CT 704 · BEI and BCT · Year IV Part I · 80 marks · 3 hours

Digital Signal Analysis and Processing

A working reader for CT 704, built around every question the Institute of Engineering has set since 2071. This subject is not memorised, it is practised: almost every mark on the paper comes from carrying out a procedure correctly, so the chapters here are written as procedures, the derivations sit behind the formulas that use them, and every theory question ever asked is answered in its own section.

past papers read

200

questions, all here

theory questions answered

marks in the paper

2071 Shrawan to 2082 Bhadra Regular and Back papers Attempt All questions Pass 32 of 80

Where the marks actually areThe same paper, every year

Every question from 19 sittings was transcribed and filed against the seven chapters of the syllabus. The result is unusual and worth knowing before you plan a single hour of revision: all seven chapters appear in all nineteen papers, and the marks each one carries are within a mark or two of the syllabus split, every time. There is no chapter to gamble on and none to drop.

The blueprint of the paper

The order of the questions barely moves. Across all nineteen papers the paper walks the syllabus from front to back, so the question number tells you the chapter before you have read a word of it.

Question	Chapter	What it asks, nearly every time	Marks
1	1	Energy or power, periodic or not, or a linearity, causality, time invariance and stability test	4 to 6
2	1	Output of an LTI system: a convolution sum, or the response to a complex exponential	5 to 6
3	2	Define the ROC, then an inverse z-transform by partial fractions, with the ROC deciding the answer	6
4	3	Plot the pole zero diagram and sketch the magnitude response, not to scale	10
5 and 6	4	Direct form I and II, cascade form, and a lattice or lattice ladder structure	10
7 and 8	5	An FIR design by window or Kaiser window, and the Remez exchange algorithm with its flow chart	14
9	6	A Butterworth IIR design by bilinear transformation, sometimes by impulse invariance, sometimes with a comparison of the two	12 to 15
10 to 12	7	An 8 point FFT by DIT or DIF, and a circular convolution	13 to 15

The consequence for revision. Nine procedures carry roughly sixty of the eighty marks: convolution, inverse z-transform, pole zero and magnitude sketch, direct form, lattice, window design, Kaiser design, bilinear Butterworth design, and the FFT butterfly. Practise those nine until they are mechanical and the paper is mostly done.

What repeats, in order

Counting how many of the nineteen papers asked for a thing, rather than how many marks it carried:

Asked in	The thing	Chapter
18 papers	A Butterworth design: order, cutoff, prototype, transform	6
17	Pole zero plot and the magnitude response sketch	3
17	Circular convolution, or a product of DFTs read back as one	7
16	Output of an LTI system by convolution	1
15	Inverse z-transform by partial fractions with a stated ROC	2
15	The Remez exchange algorithm and the optimum filter	5
14	The bilinear transformation, warping and prewarping	6
11	FIR design by a fixed window, and by the Kaiser window	5
11	An 8 point FFT, DIT or DIF	7
10	Lattice or lattice ladder structure and its coefficients	4

The Remez exchange algorithm is the single best marks per hour on the paper. It has been asked in fifteen of nineteen sittings, it is always worth 5 to 9 marks, it never changes, and it is pure bookwork: the alternation theorem, the matrix equation, and the flow chart.

How this reader is put together

Seven chapters, one per syllabus chapter, each opening with the procedures that chapter is examined on and the derivation behind each one.
The question bank holds all 200 questions verbatim, readable by paper or by chapter with the repeats merged, and every one is linked to the card that answers it.
Theory solutions answers every theory question the paper has ever set, once, most asked first, filterable by chapter. Derivations are worked in full there.
The formula sheet is every formula the paper needs, by chapter and in one scroll, and each formula opens to show where it comes from, because a formula you cannot derive is a formula you will misremember under pressure.
Numerical solutions are not here. They are their own section, and this reader does not shorten them into a corner of another page.

Five things that quietly cost marks

The ROC decides the answer, not the algebra. Two questions with the same $X (z)$ and different ROCs have different $x [n]$ , one causal and one anticausal. Write the ROC down before you invert anything.
Not to scale still means labelled. A magnitude sketch earns its marks from the peaks at the poles, the nulls at the zeros, the value at $ω = 0$ and at $ω = π$ , and the symmetry about $π$ . A smooth unlabelled curve earns almost nothing.
Prewarp before you design, in bilinear questions. Designing the analog prototype at the unwarped frequency is the commonest lost block of marks on question 9.
State the window you chose and why. A window design answer that does not name the stopband attenuation the window gives has skipped the step the question is testing.
Draw the butterfly, do not just list the numbers. In an FFT question the diagram with its twiddle factors and its bit reversed order is most of the mark.

Notation used here

Symbol	Means
$x [n]$ , $x (n)$	A discrete time signal. The papers use both brackets for the same thing.
$ω$	Digital frequency in radians per sample, from $- π$ to $π$ . The papers print it as `W`, `w` or `omega`, all meaning this.
$Ω$	Analog frequency in radians per second, used for the analog prototype in chapter 6.
$T$ , $F_{s}$	Sampling period and sampling frequency, with $ω = Ω T$ and $T = 1 / F_{s}$ .
$u [n]$ , $δ [n]$	Unit step and unit impulse.
M, N	Filter length or order in chapters 5 and 6; DFT length in chapter 7.
$W_{N}$	The twiddle factor $e^{- j 2 π / N}$ of the DFT.

The whole subject on one page

Seven chapters, and the topics each one is examined on. The number beside a leaf is how many of the nineteen sittings asked it, so this is not a contents list: it is a map of where the marks are.

Every branch is a chapter and every leaf a topic the examiner has actually set, with the number of the nineteen sittings that asked it. A bold leaf was asked in eight sittings or more. The three chapters on the left carry 45 of the 80 marks between them.

What the picture says

The right hand chapters are the foundations and the left hand ones are the money. Chapters 5, 6 and 7 carry 45 of the 80 marks and every one of their headline topics is a bold leaf.
Nine leaves carry most of the paper. The Butterworth design, the pole zero sketch, circular convolution, convolution, the inverse z-transform, the Remez algorithm, the bilinear transformation, the Kaiser window and the window method are each set in eight sittings or more.
Nothing is droppable. All seven chapters appear in all nineteen sittings, and the thinnest branch, chapter 2, still carries six marks every year for what is essentially one procedure.

How to read this reader

Two marks appear under headings all through the chapters, and they are the whole point of building this from the papers rather than from a textbook.

Chip	Means
TOP n/19	Asked in 8 or more of the nineteen sittings. There is no version of this paper that does not want it.
HOT n/19	Asked in 4 to 7 sittings.
PIN n/19	Asked in 1 to 3 sittings. Worth a read, not worth a night.
15	Marks the question has actually carried. Several chips mean it has been set at different marks in different years.

Underneath each chip is the list of sittings that asked it, in the paper's own shorthand. Bold is a Regular sitting, plain is a Back sitting.

Code	Month	Code	Month	Code	Month
Ba	Baishakh	Shr	Shrawan	Ka	Kartik
Jth	Jestha	Bh	Bhadra	Mng	Mangsir
Asa	Ashad	Ash	Ashwin	Ch	Chaitra

82 Bh is 2082 Bhadra, a Regular sitting; 81 Ba is 2081 Baishakh, a Back sitting.
Ash and Asa are different months. 2076 Ashwin and 2070 Ashad would be separate papers.
The count is over the 19 CT 704 papers in this reader, 2071 Shrawan to 2082 Bhadra. It does not include the BEX paper EX 753, which sets much of the same subject and would push every count higher.

The four sections behind the chapters

Theory answers answers every theory question the paper has set, once, most asked first, with the derivation written out.
Numerical solutions works all 170 questions that carry a computation, grouped by method, each with a note saying what is different about that one.
Question bank reproduces all 200 questions verbatim, by paper or by chapter.
The chapters themselves carry no questions at all: they are the study content, in the order the syllabus teaches it.

Chapter 1 · 8 hours · 9 marks · questions 1 and 2, every paper

Discrete-time signals and systems

What a discrete time signal is, the handful of sequences everything is built from, how to bend one in time, how to classify a system by testing it, and the one operation the whole subject rests on: the convolution sum.

What this chapter is about

Signals: the elementary sequences, the time operations, and the two classifications the paper asks for, energy against power and periodic against aperiodic.
Systems: five properties, each with a test you can carry out in three lines: linear, time invariant, causal, stable, static.
LTI systems: why the impulse response is enough to know everything, and the convolution sum that turns it into an output.
Difference equations: the other description of the same system, and the split into zero input and zero state response.

Where it fits

This chapter defines the objects. Chapter 2 gives them a transform, chapter 3 gives them a frequency response, chapter 4 gives them a circuit.
Convolution returns twice: as multiplication in chapter 2, and as circular convolution in chapter 7.
The system tests return whenever a filter has to be shown stable.

What you will learn

1.1 Discrete time signals and the elementary sequences, Even and odd parts of a signal
1.2 Energy signals and power signals
1.3 Testing a signal for periodicity
1.4 Shifting, folding and scaling a sequence
1.5 The discrete time Fourier series, and its properties
1.6 The discrete time Fourier transform, and its properties
1.7 The five system properties, and how to test each one
1.8 The convolution sum, and four ways to do it
1.9 The frequency response of an LTI system
1.10 Sampling a continuous time signal, and the spectrum of the samples
1.11 Last minute recall, chapter 1

How it is examined

Question 1 is always from here, worth 4 to 6 marks: energy or power, periodic or not, or a linearity and time invariance test.
Question 2 is always the convolution, worth 5 to 6 marks.
Both halves are procedures. Nothing in this chapter needs an essay.

1.1Discrete time signal, basic signal types

Discrete time signals and the elementary sequences PIN 2/19

76 Ch · 74 Ch2+33+2

Before the formulas

This is the first page of the subject, so start with what a discrete time signal actually is: a list of numbers, one for each sample instant, with nothing at all in between. Everything in the seven chapters is done to lists like that.

The four sequences on this card are the alphabet. An impulse, a step, a ramp and an exponential are the pieces every other signal in every question gets written in terms of, so being able to say a sequence in those terms is worth much more than it looks.

By the end of the card you should be able to read a sequence written with an arrow under one sample and say without hesitating which number sits at which n.

Discrete time signal A signal defined only at discrete instants, written as a sequence of numbers

x [n]

where n is an integer. It is undefined between samples, not zero.

Where it comes from: sampling an analog signal, $x [n]$ = xa(nT), where T is the sampling period and Fs = $\frac{1}{T}$ the sampling frequency.
Digital frequency: a sinusoid $\cos (Ω t)$ sampled at T becomes $\cos (Ω T n)$ , so the digital frequency is $ω = Ω$ T = $2 π$ F / Fs, in radians per sample, and only the range $- π$ to $π$ is distinguishable.
Three ways to write one: a closed form such as $x [n] = {(\frac{1}{2})}^{n} u [n]$ ; a list with the origin marked, $x [n] = {1, 2, 3, 1}$ with an arrow or an underline under $n = 0$ ; or a stem plot.

δ [n] & = \begin{matrix}  \end{matrix} 1 at n = 0, 0 otherwise & & the unit impulse u [n] & = 1 for n \geq 0, 0 for n < 0 & & the unit step r [n] & = n u [n] & & the unit ramp x [n] & = a^{n} u [n] & & the real exponential x [n] & = A \cos (ω_{0} n + ϕ) & & the sinusoid x [n] & = e^{j ω_{0} n} & & the complex exponential

The two relations that get used constantly: $u [n]$ = sum of $δ [n - k]$ for k from 0 to infinity, and $δ [n] = u [n] - u [n - 1]$ .
The sifting property: any sequence is a sum of shifted impulses, $x [n] = \sum_{k} x [k] δ [n - k]$ . This one line is what makes convolution possible, so it is the first line of the convolution derivation.
A gated exponential such as ${(\frac{1}{2})}^{n} {u [n] - u [n - 3]}$ is the exponential kept for $n = 0$ , 1, 2 and zero elsewhere. The paper writes impulse responses this way constantly: read off the first and last index before doing anything else.

In the exam

Read the gate first. $u [n + 2] - u [n - 2]$ runs from $n = - 2$ to $n = 1$ , four samples, not five. An off by one here loses the whole convolution.
Mark the origin in every sequence you write down. An unmarked ${2, 1, 0.5, - 1}$ is ambiguous and the marker will assume $n = 0$ at the first entry.

The sifting property at the end of this card is the first line of the convolution derivation, and the impulse train it describes is what makes sampling repeat a spectrum.

See it drawnThe four sequences everything is built from

Every signal in this subject is a sum of shifted, scaled copies of these.

Practise thisEvery question the paper has set on this topic, worked in full2

Practise this +1 like it2076 Chaitra · Q12+3 marksDefine even and odd type discrete time signals with suitable example. Plot the signal

x [- 2 n + 3]

where

x [n] = {1, 2, 0, - 1, - 3, - 4}

What is different in this one

The answer runs from $n = - 1$ to 1, 3 samples, and it does not start at the origin, so the $n = 0$ sample must be marked.
Work the index arithmetic, never the picture: solve for the n that put the argument inside the range where the original is non zero, then evaluate sample by sample.

Work the index arithmetic, never the picture. x is non zero for $0 \leq - 2 n + 3 \leq 5$ , that is $- 1 \leq n \leq 1.5$ , so only $n = - 1$ , 0, 1 survive.

\begin{aligned} n & = - 1 \to x [- 2 (- 1) + 3] = x [5] = - 4 \\ n & = 0 \to x [3] = - 1 \\ n & = 1 \to x [1] = 2 \\ n & = 2 \to x [- 1], outside  the  sequence, so  nothing \end{aligned}

y [n] = x [- 2 n + 3]

, drawn from the computed samples.

$y [n] = x [- 2 n + 3]$ = {-4, -1, 2} for $n = - 1$ to 1. Folding and decimating by 2 keeps only the odd indexed samples of the original, which is why three of the six survive.

Same as 2076 Chaitra Q12074 Chaitra · Q13+2 marksPlot the sequence

x [n] = u [n] - u [n - 3] + 5 δ [n - 4] = n u [n - 6]

. List out the properties of LTI system.

What is different in this one

The answer runs from $n = 0$ to 8, 9 samples, and it starts at the origin.
Work the index arithmetic, never the picture: solve for the n that put the argument inside the range where the original is non zero, then evaluate sample by sample.

Draw each piece on the same axis and add. Note the printed equation uses an equals sign where the last plus belongs.

\begin{aligned} u [n] - u [n - 3] a rectangle  of  height 1 over n & = 0, 1, 2 \\ 5 δ [n - 4] a single  sample  of  height 5 at n & = 4 \\ n u [n - 6] a ramp  starting  at n & = 6, so 6, 7, 8 \dots \end{aligned}

x [n]

, drawn from the computed samples.

$x [n]$ = {1, 1, 1, 0, 5, 0, 6, 7, 8} for $n = 0$ to 8. The three pieces never overlap, so no sample is the sum of two of them.

The same questions grouped by how often each topic is set: plotting a shifted, folded or scaled sequence

Even and odd parts of a signal PIN 2/19

76 Ch · 71 Shr2+34+5

Before the formulas

You have the alphabet. The next thing anyone asks about a signal is whether it has any symmetry, because symmetry saves work later: a symmetric signal has no sine terms, an antisymmetric one has no cosine terms, and half of a sum or an integral disappears before you start.

The useful fact is that every signal, however lopsided, splits into a symmetric half and an antisymmetric half that add back to the original exactly. That is the whole section: one pair of formulas, and one habit, which is to fold the signal about $n = 0$ and compare it with itself.

The questions underneath are all that same shape, so once two of them come out right you have done all of them.

Even and odd A signal is even if

x [- n] = x [n]

, symmetric about the vertical axis, and odd if

x [- n] = - x [n]

, antisymmetric, which forces

x [0] = 0

Every signal splits into one of each, and the split is unique.

\begin{aligned} x_{e} [n] & = \frac{x [n] + x [- n]}{2} & the even part \\ x_{o} [n] & = \frac{x [n] - x [- n]}{2} & the odd part \\ x [n] & = x_{e} [n] + x_{o} [n] & they add back to the signal \end{aligned}

The procedure: tabulate $x [n]$ ; write $x [- n]$ underneath it by reversing the table about $n = 0$ ; add the two rows and halve for the even part; subtract and halve for the odd part. Check that $x_{o} [0]$ came out zero and that the two parts add back.

In the exam

Line the tables up on $n = 0$ , not on the first entry. Reversing a list without tracking the origin is the only way to get this wrong.
Check $x [0]$ . The odd part must be zero there, every time. It is a free check and it catches a reversal error immediately.

The same split explains the conjugate symmetry of every spectrum in the subject: see the DTFT properties, and the free check it gives you in every FFT question.

See it drawnA signal is its even part plus its odd part

Fold the signal about

n = 0

; the half sum is even, the half difference is odd, and adding them back returns the original.

Practise thisEvery question the paper has set on this topic, worked in full1

Practise this2071 Shrawan · Q14+5 marksFind the odd and even part of the following signal: [a stem plot of

x [n]

with values 1 at

n = - 4

, -3, -2, -1 and 0, and 2 at

n = 1

, 2, 3, 4]. A discrete time LTI system has input signal and impulse response as,

x [n] = 1

for

- 1 \leq n \leq 1

and 0 elsewhere, and

h [n] = 1

for

- 1 \leq n \leq 1

and 0 elsewhere. Find the output of the system using graphical method.

How this is readThe stem plot is $x [n] = 1$ for $n = - 4$ to 0 and 2 for $n = 1$ to 4. The second half convolves two identical three sample rectangles, each from $n = - 1$ to 1.

The even and odd parts

Tabulate $x [n]$ against n, write $x [- n]$ under it by reversing about $n = 0$ , then halve the sum and the difference.

n	-4	-3	-2	-1	0	1	2	3	4
$x [n]$	1	1	1	1	1	2	2	2	2
$x [- n]$	2	2	2	2	1	1	1	1	1
$x_{e} [n]$	1.5	1.5	1.5	1.5	1	1.5	1.5	1.5	1.5
$x_{o} [n]$	-0.5	-0.5	-0.5	-0.5	0	0.5	0.5	0.5	0.5

$x_{e} [n]$ = {1.5, 1.5, 1.5, 1.5, 1, 1.5, 1.5, 1.5, 1.5} and $x_{o} [n]$ = {-0.5, -0.5, -0.5, -0.5, 0, 0.5, 0.5, 0.5, 0.5}, both for $n = - 4$ to 4. The two checks hold: $x_{o} [0] = 0$ , and $x_{e} [n] + x_{o} [n]$ returns $x [n]$ at every n.

The rest of this questionThis topic works the part that belongs to it. The other part is worked under Convolution (The output, by the graphical method), where it belongs.

The same questions grouped by how often each topic is set: even and odd parts

1.2Energy signal, power signal

Energy signals and power signals HOT 6/19

82 Bh · 79 Bh · 79 Ba · 75 Ch · 74 Ash · 72 Ka2+22+33+4

Before the formulas

So far a signal has been a shape. This section asks how big it is, and there are two different answers depending on whether the signal ever stops. A pulse that dies away has a finite total, and we call that total its energy.

A signal that runs for ever has an infinite total, so the total tells you nothing and we ask for the average instead, which is its power. Nothing here is deep. It is two sums and one question: does the signal die away or not.

The trap the paper likes is that a signal is never both at once, and some signals are neither.

The two definitions The energy of

x [n]

is E =

\sum_{n = - \infty}^{\infty} | x [n] |^{2}

. The average power is P = limit as N goes to infinity of

\frac{1}{2 N + 1}

times the sum of

| x [n] |^{2}

from -N to N.

\begin{aligned} E & = \sum_{n = - \infty}^{\infty} | x [n] |^{2} \\ P & = lim_{N \to \infty} \frac{1}{2 N + 1} \sum_{n = - N}^{N} | x [n] |^{2} \\ energy signal: 0 < E < \infty, and then P = 0 \\ power signal: 0 < P < \infty, and then E = \infty \end{aligned}

	Energy signal	Power signal
Energy	Finite and non zero	Infinite
Power	Zero	Finite and non zero
Typical shape	Dies away, or lasts a finite time	Goes on for ever without dying: periodic signals, the unit step
Examples	$δ [n]$ ; $a^{n} u [n]$ with $\| a \| < 1$ ; any finite length sequence	$u [n]$ , with P = $\frac{1}{2}$ ; A cos( $ω_{0}$ n), with P = $A^{2} / 2$ ; $e^{j ω_{0} n}$ , with $P = 1$
Can it be both	No. A signal is one, the other, or neither. Neither happens: n $u [n]$ has infinite energy and infinite power.

The procedure

Try energy first. Form sum $| x [n] |^{2}$ over the range where x is non zero. If it converges to a finite number, it is an energy signal and you are done.
If it diverges, try power. For a periodic signal the limit collapses to the average over one period, P = $\frac{1}{N}$ sum over one period of $| x [n] |^{2}$ , which is much easier than the limit.
For any complex exponential of the form $e^{j (ω_{0} n + ϕ)}$ , the magnitude is 1 at every n, so E is infinite and $P = 1$ : it is a power signal. This is the question the paper actually asks, in five sittings.

The two the paper keeps setting $x [n] = e^{j (π n / 2 + 4 p i / 7)}$ .

| x [n] |^{2} = 1

for every n, so E = sum of 1 = infinity, and P = lim

\frac{1}{2 N + 1}

times (2N+1) = 1. Finite non zero power: a power signal.
$x [n] = u [n]$ . E = sum from 0 to infinity of 1 = infinity; P = lim

(N + 1) / (2 N + 1) = \frac{1}{2}

. A power signal.
$x [n] = δ [n]$ .

E = 1

, finite, so P = 0. An energy signal.

Energy signals are the ones with a DTFT; power signals are the ones with a Fourier series. The same absolute summability condition returns as BIBO stability.

See it drawnWhich one a signal has, and why never both

An energy signal dies away, so its total is finite and its average is zero. A power signal never dies, so its total is infinite and its average is not.

Practise thisEvery question the paper has set on this topic, worked in full3

Practise this2082 Bhadra · Q12+2 marksCompare between energy signal and power signal. Determine whether the signal

x [n] = e^{j (π / 2 n + 4 π / 7)}

is energy signal or power signal.What is different hereA complex exponential has magnitude 1 at every n, so no summation is needed at all: E is infinite and P is 1. That is the whole answer, in one line.

What is different in this one

Try the energy first. If the sum settles the signal is an energy signal and its power is zero; if it grows without bound, form the power instead. Every question in this topic is that one decision.

Form the energy first. If it settles on a finite value the signal is an energy signal and its power is zero; if it grows without bound, form the power instead.

\begin{aligned} inf                                 1        N \\ E & = \sum | x [n] |^{2} P = lim - - - - - - \sum | x [n] |^{2} \\ n & = - inf N \to inf 2 N + 1 n = - N \end{aligned}

Carried out over windows of growing width, with $x [n] = e^{j (π n / 2 + 4 π / 7)}$ :

N	E over -N to N	$P = E / (2 N + 1)$
50	101	1
200	401	1
1000	2001	1
5000	10001	1

$x [n] = e^{j (π n / 2 + 4 π / 7)}$ is a power signal: the energy grows without bound while the power settles at 1.

Practise this2079 Baishakh · Q12+2 marksCompare between energy signal and power signal. Determine whether the signal

x [n] = e^{j (π / 2 n + 4 π / 7)}

is energy signal or power signal.

What is different in this one

Try the energy first. If the sum settles the signal is an energy signal and its power is zero; if it grows without bound, form the power instead. Every question in this topic is that one decision.

Form the energy first. If it settles on a finite value the signal is an energy signal and its power is zero; if it grows without bound, form the power instead.

\begin{aligned} inf                                 1        N \\ E & = \sum | x [n] |^{2} P = lim - - - - - - \sum | x [n] |^{2} \\ n & = - inf N \to inf 2 N + 1 n = - N \end{aligned}

Carried out over windows of growing width, with $x [n] = e^{j (π n / 2 + 4 π / 7)}$ :

N	E over -N to N	$P = E / (2 N + 1)$
50	101	1
200	401	1
1000	2001	1
5000	10001	1

$x [n] = e^{j (π n / 2 + 4 π / 7)}$ is a power signal: the energy grows without bound while the power settles at 1.

Practise this2072 Kartik · Q12+3 marksDefine energy and power signal. Check the signal

x [n] = u [n]

and

x [n] = δ [n]

is Energy or Power type.

How this is readThe step is the standard power signal with P = $\frac{1}{2}$ , not 1: the window holds $2 N + 1$ samples of which only $N + 1$ are non zero. The impulse is the standard energy signal with E = 1.

a) $x [n] = u [n]$

What is different in this one

Try the energy first. If the sum settles the signal is an energy signal and its power is zero; if it grows without bound, form the power instead. Every question in this topic is that one decision.

Form the energy first. If it settles on a finite value the signal is an energy signal and its power is zero; if it grows without bound, form the power instead.

\begin{aligned} inf                                 1        N \\ E & = \sum | x [n] |^{2} P = lim - - - - - - \sum | x [n] |^{2} \\ n & = - inf N \to inf 2 N + 1 n = - N \end{aligned}

Carried out over windows of growing width, with $x [n] = u [n]$ :

N	E over -N to N	$P = E / (2 N + 1)$
50	51	0.505
200	201	0.5012
1000	1001	0.5002
5000	5001	0.5

$x [n] = u [n]$ is a power signal: the energy grows without bound while the power settles at 0.5.

b) $x [n] = δ [n]$

What is different in this one

Try the energy first. If the sum settles the signal is an energy signal and its power is zero; if it grows without bound, form the power instead. Every question in this topic is that one decision.

Form the energy first. If it settles on a finite value the signal is an energy signal and its power is zero; if it grows without bound, form the power instead.

\begin{aligned} inf                                 1        N \\ E & = \sum | x [n] |^{2} P = lim - - - - - - \sum | x [n] |^{2} \\ n & = - inf N \to inf 2 N + 1 n = - N \end{aligned}

Carried out over windows of growing width, with $x [n] = δ [n]$ :

N	E over -N to N	$P = E / (2 N + 1)$
50	1	0.0099
200	1	0.0025
1000	1	0.0005
5000	1	0.0001

$x [n] = δ [n]$ is a energy signal: E settles at 1 and the power falls to zero.

The same questions grouped by how often each topic is set: energy signal or power signal

1.3Periodicity of a discrete time signal

Testing a signal for periodicity HOT 6/19

81 Bh · 80 Bh · 80 Ba · 79 Bh · 78 Bh · 74 Ash2+245

Before the formulas

In continuous time every sinusoid repeats, so nobody bothers asking. In discrete time you only see the signal at whole numbered instants, and those samples do not have to line up again the same way. A cosine with an awkward frequency can be sampled for ever and never once repeat its pattern, even though the smooth curve it came from repeats perfectly well.

This section is the test for that, and it is one line of arithmetic: write the frequency over two $π$ and ask whether it is a ratio of two whole numbers. The questions are that one test applied to sums of two or three sinusoids, plus the rule for putting their periods back together.

Periodic

x [n]

is periodic if $x [n + N] = x [n]$ for all n, for some positive integer N. The smallest such N is the fundamental period. The integer requirement is the whole difference from continuous time, where any period is allowed.

\begin{aligned} for x [n] = \cos (ω_{0} n + ϕ) or e^{j ω_{0} n} : \\ periodic ⟺ \frac{ω_{0}}{2 π} is rational \\ \frac{ω_{0}}{2 π} & = \frac{k}{N} in lowest terms ⟹ fundamental period N \end{aligned}

The procedure

Read off $ω_{0}$ , the coefficient of n inside the cosine or the exponent. In cos( $2 π$ n/5) it is $2 π$ /5; in $e^{j π n / 16}$ it is $\frac{π}{16}$ .
Form $ω_{0}$ /( $2 π$ ) and reduce the fraction to lowest terms. If it does not come out as a ratio of integers, the signal is aperiodic and you stop.
The denominator is N, the fundamental period of that component.
For a sum or a product of periodic components, the whole signal is periodic with period N = LCM( $N_{1}$ , $N_{2}$ , ...). If any one component is aperiodic, the whole thing is aperiodic.

Worked, the ones actually set $x [n]$ = cos( $2 π$ n/5) + $\sin (π n / 3)$ . First:

ω_{0} / (2 π) = \frac{1}{5}

, so

N_{1} = 5

. Second:

ω_{0} = \frac{π}{3}

, so

ω_{0} / (2 π) = \frac{1}{6}

N_{2} = 6

. N = LCM(5, 6) = 30.
$x [n] = e^{j π n / 16} \cos (n π / 17)$ . First:

(\frac{π}{16}) / (2 π) = \frac{1}{32}

N_{1} = 32

. Second:

(\frac{π}{17}) / (2 π) = \frac{1}{34}

N_{2} = 34

. N = LCM(32, 34) = 544.
$x [n] = \cos (π n / 2) \cos (π n / 4)$ .

N_{1} = 4

N_{2} = 8

, N = 8.
$x [n] = e^{j (π n / 3 + π / 4)}$ .

(\frac{π}{3}) / (2 π) = \frac{1}{6}

, so N = 6. The phase

π / 4

shifts the signal and cannot change its period.

In the exam

The phase never matters. Only the coefficient of n decides periodicity.
Reduce the fraction. $ω_{0} = \frac{4 π}{6}$ gives 1/3 after reduction, so $N = 3$ , not 6.
Write the conclusion as a sentence. The marks are for the value of N and for saying periodic or aperiodic, not for the arithmetic in between.

The rule that $ω_{0} / 2 π$ must be rational is the reason the DFT assumes a period of N, and the reason a signal that does not fit the window leaks across the whole spectrum.

See it drawnWhy a discrete sinusoid is often not periodic

The samples have to line up again exactly. They only do when the frequency divided by two

π

is a ratio of whole numbers.

Practise thisEvery question the paper has set on this topic, worked in full6

Practise this +3 like it2081 Bhadra · Q14 marksDetermine if the signal

x [n] = \cos (2 π n / 5) + \sin (π n / 3)

is periodic or not. If the signal is periodic, find its fundamental period.

What is different in this one

2 components, so each period is found separately and the answer is their LCM.
The phase plays no part: only the coefficient of n decides periodicity.

A discrete time sinusoid is periodic only if $ω_{0} / (2 π)$ is rational. Reduce it to k/N in lowest terms and N is the period of that component; for several components take the LCM.

Component	$ω_{0}$	$ω_{0} / 2 π$ , lowest terms	Period
$\cos (2 π n / 5)$	$\frac{2 π}{5}$	1 / 5	5
$\sin (π n / 3)$	$\frac{π}{3}$	1 / 6	6

N = LCM(5, 6) = 30.

$x [n] = \cos (2 π n / 5) + \sin (π n / 3)$ is periodic with fundamental period $N = 30$ samples.

Same as 2081 Bhadra Q12080 Bhadra · Q25 marksDetermine whether the given signal is periodic or not. If the signal is periodic, determine the fundamental period

x [n] = e^{j π n / 16} \cos (n π / 17)

What is different in this one

2 components, so each period is found separately and the answer is their LCM.
The phase plays no part: only the coefficient of n decides periodicity.

A discrete time sinusoid is periodic only if $ω_{0} / (2 π)$ is rational. Reduce it to k/N in lowest terms and N is the period of that component; for several components take the LCM.

Component	$ω_{0}$	$ω_{0} / 2 π$ , lowest terms	Period
$e^{j π n / 16}$	$\frac{π}{16}$	1 / 32	32
$\cos (n π / 17)$	$\frac{π}{17}$	1 / 34	34

N = LCM(32, 34) = 544.

$x [n] = e^{j π n / 16} \cos (n π / 17)$ is periodic with fundamental period $N = 544$ samples.

Practise this2080 Baishakh · Q12+2 marksCheck whether following signals are periodic or not. If yes, state their periodic time. a)

x [n]

= Sin(n

π

) + Cos(n

π

) b)

x [n]

= Sin(3n

\frac{π}{5}

) + Cos(4n

\frac{π}{7}

)

a) $\sin (n π) + \cos (n π)$

What is different in this one

2 components, so each period is found separately and the answer is their LCM.
The phase plays no part: only the coefficient of n decides periodicity.

A discrete time sinusoid is periodic only if $ω_{0} / (2 π)$ is rational. Reduce it to k/N in lowest terms and N is the period of that component; for several components take the LCM.

Component	$ω_{0}$	$ω_{0} / 2 π$ , lowest terms	Period
$\sin (n π)$	$π$	1 / 2	2
$\cos (n π)$	$π$	1 / 2	2

N = LCM(2, 2) = 2.

$x [n] = \sin (n π) + \cos (n π)$ is periodic with fundamental period $N = 2$ samples.

b) $\sin (3 n π / 5) + \cos (4 n π / 7)$

What is different in this one

2 components, so each period is found separately and the answer is their LCM.
The phase plays no part: only the coefficient of n decides periodicity.

A discrete time sinusoid is periodic only if $ω_{0} / (2 π)$ is rational. Reduce it to k/N in lowest terms and N is the period of that component; for several components take the LCM.

Component	$ω_{0}$	$ω_{0} / 2 π$ , lowest terms	Period
$\sin (3 π n / 5)$	$\frac{3 π}{5}$	3 / 10	10
$\cos (4 π n / 7)$	$\frac{4 π}{7}$	2 / 7	7

N = LCM(10, 7) = 70.

$x [n] = \sin (3 n π / 5) + \cos (4 n π / 7)$ is periodic with fundamental period $N = 70$ samples.

Same as 2081 Bhadra Q12079 Bhadra · Q12+2 marksDefine energy and power signal. Determine whether the signal

x [n] = \cos (2 π n / 5) + \sin (π n / 3)

is periodic or non-periodic and if it is periodic, find its fundamental period.

What is different in this one

2 components, so each period is found separately and the answer is their LCM.
The phase plays no part: only the coefficient of n decides periodicity.

A discrete time sinusoid is periodic only if $ω_{0} / (2 π)$ is rational. Reduce it to k/N in lowest terms and N is the period of that component; for several components take the LCM.

Component	$ω_{0}$	$ω_{0} / 2 π$ , lowest terms	Period
$\cos (2 π n / 5)$	$\frac{2 π}{5}$	1 / 5	5
$\sin (π n / 3)$	$\frac{π}{3}$	1 / 6	6

N = LCM(5, 6) = 30.

$x [n] = \cos (2 π n / 5) + \sin (π n / 3)$ is periodic with fundamental period $N = 30$ samples.

Same as 2081 Bhadra Q12078 Bhadra · Q14 marksDetermine whether the signal

x [n] = \cos (π n / 2) \cos (π n / 4)

is periodic or non periodic and if it is periodic, find its fundamental period.

What is different in this one

2 components, so each period is found separately and the answer is their LCM.
The phase plays no part: only the coefficient of n decides periodicity.

A discrete time sinusoid is periodic only if $ω_{0} / (2 π)$ is rational. Reduce it to k/N in lowest terms and N is the period of that component; for several components take the LCM.

Component	$ω_{0}$	$ω_{0} / 2 π$ , lowest terms	Period
$\cos (π n / 2)$	$\frac{π}{2}$	1 / 4	4
$\cos (π n / 4)$	$\frac{π}{4}$	1 / 8	8

N = LCM(4, 8) = 8.

$x [n] = \cos (π n / 2) \cos (π n / 4)$ is periodic with fundamental period $N = 8$ samples.

Practise this2074 Ashwin · Q12+2 marksDefine Energy and Power type discrete time signal. Check whether signal

x [n] = e^{j (π n / 3 + π / 4)}

is periodic or not. If it is periodic, state its periodic time.What is different hereThe phase

\frac{π}{4}

plays no part: only the coefficient of n decides periodicity.

What is different in this one

1 component, so the period is read straight off.
The phase plays no part: only the coefficient of n decides periodicity.

A discrete time sinusoid is periodic only if $ω_{0} / (2 π)$ is rational. Reduce it to k/N in lowest terms and N is the period of that component; for several components take the LCM.

Component	$ω_{0}$	$ω_{0} / 2 π$ , lowest terms	Period
$e^{j (π n / 3 + π / 4)}$	$\frac{π}{3}$	1 / 6	6

$x [n] = e^{j (π n / 3 + π / 4)}$ is periodic with fundamental period $N = 6$ samples.

The same questions grouped by how often each topic is set: periodicity and the fundamental period

1.4Transformation of the independent variable

Shifting, folding and scaling a sequence PIN 2/19

76 Ch · 74 Ch2+33+2

Before the formulas

You can describe a signal and you can measure it. Now you move it about. Shifting, folding and squashing a sequence is the only thing this section does, and it is worth doing carefully because the operations do not commute: shifting and then folding gives a different answer from folding and then shifting, and both of them look plausible on paper.

The recipe on this card does the index arithmetic instead of the picture, which is what makes the order stop mattering. This is not a side topic. Convolution, the most asked thing in the whole paper, is these same moves done over and over, so getting them automatic here pays for itself in 1.8.

The paper asks this as plot $x [- 2 n + 3]$ where $x [n]$ = {...}. It is worth doing by a fixed recipe, because doing the operations in the wrong order gives a wrong plot that looks plausible.

\begin{aligned} x [n - k] & delay by k & shifts the sequence right \\ x [n + k] & advance by k & shifts the sequence left \\ x [- n] & folding & reflects about n = 0 \\ x [a n] & decimation, a > 1 & keeps every a -th sample \\ x [n / a] & interpolation & spreads the samples out \end{aligned}

The safe procedure, for anything of the form $x [a n + b]$

Tabulate. Write n against $x [n]$ for every n where $x [n]$ is non zero.
Work out which n you need. The new sequence $y [n] = x [a n + b]$ is non zero when an + b lands inside the range where x is non zero. Solve for n at both ends.
Evaluate sample by sample. For each of those n, compute an + b and read x at that index. Do not try to shift and fold the picture in your head.
Mark the new origin and plot.

Worked, 2076 Chaitra

x [n] = {1, 2, 0, - 1, - 3, - 4}

starting at

n = 0

. For

y [n] = x [- 2 n + 3]

: x is non zero for

0 \leq - 2 n + 3 \leq 5

, that is

- 1 \leq n \leq 1.5

, so

n = - 1

, 0, 1.

n = - 1

gives

x [5] = - 4

;

n = 0

gives

x [3] = - 1

;

n = 1

gives

x [1] = 2

y [n] = {- 4, - 1, 2}

for

n = - 1

, 0, 1.

In the exam

Never shift then scale by eye. The index arithmetic above cannot go wrong; the picture method can.
Say how many samples survive. Decimation by 2 throws half of them away, and the marker is checking that you noticed.

Folding and shifting are exactly the two operations inside the graphical convolution, so get them right here and that question becomes bookkeeping.

See it drawnWhat each operation does to the picture

A minus sign inside the bracket folds, a plus shifts left, a multiplier decimates. Do them one at a time and in the order written.

Practise thisEvery question the paper has set on this topic, worked in full2

Practise this +1 like it2076 Chaitra · Q12+3 marksDefine even and odd type discrete time signals with suitable example. Plot the signal

x [- 2 n + 3]

where

x [n] = {1, 2, 0, - 1, - 3, - 4}

What is different in this one

The answer runs from $n = - 1$ to 1, 3 samples, and it does not start at the origin, so the $n = 0$ sample must be marked.
Work the index arithmetic, never the picture: solve for the n that put the argument inside the range where the original is non zero, then evaluate sample by sample.

Work the index arithmetic, never the picture. x is non zero for $0 \leq - 2 n + 3 \leq 5$ , that is $- 1 \leq n \leq 1.5$ , so only $n = - 1$ , 0, 1 survive.

\begin{aligned} n & = - 1 \to x [- 2 (- 1) + 3] = x [5] = - 4 \\ n & = 0 \to x [3] = - 1 \\ n & = 1 \to x [1] = 2 \\ n & = 2 \to x [- 1], outside  the  sequence, so  nothing \end{aligned}

y [n] = x [- 2 n + 3]

, drawn from the computed samples.

$y [n] = x [- 2 n + 3]$ = {-4, -1, 2} for $n = - 1$ to 1. Folding and decimating by 2 keeps only the odd indexed samples of the original, which is why three of the six survive.

Same as 2076 Chaitra Q12074 Chaitra · Q13+2 marksPlot the sequence

x [n] = u [n] - u [n - 3] + 5 δ [n - 4] = n u [n - 6]

. List out the properties of LTI system.

What is different in this one

The answer runs from $n = 0$ to 8, 9 samples, and it starts at the origin.
Work the index arithmetic, never the picture: solve for the n that put the argument inside the range where the original is non zero, then evaluate sample by sample.

Draw each piece on the same axis and add. Note the printed equation uses an equals sign where the last plus belongs.

\begin{aligned} u [n] - u [n - 3] a rectangle  of  height 1 over n & = 0, 1, 2 \\ 5 δ [n - 4] a single  sample  of  height 5 at n & = 4 \\ n u [n - 6] a ramp  starting  at n & = 6, so 6, 7, 8 \dots \end{aligned}

x [n]

, drawn from the computed samples.

$x [n]$ = {1, 1, 1, 0, 5, 0, 6, 7, 8} for $n = 0$ to 8. The three pieces never overlap, so no sample is the sum of two of them.

The same questions grouped by how often each topic is set: plotting a shifted, folded or scaled sequence

1.5Discrete time Fourier series and properties

The discrete time Fourier series, and its properties PIN 3/19

76 Ash · 73 Shr · 72 Ch4+334

Before the formulas

In 1.4 you were moving a signal about. Nothing you did there told you what the signal is made of, and that is the question this section answers, for a signal that repeats. The idea is one everybody already half knows: any repeating shape can be built out of sinusoids, and if you know how much of each one you have, you know the signal.

What is different in discrete time is that the supply of sinusoids runs out. A sequence of period N is built from exactly N of them, never more, because the next one round turns out to be one you already have.

So the Fourier series here is a finite sum of N terms with N coefficients, and the card is really two formulas: one that goes from the coefficients back to the signal, and one that goes from the signal to the coefficients.

Discrete time Fourier series A periodic sequence of period N is written as a sum of N harmonically related complex exponentials. Only N of them are distinct, because

e^{j 2 π (k + N) n / N} = e^{j 2 π k n / N}

: in discrete time the harmonics run out, which is the one real difference from the continuous case.

\begin{aligned} x [n] & = \sum_{k = 0}^{N - 1} c_{k} e^{j 2 π k n / N} & synthesis \\ c_{k} & = \frac{1}{N} \sum_{n = 0}^{N - 1} x [n] e^{- j 2 π k n / N} & analysis \\ c_{k + N} & = c_{k} & the coefficients are periodic too \end{aligned}

Where the analysis equation comes from. Multiply the synthesis equation by $e^{- j 2 π m n / N}$ , sum over one period, and swap the two sums. The inner sum is $\frac{1}{N} \sum_{n} e^{j 2 π (k - m) n / N}$ , which is 1 when k = m and 0 otherwise. That is orthogonality, and it kills every term but one, leaving c[m].
Why the inner sum is zero. It is a geometric series with ratio $e^{j 2 π (k - m) / N}$ , and for k not equal to m the ratio is not 1, so the sum is (1 $- e^{j 2 π (k - m)}$ )/(1 - ratio), whose numerator is zero.
For a real $x [n]$ , c[-k] = c*[k], so | $c_{k}$ | is even and the phase is odd. Only N/2 + 1 coefficients need computing.
The power spectrum is the set | $c_{k}$ |^2, and Parseval's relation says $\frac{1}{N}$ sum of $| x [n] |^{2}$ over one period equals sum of | $c_{k}$ |^2: the average power is the sum of the powers of the harmonics.

Property	$x [n]$ , period N	$c_{k}$
Linearity	a $x_{1} [n] + b x_{2} [n]$	a $c_{1}$ [k] + b $c_{2}$ [k]
Time shift	$x [n - m]$	$c_{k} e^{- j 2 π k m / N}$
Frequency shift	$x [n] e^{j 2 π l n / N}$	c[k - l]
Time reversal	$x [- n]$	c[-k]
Conjugation	$x^{*} [n]$	c*[-k]
Periodic convolution	sum over one period of $x_{1} [m] x_{2}$ [n-m]	N $c_{1}$ [k] $c_{2}$ [k]
Multiplication	$x_{1} [n] x_{2} [n]$	periodic convolution of $c_{1}$ and $c_{2}$
Parseval	$\frac{1}{N}$ sum $\| x [n] \|^{2}$	sum \| $c_{k}$ \|^2

Dirichlet's conditions Sufficient conditions for the series to converge to the signal, over one period: absolutely summable, or absolutely integrable in continuous time; a finite number of maxima and minima; and a finite number of finite discontinuities. At a jump the series converges to the midpoint, and the partial sums overshoot it by about 9 percent however many terms are taken. That overshoot is Gibbs' phenomenon, and it is the reason a truncated ideal filter ripples.

Worked example

Find the Fourier series coefficients of $x [n] = {1, 1, 0, 0}$ repeated with period N = 4.

\begin{aligned} c_{k} & = \frac{1}{4} \sum_{n = 0}^{3} x [n] e^{- j 2 π k n / 4} \\ = \frac{1}{4} (1 + e^{- j π k / 2}) & only x [0] and x [1] are non zero \end{aligned}

c[0] = (1/4)(1 + 1) = 0.5
c[1] = (1/4)(1 - j) = 0.25 - j0.25, magnitude 0.3536, angle -45 degrees
c[2] = (1/4)(1 - 1) = 0
c[3] = (1/4)(1 + j) = 0.25 + j0.25, magnitude 0.3536, angle +45 degrees

So c = {0.5, 0.25 - j0.25, 0, 0.25 + j0.25}. Two checks: c[3] = c*[1] as a real signal demands, and Parseval holds, since (1/4)(1 + 1) = 0.5 and 0.25 + 0.125 + 0 + 0.125 = 0.5.

In the exam

Say orthogonality by name when the question asks you to explain the process of finding the coefficients. That word is the answer; the algebra is the support.
N coefficients, not infinitely many. Saying that the discrete series has a finite number of distinct harmonics is worth a mark on its own.

Dirichlet's conditions here are where Gibbs' phenomenon comes from, and the orthogonality argument returns in the DFT, which is this series with the sum truncated to N points.

See it drawnA periodic signal is a sum of N harmonics, and only N

In discrete time the harmonics run out: the (k+N)th is the same wave as the kth, which is the one real difference from the continuous case.

1.6Discrete time Fourier transform and properties

The discrete time Fourier transform, and its properties PIN 2/19

76 Ash · 75 Ch3+44+3

Before the formulas

The series in 1.5 only works if the signal repeats. Most signals do not. Take a single pulse and pretend its period is enormous. The harmonics crowd closer and closer together, and in the limit they stop being separate lines at all: the list of coefficients turns into a continuous curve.

That curve is what this section is about. It says how much of every frequency is in the signal, not just how much of N of them. Two things to carry away, both of which the exam leans on: the answer is a function of a continuous frequency, and it repeats every two $π$ . That repeat is a fact about sampling, not a slip in your algebra.

Discrete time Fourier transform For an aperiodic sequence,

X (e^{j ω}) = \sum_{n = - \infty}^{\infty} x [n] e^{- j ω n}

. It is a continuous function of

ω

and periodic in $ω$ with period $2 π$ , because

e^{- j (ω + 2 p i) n} = e^{- j ω n}

for integer n. All the information is therefore in one period, and

- π

π

is the one usually drawn.

\begin{aligned} X (e^{j ω}) & = \sum_{n = - \infty}^{\infty} x [n] e^{- j ω n} & analysis \\ x [n] & = \frac{1}{2 π} \int_{- π}^{π} X (e^{j ω}) e^{j ω n} d ω & synthesis \\ exists when \sum_{n} | x [n] | < \infty & absolutely summable \end{aligned}

Property	$x [n]$	$X (e^{j ω})$
Linearity	a $x_{1} [n] + b x_{2} [n]$	a X1 + b X2
Time shift	$x [n - k]$	$e^{- j w k} X (e^{j ω})$
Frequency shift	$e^{j ω_{0} n} x [n]$	X( $e^{j}$ ( $ω - ω_{0}$ ))
Time reversal	$x [- n]$	$X (e^{- j ω})$
Conjugation	$x^{*} [n]$	$X^{*} (e^{- j ω})$
Convolution	$x_{1} [n] * x_{2} [n]$	X1 . X2
Multiplication	$x_{1} [n]$ . $x_{2} [n]$	periodic convolution of X1 and X2, divided by $2 π$
Differentiation in $ω$	n $x [n]$	j dX( $e^{j ω}$ )/dw
Parseval	sum $\| x [n] \|^{2}$	(1/ $2 π$ ) $\int$ \| $X (e^{j ω})$ \|^2 dw
Real $x [n]$	$X (e^{- j ω}) = X$ *( $e^{j ω}$ ): magnitude even, phase odd

\begin{aligned} x_{1} [n] x_{2} [n] & ⟷ \frac{1}{2 π} \int_{- π}^{π} X_{1} (e^{j θ}) X_{2} (e^{j (ω - θ)}) d θ \end{aligned}

The derivation, in one move.

\begin{aligned} X (e^{j ω}) & = \sum_{n} x_{1} [n] x_{2} [n] e^{- j ω n} \\ = \frac{1}{2 π} \int_{- π}^{π} X_{1} (e^{j θ}) [\sum_{n} x_{2} [n] e^{- j (ω - θ) n}] d θ \\ = \frac{1}{2 π} \int_{- π}^{π} X_{1} (e^{j θ}) X_{2} (e^{j (ω - θ)}) d θ \end{aligned}

Why this property matters more than it looks. The FIR window method is exactly this: $h [n] = h_{d} [n] w [n]$ in time, so the designed response is the ideal brick wall convolved with the window's own spectrum. The window's main lobe becomes the transition band and its side lobes become the ripple. Every row of the window table in chapter 5 is a consequence of this one line.
The convolution is periodic, over one period of $2 π$ , because both spectra are periodic. In the DFT the same argument gives circular convolution.

	Fourier series	Fourier transform
Applies to	Periodic signals	Aperiodic signals of finite energy
Spectrum	Discrete lines at multiples of the fundamental	Continuous in frequency
Produces	A set of numbers $c_{k}$	A function $X (e^{j ω})$
Operation	An average over one period	A sum over all time
Describes	A power signal	An energy signal
Relation	Let the period grow without bound: the lines close up and the series becomes the transform

Worked example

Find the DTFT of $x [n] = {(\frac{1}{2})}^{n} u [n]$ , and its magnitude at $ω = 0$ and $ω = π$ .

\begin{aligned} X (e^{j ω}) & = \sum_{n = 0}^{\infty} {(\frac{1}{2})}^{n} e^{- j ω n} \\ = \sum_{n = 0}^{\infty} {(0.5 e^{- j ω})}^{n} \\ = \frac{1}{1 - 0.5 e^{- j ω}} & a geometric series, | 0.5 e^{- j ω} | = 0.5 < 1 \end{aligned}

At $ω = 0$ : $X = 1$ /(1 - 0.5) = 2, the DC gain.
At $ω = π$ : $e^{- j π} = - 1$ , so $X = 1$ /(1 + 0.5) = 0.667.
At $ω = π$ /3: 0.5 $e^{- j π / 3} = 0.25$ - j0.433, so $X = 1$ /(0.75 + j0.433) = 1.1547 angle -30 degrees.

The response falls from 2 at DC to 0.667 at Nyquist, so this is a low pass system, which is what a decaying exponential impulse response should give.

The multiplication property on this card is the whole explanation of the window method and of Gibbs oscillation. Evaluated on the unit circle it is the z-transform; sampled at N points it is the DFT.

See it drawnThe transform of a sequence is continuous, and repeats every

2 π

The sequence is discrete, so its spectrum is continuous and periodic. That periodicity is why only minus

π

π

is ever drawn.

1.7Discrete time system properties

The five system properties, and how to test each one HOT 5/19

82 Ba · 80 Bh · 76 Ch · 75 Ash · 74 Ch3+32+2+22+2

Before the formulas

The signals are finished. From here to the end of the course the subject is systems: something you put a signal into and get another signal out. Before you can do anything useful with one you have to know what kind it is, because nearly every technique that follows is only legal for one particular kind, the linear time invariant one.

Convolution, transfer functions, frequency response: all of them assume it. So this section is five tests, one per property, each a short calculation. The exam asks them as a list. Do them in the order on the card and three of the five are usually settled in a line each.

The paper gives a relation between $x [n]$ and $y [n]$ and asks for two or three of these. Each test is three lines. Do them in this order and use the same layout every time.

Property	Definition	The test
Static or dynamic	Static, or memoryless, if $y [n]$ depends only on $x [n]$ at the same n	Look for any $x [n - k]$ , $x [n + k]$ or a sum over past values. Any of them makes it dynamic.
Linear	Obeys superposition: the response to $a_{1} x_{1} [n] + a_{2} x_{2} [n]$ is $a_{1} y_{1} [n] + a_{2} y_{2} [n]$	Compute the output for the weighted sum of two inputs, compute the weighted sum of the two separate outputs, and compare.
Time invariant	A shift of the input produces the same shift of the output	Compute y[n, k], the response to $x [n - k]$ . Compute $y [n - k]$ by replacing n with n-k in the output expression. They must be identical.
Causal	$y [n]$ depends only on present and past inputs	Look for any $x [n + k]$ with k positive, or any index that runs ahead of n.
Stable (BIBO)	Every bounded input gives a bounded output	Assume $\| x [n] \|$ ≤ Mx and bound \| $y [n]$ \|. For an LTI system the test is the much simpler one below.

\begin{aligned} linearity & : T {a_{1} x_{1} [n] + a_{2} x_{2} [n]} = a_{1} T {x_{1} [n]} + a_{2} T {x_{2} [n]} \\ time invariance & : y [n, k] = y [n - k], y [n, k] = T {x [n - k]} \\ causality & : h [n] = 0 for n < 0 (LTI form) \\ BIBO stability & : \sum_{n} | h [n] | < \infty (LTI form) \end{aligned}

The five relations the paper keeps using

$y [n] = x^{2} [n]$ . Non linear, because the square of a sum is not the sum of the squares. Time invariant, static, causal, and stable.
$y [n] = x [- n]$ . Linear, and not time invariant: shifting the input shifts the output the other way. Not causal, because $y [- 1]$ needs $x [1]$ .
$y [n] = x [n] + x [- n]$ . Linear, not time invariant, and not causal, for the same reason.
$y [n] = x [n^{2}]$ . Linear, because it only re-indexes, and not time invariant.
$y [n]$ = sum from $k = 0$ to n of $x [k]$ . The accumulator. It has memory, it is linear, it is causal, it is time variant because the lower limit is fixed at 0 rather than moving with n, and it is unstable: a bounded input such as $u [n]$ gives $y [n] = n$ + 1, which grows without bound.
$y [n]$ = cos( $5 π$ n/8 + $π / 4$ ). There is no $x [n]$ in it at all, so the output does not respond to the input: the zero input gives a non zero output, which breaks superposition. Non linear.

In the exam

Show both sides. A linearity answer that only states the conclusion scores about half. Write T{a1x1 + a2x2}, write a1y1 + a2y2, then say whether they match.
For time invariance, keep the two things apart on the page: y[n,k] on one line, $y [n - k]$ on the next. Mixing them is the usual error.
An accumulator is the standard trap. It looks causal and harmless and it is both time variant and unstable.

For an LTI system all five tests collapse into statements about $h [n]$ : the convolution card gives them, and chapter 3 turns them into statements about the ROC.

See it drawnThe five tests, and the order to do them in

Each is a separate mark. Do them one at a time and show the test, not just the verdict.

Practise thisEvery question the paper has set on this topic, worked in full5

Practise this2082 Baishakh · Q12+2 marksA discrete time system has input

x [n]

and output

y [n]

and its input-output relation is

y [n] = x [n] + x [- n]

. Is the system causal, time invariant or not?What is different herea different system:

y [n] = x [n] + x [- n]

. The fold is what breaks both time invariance and causality: at

n = - 2

the output needs

x [2]

, which has not arrived, and delaying the input does not delay the folded half with it.

What is different in this one

4 properties asked here: linear, time invariant, causal, stable. Marks are given per property, so answer every one even when it is obvious.
The linear, time invariant, stable verdicts below were checked by RUNNING the system, not asserted: on a weighted sum of two random inputs, on a shifted input, and on a bounded step.
The fold is what breaks both time invariance and causality: at $n = - 2$ the output needs $x [2]$ , which has not arrived, and delaying the input does not delay the folded half with it.

Take the properties one at a time and show the test for each. A bare yes or no with no working scores nothing.

\begin{aligned} y [n] & = x [n] + x [- n] \end{aligned}

Property	The test	Verdict
Linear	is T{a $x_{1} + b x_{2}$ } the same as a T{ $x_{1}$ } + b T{ $x_{2}$ }?	yes [tested]
Time invariant	is T{ $x [n - k]$ } the same as $y [n - k]$ , the explicit n included?	no [tested]
Causal	does $y [n]$ ever need an input from the future?	no
Stable	does every bounded input give a bounded output?	yes [tested]

Linear?

\begin{aligned} T (a x_{1} [n] + b x_{2} [n]) \\ = (a x_{1} [n] + b x_{2} [n]) + (a x_{1} [- n] + b x_{2} [- n]) \\ = a (x_{1} [n] + x_{1} [- n]) + b (x_{2} [n] + x_{2} [- n]) \\ = a y_{1} [n] + b y_{2} [n] \end{aligned}

The two sides are identical for every a and b, so it is LINEAR.

Time invariant?

\begin{aligned} delay the input \\ T (x [n - k]) [n] & = x [n - k] + x [- n - k] \\ delay the output \\ y [n - k] & = x [n - k] + x [- (n - k)] = x [n - k] + x [- n + k] \\ compare the second terms \\ x [- n - k] \\ x [- n + k] \end{aligned}

They agree only when $k = 0$ , so it is NOT time invariant. The fold reverses the shift instead of carrying it.

Causal?

\begin{aligned} y [- 2] & = x [- 2] + x [- (- 2)] = x [- 2] + x [2] \end{aligned}

The output at $n = - 2$ needs $x [2]$ , two samples into the future, so it is NOT causal.

Stable?

\begin{aligned} bound the input \\ | x [n] | \leq B & for every n \\ then \\ | y [n] | & = | x [n] + x [- n] | \leq | x [n] | + | x [- n] | \leq 2 B \end{aligned}

A bound in gives a bound out, so it is STABLE.

The fold is what breaks both time invariance and causality: at $n = - 2$ the output needs $x [2]$ , which has not arrived, and delaying the input does not delay the folded half with it.

The system is linear, not time invariant, not causal, stable. Each verdict is worked above, and each one was also checked by RUNNING the system: a weighted sum of two random inputs for linearity, a shifted input for time invariance, and a bounded input for stability. The algebra is what earns the marks; the run is what stops a wrong verdict reaching this page.

Practise this2080 Bhadra · Q12+2+2 marksA discrete time system has input

x (n)

and output

y (n)

. The input output relation of the system is given by

y (n)

= sum from k=0 to n of

x (k)

. Check whether the system is memory less, time invariant and stable or not?What is different hereThe lower limit is pinned at 0 rather than moving with n, and that alone makes the accumulator time VARIANT. It is unstable too: the bounded input

u [n]

gives

y [n] = n

+ 1, which is the disproof to write down.

What is different in this one

5 properties asked here: static, linear, time invariant, causal, stable. Marks are given per property, so answer every one even when it is obvious.
The linear, time invariant, stable verdicts below were checked by RUNNING the system, not asserted: on a weighted sum of two random inputs, on a shifted input, and on a bounded step.

Take the properties one at a time and show the test for each. A bare yes or no with no working scores nothing.

\begin{aligned} y [n] & = sum  from k = 0 to n of x [k] & the accumulator \end{aligned}

Property	The test	Verdict
Static	does $y [n]$ use x at any other index, or a past output?	no
Linear	is T{a $x_{1} + b x_{2}$ } the same as a T{ $x_{1}$ } + b T{ $x_{2}$ }?	yes [tested]
Time invariant	is T{ $x [n - k]$ } the same as $y [n - k]$ , the explicit n included?	no [tested]
Causal	does $y [n]$ ever need an input from the future?	yes
Stable	does every bounded input give a bounded output?	no [tested]

Static?

\begin{aligned} y [3] & = x [0] + x [1] + x [2] + x [3] \end{aligned}

It needs every sample since 0, not $x [3]$ alone, so it has MEMORY and is not static.

Linear?

\begin{aligned} T (a x_{1} + b x_{2}) [n] \\ = \sum over k = 0 to n of (a x_{1} [k] + b x_{2} [k]) \\ = a \sumx_{1} [k] + b \sumx_{2} [k] \\ = a y_{1} [n] + b y_{2} [n] \end{aligned}

A sum splits over addition, so it is LINEAR.

Time invariant?

\begin{aligned} delay the input \\ T (x [n - d]) [n] & = \sum over k = 0 to n of x [k - d] \\ = \sum over m = - d to n - d of x [m] \\ delay the output \\ y [n - d] & = \sum over k = 0 to n - d of x [k] \\ compare the lower limits \\ - d \\ 0 \end{aligned}

The delayed input drags d extra samples into the sum, so it is NOT time invariant. The lower limit pinned at 0 is what breaks it.

Causal?

\begin{aligned} y [n] & = x [0] + x [1] + \dots + x [n] \\ the largest index used is n \end{aligned}

Nothing beyond n is ever read, so it is CAUSAL.

Stable?

\begin{aligned} take  the  bounded  input x [n] & = u [n], so | x [n] | \leq 1 \\ y [0] & = 1, y [1] = 2, y [2] = 3, \dots, y [n] = n + 1 \end{aligned}

A bounded input gives an output that grows without bound, so it is NOT stable.

The system is not static, linear, not time invariant, causal, not stable. Each verdict is worked above, and each one was also checked by RUNNING the system: a weighted sum of two random inputs for linearity, a shifted input for time invariance, and a bounded input for stability. The algebra is what earns the marks; the run is what stops a wrong verdict reaching this page.

Practise this2076 Chaitra · Q25 marksDetermine whether the following system are: a)

y [n] = x [- n]

is time-invariant or not. b)

y [n]

= x[

n^{2}

] is linear or not.

a) is $y [n] = x [- n]$ time invariant?

What is different in this one

4 properties asked here: linear, time invariant, causal, stable. Marks are given per property, so answer every one even when it is obvious.
The linear, time invariant, stable verdicts below were checked by RUNNING the system, not asserted: on a weighted sum of two random inputs, on a shifted input, and on a bounded step.

Take the properties one at a time and show the test for each. A bare yes or no with no working scores nothing.

\begin{aligned} y [n] & = x [- n] \end{aligned}

Property	The test	Verdict
Linear	is T{a $x_{1} + b x_{2}$ } the same as a T{ $x_{1}$ } + b T{ $x_{2}$ }?	yes [tested]
Time invariant	is T{ $x [n - k]$ } the same as $y [n - k]$ , the explicit n included?	no [tested]
Causal	does $y [n]$ ever need an input from the future?	no
Stable	does every bounded input give a bounded output?	yes [tested]

Linear?

\begin{aligned} T (a x_{1} [n] + b x_{2} [n]) & = a x_{1} [- n] + b x_{2} [- n] \\ = a y_{1} [n] + b y_{2} [n] \end{aligned}

Identical for every a and b, so it is LINEAR.

Time invariant?

\begin{aligned} delay the input \\ T (x [n - k]) [n] & = x [- n - k] \\ delay the output \\ y [n - k] & = x [- (n - k)] = x [- n + k] \end{aligned}

$X [- n - k]$ against $x [- n + k]$ : equal only when $k = 0$ , so it is NOT time invariant.

Causal?

\begin{aligned} y [- 1] & = x [- (- 1)] = x [1] \end{aligned}

The output at $n = - 1$ needs the input one sample ahead, so it is NOT causal.

Stable?

\begin{aligned} bound the input \\ | x [n] | \leq B & for every n \\ then \\ | y [n] | & = | x [- n] | \leq B \end{aligned}

The fold moves samples about and never makes one larger, so it is STABLE.

b) is $y [n] = x [n^{2}]$ linear?

What is different in this one

2 properties asked here: linear, time invariant. Marks are given per property, so answer every one even when it is obvious.
The linear, time invariant verdicts below were checked by RUNNING the system, not asserted: on a weighted sum of two random inputs, on a shifted input, and on a bounded step.

Take the properties one at a time and show the test for each. A bare yes or no with no working scores nothing.

\begin{aligned} y [n] & = x [n^{2}] \end{aligned}

Property	The test	Verdict
Linear	is T{a $x_{1} + b x_{2}$ } the same as a T{ $x_{1}$ } + b T{ $x_{2}$ }?	yes [tested]
Time invariant	is T{ $x [n - k]$ } the same as $y [n - k]$ , the explicit n included?	no [tested]

Linear?

\begin{aligned} T (a x_{1} [n] + b x_{2} [n]) & = a x_{1} [n^{2}] + b x_{2} [n^{2}] \\ = a y_{1} [n] + b y_{2} [n] \end{aligned}

Reading the input at a different index is still linear in the input, so it is LINEAR.

Time invariant?

\begin{aligned} delay the input \\ T (x [n - k]) [n] & = x [n^{2} - k] \\ delay the output \\ y [n - k] & = x [{(n - k)}^{2}] = x [n^{2} - 2 kn + k^{2}] \end{aligned}

The two indices differ by -2kn + $k^{2}$ , which depends on n, so it is NOT time invariant.

The system is linear, not time invariant. Each verdict is worked above, and each one was also checked by RUNNING the system: a weighted sum of two random inputs for linearity, a shifted input for time invariance, and a bounded input for stability. The algebra is what earns the marks; the run is what stops a wrong verdict reaching this page.

Practise this2075 Ashwin · Q13+3 marksDetermine whether the following sequences are linear or not: a)

y [n] = x^{2} [n]

y [n] = \cos (5 π n / 8 + π / 4)

What is different herea different system:

y [n] = x^{2} [n]

; a different system:

y [n] = \cos (5 π n / 8 + π / 4)

How this is readThe second has no $x [n]$ in it at all, so a zero input gives a non zero output and superposition fails at the first line. That is the whole answer.

a) is $y [n] = x^{2} [n]$ linear?

What is different in this one

4 properties asked here: linear, time invariant, causal, stable. Marks are given per property, so answer every one even when it is obvious.
The linear, time invariant, stable verdicts below were checked by RUNNING the system, not asserted: on a weighted sum of two random inputs, on a shifted input, and on a bounded step.

Take the properties one at a time and show the test for each. A bare yes or no with no working scores nothing.

\begin{aligned} y [n] & = x^{2} [n] \end{aligned}

Property	The test	Verdict
Linear	is T{a $x_{1} + b x_{2}$ } the same as a T{ $x_{1}$ } + b T{ $x_{2}$ }?	no [tested]
Time invariant	is T{ $x [n - k]$ } the same as $y [n - k]$ , the explicit n included?	yes [tested]
Causal	does $y [n]$ ever need an input from the future?	yes
Stable	does every bounded input give a bounded output?	yes [tested]

Linear?

\begin{aligned} T (a x_{1} [n] + b x_{2} [n]) & = {(a x 1 [n] + b x 2 [n])}^{2} \\ = a^{2} x_{1}^{2} + 2 ab x_{1} x_{2} + b^{2} x_{2}^{2} \\ a y_{1} [n] + b y_{2} [n] & = a x_{1}^{2} + b x_{2}^{2} \\ one  counterexample  settles  it, with x_{1} & = x_{2} = 1 and a = b = 1 \\ T (1 + 1) & = 2^{2} = 4 T (1) + T (1) = 1 + 1 = 2 \end{aligned}

4 is not 2, so it is NOT linear. Squaring is the classic nonlinearity.

Time invariant?

\begin{aligned} delay the input \\ T (x [n - k]) [n] & = x^{2} [n - k] \\ delay the output \\ y [n - k] & = x^{2} [n - k] \end{aligned}

The same expression on both sides, so it is TIME INVARIANT. There is no explicit n anywhere in it.

Causal?

\begin{aligned} y [n] & = x^{2} [n] \\ the only index used is n \end{aligned}

It uses the present input alone, so it is CAUSAL, and static as well.

Stable?

\begin{aligned} bound the input \\ | x [n] | \leq B & for every n \\ then \\ | y [n] | & = | x [n] |^{2} \leq B^{2} \end{aligned}

A finite bound squared is still finite, so it is STABLE.

The system is not linear, time invariant, causal, stable. Each verdict is worked above, and each one was also checked by RUNNING the system: a weighted sum of two random inputs for linearity, a shifted input for time invariance, and a bounded input for stability. The algebra is what earns the marks; the run is what stops a wrong verdict reaching this page.

b) is $y [n] = \cos (5 π n / 8 + π / 4)$ linear?

What is different in this one

2 properties asked here: linear, time invariant. Marks are given per property, so answer every one even when it is obvious.
The linear, time invariant verdicts below were checked by RUNNING the system, not asserted: on a weighted sum of two random inputs, on a shifted input, and on a bounded step.

Take the properties one at a time and show the test for each. A bare yes or no with no working scores nothing.

\begin{aligned} y [n] & = \cos (5 π n / 8 + π / 4) \end{aligned}

Property	The test	Verdict
Linear	is T{a $x_{1} + b x_{2}$ } the same as a T{ $x_{1}$ } + b T{ $x_{2}$ }?	no [tested]
Time invariant	is T{ $x [n - k]$ } the same as $y [n - k]$ , the explicit n included?	no [tested]

Linear?

\begin{aligned} the input never appears on the right hand side at all \\ put x [n] & = 0 \\ y [n] & = \cos (5 π n / 8 + π / 4), which  is  not 0 \end{aligned}

A linear system must give zero out for zero in, and this one does not, so it is NOT linear.

Time invariant?

\begin{aligned} delay the input, and nothing happens, since x is ignored \\ T (x [n - k]) [n] & = \cos (5 π n / 8 + π / 4) \\ delay the output, and the cosine moves \\ y [n - k] & = \cos (5 π (n - k) / 8 + π / 4) \end{aligned}

Delaying the input does nothing while delaying the output shifts the cosine, so it is NOT time invariant. The explicit n is what breaks it.

The system is not linear, not time invariant. Each verdict is worked above, and each one was also checked by RUNNING the system: a weighted sum of two random inputs for linearity, a shifted input for time invariance, and a bounded input for stability. The algebra is what earns the marks; the run is what stops a wrong verdict reaching this page.

Practise this2074 Chaitra · Q23+3 marksDetermine whether the following system are: a)

y [n] = y [n - 4] + x [n - 4]

is Time-invariant or not b)

y [n] = x^{2} [n]

is Linear or Non-linearWhat is different herea different system:

y [n] = y [n - 4] + x [n - 4]

; a different system:

y [n] = x^{2} [n]

How this is readThe recursion $y [n] = y [n - 4] + x [n - 4]$ with zero initial conditions accumulates every fourth past input, which is how it is written for the test.

a) is $y [n] = y [n - 4] + x [n - 4]$ time invariant?

What is different in this one

4 properties asked here: linear, time invariant, causal, stable. Marks are given per property, so answer every one even when it is obvious.
The linear, time invariant, stable verdicts below were checked by RUNNING the system, not asserted: on a weighted sum of two random inputs, on a shifted input, and on a bounded step.

Take the properties one at a time and show the test for each. A bare yes or no with no working scores nothing.

\begin{aligned} y [n] & = y [n - 4] + x [n - 4] \end{aligned}

Property	The test	Verdict
Linear	is T{a $x_{1} + b x_{2}$ } the same as a T{ $x_{1}$ } + b T{ $x_{2}$ }?	yes [tested]
Time invariant	is T{ $x [n - k]$ } the same as $y [n - k]$ , the explicit n included?	yes [tested]
Causal	does $y [n]$ ever need an input from the future?	yes
Stable	does every bounded input give a bounded output?	no [tested]

Linear?

\begin{aligned} every term is a constant times one sample of x or of y \\ nothing is squared, multiplied together, or passed \\ through a function, and the system starts at rest \end{aligned}

A constant coefficient difference equation at rest is LINEAR.

Time invariant?

\begin{aligned} the coefficients are constants and both delays are \\ fixed at 4, and there is no explicit n anywhere \end{aligned}

Delaying the input delays the whole recursion with it, so it is TIME INVARIANT.

Causal?

\begin{aligned} y [n] & = y [n - 4] + x [n - 4] \\ the  largest  index  used  is n - 4 \end{aligned}

It reads only the past, so it is CAUSAL.

Stable?

\begin{aligned} take  the  bounded  input x [n] & = δ [n] \\ y [4] & = 1, y [8] = 1, y [12] = 1, \dots & for ever \\ \sum | h [n] | & = 1 + 1 + 1 + \dots = infinity \end{aligned}

The impulse response never decays, so it is NOT stable. The pole sits ON the unit circle.

The system is linear, time invariant, causal, not stable. Each verdict is worked above, and each one was also checked by RUNNING the system: a weighted sum of two random inputs for linearity, a shifted input for time invariance, and a bounded input for stability. The algebra is what earns the marks; the run is what stops a wrong verdict reaching this page.

b) is $y [n] = x^{2} [n]$ linear?

What is different in this one

2 properties asked here: linear, time invariant. Marks are given per property, so answer every one even when it is obvious.
The linear, time invariant verdicts below were checked by RUNNING the system, not asserted: on a weighted sum of two random inputs, on a shifted input, and on a bounded step.

Take the properties one at a time and show the test for each. A bare yes or no with no working scores nothing.

\begin{aligned} y [n] & = x^{2} [n] \end{aligned}

Property	The test	Verdict
Linear	is T{a $x_{1} + b x_{2}$ } the same as a T{ $x_{1}$ } + b T{ $x_{2}$ }?	no [tested]
Time invariant	is T{ $x [n - k]$ } the same as $y [n - k]$ , the explicit n included?	yes [tested]

Linear?

\begin{aligned} T (a x_{1} [n] + b x_{2} [n]) & = {(a x 1 [n] + b x 2 [n])}^{2} \\ = a^{2} x_{1}^{2} + 2 ab x_{1} x_{2} + b^{2} x_{2}^{2} \\ a y_{1} [n] + b y_{2} [n] & = a x_{1}^{2} + b x_{2}^{2} \\ one  counterexample  settles  it, with x_{1} & = x_{2} = 1 and a = b = 1 \\ T (1 + 1) & = 2^{2} = 4 T (1) + T (1) = 1 + 1 = 2 \end{aligned}

4 is not 2, so it is NOT linear. Squaring is the classic nonlinearity.

Time invariant?

\begin{aligned} delay the input \\ T (x [n - k]) [n] & = x^{2} [n - k] \\ delay the output \\ y [n - k] & = x^{2} [n - k] \end{aligned}

The same expression on both sides, so it is TIME INVARIANT. There is no explicit n anywhere in it.

The system is not linear, time invariant. Each verdict is worked above, and each one was also checked by RUNNING the system: a weighted sum of two random inputs for linearity, a shifted input for time invariance, and a bounded input for stability. The algebra is what earns the marks; the run is what stops a wrong verdict reaching this page.

The same questions grouped by how often each topic is set: testing a system for its properties

1.8LTI systems: the convolution sum and its properties

The convolution sum, and four ways to do it TOP 18/19

82 Bh · 82 Ba · 81 Bh · 81 Ba · 80 Ba · 79 Bh · 79 Ba · 78 Bh · 76 Ch · 76 Ash · 75 Ch · 75 Ash · 74 Ch · 74 Ash · 73 Shr · 72 Ch · 72 Ka · 71 Shr563+2

Before the formulas

This is the payoff for 1.7. If a system is linear and time invariant, then knowing what it does to a single impulse tells you what it does to everything, because every signal is a row of impulses with different heights.

The sum that assembles that answer is convolution, and it is the single most asked thing in the paper. It is not hard: flip one of the two signals, slide it along, multiply and add, once for each output sample.

The work is bookkeeping, not thinking, which is exactly why a method matters. The card gives four ways to carry it out. Pick the one that suits the numbers in front of you, and practise all four at least once, because the paper sometimes names the method it wants.

Why the impulse response is enough Write the input as a sum of shifted impulses,

x [n] = \sum_{k} x [k] δ [n - k]

. Apply a linear system: the response is the sum of the responses,

\sum_{k} x [k]

δ [n - k]}

. Apply time invariance: T{

δ [n - k]}

is just

h [n - k]

. That gives the convolution sum, and it is why

h [n]

alone describes an LTI system completely.

\begin{aligned} y [n] & = \sum_{k = - \infty}^{\infty} x [k] h [n - k] = x [n] * h [n] \\ = \sum_{k = - \infty}^{\infty} h [k] x [n - k] & commutative: fold whichever is shorter \end{aligned}

The four methods, and when each is quickest

Tabular, or the multiplication method. Write x along the top and h down the side, fill the grid with products, and add along the anti diagonals. Fastest for two short finite sequences, which is what the exam gives. The output starts at n = (first index of x) + (first index of h) and has length $N_{x} + N_{h} - 1$ .
Graphical. Fold $h [k]$ to $h [- k]$ , slide it by n, multiply overlapping samples, add. Slow, but it is what the question means when it says using graphical method, and you must draw the overlap at each n.
Analytical. For infinite sequences such as $a^{n} u [n]$ against $u [n]$ , write the sum, fix the limits from where both sequences are non zero, and sum the geometric series. This is the method for anything with $u [n]$ in it.
By transform. $Y (z) = X (z) H (z)$ , then invert. Usually slower in an exam unless the z-transforms are already in front of you.

The length and position rule If x runs from n1 to n2 and h from m1 to m2, then y runs from n1 + m1 to n2 + m2, and its length is (n2 - n1 + 1) + (m2 - m1 + 1) - 1.
The free check: sum of y = (sum of x) times (sum of h). Do this every time; it catches an arithmetic slip in one line.

Properties of convolution, and therefore of LTI systems

Property	Statement	What it means for systems
Commutative	x * h = h * x	Which block you call the input does not matter
Associative	(x * $h_{1}$ ) * $h_{2} = x$ * ( $h_{1} * h_{2}$ )	Two systems in cascade have impulse response $h_{1} * h_{2}$
Distributive	x * ( $h_{1} + h_{2}$ ) = x * $h_{1}$ + x * $h_{2}$	Two systems in parallel have impulse response $h_{1} + h_{2}$
Identity	x * $δ [n] = x [n]$	The impulse is the do nothing system
Shift	$x [n] * δ [n - k] = x [n - k]$	Convolving with a shifted impulse is a delay

In the exam

Write both index ranges down before you start. Most lost marks here are an output placed at the wrong n, not a wrong number.
Use the sum check. It takes five seconds and the question often says also check the answer, which is asking for exactly this.
When the input is a complex exponential, do not convolve at all: use the eigenfunction result in chapter 3, $y [n] = H (e^{j ω_{0}}) x [n]$ . Three papers have set that as a convolution question to see whether you notice.

Convolution is the thread of the subject. It becomes a product under the z-transform, a product again under the DTFT, and a circular convolution under the DFT. Its cascade and parallel properties are what the cascade and parallel structures realize.

See it drawnFold, slide, multiply, add

The impulse response is folded once and then slid one sample at a time. Each position of the slide gives one output sample.

Practise thisEvery question the paper has set on this topic, worked in full20

Practise this +2 like it2082 Bhadra · Q25 marksFind the output of an LTI system whose impulse response is given by

{0.5}^{n} {u [n] - u [n - 3]}

and input is given by

{2, 1, 0.5, - 1}

.What is different here

h [n]

starts at

n = 0

, so the answer starts where

x [n]

does

How this is read $h [n] = {0.5}^{n} {u [n] - u [n - 3]}$ is the exponential kept for $n = 0$ , 1, 2 only, so $h [n] = {1, 0.5, 0.25}$ .

What is different in this one

Two finite sequences, 4 samples and 3, so the answer has 4 + 3 - 1 = 6 samples and the table is 4 by 3.

$x [n]$ runs from $n = 0$ to 3, 4 samples; $h [n]$ runs from $n = 0$ to 2, 3 samples. So $y [n]$ runs from $n = 0$ to 5, which is 4 + 3 - 1 = 6 samples.

Multiply every sample of one by every sample of the other and add along the anti diagonals, which is the tabular method:

	$h [0] = 1$	$h [1] = 0.5$	$h [2] = 0.25$
$x [0] = 2$	2	1	0.5
$x [1] = 1$	1	0.5	0.25
$x [2] = 0.5$	0.5	0.25	0.125
$x [3] = - 1$	-1	-0.5	-0.25

y [0] = 2

= 2

y [1] = 1

+ 1 = 2

y [2] = 0.5

+ 0.5 + 0.5 = 1.5

y [3] = 0.25

+ 0.25 + -1 = -0.5

y [4] = 0.125

+ -0.5 = -0.375

y [5] = - 0.25

= -0.25

$y [n]$ = {2, 2, 1.5, -0.5, -0.375, -0.25} for $n = 0$ to 5. The check holds: the samples of x add to 2.5, those of h add to 1.75, and 2.5 × 1.75 = 4.375, which is the sum of y.

Also worked under The output for an exponential inputPractise this +2 like it2082 Baishakh · Q25 marksFind the output of an LTI system having impulse response

h [n] = {(1 / 2)}^{n} u [n]

excited by an input

x [n] = 5 e^{j π n / 3}

What is different in this one

1 component at one frequency, so H is evaluated once and the results added: superposition.
Do not convolve. The input runs over all n, so it is not a finite sequence and the graphical method is the wrong tool.

A complex exponential is an eigenfunction of an LTI system: it comes out as itself, scaled by one complex number. A sine or a cosine is the sum of two of them, so the same rule applies to it, and the whole method is to evaluate H once per frequency and add.

x [n] = e^{j ω_{0} n} ⟶ y [n] = H (e^{j ω_{0}}) e^{j ω_{0} n}, A \cos (ω_{0} n) ⟶ A | H | \cos (ω_{0} n + ∠ H)

\begin{aligned} H (e^{j ω}) & = 1 / (1 - 0.5 e^{- j ω}), the  transform  of h [n] = {(1 / 2)}^{n} u [n] \end{aligned}

Component	$ω_{0}$	$H (e^{j ω_{0}})$	in polar form	Output amplitude
$x [n] = 5 e^{j π n / 3}$	$0.3333 π$	1 - j0.5774	1.1547 angle -30 deg	5.7735

The check that costs nothing. The whole method rests on not convolving, so confirm the closed form against the sum it replaced: $y [n] = \sum_{k} h [k] x [n - k]$ , carried far enough that the tail is negligible.

Where	By convolution	By the formula
$n = 0$	5	5
$n = 1$	5	5
$n = 2$	0	0
$n = 5$	0	0

$y [n] = 5.7735 e^{j (0.3333 π n - 30 d e g)}$ .

Practise this +4 like it2081 Bhadra · Q25 marksFind the output of an LTI System with impulse response

h [n] = 2 δ [n + 1] + 2 δ [n - 1]

when an input of

x [n] = δ [n] + 2 δ [n - 1] - δ [n - 3]

is applied to it.

How this is read $h [n] = 2 δ [n + 1] + 2 δ [n - 1]$ is ${2, 0, 2}$ starting at $n = - 1$ ; $x [n] = δ [n] + 2 δ [n - 1] - δ [n - 3]$ is ${1, 2, 0, - 1}$ from $n = 0$ .

What is different in this one

Two finite sequences, 4 samples and 3, so the answer has 4 + 3 - 1 = 6 samples and the table is 4 by 3.
One of the sequences starts at a negative n, so the output starts at $n = - 1$ , not at zero. Mark the origin on the answer or the plot is wrong even when the numbers are right.
There is a zero sample in the middle of one sequence. It still occupies an index: dropping it shortens the answer and shifts everything after it.

$x [n]$ runs from $n = 0$ to 3, 4 samples; $h [n]$ runs from $n = - 1$ to 1, 3 samples. So $y [n]$ runs from $n = - 1$ to 4, which is 4 + 3 - 1 = 6 samples.

Multiply every sample of one by every sample of the other and add along the anti diagonals, which is the tabular method:

	$h [- 1] = 2$	$h [1] = 2$
$x [0] = 1$	2	2
$x [1] = 2$	4	4
$x [2] = 0$	0	0
$x [3] = - 1$	-2	-2

y [- 1] = 2

= 2

y [0] = 4

= 4

y [1] = 2

= 2

y [2] = 4

+ -2 = 2

y [3] = 0

= 0

y [4] = - 2

= -2

$y [n]$ = {2, 4, 2, 2, 0, -2} for $n = - 1$ to 4. The check holds: the samples of x add to 2, those of h add to 4, and 2 × 4 = 8, which is the sum of y.

Practise this +3 like it2081 Baishakh · Q26 marksFind the output

y (n)

of LTI system having impulse response

h (n) = {(1 / 3)}^{n} u (n - 3)

and input

x (n) = {(1 / 6)}^{n - 6} u (n)

What is different in this one

At least one of the sequences is infinite, so this is an analytical convolution and not a table. Fix the limits of the sum from where BOTH are non zero, then sum the geometric series.
The answer comes out in regions: one while the overlap is still growing and one once it is complete. They must agree at the join, and that is the check.

Neither sequence is finite, so this is an analytical convolution, not a table.

h [k] = {(1 / 3)}^{k}

for k ≥ 3,

x [m] = {(1 / 6)}^{m - 6} u [m] = 6^{6} {(1 / 6)}^{m}

for m ≥ 0

y [n] = \sum_{k} h [k] x [n - k]

h needs k ≥ 3 and x needs n - k ≥ 0, so k runs from 3 to n, and

y [n] = 0

for n < 3. n

y [n] = 6^{6} \sum {(1 / 3)}^{k} {(1 / 6)}^{n - k}

k=3 n =

6^{6} {(1 / 6)}^{n} \sum {(6 / 3)}^{k}

k=3 n = 46656

{(1 / 6)}^{n}

[

2^{n + 1} - 2^{3}

] since

\sum 2^{k} = 2^{n + 1}

- 8 k=3

Check the closed form against the convolution sum carried out term by term, at several n. They agree, which is what makes the closed form an answer rather than a guess.

n	$\sum_{k} h [k] x [n - k]$	the closed form
3	1728	1728
4	864	864
5	336	336
6	120	120
8	14	14
10	1.57407	1.57407

$y [n] = 46656 {(1 / 6)}^{n}$ [ $2^{n + 1}$ - 8 ] for n ≥ 3, and zero before that. It starts at $y [3] = 1728$ and decays like ${(1 / 3)}^{n}$ , since the ${(1 / 6)}^{n}$ and the $2^{n}$ combine to ${(1 / 3)}^{n}$ .

Also worked under The output for an exponential inputSame as 2082 Baishakh Q22080 Baishakh · Q25 marksFind the output of LTI system having impulse response

h [n] = {(1 / 2)}^{n} u [n]

and input

x [n] = 5 e^{j π n / 3}

for -infinity < n < infinity.

What is different in this one

1 component at one frequency, so H is evaluated once and the results added: superposition.
Do not convolve. The input runs over all n, so it is not a finite sequence and the graphical method is the wrong tool.

x [n] = e^{j ω_{0} n} ⟶ y [n] = H (e^{j ω_{0}}) e^{j ω_{0} n}, A \cos (ω_{0} n) ⟶ A | H | \cos (ω_{0} n + ∠ H)

\begin{aligned} H (e^{j ω}) & = 1 / (1 - 0.5 e^{- j ω}), the  transform  of h [n] = {(1 / 2)}^{n} u [n] \end{aligned}

Component	$ω_{0}$	$H (e^{j ω_{0}})$	in polar form	Output amplitude
$x [n] = 5 e^{j π n / 3}$	$0.3333 π$	1 - j0.5774	1.1547 angle -30 deg	5.7735

Where	By convolution	By the formula
$n = 0$	5	5
$n = 1$	5	5
$n = 2$	0	0
$n = 5$	0	0

$y [n] = 5.7735 e^{j (0.3333 π n - 30 d e g)}$ .

Practise this2079 Bhadra · Q25 marksFind the output of LTI system having input signal

x [n] = δ [n] + 2 δ [n - 1] - δ [n - 3]

and

h [n] = 2 δ [n + 1] + 2 δ [n - 1]

.What is different hereTwo transcriptions of this paper disagree about the input, and the past paper itself is not on this machine to settle it. This one reads

x [n] = δ [n] + 2 δ [n - 1] - δ [n - 3]

, so x =

{1, 2, 0, - 1}

from

n = 0

. The standard text works the same paper as its Example 1.26 and reads

x [n] = δ [n + 2] + δ [n - 1] - δ [n - 3]

, so x =

{1, 0, 0, 1, 0, - 1}

from n = -2.

h [n] = 2 δ [n + 1] + 2 δ [n - 1]

either way. The method is identical and so is the shape of the answer,

y [n] = 2 x [n + 1] + 2 x [n - 1]

: only which samples are non zero moves. Work it with whichever your paper shows; practising one gives you the other.

What is different in this one

Two finite sequences, 4 samples and 3, so the answer has 4 + 3 - 1 = 6 samples and the table is 4 by 3.
One of the sequences starts at a negative n, so the output starts at $n = - 1$ , not at zero. Mark the origin on the answer or the plot is wrong even when the numbers are right.
There is a zero sample in the middle of one sequence. It still occupies an index: dropping it shortens the answer and shifts everything after it.

$x [n]$ runs from $n = 0$ to 3, 4 samples; $h [n]$ runs from $n = - 1$ to 1, 3 samples. So $y [n]$ runs from $n = - 1$ to 4, which is 4 + 3 - 1 = 6 samples.

Multiply every sample of one by every sample of the other and add along the anti diagonals, which is the tabular method:

	$h [- 1] = 2$	$h [1] = 2$
$x [0] = 1$	2	2
$x [1] = 2$	4	4
$x [2] = 0$	0	0
$x [3] = - 1$	-2	-2

y [- 1] = 2

= 2

y [0] = 4

= 4

y [1] = 2

= 2

y [2] = 4

+ -2 = 2

y [3] = 0

= 0

y [4] = - 2

= -2

$y [n]$ = {2, 4, 2, 2, 0, -2} for $n = - 1$ to 4. The check holds: the samples of x add to 2, those of h add to 4, and 2 × 4 = 8, which is the sum of y.

Same as 2081 Bhadra Q22079 Baishakh · Q25 marksFind the output of LTI system having impulse response

h [n] = {(1 / 2)}^{n} {u [n + 2] - u [n - 2]}

to the input

x [n] = {2, 1, 0, - 1, 4}

How this is read $h [n] = {(1 / 2)}^{n} {u [n + 2] - u [n - 2]}$ runs over $n = - 2$ , -1, 0, 1, taking the values 4, 2, 1 and 0.5.

What is different in this one

Two finite sequences, 5 samples and 4, so the answer has 5 + 4 - 1 = 8 samples and the table is 5 by 4.
One of the sequences starts at a negative n, so the output starts at $n = - 2$ , not at zero. Mark the origin on the answer or the plot is wrong even when the numbers are right.
There is a zero sample in the middle of one sequence. It still occupies an index: dropping it shortens the answer and shifts everything after it.

$x [n]$ runs from $n = 0$ to 4, 5 samples; $h [n]$ runs from $n = - 2$ to 1, 4 samples. So $y [n]$ runs from $n = - 2$ to 5, which is 5 + 4 - 1 = 8 samples.

Multiply every sample of one by every sample of the other and add along the anti diagonals, which is the tabular method:

	$h [- 2] = 4$	$h [- 1] = 2$	$h [0] = 1$	$h [1] = 0.5$
$x [0] = 2$	8	4	2	1
$x [1] = 1$	4	2	1	0.5
$x [2] = 0$	0	0	0	0
$x [3] = - 1$	-4	-2	-1	-0.5
$x [4] = 4$	16	8	4	2

y [- 2] = 8

= 8

y [- 1] = 4

+ 4 = 8

y [0] = 2

+ 2 = 4

y [1] = 1

+ 1 + -4 = -2

y [2] = 0.5

+ -2 + 16 = 14.5

y [3] = - 1

+ 8 = 7

y [4] = - 0.5

+ 4 = 3.5

y [5] = 2

= 2

$y [n]$ = {8, 8, 4, -2, 14.5, 7, 3.5, 2} for $n = - 2$ to 5. The check holds: the samples of x add to 6, those of h add to 7.5, and 6 × 7.5 = 45, which is the sum of y.

Same as 2081 Baishakh Q22078 Bhadra · Q25 marksFind the output of LTI system having impulse response

h [n] = u [n] - u [n - 4]

and input signal

x [n] = {(1 / 2)}^{n} u [n]

What is different in this one

At least one of the sequences is infinite, so this is an analytical convolution and not a table. Fix the limits of the sum from where BOTH are non zero, then sum the geometric series.
The answer comes out in regions: one while the overlap is still growing and one once it is complete. They must agree at the join, and that is the check.

$h [n]$ is a four sample rectangle and $x [n]$ is an infinite exponential, so the answer has two regions: while the rectangle is still sliding onto the input, and after it is fully overlapped.

\begin{aligned} h [n] & = u [n] - u [n - 4] = 1 for n = 0, 1, 2, 3 \\ x [n] & = {(1 / 2)}^{n} u [n] \\ 3 \\ y [n] & = sum h [k] x [n - k] = \sum {(1 / 2)}^{n - k}, keeping  only n - k \geq 0 \\ k & = 0 \\ PARTIAL  OVERLAP, 0 \leq n \leq 3 : k runs 0 \dots n \\ y [n] & = sum  from j = 0 to n of {(1 / 2)}^{j} = 2 - {(1 / 2)}^{n} \\ FULL  OVERLAP, n \geq 3 : k runs 0 \dots 3 \\ y [n] & = {(1 / 2)}^{n} sum  from k = 0 to 3 of 2^{k} = 15 {(1 / 2)}^{n} \end{aligned}

Check the closed form against the convolution sum carried out term by term, at several n. They agree, which is what makes the closed form an answer rather than a guess.

n	$\sum_{k} h [k] x [n - k]$	the closed form
0	1	1
1	1.5	1.5
2	1.75	1.75
3	1.875	1.875
4	0.9375	0.9375
5	0.46875	0.46875
8	0.05859	0.05859

$y [n] = 2 - {(1 / 2)}^{n}$ for $0 \leq n \leq 3$ , and 15 ${(1 / 2)}^{n}$ for n ≥ 3, zero before. The two expressions agree at $n = 3$ , where both give 1.875, which is the check worth writing.

Same as 2081 Baishakh Q22076 Chaitra · Q36 marksFind the output of LTI system having input signal

x [n] = u [n + 1] - u [n - 4]

and impulse response

h [n] = {(1 / 2)}^{n} u [n - 1]

What is different in this one

At least one of the sequences is infinite, so this is an analytical convolution and not a table. Fix the limits of the sum from where BOTH are non zero, then sum the geometric series.
The answer comes out in regions: one while the overlap is still growing and one once it is complete. They must agree at the join, and that is the check.

$x [n]$ is a five sample rectangle from $n = - 1$ and $h [n]$ is an infinite exponential that starts at $n = 1$ , so again there are two regions.

x [n] = u [n + 1] - u [n - 4] = 1

for

n = - 1

.. 3

h [n] = {(1 / 2)}^{n} u [n - 1] = {(1 / 2)}^{n}

for n ≥ 1

y [n] = \sum_{k} x [k] h [n - k]

, needing -1 ≤ k ≤ 3 and n - k ≥ 1 so k runs from -1 to min(3, n-1), and

y [n] = 0

for n < 0 K

y [n] = {(1 / 2)}^{n} \sum 2^{k}

with K = min(3, n-1) k=-1 RISING, 0 ≤ n ≤ 4: K = n-1, sum =

2^{n} - 0.5 y [n] = 1 - {(1 / 2)}^{n + 1}

FALLING, n ≥ 4:

K = 3

, sum = 16 - 0.5 = 15.5

y [n] = 15.5 {(1 / 2)}^{n}

Check the closed form against the convolution sum carried out term by term, at several n. They agree, which is what makes the closed form an answer rather than a guess.

n	$\sum_{k} h [k] x [n - k]$	the closed form
0	0.5	0.5
1	0.75	0.75
2	0.875	0.875
3	0.9375	0.9375
4	0.96875	0.96875
5	0.48438	0.48438
7	0.12109	0.12109

$y [n] = 1 - {(1 / 2)}^{n + 1}$ for $0 \leq n \leq 4$ , and 15.5 ${(1 / 2)}^{n}$ for n ≥ 4. Both give 0.96875 at $n = 4$ , which is the check.

Same as 2081 Baishakh Q22076 Ashwin · Q26 marksFind convolution between two signals

x [n] = 2^{n}

4[-n],

0 < a < 1

and

h [n]

= 4[n]

What is different in this one

At least one of the sequences is infinite, so this is an analytical convolution and not a table. Fix the limits of the sum from where BOTH are non zero, then sum the geometric series.
The answer comes out in regions: one while the overlap is still growing and one once it is complete. They must agree at the join, and that is the check.

The scan has mangled this one: the u has come out as a 4, and the base as a 2. It reads $x [n] = a^{n} u [- n]$ with $0 < a < 1$ , and $h [n] = u [n]$ .

As printed the convolution diverges. $x [n] = a^{n} u [- n]$ is non zero for n ≤ 0, where $a^{n}$ = (1/a)^ $| n |$ grows without bound for a below 1, so the sum has no finite value. The reading that makes the question work, and the one every textbook sets, is the decaying left sided exponential $x [n] = a$ ^ $| n | u [- n]$ . Solved that way:

x [k] = a^{- k}

for k ≤ 0,

h [m] = 1

for m ≥ 0

y [n] = \sum_{k} x [k] h [n - k]

, needing k ≤ 0 and k ≤ n n ≥ 0: k runs from -inf to 0

y [n]

= sum from m=0 to inf of

a^{m} = 1

/ (1 - a) n < 0: k runs from -inf to n

y [n]

= sum from m=

| n |

to inf of

a^{m} = a

| n |

/ (1 - a)

Check the closed form against the convolution sum carried out term by term, at several n. They agree, which is what makes the closed form an answer rather than a guess.

n	$\sum_{k} h [k] x [n - k]$	the closed form
-4	0.125	0.125
-2	0.5	0.5
-1	1	1
0	2	2
1	2	2
3	2	2

$y [n] = a$ ^ $| n |$ / (1 - a) for n < 0, and $1 / (1 - a)$ for n ≥ 0: it rises like the exponential and then holds flat, which is what convolving anything with a unit step does, since the step accumulates. The check above uses $a = 0.5$ , giving a plateau of 2.

Practise this2075 Chaitra · Q25+2 marksFind the output of LTI system having impulse response

h [n]

with

h [- 2] = 3

h [0] = 2

h [1] = 1

and input signal

x [n] = {(2)}^{n}

, for

- 1 \leq n \leq 3

. Also check the answer.What is different here

x [n]

starts before

n = 0

How this is readh has $h [- 2] = 3$ , $h [0] = 2$ , $h [1] = 1$ and nothing else, so h = ${3, 0, 2, 1}$ from n = -2. $x [n] = 2^{n}$ over $- 1 \leq n \leq 3$ is ${0.5, 1, 2, 4, 8}$ from n = -1.

What is different in this one

Two finite sequences, 5 samples and 4, so the answer has 5 + 4 - 1 = 8 samples and the table is 5 by 4.
One of the sequences starts at a negative n, so the output starts at $n = - 3$ , not at zero. Mark the origin on the answer or the plot is wrong even when the numbers are right.
There is a zero sample in the middle of one sequence. It still occupies an index: dropping it shortens the answer and shifts everything after it.

$x [n]$ runs from $n = - 1$ to 3, 5 samples; $h [n]$ runs from $n = - 2$ to 1, 4 samples. So $y [n]$ runs from $n = - 3$ to 4, which is 5 + 4 - 1 = 8 samples.

Multiply every sample of one by every sample of the other and add along the anti diagonals, which is the tabular method:

	$h [- 2] = 3$	$h [0] = 2$	$h [1] = 1$
$x [- 1] = 0.5$	1.5	1	0.5
$x [0] = 1$	3	2	1
$x [1] = 2$	6	4	2
$x [2] = 4$	12	8	4
$x [3] = 8$	24	16	8

y [- 3] = 1.5

= 1.5

y [- 2] = 3

= 3

y [- 1] = 1

+ 6 = 7

y [0] = 0.5

+ 2 + 12 = 14.5

y [1] = 1

+ 4 + 24 = 29

y [2] = 2

+ 8 = 10

y [3] = 4

+ 16 = 20

y [4] = 8

= 8

$y [n]$ = {1.5, 3, 7, 14.5, 29, 10, 20, 8} for $n = - 3$ to 4. The check holds: the samples of x add to 15.5, those of h add to 6, and 15.5 × 6 = 93, which is the sum of y.

Same as 2082 Bhadra Q22075 Ashwin · Q25 marksFind the output of LTI system having impulse response

h [n] = 2^{n} {u [n] - u [n - 3]}

and input signal

x [n] = δ [n] + δ [n - 1] + δ [n - 2]

How this is read $h [n] = 2^{n} {u [n] - u [n - 3]}$ is ${1, 2, 4}$ over $n = 0$ , 1, 2, and $x [n] = δ [n] + δ [n - 1] + δ [n - 2]$ is ${1, 1, 1}$ .

What is different in this one

Two finite sequences, 3 samples and 3, so the answer has 3 + 3 - 1 = 5 samples and the table is 3 by 3.
Both sequences are the same length, so the table is square and the answer is symmetric in the two if the sequences are.

$x [n]$ runs from $n = 0$ to 2, 3 samples; $h [n]$ runs from $n = 0$ to 2, 3 samples. So $y [n]$ runs from $n = 0$ to 4, which is 3 + 3 - 1 = 5 samples.

Multiply every sample of one by every sample of the other and add along the anti diagonals, which is the tabular method:

	$h [0] = 1$	$h [1] = 2$	$h [2] = 4$
$x [0] = 1$	1	2	4
$x [1] = 1$	1	2	4
$x [2] = 1$	1	2	4

y [0] = 1

= 1

y [1] = 2

+ 1 = 3

y [2] = 4

+ 2 + 1 = 7

y [3] = 4

+ 2 = 6

y [4] = 4

= 4

$y [n]$ = {1, 3, 7, 6, 4} for $n = 0$ to 4. The check holds: the samples of x add to 3, those of h add to 7, and 3 × 7 = 21, which is the sum of y.

Same as 2081 Bhadra Q22074 Ashwin · Q23+2 marksFind the output of LTI system having impulse response

h [n] = {(1 / 2)}^{n} {u [n + 2] - u [n - 2]}

and input signal

x [n] = {2, 1, 0.5, - 1}

. Also check the answer.

What is different in this one

Two finite sequences, 4 samples and 4, so the answer has 4 + 4 - 1 = 7 samples and the table is 4 by 4.
One of the sequences starts at a negative n, so the output starts at $n = - 2$ , not at zero. Mark the origin on the answer or the plot is wrong even when the numbers are right.
Both sequences are the same length, so the table is square and the answer is symmetric in the two if the sequences are.

$x [n]$ runs from $n = 0$ to 3, 4 samples; $h [n]$ runs from $n = - 2$ to 1, 4 samples. So $y [n]$ runs from $n = - 2$ to 4, which is 4 + 4 - 1 = 7 samples.

Multiply every sample of one by every sample of the other and add along the anti diagonals, which is the tabular method:

	$h [- 2] = 4$	$h [- 1] = 2$	$h [0] = 1$	$h [1] = 0.5$
$x [0] = 2$	8	4	2	1
$x [1] = 1$	4	2	1	0.5
$x [2] = 0.5$	2	1	0.5	0.25
$x [3] = - 1$	-4	-2	-1	-0.5

y [- 2] = 8

= 8

y [- 1] = 4

+ 4 = 8

y [0] = 2

+ 2 + 2 = 6

y [1] = 1

+ 1 + 1 + -4 = -1

y [2] = 0.5

+ 0.5 + -2 = -1

y [3] = 0.25

+ -1 = -0.75

y [4] = - 0.5

= -0.5

$y [n]$ = {8, 8, 6, -1, -1, -0.75, -0.5} for $n = - 2$ to 4. The check holds: the samples of x add to 2.5, those of h add to 7.5, and 2.5 × 7.5 = 18.75, which is the sum of y.

Same as 2082 Bhadra Q22073 Shrawan · Q26 marksDetermine the system output

y (n)

of the following signals:

h (n) = {1, 1, 1}

and

x (n) = {1, 1, 1, 1}

What is different in this one

Two finite sequences, 4 samples and 3, so the answer has 4 + 3 - 1 = 6 samples and the table is 4 by 3.

$x [n]$ runs from $n = 0$ to 3, 4 samples; $h [n]$ runs from $n = 0$ to 2, 3 samples. So $y [n]$ runs from $n = 0$ to 5, which is 4 + 3 - 1 = 6 samples.

Multiply every sample of one by every sample of the other and add along the anti diagonals, which is the tabular method:

	$h [0] = 1$	$h [1] = 1$	$h [2] = 1$
$x [0] = 1$	1	1	1
$x [1] = 1$	1	1	1
$x [2] = 1$	1	1	1
$x [3] = 1$	1	1	1

y [0] = 1

= 1

y [1] = 1

+ 1 = 2

y [2] = 1

+ 1 + 1 = 3

y [3] = 1

+ 1 + 1 = 3

y [4] = 1

+ 1 = 2

y [5] = 1

= 1

$y [n]$ = {1, 2, 3, 3, 2, 1} for $n = 0$ to 5. The check holds: the samples of x add to 4, those of h add to 3, and 4 × 3 = 12, which is the sum of y.

Also worked under The output for an exponential inputPractise this2073 Shrawan · Q43+7 marksState linear constant coefficient difference equation and corresponding system function. Determine the output sequence of the system with impulse response

h [n] = {(1 / 2)}^{n} u [n]

when the input signal is

x [n] = 10

- 5sin(

π

n/2) + 20cos(

π

n), -infinity < n < infinity.What is different hereThe only sitting where the input is a REAL sum rather than a single complex exponential, so the answer has to come back out real: a constant, a sine and a cosine, each with its own gain and phase. Leaving it as complex exponentials is not finishing the question.

What is different in this one

3 components at 3 different frequencies, so H is evaluated 3 times and the results added: superposition.
One of them is a constant, which is simply the $ω = 0$ case: the input times the DC gain.
The input is real, so the answer must be real. A sine goes in and a sine of the same frequency comes out, scaled by $| H |$ and shifted by the angle of H. Leaving the answer as complex exponentials is not finishing the question.
Do not convolve. The input runs over all n, so it is not a finite sequence and the graphical method is the wrong tool.

x [n] = e^{j ω_{0} n} ⟶ y [n] = H (e^{j ω_{0}}) e^{j ω_{0} n}, A \cos (ω_{0} n) ⟶ A | H | \cos (ω_{0} n + ∠ H)

\begin{aligned} H (e^{j ω}) & = 1 / (1 - 0.5 e^{- j ω}), the  transform  of h [n] = {(1 / 2)}^{n} u [n] \end{aligned}

Component	$ω_{0}$	$H (e^{j ω_{0}})$	in polar form	Output amplitude
the constant 10, which is the $ω = 0$ case	$0 π$	2	2 angle 0 deg	20
-5 $\sin (π n / 2)$ , taken at $ω = \frac{π}{2}$	$0.5 π$	0.8 - j0.4	0.8944 angle -26.57 deg	4.4721
20 $\cos (π n)$ , taken at $ω = π$	$1 π$	0.6667	0.6667 angle 0 deg	13.3333

Where	By convolution	By the formula
$n = 0$	35.3333	35.3333
$n = 1$	2.6667	2.6667
$n = 2$	31.3333	31.3333
$n = 5$	2.6667	2.6667

$y [n] = 20 - 4.4721 \sin (0.5 π n - 26.57 d e g) + 13.3333 \cos (π n)$ . Real in, real out: every term is a signal of the same frequency as the one that produced it, moved in amplitude and in phase and in nothing else.

Same as 2081 Bhadra Q22072 Chaitra · Q23+2 marksFind the output of LTI system having impulse response

h [n]

with

h [- 2] = 1

h [0] = 2

h [1] = 3

and input signal

x [n]

with

x [0] = \frac{1}{2}

x [2] = 2

x [3] = 3

. Also check the answer.

How this is read $h [- 2] = 1$ , $h [0] = 2$ , $h [1] = 3$ gives h = ${1, 0, 2, 3}$ from $n = - 2$ ; $x [0] = \frac{1}{2}$ , $x [2] = 2$ , $x [3] = 3$ gives x = ${0.5, 0, 2, 3}$ from $n = 0$ .

What is different in this one

Two finite sequences, 4 samples and 4, so the answer has 4 + 4 - 1 = 7 samples and the table is 4 by 4.
One of the sequences starts at a negative n, so the output starts at $n = - 2$ , not at zero. Mark the origin on the answer or the plot is wrong even when the numbers are right.
Both sequences are the same length, so the table is square and the answer is symmetric in the two if the sequences are.
There is a zero sample in the middle of one sequence. It still occupies an index: dropping it shortens the answer and shifts everything after it.

$x [n]$ runs from $n = 0$ to 3, 4 samples; $h [n]$ runs from $n = - 2$ to 1, 4 samples. So $y [n]$ runs from $n = - 2$ to 4, which is 4 + 4 - 1 = 7 samples.

Multiply every sample of one by every sample of the other and add along the anti diagonals, which is the tabular method:

	$h [- 2] = 1$	$h [0] = 2$	$h [1] = 3$
$x [0] = 0.5$	0.5	1	1.5
$x [1] = 0$	0	0	0
$x [2] = 2$	2	4	6
$x [3] = 3$	3	6	9

y [- 2] = 0.5

= 0.5

y [- 1] = 0

= 0

y [0] = 1

+ 2 = 3

y [1] = 1.5

+ 3 = 4.5

y [2] = 4

= 4

y [3] = 6

+ 6 = 12

y [4] = 9

= 9

$y [n]$ = {0.5, 0, 3, 4.5, 4, 12, 9} for $n = - 2$ to 4. The check holds: the samples of x add to 5.5, those of h add to 6, and 5.5 × 6 = 33, which is the sum of y.

Same as 2081 Bhadra Q22072 Kartik · Q25 marksFind the output of LTI system having impulse response

h [n] = {(1 / 3)}^{n} {u [n + 1] - u [n - 2]}

and input signal

x [n] = {2, 1, 0.5, 3}

How this is read $h [n] = {(1 / 3)}^{n} {u [n + 1] - u [n - 2]}$ runs over $n = - 1$ , 0, 1 with the values 3, 1 and 1/3.

What is different in this one

Two finite sequences, 4 samples and 3, so the answer has 4 + 3 - 1 = 6 samples and the table is 4 by 3.
One of the sequences starts at a negative n, so the output starts at $n = - 1$ , not at zero. Mark the origin on the answer or the plot is wrong even when the numbers are right.

$x [n]$ runs from $n = 0$ to 3, 4 samples; $h [n]$ runs from $n = - 1$ to 1, 3 samples. So $y [n]$ runs from $n = - 1$ to 4, which is 4 + 3 - 1 = 6 samples.

Multiply every sample of one by every sample of the other and add along the anti diagonals, which is the tabular method:

	$h [- 1] = 3$	$h [0] = 1$	$h [1] = 0.3333$
$x [0] = 2$	6	2	0.6667
$x [1] = 1$	3	1	0.3333
$x [2] = 0.5$	1.5	0.5	0.1667
$x [3] = 3$	9	3	1

y [- 1] = 6

= 6

y [0] = 2

+ 3 = 5

y [1] = 0.6667

+ 1 + 1.5 = 3.1667

y [2] = 0.3333

+ 0.5 + 9 = 9.8333

y [3] = 0.1667

+ 3 = 3.1667

y [4] = 1

= 1

$y [n]$ = {6, 5, 3.1667, 9.8333, 3.1667, 1} for $n = - 1$ to 4. The check holds: the samples of x add to 6.5, those of h add to 4.3333, and 6.5 × 4.3333 = 28.1667, which is the sum of y.

Also worked under The output for an exponential inputSame as 2082 Baishakh Q22072 Kartik · Q44 marksFind the output of LTI System having impulse response

h [n] = {(1 / 2)}^{n} u [n]

and input signal

x [n] = 5 e^{j π n / 3}

for -infinity < n < infinity.

What is different in this one

1 component at one frequency, so H is evaluated once and the results added: superposition.
Do not convolve. The input runs over all n, so it is not a finite sequence and the graphical method is the wrong tool.

x [n] = e^{j ω_{0} n} ⟶ y [n] = H (e^{j ω_{0}}) e^{j ω_{0} n}, A \cos (ω_{0} n) ⟶ A | H | \cos (ω_{0} n + ∠ H)

\begin{aligned} H (e^{j ω}) & = 1 / (1 - 0.5 e^{- j ω}), the  transform  of h [n] = {(1 / 2)}^{n} u [n] \end{aligned}

Component	$ω_{0}$	$H (e^{j ω_{0}})$	in polar form	Output amplitude
$x [n] = 5 e^{j π n / 3}$	$0.3333 π$	1 - j0.5774	1.1547 angle -30 deg	5.7735

Where	By convolution	By the formula
$n = 0$	5	5
$n = 1$	5	5
$n = 2$	0	0
$n = 5$	0	0

$y [n] = 5.7735 e^{j (0.3333 π n - 30 d e g)}$ .

Also worked under Even and odd partsPractise this2071 Shrawan · Q14+5 marksFind the odd and even part of the following signal: [a stem plot of

x [n]

with values 1 at

n = - 4

, -3, -2, -1 and 0, and 2 at

n = 1

, 2, 3, 4]. A discrete time LTI system has input signal and impulse response as,

x [n] = 1

for

- 1 \leq n \leq 1

and 0 elsewhere, and

h [n] = 1

for

- 1 \leq n \leq 1

and 0 elsewhere. Find the output of the system using graphical method.

How this is readThe stem plot is $x [n] = 1$ for $n = - 4$ to 0 and 2 for $n = 1$ to 4. The second half convolves two identical three sample rectangles, each from $n = - 1$ to 1.

The output, by the graphical method

What is different in this one

Two finite sequences, 3 samples and 3, so the answer has 3 + 3 - 1 = 5 samples and the table is 3 by 3.
One of the sequences starts at a negative n, so the output starts at $n = - 2$ , not at zero. Mark the origin on the answer or the plot is wrong even when the numbers are right.
Both sequences are the same length, so the table is square and the answer is symmetric in the two if the sequences are.

$x [n]$ runs from $n = - 1$ to 1, 3 samples; $h [n]$ runs from $n = - 1$ to 1, 3 samples. So $y [n]$ runs from $n = - 2$ to 2, which is 3 + 3 - 1 = 5 samples.

Multiply every sample of one by every sample of the other and add along the anti diagonals, which is the tabular method:

	$h [- 1] = 1$	$h [0] = 1$	$h [1] = 1$
$x [- 1] = 1$	1	1	1
$x [0] = 1$	1	1	1
$x [1] = 1$	1	1	1

y [- 2] = 1

= 1

y [- 1] = 1

+ 1 = 2

y [0] = 1

+ 1 + 1 = 3

y [1] = 1

+ 1 = 2

y [2] = 1

= 1

$y [n]$ = {1, 2, 3, 2, 1} for $n = - 2$ to 2. The check holds: the samples of x add to 3, those of h add to 3, and 3 × 3 = 9, which is the sum of y.

The rest of this questionThis topic works the part that belongs to it. The other part is worked under Even and odd parts (The even and odd parts), where it belongs.

Also worked under The output for an exponential inputPractise this2071 Shrawan · Q32+3+5 marksWhy do we need difference equation? State linear constant coefficient difference equation and corresponding system function. Consider an LTI system with impulse response

h [n] = {(1 / 2)}^{n} u [n]

. Determine

y [n]

, if the input is

x [n] = A e^{j π n}

.What is different hereAt

ω = π

the exponential

e^{- j ω}

is exactly -1, so H is

1 / (1 + 0.5) = 2 / 3

with no complex arithmetic at all. The answer is

y [n]

= (2A/3)

e^{j π n}

What is different in this one

1 component at one frequency, so H is evaluated once and the results added: superposition.
Do not convolve. The input runs over all n, so it is not a finite sequence and the graphical method is the wrong tool.

x [n] = e^{j ω_{0} n} ⟶ y [n] = H (e^{j ω_{0}}) e^{j ω_{0} n}, A \cos (ω_{0} n) ⟶ A | H | \cos (ω_{0} n + ∠ H)

\begin{aligned} H (e^{j ω}) & = 1 / (1 - 0.5 e^{- j ω}), the  transform  of h [n] = {(1 / 2)}^{n} u [n] \end{aligned}

Component	$ω_{0}$	$H (e^{j ω_{0}})$	in polar form	Output amplitude
$x [n] = A e^{j π n}$ , with A factored out	$1 π$	0.6667	0.6667 angle 0 deg	0.6667

Where	By convolution	By the formula
$n = 0$	0.6667	0.6667
$n = 1$	-0.6667	-0.6667
$n = 2$	0.6667	0.6667
$n = 5$	-0.6667	-0.6667

$y [n] = 0.6667 e^{j (π n)}$ .

The same questions grouped by how often each topic is set: convolution

1.9Frequency response of an LTI system

The frequency response of an LTI system

Before the formulas

Convolution gives you the output for any input at all, which is more than you usually need and more work than you want. There is one family of inputs where the answer is almost free. Put a complex exponential into an LTI system and the same exponential comes out, only scaled.

Nothing else has that property, and the scale factor depends on the frequency you put in. The function that gives it is the frequency response. That one fact is why the rest of the subject lives in frequency: an entire convolution collapses into a single multiplication.

Frequency response

H (e^{j ω}) = \sum_{n} h [n] e^{- j ω n}

, the DTFT of the impulse response. It is what the system does to a complex exponential of frequency

ω

, and like every DTFT it is periodic in $2 π$ .

\begin{aligned} x [n] = A e^{j ω_{0} n} & ⟹ y [n] = A H (e^{j ω_{0}}) e^{j ω_{0} n} \\ x [n] = A \cos (ω_{0} n + ϕ) & ⟹ y [n] = A | H (e^{j ω_{0}}) | \cos (ω_{0} n + ϕ + ∠ H (e^{j ω_{0}})) \end{aligned}

An exponential comes out as the same exponential, scaled and phase shifted. That makes it an eigenfunction of every LTI system, with $H (e^{j ω_{0}})$ the eigenvalue.

The proof is three lines: put A $e^{j ω_{0} (n - k)}$ into the convolution sum, pull A $e^{j ω_{0} n}$ out of it, and what is left is $\sum_{k} h [k] e^{- j ω_{0} k}$ , which is $H (e^{j ω_{0}})$ .
What a filter can and cannot do. It scales the amplitude and shifts the phase. It cannot change the frequency and it cannot create a frequency that was not in the input. Every filter in this subject rests on that sentence.
Do not convolve when the input is an exponential or a sinusoid. Evaluate $H (e^{j ω})$ at that one frequency instead. Several papers set this as though it were a convolution question.
A constant input is the $ω = 0$ case, so its output is the input times the DC gain $H (e^{j 0})$ = sum of $h [n]$ .

In the exam

This is question 2 in several sittings, usually with $h [n] = {(\frac{1}{2})}^{n} u [n]$ and an input of 5 $e^{j π n / 3}$ , sometimes with a constant added. Answer it term by term: each input frequency is scaled by H at that frequency, and the results are added. The full treatment of $H (e^{j ω})$ , including the pole zero sketch, is chapter 3.

The full treatment, including the pole zero sketch that seventeen sittings ask for, is chapter 3.

See it drawnAn exponential goes through unchanged, except for a gain

That is the whole reason the frequency response exists: the shape survives, and only the size and the phase change.

1.10Sampling a continuous time signal, spectral properties

Sampling a continuous time signal, and the spectrum of the samples

Before the formulas

The last section of the chapter is where all the discrete signals came from in the first place. Sampling means keeping a continuous signal's value every T seconds and throwing away everything in between. What happens to the spectrum is the thing to hold on to: it gets copied, for ever, along the frequency axis.

If the copies are far enough apart nothing is lost and the original can be rebuilt exactly. If they overlap, frequencies that were never there appear in the samples, and no processing afterwards can separate them out again.

That overlap is aliasing, it is why the sampling rate has a lower limit, and that limit is all the Nyquist theorem says.

Sampling Taking

x [n]

= xa(nT) at a uniform spacing T. The sampling frequency is Fs =

\frac{1}{T}

. The spectrum of the sequence is the spectrum of the analog signal repeated every Fs and summed, which is the single fact the whole of this section rests on.

\begin{aligned} 1 inf \\ X (e^{j ω}) & = - - - \sum Xa (j (ω - 2 \pik) / T) \\ T k & = - inf \\ or, in analog frequency, \\ 1 inf \\ Xs (jW) & = - - - \sum Xa (j (W - k Ω_{s})) Ω_{s} = 2 π / T \\ T k & = - inf \\ FREQUENCY  MAPPING ω & = W T = 2 \piF / Fs & radians per sample \\ F & = 0 \to ω = 0 \\ F & = Fs / 2 \to ω = π & the folding frequency \\ F & = Fs \to ω = 2 π which  is  the  same  as ω = 0 \end{aligned}

The sampling theorem A signal band limited to Fmax is recovered exactly from its samples if $F s > 2$ Fmax. The rate 2 Fmax is the Nyquist rate, and Fs/2 is the folding frequency.

Why the repeats appear. Sampling is multiplication by an impulse train, whose own spectrum is an impulse train. Multiplying in time convolves in frequency, and convolving with an impulse train makes copies. The multiplication property from the last card is doing all the work.
Aliasing is what happens when Fs is too low: the copies overlap, high frequencies fold back and appear as low ones, and no processing afterwards can undo it. A 6 kHz tone sampled at 8 kHz comes back as 2 kHz, because 8 - 6 = 2.
The defence is an analog anti alias filter before the ADC, cutting everything above Fs/2. This is why a digital filter always has an analog filter in front of it.
Reconstruction is an ideal low pass filter of gain T and cutoff Fs/2, which in the time domain interpolates with sinc functions, xa(t) = sum of $x [n] sinc ((t - n T) / T)$ . Real converters use a hold followed by a smoothing filter, which is why a reconstruction filter is needed at the output too.
Only $- π$ to $π$ is distinguishable. Two analog frequencies that differ by a multiple of Fs land on the same $ω$ , which is another way of saying the same thing.

Worked example

An analog signal xa(t) = 3 cos( $2 π$ 2000 t) + 5 sin( $2 π$ 6000 t) is sampled at Fs = 8 kHz. What digital frequencies appear, and is anything aliased?

The folding frequency is Fs/2 = 4 kHz.
The 2 kHz component: $ω = 2 π$ (2000)/8000 = $0.5 π$ , below folding, so it survives untouched.
The 6 kHz component: $ω = 2 π$ (6000)/8000 = $1.5 π$ , which is above $π$ . It folds to $1.5 π - 2 π = - 0.5 π$ , and a negative frequency in a real signal is read as $0.5 π$ with the sign of the sine reversed.

Both components land on $ω = 0$ .5 $π$ , so the 6 kHz tone is aliased and appears as a 2 kHz tone. The sampled sequence is $x [n] = 3 \cos (0.5 π n) - 5 \sin (0.5 π n)$ , and the two tones can never be separated again. To keep them apart, Fs would have to exceed 12 kHz.

The spectrum repeating every $2 π$ is the same overlap that impulse invariance suffers, and the reason the bilinear transformation is preferred. The $2 π$ periodicity of every plot in chapter 3 comes from here.

See it drawnTwo tones, one sequence: what aliasing is

Sample too slowly and a fast tone lands on the same samples as a slow one. After that no filter can tell them apart.

1.11Last minute recall

Last minute recall, chapter 1

Must memorise

xe = ( $x [n] + x [- n]$ )/2, xo = ( $x [n] - x [- n]$ )/2, and $x_{o} [0] = 0$ always.
Periodic if and only if $ω_{0} / 2 π$ is rational, and then N is its denominator in lowest terms. No $π$ in $ω_{0}$ means aperiodic. Several components: take the LCM.
Energy signal: 0 < E < infinity and P = 0. Power signal: 0 < P < infinity and E = infinity. Periodic means power. | $e^{j θ}$ | = 1 means power, with P = 1.
E for $a^{n} u [n]$ is 1/(1 $- a^{2}$ ); P for $u [n]$ is $\frac{1}{2}$ , not 1.
$y [n]$ = sum of $x [k] h [n - k]$ . The output starts at nx + nh and has Lx + Lh - 1 samples.
LTI causal if and only if $h [n] = 0$ for n < 0. LTI stable if and only if sum $| h [n] |$ is finite.
$e^{j ω n}$ in gives $H (e^{j ω}) e^{j ω n}$ out. Never convolve these.
$X (e^{j ω})$ = ( $\frac{1}{T}$ ) sum of Xa( j( $ω - 2 π$ k)/T ), and $F s > 2$ Fmax.
$c_{k} = \frac{1}{N}$ sum of $x [n] e^{- j 2 π k n / N}$ , and only N of them are distinct.

Most repeated in this chapter, in order

Convolution, find $y [n]$ TOP 16/19
Energy against power HOT 6/19 and the fundamental period HOT 6/19, usually as the two halves of one question
Test a system for linearity, time invariance, causality and stability HOT 5/19
Even and odd parts PIN 2/19 and plotting a transformed sequence PIN 2/19
The Fourier series coefficients PIN 2/19

Questions 1 and 2 come from this chapter in almost every sitting. If time is short, learn the convolution sum and the energy against power test, and leave the Fourier series: it has been asked twice in nineteen papers and never for more than five marks.

Chapter 2 · 4 hours · 6 marks · question 3, every paper

The z-transform

The transform that turns convolution into multiplication and a difference equation into algebra. The examinable core is small and fixed: define it, state the ROC and its properties, and invert a rational $X (z)$ by partial fractions with the ROC deciding which answer is right.

What this chapter is about

The definition, bilateral and unilateral, and why the ROC has to be quoted with it.
The ROC: its properties, and the four shapes it can take.
The properties, above all the convolution property, which is the reason the transform exists.
The inverse by partial fractions, which is question 3 in every single paper.

Where it fits

Chapter 1 gave convolution; here it becomes multiplication.
Chapter 3 evaluates $H (z)$ on the unit circle to get the frequency response.
Chapters 4, 5 and 6 all start from an $H (z)$ written as a ratio of polynomials.

What you will learn

2.1 The definition, The region of convergence, and its properties, The standard pairs
2.2 The properties, The convolution property, derived
2.3 Inverting $X (z)$ by long division, Inverting $X (z)$ by partial fractions
2.4 Last minute recall, chapter 2

How it is examined

Always question 3, and always in two parts: 1 to 3 marks of bookwork, then 5 or 6 marks of inversion.
The bookwork half is one of three things: define the z-transform, define the ROC, or list the properties of the ROC.
The inversion half always states an ROC, and the ROC is the point of the question.

2.1Definition, convergence and the region of convergence

The definition PIN 2/19

82 Bh · 79 Ba1+5

Before the formulas

Chapter 1 ended with a transform that works in frequency alone. It has a weakness: it only exists if the sequence's sum settles down, and plenty of useful sequences grow instead. The z-transform fixes that by attaching a knob.

Write z as a radius times an angle, and the radius supplies a shrinking factor that can drag a growing sequence back into a finite sum. Turn the radius to one and you are standing exactly where chapter 1 left you.

So this is not a new idea to learn from scratch. It is the transform you already have, with room to move, and that room is what makes the whole of filter design possible.

The z-transform The z-transform of a discrete time signal

x [n]

is the power series $X (z) = \sum_{n = - \infty}^{\infty} x [n] z^{- n}$ , where z is a complex variable. It converts a sequence into a function of z, and it exists only for the values of z that make the series converge; that set of values is the region of convergence.

\begin{aligned} bilateral X (z) & = \sum_{n = - \infty}^{\infty} x [n] z^{- n} \\ unilateral X (z) & = \sum_{n = 0}^{\infty} x [n] z^{- n} \\ z = r e^{j ω} ⟹ X (z) & = \sum_{n} x [n] r^{- n} e^{- j ω n} \\ r = 1 ⟹ X (z) |_{z = e^{j ω}} & = X (e^{j ω}) = the DTFT \end{aligned}

The transform is not complete without its ROC. Two different sequences, one right sided and one left sided, can produce exactly the same algebraic $X (z)$ ; only the ROC tells them apart. This is why every exam question states one.
The bilateral form is the one meant unless the question says otherwise. The unilateral form ignores n < 0 and is used for difference equations with initial conditions.
It generalises the DTFT. Writing z = r $e^{j ω}$ shows the z-transform is the DTFT of $x [n] r^{- n}$ . The extra factor $r^{- n}$ is what lets the sum converge for sequences whose DTFT does not exist, such as $u [n]$ or a growing exponential.
Why it is useful: convolution becomes multiplication, a difference equation becomes a ratio of polynomials, and stability and causality become statements about where the poles sit.

Set z = $e^{j ω}$ and this is the DTFT; sample that at N points and it is the DFT. The three transforms are one transform looked at from three distances.

See it drawnThe z-transform is the DTFT with a convergence factor

Writing z as r times a complex exponential shows what the transform adds: an r to the minus n that can drag a growing sequence back into convergence.

Practise thisEvery question the paper has set on this topic, worked in full2

Practise this +1 like it2082 Bhadra · Q31+5 marksDefine z-transform for a discrete time signal. Find the inverse z-transform for

H (z) = z / (3 z^{2} - 4 z + 1)

using partial fraction method for

1 / 3 < | z | < 1

How this is read $H (z) = z / (3 z^{2} - 4 z + 1)$ . Dividing numerator and denominator by $z^{2}$ gives $z^{- 1}$ over $3 - 4 z^{- 1} + z^{- 2}$ , which is the form the partial fractions are taken from.

What is different in this one

The numerator is of lower degree than the denominator, so it is already proper: no division, straight to F(z) = $X (z)$ /z.
The ROC is a ring, so the answer is two sided: the poles inside the ring give right sided terms and the ones outside it give left sided terms. This is the case that separates a real answer from a memorised one.

Six steps, and the order matters: write it in positive powers of z, divide out any excess degree, take $X (z)$ /z, split that into partial fractions, multiply the z back in, then let the ROC decide which side each pole sits on.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{z}{3 z^{2} - 4 z + 1}

2. Nothing to divide. Degree 1 over degree 2, so the fraction is already low enough on top.

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = \frac{1}{3 z^{2} - 4 z + 1}

The numerator carried a z of its own, and it cancels the z we just divided by, so there is no term at the origin at all.

3 z^{2} - 4 z + 1 = 3 (z - 0.3333) (z - 1) = 0

So the poles are $0.3333$ and $1$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{1}{3 (z - 0.3333) (z - 1)} \\ = \frac{A}{(z - 0.3333)} + \frac{B}{(z - 1)} \end{aligned}

Put them back over the common denominator $3 (z - 0.3333) (z - 1)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z - 1) + B (z - 0.3333) \\ A (z - 1) & = A z - A \\ B (z - 0.3333) & = B z - 0.3333 B \end{aligned}

Now collect the powers of z and set each one equal to the same power in $0.3333$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + B & = 0 & from z \\ - A - 0.3333 B & = 0.3333 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = - 0.5 \\ B & = 0.5 \end{aligned}

Putting those back and adding the terms returns $1$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = \frac{- 0.5 z}{(z - 0.3333)} + \frac{0.5 z}{(z - 1)}

6. Invert term by term. The ROC $0.3333 < | z | < 1$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
0.3333	0.3333	-0.5	the ROC lies outside it	-0.5 ${(0.3333)}^{n} u [n]$
1	1	0.5	the ROC lies inside it	-0.5 ${(1)}^{n} u [- n - 1]$

With the ROC $0.3333 < | z | < 1$ , $x [n] = - 0.5 {(0.3333)}^{n} u [n] - 0.5 {(1)}^{n} u [- n - 1]$ .

Same as 2082 Bhadra Q32079 Baishakh · Q31+5 marksDefine z-transform for a discrete time signal. Find the inverse z-transform for

H (z) = z / (3 z^{2} - 4 z + 1)

using partial fraction method for

1 / 3 < | z | < 1

What is different in this one

The numerator is of lower degree than the denominator, so it is already proper: no division, straight to F(z) = $X (z)$ /z.
The ROC is a ring, so the answer is two sided: the poles inside the ring give right sided terms and the ones outside it give left sided terms. This is the case that separates a real answer from a memorised one.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{z}{3 z^{2} - 4 z + 1}

2. Nothing to divide. Degree 1 over degree 2, so the fraction is already low enough on top.

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = \frac{1}{3 z^{2} - 4 z + 1}

The numerator carried a z of its own, and it cancels the z we just divided by, so there is no term at the origin at all.

3 z^{2} - 4 z + 1 = 3 (z - 0.3333) (z - 1) = 0

So the poles are $0.3333$ and $1$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{1}{3 (z - 0.3333) (z - 1)} \\ = \frac{A}{(z - 0.3333)} + \frac{B}{(z - 1)} \end{aligned}

Put them back over the common denominator $3 (z - 0.3333) (z - 1)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z - 1) + B (z - 0.3333) \\ A (z - 1) & = A z - A \\ B (z - 0.3333) & = B z - 0.3333 B \end{aligned}

Now collect the powers of z and set each one equal to the same power in $0.3333$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + B & = 0 & from z \\ - A - 0.3333 B & = 0.3333 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = - 0.5 \\ B & = 0.5 \end{aligned}

Putting those back and adding the terms returns $1$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = \frac{- 0.5 z}{(z - 0.3333)} + \frac{0.5 z}{(z - 1)}

6. Invert term by term. The ROC $0.3333 < | z | < 1$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
0.3333	0.3333	-0.5	the ROC lies outside it	-0.5 ${(0.3333)}^{n} u [n]$
1	1	0.5	the ROC lies inside it	-0.5 ${(1)}^{n} u [- n - 1]$

With the ROC $0.3333 < | z | < 1$ , $x [n] = - 0.5 {(0.3333)}^{n} u [n] - 0.5 {(1)}^{n} u [- n - 1]$ .

The region of convergence, and its properties TOP 14/19

82 Ba · 81 Bh · 81 Ba · 80 Bh · 80 Ba · 78 Bh · 76 Ch · 75 Ch · 75 Ash · 74 Ch · 74 Ash · 73 Shr · 72 Ch · 72 Ka1+53+31+3+5

Before the formulas

Because the radius is a knob, the question "does this sum converge" has no single answer. It converges for some radii and not for others, and the set that works is the region of convergence. This is not a footnote.

The same algebraic expression is two completely different sequences depending on which region came with it, so an answer without its region is not an answer. Most of the marks in this chapter are for saying which region and why.

The four shapes it can take are decided by the sequence itself, not by your algebra, and learning to say which shape from the look of the sequence is the skill this card is for.

ROC The region of convergence is the set of values of z in the complex plane for which the z-transform sum converges to a finite value. It is always an annulus centred on the origin,

r 1 < | z | < r 2

, because convergence depends only on

| z |

The properties, which is the list the paper asks for

It is a ring or a disc centred at the origin. Convergence depends on $| z |$ alone, so the region can only be bounded by circles.
The ROC contains no poles. $X (z)$ is infinite at a pole, so the series cannot converge there. The ROC is therefore always bounded by poles.
A finite length sequence converges everywhere except possibly $z = 0$ (if it has positive n terms) and z = infinity (if it has negative n terms).
A right sided sequence has an ROC outside the outermost pole: $| z |$ > rmax. If it is also causal, the ROC includes z = infinity.
A left sided sequence has an ROC inside the innermost pole: $| z |$ < rmin.
A two sided sequence has an ROC that is a ring between two poles, $r 1 < | z | < r 2$ , and it may be empty, in which case the transform does not exist.
The ROC must be a connected region. It cannot be two separate rings.
For an LTI system: the system is stable if and only if the ROC contains the unit circle, and causal if and only if the ROC is the outside of the outermost pole. Both at once means every pole is inside the unit circle.

Sequence	ROC	Example
Finite length	The whole plane, except perhaps 0 or infinity	${1, 2, 3}$
Right sided, causal	$\| z \| > a$ , outside the outermost pole	$a^{n} u [n]$
Left sided, anticausal	$\| z \| < a$ , inside the innermost pole	- $a^{n} u [- n - 1]$
Two sided	$a < \| z \| < b$ , a ring	$a^{n} u [n] + b^{n} u [- n - 1]$

Locating the ROC, worked $x [n] = {(0.1)}^{n} u [n] + {(0.3)}^{n} u [- n - 1]$ . The first term is right sided with a pole at 0.1, so it needs

| z | > 0.1

. The second is left sided with a pole at 0.3, so it needs

| z | < 0.3

. Both converge on the ring $0.1 < | z | < 0.3$ , and the transform exists.
$x [n] = {(0.6)}^{n} u [n] + {(0.25)}^{n} u [n]$ . Both terms are right sided, with poles at 0.6 and 0.25, so the ROC is outside the outermost: $| z | > 0.6$ . Because that region contains the unit circle, the signal is absolutely summable.

In the exam

Draw it. A pole zero plot with the annulus shaded costs thirty seconds and carries a mark on its own.
Name the three cases when the question gives one $X (z)$ and three ROCs, which is the 2082 Baishakh and 2081 Baishakh pattern: the outer ROC is causal, the inner one is anticausal, the ring is two sided.

The ROC is where stability and causality live, and it is what decides every term of the inverse transform.

See it drawnThe four shapes a region of convergence can take

The ROC is always bounded by pole magnitudes and never contains a pole. Which of the four you get is decided by the sequence, not by the algebra.

Practise thisEvery question the paper has set on this topic, worked in full2

Practise this2075 Chaitra · Q64+6 marksList out the properties of Region of convergence and locate the ROC of the following signal

x [n] = {(0.1)}^{n} u [n] + {(0.3)}^{n} u [- n - 1]

What is different in this one

2 terms, 1 right sided and 1 left sided, so the ROC will be a ring if it exists at all.
Watch the sign on the left sided term: the standard pair is $- b^{n} u [- n - 1]$ , so a plus in the question picks up a minus in the transform.

Transform each term with its own ROC, then intersect them.

Term	Transform	Its own ROC
${(0.1)}^{n} u [n]$	$z / (z - 0.1)$	$\| z \| > 0.1$
${(0.3)}^{n} u [- n - 1]$	$- z / (z - 0.3)$	$\| z \| < 0.3$

For $x [n] = {(0.1)}^{n} u [n] + {(0.3)}^{n} u [- n - 1]$ the ROC is $0.1 < | z | < 0.3$ , a ring, non empty because the right sided pole is the inner one. It does not contain the unit circle, so the signal has no DTFT.

Practise this2075 Ashwin · Q33+6 marksList out the properties of Region of convergence and locate the ROC of the following signal.

x [n] = {(0.6)}^{n} u [n] + {(0.25)}^{n} u [n]

What is different hereBoth terms are right sided here, so the ROC is the outside of the OUTERMOST pole and there is no ring. Contrast the 2075 Chaitra question, where one term is left sided and the answer is a ring.

What is different in this one

2 terms, 2 right sided and 0 left sided, so the ROC is simply the outside of the outermost pole.
Both terms are right sided, so no sign trap here.

Transform each term with its own ROC, then intersect them.

Term	Transform	Its own ROC
${(0.6)}^{n} u [n]$	$z / (z - 0.6)$	$\| z \| > 0.6$
${(0.25)}^{n} u [n]$	$z / (z - 0.25)$	$\| z \| > 0.25$

For $x [n] = {(0.6)}^{n} u [n] + {(0.25)}^{n} u [n]$ the ROC is every term is right sided, so the ROC is the outside of the outermost pole: $| z | > 0.6$ . It contains the unit circle, so the signal is absolutely summable and its DTFT exists.

The same questions grouped by how often each topic is set: locating the region of convergence

The standard pairs PIN 2/19

82 Bh · 79 Ba1+5

Before the formulas

Every inverse transform question in the paper is finished by recognising a standard form, so the short list is worth knowing by heart rather than deriving under time pressure. There are only a handful, and two of them are the same expression with different regions of convergence, which is the exam's favourite trap: identical algebra, one causal answer and one anticausal answer. Read this card as a table to memorise, and read the two look alike pairs twice.

Every inversion question reduces to recognising one of these. They all come from summing a geometric series, so if you forget one you can rebuild it in two lines.

$x [n]$	$X (z)$	ROC
$δ [n]$	1	All z
$δ [n - k]$	$z^{- k}$	All z except 0
$u [n]$	1/(1 $- z^{- 1}$ )	$\| z \| > 1$
$a^{n} u [n]$	1/(1 - a $z^{- 1}$ )	$\| z \| > \| a \|$
- $a^{n} u [- n - 1]$	1/(1 - a $z^{- 1}$ )	$\| z \| < \| a \|$
n $a^{n} u [n]$	a $z^{- 1}$ / (1 - a $z^{- 1}$ )^2	$\| z \| > \| a \|$
(n+1) $a^{n} u [n]$	1 / (1 - a $z^{- 1}$ )^2	$\| z \| > \| a \|$
cos( $ω_{0}$ n) $u [n]$	(1 - cos( $ω_{0}$ ) $z^{- 1}$ )/(1 - 2cos( $ω_{0}$ ) $z^{- 1} + z^{- 2}$ )	$\| z \| > 1$
sin( $ω_{0}$ n) $u [n]$	sin( $ω_{0}$ ) $z^{- 1}$ /(1 - 2cos( $ω_{0}$ ) $z^{- 1} + z^{- 2}$ )	$\| z \| > 1$
$a^{n}$ cos( $ω_{0}$ n) $u [n]$	(1 - a cos( $ω_{0}$ ) $z^{- 1}$ )/(1 - 2a cos( $ω_{0}$ ) $z^{- 1} + a^{2} z^{- 2}$ )	$\| z \| > \| a \|$

The one line behind the whole table:

\sum_{n = 0}^{\infty} {(a z^{- 1})}^{n} = \frac{1}{1 - a z^{- 1}} for | a z^{- 1} | < 1, that is | z | > | a |

The condition is the ROC.

The pair that catches people: $a^{n} u [n]$ and $- a^{n} u [- n - 1]$ have the same $X (z)$ . Only the ROC separates them, and that is exactly the fact the exam question is built on.

See it drawnThe two pairs that differ only in their ROC

The same algebraic

X (z)

is two different sequences. Only the ROC says which, and the question always gives you one.

Practise thisEvery question the paper has set on this topic, worked in full2

Practise this +1 like it2082 Bhadra · Q31+5 marksDefine z-transform for a discrete time signal. Find the inverse z-transform for

H (z) = z / (3 z^{2} - 4 z + 1)

using partial fraction method for

1 / 3 < | z | < 1

What is different in this one

The numerator is of lower degree than the denominator, so it is already proper: no division, straight to F(z) = $X (z)$ /z.
The ROC is a ring, so the answer is two sided: the poles inside the ring give right sided terms and the ones outside it give left sided terms. This is the case that separates a real answer from a memorised one.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{z}{3 z^{2} - 4 z + 1}

2. Nothing to divide. Degree 1 over degree 2, so the fraction is already low enough on top.

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = \frac{1}{3 z^{2} - 4 z + 1}

The numerator carried a z of its own, and it cancels the z we just divided by, so there is no term at the origin at all.

3 z^{2} - 4 z + 1 = 3 (z - 0.3333) (z - 1) = 0

So the poles are $0.3333$ and $1$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{1}{3 (z - 0.3333) (z - 1)} \\ = \frac{A}{(z - 0.3333)} + \frac{B}{(z - 1)} \end{aligned}

Put them back over the common denominator $3 (z - 0.3333) (z - 1)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z - 1) + B (z - 0.3333) \\ A (z - 1) & = A z - A \\ B (z - 0.3333) & = B z - 0.3333 B \end{aligned}

Now collect the powers of z and set each one equal to the same power in $0.3333$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + B & = 0 & from z \\ - A - 0.3333 B & = 0.3333 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = - 0.5 \\ B & = 0.5 \end{aligned}

Putting those back and adding the terms returns $1$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = \frac{- 0.5 z}{(z - 0.3333)} + \frac{0.5 z}{(z - 1)}

6. Invert term by term. The ROC $0.3333 < | z | < 1$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
0.3333	0.3333	-0.5	the ROC lies outside it	-0.5 ${(0.3333)}^{n} u [n]$
1	1	0.5	the ROC lies inside it	-0.5 ${(1)}^{n} u [- n - 1]$

With the ROC $0.3333 < | z | < 1$ , $x [n] = - 0.5 {(0.3333)}^{n} u [n] - 0.5 {(1)}^{n} u [- n - 1]$ .

Same as 2082 Bhadra Q32079 Baishakh · Q31+5 marksDefine z-transform for a discrete time signal. Find the inverse z-transform for

H (z) = z / (3 z^{2} - 4 z + 1)

using partial fraction method for

1 / 3 < | z | < 1

What is different in this one

The numerator is of lower degree than the denominator, so it is already proper: no division, straight to F(z) = $X (z)$ /z.
The ROC is a ring, so the answer is two sided: the poles inside the ring give right sided terms and the ones outside it give left sided terms. This is the case that separates a real answer from a memorised one.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{z}{3 z^{2} - 4 z + 1}

2. Nothing to divide. Degree 1 over degree 2, so the fraction is already low enough on top.

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = \frac{1}{3 z^{2} - 4 z + 1}

The numerator carried a z of its own, and it cancels the z we just divided by, so there is no term at the origin at all.

3 z^{2} - 4 z + 1 = 3 (z - 0.3333) (z - 1) = 0

So the poles are $0.3333$ and $1$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{1}{3 (z - 0.3333) (z - 1)} \\ = \frac{A}{(z - 0.3333)} + \frac{B}{(z - 1)} \end{aligned}

Put them back over the common denominator $3 (z - 0.3333) (z - 1)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z - 1) + B (z - 0.3333) \\ A (z - 1) & = A z - A \\ B (z - 0.3333) & = B z - 0.3333 B \end{aligned}

Now collect the powers of z and set each one equal to the same power in $0.3333$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + B & = 0 & from z \\ - A - 0.3333 B & = 0.3333 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = - 0.5 \\ B & = 0.5 \end{aligned}

Putting those back and adding the terms returns $1$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = \frac{- 0.5 z}{(z - 0.3333)} + \frac{0.5 z}{(z - 1)}

6. Invert term by term. The ROC $0.3333 < | z | < 1$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
0.3333	0.3333	-0.5	the ROC lies outside it	-0.5 ${(0.3333)}^{n} u [n]$
1	1	0.5	the ROC lies inside it	-0.5 ${(1)}^{n} u [- n - 1]$

With the ROC $0.3333 < | z | < 1$ , $x [n] = - 0.5 {(0.3333)}^{n} u [n] - 0.5 {(1)}^{n} u [- n - 1]$ .

2.2Properties of the z-transform

The properties HOT 7/19

81 Ba · 75 Ch · 75 Ash · 74 Ch · 74 Ash · 72 Ch · 72 Ka3+31+3+52+2+3

Before the formulas

You have the transform and you have its region. The properties are what make it worth having. A shift becomes a power of z. A fold becomes z to the minus one. Multiplying by an exponential moves the whole picture.

Each property turns an awkward operation on the sequence into ordinary algebra on the transform, which is the only reason anyone works in the z domain at all. There is nothing deep to understand here. There is a table to know, plus what each property does to the region of convergence, which is the half that gets forgotten and the half that is marked.

Property	$x [n]$	$X (z)$	ROC
Linearity	a $x_{1} [n] + b x_{2} [n]$	a X1(z) + b X2(z)	At least the intersection
Time shift	$x [n - k]$	$z^{- k} X (z)$	Same, except 0 or infinity
Scaling in z	$a^{n} x [n]$	X(z/a)	$\| a \|$ times the original
Time reversal	$x [- n]$	X(1/z)	Inverted: $1 / r 2 < \| z \| < 1$ /r1
Differentiation	n $x [n]$	-z dX(z)/dz	Same
Convolution	$x_{1} [n] * x_{2} [n]$	X1(z) X2(z)	At least the intersection
Conjugation	$x^{*} [n]$	X(z)	Same
Initial value	$x [0]$ for causal x	limit of $X (z)$ as z goes to infinity
Final value	limit of $x [n]$ as n goes to infinity	limit of (z-1) $X (z)$ as z goes to 1	Poles of (z-1) $X (z)$ inside the unit circle

The time shift property is what turns a difference equation into $H (z)$ in one line, and therefore what makes every structure in chapter 4 drawable.

See it drawnWhat a shift, a scale and a fold do to the transform

Each property trades work in one domain for work in the other, which is the whole reason the transform is worth taking.

Practise thisEvery question the paper has set on this topic, worked in full2

Practise this2075 Chaitra · Q64+6 marksList out the properties of Region of convergence and locate the ROC of the following signal

x [n] = {(0.1)}^{n} u [n] + {(0.3)}^{n} u [- n - 1]

What is different in this one

2 terms, 1 right sided and 1 left sided, so the ROC will be a ring if it exists at all.
Watch the sign on the left sided term: the standard pair is $- b^{n} u [- n - 1]$ , so a plus in the question picks up a minus in the transform.

Transform each term with its own ROC, then intersect them.

Term	Transform	Its own ROC
${(0.1)}^{n} u [n]$	$z / (z - 0.1)$	$\| z \| > 0.1$
${(0.3)}^{n} u [- n - 1]$	$- z / (z - 0.3)$	$\| z \| < 0.3$

Practise this2075 Ashwin · Q33+6 marksList out the properties of Region of convergence and locate the ROC of the following signal.

x [n] = {(0.6)}^{n} u [n] + {(0.25)}^{n} u [n]

What is different in this one

2 terms, 2 right sided and 0 left sided, so the ROC is simply the outside of the outermost pole.
Both terms are right sided, so no sign trap here.

Transform each term with its own ROC, then intersect them.

Term	Transform	Its own ROC
${(0.6)}^{n} u [n]$	$z / (z - 0.6)$	$\| z \| > 0.6$
${(0.25)}^{n} u [n]$	$z / (z - 0.25)$	$\| z \| > 0.25$

The same questions grouped by how often each topic is set: locating the region of convergence

The convolution property, derived PIN 2/19

76 Ash · 72 Ka3+33+6

Before the formulas

This is the property the rest of the subject is built on. Convolution took you a page of flipping and sliding in chapter 1. After transforming, it is a plain multiplication. That single fact is why filters are designed in the z domain and then brought back, rather than designed in time.

The derivation is four lines and the paper does ask for it, so learn it as a derivation and not only as a result.

Asked as derive the convolution property of z-transform and as state the convolution property. It is five lines and worth writing out in full.

\begin{aligned} y [n] & = x_{1} [n] * x_{2} [n] = \sum_{k} x_{1} [k] x_{2} [n - k] \\ Y (z) & = \sum_{n} y [n] z^{- n} = \sum_{n} [\sum_{k} x_{1} [k] x_{2} [n - k]] z^{- n} \\ = \sum_{k} x_{1} [k] [\sum_{n} x_{2} [n - k] z^{- n}] & interchange, valid in the common ROC \\ = \sum_{k} x_{1} [k] z^{- k} [\sum_{m} x_{2} [m] z^{- m}] & put m = n - k \\ = X_{1} (z) X_{2} (z) \end{aligned}

The ROC of the product is at least the intersection of the two ROCs, and can be larger if a pole of one is cancelled by a zero of the other.
Why this property is the whole point. It turns the convolution sum, which costs a sum per output sample, into one multiplication, and it is what makes $Y (z) = H (z) X (z)$ , the definition of the system function.
The same argument in chapter 7 gives circular convolution for the DFT, with the shift being circular instead of linear.

This single property gives $H (z) = Y (z) / X (z)$ , which is the object every remaining chapter manipulates: its poles and zeros, its realization, and its design.

See it drawnConvolution in time is multiplication in z

The long way round is shorter: transform, multiply, invert. That is the whole reason chapter 2 exists.

Practise thisEvery question the paper has set on this topic, worked in full1

Practise this2076 Ashwin · Q33+6 marksState Convolution property of Z-transform. Find inverse Z-transform of

X (z) = z

(z - 0.6) (z + 0.5)^{2}

}, ROC:

| z | > 0.6

What is different herethe region of convergence is outside the poles, so every term inverts to a right sided, causal sequence; a pole is REPEATED, so it needs a term for every power up to its order, the lower constant cannot be found by covering up, and the extra term inverts with an n in front of it

How this is read $X (z) = z / ((z - 0.6) (z + 0.5)^{2})$ , whose denominator multiplies out to $z^{3} + 0.4 z^{2} - 0.35 z - 0.15$ . The z on top cancels the z in F(z) = $X (z)$ /z, so this one never gets a term at the origin.

What is different in this one

The numerator is of lower degree than the denominator, so it is already proper: no division, straight to F(z) = $X (z)$ /z.
There is a repeated pole at -0.5. It needs a term for every power up to its order, and the extra one inverts with an n in front: $A z / (z - p)^{2}$ is (A/p) n $p^{n} u [n]$ , not A $p^{n} u [n]$ . Cover up cannot reach the lower constant at all, which is why the constants are found by comparing coefficients here.
The ROC is the outside of the outermost pole, so every term is right sided and the answer is causal.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{z}{z^{3} + 0.4 z^{2} - 0.35 z - 0.15}

2. Nothing to divide. Degree 1 over degree 3, so the fraction is already low enough on top.

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = \frac{1}{z^{3} + 0.4 z^{2} - 0.35 z - 0.15}

The numerator carried a z of its own, and it cancels the z we just divided by, so there is no term at the origin at all.

z^{3} + 0.4 z^{2} - 0.35 z - 0.15 = {(z + 0.5)}^{2} (z - 0.6) = 0

So the poles are $- 0.5$ , repeated 2 times and $0.6$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{1}{{(z + 0.5)}^{2} (z - 0.6)} \\ = \frac{A}{(z + 0.5)} + \frac{B}{{(z + 0.5)}^{2}} + \frac{C}{(z - 0.6)} \end{aligned}

Put them back over the common denominator ${(z + 0.5)}^{2} (z - 0.6)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z + 0.5) (z - 0.6) + B (z - 0.6) + C {(z + 0.5)}^{2} \\ A (z + 0.5) (z - 0.6) & = A z^{2} - 0.1 A z - 0.3 A \\ B (z - 0.6) & = B z - 0.6 B \\ C {(z + 0.5)}^{2} & = C z^{2} + C z + 0.25 C \end{aligned}

Now collect the powers of z and set each one equal to the same power in $1$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + C & = 0 & from z^{2} \\ - 0.1 A + B + C & = 0 & from z \\ - 0.3 A - 0.6 B + 0.25 C & = 1 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = - 0.8264 \\ B & = - 0.9091 \\ C & = 0.8264 \end{aligned}

Putting those back and adding the terms returns $1$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = \frac{- 0.8264 z}{(z + 0.5)} + \frac{- 0.9091 z}{{(z + 0.5)}^{2}} + \frac{0.8264 z}{(z - 0.6)}

6. Invert term by term. The ROC $| z | > 0.6$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
-0.5	0.5	-0.8264	the ROC lies outside it	-0.8264 ${(- 0.5)}^{n} u [n]$
-0.5, order 2	0.5	-0.9091	the ROC lies outside it	1.8182 n ${(- 0.5)}^{n} u [n]$
0.6	0.6	0.8264	the ROC lies outside it	0.8264 ${(0.6)}^{n} u [n]$

With the ROC $| z | > 0.6$ , $x [n] = - 0.8264 {(- 0.5)}^{n} u [n]$ + 1.8182 n ${(- 0.5)}^{n} u [n] + 0.8264 {(0.6)}^{n} u [n]$ .

Check it without trusting any of the algebra: run the difference equation the transform encodes and compare the first samples. Both routes give ${0, 0, 1, - 0.4, 0.51, - 0.194}$ .

2.3Inverse z-transform by long division and partial fractions

Inverting $X (z)$ by long division TOP 15/19

82 Bh · 82 Ba · 81 Bh · 81 Ba · 80 Bh · 80 Ba · 79 Bh · 79 Ba · 78 Bh · 76 Ch · 76 Ash · 74 Ch · 74 Ash · 73 Shr · 71 Shr1+561+3+5

Before the formulas

Having a transform is only useful if you can get back out of it. There are two ways, and this is the blunt one. Divide the numerator by the denominator the way you would divide numbers, and the coefficients of the quotient are the samples in order, one at a time.

It gives you the first few values quickly, it confirms a causal answer, and it never gives you a closed form for the general term. Use it when the question asks for a few values, and say which way you arranged the polynomials, because that is what decides whether you get the causal sequence or the other one.

The power series, or long division, method divides the numerator by the denominator until a pattern appears. The z-transform is a power series in $z^{- 1}$ by definition, so the coefficient of $z^{- n}$ is $x [n]$ , read straight off.

X (z) = \sum_{n} x [n] z^{- n} = x [0] + x [1] z^{- 1} + x [2] z^{- 2} + \dots

ROC	Arrange both polynomials in	The quotient is
causal, $\| z \| > r$	ascending powers of $z^{- 1}$	$x [0]$ , $x [1]$ , $x [2]$ ...
anticausal, $\| z \| < r$	ascending powers of z	$x [- 1]$ , $x [- 2]$ ...

The ROC chooses the direction of the division, exactly as it chooses $u [n]$ against $- u [- n - 1]$ in the partial fraction method. Dividing the wrong way gives a valid series that converges nowhere near the given ROC.
What it is good for: getting the first few samples quickly, and checking a partial fraction answer. It is the method to reach for when the question says find the first four samples.
What it is bad for: a closed form. Long division gives numbers, not a formula, unless the pattern is obvious. For a closed form, use partial fractions.

Worked example

Find the first four samples of $x [n]$ for $X (z) = 1 / (1 - 1.5 z^{- 1} + 0.5 z^{- 2})$ , with the ROC $| z | > 1$ , so the sequence is causal.

1 + 1.5 z^-1 + 1.75 z^-2 + 1.875 z^-3 ... ______________________________________________ 1 - 1.5z^-1 + 0.5z^-2 ) 1 1 - 1.5 z^-1 + 0.5 z^-2 --------------------------- 1.5 z^-1 - 0.5 z^-2 1.5 z^-1 - 2.25 z^-2 + 0.75 z^-3 --------------------------------- 1.75 z^-2 - 0.75 z^-3 1.75 z^-2 - 2.625 z^-3 + 0.875 z^-4 ------------------------------------ 1.875 z^-3 ...

The same numbers come out of the difference equation the transform encodes, $x [n] = 1$ . $5 x [n - 1]$ - 0. $5 x [n - 2] + δ [n]$ , which is the quick way to carry the division on: 1, 1.5, 1.75, 1.875, 1.9375, and so on.

$x [n]$ = {1, 1.5, 1.75, 1.875, ...} for $n = 0$ , 1, 2, 3. Partial fractions confirm it in closed form: $X (z) = 2 z / (z - 1) - z / (z - 0.5)$ , so $x [n] = 2 - {(0.5)}^{n}$ for n ≥ 0, which gives 1, 1.5, 1.75, 1.875 and tends to 2.

The same division, run the other way, is how partial fractions start when the numerator is the larger.

See it drawnLong division gives the samples one at a time

It never gives a closed form, so it answers "find the first four samples" and nothing more. That is exactly what the paper asks.

Practise thisEvery question the paper has set on this topic, worked in full15

Practise this +1 like it2082 Bhadra · Q31+5 marksDefine z-transform for a discrete time signal. Find the inverse z-transform for

H (z) = z / (3 z^{2} - 4 z + 1)

using partial fraction method for

1 / 3 < | z | < 1

What is different in this one

The numerator is of lower degree than the denominator, so it is already proper: no division, straight to F(z) = $X (z)$ /z.
The ROC is a ring, so the answer is two sided: the poles inside the ring give right sided terms and the ones outside it give left sided terms. This is the case that separates a real answer from a memorised one.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{z}{3 z^{2} - 4 z + 1}

2. Nothing to divide. Degree 1 over degree 2, so the fraction is already low enough on top.

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = \frac{1}{3 z^{2} - 4 z + 1}

The numerator carried a z of its own, and it cancels the z we just divided by, so there is no term at the origin at all.

3 z^{2} - 4 z + 1 = 3 (z - 0.3333) (z - 1) = 0

So the poles are $0.3333$ and $1$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{1}{3 (z - 0.3333) (z - 1)} \\ = \frac{A}{(z - 0.3333)} + \frac{B}{(z - 1)} \end{aligned}

Put them back over the common denominator $3 (z - 0.3333) (z - 1)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z - 1) + B (z - 0.3333) \\ A (z - 1) & = A z - A \\ B (z - 0.3333) & = B z - 0.3333 B \end{aligned}

Now collect the powers of z and set each one equal to the same power in $0.3333$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + B & = 0 & from z \\ - A - 0.3333 B & = 0.3333 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = - 0.5 \\ B & = 0.5 \end{aligned}

Putting those back and adding the terms returns $1$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = \frac{- 0.5 z}{(z - 0.3333)} + \frac{0.5 z}{(z - 1)}

6. Invert term by term. The ROC $0.3333 < | z | < 1$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
0.3333	0.3333	-0.5	the ROC lies outside it	-0.5 ${(0.3333)}^{n} u [n]$
1	1	0.5	the ROC lies inside it	-0.5 ${(1)}^{n} u [- n - 1]$

With the ROC $0.3333 < | z | < 1$ , $x [n] = - 0.5 {(0.3333)}^{n} u [n] - 0.5 {(1)}^{n} u [- n - 1]$ .

Practise this2082 Baishakh · Q31+5 marksDefine ROC of a Z-transform. Determine

x [n]

for

X (z) = 1 / (1 - 1.5 z^{- 1} + 0.5 z^{- 2})

if: a)

| z | > 1

| z | < 0.5

0.5 < | z | < 1

using partial fraction method.What is different hereOne

X (z)

, three answers. This is the question the whole chapter exists for, and the partial fractions are done ONCE: only the side each pole falls on changes.

What is different in this one

The same $X (z)$ is given with 3 different regions, so it stands for 3 different signals. The partial fractions are worked once: the poles and the constants do not know about the ROC, and only the very last step, which side each pole falls on, changes between the parts.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{z^{2}}{z^{2} - 1.5 z + 0.5}

2. Nothing to divide. Degree 2 over degree 2, so the fraction is already low enough on top.

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = \frac{z}{z^{2} - 1.5 z + 0.5}

The numerator carried a z of its own, and it cancels the z we just divided by, so there is no term at the origin at all.

z^{2} - 1.5 z + 0.5 = (z - 0.5) (z - 1) = 0

So the poles are $0.5$ and $1$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{z}{(z - 0.5) (z - 1)} \\ = \frac{A}{(z - 0.5)} + \frac{B}{(z - 1)} \end{aligned}

Put them back over the common denominator $(z - 0.5) (z - 1)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z - 1) + B (z - 0.5) \\ A (z - 1) & = A z - A \\ B (z - 0.5) & = B z - 0.5 B \end{aligned}

Now collect the powers of z and set each one equal to the same power in $z$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + B & = 1 & from z \\ - A - 0.5 B & = 0 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = - 1 \\ B & = 2 \end{aligned}

Putting those back and adding the terms returns $z$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = \frac{- 1 z}{(z - 0.5)} + \frac{2 z}{(z - 1)}

6. Now the regions. Everything above is common to all of them. Each part below changes one thing only: whether the region lies outside a pole, which makes it causal, or inside it, which makes it anticausal.

$| z | > 1$

Pole	$\| p \|$	Constant	Which side	Inverts to
0.5	0.5	-1	the ROC lies outside it	- ${(0.5)}^{n} u [n]$
1	1	2	the ROC lies outside it	2 ${(1)}^{n} u [n]$

With the ROC $| z | > 1$ , $x [n] = - {(0.5)}^{n} u [n] + 2 {(1)}^{n} u [n]$ .

$| z | < 0.5$

Pole	$\| p \|$	Constant	Which side	Inverts to
0.5	0.5	-1	the ROC lies inside it	${(0.5)}^{n} u [- n - 1]$
1	1	2	the ROC lies inside it	-2 ${(1)}^{n} u [- n - 1]$

With the ROC $| z | < 0.5$ , $x [n] = {(0.5)}^{n} u [- n - 1] - 2 {(1)}^{n} u [- n - 1]$ .

$0.5 < | z | < 1$

Pole	$\| p \|$	Constant	Which side	Inverts to
0.5	0.5	-1	the ROC lies outside it	- ${(0.5)}^{n} u [n]$
1	1	2	the ROC lies inside it	-2 ${(1)}^{n} u [- n - 1]$

With the ROC $0.5 < | z | < 1$ , $x [n] = - {(0.5)}^{n} u [n] - 2 {(1)}^{n} u [- n - 1]$ .

Side by side, the same transform read 3 ways:

ROC	Which side each pole falls	$x [n]$	Stable?
$\| z \| > 1$	0.5 is right sided, 1 is right sided	- ${(0.5)}^{n} u [n] + 2 {(1)}^{n} u [n]$	no
$\| z \| < 0.5$	0.5 is left sided, 1 is left sided	${(0.5)}^{n} u [- n - 1] - 2 {(1)}^{n} u [- n - 1]$	no
$0.5 < \| z \| < 1$	0.5 is right sided, 1 is left sided	- ${(0.5)}^{n} u [n] - 2 {(1)}^{n} u [- n - 1]$	no

Practise this +1 like it2081 Bhadra · Q31+5 marksDefine ROC of z-transform. Find the inverse z-transform using partial fraction expansion of

X (z) = (2 z^{3} - 5 z^{2} + z + 3) / ((z - 1) (z - 2))

for ROC

| z | < 1

What is different herethe region of convergence is inside the pole, so every term inverts to a left sided sequence; the numerator is ONE degree higher, so a single long division step comes first, and its quotient is an impulse at a negative n; the numerator reaches the degree of the denominator, so dividing by z leaves a pole AT THE ORIGIN, and that term inverts to a plain impulse on top of the exponentials

How this is read $X (z) = (2 z^{3} - 5 z^{2} + z + 3) / ((z - 1) (z - 2))$ , and the denominator multiplies out to $z^{2} - 3 z + 2$ . The numerator is one degree higher, so it is long divided once before anything else.

What is different in this one

The numerator is of higher degree than the denominator (3 against 2), so it has to be long divided first. The quotient is a polynomial in z, which inverts to impulses at negative n, and only the remainder goes through partial fractions. Divide until the remainder reaches the degree of the divisor and stop there, not further.
The ROC is the inside of the innermost pole, so every term is left sided and the whole answer carries $u [- n - 1]$ with a sign change.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{2 z^{3} - 5 z^{2} + z + 3}{z^{2} - 3 z + 2}

2. The numerator is of higher degree, so divide first. Degree 3 over degree 2. Divide until the remainder has the same degree as the divisor and then stop: that is exactly far enough to make the leftover proper once it is divided by z, and it leaves the quotient with no constant term.

			$2 z$
$z^{2} - 3 z + 2$ )	$2 z^{3}$	$- 5 z^{2}$	$+ z$	$+ 3$
	$2 z^{3}$	$- 6 z^{2}$	$+ 4 z$
		$z^{2}$	$- 3 z$	$+ 3$

X (z) = 2 z + \frac{z^{2} - 3 z + 3}{z^{2} - 3 z + 2}

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = 2 + \frac{z^{2} - 3 z + 3}{z^{3} - 3 z^{2} + 2 z}

The poles are the roots of that denominator. Note the z we just divided by is one of them, sitting at the origin.

z^{3} - 3 z^{2} + 2 z = z (z - 1) (z - 2) = 0

So the poles are $0$ , $1$ and $2$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{z^{2} - 3 z + 3}{z (z - 1) (z - 2)} \\ = \frac{A}{z} + \frac{B}{(z - 1)} + \frac{C}{(z - 2)} \end{aligned}

Put them back over the common denominator $z (z - 1) (z - 2)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z - 1) (z - 2) + B z (z - 2) + C z (z - 1) \\ A (z - 1) (z - 2) & = A z^{2} - 3 A z + 2 A \\ B z (z - 2) & = B z^{2} - 2 B z \\ C z (z - 1) & = C z^{2} - C z \end{aligned}

Now collect the powers of z and set each one equal to the same power in $z^{2} - 3 z + 3$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + B + C & = 1 & from z^{2} \\ - 3 A - 2 B - C & = - 3 & from z \\ 2 A & = 3 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = 1.5 \\ B & = - 1 \\ C & = 0.5 \end{aligned}

Putting those back and adding the terms returns $z^{2} - 3 z + 3$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = 2 z + 1.5 + \frac{- 1 z}{(z - 1)} + \frac{0.5 z}{(z - 2)}

6. Invert term by term. The ROC $| z | < 1$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
none, from the division		2	a polynomial term	2 $δ [n + 1]$
0, the origin	0	1.5	a polynomial term	1.5 $δ [n]$
1	1	-1	the ROC lies inside it	${(1)}^{n} u [- n - 1]$
2	2	0.5	the ROC lies inside it	-0.5 ${(2)}^{n} u [- n - 1]$

With the ROC $| z | < 1$ , $x [n] = 2 δ [n + 1] + 1.5 δ [n] + {(1)}^{n} u [- n - 1] - 0.5 {(2)}^{n} u [- n - 1]$ .

Practise this2081 Baishakh · Q32+2+3 marksDetermine the inverse z-transform of

X (z) = 1 / (1 - 0.8 z^{- 1} + 0.12 z^{- 2}) . (i)

if ROC is

| z | > 0.6

(ii) if ROC is

| z | < 0.2

(iii) if ROC is

0.2 < | z | < 0.6

What is different hereSame shape as the 2082 Baishakh question with different poles: 0.6 and 0.2 rather than 1 and 0.5. Note that here the middle ROC does NOT contain the unit circle, so none of the three is stable. This paper does not name a method, so the long division is worked underneath as a second route; the 2082 Baishakh one says partial fraction and gets partial fractions only.

What is different in this one

The same $X (z)$ is given with 3 different regions, so it stands for 3 different signals. The partial fractions are worked once: the poles and the constants do not know about the ROC, and only the very last step, which side each pole falls on, changes between the parts.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{z^{2}}{z^{2} - 0.8 z + 0.12}

2. Nothing to divide. Degree 2 over degree 2, so the fraction is already low enough on top.

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = \frac{z}{z^{2} - 0.8 z + 0.12}

The numerator carried a z of its own, and it cancels the z we just divided by, so there is no term at the origin at all.

z^{2} - 0.8 z + 0.12 = (z - 0.2) (z - 0.6) = 0

So the poles are $0.2$ and $0.6$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{z}{(z - 0.2) (z - 0.6)} \\ = \frac{A}{(z - 0.2)} + \frac{B}{(z - 0.6)} \end{aligned}

Put them back over the common denominator $(z - 0.2) (z - 0.6)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z - 0.6) + B (z - 0.2) \\ A (z - 0.6) & = A z - 0.6 A \\ B (z - 0.2) & = B z - 0.2 B \end{aligned}

Now collect the powers of z and set each one equal to the same power in $z$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + B & = 1 & from z \\ - 0.6 A - 0.2 B & = 0 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = - 0.5 \\ B & = 1.5 \end{aligned}

Putting those back and adding the terms returns $z$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = \frac{- 0.5 z}{(z - 0.2)} + \frac{1.5 z}{(z - 0.6)}

$| z | > 0.6$

Pole	$\| p \|$	Constant	Which side	Inverts to
0.2	0.2	-0.5	the ROC lies outside it	-0.5 ${(0.2)}^{n} u [n]$
0.6	0.6	1.5	the ROC lies outside it	1.5 ${(0.6)}^{n} u [n]$

With the ROC $| z | > 0.6$ , $x [n] = - 0.5 {(0.2)}^{n} u [n] + 1.5 {(0.6)}^{n} u [n]$ .

$| z | < 0.2$

Pole	$\| p \|$	Constant	Which side	Inverts to
0.2	0.2	-0.5	the ROC lies inside it	0.5 ${(0.2)}^{n} u [- n - 1]$
0.6	0.6	1.5	the ROC lies inside it	-1.5 ${(0.6)}^{n} u [- n - 1]$

With the ROC $| z | < 0.2$ , $x [n] = 0.5 {(0.2)}^{n} u [- n - 1] - 1.5 {(0.6)}^{n} u [- n - 1]$ .

$0.2 < | z | < 0.6$

Pole	$\| p \|$	Constant	Which side	Inverts to
0.2	0.2	-0.5	the ROC lies outside it	-0.5 ${(0.2)}^{n} u [n]$
0.6	0.6	1.5	the ROC lies inside it	-1.5 ${(0.6)}^{n} u [- n - 1]$

With the ROC $0.2 < | z | < 0.6$ , $x [n] = - 0.5 {(0.2)}^{n} u [n] - 1.5 {(0.6)}^{n} u [- n - 1]$ .

Side by side, the same transform read 3 ways:

ROC	Which side each pole falls	$x [n]$	Stable?
$\| z \| > 0.6$	0.2 is right sided, 0.6 is right sided	-0.5 ${(0.2)}^{n} u [n] + 1.5 {(0.6)}^{n} u [n]$	yes
$\| z \| < 0.2$	0.2 is left sided, 0.6 is left sided	0.5 ${(0.2)}^{n} u [- n - 1] - 1.5 {(0.6)}^{n} u [- n - 1]$	no
$0.2 < \| z \| < 0.6$	0.2 is right sided, 0.6 is left sided	-0.5 ${(0.2)}^{n} u [n] - 1.5 {(0.6)}^{n} u [- n - 1]$	no

Because this paper does not name a method, here is the other route as a check on the first few samples. It gives the numbers but never the general term, which is why partial fractions is the answer above.

$| z | > 0.6$ , so start the divisor at its highest term.

	$1$	$+ 0.8 z^{- 1}$	$+ 0.52 z^{- 2}$	$+ 0.32 z^{- 3}$	$+ 0.1936 z^{- 4}$	$+ 0.1165 z^{- 5}$	$+ 0.07 z^{- 6}$	$+ 0.042 z^{- 7}$
$1 - 0.8 z^{- 1} + 0.12 z^{- 2}$ )	$1$
	$1$	$- 0.8 z^{- 1}$	$+ 0.12 z^{- 2}$
		$0.8 z^{- 1}$	$- 0.12 z^{- 2}$
		$0.8 z^{- 1}$	$- 0.64 z^{- 2}$	$+ 0.096 z^{- 3}$
			$0.52 z^{- 2}$	$- 0.096 z^{- 3}$
			$0.52 z^{- 2}$	$- 0.416 z^{- 3}$	$+ 0.0624 z^{- 4}$
				$0.32 z^{- 3}$	$- 0.0624 z^{- 4}$
				$0.32 z^{- 3}$	$- 0.256 z^{- 4}$	$+ 0.0384 z^{- 5}$
					$0.1936 z^{- 4}$	$- 0.0384 z^{- 5}$
					$0.1936 z^{- 4}$	$- 0.1549 z^{- 5}$	$+ 0.0232 z^{- 6}$
						$0.1165 z^{- 5}$	$- 0.0232 z^{- 6}$
						$0.1165 z^{- 5}$	$- 0.0932 z^{- 6}$	$+ 0.014 z^{- 7}$
							$0.07 z^{- 6}$	$- 0.014 z^{- 7}$
							$0.07 z^{- 6}$	$- 0.056 z^{- 7}$	$+ 0.0084 z^{- 8}$
								$0.042 z^{- 7}$	$- 0.0084 z^{- 8}$
								$0.042 z^{- 7}$	$- 0.0336 z^{- 8}$	$+ 0.005 z^{- 9}$
									$0.0252 z^{- 8}$	$- 0.005 z^{- 9}$

$| z | < 0.2$ , so start the divisor at its lowest term.

	$8.3333 z^{2}$	$+ 55.5556 z^{3}$	$+ 300.9259 z^{4}$	$+ 1543.2099 z^{5}$	$+ 7780.3498 z^{6}$	$+ 39008.9163 z^{7}$
$0.12 - 0.8 z + z^{2}$ )	$z^{2}$
	$z^{2}$	$- 6.6667 z^{3}$	$+ 8.3333 z^{4}$
		$6.6667 z^{3}$	$- 8.3333 z^{4}$
		$6.6667 z^{3}$	$- 44.4444 z^{4}$	$+ 55.5556 z^{5}$
			$36.1111 z^{4}$	$- 55.5556 z^{5}$
			$36.1111 z^{4}$	$- 240.7407 z^{5}$	$+ 300.9259 z^{6}$
				$185.1852 z^{5}$	$- 300.9259 z^{6}$
				$185.1852 z^{5}$	$- 1234.5679 z^{6}$	$+ 1543.2099 z^{7}$
					$933.642 z^{6}$	$- 1543.2099 z^{7}$
					$933.642 z^{6}$	$- 6224.2798 z^{7}$	$+ 7780.3498 z^{8}$
						$4681.07 z^{7}$	$- 7780.3498 z^{8}$
						$4681.07 z^{7}$	$- 31207.1331 z^{8}$	$+ 39008.9163 z^{9}$
							$23426.7833 z^{8}$	$- 39008.9163 z^{9}$

Practise this2080 Bhadra · Q31+6 marksDefine the Region of Convergence (ROC). Find the inverse of

H (z) = (1 + 2 z^{- 1} + z^{- 2}) / (1 - 0.75 z^{- 1} + 0.125 z^{- 2})

; ROC

0.25 < | z | < 0.5

What is different herethe numerator reaches the degree of the denominator, so dividing by z leaves a pole AT THE ORIGIN, and that term inverts to a plain impulse on top of the exponentials

What is different in this one

The numerator reaches the degree of the denominator. Dividing by z is enough to make it proper, so no long division is needed, but the expansion will carry a term at the origin that inverts to an impulse.
The ROC is a ring, so the answer is two sided: the poles inside the ring give right sided terms and the ones outside it give left sided terms. This is the case that separates a real answer from a memorised one.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{z^{2} + 2 z + 1}{z^{2} - 0.75 z + 0.125}

2. Nothing to divide. Degree 2 over degree 2, so the fraction is already low enough on top.

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = \frac{z^{2} + 2 z + 1}{z^{3} - 0.75 z^{2} + 0.125 z}

The poles are the roots of that denominator. Note the z we just divided by is one of them, sitting at the origin.

z^{3} - 0.75 z^{2} + 0.125 z = z (z - 0.25) (z - 0.5) = 0

So the poles are $0$ , $0.25$ and $0.5$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{z^{2} + 2 z + 1}{z (z - 0.25) (z - 0.5)} \\ = \frac{A}{z} + \frac{B}{(z - 0.25)} + \frac{C}{(z - 0.5)} \end{aligned}

Put them back over the common denominator $z (z - 0.25) (z - 0.5)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z - 0.25) (z - 0.5) + B z (z - 0.5) + C z (z - 0.25) \\ A (z - 0.25) (z - 0.5) & = A z^{2} - 0.75 A z + 0.125 A \\ B z (z - 0.5) & = B z^{2} - 0.5 B z \\ C z (z - 0.25) & = C z^{2} - 0.25 C z \end{aligned}

Now collect the powers of z and set each one equal to the same power in $z^{2} + 2 z + 1$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + B + C & = 1 & from z^{2} \\ - 0.75 A - 0.5 B - 0.25 C & = 2 & from z \\ 0.125 A & = 1 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = 8 \\ B & = - 25 \\ C & = 18 \end{aligned}

Putting those back and adding the terms returns $z^{2} + 2 z + 1$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = 8 + \frac{- 25 z}{(z - 0.25)} + \frac{18 z}{(z - 0.5)}

6. Invert term by term. The ROC $0.25 < | z | < 0.5$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
0, the origin	0	8	a polynomial term	8 $δ [n]$
0.25	0.25	-25	the ROC lies outside it	-25 ${(0.25)}^{n} u [n]$
0.5	0.5	18	the ROC lies inside it	-18 ${(0.5)}^{n} u [- n - 1]$

With the ROC $0.25 < | z | < 0.5$ , $x [n] = 8 δ [n] - 25 {(0.25)}^{n} u [n] - 18 {(0.5)}^{n} u [- n - 1]$ .

Practise this +1 like it2080 Baishakh · Q31+5 marksDefine ROC. Find inverse z-transform of

X (z) = (1 + 2 z^{- 1} + z^{- 2}) / (1 - 1.5 z^{- 1} + 0.5 z^{- 2})

, ROC:

| z | > 1

.What is different herethe region of convergence is outside the poles, so every term inverts to a right sided, causal sequence; the numerator reaches the degree of the denominator, so dividing by z leaves a pole AT THE ORIGIN, and that term inverts to a plain impulse on top of the exponentials

What is different in this one

The numerator reaches the degree of the denominator. Dividing by z is enough to make it proper, so no long division is needed, but the expansion will carry a term at the origin that inverts to an impulse.
The ROC is the outside of the outermost pole, so every term is right sided and the answer is causal.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{z^{2} + 2 z + 1}{z^{2} - 1.5 z + 0.5}

2. Nothing to divide. Degree 2 over degree 2, so the fraction is already low enough on top.

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = \frac{z^{2} + 2 z + 1}{z^{3} - 1.5 z^{2} + 0.5 z}

The poles are the roots of that denominator. Note the z we just divided by is one of them, sitting at the origin.

z^{3} - 1.5 z^{2} + 0.5 z = z (z - 0.5) (z - 1) = 0

So the poles are $0$ , $0.5$ and $1$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{z^{2} + 2 z + 1}{z (z - 0.5) (z - 1)} \\ = \frac{A}{z} + \frac{B}{(z - 0.5)} + \frac{C}{(z - 1)} \end{aligned}

Put them back over the common denominator $z (z - 0.5) (z - 1)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z - 0.5) (z - 1) + B z (z - 1) + C z (z - 0.5) \\ A (z - 0.5) (z - 1) & = A z^{2} - 1.5 A z + 0.5 A \\ B z (z - 1) & = B z^{2} - B z \\ C z (z - 0.5) & = C z^{2} - 0.5 C z \end{aligned}

Now collect the powers of z and set each one equal to the same power in $z^{2} + 2 z + 1$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + B + C & = 1 & from z^{2} \\ - 1.5 A - B - 0.5 C & = 2 & from z \\ 0.5 A & = 1 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = 2 \\ B & = - 9 \\ C & = 8 \end{aligned}

Putting those back and adding the terms returns $z^{2} + 2 z + 1$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = 2 + \frac{- 9 z}{(z - 0.5)} + \frac{8 z}{(z - 1)}

6. Invert term by term. The ROC $| z | > 1$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
0, the origin	0	2	a polynomial term	2 $δ [n]$
0.5	0.5	-9	the ROC lies outside it	-9 ${(0.5)}^{n} u [n]$
1	1	8	the ROC lies outside it	8 ${(1)}^{n} u [n]$

With the ROC $| z | > 1$ , $x [n] = 2 δ [n] - 9 {(0.5)}^{n} u [n] + 8 {(1)}^{n} u [n]$ .

Check it without trusting any of the algebra: run the difference equation the transform encodes and compare the first samples. Both routes give ${1, 3.5, 5.75, 6.875, 7.4375, 7.71875}$ .

Practise this +1 like it2079 Bhadra · Q36 marksFind inverses Z-transform of

X (z) = (2 z^{4} + 2 z^{3} - 3 z + 2) / (z^{2} - 1.5 z - 1)

, ROC:

| z | < 0.5

, using partial fraction expansion method.What is different herethe region of convergence is inside the pole, so every term inverts to a left sided sequence; the numerator is TWO degrees higher, so the long division runs for two steps before the partial fractions even start; the numerator reaches the degree of the denominator, so dividing by z leaves a pole AT THE ORIGIN, and that term inverts to a plain impulse on top of the exponentials

How this is read $X (z) = (2 z^{4} + 2 z^{3} - 3 z + 2) / (z^{2} - 1.5 z - 1)$ . The denominator factors as (z - 2)(z + 0.5), so the poles are at 2 and -0.5 and the printed ROC $| z | < 0.5$ is exactly the innermost region, the anticausal one.

What is different in this one

The numerator is of higher degree than the denominator (4 against 2), so it has to be long divided first. The quotient is a polynomial in z, which inverts to impulses at negative n, and only the remainder goes through partial fractions. Divide until the remainder reaches the degree of the divisor and stop there, not further.
The ROC is the inside of the innermost pole, so every term is left sided and the whole answer carries $u [- n - 1]$ with a sign change.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{2 z^{4} + 2 z^{3} - 3 z + 2}{z^{2} - 1.5 z - 1}

2. The numerator is of higher degree, so divide first. Degree 4 over degree 2. Divide until the remainder has the same degree as the divisor and then stop: that is exactly far enough to make the leftover proper once it is divided by z, and it leaves the quotient with no constant term.

			$2 z^{2}$	$+ 5 z$
$z^{2} - 1.5 z - 1$ )	$2 z^{4}$	$+ 2 z^{3}$	$+ 0 z^{2}$	$- 3 z$	$+ 2$
	$2 z^{4}$	$- 3 z^{3}$	$- 2 z^{2}$
		$5 z^{3}$	$+ 2 z^{2}$	$- 3 z$	$+ 2$
		$5 z^{3}$	$- 7.5 z^{2}$	$- 5 z$
			$9.5 z^{2}$	$+ 2 z$	$+ 2$

X (z) = 2 z^{2} + 5 z + \frac{9.5 z^{2} + 2 z + 2}{z^{2} - 1.5 z - 1}

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = 2 z + 5 + \frac{9.5 z^{2} + 2 z + 2}{z^{3} - 1.5 z^{2} - z}

The poles are the roots of that denominator. Note the z we just divided by is one of them, sitting at the origin.

z^{3} - 1.5 z^{2} - z = z (z + 0.5) (z - 2) = 0

So the poles are $0$ , $- 0.5$ and $2$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{9.5 z^{2} + 2 z + 2}{z (z + 0.5) (z - 2)} \\ = \frac{A}{z} + \frac{B}{(z + 0.5)} + \frac{C}{(z - 2)} \end{aligned}

Put them back over the common denominator $z (z + 0.5) (z - 2)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z + 0.5) (z - 2) + B z (z - 2) + C z (z + 0.5) \\ A (z + 0.5) (z - 2) & = A z^{2} - 1.5 A z - A \\ B z (z - 2) & = B z^{2} - 2 B z \\ C z (z + 0.5) & = C z^{2} + 0.5 C z \end{aligned}

Now collect the powers of z and set each one equal to the same power in $9.5 z^{2} + 2 z + 2$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + B + C & = 9.5 & from z^{2} \\ - 1.5 A - 2 B + 0.5 C & = 2 & from z \\ - A & = 2 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = - 2 \\ B & = 2.7 \\ C & = 8.8 \end{aligned}

Putting those back and adding the terms returns $9.5 z^{2} + 2 z + 2$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = 2 z^{2} + 5 z - 2 + \frac{2.7 z}{(z + 0.5)} + \frac{8.8 z}{(z - 2)}

6. Invert term by term. The ROC $| z | < 0.5$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
none, from the division		2	a polynomial term	2 $δ [n + 2]$
none, from the division		5	a polynomial term	5 $δ [n + 1]$
0, the origin	0	-2	a polynomial term	-2 $δ [n]$
-0.5	0.5	2.7	the ROC lies inside it	-2.7 ${(- 0.5)}^{n} u [- n - 1]$
2	2	8.8	the ROC lies inside it	-8.8 ${(2)}^{n} u [- n - 1]$

With the ROC $| z | < 0.5$ , $x [n] = 2 δ [n + 2] + 5 δ [n + 1] - 2 δ [n] - 2.7 {(- 0.5)}^{n} u [- n - 1] - 8.8 {(2)}^{n} u [- n - 1]$ .

Same as 2082 Bhadra Q32079 Baishakh · Q31+5 marksDefine z-transform for a discrete time signal. Find the inverse z-transform for

H (z) = z / (3 z^{2} - 4 z + 1)

using partial fraction method for

1 / 3 < | z | < 1

What is different in this one

The numerator is of lower degree than the denominator, so it is already proper: no division, straight to F(z) = $X (z)$ /z.
The ROC is a ring, so the answer is two sided: the poles inside the ring give right sided terms and the ones outside it give left sided terms. This is the case that separates a real answer from a memorised one.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{z}{3 z^{2} - 4 z + 1}

2. Nothing to divide. Degree 1 over degree 2, so the fraction is already low enough on top.

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = \frac{1}{3 z^{2} - 4 z + 1}

The numerator carried a z of its own, and it cancels the z we just divided by, so there is no term at the origin at all.

3 z^{2} - 4 z + 1 = 3 (z - 0.3333) (z - 1) = 0

So the poles are $0.3333$ and $1$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{1}{3 (z - 0.3333) (z - 1)} \\ = \frac{A}{(z - 0.3333)} + \frac{B}{(z - 1)} \end{aligned}

Put them back over the common denominator $3 (z - 0.3333) (z - 1)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z - 1) + B (z - 0.3333) \\ A (z - 1) & = A z - A \\ B (z - 0.3333) & = B z - 0.3333 B \end{aligned}

Now collect the powers of z and set each one equal to the same power in $0.3333$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + B & = 0 & from z \\ - A - 0.3333 B & = 0.3333 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = - 0.5 \\ B & = 0.5 \end{aligned}

Putting those back and adding the terms returns $1$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = \frac{- 0.5 z}{(z - 0.3333)} + \frac{0.5 z}{(z - 1)}

6. Invert term by term. The ROC $0.3333 < | z | < 1$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
0.3333	0.3333	-0.5	the ROC lies outside it	-0.5 ${(0.3333)}^{n} u [n]$
1	1	0.5	the ROC lies inside it	-0.5 ${(1)}^{n} u [- n - 1]$

With the ROC $0.3333 < | z | < 1$ , $x [n] = - 0.5 {(0.3333)}^{n} u [n] - 0.5 {(1)}^{n} u [- n - 1]$ .

Practise this2078 Bhadra · Q31+5 marksDefine ROC. Find inverse z-transform of

X (z) = (z^{3} + z^{2} + 1.5 z + 0.5) / (z^{3} + 1.5 z^{2} + 0.5 z)

, ROC:

| z | < 1 / 2

.What is different herethe region of convergence is inside the pole, so every term inverts to a left sided sequence; the numerator reaches the degree of the denominator, so dividing by z leaves a pole AT THE ORIGIN, and that term inverts to a plain impulse on top of the exponentials; a pole is REPEATED, so it needs a term for every power up to its order, the lower constant cannot be found by covering up, and the extra term inverts with an n in front of it

How this is read $X (z) = (z^{3} + z^{2} + 1.5 z + 0.5) / (z^{3} + 1.5 z^{2} + 0.5 z)$ . The denominator is $z (z^{2} + 1.5 z + 0.5)$ , and dividing both by $z^{3}$ gives the proper form in $z^{- 1}$ with poles at -1 and -0.5.

What is different in this one

The numerator reaches the degree of the denominator. Dividing by z is enough to make it proper, so no long division is needed, but the expansion will carry a term at the origin that inverts to an impulse.
There is a repeated pole at 0. It needs a term for every power up to its order, and the extra one inverts with an n in front: $A z / (z - p)^{2}$ is (A/p) n $p^{n} u [n]$ , not A $p^{n} u [n]$ . Cover up cannot reach the lower constant at all, which is why the constants are found by comparing coefficients here.
The ROC is the inside of the innermost pole, so every term is left sided and the whole answer carries $u [- n - 1]$ with a sign change.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{z^{3} + z^{2} + 1.5 z + 0.5}{z^{3} + 1.5 z^{2} + 0.5 z}

2. Nothing to divide. Degree 3 over degree 3, so the fraction is already low enough on top.

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = \frac{z^{3} + z^{2} + 1.5 z + 0.5}{z^{4} + 1.5 z^{3} + 0.5 z^{2}}

The poles are the roots of that denominator. Note the z we just divided by is one of them, sitting at the origin.

z^{4} + 1.5 z^{3} + 0.5 z^{2} = z^{2} (z + 0.5) (z + 1) = 0

So the poles are $0$ , repeated 2 times, $- 0.5$ and $- 1$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{z^{3} + z^{2} + 1.5 z + 0.5}{z^{2} (z + 0.5) (z + 1)} \\ = \frac{A}{z} + \frac{B}{z^{2}} + \frac{C}{(z + 0.5)} + \frac{D}{(z + 1)} \end{aligned}

Put them back over the common denominator $z^{2} (z + 0.5) (z + 1)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A z (z + 0.5) (z + 1) + B (z + 0.5) (z + 1) + C z^{2} (z + 1) + D z^{2} (z + 0.5) \\ A z (z + 0.5) (z + 1) & = A z^{3} + 1.5 A z^{2} + 0.5 A z \\ B (z + 0.5) (z + 1) & = B z^{2} + 1.5 B z + 0.5 B \\ C z^{2} (z + 1) & = C z^{3} + C z^{2} \\ D z^{2} (z + 0.5) & = D z^{3} + 0.5 D z^{2} \end{aligned}

Now collect the powers of z and set each one equal to the same power in $z^{3} + z^{2} + 1.5 z + 0.5$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + C + D & = 1 & from z^{3} \\ 1.5 A + B + C + 0.5 D & = 1 & from z^{2} \\ 0.5 A + 1.5 B & = 1.5 & from z \\ 0.5 B & = 0.5 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = 0 \\ B & = 1 \\ C & = - 1 \\ D & = 2 \end{aligned}

Putting those back and adding the terms returns $z^{3} + z^{2} + 1.5 z + 0.5$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = \frac{1}{z} + \frac{- 1 z}{(z + 0.5)} + \frac{2 z}{(z + 1)}

6. Invert term by term. The ROC $| z | < 0.5$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
0, the origin	0	1	a polynomial term	$δ [n - 1]$
-0.5	0.5	-1	the ROC lies inside it	${(- 0.5)}^{n} u [- n - 1]$
-1	1	2	the ROC lies inside it	-2 ${(- 1)}^{n} u [- n - 1]$

The constant at 0 came out zero: a zero of the numerator sits on that pole and cancels it, so it contributes nothing.

With the ROC $| z | < 0.5$ , $x [n] = δ [n - 1] + {(- 0.5)}^{n} u [- n - 1] - 2 {(- 1)}^{n} u [- n - 1]$ .

Same as 2079 Bhadra Q32076 Chaitra · Q42+6 marksDefine ROC of z-transform. Find inverse z-transform using partial fraction expansion of

X (z) = (z^{4} + 5 z^{3} - 3 z + 4) / (z^{2} - 1.5 z - 1)

, ROC:

| z | < 0.5

How this is read $X (z) = (z^{4} + 5 z^{3} - 3 z + 4) / (z^{2} - 1.5 z - 1)$ . The numerator is two degrees higher, so the long division runs for two terms before the partial fractions start.

What is different in this one

The numerator is of higher degree than the denominator (4 against 2), so it has to be long divided first. The quotient is a polynomial in z, which inverts to impulses at negative n, and only the remainder goes through partial fractions. Divide until the remainder reaches the degree of the divisor and stop there, not further.
The ROC is the inside of the innermost pole, so every term is left sided and the whole answer carries $u [- n - 1]$ with a sign change.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{z^{4} + 5 z^{3} - 3 z + 4}{z^{2} - 1.5 z - 1}

			$z^{2}$	$+ 6.5 z$
$z^{2} - 1.5 z - 1$ )	$z^{4}$	$+ 5 z^{3}$	$+ 0 z^{2}$	$- 3 z$	$+ 4$
	$z^{4}$	$- 1.5 z^{3}$	$- z^{2}$
		$6.5 z^{3}$	$+ z^{2}$	$- 3 z$	$+ 4$
		$6.5 z^{3}$	$- 9.75 z^{2}$	$- 6.5 z$
			$10.75 z^{2}$	$+ 3.5 z$	$+ 4$

X (z) = z^{2} + 6.5 z + \frac{10.75 z^{2} + 3.5 z + 4}{z^{2} - 1.5 z - 1}

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = z + 6.5 + \frac{10.75 z^{2} + 3.5 z + 4}{z^{3} - 1.5 z^{2} - z}

The poles are the roots of that denominator. Note the z we just divided by is one of them, sitting at the origin.

z^{3} - 1.5 z^{2} - z = z (z + 0.5) (z - 2) = 0

So the poles are $0$ , $- 0.5$ and $2$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{10.75 z^{2} + 3.5 z + 4}{z (z + 0.5) (z - 2)} \\ = \frac{A}{z} + \frac{B}{(z + 0.5)} + \frac{C}{(z - 2)} \end{aligned}

Put them back over the common denominator $z (z + 0.5) (z - 2)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z + 0.5) (z - 2) + B z (z - 2) + C z (z + 0.5) \\ A (z + 0.5) (z - 2) & = A z^{2} - 1.5 A z - A \\ B z (z - 2) & = B z^{2} - 2 B z \\ C z (z + 0.5) & = C z^{2} + 0.5 C z \end{aligned}

Now collect the powers of z and set each one equal to the same power in $10.75 z^{2} + 3.5 z + 4$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + B + C & = 10.75 & from z^{2} \\ - 1.5 A - 2 B + 0.5 C & = 3.5 & from z \\ - A & = 4 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = - 4 \\ B & = 3.95 \\ C & = 10.8 \end{aligned}

Putting those back and adding the terms returns $10.75 z^{2} + 3.5 z + 4$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = z^{2} + 6.5 z - 4 + \frac{3.95 z}{(z + 0.5)} + \frac{10.8 z}{(z - 2)}

6. Invert term by term. The ROC $| z | < 0.5$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
none, from the division		1	a polynomial term	$δ [n + 2]$
none, from the division		6.5	a polynomial term	6.5 $δ [n + 1]$
0, the origin	0	-4	a polynomial term	-4 $δ [n]$
-0.5	0.5	3.95	the ROC lies inside it	-3.95 ${(- 0.5)}^{n} u [- n - 1]$
2	2	10.8	the ROC lies inside it	-10.8 ${(2)}^{n} u [- n - 1]$

With the ROC $| z | < 0.5$ , $x [n] = δ [n + 2] + 6.5 δ [n + 1] - 4 δ [n] - 3.95 {(- 0.5)}^{n} u [- n - 1] - 10.8 {(2)}^{n} u [- n - 1]$ .

Practise this2076 Ashwin · Q33+6 marksState Convolution property of Z-transform. Find inverse Z-transform of

X (z) = z

(z - 0.6) (z + 0.5)^{2}

}, ROC:

| z | > 0.6

What is different in this one

The numerator is of lower degree than the denominator, so it is already proper: no division, straight to F(z) = $X (z)$ /z.
There is a repeated pole at -0.5. It needs a term for every power up to its order, and the extra one inverts with an n in front: $A z / (z - p)^{2}$ is (A/p) n $p^{n} u [n]$ , not A $p^{n} u [n]$ . Cover up cannot reach the lower constant at all, which is why the constants are found by comparing coefficients here.
The ROC is the outside of the outermost pole, so every term is right sided and the answer is causal.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{z}{z^{3} + 0.4 z^{2} - 0.35 z - 0.15}

2. Nothing to divide. Degree 1 over degree 3, so the fraction is already low enough on top.

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = \frac{1}{z^{3} + 0.4 z^{2} - 0.35 z - 0.15}

The numerator carried a z of its own, and it cancels the z we just divided by, so there is no term at the origin at all.

z^{3} + 0.4 z^{2} - 0.35 z - 0.15 = {(z + 0.5)}^{2} (z - 0.6) = 0

So the poles are $- 0.5$ , repeated 2 times and $0.6$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{1}{{(z + 0.5)}^{2} (z - 0.6)} \\ = \frac{A}{(z + 0.5)} + \frac{B}{{(z + 0.5)}^{2}} + \frac{C}{(z - 0.6)} \end{aligned}

Put them back over the common denominator ${(z + 0.5)}^{2} (z - 0.6)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z + 0.5) (z - 0.6) + B (z - 0.6) + C {(z + 0.5)}^{2} \\ A (z + 0.5) (z - 0.6) & = A z^{2} - 0.1 A z - 0.3 A \\ B (z - 0.6) & = B z - 0.6 B \\ C {(z + 0.5)}^{2} & = C z^{2} + C z + 0.25 C \end{aligned}

Now collect the powers of z and set each one equal to the same power in $1$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + C & = 0 & from z^{2} \\ - 0.1 A + B + C & = 0 & from z \\ - 0.3 A - 0.6 B + 0.25 C & = 1 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = - 0.8264 \\ B & = - 0.9091 \\ C & = 0.8264 \end{aligned}

Putting those back and adding the terms returns $1$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = \frac{- 0.8264 z}{(z + 0.5)} + \frac{- 0.9091 z}{{(z + 0.5)}^{2}} + \frac{0.8264 z}{(z - 0.6)}

6. Invert term by term. The ROC $| z | > 0.6$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
-0.5	0.5	-0.8264	the ROC lies outside it	-0.8264 ${(- 0.5)}^{n} u [n]$
-0.5, order 2	0.5	-0.9091	the ROC lies outside it	1.8182 n ${(- 0.5)}^{n} u [n]$
0.6	0.6	0.8264	the ROC lies outside it	0.8264 ${(0.6)}^{n} u [n]$

With the ROC $| z | > 0.6$ , $x [n] = - 0.8264 {(- 0.5)}^{n} u [n]$ + 1.8182 n ${(- 0.5)}^{n} u [n] + 0.8264 {(0.6)}^{n} u [n]$ .

Check it without trusting any of the algebra: run the difference equation the transform encodes and compare the first samples. Both routes give ${0, 0, 1, - 0.4, 0.51, - 0.194}$ .

Same as 2081 Bhadra Q32074 Chaitra · Q31+3+5 marksDefine a ROC. What are the properties of ROC of z-transform? Find the inverse Z-transform of

X (z) = (2 z^{2} + 2 z^{2} + 3 z + 5) / (z^{2} - 0.1 z - 0.2)

, ROC:

| z | < 0.4

How this is readThe paper prints " $2 z^{2} + 2 z^{2}$ " in the numerator, which cannot be right. Read as $2 z^{3} + 2 z^{2} + 3 z + 5$ , which is the only reading that leaves the question non trivial, and is the reading the standard text uses for this same paper.

What is different in this one

The numerator is of higher degree than the denominator (3 against 2), so it has to be long divided first. The quotient is a polynomial in z, which inverts to impulses at negative n, and only the remainder goes through partial fractions. Divide until the remainder reaches the degree of the divisor and stop there, not further.
The ROC is the inside of the innermost pole, so every term is left sided and the whole answer carries $u [- n - 1]$ with a sign change.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{2 z^{3} + 2 z^{2} + 3 z + 5}{z^{2} - 0.1 z - 0.2}

			$2 z$
$z^{2} - 0.1 z - 0.2$ )	$2 z^{3}$	$+ 2 z^{2}$	$+ 3 z$	$+ 5$
	$2 z^{3}$	$- 0.2 z^{2}$	$- 0.4 z$
		$2.2 z^{2}$	$+ 3.4 z$	$+ 5$

X (z) = 2 z + \frac{2.2 z^{2} + 3.4 z + 5}{z^{2} - 0.1 z - 0.2}

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = 2 + \frac{2.2 z^{2} + 3.4 z + 5}{z^{3} - 0.1 z^{2} - 0.2 z}

The poles are the roots of that denominator. Note the z we just divided by is one of them, sitting at the origin.

z^{3} - 0.1 z^{2} - 0.2 z = z (z + 0.4) (z - 0.5) = 0

So the poles are $0$ , $- 0.4$ and $0.5$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{2.2 z^{2} + 3.4 z + 5}{z (z + 0.4) (z - 0.5)} \\ = \frac{A}{z} + \frac{B}{(z + 0.4)} + \frac{C}{(z - 0.5)} \end{aligned}

Put them back over the common denominator $z (z + 0.4) (z - 0.5)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z + 0.4) (z - 0.5) + B z (z - 0.5) + C z (z + 0.4) \\ A (z + 0.4) (z - 0.5) & = A z^{2} - 0.1 A z - 0.2 A \\ B z (z - 0.5) & = B z^{2} - 0.5 B z \\ C z (z + 0.4) & = C z^{2} + 0.4 C z \end{aligned}

Now collect the powers of z and set each one equal to the same power in $2.2 z^{2} + 3.4 z + 5$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + B + C & = 2.2 & from z^{2} \\ - 0.1 A - 0.5 B + 0.4 C & = 3.4 & from z \\ - 0.2 A & = 5 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = - 25 \\ B & = 11.0889 \\ C & = 16.1111 \end{aligned}

Putting those back and adding the terms returns $2.2 z^{2} + 3.4 z + 5$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = 2 z - 25 + \frac{11.0889 z}{(z + 0.4)} + \frac{16.1111 z}{(z - 0.5)}

6. Invert term by term. The ROC $| z | < 0.4$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
none, from the division		2	a polynomial term	2 $δ [n + 1]$
0, the origin	0	-25	a polynomial term	-25 $δ [n]$
-0.4	0.4	11.0889	the ROC lies inside it	-11.0889 ${(- 0.4)}^{n} u [- n - 1]$
0.5	0.5	16.1111	the ROC lies inside it	-16.1111 ${(0.5)}^{n} u [- n - 1]$

With the ROC $| z | < 0.4$ , $x [n] = 2 δ [n + 1] - 25 δ [n] - 11.0889 {(- 0.4)}^{n} u [- n - 1] - 16.1111 {(0.5)}^{n} u [- n - 1]$ .

Practise this2074 Ashwin · Q33+3 marksState and explain the properties of a Region of Convergence (ROC). Find the inverse z-transform of

X (z) = z^{2} [1 - (3 / 2) z^{- 1}] (1 + z^{- 1}) (1 - z^{- 1})

What is different hereNo ROC is given because none is needed: a finite length sequence converges everywhere except possibly at

z = 0

and z = infinity.

What is different in this one

There is nothing to invert: no partial fractions, no ROC decision. $X (z)$ is already a finite polynomial, so its coefficients are the sequence. Thirty seconds if you spot it.
A finite length sequence converges everywhere except possibly at $z = 0$ and z = infinity, which is why the question does not give you an ROC.
The factor $z^{2}$ is an advance, so the sequence starts at $n = - 2$ , before the origin.

Multiply the factors out, then read each coefficient as an impulse: a power of $z^{- k}$ is a delay of k and a power of $z^{k}$ an advance.

\begin{aligned} X (z) & = z^{2} (1 - 1.5 z^{- 1}) (1 + z^{- 1}) (1 - z^{- 1}) \\ = z^{2} (1 - 1.5 z^{- 1}) \\ = z^{2} (1 - 0.5 z^{- 1} - 1.5 z^{- 2}) \\ = z^{2} (1 - 1.5 z^{- 1} - z^{- 2} + 1.5 z^{- 3}) \end{aligned}

$x [n] = 1 δ [n + 2] - 1.5 δ [n + 1] - 1 δ [n] + 1.5 δ [n - 1]$ , that is {1, -1.5, -1, 1.5} for $n = - 2$ to 1.

Practise this2073 Shrawan · Q31+5 marksDefine a ROC. Find inverse Z-transform of

X (z) = z

(z - 0.4) (z + 1.5)^{2}

}, ROC:

| z | < 0.4

What is different herethe region of convergence is inside the pole, so every term inverts to a left sided sequence; a pole is REPEATED, so it needs a term for every power up to its order, the lower constant cannot be found by covering up, and the extra term inverts with an n in front of it

How this is read $X (z) = z / ((z - 0.4) (z + 1.5)^{2})$ , whose denominator multiplies out to $z^{3} + 2.6 z^{2} + 1.05 z - 0.9$ .

What is different in this one

The numerator is of lower degree than the denominator, so it is already proper: no division, straight to F(z) = $X (z)$ /z.
There is a repeated pole at -1.5. It needs a term for every power up to its order, and the extra one inverts with an n in front: $A z / (z - p)^{2}$ is (A/p) n $p^{n} u [n]$ , not A $p^{n} u [n]$ . Cover up cannot reach the lower constant at all, which is why the constants are found by comparing coefficients here.
The ROC is the inside of the innermost pole, so every term is left sided and the whole answer carries $u [- n - 1]$ with a sign change.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{z}{z^{3} + 2.6 z^{2} + 1.05 z - 0.9}

2. Nothing to divide. Degree 1 over degree 3, so the fraction is already low enough on top.

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = \frac{1}{z^{3} + 2.6 z^{2} + 1.05 z - 0.9}

The numerator carried a z of its own, and it cancels the z we just divided by, so there is no term at the origin at all.

z^{3} + 2.6 z^{2} + 1.05 z - 0.9 = (z - 0.4) {(z + 1.5)}^{2} = 0

So the poles are $0.4$ and $- 1.5$ , repeated 2 times.

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{1}{(z - 0.4) {(z + 1.5)}^{2}} \\ = \frac{A}{(z - 0.4)} + \frac{B}{(z + 1.5)} + \frac{C}{{(z + 1.5)}^{2}} \end{aligned}

Put them back over the common denominator $(z - 0.4) {(z + 1.5)}^{2}$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A {(z + 1.5)}^{2} + B (z - 0.4) (z + 1.5) + C (z - 0.4) \\ A {(z + 1.5)}^{2} & = A z^{2} + 3 A z + 2.25 A \\ B (z - 0.4) (z + 1.5) & = B z^{2} + 1.1 B z - 0.6 B \\ C (z - 0.4) & = C z - 0.4 C \end{aligned}

Now collect the powers of z and set each one equal to the same power in $1$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + B & = 0 & from z^{2} \\ 3 A + 1.1 B + C & = 0 & from z \\ 2.25 A - 0.6 B - 0.4 C & = 1 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = 0.277 \\ B & = - 0.277 \\ C & = - 0.5263 \end{aligned}

Putting those back and adding the terms returns $1$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = \frac{0.277 z}{(z - 0.4)} + \frac{- 0.277 z}{(z + 1.5)} + \frac{- 0.5263 z}{{(z + 1.5)}^{2}}

6. Invert term by term. The ROC $| z | < 0.4$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
0.4	0.4	0.277	the ROC lies inside it	-0.277 ${(0.4)}^{n} u [- n - 1]$
-1.5	1.5	-0.277	the ROC lies inside it	0.277 ${(- 1.5)}^{n} u [- n - 1]$
-1.5, order 2	1.5	-0.5263	the ROC lies inside it	-0.3509 n ${(- 1.5)}^{n} u [- n - 1]$

With the ROC $| z | < 0.4$ , $x [n] = - 0.277 {(0.4)}^{n} u [- n - 1] + 0.277 {(- 1.5)}^{n} u [- n - 1]$ - 0.3509 n ${(- 1.5)}^{n} u [- n - 1]$ .

Same as 2080 Baishakh Q32071 Shrawan · Q26 marksFind the inverse z transform of:

X (Z) = (1 + 2 z^{- 1} + z^{- 2}) / (1 + 1.5 z^{- 1} + 0.5 z^{- 2})

| z | > 1

using partial fraction method.

What is different in this one

The numerator reaches the degree of the denominator. Dividing by z is enough to make it proper, so no long division is needed, but the expansion will carry a term at the origin that inverts to an impulse.
The ROC is the outside of the outermost pole, so every term is right sided and the answer is causal.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{z^{2} + 2 z + 1}{z^{2} + 1.5 z + 0.5}

2. Nothing to divide. Degree 2 over degree 2, so the fraction is already low enough on top.

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = \frac{z^{2} + 2 z + 1}{z^{3} + 1.5 z^{2} + 0.5 z}

The poles are the roots of that denominator. Note the z we just divided by is one of them, sitting at the origin.

z^{3} + 1.5 z^{2} + 0.5 z = z (z + 0.5) (z + 1) = 0

So the poles are $0$ , $- 0.5$ and $- 1$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{z^{2} + 2 z + 1}{z (z + 0.5) (z + 1)} \\ = \frac{A}{z} + \frac{B}{(z + 0.5)} + \frac{C}{(z + 1)} \end{aligned}

Put them back over the common denominator $z (z + 0.5) (z + 1)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z + 0.5) (z + 1) + B z (z + 1) + C z (z + 0.5) \\ A (z + 0.5) (z + 1) & = A z^{2} + 1.5 A z + 0.5 A \\ B z (z + 1) & = B z^{2} + B z \\ C z (z + 0.5) & = C z^{2} + 0.5 C z \end{aligned}

Now collect the powers of z and set each one equal to the same power in $z^{2} + 2 z + 1$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + B + C & = 1 & from z^{2} \\ 1.5 A + B + 0.5 C & = 2 & from z \\ 0.5 A & = 1 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = 2 \\ B & = - 1 \\ C & = 0 \end{aligned}

Putting those back and adding the terms returns $z^{2} + 2 z + 1$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = 2 + \frac{- 1 z}{(z + 0.5)}

6. Invert term by term. The ROC $| z | > 1$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
0, the origin	0	2	a polynomial term	2 $δ [n]$
-0.5	0.5	-1	the ROC lies outside it	- ${(- 0.5)}^{n} u [n]$

The constant at -1 came out zero: a zero of the numerator sits on that pole and cancels it, so it contributes nothing.

With the ROC $| z | > 1$ , $x [n] = 2 δ [n] - {(- 0.5)}^{n} u [n]$ .

Check it without trusting any of the algebra: run the difference equation the transform encodes and compare the first samples. Both routes give ${1, 0.5, - 0.25, 0.125, - 0.0625, 0.03125}$ .

The same questions grouped by how often each topic is set: the inverse z-transform

Inverting $X (z)$ by partial fractions TOP 15/19

82 Bh · 82 Ba · 81 Bh · 81 Ba · 80 Bh · 80 Ba · 79 Bh · 79 Ba · 78 Bh · 76 Ch · 76 Ash · 74 Ch · 74 Ash · 73 Shr · 71 Shr1+561+3+5

Before the formulas

The other way back, and the one that gives a formula rather than a list. Split the expression into one term per pole, look each term up in the short table of pairs, and add the results. The only real work is finding the coefficients; the only real thinking is deciding which standard sequence each term becomes, and that is decided by the region of convergence, not by the term itself.

Get in the habit of writing the region down before you invert anything, because by the last line it is too late.

This is question 3 in nineteen papers out of nineteen. The method never changes, so learn it as a fixed sequence of steps.

\begin{aligned} 1. if \deg (num) \geq \deg (den), long divide first \\ 2. expand \frac{X (z)}{z}, not X (z) \\ \frac{X (z)}{z} = \frac{A_{1}}{z - p_{1}} + \frac{A_{2}}{z - p_{2}} + \dots, A_{k} = {[(z - p_{k}) \frac{X (z)}{z}]}_{z = p_{k}} \\ 3. multiply back by z : X (z) = \frac{A_{1} z}{z - p_{1}} + \frac{A_{2} z}{z - p_{2}} + \dots \\ 4. invert TERM BY TERM against the ROC \\ ROC outside p_{k} : A_{k} p_{k}^{n} u [n] ROC inside p_{k} : - A_{k} p_{k}^{n} u [- n - 1] \end{aligned}

Step 5 is the whole question. Everything before it is algebra that most candidates get right. The marks are in reading the ROC against each pole separately.

How to read the ROC, pole by pole Draw the poles on a line by magnitude. Mark the given ROC on it. For each pole ask: is this pole inside the ROC boundary, meaning the ROC lies outside it? Then that term is right sided, with

u [n]

. Is the ROC inside this pole? Then that term is left sided, with

- u [- n - 1]

.
A two sided ROC such as

1 / 3 < | z | < 1

gives one term of each kind, which is what makes those questions worth five marks.

Worked, the 2082 Bhadra and 2079 Baishakh question

H (z) = z / (3 z^{2} - 4 z + 1)

for

1 / 3 < | z | < 1

.
Factor:

3 z^{2} - 4 z + 1 = 3 (z - 1) (z - 1 / 3)

, so

H (z) = z

/ [3(z-1)(z-1/3)].

H (z) / z = 1 / [3 (z - 1) (z - 1 / 3)] = A / (z - 1) + B / (z - 1 / 3)

A = 1

/[3(1 - 1/3)] =

\frac{1}{2}

B = 1

/[3(1/3 - 1)] =

- \frac{1}{2}

H (z)

= (

\frac{1}{2}) z / (z - 1) - (\frac{1}{2}) z / (z - 1 / 3

).
Now the ROC,

1 / 3 < | z | < 1

. The pole at 1/3 has the ROC outside it, so that term is right sided. The pole at 1 has the ROC inside it, so that term is left sided.
$h [n]$ = -( $\frac{1}{2}$ ) $u [- n - 1]$ - ( $\frac{1}{2}$ ) ${(1 / 3)}^{n} u [n]$ .

The other two methods, and when to use them

Long division, or the power series method. Divide numerator by denominator to get a series in $z^{- 1}$ ; the coefficients are $x [0]$ , $x [1]$ , $x [2]$ and so on. Use it when the question wants only the first few samples, or when the denominator does not factor nicely. Divide in ascending powers of $z^{- 1}$ for a causal answer, in ascending powers of z for an anticausal one.
The residue method. $x [n]$ = sum of residues of $X (z) z^{n - 1}$ at the poles inside the contour. Correct, and slower than partial fractions for exam sized problems.

Repeated poles. For a pole of order m at p, the expansion needs terms $A 1 / (z - p) + A 2 / {(z - p)}^{2}$ + ... + Am/ ${(z - p)}^{m}$ , and the constants come from differentiating: A(m-k) = (1/k!) times the k-th derivative of [ ${(z - p)}^{m} X (z)$ /z] evaluated at z = p. The inverse of z/ ${(z - p)}^{2}$ is n $p^{n - 1} u [n]$ .

In the exam

Long divide first if the degrees demand it. Skipping that step on $X (z)$ = ( $2 z^{4} + 2 z^{3} - 3 z + 2) / (z^{2}$ - 1.5z - 1) makes the partial fractions come out wrong, and that exact fraction has been set twice.
Expand $X (z)$ /z, not $X (z)$ . Expanding $X (z)$ directly gives terms in $1 / (z - p)$ whose inverse carries an awkward shift.
Check $x [0]$ against the initial value theorem when the sequence is causal. It is one line and it catches a wrong residue.

The identical partial fraction expansion is the first step of impulse invariance, where each term is mapped separately from s to z.

See it drawnPartial fractions: one pole, one term, one sequence

Split

X (z)

/z into one term per pole, then invert each term on its own. The ROC decides, pole by pole, which of the two standard pairs each one becomes.

Practise thisEvery question the paper has set on this topic, worked in full15

Practise this +1 like it2082 Bhadra · Q31+5 marksDefine z-transform for a discrete time signal. Find the inverse z-transform for

H (z) = z / (3 z^{2} - 4 z + 1)

using partial fraction method for

1 / 3 < | z | < 1

What is different in this one

The numerator is of lower degree than the denominator, so it is already proper: no division, straight to F(z) = $X (z)$ /z.
The ROC is a ring, so the answer is two sided: the poles inside the ring give right sided terms and the ones outside it give left sided terms. This is the case that separates a real answer from a memorised one.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{z}{3 z^{2} - 4 z + 1}

2. Nothing to divide. Degree 1 over degree 2, so the fraction is already low enough on top.

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = \frac{1}{3 z^{2} - 4 z + 1}

The numerator carried a z of its own, and it cancels the z we just divided by, so there is no term at the origin at all.

3 z^{2} - 4 z + 1 = 3 (z - 0.3333) (z - 1) = 0

So the poles are $0.3333$ and $1$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{1}{3 (z - 0.3333) (z - 1)} \\ = \frac{A}{(z - 0.3333)} + \frac{B}{(z - 1)} \end{aligned}

Put them back over the common denominator $3 (z - 0.3333) (z - 1)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z - 1) + B (z - 0.3333) \\ A (z - 1) & = A z - A \\ B (z - 0.3333) & = B z - 0.3333 B \end{aligned}

Now collect the powers of z and set each one equal to the same power in $0.3333$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + B & = 0 & from z \\ - A - 0.3333 B & = 0.3333 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = - 0.5 \\ B & = 0.5 \end{aligned}

Putting those back and adding the terms returns $1$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = \frac{- 0.5 z}{(z - 0.3333)} + \frac{0.5 z}{(z - 1)}

6. Invert term by term. The ROC $0.3333 < | z | < 1$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
0.3333	0.3333	-0.5	the ROC lies outside it	-0.5 ${(0.3333)}^{n} u [n]$
1	1	0.5	the ROC lies inside it	-0.5 ${(1)}^{n} u [- n - 1]$

With the ROC $0.3333 < | z | < 1$ , $x [n] = - 0.5 {(0.3333)}^{n} u [n] - 0.5 {(1)}^{n} u [- n - 1]$ .

Practise this2082 Baishakh · Q31+5 marksDefine ROC of a Z-transform. Determine

x [n]

for

X (z) = 1 / (1 - 1.5 z^{- 1} + 0.5 z^{- 2})

if: a)

| z | > 1

| z | < 0.5

0.5 < | z | < 1

using partial fraction method.What is different hereOne

X (z)

, three answers. This is the question the whole chapter exists for, and the partial fractions are done ONCE: only the side each pole falls on changes.

What is different in this one

The same $X (z)$ is given with 3 different regions, so it stands for 3 different signals. The partial fractions are worked once: the poles and the constants do not know about the ROC, and only the very last step, which side each pole falls on, changes between the parts.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{z^{2}}{z^{2} - 1.5 z + 0.5}

2. Nothing to divide. Degree 2 over degree 2, so the fraction is already low enough on top.

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = \frac{z}{z^{2} - 1.5 z + 0.5}

The numerator carried a z of its own, and it cancels the z we just divided by, so there is no term at the origin at all.

z^{2} - 1.5 z + 0.5 = (z - 0.5) (z - 1) = 0

So the poles are $0.5$ and $1$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{z}{(z - 0.5) (z - 1)} \\ = \frac{A}{(z - 0.5)} + \frac{B}{(z - 1)} \end{aligned}

Put them back over the common denominator $(z - 0.5) (z - 1)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z - 1) + B (z - 0.5) \\ A (z - 1) & = A z - A \\ B (z - 0.5) & = B z - 0.5 B \end{aligned}

Now collect the powers of z and set each one equal to the same power in $z$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + B & = 1 & from z \\ - A - 0.5 B & = 0 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = - 1 \\ B & = 2 \end{aligned}

Putting those back and adding the terms returns $z$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = \frac{- 1 z}{(z - 0.5)} + \frac{2 z}{(z - 1)}

$| z | > 1$

Pole	$\| p \|$	Constant	Which side	Inverts to
0.5	0.5	-1	the ROC lies outside it	- ${(0.5)}^{n} u [n]$
1	1	2	the ROC lies outside it	2 ${(1)}^{n} u [n]$

With the ROC $| z | > 1$ , $x [n] = - {(0.5)}^{n} u [n] + 2 {(1)}^{n} u [n]$ .

$| z | < 0.5$

Pole	$\| p \|$	Constant	Which side	Inverts to
0.5	0.5	-1	the ROC lies inside it	${(0.5)}^{n} u [- n - 1]$
1	1	2	the ROC lies inside it	-2 ${(1)}^{n} u [- n - 1]$

With the ROC $| z | < 0.5$ , $x [n] = {(0.5)}^{n} u [- n - 1] - 2 {(1)}^{n} u [- n - 1]$ .

$0.5 < | z | < 1$

Pole	$\| p \|$	Constant	Which side	Inverts to
0.5	0.5	-1	the ROC lies outside it	- ${(0.5)}^{n} u [n]$
1	1	2	the ROC lies inside it	-2 ${(1)}^{n} u [- n - 1]$

With the ROC $0.5 < | z | < 1$ , $x [n] = - {(0.5)}^{n} u [n] - 2 {(1)}^{n} u [- n - 1]$ .

Side by side, the same transform read 3 ways:

ROC	Which side each pole falls	$x [n]$	Stable?
$\| z \| > 1$	0.5 is right sided, 1 is right sided	- ${(0.5)}^{n} u [n] + 2 {(1)}^{n} u [n]$	no
$\| z \| < 0.5$	0.5 is left sided, 1 is left sided	${(0.5)}^{n} u [- n - 1] - 2 {(1)}^{n} u [- n - 1]$	no
$0.5 < \| z \| < 1$	0.5 is right sided, 1 is left sided	- ${(0.5)}^{n} u [n] - 2 {(1)}^{n} u [- n - 1]$	no

Practise this +1 like it2081 Bhadra · Q31+5 marksDefine ROC of z-transform. Find the inverse z-transform using partial fraction expansion of

X (z) = (2 z^{3} - 5 z^{2} + z + 3) / ((z - 1) (z - 2))

for ROC

| z | < 1

What is different in this one

The numerator is of higher degree than the denominator (3 against 2), so it has to be long divided first. The quotient is a polynomial in z, which inverts to impulses at negative n, and only the remainder goes through partial fractions. Divide until the remainder reaches the degree of the divisor and stop there, not further.
The ROC is the inside of the innermost pole, so every term is left sided and the whole answer carries $u [- n - 1]$ with a sign change.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{2 z^{3} - 5 z^{2} + z + 3}{z^{2} - 3 z + 2}

			$2 z$
$z^{2} - 3 z + 2$ )	$2 z^{3}$	$- 5 z^{2}$	$+ z$	$+ 3$
	$2 z^{3}$	$- 6 z^{2}$	$+ 4 z$
		$z^{2}$	$- 3 z$	$+ 3$

X (z) = 2 z + \frac{z^{2} - 3 z + 3}{z^{2} - 3 z + 2}

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = 2 + \frac{z^{2} - 3 z + 3}{z^{3} - 3 z^{2} + 2 z}

The poles are the roots of that denominator. Note the z we just divided by is one of them, sitting at the origin.

z^{3} - 3 z^{2} + 2 z = z (z - 1) (z - 2) = 0

So the poles are $0$ , $1$ and $2$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{z^{2} - 3 z + 3}{z (z - 1) (z - 2)} \\ = \frac{A}{z} + \frac{B}{(z - 1)} + \frac{C}{(z - 2)} \end{aligned}

Put them back over the common denominator $z (z - 1) (z - 2)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z - 1) (z - 2) + B z (z - 2) + C z (z - 1) \\ A (z - 1) (z - 2) & = A z^{2} - 3 A z + 2 A \\ B z (z - 2) & = B z^{2} - 2 B z \\ C z (z - 1) & = C z^{2} - C z \end{aligned}

Now collect the powers of z and set each one equal to the same power in $z^{2} - 3 z + 3$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + B + C & = 1 & from z^{2} \\ - 3 A - 2 B - C & = - 3 & from z \\ 2 A & = 3 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = 1.5 \\ B & = - 1 \\ C & = 0.5 \end{aligned}

Putting those back and adding the terms returns $z^{2} - 3 z + 3$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = 2 z + 1.5 + \frac{- 1 z}{(z - 1)} + \frac{0.5 z}{(z - 2)}

6. Invert term by term. The ROC $| z | < 1$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
none, from the division		2	a polynomial term	2 $δ [n + 1]$
0, the origin	0	1.5	a polynomial term	1.5 $δ [n]$
1	1	-1	the ROC lies inside it	${(1)}^{n} u [- n - 1]$
2	2	0.5	the ROC lies inside it	-0.5 ${(2)}^{n} u [- n - 1]$

With the ROC $| z | < 1$ , $x [n] = 2 δ [n + 1] + 1.5 δ [n] + {(1)}^{n} u [- n - 1] - 0.5 {(2)}^{n} u [- n - 1]$ .

Practise this2081 Baishakh · Q32+2+3 marksDetermine the inverse z-transform of

X (z) = 1 / (1 - 0.8 z^{- 1} + 0.12 z^{- 2}) . (i)

if ROC is

| z | > 0.6

(ii) if ROC is

| z | < 0.2

(iii) if ROC is

0.2 < | z | < 0.6

What is different in this one

The same $X (z)$ is given with 3 different regions, so it stands for 3 different signals. The partial fractions are worked once: the poles and the constants do not know about the ROC, and only the very last step, which side each pole falls on, changes between the parts.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{z^{2}}{z^{2} - 0.8 z + 0.12}

2. Nothing to divide. Degree 2 over degree 2, so the fraction is already low enough on top.

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = \frac{z}{z^{2} - 0.8 z + 0.12}

The numerator carried a z of its own, and it cancels the z we just divided by, so there is no term at the origin at all.

z^{2} - 0.8 z + 0.12 = (z - 0.2) (z - 0.6) = 0

So the poles are $0.2$ and $0.6$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{z}{(z - 0.2) (z - 0.6)} \\ = \frac{A}{(z - 0.2)} + \frac{B}{(z - 0.6)} \end{aligned}

Put them back over the common denominator $(z - 0.2) (z - 0.6)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z - 0.6) + B (z - 0.2) \\ A (z - 0.6) & = A z - 0.6 A \\ B (z - 0.2) & = B z - 0.2 B \end{aligned}

Now collect the powers of z and set each one equal to the same power in $z$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + B & = 1 & from z \\ - 0.6 A - 0.2 B & = 0 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = - 0.5 \\ B & = 1.5 \end{aligned}

Putting those back and adding the terms returns $z$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = \frac{- 0.5 z}{(z - 0.2)} + \frac{1.5 z}{(z - 0.6)}

$| z | > 0.6$

Pole	$\| p \|$	Constant	Which side	Inverts to
0.2	0.2	-0.5	the ROC lies outside it	-0.5 ${(0.2)}^{n} u [n]$
0.6	0.6	1.5	the ROC lies outside it	1.5 ${(0.6)}^{n} u [n]$

With the ROC $| z | > 0.6$ , $x [n] = - 0.5 {(0.2)}^{n} u [n] + 1.5 {(0.6)}^{n} u [n]$ .

$| z | < 0.2$

Pole	$\| p \|$	Constant	Which side	Inverts to
0.2	0.2	-0.5	the ROC lies inside it	0.5 ${(0.2)}^{n} u [- n - 1]$
0.6	0.6	1.5	the ROC lies inside it	-1.5 ${(0.6)}^{n} u [- n - 1]$

With the ROC $| z | < 0.2$ , $x [n] = 0.5 {(0.2)}^{n} u [- n - 1] - 1.5 {(0.6)}^{n} u [- n - 1]$ .

$0.2 < | z | < 0.6$

Pole	$\| p \|$	Constant	Which side	Inverts to
0.2	0.2	-0.5	the ROC lies outside it	-0.5 ${(0.2)}^{n} u [n]$
0.6	0.6	1.5	the ROC lies inside it	-1.5 ${(0.6)}^{n} u [- n - 1]$

With the ROC $0.2 < | z | < 0.6$ , $x [n] = - 0.5 {(0.2)}^{n} u [n] - 1.5 {(0.6)}^{n} u [- n - 1]$ .

Side by side, the same transform read 3 ways:

ROC	Which side each pole falls	$x [n]$	Stable?
$\| z \| > 0.6$	0.2 is right sided, 0.6 is right sided	-0.5 ${(0.2)}^{n} u [n] + 1.5 {(0.6)}^{n} u [n]$	yes
$\| z \| < 0.2$	0.2 is left sided, 0.6 is left sided	0.5 ${(0.2)}^{n} u [- n - 1] - 1.5 {(0.6)}^{n} u [- n - 1]$	no
$0.2 < \| z \| < 0.6$	0.2 is right sided, 0.6 is left sided	-0.5 ${(0.2)}^{n} u [n] - 1.5 {(0.6)}^{n} u [- n - 1]$	no

$| z | > 0.6$ , so start the divisor at its highest term.

	$1$	$+ 0.8 z^{- 1}$	$+ 0.52 z^{- 2}$	$+ 0.32 z^{- 3}$	$+ 0.1936 z^{- 4}$	$+ 0.1165 z^{- 5}$	$+ 0.07 z^{- 6}$	$+ 0.042 z^{- 7}$
$1 - 0.8 z^{- 1} + 0.12 z^{- 2}$ )	$1$
	$1$	$- 0.8 z^{- 1}$	$+ 0.12 z^{- 2}$
		$0.8 z^{- 1}$	$- 0.12 z^{- 2}$
		$0.8 z^{- 1}$	$- 0.64 z^{- 2}$	$+ 0.096 z^{- 3}$
			$0.52 z^{- 2}$	$- 0.096 z^{- 3}$
			$0.52 z^{- 2}$	$- 0.416 z^{- 3}$	$+ 0.0624 z^{- 4}$
				$0.32 z^{- 3}$	$- 0.0624 z^{- 4}$
				$0.32 z^{- 3}$	$- 0.256 z^{- 4}$	$+ 0.0384 z^{- 5}$
					$0.1936 z^{- 4}$	$- 0.0384 z^{- 5}$
					$0.1936 z^{- 4}$	$- 0.1549 z^{- 5}$	$+ 0.0232 z^{- 6}$
						$0.1165 z^{- 5}$	$- 0.0232 z^{- 6}$
						$0.1165 z^{- 5}$	$- 0.0932 z^{- 6}$	$+ 0.014 z^{- 7}$
							$0.07 z^{- 6}$	$- 0.014 z^{- 7}$
							$0.07 z^{- 6}$	$- 0.056 z^{- 7}$	$+ 0.0084 z^{- 8}$
								$0.042 z^{- 7}$	$- 0.0084 z^{- 8}$
								$0.042 z^{- 7}$	$- 0.0336 z^{- 8}$	$+ 0.005 z^{- 9}$
									$0.0252 z^{- 8}$	$- 0.005 z^{- 9}$

$| z | < 0.2$ , so start the divisor at its lowest term.

	$8.3333 z^{2}$	$+ 55.5556 z^{3}$	$+ 300.9259 z^{4}$	$+ 1543.2099 z^{5}$	$+ 7780.3498 z^{6}$	$+ 39008.9163 z^{7}$
$0.12 - 0.8 z + z^{2}$ )	$z^{2}$
	$z^{2}$	$- 6.6667 z^{3}$	$+ 8.3333 z^{4}$
		$6.6667 z^{3}$	$- 8.3333 z^{4}$
		$6.6667 z^{3}$	$- 44.4444 z^{4}$	$+ 55.5556 z^{5}$
			$36.1111 z^{4}$	$- 55.5556 z^{5}$
			$36.1111 z^{4}$	$- 240.7407 z^{5}$	$+ 300.9259 z^{6}$
				$185.1852 z^{5}$	$- 300.9259 z^{6}$
				$185.1852 z^{5}$	$- 1234.5679 z^{6}$	$+ 1543.2099 z^{7}$
					$933.642 z^{6}$	$- 1543.2099 z^{7}$
					$933.642 z^{6}$	$- 6224.2798 z^{7}$	$+ 7780.3498 z^{8}$
						$4681.07 z^{7}$	$- 7780.3498 z^{8}$
						$4681.07 z^{7}$	$- 31207.1331 z^{8}$	$+ 39008.9163 z^{9}$
							$23426.7833 z^{8}$	$- 39008.9163 z^{9}$

Practise this2080 Bhadra · Q31+6 marksDefine the Region of Convergence (ROC). Find the inverse of

H (z) = (1 + 2 z^{- 1} + z^{- 2}) / (1 - 0.75 z^{- 1} + 0.125 z^{- 2})

; ROC

0.25 < | z | < 0.5

What is different herethe numerator reaches the degree of the denominator, so dividing by z leaves a pole AT THE ORIGIN, and that term inverts to a plain impulse on top of the exponentials

What is different in this one

The numerator reaches the degree of the denominator. Dividing by z is enough to make it proper, so no long division is needed, but the expansion will carry a term at the origin that inverts to an impulse.
The ROC is a ring, so the answer is two sided: the poles inside the ring give right sided terms and the ones outside it give left sided terms. This is the case that separates a real answer from a memorised one.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{z^{2} + 2 z + 1}{z^{2} - 0.75 z + 0.125}

2. Nothing to divide. Degree 2 over degree 2, so the fraction is already low enough on top.

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = \frac{z^{2} + 2 z + 1}{z^{3} - 0.75 z^{2} + 0.125 z}

The poles are the roots of that denominator. Note the z we just divided by is one of them, sitting at the origin.

z^{3} - 0.75 z^{2} + 0.125 z = z (z - 0.25) (z - 0.5) = 0

So the poles are $0$ , $0.25$ and $0.5$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{z^{2} + 2 z + 1}{z (z - 0.25) (z - 0.5)} \\ = \frac{A}{z} + \frac{B}{(z - 0.25)} + \frac{C}{(z - 0.5)} \end{aligned}

Put them back over the common denominator $z (z - 0.25) (z - 0.5)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z - 0.25) (z - 0.5) + B z (z - 0.5) + C z (z - 0.25) \\ A (z - 0.25) (z - 0.5) & = A z^{2} - 0.75 A z + 0.125 A \\ B z (z - 0.5) & = B z^{2} - 0.5 B z \\ C z (z - 0.25) & = C z^{2} - 0.25 C z \end{aligned}

Now collect the powers of z and set each one equal to the same power in $z^{2} + 2 z + 1$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + B + C & = 1 & from z^{2} \\ - 0.75 A - 0.5 B - 0.25 C & = 2 & from z \\ 0.125 A & = 1 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = 8 \\ B & = - 25 \\ C & = 18 \end{aligned}

Putting those back and adding the terms returns $z^{2} + 2 z + 1$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = 8 + \frac{- 25 z}{(z - 0.25)} + \frac{18 z}{(z - 0.5)}

6. Invert term by term. The ROC $0.25 < | z | < 0.5$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
0, the origin	0	8	a polynomial term	8 $δ [n]$
0.25	0.25	-25	the ROC lies outside it	-25 ${(0.25)}^{n} u [n]$
0.5	0.5	18	the ROC lies inside it	-18 ${(0.5)}^{n} u [- n - 1]$

With the ROC $0.25 < | z | < 0.5$ , $x [n] = 8 δ [n] - 25 {(0.25)}^{n} u [n] - 18 {(0.5)}^{n} u [- n - 1]$ .

Practise this +1 like it2080 Baishakh · Q31+5 marksDefine ROC. Find inverse z-transform of

X (z) = (1 + 2 z^{- 1} + z^{- 2}) / (1 - 1.5 z^{- 1} + 0.5 z^{- 2})

, ROC:

| z | > 1

What is different in this one

The numerator reaches the degree of the denominator. Dividing by z is enough to make it proper, so no long division is needed, but the expansion will carry a term at the origin that inverts to an impulse.
The ROC is the outside of the outermost pole, so every term is right sided and the answer is causal.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{z^{2} + 2 z + 1}{z^{2} - 1.5 z + 0.5}

2. Nothing to divide. Degree 2 over degree 2, so the fraction is already low enough on top.

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = \frac{z^{2} + 2 z + 1}{z^{3} - 1.5 z^{2} + 0.5 z}

The poles are the roots of that denominator. Note the z we just divided by is one of them, sitting at the origin.

z^{3} - 1.5 z^{2} + 0.5 z = z (z - 0.5) (z - 1) = 0

So the poles are $0$ , $0.5$ and $1$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{z^{2} + 2 z + 1}{z (z - 0.5) (z - 1)} \\ = \frac{A}{z} + \frac{B}{(z - 0.5)} + \frac{C}{(z - 1)} \end{aligned}

Put them back over the common denominator $z (z - 0.5) (z - 1)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z - 0.5) (z - 1) + B z (z - 1) + C z (z - 0.5) \\ A (z - 0.5) (z - 1) & = A z^{2} - 1.5 A z + 0.5 A \\ B z (z - 1) & = B z^{2} - B z \\ C z (z - 0.5) & = C z^{2} - 0.5 C z \end{aligned}

Now collect the powers of z and set each one equal to the same power in $z^{2} + 2 z + 1$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + B + C & = 1 & from z^{2} \\ - 1.5 A - B - 0.5 C & = 2 & from z \\ 0.5 A & = 1 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = 2 \\ B & = - 9 \\ C & = 8 \end{aligned}

Putting those back and adding the terms returns $z^{2} + 2 z + 1$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = 2 + \frac{- 9 z}{(z - 0.5)} + \frac{8 z}{(z - 1)}

6. Invert term by term. The ROC $| z | > 1$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
0, the origin	0	2	a polynomial term	2 $δ [n]$
0.5	0.5	-9	the ROC lies outside it	-9 ${(0.5)}^{n} u [n]$
1	1	8	the ROC lies outside it	8 ${(1)}^{n} u [n]$

With the ROC $| z | > 1$ , $x [n] = 2 δ [n] - 9 {(0.5)}^{n} u [n] + 8 {(1)}^{n} u [n]$ .

Check it without trusting any of the algebra: run the difference equation the transform encodes and compare the first samples. Both routes give ${1, 3.5, 5.75, 6.875, 7.4375, 7.71875}$ .

Practise this +1 like it2079 Bhadra · Q36 marksFind inverses Z-transform of

X (z) = (2 z^{4} + 2 z^{3} - 3 z + 2) / (z^{2} - 1.5 z - 1)

, ROC:

| z | < 0.5

What is different in this one

The numerator is of higher degree than the denominator (4 against 2), so it has to be long divided first. The quotient is a polynomial in z, which inverts to impulses at negative n, and only the remainder goes through partial fractions. Divide until the remainder reaches the degree of the divisor and stop there, not further.
The ROC is the inside of the innermost pole, so every term is left sided and the whole answer carries $u [- n - 1]$ with a sign change.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{2 z^{4} + 2 z^{3} - 3 z + 2}{z^{2} - 1.5 z - 1}

			$2 z^{2}$	$+ 5 z$
$z^{2} - 1.5 z - 1$ )	$2 z^{4}$	$+ 2 z^{3}$	$+ 0 z^{2}$	$- 3 z$	$+ 2$
	$2 z^{4}$	$- 3 z^{3}$	$- 2 z^{2}$
		$5 z^{3}$	$+ 2 z^{2}$	$- 3 z$	$+ 2$
		$5 z^{3}$	$- 7.5 z^{2}$	$- 5 z$
			$9.5 z^{2}$	$+ 2 z$	$+ 2$

X (z) = 2 z^{2} + 5 z + \frac{9.5 z^{2} + 2 z + 2}{z^{2} - 1.5 z - 1}

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = 2 z + 5 + \frac{9.5 z^{2} + 2 z + 2}{z^{3} - 1.5 z^{2} - z}

The poles are the roots of that denominator. Note the z we just divided by is one of them, sitting at the origin.

z^{3} - 1.5 z^{2} - z = z (z + 0.5) (z - 2) = 0

So the poles are $0$ , $- 0.5$ and $2$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{9.5 z^{2} + 2 z + 2}{z (z + 0.5) (z - 2)} \\ = \frac{A}{z} + \frac{B}{(z + 0.5)} + \frac{C}{(z - 2)} \end{aligned}

Put them back over the common denominator $z (z + 0.5) (z - 2)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z + 0.5) (z - 2) + B z (z - 2) + C z (z + 0.5) \\ A (z + 0.5) (z - 2) & = A z^{2} - 1.5 A z - A \\ B z (z - 2) & = B z^{2} - 2 B z \\ C z (z + 0.5) & = C z^{2} + 0.5 C z \end{aligned}

Now collect the powers of z and set each one equal to the same power in $9.5 z^{2} + 2 z + 2$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + B + C & = 9.5 & from z^{2} \\ - 1.5 A - 2 B + 0.5 C & = 2 & from z \\ - A & = 2 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = - 2 \\ B & = 2.7 \\ C & = 8.8 \end{aligned}

Putting those back and adding the terms returns $9.5 z^{2} + 2 z + 2$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = 2 z^{2} + 5 z - 2 + \frac{2.7 z}{(z + 0.5)} + \frac{8.8 z}{(z - 2)}

6. Invert term by term. The ROC $| z | < 0.5$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
none, from the division		2	a polynomial term	2 $δ [n + 2]$
none, from the division		5	a polynomial term	5 $δ [n + 1]$
0, the origin	0	-2	a polynomial term	-2 $δ [n]$
-0.5	0.5	2.7	the ROC lies inside it	-2.7 ${(- 0.5)}^{n} u [- n - 1]$
2	2	8.8	the ROC lies inside it	-8.8 ${(2)}^{n} u [- n - 1]$

With the ROC $| z | < 0.5$ , $x [n] = 2 δ [n + 2] + 5 δ [n + 1] - 2 δ [n] - 2.7 {(- 0.5)}^{n} u [- n - 1] - 8.8 {(2)}^{n} u [- n - 1]$ .

Same as 2082 Bhadra Q32079 Baishakh · Q31+5 marksDefine z-transform for a discrete time signal. Find the inverse z-transform for

H (z) = z / (3 z^{2} - 4 z + 1)

using partial fraction method for

1 / 3 < | z | < 1

What is different in this one

The numerator is of lower degree than the denominator, so it is already proper: no division, straight to F(z) = $X (z)$ /z.
The ROC is a ring, so the answer is two sided: the poles inside the ring give right sided terms and the ones outside it give left sided terms. This is the case that separates a real answer from a memorised one.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{z}{3 z^{2} - 4 z + 1}

2. Nothing to divide. Degree 1 over degree 2, so the fraction is already low enough on top.

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = \frac{1}{3 z^{2} - 4 z + 1}

The numerator carried a z of its own, and it cancels the z we just divided by, so there is no term at the origin at all.

3 z^{2} - 4 z + 1 = 3 (z - 0.3333) (z - 1) = 0

So the poles are $0.3333$ and $1$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{1}{3 (z - 0.3333) (z - 1)} \\ = \frac{A}{(z - 0.3333)} + \frac{B}{(z - 1)} \end{aligned}

Put them back over the common denominator $3 (z - 0.3333) (z - 1)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z - 1) + B (z - 0.3333) \\ A (z - 1) & = A z - A \\ B (z - 0.3333) & = B z - 0.3333 B \end{aligned}

Now collect the powers of z and set each one equal to the same power in $0.3333$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + B & = 0 & from z \\ - A - 0.3333 B & = 0.3333 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = - 0.5 \\ B & = 0.5 \end{aligned}

Putting those back and adding the terms returns $1$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = \frac{- 0.5 z}{(z - 0.3333)} + \frac{0.5 z}{(z - 1)}

6. Invert term by term. The ROC $0.3333 < | z | < 1$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
0.3333	0.3333	-0.5	the ROC lies outside it	-0.5 ${(0.3333)}^{n} u [n]$
1	1	0.5	the ROC lies inside it	-0.5 ${(1)}^{n} u [- n - 1]$

With the ROC $0.3333 < | z | < 1$ , $x [n] = - 0.5 {(0.3333)}^{n} u [n] - 0.5 {(1)}^{n} u [- n - 1]$ .

Practise this2078 Bhadra · Q31+5 marksDefine ROC. Find inverse z-transform of

X (z) = (z^{3} + z^{2} + 1.5 z + 0.5) / (z^{3} + 1.5 z^{2} + 0.5 z)

, ROC:

| z | < 1 / 2

What is different in this one

The numerator reaches the degree of the denominator. Dividing by z is enough to make it proper, so no long division is needed, but the expansion will carry a term at the origin that inverts to an impulse.
There is a repeated pole at 0. It needs a term for every power up to its order, and the extra one inverts with an n in front: $A z / (z - p)^{2}$ is (A/p) n $p^{n} u [n]$ , not A $p^{n} u [n]$ . Cover up cannot reach the lower constant at all, which is why the constants are found by comparing coefficients here.
The ROC is the inside of the innermost pole, so every term is left sided and the whole answer carries $u [- n - 1]$ with a sign change.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{z^{3} + z^{2} + 1.5 z + 0.5}{z^{3} + 1.5 z^{2} + 0.5 z}

2. Nothing to divide. Degree 3 over degree 3, so the fraction is already low enough on top.

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = \frac{z^{3} + z^{2} + 1.5 z + 0.5}{z^{4} + 1.5 z^{3} + 0.5 z^{2}}

The poles are the roots of that denominator. Note the z we just divided by is one of them, sitting at the origin.

z^{4} + 1.5 z^{3} + 0.5 z^{2} = z^{2} (z + 0.5) (z + 1) = 0

So the poles are $0$ , repeated 2 times, $- 0.5$ and $- 1$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{z^{3} + z^{2} + 1.5 z + 0.5}{z^{2} (z + 0.5) (z + 1)} \\ = \frac{A}{z} + \frac{B}{z^{2}} + \frac{C}{(z + 0.5)} + \frac{D}{(z + 1)} \end{aligned}

Put them back over the common denominator $z^{2} (z + 0.5) (z + 1)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A z (z + 0.5) (z + 1) + B (z + 0.5) (z + 1) + C z^{2} (z + 1) + D z^{2} (z + 0.5) \\ A z (z + 0.5) (z + 1) & = A z^{3} + 1.5 A z^{2} + 0.5 A z \\ B (z + 0.5) (z + 1) & = B z^{2} + 1.5 B z + 0.5 B \\ C z^{2} (z + 1) & = C z^{3} + C z^{2} \\ D z^{2} (z + 0.5) & = D z^{3} + 0.5 D z^{2} \end{aligned}

Now collect the powers of z and set each one equal to the same power in $z^{3} + z^{2} + 1.5 z + 0.5$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + C + D & = 1 & from z^{3} \\ 1.5 A + B + C + 0.5 D & = 1 & from z^{2} \\ 0.5 A + 1.5 B & = 1.5 & from z \\ 0.5 B & = 0.5 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = 0 \\ B & = 1 \\ C & = - 1 \\ D & = 2 \end{aligned}

Putting those back and adding the terms returns $z^{3} + z^{2} + 1.5 z + 0.5$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = \frac{1}{z} + \frac{- 1 z}{(z + 0.5)} + \frac{2 z}{(z + 1)}

6. Invert term by term. The ROC $| z | < 0.5$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
0, the origin	0	1	a polynomial term	$δ [n - 1]$
-0.5	0.5	-1	the ROC lies inside it	${(- 0.5)}^{n} u [- n - 1]$
-1	1	2	the ROC lies inside it	-2 ${(- 1)}^{n} u [- n - 1]$

The constant at 0 came out zero: a zero of the numerator sits on that pole and cancels it, so it contributes nothing.

With the ROC $| z | < 0.5$ , $x [n] = δ [n - 1] + {(- 0.5)}^{n} u [- n - 1] - 2 {(- 1)}^{n} u [- n - 1]$ .

Same as 2079 Bhadra Q32076 Chaitra · Q42+6 marksDefine ROC of z-transform. Find inverse z-transform using partial fraction expansion of

X (z) = (z^{4} + 5 z^{3} - 3 z + 4) / (z^{2} - 1.5 z - 1)

, ROC:

| z | < 0.5

How this is read $X (z) = (z^{4} + 5 z^{3} - 3 z + 4) / (z^{2} - 1.5 z - 1)$ . The numerator is two degrees higher, so the long division runs for two terms before the partial fractions start.

What is different in this one

The numerator is of higher degree than the denominator (4 against 2), so it has to be long divided first. The quotient is a polynomial in z, which inverts to impulses at negative n, and only the remainder goes through partial fractions. Divide until the remainder reaches the degree of the divisor and stop there, not further.
The ROC is the inside of the innermost pole, so every term is left sided and the whole answer carries $u [- n - 1]$ with a sign change.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{z^{4} + 5 z^{3} - 3 z + 4}{z^{2} - 1.5 z - 1}

			$z^{2}$	$+ 6.5 z$
$z^{2} - 1.5 z - 1$ )	$z^{4}$	$+ 5 z^{3}$	$+ 0 z^{2}$	$- 3 z$	$+ 4$
	$z^{4}$	$- 1.5 z^{3}$	$- z^{2}$
		$6.5 z^{3}$	$+ z^{2}$	$- 3 z$	$+ 4$
		$6.5 z^{3}$	$- 9.75 z^{2}$	$- 6.5 z$
			$10.75 z^{2}$	$+ 3.5 z$	$+ 4$

X (z) = z^{2} + 6.5 z + \frac{10.75 z^{2} + 3.5 z + 4}{z^{2} - 1.5 z - 1}

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = z + 6.5 + \frac{10.75 z^{2} + 3.5 z + 4}{z^{3} - 1.5 z^{2} - z}

The poles are the roots of that denominator. Note the z we just divided by is one of them, sitting at the origin.

z^{3} - 1.5 z^{2} - z = z (z + 0.5) (z - 2) = 0

So the poles are $0$ , $- 0.5$ and $2$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{10.75 z^{2} + 3.5 z + 4}{z (z + 0.5) (z - 2)} \\ = \frac{A}{z} + \frac{B}{(z + 0.5)} + \frac{C}{(z - 2)} \end{aligned}

Put them back over the common denominator $z (z + 0.5) (z - 2)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z + 0.5) (z - 2) + B z (z - 2) + C z (z + 0.5) \\ A (z + 0.5) (z - 2) & = A z^{2} - 1.5 A z - A \\ B z (z - 2) & = B z^{2} - 2 B z \\ C z (z + 0.5) & = C z^{2} + 0.5 C z \end{aligned}

Now collect the powers of z and set each one equal to the same power in $10.75 z^{2} + 3.5 z + 4$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + B + C & = 10.75 & from z^{2} \\ - 1.5 A - 2 B + 0.5 C & = 3.5 & from z \\ - A & = 4 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = - 4 \\ B & = 3.95 \\ C & = 10.8 \end{aligned}

Putting those back and adding the terms returns $10.75 z^{2} + 3.5 z + 4$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = z^{2} + 6.5 z - 4 + \frac{3.95 z}{(z + 0.5)} + \frac{10.8 z}{(z - 2)}

6. Invert term by term. The ROC $| z | < 0.5$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
none, from the division		1	a polynomial term	$δ [n + 2]$
none, from the division		6.5	a polynomial term	6.5 $δ [n + 1]$
0, the origin	0	-4	a polynomial term	-4 $δ [n]$
-0.5	0.5	3.95	the ROC lies inside it	-3.95 ${(- 0.5)}^{n} u [- n - 1]$
2	2	10.8	the ROC lies inside it	-10.8 ${(2)}^{n} u [- n - 1]$

With the ROC $| z | < 0.5$ , $x [n] = δ [n + 2] + 6.5 δ [n + 1] - 4 δ [n] - 3.95 {(- 0.5)}^{n} u [- n - 1] - 10.8 {(2)}^{n} u [- n - 1]$ .

Practise this2076 Ashwin · Q33+6 marksState Convolution property of Z-transform. Find inverse Z-transform of

X (z) = z

(z - 0.6) (z + 0.5)^{2}

}, ROC:

| z | > 0.6

What is different in this one

The numerator is of lower degree than the denominator, so it is already proper: no division, straight to F(z) = $X (z)$ /z.
There is a repeated pole at -0.5. It needs a term for every power up to its order, and the extra one inverts with an n in front: $A z / (z - p)^{2}$ is (A/p) n $p^{n} u [n]$ , not A $p^{n} u [n]$ . Cover up cannot reach the lower constant at all, which is why the constants are found by comparing coefficients here.
The ROC is the outside of the outermost pole, so every term is right sided and the answer is causal.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{z}{z^{3} + 0.4 z^{2} - 0.35 z - 0.15}

2. Nothing to divide. Degree 1 over degree 3, so the fraction is already low enough on top.

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = \frac{1}{z^{3} + 0.4 z^{2} - 0.35 z - 0.15}

The numerator carried a z of its own, and it cancels the z we just divided by, so there is no term at the origin at all.

z^{3} + 0.4 z^{2} - 0.35 z - 0.15 = {(z + 0.5)}^{2} (z - 0.6) = 0

So the poles are $- 0.5$ , repeated 2 times and $0.6$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{1}{{(z + 0.5)}^{2} (z - 0.6)} \\ = \frac{A}{(z + 0.5)} + \frac{B}{{(z + 0.5)}^{2}} + \frac{C}{(z - 0.6)} \end{aligned}

Put them back over the common denominator ${(z + 0.5)}^{2} (z - 0.6)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z + 0.5) (z - 0.6) + B (z - 0.6) + C {(z + 0.5)}^{2} \\ A (z + 0.5) (z - 0.6) & = A z^{2} - 0.1 A z - 0.3 A \\ B (z - 0.6) & = B z - 0.6 B \\ C {(z + 0.5)}^{2} & = C z^{2} + C z + 0.25 C \end{aligned}

Now collect the powers of z and set each one equal to the same power in $1$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + C & = 0 & from z^{2} \\ - 0.1 A + B + C & = 0 & from z \\ - 0.3 A - 0.6 B + 0.25 C & = 1 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = - 0.8264 \\ B & = - 0.9091 \\ C & = 0.8264 \end{aligned}

Putting those back and adding the terms returns $1$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = \frac{- 0.8264 z}{(z + 0.5)} + \frac{- 0.9091 z}{{(z + 0.5)}^{2}} + \frac{0.8264 z}{(z - 0.6)}

6. Invert term by term. The ROC $| z | > 0.6$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
-0.5	0.5	-0.8264	the ROC lies outside it	-0.8264 ${(- 0.5)}^{n} u [n]$
-0.5, order 2	0.5	-0.9091	the ROC lies outside it	1.8182 n ${(- 0.5)}^{n} u [n]$
0.6	0.6	0.8264	the ROC lies outside it	0.8264 ${(0.6)}^{n} u [n]$

With the ROC $| z | > 0.6$ , $x [n] = - 0.8264 {(- 0.5)}^{n} u [n]$ + 1.8182 n ${(- 0.5)}^{n} u [n] + 0.8264 {(0.6)}^{n} u [n]$ .

Check it without trusting any of the algebra: run the difference equation the transform encodes and compare the first samples. Both routes give ${0, 0, 1, - 0.4, 0.51, - 0.194}$ .

Same as 2081 Bhadra Q32074 Chaitra · Q31+3+5 marksDefine a ROC. What are the properties of ROC of z-transform? Find the inverse Z-transform of

X (z) = (2 z^{2} + 2 z^{2} + 3 z + 5) / (z^{2} - 0.1 z - 0.2)

, ROC:

| z | < 0.4

What is different in this one

The numerator is of higher degree than the denominator (3 against 2), so it has to be long divided first. The quotient is a polynomial in z, which inverts to impulses at negative n, and only the remainder goes through partial fractions. Divide until the remainder reaches the degree of the divisor and stop there, not further.
The ROC is the inside of the innermost pole, so every term is left sided and the whole answer carries $u [- n - 1]$ with a sign change.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{2 z^{3} + 2 z^{2} + 3 z + 5}{z^{2} - 0.1 z - 0.2}

			$2 z$
$z^{2} - 0.1 z - 0.2$ )	$2 z^{3}$	$+ 2 z^{2}$	$+ 3 z$	$+ 5$
	$2 z^{3}$	$- 0.2 z^{2}$	$- 0.4 z$
		$2.2 z^{2}$	$+ 3.4 z$	$+ 5$

X (z) = 2 z + \frac{2.2 z^{2} + 3.4 z + 5}{z^{2} - 0.1 z - 0.2}

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = 2 + \frac{2.2 z^{2} + 3.4 z + 5}{z^{3} - 0.1 z^{2} - 0.2 z}

The poles are the roots of that denominator. Note the z we just divided by is one of them, sitting at the origin.

z^{3} - 0.1 z^{2} - 0.2 z = z (z + 0.4) (z - 0.5) = 0

So the poles are $0$ , $- 0.4$ and $0.5$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{2.2 z^{2} + 3.4 z + 5}{z (z + 0.4) (z - 0.5)} \\ = \frac{A}{z} + \frac{B}{(z + 0.4)} + \frac{C}{(z - 0.5)} \end{aligned}

Put them back over the common denominator $z (z + 0.4) (z - 0.5)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z + 0.4) (z - 0.5) + B z (z - 0.5) + C z (z + 0.4) \\ A (z + 0.4) (z - 0.5) & = A z^{2} - 0.1 A z - 0.2 A \\ B z (z - 0.5) & = B z^{2} - 0.5 B z \\ C z (z + 0.4) & = C z^{2} + 0.4 C z \end{aligned}

Now collect the powers of z and set each one equal to the same power in $2.2 z^{2} + 3.4 z + 5$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + B + C & = 2.2 & from z^{2} \\ - 0.1 A - 0.5 B + 0.4 C & = 3.4 & from z \\ - 0.2 A & = 5 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = - 25 \\ B & = 11.0889 \\ C & = 16.1111 \end{aligned}

Putting those back and adding the terms returns $2.2 z^{2} + 3.4 z + 5$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = 2 z - 25 + \frac{11.0889 z}{(z + 0.4)} + \frac{16.1111 z}{(z - 0.5)}

6. Invert term by term. The ROC $| z | < 0.4$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
none, from the division		2	a polynomial term	2 $δ [n + 1]$
0, the origin	0	-25	a polynomial term	-25 $δ [n]$
-0.4	0.4	11.0889	the ROC lies inside it	-11.0889 ${(- 0.4)}^{n} u [- n - 1]$
0.5	0.5	16.1111	the ROC lies inside it	-16.1111 ${(0.5)}^{n} u [- n - 1]$

With the ROC $| z | < 0.4$ , $x [n] = 2 δ [n + 1] - 25 δ [n] - 11.0889 {(- 0.4)}^{n} u [- n - 1] - 16.1111 {(0.5)}^{n} u [- n - 1]$ .

Practise this2074 Ashwin · Q33+3 marksState and explain the properties of a Region of Convergence (ROC). Find the inverse z-transform of

X (z) = z^{2} [1 - (3 / 2) z^{- 1}] (1 + z^{- 1}) (1 - z^{- 1})

What is different hereNo ROC is given because none is needed: a finite length sequence converges everywhere except possibly at

z = 0

and z = infinity.

What is different in this one

There is nothing to invert: no partial fractions, no ROC decision. $X (z)$ is already a finite polynomial, so its coefficients are the sequence. Thirty seconds if you spot it.
A finite length sequence converges everywhere except possibly at $z = 0$ and z = infinity, which is why the question does not give you an ROC.
The factor $z^{2}$ is an advance, so the sequence starts at $n = - 2$ , before the origin.

Multiply the factors out, then read each coefficient as an impulse: a power of $z^{- k}$ is a delay of k and a power of $z^{k}$ an advance.

\begin{aligned} X (z) & = z^{2} (1 - 1.5 z^{- 1}) (1 + z^{- 1}) (1 - z^{- 1}) \\ = z^{2} (1 - 1.5 z^{- 1}) \\ = z^{2} (1 - 0.5 z^{- 1} - 1.5 z^{- 2}) \\ = z^{2} (1 - 1.5 z^{- 1} - z^{- 2} + 1.5 z^{- 3}) \end{aligned}

$x [n] = 1 δ [n + 2] - 1.5 δ [n + 1] - 1 δ [n] + 1.5 δ [n - 1]$ , that is {1, -1.5, -1, 1.5} for $n = - 2$ to 1.

Practise this2073 Shrawan · Q31+5 marksDefine a ROC. Find inverse Z-transform of

X (z) = z

(z - 0.4) (z + 1.5)^{2}

}, ROC:

| z | < 0.4

How this is read $X (z) = z / ((z - 0.4) (z + 1.5)^{2})$ , whose denominator multiplies out to $z^{3} + 2.6 z^{2} + 1.05 z - 0.9$ .

What is different in this one

The numerator is of lower degree than the denominator, so it is already proper: no division, straight to F(z) = $X (z)$ /z.
There is a repeated pole at -1.5. It needs a term for every power up to its order, and the extra one inverts with an n in front: $A z / (z - p)^{2}$ is (A/p) n $p^{n} u [n]$ , not A $p^{n} u [n]$ . Cover up cannot reach the lower constant at all, which is why the constants are found by comparing coefficients here.
The ROC is the inside of the innermost pole, so every term is left sided and the whole answer carries $u [- n - 1]$ with a sign change.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{z}{z^{3} + 2.6 z^{2} + 1.05 z - 0.9}

2. Nothing to divide. Degree 1 over degree 3, so the fraction is already low enough on top.

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = \frac{1}{z^{3} + 2.6 z^{2} + 1.05 z - 0.9}

The numerator carried a z of its own, and it cancels the z we just divided by, so there is no term at the origin at all.

z^{3} + 2.6 z^{2} + 1.05 z - 0.9 = (z - 0.4) {(z + 1.5)}^{2} = 0

So the poles are $0.4$ and $- 1.5$ , repeated 2 times.

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{1}{(z - 0.4) {(z + 1.5)}^{2}} \\ = \frac{A}{(z - 0.4)} + \frac{B}{(z + 1.5)} + \frac{C}{{(z + 1.5)}^{2}} \end{aligned}

Put them back over the common denominator $(z - 0.4) {(z + 1.5)}^{2}$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A {(z + 1.5)}^{2} + B (z - 0.4) (z + 1.5) + C (z - 0.4) \\ A {(z + 1.5)}^{2} & = A z^{2} + 3 A z + 2.25 A \\ B (z - 0.4) (z + 1.5) & = B z^{2} + 1.1 B z - 0.6 B \\ C (z - 0.4) & = C z - 0.4 C \end{aligned}

Now collect the powers of z and set each one equal to the same power in $1$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + B & = 0 & from z^{2} \\ 3 A + 1.1 B + C & = 0 & from z \\ 2.25 A - 0.6 B - 0.4 C & = 1 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = 0.277 \\ B & = - 0.277 \\ C & = - 0.5263 \end{aligned}

Putting those back and adding the terms returns $1$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = \frac{0.277 z}{(z - 0.4)} + \frac{- 0.277 z}{(z + 1.5)} + \frac{- 0.5263 z}{{(z + 1.5)}^{2}}

6. Invert term by term. The ROC $| z | < 0.4$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
0.4	0.4	0.277	the ROC lies inside it	-0.277 ${(0.4)}^{n} u [- n - 1]$
-1.5	1.5	-0.277	the ROC lies inside it	0.277 ${(- 1.5)}^{n} u [- n - 1]$
-1.5, order 2	1.5	-0.5263	the ROC lies inside it	-0.3509 n ${(- 1.5)}^{n} u [- n - 1]$

With the ROC $| z | < 0.4$ , $x [n] = - 0.277 {(0.4)}^{n} u [- n - 1] + 0.277 {(- 1.5)}^{n} u [- n - 1]$ - 0.3509 n ${(- 1.5)}^{n} u [- n - 1]$ .

Same as 2080 Baishakh Q32071 Shrawan · Q26 marksFind the inverse z transform of:

X (Z) = (1 + 2 z^{- 1} + z^{- 2}) / (1 + 1.5 z^{- 1} + 0.5 z^{- 2})

| z | > 1

using partial fraction method.

What is different in this one

The numerator reaches the degree of the denominator. Dividing by z is enough to make it proper, so no long division is needed, but the expansion will carry a term at the origin that inverts to an impulse.
The ROC is the outside of the outermost pole, so every term is right sided and the answer is causal.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{z^{2} + 2 z + 1}{z^{2} + 1.5 z + 0.5}

2. Nothing to divide. Degree 2 over degree 2, so the fraction is already low enough on top.

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = \frac{z^{2} + 2 z + 1}{z^{3} + 1.5 z^{2} + 0.5 z}

The poles are the roots of that denominator. Note the z we just divided by is one of them, sitting at the origin.

z^{3} + 1.5 z^{2} + 0.5 z = z (z + 0.5) (z + 1) = 0

So the poles are $0$ , $- 0.5$ and $- 1$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{z^{2} + 2 z + 1}{z (z + 0.5) (z + 1)} \\ = \frac{A}{z} + \frac{B}{(z + 0.5)} + \frac{C}{(z + 1)} \end{aligned}

Put them back over the common denominator $z (z + 0.5) (z + 1)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z + 0.5) (z + 1) + B z (z + 1) + C z (z + 0.5) \\ A (z + 0.5) (z + 1) & = A z^{2} + 1.5 A z + 0.5 A \\ B z (z + 1) & = B z^{2} + B z \\ C z (z + 0.5) & = C z^{2} + 0.5 C z \end{aligned}

Now collect the powers of z and set each one equal to the same power in $z^{2} + 2 z + 1$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + B + C & = 1 & from z^{2} \\ 1.5 A + B + 0.5 C & = 2 & from z \\ 0.5 A & = 1 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = 2 \\ B & = - 1 \\ C & = 0 \end{aligned}

Putting those back and adding the terms returns $z^{2} + 2 z + 1$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = 2 + \frac{- 1 z}{(z + 0.5)}

6. Invert term by term. The ROC $| z | > 1$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
0, the origin	0	2	a polynomial term	2 $δ [n]$
-0.5	0.5	-1	the ROC lies outside it	- ${(- 0.5)}^{n} u [n]$

The constant at -1 came out zero: a zero of the numerator sits on that pole and cancels it, so it contributes nothing.

With the ROC $| z | > 1$ , $x [n] = 2 δ [n] - {(- 0.5)}^{n} u [n]$ .

Check it without trusting any of the algebra: run the difference equation the transform encodes and compare the first samples. Both routes give ${1, 0.5, - 0.25, 0.125, - 0.0625, 0.03125}$ .

The same questions grouped by how often each topic is set: the inverse z-transform

2.4Last minute recall

Last minute recall, chapter 2

Must memorise

$X (z)$ = sum of $x [n] z^{- n}$ , and the ROC must always be quoted with it: the same algebra with two ROCs is two different sequences.
The ROC is a ring centred on the origin and contains no pole. Right sided means outside the outermost pole; left sided means inside the innermost.
Stable if and only if the ROC contains the unit circle. Causal if and only if the ROC is the outside of the outermost pole, and both together means every pole is inside the unit circle.
Expand $X (z)$ /z, not $X (z)$ , so every term returns as $z / (z - p)$ .
ROC outside the pole gives A $p^{n} u [n]$ ; ROC inside gives -A $p^{n} u [- n - 1]$ .
Long divide first if the numerator's degree reaches the denominator's: the quotient becomes impulses.
The pairs: $z / (z - a)$ to $a^{n} u [n]$ , z/ ${(z - a)}^{2}$ to n $a^{n - 1} u [n]$ , a constant to an impulse, $z^{k}$ to $δ [n + k]$ .
Convolution becomes multiplication: $Y (z) = H (z) X (z)$ . That is the whole reason the transform exists.

Most repeated in this chapter, in order

Inverse z-transform by partial fractions, with the ROC deciding TOP 15/19
Define the ROC TOP 8/19, almost always as the opening two marks of the same question
Properties of the ROC, and locating it HOT 7/19
Define the z-transform PIN 2/19 and derive the convolution property PIN 2/19

Question 3 is this chapter, every year, and it is almost always the same question: define the ROC for two marks, then invert an $X (z)$ for five or six. The marks are in the ROC assignment, not the algebra.

Chapter 3 · 6 hours · 10 marks · question 4, every paper

LTI systems in the frequency domain

What an LTI system does to a sinusoid, how to read that straight off a pole zero diagram without computing anything, and how stability and causality are written in the same picture. Question 4 is this chapter in every paper, and it is ten marks.

What this chapter is about

The frequency response: why a complex exponential goes through an LTI system unchanged except for a complex gain.
The geometric method: reading magnitude and phase off the distances from a point on the unit circle to the poles and zeros.
Stability and causality stated three ways: on $h [n]$ , on the ROC, and on the pole positions.
FIR against IIR, which the paper asks as a straight comparison.

Where it fits

Chapter 2 gave $H (z)$ ; this chapter walks around the unit circle in it.
Chapters 5 and 6 design $H (z)$ so that this magnitude response comes out the way the specification demands.
Chapter 7 samples this same response at N points.

What you will learn

3.1 The frequency response, and why exponentials pass straight through
3.2 Difference equations, and the system function, The zero input response
3.3 Plotting poles and zeros and sketching the magnitude response, The phase response
3.4 Linear phase, and its relationship with causality, Stability and causality, three ways of saying the same thing, The BIBO stability test, FIR against IIR
3.5 Last minute recall, chapter 3

How it is examined

Question 4 is always: plot the pole zero in the z-plane and draw the magnitude response, not to scale. Ten marks, seventeen papers.
The question is given either as a difference equation or as a list of pole and zero locations. Both reduce to the same picture.
Not to scale means labelled, not vague. Values at $ω = 0$ and $ω = π$ , peaks at the poles, nulls at the zeros.

3.1Frequency response of an LTI system, response to a complex exponential

The frequency response, and why exponentials pass straight through HOT 5/19

82 Ba · 80 Ba · 73 Shr · 72 Ka · 71 Shr52+3+53+7

Before the formulas

Chapter 2 gave you $H (z)$ , which is algebra. This chapter asks what that algebra sounds like: what does the system do to each frequency that goes through it. The bridge is short. Set z to a point on the unit circle and the transfer function becomes the gain and phase at that frequency, so walking once round the circle traces out the entire response.

Chapter 1 showed you why exponentials pass through unchanged. Here that fact turns into something you can draw, and drawing it is most of this chapter.

The eigenfunction property Feed

x [n] = e^{j ω_{0} n}

into an LTI system with impulse response

h [n]

. The convolution sum gives back the same exponential, multiplied by a complex number that depends only on

ω_{0}

. The exponential is an eigenfunction of every LTI system, and the complex number is the frequency response.

\begin{aligned} y [n] & = \sum_{k} h [k] x [n - k] \\ = \sum_{k} h [k] e^{j ω_{0} (n - k)} \\ = e^{j ω_{0} n} \sum_{k} h [k] e^{- j ω_{0} k} \\ = e^{j ω_{0} n} H (e^{j ω_{0}}) \\ H (e^{j ω}) & = \sum_{k = - \infty}^{\infty} h [k] e^{- j ω k} = H (z) |_{z = e^{j ω}} \\ = | H (e^{j ω}) | e^{j θ (ω)} & magnitude and phase \end{aligned}

What this buys you

No convolution needed. When the input is a complex exponential or a sinusoid, the output is the input scaled by $| H |$ and delayed in phase by $θ$ . Three papers set this as a convolution question; recognising it saves ten minutes.
For a real sinusoid: if $x [n] = A$ cos( $ω_{0}$ n + $ϕ$ ), then $y [n] = A$ | $H (e^{j ω_{0}})$ | cos( $ω_{0}$ n + $ϕ + θ$ ( $ω_{0}$ )).
For a sum of sinusoids, do each one separately and add, which is what the question with $x [n] = 10 - 5 \sin (π n / 2) + 20 \cos (π n)$ is testing. Note that the constant 10 is the $ω = 0$ component, so it is scaled by $H (e^{j 0})$ .

Worked, the one set four times

h [n] = {(\frac{1}{2})}^{n} u [n]

and

x [n] = 5 e^{j π n / 3}

H (e^{j ω}) = 1

/(1 - 0.5

e^{- j ω}

). At

ω = π

/3:

e^{- j π / 3} = 0.5

- j0.866, so 1 - 0.5(0.5 - j0.866) = 0.75 + j0.433, whose magnitude is 0.866 and angle 30 degrees.

H = 1

/(0.866 angle 30 deg) = 1.1547 angle -30 deg.
$y [n] = 5$ (1.1547) $e^{j (π n / 3 - π / 6)} = 5.7735 e^{j (π n / 3 - π / 6)}$ .

Two properties of $H (e^{j ω})$ worth stating in any answer

Periodic in $2 π$ . $H (e^{j (ω + 2 p i)}) = H (e^{j ω})$ , because $e^{- j 2 p i k} = 1$ . Only one period of the response exists, which is why every plot runs from 0 to $2 π$ , or $- π$ to $π$ .
Conjugate symmetric for real $h [n]$ . $H (e^{- j ω}) = H$ *( $e^{j ω}$ ), so the magnitude is even and the phase is odd. That is why the magnitude is always sketched over 0 to $π$ and then mirrored.

This is the DTFT of $h [n]$ under another name, and FIR design and IIR design are both nothing but choosing the $| H (e^{j ω}) |$ this card evaluates.

See it drawnThe frequency response is

H (z)

walked round the unit circle

Set z to a point on the circle and the transfer function becomes the gain at that frequency. Walking once round is the whole response.

Practise thisEvery question the paper has set on this topic, worked in full5

Practise this +2 like it2082 Baishakh · Q25 marksFind the output of an LTI system having impulse response

h [n] = {(1 / 2)}^{n} u [n]

excited by an input

x [n] = 5 e^{j π n / 3}

What is different in this one

1 component at one frequency, so H is evaluated once and the results added: superposition.
Do not convolve. The input runs over all n, so it is not a finite sequence and the graphical method is the wrong tool.

x [n] = e^{j ω_{0} n} ⟶ y [n] = H (e^{j ω_{0}}) e^{j ω_{0} n}, A \cos (ω_{0} n) ⟶ A | H | \cos (ω_{0} n + ∠ H)

\begin{aligned} H (e^{j ω}) & = 1 / (1 - 0.5 e^{- j ω}), the  transform  of h [n] = {(1 / 2)}^{n} u [n] \end{aligned}

Component	$ω_{0}$	$H (e^{j ω_{0}})$	in polar form	Output amplitude
$x [n] = 5 e^{j π n / 3}$	$0.3333 π$	1 - j0.5774	1.1547 angle -30 deg	5.7735

Where	By convolution	By the formula
$n = 0$	5	5
$n = 1$	5	5
$n = 2$	0	0
$n = 5$	0	0

$y [n] = 5.7735 e^{j (0.3333 π n - 30 d e g)}$ .

Same as 2082 Baishakh Q22080 Baishakh · Q25 marksFind the output of LTI system having impulse response

h [n] = {(1 / 2)}^{n} u [n]

and input

x [n] = 5 e^{j π n / 3}

for -infinity < n < infinity.

What is different in this one

1 component at one frequency, so H is evaluated once and the results added: superposition.
Do not convolve. The input runs over all n, so it is not a finite sequence and the graphical method is the wrong tool.

x [n] = e^{j ω_{0} n} ⟶ y [n] = H (e^{j ω_{0}}) e^{j ω_{0} n}, A \cos (ω_{0} n) ⟶ A | H | \cos (ω_{0} n + ∠ H)

\begin{aligned} H (e^{j ω}) & = 1 / (1 - 0.5 e^{- j ω}), the  transform  of h [n] = {(1 / 2)}^{n} u [n] \end{aligned}

Component	$ω_{0}$	$H (e^{j ω_{0}})$	in polar form	Output amplitude
$x [n] = 5 e^{j π n / 3}$	$0.3333 π$	1 - j0.5774	1.1547 angle -30 deg	5.7735

Where	By convolution	By the formula
$n = 0$	5	5
$n = 1$	5	5
$n = 2$	0	0
$n = 5$	0	0

$y [n] = 5.7735 e^{j (0.3333 π n - 30 d e g)}$ .

Practise this2073 Shrawan · Q43+7 marksState linear constant coefficient difference equation and corresponding system function. Determine the output sequence of the system with impulse response

h [n] = {(1 / 2)}^{n} u [n]

when the input signal is

x [n] = 10

- 5sin(

π

n/2) + 20cos(

π

What is different in this one

3 components at 3 different frequencies, so H is evaluated 3 times and the results added: superposition.
One of them is a constant, which is simply the $ω = 0$ case: the input times the DC gain.
The input is real, so the answer must be real. A sine goes in and a sine of the same frequency comes out, scaled by $| H |$ and shifted by the angle of H. Leaving the answer as complex exponentials is not finishing the question.
Do not convolve. The input runs over all n, so it is not a finite sequence and the graphical method is the wrong tool.

x [n] = e^{j ω_{0} n} ⟶ y [n] = H (e^{j ω_{0}}) e^{j ω_{0} n}, A \cos (ω_{0} n) ⟶ A | H | \cos (ω_{0} n + ∠ H)

\begin{aligned} H (e^{j ω}) & = 1 / (1 - 0.5 e^{- j ω}), the  transform  of h [n] = {(1 / 2)}^{n} u [n] \end{aligned}

Component	$ω_{0}$	$H (e^{j ω_{0}})$	in polar form	Output amplitude
the constant 10, which is the $ω = 0$ case	$0 π$	2	2 angle 0 deg	20
-5 $\sin (π n / 2)$ , taken at $ω = \frac{π}{2}$	$0.5 π$	0.8 - j0.4	0.8944 angle -26.57 deg	4.4721
20 $\cos (π n)$ , taken at $ω = π$	$1 π$	0.6667	0.6667 angle 0 deg	13.3333

Where	By convolution	By the formula
$n = 0$	35.3333	35.3333
$n = 1$	2.6667	2.6667
$n = 2$	31.3333	31.3333
$n = 5$	2.6667	2.6667

Same as 2082 Baishakh Q22072 Kartik · Q44 marksFind the output of LTI System having impulse response

h [n] = {(1 / 2)}^{n} u [n]

and input signal

x [n] = 5 e^{j π n / 3}

for -infinity < n < infinity.

What is different in this one

1 component at one frequency, so H is evaluated once and the results added: superposition.
Do not convolve. The input runs over all n, so it is not a finite sequence and the graphical method is the wrong tool.

x [n] = e^{j ω_{0} n} ⟶ y [n] = H (e^{j ω_{0}}) e^{j ω_{0} n}, A \cos (ω_{0} n) ⟶ A | H | \cos (ω_{0} n + ∠ H)

\begin{aligned} H (e^{j ω}) & = 1 / (1 - 0.5 e^{- j ω}), the  transform  of h [n] = {(1 / 2)}^{n} u [n] \end{aligned}

Component	$ω_{0}$	$H (e^{j ω_{0}})$	in polar form	Output amplitude
$x [n] = 5 e^{j π n / 3}$	$0.3333 π$	1 - j0.5774	1.1547 angle -30 deg	5.7735

Where	By convolution	By the formula
$n = 0$	5	5
$n = 1$	5	5
$n = 2$	0	0
$n = 5$	0	0

$y [n] = 5.7735 e^{j (0.3333 π n - 30 d e g)}$ .

Practise this2071 Shrawan · Q32+3+5 marksWhy do we need difference equation? State linear constant coefficient difference equation and corresponding system function. Consider an LTI system with impulse response

h [n] = {(1 / 2)}^{n} u [n]

. Determine

y [n]

, if the input is

x [n] = A e^{j π n}

.What is different hereAt

ω = π

the exponential

e^{- j ω}

is exactly -1, so H is

1 / (1 + 0.5) = 2 / 3

with no complex arithmetic at all. The answer is

y [n]

= (2A/3)

e^{j π n}

What is different in this one

1 component at one frequency, so H is evaluated once and the results added: superposition.
Do not convolve. The input runs over all n, so it is not a finite sequence and the graphical method is the wrong tool.

x [n] = e^{j ω_{0} n} ⟶ y [n] = H (e^{j ω_{0}}) e^{j ω_{0} n}, A \cos (ω_{0} n) ⟶ A | H | \cos (ω_{0} n + ∠ H)

\begin{aligned} H (e^{j ω}) & = 1 / (1 - 0.5 e^{- j ω}), the  transform  of h [n] = {(1 / 2)}^{n} u [n] \end{aligned}

Component	$ω_{0}$	$H (e^{j ω_{0}})$	in polar form	Output amplitude
$x [n] = A e^{j π n}$ , with A factored out	$1 π$	0.6667	0.6667 angle 0 deg	0.6667

Where	By convolution	By the formula
$n = 0$	0.6667	0.6667
$n = 1$	-0.6667	-0.6667
$n = 2$	0.6667	0.6667
$n = 5$	-0.6667	-0.6667

$y [n] = 0.6667 e^{j (π n)}$ .

The same questions grouped by how often each topic is set: the output for an exponential input

3.2Linear constant coefficient difference equation and the system function

Difference equations, and the system function PIN 2/19

73 Shr · 71 Shr2+3+53+7

Before the formulas

A transfer function is a description of a filter. A difference equation is a thing you can build: every term in it is a delay, a multiplication or an addition, which is exactly what hardware has and nothing more.

This section moves between the two, and it is a move you will make in almost every numerical for the rest of the course. The system function is nothing more than the difference equation transformed and rearranged.

The one place marks go missing is the sign: the feedback terms change sign as they cross the equals sign, every time.

Linear constant coefficient difference equation A relation between the output and the input of the form below, with coefficients that do not depend on n. It is the discrete time counterpart of a differential equation, and it is how a system is actually implemented.

\begin{aligned} \sum_{k = 0}^{N} a_{k} y [n - k] & = \sum_{k = 0}^{M} b_{k} x [n - k], a_{0} = 1 \\ y [n] & = \sum_{k = 0}^{M} b_{k} x [n - k] - \sum_{k = 1}^{N} a_{k} y [n - k] \\ H (z) & = \frac{B (z)}{A (z)} = \frac{b_{0} + b_{1} z^{- 1} + \dots + b_{M} z^{- M}}{1 + a_{1} z^{- 1} + \dots + a_{N} z^{- N}} \end{aligned}

Why we need it

It is the implementation. Convolution with an infinite $h [n]$ cannot be programmed; a difference equation with a handful of coefficients can, and it runs in real time.
It is finite memory. Only N past outputs and M past inputs are stored, whatever the length of $h [n]$ .
It carries the structure. The b coefficients are the feed forward path and the a coefficients the feedback path, which is exactly what chapter 4 draws.
It classifies the filter. No a terms means FIR, non recursive, $h [n]$ of finite length. Any a term means IIR, recursive, $h [n]$ of infinite length.

The two responses. The total solution of a difference equation splits in two, and the paper asks for the first half on its own.

	Zero input response	Zero state response
Also called	Natural or free response	Forced response
Caused by	The initial conditions, with $x [n] = 0$	The input, with all initial conditions zero
Found from	The roots of the characteristic equation	Convolution, or the z-transform
Shape	$\sum_{k} C_{k} λ_{k}^{n}$	Depends on the input

Taking the z-transform of this equation, with the time shift property, gives $H (z)$ directly; the coefficients are then read straight onto a direct form structure.

See it drawnA difference equation is the filter you can actually build

Every term is a delay, a multiply or an add. That is why this form, and not the convolution sum, is what gets implemented.

Practise thisEvery question the paper has set on this topic, worked in full2

Practise this2073 Shrawan · Q43+7 marksState linear constant coefficient difference equation and corresponding system function. Determine the output sequence of the system with impulse response

h [n] = {(1 / 2)}^{n} u [n]

when the input signal is

x [n] = 10

- 5sin(

π

n/2) + 20cos(

π

What is different in this one

3 components at 3 different frequencies, so H is evaluated 3 times and the results added: superposition.
One of them is a constant, which is simply the $ω = 0$ case: the input times the DC gain.
The input is real, so the answer must be real. A sine goes in and a sine of the same frequency comes out, scaled by $| H |$ and shifted by the angle of H. Leaving the answer as complex exponentials is not finishing the question.
Do not convolve. The input runs over all n, so it is not a finite sequence and the graphical method is the wrong tool.

x [n] = e^{j ω_{0} n} ⟶ y [n] = H (e^{j ω_{0}}) e^{j ω_{0} n}, A \cos (ω_{0} n) ⟶ A | H | \cos (ω_{0} n + ∠ H)

\begin{aligned} H (e^{j ω}) & = 1 / (1 - 0.5 e^{- j ω}), the  transform  of h [n] = {(1 / 2)}^{n} u [n] \end{aligned}

Component	$ω_{0}$	$H (e^{j ω_{0}})$	in polar form	Output amplitude
the constant 10, which is the $ω = 0$ case	$0 π$	2	2 angle 0 deg	20
-5 $\sin (π n / 2)$ , taken at $ω = \frac{π}{2}$	$0.5 π$	0.8 - j0.4	0.8944 angle -26.57 deg	4.4721
20 $\cos (π n)$ , taken at $ω = π$	$1 π$	0.6667	0.6667 angle 0 deg	13.3333

Where	By convolution	By the formula
$n = 0$	35.3333	35.3333
$n = 1$	2.6667	2.6667
$n = 2$	31.3333	31.3333
$n = 5$	2.6667	2.6667

h [n] = {(1 / 2)}^{n} u [n]

. Determine

y [n]

, if the input is

x [n] = A e^{j π n}

.What is different hereAt

ω = π

the exponential

e^{- j ω}

is exactly -1, so H is

1 / (1 + 0.5) = 2 / 3

with no complex arithmetic at all. The answer is

y [n]

= (2A/3)

e^{j π n}

What is different in this one

1 component at one frequency, so H is evaluated once and the results added: superposition.
Do not convolve. The input runs over all n, so it is not a finite sequence and the graphical method is the wrong tool.

x [n] = e^{j ω_{0} n} ⟶ y [n] = H (e^{j ω_{0}}) e^{j ω_{0} n}, A \cos (ω_{0} n) ⟶ A | H | \cos (ω_{0} n + ∠ H)

\begin{aligned} H (e^{j ω}) & = 1 / (1 - 0.5 e^{- j ω}), the  transform  of h [n] = {(1 / 2)}^{n} u [n] \end{aligned}

Component	$ω_{0}$	$H (e^{j ω_{0}})$	in polar form	Output amplitude
$x [n] = A e^{j π n}$ , with A factored out	$1 π$	0.6667	0.6667 angle 0 deg	0.6667

Where	By convolution	By the formula
$n = 0$	0.6667	0.6667
$n = 1$	-0.6667	-0.6667
$n = 2$	0.6667	0.6667
$n = 5$	-0.6667	-0.6667

$y [n] = 0.6667 e^{j (π n)}$ .

The zero input response PIN 1/19

78 Bh4

Before the formulas

Set the input to zero and the system still does something, because it has memory. What comes out is the system running down whatever was stored in it, and its shape is decided entirely by the roots of the characteristic equation: real roots decay smoothly, a complex pair rings, a repeated root decays with a bend in it.

It matters because the full response is this part plus the part the input drives, and the paper separates them on purpose to see whether you know which is which.

Asked as determine the zero input response for a second order system. It is a four step procedure and nothing else.

Set $x [n] = 0$ and write the homogeneous equation: $y [n] + a_{1} y [n - 1] + a_{2} y [n - 2] = 0$ .
Assume $y [n] = λ$ ^n and substitute. Dividing through by $λ$ ^(n-2) gives the characteristic equation $λ$ ^2 + $a_{1} λ + a_{2} = 0$ .
Solve for the roots. Distinct real roots give $y [n] = C_{1} λ_{1}^{n} + C_{2} λ_{2}^{n}$ . A repeated root $λ$ gives $y [n]$ = ( $C_{1} + C_{2}$ n) $λ$ ^n. Complex conjugate roots give a damped sinusoid, $y [n] = r^{n}$ ( $C_{1}$ cos( $θ$ n) + $C_{2}$ sin( $θ$ n)).
Fit the constants to the given initial conditions $y [- 1]$ , $y [- 2]$ and state the answer.

Worked, 2078 Bhadra

y [n] - 3 y [n - 1] - 4 y [n - 2] = x [n]

. With

x [n] = 0

the characteristic equation is

λ

^2 - 3

λ

- 4 = 0, so (

λ

- 4)(

λ

+ 1) = 0 and

λ = 4

, -1.
The zero input response is $y [n] = C_{1} {(4)}^{n} + C_{2} {(- 1)}^{n}$ , with

C_{1}

and

C_{2}

fixed by the two initial conditions.
Read the stability off it: the root at 4 lies outside the unit circle, so the natural response grows without bound and the system is unstable.

The roots of the characteristic equation are the poles of $H (z)$ , which is why a root outside the unit circle and an unstable filter are the same statement.

See it drawnThe response with nothing going in

Set the input to zero and what is left is the system running on its stored energy. The roots of the characteristic equation decide its shape.

Practise thisEvery question the paper has set on this topic, worked in full1

Practise this2078 Bhadra · Q44 marksDetermine the zero-input response for a second order system given by:

y [n] - 3 y [n - 1] - 4 y [n - 2] = x [n]

What is different in this one

An order 2 system, so the characteristic equation is a quadratic and there are 2 constants to fit to the initial conditions.
The roots are real and distinct, so the response is a plain sum of two exponentials.

1. The characteristic equation. Set the input to zero and assume $y [n] = λ^{n}$ , which is the one shape that survives a constant coefficient recursion. Divide through by $λ^{n - N}$ .

\begin{aligned} with x [n] & = 0 : 1 - 3 λ^- 1 - 4 λ^- 2 = 0 \\ multiply  through  by λ^{2} : \\ λ^{2} - 3 λ - 4 & = 0 \end{aligned}

2. Solve it. Factorise, or use the quadratic formula.

λ

= ( 3 +/-

\sqrt{< m > {(- 3)}^{2}}

- 4(-4) } ) / 2 = ( 3 +/-

\sqrt{25}

) / 2 = ( 3 +/- 5 ) / 2

λ = 4

, -1

3. The form of the response. One term per root, each with its own constant.

\begin{aligned} y [n] & = C_{1} {(4)}^{n} + C_{2} {(- 1)}^{n} \end{aligned}

4. Fix the constants. Write the response at the 2 instants the initial conditions give and solve the pair. This paper gives none, so the answer is the general form above. The arithmetic below shows what to do the moment they are given, on the usual pair $y [- 1] = 1$ , $y [- 2] = 0$ .

\begin{aligned} y [- 2] & = {(4)}^{- 2} C_{1} + {(- 1)}^{- 2} C_{2} \\ = 0.0625 C_{1} + 1 C_{2} \\ y [- 1] & = {(4)}^{- 1} C_{1} + {(- 1)}^{- 1} C_{2} \\ = 0.25 C_{1} - 1 C_{2} \\ with y [- 2] & = 0, y [- 1] = 1 \end{aligned}

\begin{aligned} solving : C_{1} & = 3.2, C_{2} = - 0.2 \end{aligned}

$y [n] = C_{1} {(4)}^{n} + C_{2} {(- 1)}^{n}$ , and the two constants are whatever the initial conditions make them. This paper states none, so the general form IS the answer: say so, and show the pair of equations you would solve. On the illustration above they come out $C_{1} = 3.2$ and $C_{2} = - 0.2$ . The root at 4 lies outside the unit circle, so the natural response grows without bound and the system is unstable.

The same questions grouped by how often each topic is set: the zero input response

3.3Relationship of the frequency response to the poles and zeros

Plotting poles and zeros and sketching the magnitude response TOP 17/19

82 Bh · 82 Ba · 81 Bh · 81 Ba · 80 Bh · 80 Ba · 79 Bh · 79 Ba · 78 Bh · 76 Ch · 76 Ash · 75 Ch · 75 Ash · 74 Ch · 74 Ash · 72 Ch · 72 Ka3+72+82+4

Before the formulas

Here is the payoff of the whole chapter, and the most asked question in it. The gain at a frequency is the product of the distances from that point on the circle to the zeros, divided by the product of the distances to the poles.

That sentence means you can sketch the response by eye, with no computing at all: a pole close to the circle pushes the response up where it sits, a zero on the circle pulls it down to nothing. So this card is drawing, not algebra, and the marks are for a sketch with the right shape at the right frequencies, labelled.

Seventeen of nineteen papers. Always ten marks, or close to it. The whole answer is a fixed six step procedure, and the sketch is read off the geometry rather than computed.

1. Get $H (z)$ from the difference equation. Take the z-transform of both sides, using $x [n - k]$ mapping to $z^{- k} X (z)$ .

\begin{aligned} y [n] + a_{1} y [n - 1] + a_{2} y [n - 2] & = b_{0} x [n] + b_{1} x [n - 1] \\ H (z) = \frac{b_{0} + b_{1} z^{- 1}}{1 + a_{1} z^{- 1} + a_{2} z^{- 2}} & = \frac{z (b_{0} z + b_{1})}{z^{2} + a_{1} z + a_{2}} \end{aligned}

2. Multiply top and bottom by $z^{N}$ so both are polynomials in z, never in $z^{- 1}$ . Zeros at $z = 0$ and poles at $z = 0$ are easy to lose otherwise.

3. Factor. Roots of the numerator are zeros, drawn o; roots of the denominator are poles, drawn x.

The four rules that give you the sketch without arithmetic Near a pole the response peaks. The closer the pole is to the unit circle, the taller and narrower the peak, and the peak sits at the angle of the pole.
Near a zero the response dips. A zero on the unit circle forces the response to exactly zero at that angle.
At $ω = 0$ the point is

z = 1

; at $ω = π$ it is

z = - 1

. Compute the two numbers H(1) and H(-1) exactly, since they anchor the sketch and are cheap.
Poles and zeros far from the unit circle barely matter. A pole at

| z | = 2

contributes a slow, gentle trend, not a feature.

Reading the filter type off the picture

Where the poles sit	Where the zeros sit	The filter is
Near z = +1, angle 0	Near $z = - 1$	Low pass
Near $z = - 1$ , angle $π$	Near z = +1	High pass
At angle $ω_{0}$ on both sides	At z = +1 and $z = - 1$	Band pass centred on $ω_{0}$
Just inside a zero at angle $ω_{0}$	On the circle at angle $ω_{0}$	Band stop or notch at $ω_{0}$
Anywhere, with each zero the reciprocal conjugate of a pole		All pass: flat magnitude

Worked, the pattern the paper repeats Poles at 0.45 +/- j1.6, zeros at 0.58 +/- j2.06. Both pairs are complex conjugates.

| p o l e |

= sqrt(0.

45^{2}

+ 1.

6^{2}

) = 1.66 and

| z e r o |

= sqrt(0.

58^{2}

+ 2.

06^{2}

) = 2.14, so both lie outside the unit circle, the poles nearer to it than the zeros.
Pole angle = arctan(1.6/0.45) = 74.3 degrees; zero angle = arctan(2.06/0.58) = 74.3 degrees as well, so the pole and the zero sit on the same ray.
Because the poles are outside the unit circle, the system is unstable, and the magnitude response has a broad rise near

ω = 74.3

degrees, which is about 1.3 radians, rather than a sharp peak. Say that: the marks are for reading the geometry, and for noticing the instability.

Worked, from a difference equation $y [n] - 0.4 y [n - 1] + 0.25 y [n - 2] = x [n] - 0.4 x [n - 1]$ .

H (z)

= (1 - 0.4

z^{- 1}) / (1 - 0.4 z^{- 1} + 0.25 z^{- 2}) = z (z - 0.4) / (z^{2}

- 0.4 z + 0.25).
Zeros:

z = 0

and

z = 0

.4. Poles:

z = (0.4 \pm \sqrt{0.16 - 1}) / 2 = 0.2 \pm j 0.458

, so

| p | = 0.5

and angle = 66.4 degrees.
Both poles are inside the unit circle, so the system is stable. The response peaks near

ω = 66.4

degrees (1.16 rad), and has a dip near

ω = 0

from the zero at 0.4.
Anchors:

H (1) = (1 - 0.4) / (1 - 0.4 + 0.25) = 0.6 / 0.85 = 0.706

H (- 1) = (1 + 0.4) / (1 + 0.4 + 0.25) = 1.4 / 1.65 = 0.848

In the exam

Draw the unit circle and label both axes. A pole zero plot without the unit circle cannot be marked, because every conclusion depends on which side of it things are.
Mark the pole angle on the frequency axis of the magnitude sketch. That correspondence is the thing being examined.
Compute H(1) and H(-1). Two numbers, two marks, and they stop the sketch floating.
Say whether it is stable even when the question does not ask. It costs one sentence and it is often a mark.
Sketch only 0 to $π$ and note that the rest is the mirror image.

The poles you plot here are the same numbers that the lattice reflection coefficients test without being found, and the same ones that a Butterworth design places deliberately.

See it drawnA pole pulls the response up, a zero pulls it down

The magnitude at a frequency is the product of the distances to the zeros over the distances to the poles. Near a pole the denominator is small.

Practise thisEvery question the paper has set on this topic, worked in full17

Practise this +4 like it2082 Bhadra · Q42+8 marksDraw the poles and zeros in the z-plane for the system with poles at 0.45 +/- j1.6 and zeros at 0.58 +/- j2.06. Also plot the magnitude response (not in scale) of the system.

The poles and zeros are given, so build $H (z)$ from them: the numerator is the product of (z - zero) and the denominator the product of (z - pole). Conjugate pairs multiply out to real quadratics, which is why the coefficients come out real.

What is different in this one

A pole lies on or outside the unit circle, at magnitude 1.6621, so this system is unstable. Say so: it is a mark, and it also tells you the magnitude response will be dominated by that pole.
The poles are a conjugate pair at an angle of 1.297 rad, so the response has a peak near that frequency and the sketch is symmetric about $ω = 0$ .

Take the z-transform of the difference equation, form $H (z)$ , then multiply above and below by $z^{2}$ so both are positive powers of z. The roots of the numerator are the zeros and the roots of the denominator are the poles.

\begin{aligned} H (z) & = \frac{1 - 1.16 z^{- 1} + 4.58 z^{- 2}}{1 - 0.9 z^{- 1} + 2.7625 z^{- 2}} & from the difference equation \\ = \frac{z^{2} - 1.16 z + 4.58}{z^{2} - 0.9 z + 2.7625} & \times z^{2} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	0.58 + j2.06	2.1401	74.28
zero, o	0.58 - j2.06	2.1401	-74.28
pole, x	0.45 + j1.6	1.6621	74.29
pole, x	0.45 - j1.6	1.6621	-74.29

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 1.6621, so for a causal system the ROC is $| z | > 1.6621$ , which does not contain the unit circle, so the system is unstable.

Now put z = $e^{j ω}$ , which walks the point round the unit circle, and work out the numerator and the denominator separately at nine frequencies across a full turn. The sketch is read off the last column.

\begin{aligned} N (e^{j ω}) & = e^{j 2 ω} - 1.16 e^ jw + 4.58 \\ = \cos (2 ω) + jsin (2 ω) - 1.16 cos (ω) - j 1.16 sin (ω) + 4.58 \\ D (e^{j ω}) & = e^{j 2 ω} - 0.9 e^ jw + 2.7625 \\ = \cos (2 ω) + jsin (2 ω) - 0.9 cos (ω) - j 0.9 sin (ω) + 2.7625 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	4.42	2.8625	1.5441	1.5441
$\frac{π}{4}$	3.7598 + j0.1798	2.1261 + j0.3636	1.7322 - j0.2117	1.7451
$\frac{π}{2}$	3.58 - j1.16	1.7625 - j0.9	1.8777 + j0.3007	1.9016
$\frac{3 π}{4}$	5.4002 - j1.8202	3.3989 - j1.6364	1.4992 + j0.1862	1.5107
$π$	6.74	4.6625	1.4456	1.4456
$\frac{5 π}{4}$	5.4002 + j1.8202	3.3989 + j1.6364	1.4992 - j0.1862	1.5107
$\frac{3 π}{2}$	3.58 + j1.16	1.7625 + j0.9	1.8777 - j0.3007	1.9016
$\frac{7 π}{4}$	3.7598 - j0.1798	2.1261 - j0.3636	1.7322 + j0.2117	1.7451
$2 π$	4.42	2.8625	1.5441	1.5441

The geometry says the same thing without arithmetic, and it is worth carrying as a check: $| H |$ at a point on the unit circle is the product of the distances to the zeros divided by the product of the distances to the poles. Near a pole the response peaks, near a zero it dips, and anything at the origin changes only the phase.

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	1.5441
$ω = 1.297$ , the angle of the poles	1.2966	2.0367
$ω = \frac{π}{2}$ , the point z = j	1.5708	1.9016
$ω = π$ , the point $z = - 1$	3.1416	1.4456

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response rises to a peak of 2.0368 near $ω = 1.291$ , with $| H |$ = 1.5441 at $ω = 0$ and 1.4456 at $ω = π$ . That makes it a low pass shape.

Practise this2082 Baishakh · Q43+7 marksThe poles of a system are located at 0.45 +/- j1.6 and the zeros at 0.58 +/- j2.06. Map the poles and zeros in the z-plane and plot the magnitude and phase response (not to scale) of the system.What is different hereboth curves are asked for, the magnitude AND the phase

What is different in this one

A pole lies on or outside the unit circle, at magnitude 1.6621, so this system is unstable. Say so: it is a mark, and it also tells you the magnitude response will be dominated by that pole.
The poles are a conjugate pair at an angle of 1.297 rad, so the response has a peak near that frequency and the sketch is symmetric about $ω = 0$ .

\begin{aligned} H (z) & = \frac{1 - 1.16 z^{- 1} + 4.58 z^{- 2}}{1 - 0.9 z^{- 1} + 2.7625 z^{- 2}} & from the difference equation \\ = \frac{z^{2} - 1.16 z + 4.58}{z^{2} - 0.9 z + 2.7625} & \times z^{2} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	0.58 + j2.06	2.1401	74.28
zero, o	0.58 - j2.06	2.1401	-74.28
pole, x	0.45 + j1.6	1.6621	74.29
pole, x	0.45 - j1.6	1.6621	-74.29

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 1.6621, so for a causal system the ROC is $| z | > 1.6621$ , which does not contain the unit circle, so the system is unstable.

\begin{aligned} N (e^{j ω}) & = e^{j 2 ω} - 1.16 e^ jw + 4.58 \\ = \cos (2 ω) + jsin (2 ω) - 1.16 cos (ω) - j 1.16 sin (ω) + 4.58 \\ D (e^{j ω}) & = e^{j 2 ω} - 0.9 e^ jw + 2.7625 \\ = \cos (2 ω) + jsin (2 ω) - 0.9 cos (ω) - j 0.9 sin (ω) + 2.7625 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	4.42	2.8625	1.5441	1.5441
$\frac{π}{4}$	3.7598 + j0.1798	2.1261 + j0.3636	1.7322 - j0.2117	1.7451
$\frac{π}{2}$	3.58 - j1.16	1.7625 - j0.9	1.8777 + j0.3007	1.9016
$\frac{3 π}{4}$	5.4002 - j1.8202	3.3989 - j1.6364	1.4992 + j0.1862	1.5107
$π$	6.74	4.6625	1.4456	1.4456
$\frac{5 π}{4}$	5.4002 + j1.8202	3.3989 + j1.6364	1.4992 - j0.1862	1.5107
$\frac{3 π}{2}$	3.58 + j1.16	1.7625 + j0.9	1.8777 - j0.3007	1.9016
$\frac{7 π}{4}$	3.7598 - j0.1798	2.1261 - j0.3636	1.7322 + j0.2117	1.7451
$2 π$	4.42	2.8625	1.5441	1.5441

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	1.5441
$ω = 1.297$ , the angle of the poles	1.2966	2.0367
$ω = \frac{π}{2}$ , the point z = j	1.5708	1.9016
$ω = π$ , the point $z = - 1$	3.1416	1.4456

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The phase. At each frequency it is the sum of the angles from that point on the circle to every zero, minus the sum of the angles to every pole. It is drawn unwrapped, because the $2 π$ jumps an arctangent produces are an artefact and not the filter.

arg

H (e^{j ω})

in radians, unwrapped, drawn from the computed response.

Where	$ω$	arg H, radians
$ω = 0$	0	0
$ω = \frac{π}{2}$	1.5708	0.159
$ω = π$	3.1416	0

The response rises to a peak of 2.0368 near $ω = 1.291$ , with $| H |$ = 1.5441 at $ω = 0$ and 1.4456 at $ω = π$ . That makes it a low pass shape. The phase runs from 0 rad at $ω = 0$ to 0 rad at $ω = π$ , rising steadily.

Practise this +5 like it2081 Bhadra · Q43+7 marksPlot the pole zero in z plane and draw magnitude response (not to the scale) of the system described by difference equation

y [n] - 0.4 y [n - 1] + 0.25 y [n - 2] = x [n] - 0.4 x [n - 1]

What is different in this one

There is a zero at the origin, which appears only after multiplying by $z^{2}$ . It changes the phase and not the magnitude, since its distance to every point of the unit circle is 1, and forgetting it is the usual lost mark.
The poles are a conjugate pair at an angle of 1.159 rad, so the response has a peak near that frequency and the sketch is symmetric about $ω = 0$ .

\begin{aligned} H (z) & = \frac{1 - 0.4 z^{- 1}}{1 - 0.4 z^{- 1} + 0.25 z^{- 2}} & from the difference equation \\ = \frac{z^{2} - 0.4 z}{z^{2} - 0.4 z + 0.25} & \times z^{2} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	0.4	0.4	0
zero, o	0	0	0
pole, x	0.2 + j0.4583	0.5	66.42
pole, x	0.2 - j0.4583	0.5	-66.42

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 0.5, so for a causal system the ROC is $| z | > 0.5$ , which contains the unit circle, so the system is stable.

\begin{aligned} N (e^{j ω}) & = e^{j 2 ω} - 0.4 e^ jw \\ = \cos (2 ω) + jsin (2 ω) - 0.4 cos (ω) - j 0.4 sin (ω) \\ D (e^{j ω}) & = e^{j 2 ω} - 0.4 e^ jw + 0.25 \\ = \cos (2 ω) + jsin (2 ω) - 0.4 cos (ω) - j 0.4 sin (ω) + 0.25 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	0.6	0.85	0.7059	0.7059
$\frac{π}{4}$	-0.2828 + j0.7172	-0.0328 + j0.7172	1.0159 + j0.3479	1.0738
$\frac{π}{2}$	-1 - j0.4	-0.75 - j0.4	1.2595 - j0.1384	1.2671
$\frac{3 π}{4}$	0.2828 - j1.2828	0.5328 - j1.2828	0.931 - j0.1662	0.9457
$π$	1.4	1.65	0.8485	0.8485
$\frac{5 π}{4}$	0.2828 + j1.2828	0.5328 + j1.2828	0.931 + j0.1662	0.9457
$\frac{3 π}{2}$	-1 + j0.4	-0.75 + j0.4	1.2595 + j0.1384	1.2671
$\frac{7 π}{4}$	-0.2828 - j0.7172	-0.0328 - j0.7172	1.0159 - j0.3479	1.0738
$2 π$	0.6	0.85	0.7059	0.7059

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	0.7059
$ω = 1.159$ , the angle of the poles	1.1593	1.3194
$ω = \frac{π}{2}$ , the point z = j	1.5708	1.2671
$ω = π$ , the point $z = - 1$	3.1416	0.8485

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response rises to a peak of 1.3405 near $ω = 1.291$ , with $| H |$ = 0.7059 at $ω = 0$ and 0.8485 at $ω = π$ . That makes it a high pass shape.

Practise this +1 like it2081 Baishakh · Q43+7 marksPlot the pole-zero in z plane and draw the magnitude response (not to the scale) of the system described by difference equation

y (n) = 0.67 x (n) - 0.3 x (n - 1) + 2.75 y (n - 1)

What is different herea first order denominator; the poles are real, so the response falls away without a peak

What is different in this one

A pole lies on or outside the unit circle, at magnitude 2.75, so this system is unstable. Say so: it is a mark, and it also tells you the magnitude response will be dominated by that pole.
Every pole is real, so the response has no resonant peak: it rises or falls monotonically between the two ends.

Take the z-transform of the difference equation, form $H (z)$ , then multiply above and below by $z^{1}$ so both are positive powers of z. The roots of the numerator are the zeros and the roots of the denominator are the poles.

\begin{aligned} H (z) & = \frac{0.67 - 0.3 z^{- 1}}{1 - 2.75 z^{- 1}} & from the difference equation \\ = \frac{0.67 z - 0.3}{z - 2.75} & \times z^{1} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	0.4478	0.4478	0
pole, x	2.75	2.75	0

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 2.75, so for a causal system the ROC is $| z | > 2.75$ , which does not contain the unit circle, so the system is unstable.

\begin{aligned} N (e^{j ω}) & = 0.67 e^ jw - 0.3 \\ = 0.67 cos (ω) + j 0.67 sin (ω) - 0.3 \\ D (e^{j ω}) & = e^{j ω} - 2.75 \\ = \cos (ω) + jsin (ω) - 2.75 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	0.37	-1.75	-0.2114	0.2114
$\frac{π}{4}$	0.1738 + j0.4738	-2.0429 + j0.7071	-0.0043 - j0.2334	0.2334
$\frac{π}{2}$	-0.3 + j0.67	-2.75 + j1	0.1746 - j0.1801	0.2509
$\frac{3 π}{4}$	-0.7738 + j0.4738	-3.4571 + j0.7071	0.2417 - j0.0876	0.2571
$π$	-0.97	-3.75	0.2587	0.2587
$\frac{5 π}{4}$	-0.7738 - j0.4738	-3.4571 - j0.7071	0.2417 + j0.0876	0.2571
$\frac{3 π}{2}$	-0.3 - j0.67	-2.75 - j1	0.1746 + j0.1801	0.2509
$\frac{7 π}{4}$	0.1738 - j0.4738	-2.0429 - j0.7071	-0.0043 + j0.2334	0.2334
$2 π$	0.37	-1.75	-0.2114	0.2114

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	0.2114
$ω = \frac{π}{2}$ , the point z = j	1.5708	0.2509
$ω = π$ , the point $z = - 1$	3.1416	0.2587

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response rises monotonically, with $| H |$ = 0.2114 at $ω = 0$ and 0.2587 at $ω = π$ . That makes it a high pass shape.

Practise this2080 Bhadra · Q43+7 marksPlot the pole-zero on the z-plane and draw Magnitude response (not to the scale) of an LTI system described by the equation,

y (n) = x (n) + 0.8 x (n - 1) + 0.8 x (n - 2) + 0.49 y (n - 2)

.What is different herethe poles are real, so the response falls away without a peak

What is different in this one

Every pole is real, so the response has no resonant peak: it rises or falls monotonically between the two ends.

\begin{aligned} H (z) & = \frac{1 + 0.8 z^{- 1} + 0.8 z^{- 2}}{1 - 0.49 z^{- 2}} & from the difference equation \\ = \frac{z^{2} + 0.8 z + 0.8}{z^{2} - 0.49} & \times z^{2} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	-0.4 + j0.8	0.8944	116.57
zero, o	-0.4 - j0.8	0.8944	-116.57
pole, x	-0.7	0.7	180
pole, x	0.7	0.7	0

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 0.7, so for a causal system the ROC is $| z | > 0.7$ , which contains the unit circle, so the system is stable.

\begin{aligned} N (e^{j ω}) & = e^{j 2 ω} + 0.8 e^ jw + 0.8 \\ = \cos (2 ω) + jsin (2 ω) + 0.8 cos (ω) + j 0.8 sin (ω) + 0.8 \\ D (e^{j ω}) & = e^{j 2 ω} - 0.49 \\ = \cos (2 ω) + jsin (2 ω) - 0.49 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	2.6	0.51	5.098	5.098
$\frac{π}{4}$	1.3657 + j1.5657	-0.49 + j1	0.7229 - j1.7199	1.8657
$\frac{π}{2}$	-0.2 + j0.8	-1.49	0.1342 - j0.5369	0.5534
$\frac{3 π}{4}$	0.2343 - j0.4343	-0.49 - j1	0.2576 + j0.3606	0.4431
$π$	1	0.51	1.9608	1.9608
$\frac{5 π}{4}$	0.2343 + j0.4343	-0.49 + j1	0.2576 - j0.3606	0.4431
$\frac{3 π}{2}$	-0.2 - j0.8	-1.49	0.1342 + j0.5369	0.5534
$\frac{7 π}{4}$	1.3657 - j1.5657	-0.49 - j1	0.7229 + j1.7199	1.8657
$2 π$	2.6	0.51	5.098	5.098

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	5.098
$ω = \frac{π}{2}$ , the point z = j	1.5708	0.5534
$ω = π$ , the point $z = - 1$	3.1416	1.9608

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response falls monotonically, with $| H |$ = 5.098 at $ω = 0$ and 1.9608 at $ω = π$ . That makes it a low pass shape.

Same as 2082 Bhadra Q42080 Baishakh · Q44+6 marksDifferentiate between FIR system and IIR System. The poles of a system are located at 0.45 +/- j1.6 and zeros at 0.58 +/- j2.06. Map the poles and zeros in the z-plane and plot the magnitude response (not in scale) of the system.

What is different in this one

A pole lies on or outside the unit circle, at magnitude 1.6621, so this system is unstable. Say so: it is a mark, and it also tells you the magnitude response will be dominated by that pole.
The poles are a conjugate pair at an angle of 1.297 rad, so the response has a peak near that frequency and the sketch is symmetric about $ω = 0$ .

\begin{aligned} H (z) & = \frac{1 - 1.16 z^{- 1} + 4.58 z^{- 2}}{1 - 0.9 z^{- 1} + 2.7625 z^{- 2}} & from the difference equation \\ = \frac{z^{2} - 1.16 z + 4.58}{z^{2} - 0.9 z + 2.7625} & \times z^{2} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	0.58 + j2.06	2.1401	74.28
zero, o	0.58 - j2.06	2.1401	-74.28
pole, x	0.45 + j1.6	1.6621	74.29
pole, x	0.45 - j1.6	1.6621	-74.29

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 1.6621, so for a causal system the ROC is $| z | > 1.6621$ , which does not contain the unit circle, so the system is unstable.

\begin{aligned} N (e^{j ω}) & = e^{j 2 ω} - 1.16 e^ jw + 4.58 \\ = \cos (2 ω) + jsin (2 ω) - 1.16 cos (ω) - j 1.16 sin (ω) + 4.58 \\ D (e^{j ω}) & = e^{j 2 ω} - 0.9 e^ jw + 2.7625 \\ = \cos (2 ω) + jsin (2 ω) - 0.9 cos (ω) - j 0.9 sin (ω) + 2.7625 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	4.42	2.8625	1.5441	1.5441
$\frac{π}{4}$	3.7598 + j0.1798	2.1261 + j0.3636	1.7322 - j0.2117	1.7451
$\frac{π}{2}$	3.58 - j1.16	1.7625 - j0.9	1.8777 + j0.3007	1.9016
$\frac{3 π}{4}$	5.4002 - j1.8202	3.3989 - j1.6364	1.4992 + j0.1862	1.5107
$π$	6.74	4.6625	1.4456	1.4456
$\frac{5 π}{4}$	5.4002 + j1.8202	3.3989 + j1.6364	1.4992 - j0.1862	1.5107
$\frac{3 π}{2}$	3.58 + j1.16	1.7625 + j0.9	1.8777 - j0.3007	1.9016
$\frac{7 π}{4}$	3.7598 - j0.1798	2.1261 - j0.3636	1.7322 + j0.2117	1.7451
$2 π$	4.42	2.8625	1.5441	1.5441

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	1.5441
$ω = 1.297$ , the angle of the poles	1.2966	2.0367
$ω = \frac{π}{2}$ , the point z = j	1.5708	1.9016
$ω = π$ , the point $z = - 1$	3.1416	1.4456

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response rises to a peak of 2.0368 near $ω = 1.291$ , with $| H |$ = 1.5441 at $ω = 0$ and 1.4456 at $ω = π$ . That makes it a low pass shape.

Practise this2079 Bhadra · Q43+7 marksPlot the pole-zero in z-plane and draw the magnitude response (not to the scale) of the equation of the system describe by difference equation:

y [n]

- 0.

35 y [n - 1]

+ 0.

25 y [n - 2] = x [n]

- 0.

75 x [n - 1]

.What is different hereThe standard text works this one as its Example 3.3 and slips in the

ω = \frac{π}{4}

row: it prints

D (e^{j ω}) = - 0.025

+ j0.75, but 0.25 - 0.2475 is <

b > 0.0025 < / b

>, not -0.025, so

D = 0.0025

+ j0.7525 and

H = 0.6218

+ j0.7068. The same slip is repeated at

\frac{7 π}{4}

. The magnitude survives it almost unchanged, 0.9414 against their 0.943, which is exactly why it is worth checking the real and imaginary parts and not only the modulus.

What is different in this one

There is a zero at the origin, which appears only after multiplying by $z^{2}$ . It changes the phase and not the magnitude, since its distance to every point of the unit circle is 1, and forgetting it is the usual lost mark.
The poles are a conjugate pair at an angle of 1.213 rad, so the response has a peak near that frequency and the sketch is symmetric about $ω = 0$ .

\begin{aligned} H (z) & = \frac{1 - 0.75 z^{- 1}}{1 - 0.35 z^{- 1} + 0.25 z^{- 2}} & from the difference equation \\ = \frac{z^{2} - 0.75 z}{z^{2} - 0.35 z + 0.25} & \times z^{2} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	0.75	0.75	0
zero, o	0	0	0
pole, x	0.175 + j0.4684	0.5	69.51
pole, x	0.175 - j0.4684	0.5	-69.51

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 0.5, so for a causal system the ROC is $| z | > 0.5$ , which contains the unit circle, so the system is stable.

\begin{aligned} N (e^{j ω}) & = e^{j 2 ω} - 0.75 e^ jw \\ = \cos (2 ω) + jsin (2 ω) - 0.75 cos (ω) - j 0.75 sin (ω) \\ D (e^{j ω}) & = e^{j 2 ω} - 0.35 e^ jw + 0.25 \\ = \cos (2 ω) + jsin (2 ω) - 0.35 cos (ω) - j 0.35 sin (ω) + 0.25 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	0.25	0.9	0.2778	0.2778
$\frac{π}{4}$	-0.5303 + j0.4697	0.0025 + j0.7525	0.6218 + j0.7068	0.9414
$\frac{π}{2}$	-1 - j0.75	-0.75 - j0.35	1.4781 + j0.3102	1.5103
$\frac{3 π}{4}$	0.5303 - j1.5303	0.4975 - j1.2475	1.2047 - j0.0553	1.2059
$π$	1.75	1.6	1.0938	1.0938
$\frac{5 π}{4}$	0.5303 + j1.5303	0.4975 + j1.2475	1.2047 + j0.0553	1.2059
$\frac{3 π}{2}$	-1 + j0.75	-0.75 + j0.35	1.4781 - j0.3102	1.5103
$\frac{7 π}{4}$	-0.5303 - j0.4697	0.0025 - j0.7525	0.6218 - j0.7068	0.9414
$2 π$	0.25	0.9	0.2778	0.2778

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	0.2778
$ω = 1.213$ , the angle of the poles	1.2132	1.4387
$ω = \frac{π}{2}$ , the point z = j	1.5708	1.5103
$ω = π$ , the point $z = - 1$	3.1416	1.0938

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response rises to a peak of 1.5251 near $ω = 1.456$ , with $| H |$ = 0.2778 at $ω = 0$ and 1.0938 at $ω = π$ . That makes it a high pass shape.

Same as 2081 Bhadra Q42079 Baishakh · Q43+7 marksPlot pole-zero in z-plane and draw magnitude response (not to the scale) of the system described by difference equation

y [n] - 0.3 y [n - 1]

+ 0.

2 y [n - 2] = x [n]

- 0.

5 x [n - 1]

What is different in this one

There is a zero at the origin, which appears only after multiplying by $z^{2}$ . It changes the phase and not the magnitude, since its distance to every point of the unit circle is 1, and forgetting it is the usual lost mark.
The poles are a conjugate pair at an angle of 1.229 rad, so the response has a peak near that frequency and the sketch is symmetric about $ω = 0$ .

\begin{aligned} H (z) & = \frac{1 - 0.5 z^{- 1}}{1 - 0.3 z^{- 1} + 0.2 z^{- 2}} & from the difference equation \\ = \frac{z^{2} - 0.5 z}{z^{2} - 0.3 z + 0.2} & \times z^{2} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	0.5	0.5	0
zero, o	0	0	0
pole, x	0.15 + j0.4213	0.4472	70.4
pole, x	0.15 - j0.4213	0.4472	-70.4

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 0.4472, so for a causal system the ROC is $| z | > 0.4472$ , which contains the unit circle, so the system is stable.

\begin{aligned} N (e^{j ω}) & = e^{j 2 ω} - 0.5 e^ jw \\ = \cos (2 ω) + jsin (2 ω) - 0.5 cos (ω) - j 0.5 sin (ω) \\ D (e^{j ω}) & = e^{j 2 ω} - 0.3 e^ jw + 0.2 \\ = \cos (2 ω) + jsin (2 ω) - 0.3 cos (ω) - j 0.3 sin (ω) + 0.2 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	0.5	0.9	0.5556	0.5556
$\frac{π}{4}$	-0.3536 + j0.6464	-0.0121 + j0.7879	0.8272 + j0.436	0.9351
$\frac{π}{2}$	-1 - j0.5	-0.8 - j0.3	1.3014 + j0.137	1.3086
$\frac{3 π}{4}$	0.3536 - j1.3536	0.4121 - j1.2121	1.0899 - j0.0789	1.0927
$π$	1.5	1.5	1	1
$\frac{5 π}{4}$	0.3536 + j1.3536	0.4121 + j1.2121	1.0899 + j0.0789	1.0927
$\frac{3 π}{2}$	-1 + j0.5	-0.8 + j0.3	1.3014 - j0.137	1.3086
$\frac{7 π}{4}$	-0.3536 - j0.6464	-0.0121 - j0.7879	0.8272 - j0.436	0.9351
$2 π$	0.5	0.9	0.5556	0.5556

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	0.5556
$ω = 1.229$ , the angle of the poles	1.2288	1.2574
$ω = \frac{π}{2}$ , the point z = j	1.5708	1.3086
$ω = π$ , the point $z = - 1$	3.1416	1

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response rises to a peak of 1.3142 near $ω = 1.482$ , with $| H |$ = 0.5556 at $ω = 0$ and 1 at $ω = π$ . That makes it a high pass shape.

Same as 2081 Bhadra Q42078 Bhadra · Q52+4 marksPlot the pole-zero in z-plane and draw magnitude response (not to the scale) of the system described by difference equation.

y [n] - 0.4 y [n - 1] + 0.25 y [n - 2] = x [n]

- 0.

4 x [n - 1]

What is different in this one

There is a zero at the origin, which appears only after multiplying by $z^{2}$ . It changes the phase and not the magnitude, since its distance to every point of the unit circle is 1, and forgetting it is the usual lost mark.
The poles are a conjugate pair at an angle of 1.159 rad, so the response has a peak near that frequency and the sketch is symmetric about $ω = 0$ .

\begin{aligned} H (z) & = \frac{1 - 0.4 z^{- 1}}{1 - 0.4 z^{- 1} + 0.25 z^{- 2}} & from the difference equation \\ = \frac{z^{2} - 0.4 z}{z^{2} - 0.4 z + 0.25} & \times z^{2} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	0.4	0.4	0
zero, o	0	0	0
pole, x	0.2 + j0.4583	0.5	66.42
pole, x	0.2 - j0.4583	0.5	-66.42

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 0.5, so for a causal system the ROC is $| z | > 0.5$ , which contains the unit circle, so the system is stable.

\begin{aligned} N (e^{j ω}) & = e^{j 2 ω} - 0.4 e^ jw \\ = \cos (2 ω) + jsin (2 ω) - 0.4 cos (ω) - j 0.4 sin (ω) \\ D (e^{j ω}) & = e^{j 2 ω} - 0.4 e^ jw + 0.25 \\ = \cos (2 ω) + jsin (2 ω) - 0.4 cos (ω) - j 0.4 sin (ω) + 0.25 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	0.6	0.85	0.7059	0.7059
$\frac{π}{4}$	-0.2828 + j0.7172	-0.0328 + j0.7172	1.0159 + j0.3479	1.0738
$\frac{π}{2}$	-1 - j0.4	-0.75 - j0.4	1.2595 - j0.1384	1.2671
$\frac{3 π}{4}$	0.2828 - j1.2828	0.5328 - j1.2828	0.931 - j0.1662	0.9457
$π$	1.4	1.65	0.8485	0.8485
$\frac{5 π}{4}$	0.2828 + j1.2828	0.5328 + j1.2828	0.931 + j0.1662	0.9457
$\frac{3 π}{2}$	-1 + j0.4	-0.75 + j0.4	1.2595 + j0.1384	1.2671
$\frac{7 π}{4}$	-0.2828 - j0.7172	-0.0328 - j0.7172	1.0159 - j0.3479	1.0738
$2 π$	0.6	0.85	0.7059	0.7059

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	0.7059
$ω = 1.159$ , the angle of the poles	1.1593	1.3194
$ω = \frac{π}{2}$ , the point z = j	1.5708	1.2671
$ω = π$ , the point $z = - 1$	3.1416	0.8485

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response rises to a peak of 1.3405 near $ω = 1.291$ , with $| H |$ = 0.7059 at $ω = 0$ and 0.8485 at $ω = π$ . That makes it a high pass shape.

Same as 2082 Bhadra Q42076 Chaitra · Q52+6 marksDraw the pole-zero in the z-plane for a system with poles at 0.45 +/- j1.06 and zeroes at 0.58 +/- j2.06. Also plot the magnitude response (not to the scale) of the system.

What is different in this one

A pole lies on or outside the unit circle, at magnitude 1.1516, so this system is unstable. Say so: it is a mark, and it also tells you the magnitude response will be dominated by that pole.
The poles are a conjugate pair at an angle of 1.169 rad, so the response has a peak near that frequency and the sketch is symmetric about $ω = 0$ .

\begin{aligned} H (z) & = \frac{1 - 1.16 z^{- 1} + 4.58 z^{- 2}}{1 - 0.9 z^{- 1} + 1.3261 z^{- 2}} & from the difference equation \\ = \frac{z^{2} - 1.16 z + 4.58}{z^{2} - 0.9 z + 1.3261} & \times z^{2} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	0.58 + j2.06	2.1401	74.28
zero, o	0.58 - j2.06	2.1401	-74.28
pole, x	0.45 + j1.06	1.1516	67
pole, x	0.45 - j1.06	1.1516	-67

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 1.1516, so for a causal system the ROC is $| z | > 1.1516$ , which does not contain the unit circle, so the system is unstable.

\begin{aligned} N (e^{j ω}) & = e^{j 2 ω} - 1.16 e^ jw + 4.58 \\ = \cos (2 ω) + jsin (2 ω) - 1.16 cos (ω) - j 1.16 sin (ω) + 4.58 \\ D (e^{j ω}) & = e^{j 2 ω} - 0.9 e^ jw + 1.3261 \\ = \cos (2 ω) + jsin (2 ω) - 0.9 cos (ω) - j 0.9 sin (ω) + 1.3261 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	4.42	1.4261	3.0994	3.0994
$\frac{π}{4}$	3.7598 + j0.1798	0.6897 + j0.3636	4.3732 - j2.0449	4.8277
$\frac{π}{2}$	3.58 - j1.16	0.3261 - j0.9	2.4133 + j3.1033	3.9313
$\frac{3 π}{4}$	5.4002 - j1.8202	1.9625 - j1.6364	2.0794 + j0.8063	2.2302
$π$	6.74	3.2261	2.0892	2.0892
$\frac{5 π}{4}$	5.4002 + j1.8202	1.9625 + j1.6364	2.0794 - j0.8063	2.2302
$\frac{3 π}{2}$	3.58 + j1.16	0.3261 + j0.9	2.4133 - j3.1033	3.9313
$\frac{7 π}{4}$	3.7598 - j0.1798	0.6897 - j0.3636	4.3732 + j2.0449	4.8277
$2 π$	4.42	1.4261	3.0994	3.0994

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	3.0994
$ω = 1.169$ , the angle of the poles	1.1693	11.4875
$ω = \frac{π}{2}$ , the point z = j	1.5708	3.9313
$ω = π$ , the point $z = - 1$	3.1416	2.0892

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response rises to a peak of 11.4954 near $ω = 1.165$ , with $| H |$ = 3.0994 at $ω = 0$ and 2.0892 at $ω = π$ . That makes it a low pass shape.

Practise this2076 Ashwin · Q62+8 marksFor the system described by the following difference equation:

y [n] = 0

67 x [n]

- 0.

3 x [n - 1]

+ 2.

75 y [n - 1]

. Map the poles and zero in the z-plane and plot the phase response of the system.What is different herea first order denominator; the poles are real, so the response falls away without a peak; the PHASE response is what this one asks for, not the magnitude: the angles to the zeros minus the angles to the poles

What is different in this one

A pole lies on or outside the unit circle, at magnitude 2.75, so this system is unstable. Say so: it is a mark, and it also tells you the magnitude response will be dominated by that pole.
Every pole is real, so the response has no resonant peak: it rises or falls monotonically between the two ends.

\begin{aligned} H (z) & = \frac{0.67 - 0.3 z^{- 1}}{1 - 2.75 z^{- 1}} & from the difference equation \\ = \frac{0.67 z - 0.3}{z - 2.75} & \times z^{1} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	0.4478	0.4478	0
pole, x	2.75	2.75	0

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 2.75, so for a causal system the ROC is $| z | > 2.75$ , which does not contain the unit circle, so the system is unstable.

\begin{aligned} N (e^{j ω}) & = 0.67 e^ jw - 0.3 \\ = 0.67 cos (ω) + j 0.67 sin (ω) - 0.3 \\ D (e^{j ω}) & = e^{j ω} - 2.75 \\ = \cos (ω) + jsin (ω) - 2.75 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	0.37	-1.75	-0.2114	0.2114
$\frac{π}{4}$	0.1738 + j0.4738	-2.0429 + j0.7071	-0.0043 - j0.2334	0.2334
$\frac{π}{2}$	-0.3 + j0.67	-2.75 + j1	0.1746 - j0.1801	0.2509
$\frac{3 π}{4}$	-0.7738 + j0.4738	-3.4571 + j0.7071	0.2417 - j0.0876	0.2571
$π$	-0.97	-3.75	0.2587	0.2587
$\frac{5 π}{4}$	-0.7738 - j0.4738	-3.4571 - j0.7071	0.2417 + j0.0876	0.2571
$\frac{3 π}{2}$	-0.3 - j0.67	-2.75 - j1	0.1746 + j0.1801	0.2509
$\frac{7 π}{4}$	0.1738 - j0.4738	-2.0429 - j0.7071	-0.0043 + j0.2334	0.2334
$2 π$	0.37	-1.75	-0.2114	0.2114

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	0.2114
$ω = \frac{π}{2}$ , the point z = j	1.5708	0.2509
$ω = π$ , the point $z = - 1$	3.1416	0.2587

arg

H (e^{j ω})

in radians, unwrapped, drawn from the computed response.

Where	$ω$	arg H, radians
$ω = 0$	0	-3.142
$ω = \frac{π}{2}$	1.5708	-0.801
$ω = π$	3.1416	0

The magnitude is not asked here, only the pole zero map and the phase. The phase runs from -3.142 rad at $ω = 0$ to 0 rad at $ω = π$ , rising steadily.

Same as 2081 Baishakh Q42075 Chaitra · Q32+7 marksPlot the pole-zero in z-plane and draw magnitude response (not to scale) of the system described by differential equation

y (n) - 0.3 y (n - 1) = 2 x (n - 2) + 0.7 x (n - 1) + 4 x (n)

What is different in this one

Every pole is real, so the response has no resonant peak: it rises or falls monotonically between the two ends.

\begin{aligned} H (z) & = \frac{4 + 0.7 z^{- 1} + 2 z^{- 2}}{1 - 0.3 z^{- 1}} & from the difference equation \\ = \frac{4 z^{2} + 0.7 z + 2}{z^{2} - 0.3 z} & \times z^{2} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	-0.0875 + j0.7017	0.7071	97.11
zero, o	-0.0875 - j0.7017	0.7071	-97.11
pole, x	0.3	0.3	0
pole, x	0	0	0

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 0.3, so for a causal system the ROC is $| z | > 0.3$ , which contains the unit circle, so the system is stable.

\begin{aligned} N (e^{j ω}) & = 4 e^j 2 ω + 0.7 e^ jw + 2 \\ = 4 cos (2 ω) + j 4 sin (2 ω) + 0.7 cos (ω) + j 0.7 sin (ω) + 2 \\ D (e^{j ω}) & = e^{j 2 ω} - 0.3 e^ jw \\ = \cos (2 ω) + jsin (2 ω) - 0.3 cos (ω) - j 0.3 sin (ω) \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	6.7	0.7	9.5714	9.5714
$\frac{π}{4}$	2.495 + j4.495	-0.2121 + j0.7879	4.5246 - j4.385	6.3008
$\frac{π}{2}$	-2 + j0.7	-1 - j0.3	1.6422 - j1.1927	2.0296
$\frac{3 π}{4}$	1.505 - j3.505	0.2121 - j1.2121	3.0165 + j0.7137	3.0998
$π$	5.3	1.3	4.0769	4.0769
$\frac{5 π}{4}$	1.505 + j3.505	0.2121 + j1.2121	3.0165 - j0.7137	3.0998
$\frac{3 π}{2}$	-2 - j0.7	-1 + j0.3	1.6422 + j1.1927	2.0296
$\frac{7 π}{4}$	2.495 - j4.495	-0.2121 - j0.7879	4.5246 + j4.385	6.3008
$2 π$	6.7	0.7	9.5714	9.5714

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	9.5714
$ω = \frac{π}{2}$ , the point z = j	1.5708	2.0296
$ω = π$ , the point $z = - 1$	3.1416	4.0769

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response falls monotonically, with $| H |$ = 9.5714 at $ω = 0$ and 4.0769 at $ω = π$ . That makes it a low pass shape.

Same as 2082 Bhadra Q42075 Ashwin · Q42+8 marksDraw the poles and zeros in the z-plane for a system with poles at 0.45 +/- j1.06 and zeros at 0.58 +/- j2.06. Also plot the magnitude response of the system.

What is different in this one

A pole lies on or outside the unit circle, at magnitude 1.1516, so this system is unstable. Say so: it is a mark, and it also tells you the magnitude response will be dominated by that pole.
The poles are a conjugate pair at an angle of 1.169 rad, so the response has a peak near that frequency and the sketch is symmetric about $ω = 0$ .

\begin{aligned} H (z) & = \frac{1 - 1.16 z^{- 1} + 4.58 z^{- 2}}{1 - 0.9 z^{- 1} + 1.3261 z^{- 2}} & from the difference equation \\ = \frac{z^{2} - 1.16 z + 4.58}{z^{2} - 0.9 z + 1.3261} & \times z^{2} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	0.58 + j2.06	2.1401	74.28
zero, o	0.58 - j2.06	2.1401	-74.28
pole, x	0.45 + j1.06	1.1516	67
pole, x	0.45 - j1.06	1.1516	-67

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 1.1516, so for a causal system the ROC is $| z | > 1.1516$ , which does not contain the unit circle, so the system is unstable.

\begin{aligned} N (e^{j ω}) & = e^{j 2 ω} - 1.16 e^ jw + 4.58 \\ = \cos (2 ω) + jsin (2 ω) - 1.16 cos (ω) - j 1.16 sin (ω) + 4.58 \\ D (e^{j ω}) & = e^{j 2 ω} - 0.9 e^ jw + 1.3261 \\ = \cos (2 ω) + jsin (2 ω) - 0.9 cos (ω) - j 0.9 sin (ω) + 1.3261 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	4.42	1.4261	3.0994	3.0994
$\frac{π}{4}$	3.7598 + j0.1798	0.6897 + j0.3636	4.3732 - j2.0449	4.8277
$\frac{π}{2}$	3.58 - j1.16	0.3261 - j0.9	2.4133 + j3.1033	3.9313
$\frac{3 π}{4}$	5.4002 - j1.8202	1.9625 - j1.6364	2.0794 + j0.8063	2.2302
$π$	6.74	3.2261	2.0892	2.0892
$\frac{5 π}{4}$	5.4002 + j1.8202	1.9625 + j1.6364	2.0794 - j0.8063	2.2302
$\frac{3 π}{2}$	3.58 + j1.16	0.3261 + j0.9	2.4133 - j3.1033	3.9313
$\frac{7 π}{4}$	3.7598 - j0.1798	0.6897 - j0.3636	4.3732 + j2.0449	4.8277
$2 π$	4.42	1.4261	3.0994	3.0994

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	3.0994
$ω = 1.169$ , the angle of the poles	1.1693	11.4875
$ω = \frac{π}{2}$ , the point z = j	1.5708	3.9313
$ω = π$ , the point $z = - 1$	3.1416	2.0892

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response rises to a peak of 11.4954 near $ω = 1.165$ , with $| H |$ = 3.0994 at $ω = 0$ and 2.0892 at $ω = π$ . That makes it a low pass shape.

Same as 2082 Bhadra Q42074 Chaitra · Q42+8 marksThe poles of a system are located at: 0.45 - 0.77i and -2 +/- 0.3i. Map the poles and zero in the z-plane and plot the magnitude response of the system.

How this is readThe paper prints one pole of the first pair only. Real coefficients force poles into conjugate pairs, so read it as 0.45 +/- j0.77 together with -2 +/- j0.3. No zeros are given, so $H (z)$ is all pole.

What is different in this one

A pole lies on or outside the unit circle, at magnitude 2.0224, so this system is unstable. Say so: it is a mark, and it also tells you the magnitude response will be dominated by that pole.
There is a zero at the origin, which appears only after multiplying by $z^{4}$ . It changes the phase and not the magnitude, since its distance to every point of the unit circle is 1, and forgetting it is the usual lost mark.
The poles are a conjugate pair at an angle of 1.042 rad, so the response has a peak near that frequency and the sketch is symmetric about $ω = 0$ .

Take the z-transform of the difference equation, form $H (z)$ , then multiply above and below by $z^{4}$ so both are positive powers of z. The roots of the numerator are the zeros and the roots of the denominator are the poles.

\begin{aligned} H (z) & = \frac{1}{1 + 3.1 z^{- 1} + 1.2854 z^{- 2} - 0.4994 z^{- 3} + 3.2532 z^{- 4}} & from the difference equation \\ = \frac{z^{4}}{z^{4} + 3.1 z^{3} + 1.2854 z^{2} - 0.4994 z + 3.2532} & \times z^{4} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	0	0	0
zero, o	0	0	0
zero, o	0	0	0
zero, o	0	0	0
pole, x	-2 + j0.3	2.0224	171.47
pole, x	-2 - j0.3	2.0224	-171.47
pole, x	0.45 + j0.77	0.8919	59.7
pole, x	0.45 - j0.77	0.8919	-59.7

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 2.0224, so for a causal system the ROC is $| z | > 2.0224$ , which does not contain the unit circle, so the system is unstable.

\begin{aligned} N (e^{j ω}) & = e^{j 4 ω} \\ = \cos (4 ω) + jsin (4 ω) \\ D (e^{j ω}) & = e^{j 4 ω} + 3.1 e^j 3 ω + 1.2854 e^j 2 ω - 0.4994 e^ jw + 3.2532 \\ = \cos (4 ω) + jsin (4 ω) + 3.1 cos (3 ω) + j 3.1 sin (3 ω) + 1.2854 cos (2 ω) + j 1.2854 sin (2 ω) - 0.4994 cos (ω) - j 0.4994 sin (ω) + 3.2532 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	1	8.1392	0.1229	0.1229
$\frac{π}{4}$	-1	-0.292 + j3.1243	0.0297 + j0.3173	0.3187
$\frac{π}{2}$	1	2.9678 - j3.5994	0.1364 + j0.1654	0.2144
$\frac{3 π}{4}$	-1	4.7983 + j0.5535	-0.2057 + j0.0237	0.207
$π$	1	2.938	0.3404	0.3404
$\frac{5 π}{4}$	-1	4.7983 - j0.5535	-0.2057 - j0.0237	0.207
$\frac{3 π}{2}$	1	2.9678 + j3.5994	0.1364 - j0.1654	0.2144
$\frac{7 π}{4}$	-1	-0.292 - j3.1243	0.0297 - j0.3173	0.3187
$2 π$	1	8.1392	0.1229	0.1229

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	0.1229
$ω = 1.042$ , the angle of the poles	1.0419	0.7981
$ω = \frac{π}{2}$ , the point z = j	1.5708	0.2144
$ω = π$ , the point $z = - 1$	3.1416	0.3404

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response rises to a peak of 0.7983 near $ω = 1.045$ , with $| H |$ = 0.1229 at $ω = 0$ and 0.3404 at $ω = π$ . That makes it a high pass shape.

Same as 2081 Bhadra Q42074 Ashwin · Q43+7 marksPlot the pole-zero in z-plane and Draw Magnitude Response (not to the scale) of the system described by difference equation

y [n]

- 0.

4 y [n - 1]

+ 0.

2 y [n - 2] = x [n]

+ 0.

5 x [n - 1]

+ 0.

6 x [n - 2]

+ 0.

8 x [n - 3]

What is different in this one

The poles are a conjugate pair at an angle of 1.107 rad, so the response has a peak near that frequency and the sketch is symmetric about $ω = 0$ .

Take the z-transform of the difference equation, form $H (z)$ , then multiply above and below by $z^{3}$ so both are positive powers of z. The roots of the numerator are the zeros and the roots of the denominator are the poles.

\begin{aligned} H (z) & = \frac{1 + 0.5 z^{- 1} + 0.6 z^{- 2} + 0.8 z^{- 3}}{1 - 0.4 z^{- 1} + 0.2 z^{- 2}} & from the difference equation \\ = \frac{z^{3} + 0.5 z^{2} + 0.6 z + 0.8}{z^{3} - 0.4 z^{2} + 0.2 z} & \times z^{3} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	0.1845 + j0.9416	0.9595	78.91
zero, o	0.1845 - j0.9416	0.9595	-78.91
zero, o	-0.869	0.869	180
pole, x	0.2 + j0.4	0.4472	63.43
pole, x	0.2 - j0.4	0.4472	-63.43
pole, x	0	0	0

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 0.4472, so for a causal system the ROC is $| z | > 0.4472$ , which contains the unit circle, so the system is stable.

\begin{aligned} N (e^{j ω}) & = e^{j 3 ω} + 0.5 e^j 2 ω + 0.6 e^ jw + 0.8 \\ = \cos (3 ω) + jsin (3 ω) + 0.5 cos (2 ω) + j 0.5 sin (2 ω) + 0.6 cos (ω) + j 0.6 sin (ω) + 0.8 \\ D (e^{j ω}) & = e^{j 3 ω} - 0.4 e^j 2 ω + 0.2 e^ jw \\ = \cos (3 ω) + jsin (3 ω) - 0.4 cos (2 ω) - j 0.4 sin (2 ω) + 0.2 cos (ω) + j 0.2 sin (ω) \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	2.9	0.8	3.625	3.625
$\frac{π}{4}$	0.5172 + j1.6314	-0.5657 + j0.4485	0.8426 - j2.2158	2.3706
$\frac{π}{2}$	0.3 - j0.4	0.4 - j0.8	0.55 + j0.1	0.559
$\frac{3 π}{4}$	1.0828 + j0.6314	0.5657 + j1.2485	0.7456 - j0.5295	0.9145
$π$	-0.3	-1.6	0.1875	0.1875
$\frac{5 π}{4}$	1.0828 - j0.6314	0.5657 - j1.2485	0.7456 + j0.5295	0.9145
$\frac{3 π}{2}$	0.3 + j0.4	0.4 + j0.8	0.55 - j0.1	0.559
$\frac{7 π}{4}$	0.5172 - j1.6314	-0.5657 - j0.4485	0.8426 + j2.2158	2.3706
$2 π$	2.9	0.8	3.625	3.625

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	3.625
$ω = 1.107$ , the angle of the poles	1.1071	1.0815
$ω = \frac{π}{2}$ , the point z = j	1.5708	0.559
$ω = π$ , the point $z = - 1$	3.1416	0.1875

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response falls monotonically, with $| H |$ = 3.625 at $ω = 0$ and 0.1875 at $ω = π$ . That makes it a low pass shape.

Same as 2081 Bhadra Q42072 Chaitra · Q52+4 marksPlot the pole-zero in z-plane and Draw Magnitude Response (not to the scale) of the system described by difference equation.

y [n]

- 0.

4 y [n - 1]

+ 0.

1 y [n - 2] = x [n]

+ 0.

6 x [n - 1]

What is different in this one

There is a zero at the origin, which appears only after multiplying by $z^{2}$ . It changes the phase and not the magnitude, since its distance to every point of the unit circle is 1, and forgetting it is the usual lost mark.
The poles are a conjugate pair at an angle of 0.886 rad, so the response has a peak near that frequency and the sketch is symmetric about $ω = 0$ .

\begin{aligned} H (z) & = \frac{1 + 0.6 z^{- 1}}{1 - 0.4 z^{- 1} + 0.1 z^{- 2}} & from the difference equation \\ = \frac{z^{2} + 0.6 z}{z^{2} - 0.4 z + 0.1} & \times z^{2} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	-0.6	0.6	180
zero, o	0	0	0
pole, x	0.2 + j0.2449	0.3162	50.77
pole, x	0.2 - j0.2449	0.3162	-50.77

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 0.3162, so for a causal system the ROC is $| z | > 0.3162$ , which contains the unit circle, so the system is stable.

\begin{aligned} N (e^{j ω}) & = e^{j 2 ω} + 0.6 e^ jw \\ = \cos (2 ω) + jsin (2 ω) + 0.6 cos (ω) + j 0.6 sin (ω) \\ D (e^{j ω}) & = e^{j 2 ω} - 0.4 e^ jw + 0.1 \\ = \cos (2 ω) + jsin (2 ω) - 0.4 cos (ω) - j 0.4 sin (ω) + 0.1 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	1.6	0.7	2.2857	2.2857
$\frac{π}{4}$	0.4243 + j1.4243	-0.1828 + j0.7172	1.7231 - j1.0309	2.008
$\frac{π}{2}$	-1 + j0.6	-0.9 - j0.4	0.6804 - j0.9691	1.1841
$\frac{3 π}{4}$	-0.4243 - j0.5757	0.3828 - j1.2828	0.3215 - j0.4267	0.5342
$π$	0.4	1.5	0.2667	0.2667
$\frac{5 π}{4}$	-0.4243 + j0.5757	0.3828 + j1.2828	0.3215 + j0.4267	0.5342
$\frac{3 π}{2}$	-1 - j0.6	-0.9 + j0.4	0.6804 + j0.9691	1.1841
$\frac{7 π}{4}$	0.4243 - j1.4243	-0.1828 - j0.7172	1.7231 + j1.0309	2.008
$2 π$	1.6	0.7	2.2857	2.2857

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	2.2857
$ω = 0.886$ , the angle of the poles	0.8861	1.9223
$ω = \frac{π}{2}$ , the point z = j	1.5708	1.1841
$ω = π$ , the point $z = - 1$	3.1416	0.2667

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response falls monotonically, with $| H |$ = 2.2857 at $ω = 0$ and 0.2667 at $ω = π$ . That makes it a low pass shape.

Same as 2081 Bhadra Q42072 Kartik · Q56 marksPlot Magnitude Response (not to the scale) of the system described by difference equation.

y [n]

- 0.

5 y [n - 1]

+ 0.

3 y [n - 2] = x [n]

+ 0.

7 x [n - 1]

What is different in this one

There is a zero at the origin, which appears only after multiplying by $z^{2}$ . It changes the phase and not the magnitude, since its distance to every point of the unit circle is 1, and forgetting it is the usual lost mark.
The poles are a conjugate pair at an angle of 1.097 rad, so the response has a peak near that frequency and the sketch is symmetric about $ω = 0$ .

\begin{aligned} H (z) & = \frac{1 + 0.7 z^{- 1}}{1 - 0.5 z^{- 1} + 0.3 z^{- 2}} & from the difference equation \\ = \frac{z^{2} + 0.7 z}{z^{2} - 0.5 z + 0.3} & \times z^{2} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	-0.7	0.7	180
zero, o	0	0	0
pole, x	0.25 + j0.4873	0.5477	62.84
pole, x	0.25 - j0.4873	0.5477	-62.84

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 0.5477, so for a causal system the ROC is $| z | > 0.5477$ , which contains the unit circle, so the system is stable.

\begin{aligned} N (e^{j ω}) & = e^{j 2 ω} + 0.7 e^ jw \\ = \cos (2 ω) + jsin (2 ω) + 0.7 cos (ω) + j 0.7 sin (ω) \\ D (e^{j ω}) & = e^{j 2 ω} - 0.5 e^ jw + 0.3 \\ = \cos (2 ω) + jsin (2 ω) - 0.5 cos (ω) - j 0.5 sin (ω) + 0.3 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	1.7	0.8	2.125	2.125
$\frac{π}{4}$	0.495 + j1.495	-0.0536 + j0.6464	2.2338 - j0.9507	2.4277
$\frac{π}{2}$	-1 + j0.7	-0.7 - j0.5	0.473 - j1.3378	1.419
$\frac{3 π}{4}$	-0.495 - j0.505	0.6536 - j1.3536	0.1594 - j0.4426	0.4705
$π$	0.3	1.8	0.1667	0.1667
$\frac{5 π}{4}$	-0.495 + j0.505	0.6536 + j1.3536	0.1594 + j0.4426	0.4705
$\frac{3 π}{2}$	-1 - j0.7	-0.7 + j0.5	0.473 + j1.3378	1.419
$\frac{7 π}{4}$	0.495 - j1.495	-0.0536 - j0.6464	2.2338 + j0.9507	2.4277
$2 π$	1.7	0.8	2.125	2.125

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	2.125
$ω = 1.097$ , the angle of the poles	1.0968	2.3168
$ω = \frac{π}{2}$ , the point z = j	1.5708	1.419
$ω = π$ , the point $z = - 1$	3.1416	0.1667

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response rises to a peak of 2.4438 near $ω = 0.877$ , with $| H |$ = 2.125 at $ω = 0$ and 0.1667 at $ω = π$ . That makes it a low pass shape.

The same questions grouped by how often each topic is set: the pole zero map and the magnitude sketch

The phase response TOP 17/19

82 Bh · 82 Ba · 81 Bh · 81 Ba · 80 Bh · 80 Ba · 79 Bh · 79 Ba · 78 Bh · 76 Ch · 76 Ash · 75 Ch · 75 Ash · 74 Ch · 74 Ash · 72 Ch · 72 Ka3+72+82+4

Before the formulas

Magnitude is only half the answer. The phase comes off the same picture: add up the angles from the point to each zero, subtract the angles to each pole, and that is it. Beyond that there is one idea worth carrying: a phase that falls as a straight line is a pure delay, which shifts the signal without changing its shape.

Anything that is not a straight line distorts the shape, and the next card is about buying that straight line deliberately.

\begin{aligned} θ (ω) & = \sum angles from e^{j ω} to each zero - \sum angles from e^{j ω} to each pole \\ - ω k for any pure delay z^{- k} \\ τ (ω) & = - \frac{d θ (ω)}{d ω} & the group delay \end{aligned}

The geometric rule is the same as for magnitude, with angles added and subtracted instead of distances multiplied and divided.
Phase is odd for a real system, so it is sketched over 0 to $π$ and mirrored with a sign change.
Linear phase means constant group delay: every frequency is delayed by the same number of samples, so the shape of the signal survives. This is the property FIR filters can have exactly and IIR filters cannot, and it is the whole argument of chapter 5.
A phase sketch is asked in 2082 Baishakh and 2076 Ashwin. Mark $θ$ (0), which is 0 or $π$ for a real system, and the jumps of $π$ that occur wherever the response passes through a zero on the unit circle.

A phase that is a straight line is the subject of the next card, and the property that every FIR design is built to have.

See it drawnPhase is the sum of the angles, and a delay is a straight line

Every zero adds its angle and every pole subtracts one. A pure delay adds a term proportional to

ω

, which is what makes the phase linear.

Practise thisEvery question the paper has set on this topic, worked in full17

What is different in this one

A pole lies on or outside the unit circle, at magnitude 1.6621, so this system is unstable. Say so: it is a mark, and it also tells you the magnitude response will be dominated by that pole.
The poles are a conjugate pair at an angle of 1.297 rad, so the response has a peak near that frequency and the sketch is symmetric about $ω = 0$ .

\begin{aligned} H (z) & = \frac{1 - 1.16 z^{- 1} + 4.58 z^{- 2}}{1 - 0.9 z^{- 1} + 2.7625 z^{- 2}} & from the difference equation \\ = \frac{z^{2} - 1.16 z + 4.58}{z^{2} - 0.9 z + 2.7625} & \times z^{2} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	0.58 + j2.06	2.1401	74.28
zero, o	0.58 - j2.06	2.1401	-74.28
pole, x	0.45 + j1.6	1.6621	74.29
pole, x	0.45 - j1.6	1.6621	-74.29

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 1.6621, so for a causal system the ROC is $| z | > 1.6621$ , which does not contain the unit circle, so the system is unstable.

\begin{aligned} N (e^{j ω}) & = e^{j 2 ω} - 1.16 e^ jw + 4.58 \\ = \cos (2 ω) + jsin (2 ω) - 1.16 cos (ω) - j 1.16 sin (ω) + 4.58 \\ D (e^{j ω}) & = e^{j 2 ω} - 0.9 e^ jw + 2.7625 \\ = \cos (2 ω) + jsin (2 ω) - 0.9 cos (ω) - j 0.9 sin (ω) + 2.7625 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	4.42	2.8625	1.5441	1.5441
$\frac{π}{4}$	3.7598 + j0.1798	2.1261 + j0.3636	1.7322 - j0.2117	1.7451
$\frac{π}{2}$	3.58 - j1.16	1.7625 - j0.9	1.8777 + j0.3007	1.9016
$\frac{3 π}{4}$	5.4002 - j1.8202	3.3989 - j1.6364	1.4992 + j0.1862	1.5107
$π$	6.74	4.6625	1.4456	1.4456
$\frac{5 π}{4}$	5.4002 + j1.8202	3.3989 + j1.6364	1.4992 - j0.1862	1.5107
$\frac{3 π}{2}$	3.58 + j1.16	1.7625 + j0.9	1.8777 - j0.3007	1.9016
$\frac{7 π}{4}$	3.7598 - j0.1798	2.1261 - j0.3636	1.7322 + j0.2117	1.7451
$2 π$	4.42	2.8625	1.5441	1.5441

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	1.5441
$ω = 1.297$ , the angle of the poles	1.2966	2.0367
$ω = \frac{π}{2}$ , the point z = j	1.5708	1.9016
$ω = π$ , the point $z = - 1$	3.1416	1.4456

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response rises to a peak of 2.0368 near $ω = 1.291$ , with $| H |$ = 1.5441 at $ω = 0$ and 1.4456 at $ω = π$ . That makes it a low pass shape.

What is different in this one

A pole lies on or outside the unit circle, at magnitude 1.6621, so this system is unstable. Say so: it is a mark, and it also tells you the magnitude response will be dominated by that pole.
The poles are a conjugate pair at an angle of 1.297 rad, so the response has a peak near that frequency and the sketch is symmetric about $ω = 0$ .

\begin{aligned} H (z) & = \frac{1 - 1.16 z^{- 1} + 4.58 z^{- 2}}{1 - 0.9 z^{- 1} + 2.7625 z^{- 2}} & from the difference equation \\ = \frac{z^{2} - 1.16 z + 4.58}{z^{2} - 0.9 z + 2.7625} & \times z^{2} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	0.58 + j2.06	2.1401	74.28
zero, o	0.58 - j2.06	2.1401	-74.28
pole, x	0.45 + j1.6	1.6621	74.29
pole, x	0.45 - j1.6	1.6621	-74.29

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 1.6621, so for a causal system the ROC is $| z | > 1.6621$ , which does not contain the unit circle, so the system is unstable.

\begin{aligned} N (e^{j ω}) & = e^{j 2 ω} - 1.16 e^ jw + 4.58 \\ = \cos (2 ω) + jsin (2 ω) - 1.16 cos (ω) - j 1.16 sin (ω) + 4.58 \\ D (e^{j ω}) & = e^{j 2 ω} - 0.9 e^ jw + 2.7625 \\ = \cos (2 ω) + jsin (2 ω) - 0.9 cos (ω) - j 0.9 sin (ω) + 2.7625 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	4.42	2.8625	1.5441	1.5441
$\frac{π}{4}$	3.7598 + j0.1798	2.1261 + j0.3636	1.7322 - j0.2117	1.7451
$\frac{π}{2}$	3.58 - j1.16	1.7625 - j0.9	1.8777 + j0.3007	1.9016
$\frac{3 π}{4}$	5.4002 - j1.8202	3.3989 - j1.6364	1.4992 + j0.1862	1.5107
$π$	6.74	4.6625	1.4456	1.4456
$\frac{5 π}{4}$	5.4002 + j1.8202	3.3989 + j1.6364	1.4992 - j0.1862	1.5107
$\frac{3 π}{2}$	3.58 + j1.16	1.7625 + j0.9	1.8777 - j0.3007	1.9016
$\frac{7 π}{4}$	3.7598 - j0.1798	2.1261 - j0.3636	1.7322 + j0.2117	1.7451
$2 π$	4.42	2.8625	1.5441	1.5441

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	1.5441
$ω = 1.297$ , the angle of the poles	1.2966	2.0367
$ω = \frac{π}{2}$ , the point z = j	1.5708	1.9016
$ω = π$ , the point $z = - 1$	3.1416	1.4456

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

arg

H (e^{j ω})

in radians, unwrapped, drawn from the computed response.

Where	$ω$	arg H, radians
$ω = 0$	0	0
$ω = \frac{π}{2}$	1.5708	0.159
$ω = π$	3.1416	0

Practise this +5 like it2081 Bhadra · Q43+7 marksPlot the pole zero in z plane and draw magnitude response (not to the scale) of the system described by difference equation

y [n] - 0.4 y [n - 1] + 0.25 y [n - 2] = x [n] - 0.4 x [n - 1]

What is different in this one

There is a zero at the origin, which appears only after multiplying by $z^{2}$ . It changes the phase and not the magnitude, since its distance to every point of the unit circle is 1, and forgetting it is the usual lost mark.
The poles are a conjugate pair at an angle of 1.159 rad, so the response has a peak near that frequency and the sketch is symmetric about $ω = 0$ .

\begin{aligned} H (z) & = \frac{1 - 0.4 z^{- 1}}{1 - 0.4 z^{- 1} + 0.25 z^{- 2}} & from the difference equation \\ = \frac{z^{2} - 0.4 z}{z^{2} - 0.4 z + 0.25} & \times z^{2} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	0.4	0.4	0
zero, o	0	0	0
pole, x	0.2 + j0.4583	0.5	66.42
pole, x	0.2 - j0.4583	0.5	-66.42

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 0.5, so for a causal system the ROC is $| z | > 0.5$ , which contains the unit circle, so the system is stable.

\begin{aligned} N (e^{j ω}) & = e^{j 2 ω} - 0.4 e^ jw \\ = \cos (2 ω) + jsin (2 ω) - 0.4 cos (ω) - j 0.4 sin (ω) \\ D (e^{j ω}) & = e^{j 2 ω} - 0.4 e^ jw + 0.25 \\ = \cos (2 ω) + jsin (2 ω) - 0.4 cos (ω) - j 0.4 sin (ω) + 0.25 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	0.6	0.85	0.7059	0.7059
$\frac{π}{4}$	-0.2828 + j0.7172	-0.0328 + j0.7172	1.0159 + j0.3479	1.0738
$\frac{π}{2}$	-1 - j0.4	-0.75 - j0.4	1.2595 - j0.1384	1.2671
$\frac{3 π}{4}$	0.2828 - j1.2828	0.5328 - j1.2828	0.931 - j0.1662	0.9457
$π$	1.4	1.65	0.8485	0.8485
$\frac{5 π}{4}$	0.2828 + j1.2828	0.5328 + j1.2828	0.931 + j0.1662	0.9457
$\frac{3 π}{2}$	-1 + j0.4	-0.75 + j0.4	1.2595 + j0.1384	1.2671
$\frac{7 π}{4}$	-0.2828 - j0.7172	-0.0328 - j0.7172	1.0159 - j0.3479	1.0738
$2 π$	0.6	0.85	0.7059	0.7059

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	0.7059
$ω = 1.159$ , the angle of the poles	1.1593	1.3194
$ω = \frac{π}{2}$ , the point z = j	1.5708	1.2671
$ω = π$ , the point $z = - 1$	3.1416	0.8485

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response rises to a peak of 1.3405 near $ω = 1.291$ , with $| H |$ = 0.7059 at $ω = 0$ and 0.8485 at $ω = π$ . That makes it a high pass shape.

Practise this +1 like it2081 Baishakh · Q43+7 marksPlot the pole-zero in z plane and draw the magnitude response (not to the scale) of the system described by difference equation

y (n) = 0.67 x (n) - 0.3 x (n - 1) + 2.75 y (n - 1)

What is different herea first order denominator; the poles are real, so the response falls away without a peak

What is different in this one

A pole lies on or outside the unit circle, at magnitude 2.75, so this system is unstable. Say so: it is a mark, and it also tells you the magnitude response will be dominated by that pole.
Every pole is real, so the response has no resonant peak: it rises or falls monotonically between the two ends.

\begin{aligned} H (z) & = \frac{0.67 - 0.3 z^{- 1}}{1 - 2.75 z^{- 1}} & from the difference equation \\ = \frac{0.67 z - 0.3}{z - 2.75} & \times z^{1} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	0.4478	0.4478	0
pole, x	2.75	2.75	0

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 2.75, so for a causal system the ROC is $| z | > 2.75$ , which does not contain the unit circle, so the system is unstable.

\begin{aligned} N (e^{j ω}) & = 0.67 e^ jw - 0.3 \\ = 0.67 cos (ω) + j 0.67 sin (ω) - 0.3 \\ D (e^{j ω}) & = e^{j ω} - 2.75 \\ = \cos (ω) + jsin (ω) - 2.75 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	0.37	-1.75	-0.2114	0.2114
$\frac{π}{4}$	0.1738 + j0.4738	-2.0429 + j0.7071	-0.0043 - j0.2334	0.2334
$\frac{π}{2}$	-0.3 + j0.67	-2.75 + j1	0.1746 - j0.1801	0.2509
$\frac{3 π}{4}$	-0.7738 + j0.4738	-3.4571 + j0.7071	0.2417 - j0.0876	0.2571
$π$	-0.97	-3.75	0.2587	0.2587
$\frac{5 π}{4}$	-0.7738 - j0.4738	-3.4571 - j0.7071	0.2417 + j0.0876	0.2571
$\frac{3 π}{2}$	-0.3 - j0.67	-2.75 - j1	0.1746 + j0.1801	0.2509
$\frac{7 π}{4}$	0.1738 - j0.4738	-2.0429 - j0.7071	-0.0043 + j0.2334	0.2334
$2 π$	0.37	-1.75	-0.2114	0.2114

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	0.2114
$ω = \frac{π}{2}$ , the point z = j	1.5708	0.2509
$ω = π$ , the point $z = - 1$	3.1416	0.2587

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response rises monotonically, with $| H |$ = 0.2114 at $ω = 0$ and 0.2587 at $ω = π$ . That makes it a high pass shape.

Practise this2080 Bhadra · Q43+7 marksPlot the pole-zero on the z-plane and draw Magnitude response (not to the scale) of an LTI system described by the equation,

y (n) = x (n) + 0.8 x (n - 1) + 0.8 x (n - 2) + 0.49 y (n - 2)

.What is different herethe poles are real, so the response falls away without a peak

What is different in this one

Every pole is real, so the response has no resonant peak: it rises or falls monotonically between the two ends.

\begin{aligned} H (z) & = \frac{1 + 0.8 z^{- 1} + 0.8 z^{- 2}}{1 - 0.49 z^{- 2}} & from the difference equation \\ = \frac{z^{2} + 0.8 z + 0.8}{z^{2} - 0.49} & \times z^{2} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	-0.4 + j0.8	0.8944	116.57
zero, o	-0.4 - j0.8	0.8944	-116.57
pole, x	-0.7	0.7	180
pole, x	0.7	0.7	0

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 0.7, so for a causal system the ROC is $| z | > 0.7$ , which contains the unit circle, so the system is stable.

\begin{aligned} N (e^{j ω}) & = e^{j 2 ω} + 0.8 e^ jw + 0.8 \\ = \cos (2 ω) + jsin (2 ω) + 0.8 cos (ω) + j 0.8 sin (ω) + 0.8 \\ D (e^{j ω}) & = e^{j 2 ω} - 0.49 \\ = \cos (2 ω) + jsin (2 ω) - 0.49 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	2.6	0.51	5.098	5.098
$\frac{π}{4}$	1.3657 + j1.5657	-0.49 + j1	0.7229 - j1.7199	1.8657
$\frac{π}{2}$	-0.2 + j0.8	-1.49	0.1342 - j0.5369	0.5534
$\frac{3 π}{4}$	0.2343 - j0.4343	-0.49 - j1	0.2576 + j0.3606	0.4431
$π$	1	0.51	1.9608	1.9608
$\frac{5 π}{4}$	0.2343 + j0.4343	-0.49 + j1	0.2576 - j0.3606	0.4431
$\frac{3 π}{2}$	-0.2 - j0.8	-1.49	0.1342 + j0.5369	0.5534
$\frac{7 π}{4}$	1.3657 - j1.5657	-0.49 - j1	0.7229 + j1.7199	1.8657
$2 π$	2.6	0.51	5.098	5.098

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	5.098
$ω = \frac{π}{2}$ , the point z = j	1.5708	0.5534
$ω = π$ , the point $z = - 1$	3.1416	1.9608

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response falls monotonically, with $| H |$ = 5.098 at $ω = 0$ and 1.9608 at $ω = π$ . That makes it a low pass shape.

What is different in this one

A pole lies on or outside the unit circle, at magnitude 1.6621, so this system is unstable. Say so: it is a mark, and it also tells you the magnitude response will be dominated by that pole.
The poles are a conjugate pair at an angle of 1.297 rad, so the response has a peak near that frequency and the sketch is symmetric about $ω = 0$ .

\begin{aligned} H (z) & = \frac{1 - 1.16 z^{- 1} + 4.58 z^{- 2}}{1 - 0.9 z^{- 1} + 2.7625 z^{- 2}} & from the difference equation \\ = \frac{z^{2} - 1.16 z + 4.58}{z^{2} - 0.9 z + 2.7625} & \times z^{2} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	0.58 + j2.06	2.1401	74.28
zero, o	0.58 - j2.06	2.1401	-74.28
pole, x	0.45 + j1.6	1.6621	74.29
pole, x	0.45 - j1.6	1.6621	-74.29

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 1.6621, so for a causal system the ROC is $| z | > 1.6621$ , which does not contain the unit circle, so the system is unstable.

\begin{aligned} N (e^{j ω}) & = e^{j 2 ω} - 1.16 e^ jw + 4.58 \\ = \cos (2 ω) + jsin (2 ω) - 1.16 cos (ω) - j 1.16 sin (ω) + 4.58 \\ D (e^{j ω}) & = e^{j 2 ω} - 0.9 e^ jw + 2.7625 \\ = \cos (2 ω) + jsin (2 ω) - 0.9 cos (ω) - j 0.9 sin (ω) + 2.7625 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	4.42	2.8625	1.5441	1.5441
$\frac{π}{4}$	3.7598 + j0.1798	2.1261 + j0.3636	1.7322 - j0.2117	1.7451
$\frac{π}{2}$	3.58 - j1.16	1.7625 - j0.9	1.8777 + j0.3007	1.9016
$\frac{3 π}{4}$	5.4002 - j1.8202	3.3989 - j1.6364	1.4992 + j0.1862	1.5107
$π$	6.74	4.6625	1.4456	1.4456
$\frac{5 π}{4}$	5.4002 + j1.8202	3.3989 + j1.6364	1.4992 - j0.1862	1.5107
$\frac{3 π}{2}$	3.58 + j1.16	1.7625 + j0.9	1.8777 - j0.3007	1.9016
$\frac{7 π}{4}$	3.7598 - j0.1798	2.1261 - j0.3636	1.7322 + j0.2117	1.7451
$2 π$	4.42	2.8625	1.5441	1.5441

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	1.5441
$ω = 1.297$ , the angle of the poles	1.2966	2.0367
$ω = \frac{π}{2}$ , the point z = j	1.5708	1.9016
$ω = π$ , the point $z = - 1$	3.1416	1.4456

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response rises to a peak of 2.0368 near $ω = 1.291$ , with $| H |$ = 1.5441 at $ω = 0$ and 1.4456 at $ω = π$ . That makes it a low pass shape.

Practise this2079 Bhadra · Q43+7 marksPlot the pole-zero in z-plane and draw the magnitude response (not to the scale) of the equation of the system describe by difference equation:

y [n]

- 0.

35 y [n - 1]

+ 0.

25 y [n - 2] = x [n]

- 0.

75 x [n - 1]

.What is different hereThe standard text works this one as its Example 3.3 and slips in the

ω = \frac{π}{4}

row: it prints

D (e^{j ω}) = - 0.025

+ j0.75, but 0.25 - 0.2475 is <

b > 0.0025 < / b

>, not -0.025, so

D = 0.0025

+ j0.7525 and

H = 0.6218

+ j0.7068. The same slip is repeated at

\frac{7 π}{4}

. The magnitude survives it almost unchanged, 0.9414 against their 0.943, which is exactly why it is worth checking the real and imaginary parts and not only the modulus.

What is different in this one

There is a zero at the origin, which appears only after multiplying by $z^{2}$ . It changes the phase and not the magnitude, since its distance to every point of the unit circle is 1, and forgetting it is the usual lost mark.
The poles are a conjugate pair at an angle of 1.213 rad, so the response has a peak near that frequency and the sketch is symmetric about $ω = 0$ .

\begin{aligned} H (z) & = \frac{1 - 0.75 z^{- 1}}{1 - 0.35 z^{- 1} + 0.25 z^{- 2}} & from the difference equation \\ = \frac{z^{2} - 0.75 z}{z^{2} - 0.35 z + 0.25} & \times z^{2} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	0.75	0.75	0
zero, o	0	0	0
pole, x	0.175 + j0.4684	0.5	69.51
pole, x	0.175 - j0.4684	0.5	-69.51

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 0.5, so for a causal system the ROC is $| z | > 0.5$ , which contains the unit circle, so the system is stable.

\begin{aligned} N (e^{j ω}) & = e^{j 2 ω} - 0.75 e^ jw \\ = \cos (2 ω) + jsin (2 ω) - 0.75 cos (ω) - j 0.75 sin (ω) \\ D (e^{j ω}) & = e^{j 2 ω} - 0.35 e^ jw + 0.25 \\ = \cos (2 ω) + jsin (2 ω) - 0.35 cos (ω) - j 0.35 sin (ω) + 0.25 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	0.25	0.9	0.2778	0.2778
$\frac{π}{4}$	-0.5303 + j0.4697	0.0025 + j0.7525	0.6218 + j0.7068	0.9414
$\frac{π}{2}$	-1 - j0.75	-0.75 - j0.35	1.4781 + j0.3102	1.5103
$\frac{3 π}{4}$	0.5303 - j1.5303	0.4975 - j1.2475	1.2047 - j0.0553	1.2059
$π$	1.75	1.6	1.0938	1.0938
$\frac{5 π}{4}$	0.5303 + j1.5303	0.4975 + j1.2475	1.2047 + j0.0553	1.2059
$\frac{3 π}{2}$	-1 + j0.75	-0.75 + j0.35	1.4781 - j0.3102	1.5103
$\frac{7 π}{4}$	-0.5303 - j0.4697	0.0025 - j0.7525	0.6218 - j0.7068	0.9414
$2 π$	0.25	0.9	0.2778	0.2778

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	0.2778
$ω = 1.213$ , the angle of the poles	1.2132	1.4387
$ω = \frac{π}{2}$ , the point z = j	1.5708	1.5103
$ω = π$ , the point $z = - 1$	3.1416	1.0938

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response rises to a peak of 1.5251 near $ω = 1.456$ , with $| H |$ = 0.2778 at $ω = 0$ and 1.0938 at $ω = π$ . That makes it a high pass shape.

Same as 2081 Bhadra Q42079 Baishakh · Q43+7 marksPlot pole-zero in z-plane and draw magnitude response (not to the scale) of the system described by difference equation

y [n] - 0.3 y [n - 1]

+ 0.

2 y [n - 2] = x [n]

- 0.

5 x [n - 1]

What is different in this one

There is a zero at the origin, which appears only after multiplying by $z^{2}$ . It changes the phase and not the magnitude, since its distance to every point of the unit circle is 1, and forgetting it is the usual lost mark.
The poles are a conjugate pair at an angle of 1.229 rad, so the response has a peak near that frequency and the sketch is symmetric about $ω = 0$ .

\begin{aligned} H (z) & = \frac{1 - 0.5 z^{- 1}}{1 - 0.3 z^{- 1} + 0.2 z^{- 2}} & from the difference equation \\ = \frac{z^{2} - 0.5 z}{z^{2} - 0.3 z + 0.2} & \times z^{2} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	0.5	0.5	0
zero, o	0	0	0
pole, x	0.15 + j0.4213	0.4472	70.4
pole, x	0.15 - j0.4213	0.4472	-70.4

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 0.4472, so for a causal system the ROC is $| z | > 0.4472$ , which contains the unit circle, so the system is stable.

\begin{aligned} N (e^{j ω}) & = e^{j 2 ω} - 0.5 e^ jw \\ = \cos (2 ω) + jsin (2 ω) - 0.5 cos (ω) - j 0.5 sin (ω) \\ D (e^{j ω}) & = e^{j 2 ω} - 0.3 e^ jw + 0.2 \\ = \cos (2 ω) + jsin (2 ω) - 0.3 cos (ω) - j 0.3 sin (ω) + 0.2 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	0.5	0.9	0.5556	0.5556
$\frac{π}{4}$	-0.3536 + j0.6464	-0.0121 + j0.7879	0.8272 + j0.436	0.9351
$\frac{π}{2}$	-1 - j0.5	-0.8 - j0.3	1.3014 + j0.137	1.3086
$\frac{3 π}{4}$	0.3536 - j1.3536	0.4121 - j1.2121	1.0899 - j0.0789	1.0927
$π$	1.5	1.5	1	1
$\frac{5 π}{4}$	0.3536 + j1.3536	0.4121 + j1.2121	1.0899 + j0.0789	1.0927
$\frac{3 π}{2}$	-1 + j0.5	-0.8 + j0.3	1.3014 - j0.137	1.3086
$\frac{7 π}{4}$	-0.3536 - j0.6464	-0.0121 - j0.7879	0.8272 - j0.436	0.9351
$2 π$	0.5	0.9	0.5556	0.5556

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	0.5556
$ω = 1.229$ , the angle of the poles	1.2288	1.2574
$ω = \frac{π}{2}$ , the point z = j	1.5708	1.3086
$ω = π$ , the point $z = - 1$	3.1416	1

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response rises to a peak of 1.3142 near $ω = 1.482$ , with $| H |$ = 0.5556 at $ω = 0$ and 1 at $ω = π$ . That makes it a high pass shape.

Same as 2081 Bhadra Q42078 Bhadra · Q52+4 marksPlot the pole-zero in z-plane and draw magnitude response (not to the scale) of the system described by difference equation.

y [n] - 0.4 y [n - 1] + 0.25 y [n - 2] = x [n]

- 0.

4 x [n - 1]

What is different in this one

There is a zero at the origin, which appears only after multiplying by $z^{2}$ . It changes the phase and not the magnitude, since its distance to every point of the unit circle is 1, and forgetting it is the usual lost mark.
The poles are a conjugate pair at an angle of 1.159 rad, so the response has a peak near that frequency and the sketch is symmetric about $ω = 0$ .

\begin{aligned} H (z) & = \frac{1 - 0.4 z^{- 1}}{1 - 0.4 z^{- 1} + 0.25 z^{- 2}} & from the difference equation \\ = \frac{z^{2} - 0.4 z}{z^{2} - 0.4 z + 0.25} & \times z^{2} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	0.4	0.4	0
zero, o	0	0	0
pole, x	0.2 + j0.4583	0.5	66.42
pole, x	0.2 - j0.4583	0.5	-66.42

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 0.5, so for a causal system the ROC is $| z | > 0.5$ , which contains the unit circle, so the system is stable.

\begin{aligned} N (e^{j ω}) & = e^{j 2 ω} - 0.4 e^ jw \\ = \cos (2 ω) + jsin (2 ω) - 0.4 cos (ω) - j 0.4 sin (ω) \\ D (e^{j ω}) & = e^{j 2 ω} - 0.4 e^ jw + 0.25 \\ = \cos (2 ω) + jsin (2 ω) - 0.4 cos (ω) - j 0.4 sin (ω) + 0.25 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	0.6	0.85	0.7059	0.7059
$\frac{π}{4}$	-0.2828 + j0.7172	-0.0328 + j0.7172	1.0159 + j0.3479	1.0738
$\frac{π}{2}$	-1 - j0.4	-0.75 - j0.4	1.2595 - j0.1384	1.2671
$\frac{3 π}{4}$	0.2828 - j1.2828	0.5328 - j1.2828	0.931 - j0.1662	0.9457
$π$	1.4	1.65	0.8485	0.8485
$\frac{5 π}{4}$	0.2828 + j1.2828	0.5328 + j1.2828	0.931 + j0.1662	0.9457
$\frac{3 π}{2}$	-1 + j0.4	-0.75 + j0.4	1.2595 + j0.1384	1.2671
$\frac{7 π}{4}$	-0.2828 - j0.7172	-0.0328 - j0.7172	1.0159 - j0.3479	1.0738
$2 π$	0.6	0.85	0.7059	0.7059

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	0.7059
$ω = 1.159$ , the angle of the poles	1.1593	1.3194
$ω = \frac{π}{2}$ , the point z = j	1.5708	1.2671
$ω = π$ , the point $z = - 1$	3.1416	0.8485

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response rises to a peak of 1.3405 near $ω = 1.291$ , with $| H |$ = 0.7059 at $ω = 0$ and 0.8485 at $ω = π$ . That makes it a high pass shape.

What is different in this one

A pole lies on or outside the unit circle, at magnitude 1.1516, so this system is unstable. Say so: it is a mark, and it also tells you the magnitude response will be dominated by that pole.
The poles are a conjugate pair at an angle of 1.169 rad, so the response has a peak near that frequency and the sketch is symmetric about $ω = 0$ .

\begin{aligned} H (z) & = \frac{1 - 1.16 z^{- 1} + 4.58 z^{- 2}}{1 - 0.9 z^{- 1} + 1.3261 z^{- 2}} & from the difference equation \\ = \frac{z^{2} - 1.16 z + 4.58}{z^{2} - 0.9 z + 1.3261} & \times z^{2} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	0.58 + j2.06	2.1401	74.28
zero, o	0.58 - j2.06	2.1401	-74.28
pole, x	0.45 + j1.06	1.1516	67
pole, x	0.45 - j1.06	1.1516	-67

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 1.1516, so for a causal system the ROC is $| z | > 1.1516$ , which does not contain the unit circle, so the system is unstable.

\begin{aligned} N (e^{j ω}) & = e^{j 2 ω} - 1.16 e^ jw + 4.58 \\ = \cos (2 ω) + jsin (2 ω) - 1.16 cos (ω) - j 1.16 sin (ω) + 4.58 \\ D (e^{j ω}) & = e^{j 2 ω} - 0.9 e^ jw + 1.3261 \\ = \cos (2 ω) + jsin (2 ω) - 0.9 cos (ω) - j 0.9 sin (ω) + 1.3261 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	4.42	1.4261	3.0994	3.0994
$\frac{π}{4}$	3.7598 + j0.1798	0.6897 + j0.3636	4.3732 - j2.0449	4.8277
$\frac{π}{2}$	3.58 - j1.16	0.3261 - j0.9	2.4133 + j3.1033	3.9313
$\frac{3 π}{4}$	5.4002 - j1.8202	1.9625 - j1.6364	2.0794 + j0.8063	2.2302
$π$	6.74	3.2261	2.0892	2.0892
$\frac{5 π}{4}$	5.4002 + j1.8202	1.9625 + j1.6364	2.0794 - j0.8063	2.2302
$\frac{3 π}{2}$	3.58 + j1.16	0.3261 + j0.9	2.4133 - j3.1033	3.9313
$\frac{7 π}{4}$	3.7598 - j0.1798	0.6897 - j0.3636	4.3732 + j2.0449	4.8277
$2 π$	4.42	1.4261	3.0994	3.0994

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	3.0994
$ω = 1.169$ , the angle of the poles	1.1693	11.4875
$ω = \frac{π}{2}$ , the point z = j	1.5708	3.9313
$ω = π$ , the point $z = - 1$	3.1416	2.0892

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response rises to a peak of 11.4954 near $ω = 1.165$ , with $| H |$ = 3.0994 at $ω = 0$ and 2.0892 at $ω = π$ . That makes it a low pass shape.

Practise this2076 Ashwin · Q62+8 marksFor the system described by the following difference equation:

y [n] = 0

67 x [n]

- 0.

3 x [n - 1]

+ 2.

75 y [n - 1]

What is different in this one

A pole lies on or outside the unit circle, at magnitude 2.75, so this system is unstable. Say so: it is a mark, and it also tells you the magnitude response will be dominated by that pole.
Every pole is real, so the response has no resonant peak: it rises or falls monotonically between the two ends.

\begin{aligned} H (z) & = \frac{0.67 - 0.3 z^{- 1}}{1 - 2.75 z^{- 1}} & from the difference equation \\ = \frac{0.67 z - 0.3}{z - 2.75} & \times z^{1} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	0.4478	0.4478	0
pole, x	2.75	2.75	0

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 2.75, so for a causal system the ROC is $| z | > 2.75$ , which does not contain the unit circle, so the system is unstable.

\begin{aligned} N (e^{j ω}) & = 0.67 e^ jw - 0.3 \\ = 0.67 cos (ω) + j 0.67 sin (ω) - 0.3 \\ D (e^{j ω}) & = e^{j ω} - 2.75 \\ = \cos (ω) + jsin (ω) - 2.75 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	0.37	-1.75	-0.2114	0.2114
$\frac{π}{4}$	0.1738 + j0.4738	-2.0429 + j0.7071	-0.0043 - j0.2334	0.2334
$\frac{π}{2}$	-0.3 + j0.67	-2.75 + j1	0.1746 - j0.1801	0.2509
$\frac{3 π}{4}$	-0.7738 + j0.4738	-3.4571 + j0.7071	0.2417 - j0.0876	0.2571
$π$	-0.97	-3.75	0.2587	0.2587
$\frac{5 π}{4}$	-0.7738 - j0.4738	-3.4571 - j0.7071	0.2417 + j0.0876	0.2571
$\frac{3 π}{2}$	-0.3 - j0.67	-2.75 - j1	0.1746 + j0.1801	0.2509
$\frac{7 π}{4}$	0.1738 - j0.4738	-2.0429 - j0.7071	-0.0043 + j0.2334	0.2334
$2 π$	0.37	-1.75	-0.2114	0.2114

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	0.2114
$ω = \frac{π}{2}$ , the point z = j	1.5708	0.2509
$ω = π$ , the point $z = - 1$	3.1416	0.2587

arg

H (e^{j ω})

in radians, unwrapped, drawn from the computed response.

Where	$ω$	arg H, radians
$ω = 0$	0	-3.142
$ω = \frac{π}{2}$	1.5708	-0.801
$ω = π$	3.1416	0

The magnitude is not asked here, only the pole zero map and the phase. The phase runs from -3.142 rad at $ω = 0$ to 0 rad at $ω = π$ , rising steadily.

Same as 2081 Baishakh Q42075 Chaitra · Q32+7 marksPlot the pole-zero in z-plane and draw magnitude response (not to scale) of the system described by differential equation

y (n) - 0.3 y (n - 1) = 2 x (n - 2) + 0.7 x (n - 1) + 4 x (n)

What is different in this one

Every pole is real, so the response has no resonant peak: it rises or falls monotonically between the two ends.

\begin{aligned} H (z) & = \frac{4 + 0.7 z^{- 1} + 2 z^{- 2}}{1 - 0.3 z^{- 1}} & from the difference equation \\ = \frac{4 z^{2} + 0.7 z + 2}{z^{2} - 0.3 z} & \times z^{2} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	-0.0875 + j0.7017	0.7071	97.11
zero, o	-0.0875 - j0.7017	0.7071	-97.11
pole, x	0.3	0.3	0
pole, x	0	0	0

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 0.3, so for a causal system the ROC is $| z | > 0.3$ , which contains the unit circle, so the system is stable.

\begin{aligned} N (e^{j ω}) & = 4 e^j 2 ω + 0.7 e^ jw + 2 \\ = 4 cos (2 ω) + j 4 sin (2 ω) + 0.7 cos (ω) + j 0.7 sin (ω) + 2 \\ D (e^{j ω}) & = e^{j 2 ω} - 0.3 e^ jw \\ = \cos (2 ω) + jsin (2 ω) - 0.3 cos (ω) - j 0.3 sin (ω) \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	6.7	0.7	9.5714	9.5714
$\frac{π}{4}$	2.495 + j4.495	-0.2121 + j0.7879	4.5246 - j4.385	6.3008
$\frac{π}{2}$	-2 + j0.7	-1 - j0.3	1.6422 - j1.1927	2.0296
$\frac{3 π}{4}$	1.505 - j3.505	0.2121 - j1.2121	3.0165 + j0.7137	3.0998
$π$	5.3	1.3	4.0769	4.0769
$\frac{5 π}{4}$	1.505 + j3.505	0.2121 + j1.2121	3.0165 - j0.7137	3.0998
$\frac{3 π}{2}$	-2 - j0.7	-1 + j0.3	1.6422 + j1.1927	2.0296
$\frac{7 π}{4}$	2.495 - j4.495	-0.2121 - j0.7879	4.5246 + j4.385	6.3008
$2 π$	6.7	0.7	9.5714	9.5714

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	9.5714
$ω = \frac{π}{2}$ , the point z = j	1.5708	2.0296
$ω = π$ , the point $z = - 1$	3.1416	4.0769

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response falls monotonically, with $| H |$ = 9.5714 at $ω = 0$ and 4.0769 at $ω = π$ . That makes it a low pass shape.

What is different in this one

A pole lies on or outside the unit circle, at magnitude 1.1516, so this system is unstable. Say so: it is a mark, and it also tells you the magnitude response will be dominated by that pole.
The poles are a conjugate pair at an angle of 1.169 rad, so the response has a peak near that frequency and the sketch is symmetric about $ω = 0$ .

\begin{aligned} H (z) & = \frac{1 - 1.16 z^{- 1} + 4.58 z^{- 2}}{1 - 0.9 z^{- 1} + 1.3261 z^{- 2}} & from the difference equation \\ = \frac{z^{2} - 1.16 z + 4.58}{z^{2} - 0.9 z + 1.3261} & \times z^{2} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	0.58 + j2.06	2.1401	74.28
zero, o	0.58 - j2.06	2.1401	-74.28
pole, x	0.45 + j1.06	1.1516	67
pole, x	0.45 - j1.06	1.1516	-67

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 1.1516, so for a causal system the ROC is $| z | > 1.1516$ , which does not contain the unit circle, so the system is unstable.

\begin{aligned} N (e^{j ω}) & = e^{j 2 ω} - 1.16 e^ jw + 4.58 \\ = \cos (2 ω) + jsin (2 ω) - 1.16 cos (ω) - j 1.16 sin (ω) + 4.58 \\ D (e^{j ω}) & = e^{j 2 ω} - 0.9 e^ jw + 1.3261 \\ = \cos (2 ω) + jsin (2 ω) - 0.9 cos (ω) - j 0.9 sin (ω) + 1.3261 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	4.42	1.4261	3.0994	3.0994
$\frac{π}{4}$	3.7598 + j0.1798	0.6897 + j0.3636	4.3732 - j2.0449	4.8277
$\frac{π}{2}$	3.58 - j1.16	0.3261 - j0.9	2.4133 + j3.1033	3.9313
$\frac{3 π}{4}$	5.4002 - j1.8202	1.9625 - j1.6364	2.0794 + j0.8063	2.2302
$π$	6.74	3.2261	2.0892	2.0892
$\frac{5 π}{4}$	5.4002 + j1.8202	1.9625 + j1.6364	2.0794 - j0.8063	2.2302
$\frac{3 π}{2}$	3.58 + j1.16	0.3261 + j0.9	2.4133 - j3.1033	3.9313
$\frac{7 π}{4}$	3.7598 - j0.1798	0.6897 - j0.3636	4.3732 + j2.0449	4.8277
$2 π$	4.42	1.4261	3.0994	3.0994

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	3.0994
$ω = 1.169$ , the angle of the poles	1.1693	11.4875
$ω = \frac{π}{2}$ , the point z = j	1.5708	3.9313
$ω = π$ , the point $z = - 1$	3.1416	2.0892

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response rises to a peak of 11.4954 near $ω = 1.165$ , with $| H |$ = 3.0994 at $ω = 0$ and 2.0892 at $ω = π$ . That makes it a low pass shape.

What is different in this one

A pole lies on or outside the unit circle, at magnitude 2.0224, so this system is unstable. Say so: it is a mark, and it also tells you the magnitude response will be dominated by that pole.
There is a zero at the origin, which appears only after multiplying by $z^{4}$ . It changes the phase and not the magnitude, since its distance to every point of the unit circle is 1, and forgetting it is the usual lost mark.
The poles are a conjugate pair at an angle of 1.042 rad, so the response has a peak near that frequency and the sketch is symmetric about $ω = 0$ .

\begin{aligned} H (z) & = \frac{1}{1 + 3.1 z^{- 1} + 1.2854 z^{- 2} - 0.4994 z^{- 3} + 3.2532 z^{- 4}} & from the difference equation \\ = \frac{z^{4}}{z^{4} + 3.1 z^{3} + 1.2854 z^{2} - 0.4994 z + 3.2532} & \times z^{4} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	0	0	0
zero, o	0	0	0
zero, o	0	0	0
zero, o	0	0	0
pole, x	-2 + j0.3	2.0224	171.47
pole, x	-2 - j0.3	2.0224	-171.47
pole, x	0.45 + j0.77	0.8919	59.7
pole, x	0.45 - j0.77	0.8919	-59.7

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 2.0224, so for a causal system the ROC is $| z | > 2.0224$ , which does not contain the unit circle, so the system is unstable.

\begin{aligned} N (e^{j ω}) & = e^{j 4 ω} \\ = \cos (4 ω) + jsin (4 ω) \\ D (e^{j ω}) & = e^{j 4 ω} + 3.1 e^j 3 ω + 1.2854 e^j 2 ω - 0.4994 e^ jw + 3.2532 \\ = \cos (4 ω) + jsin (4 ω) + 3.1 cos (3 ω) + j 3.1 sin (3 ω) + 1.2854 cos (2 ω) + j 1.2854 sin (2 ω) - 0.4994 cos (ω) - j 0.4994 sin (ω) + 3.2532 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	1	8.1392	0.1229	0.1229
$\frac{π}{4}$	-1	-0.292 + j3.1243	0.0297 + j0.3173	0.3187
$\frac{π}{2}$	1	2.9678 - j3.5994	0.1364 + j0.1654	0.2144
$\frac{3 π}{4}$	-1	4.7983 + j0.5535	-0.2057 + j0.0237	0.207
$π$	1	2.938	0.3404	0.3404
$\frac{5 π}{4}$	-1	4.7983 - j0.5535	-0.2057 - j0.0237	0.207
$\frac{3 π}{2}$	1	2.9678 + j3.5994	0.1364 - j0.1654	0.2144
$\frac{7 π}{4}$	-1	-0.292 - j3.1243	0.0297 - j0.3173	0.3187
$2 π$	1	8.1392	0.1229	0.1229

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	0.1229
$ω = 1.042$ , the angle of the poles	1.0419	0.7981
$ω = \frac{π}{2}$ , the point z = j	1.5708	0.2144
$ω = π$ , the point $z = - 1$	3.1416	0.3404

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response rises to a peak of 0.7983 near $ω = 1.045$ , with $| H |$ = 0.1229 at $ω = 0$ and 0.3404 at $ω = π$ . That makes it a high pass shape.

Same as 2081 Bhadra Q42074 Ashwin · Q43+7 marksPlot the pole-zero in z-plane and Draw Magnitude Response (not to the scale) of the system described by difference equation

y [n]

- 0.

4 y [n - 1]

+ 0.

2 y [n - 2] = x [n]

+ 0.

5 x [n - 1]

+ 0.

6 x [n - 2]

+ 0.

8 x [n - 3]

What is different in this one

The poles are a conjugate pair at an angle of 1.107 rad, so the response has a peak near that frequency and the sketch is symmetric about $ω = 0$ .

\begin{aligned} H (z) & = \frac{1 + 0.5 z^{- 1} + 0.6 z^{- 2} + 0.8 z^{- 3}}{1 - 0.4 z^{- 1} + 0.2 z^{- 2}} & from the difference equation \\ = \frac{z^{3} + 0.5 z^{2} + 0.6 z + 0.8}{z^{3} - 0.4 z^{2} + 0.2 z} & \times z^{3} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	0.1845 + j0.9416	0.9595	78.91
zero, o	0.1845 - j0.9416	0.9595	-78.91
zero, o	-0.869	0.869	180
pole, x	0.2 + j0.4	0.4472	63.43
pole, x	0.2 - j0.4	0.4472	-63.43
pole, x	0	0	0

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 0.4472, so for a causal system the ROC is $| z | > 0.4472$ , which contains the unit circle, so the system is stable.

\begin{aligned} N (e^{j ω}) & = e^{j 3 ω} + 0.5 e^j 2 ω + 0.6 e^ jw + 0.8 \\ = \cos (3 ω) + jsin (3 ω) + 0.5 cos (2 ω) + j 0.5 sin (2 ω) + 0.6 cos (ω) + j 0.6 sin (ω) + 0.8 \\ D (e^{j ω}) & = e^{j 3 ω} - 0.4 e^j 2 ω + 0.2 e^ jw \\ = \cos (3 ω) + jsin (3 ω) - 0.4 cos (2 ω) - j 0.4 sin (2 ω) + 0.2 cos (ω) + j 0.2 sin (ω) \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	2.9	0.8	3.625	3.625
$\frac{π}{4}$	0.5172 + j1.6314	-0.5657 + j0.4485	0.8426 - j2.2158	2.3706
$\frac{π}{2}$	0.3 - j0.4	0.4 - j0.8	0.55 + j0.1	0.559
$\frac{3 π}{4}$	1.0828 + j0.6314	0.5657 + j1.2485	0.7456 - j0.5295	0.9145
$π$	-0.3	-1.6	0.1875	0.1875
$\frac{5 π}{4}$	1.0828 - j0.6314	0.5657 - j1.2485	0.7456 + j0.5295	0.9145
$\frac{3 π}{2}$	0.3 + j0.4	0.4 + j0.8	0.55 - j0.1	0.559
$\frac{7 π}{4}$	0.5172 - j1.6314	-0.5657 - j0.4485	0.8426 + j2.2158	2.3706
$2 π$	2.9	0.8	3.625	3.625

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	3.625
$ω = 1.107$ , the angle of the poles	1.1071	1.0815
$ω = \frac{π}{2}$ , the point z = j	1.5708	0.559
$ω = π$ , the point $z = - 1$	3.1416	0.1875

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response falls monotonically, with $| H |$ = 3.625 at $ω = 0$ and 0.1875 at $ω = π$ . That makes it a low pass shape.

Same as 2081 Bhadra Q42072 Chaitra · Q52+4 marksPlot the pole-zero in z-plane and Draw Magnitude Response (not to the scale) of the system described by difference equation.

y [n]

- 0.

4 y [n - 1]

+ 0.

1 y [n - 2] = x [n]

+ 0.

6 x [n - 1]

What is different in this one

There is a zero at the origin, which appears only after multiplying by $z^{2}$ . It changes the phase and not the magnitude, since its distance to every point of the unit circle is 1, and forgetting it is the usual lost mark.
The poles are a conjugate pair at an angle of 0.886 rad, so the response has a peak near that frequency and the sketch is symmetric about $ω = 0$ .

\begin{aligned} H (z) & = \frac{1 + 0.6 z^{- 1}}{1 - 0.4 z^{- 1} + 0.1 z^{- 2}} & from the difference equation \\ = \frac{z^{2} + 0.6 z}{z^{2} - 0.4 z + 0.1} & \times z^{2} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	-0.6	0.6	180
zero, o	0	0	0
pole, x	0.2 + j0.2449	0.3162	50.77
pole, x	0.2 - j0.2449	0.3162	-50.77

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 0.3162, so for a causal system the ROC is $| z | > 0.3162$ , which contains the unit circle, so the system is stable.

\begin{aligned} N (e^{j ω}) & = e^{j 2 ω} + 0.6 e^ jw \\ = \cos (2 ω) + jsin (2 ω) + 0.6 cos (ω) + j 0.6 sin (ω) \\ D (e^{j ω}) & = e^{j 2 ω} - 0.4 e^ jw + 0.1 \\ = \cos (2 ω) + jsin (2 ω) - 0.4 cos (ω) - j 0.4 sin (ω) + 0.1 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	1.6	0.7	2.2857	2.2857
$\frac{π}{4}$	0.4243 + j1.4243	-0.1828 + j0.7172	1.7231 - j1.0309	2.008
$\frac{π}{2}$	-1 + j0.6	-0.9 - j0.4	0.6804 - j0.9691	1.1841
$\frac{3 π}{4}$	-0.4243 - j0.5757	0.3828 - j1.2828	0.3215 - j0.4267	0.5342
$π$	0.4	1.5	0.2667	0.2667
$\frac{5 π}{4}$	-0.4243 + j0.5757	0.3828 + j1.2828	0.3215 + j0.4267	0.5342
$\frac{3 π}{2}$	-1 - j0.6	-0.9 + j0.4	0.6804 + j0.9691	1.1841
$\frac{7 π}{4}$	0.4243 - j1.4243	-0.1828 - j0.7172	1.7231 + j1.0309	2.008
$2 π$	1.6	0.7	2.2857	2.2857

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	2.2857
$ω = 0.886$ , the angle of the poles	0.8861	1.9223
$ω = \frac{π}{2}$ , the point z = j	1.5708	1.1841
$ω = π$ , the point $z = - 1$	3.1416	0.2667

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response falls monotonically, with $| H |$ = 2.2857 at $ω = 0$ and 0.2667 at $ω = π$ . That makes it a low pass shape.

Same as 2081 Bhadra Q42072 Kartik · Q56 marksPlot Magnitude Response (not to the scale) of the system described by difference equation.

y [n]

- 0.

5 y [n - 1]

+ 0.

3 y [n - 2] = x [n]

+ 0.

7 x [n - 1]

What is different in this one

There is a zero at the origin, which appears only after multiplying by $z^{2}$ . It changes the phase and not the magnitude, since its distance to every point of the unit circle is 1, and forgetting it is the usual lost mark.
The poles are a conjugate pair at an angle of 1.097 rad, so the response has a peak near that frequency and the sketch is symmetric about $ω = 0$ .

\begin{aligned} H (z) & = \frac{1 + 0.7 z^{- 1}}{1 - 0.5 z^{- 1} + 0.3 z^{- 2}} & from the difference equation \\ = \frac{z^{2} + 0.7 z}{z^{2} - 0.5 z + 0.3} & \times z^{2} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	-0.7	0.7	180
zero, o	0	0	0
pole, x	0.25 + j0.4873	0.5477	62.84
pole, x	0.25 - j0.4873	0.5477	-62.84

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 0.5477, so for a causal system the ROC is $| z | > 0.5477$ , which contains the unit circle, so the system is stable.

\begin{aligned} N (e^{j ω}) & = e^{j 2 ω} + 0.7 e^ jw \\ = \cos (2 ω) + jsin (2 ω) + 0.7 cos (ω) + j 0.7 sin (ω) \\ D (e^{j ω}) & = e^{j 2 ω} - 0.5 e^ jw + 0.3 \\ = \cos (2 ω) + jsin (2 ω) - 0.5 cos (ω) - j 0.5 sin (ω) + 0.3 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	1.7	0.8	2.125	2.125
$\frac{π}{4}$	0.495 + j1.495	-0.0536 + j0.6464	2.2338 - j0.9507	2.4277
$\frac{π}{2}$	-1 + j0.7	-0.7 - j0.5	0.473 - j1.3378	1.419
$\frac{3 π}{4}$	-0.495 - j0.505	0.6536 - j1.3536	0.1594 - j0.4426	0.4705
$π$	0.3	1.8	0.1667	0.1667
$\frac{5 π}{4}$	-0.495 + j0.505	0.6536 + j1.3536	0.1594 + j0.4426	0.4705
$\frac{3 π}{2}$	-1 - j0.7	-0.7 + j0.5	0.473 + j1.3378	1.419
$\frac{7 π}{4}$	0.495 - j1.495	-0.0536 - j0.6464	2.2338 + j0.9507	2.4277
$2 π$	1.7	0.8	2.125	2.125

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	2.125
$ω = 1.097$ , the angle of the poles	1.0968	2.3168
$ω = \frac{π}{2}$ , the point z = j	1.5708	1.419
$ω = π$ , the point $z = - 1$	3.1416	0.1667

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response rises to a peak of 2.4438 near $ω = 0.877$ , with $| H |$ = 2.125 at $ω = 0$ and 0.1667 at $ω = π$ . That makes it a low pass shape.

The same questions grouped by how often each topic is set: the pole zero map and the magnitude sketch

3.4Linear phase, causality and stability

Linear phase, and its relationship with causality

Before the formulas

A filter that delays every frequency by the same amount does not bend the shape of what goes through it, it only moves it later. That is linear phase, and for anything carrying a waveform, a pulse, a picture, a heartbeat, it is the property you care about most.

It is bought with symmetry in the impulse response, and it is not free: the symmetry forces a delay of half the filter length, and a filter with exactly linear phase cannot be causal unless you accept that delay.

This is a short card built around one clean trade, and the trade is what the exam asks for.

Linear phase A system has linear phase if its phase response is a straight line in

ω

, angle $H (e^{j ω}) = - α ω$ . Every frequency is then delayed by the same

α

samples, so the shape of the input survives: no component overtakes another.

\begin{aligned} τ (ω) & = - \frac{d}{d ω} [∠ H (e^{j ω})] \\ linear phase & ⟺ τ (ω) = α, constant at every ω \\ h [n] & = h [M - n] & symmetric, M = N - 1 \\ h [n] & = - h [M - n] & antisymmetric \\ H (e^{j ω}) & = e^{- j ω M / 2} H_{r} (ω), H_{r} real \end{aligned}

The relationship with causality, which is the point of the question

An ideal filter is not causal. The ideal low pass response has zero phase and a brick wall magnitude, and its impulse response sin( $ω_{c}$ n)/( $π$ n) is non zero for every negative n and never ends. It cannot be built, because an output would be needed before its input arrived.
A delay is what buys causality. Shift the ideal response right by $α = (N - 1) / 2$ samples and truncate. The shift is exactly the factor $e^{- j ω a l p h a}$ , which is linear phase. So linear phase is not an extra requirement bolted on: it is the price and the proof of having made the filter causal.
The cost is latency. A causal linear phase FIR filter of length N delays the signal by $(N - 1) / 2$ samples. Longer filter, sharper response, more delay. That trade is why linear phase FIR filters are avoided in a control loop.
A causal IIR filter cannot have exactly linear phase. Symmetry of $h [n]$ about a point demands a finite, symmetric impulse response; an infinite one that is also symmetric would have to extend to n negative, which is non causal. So exact linear phase belongs to FIR filters alone.
Zero phase is possible offline. Filtering forwards and then backwards through the same IIR filter gives $| H |$ ^2 and zero phase, at the cost of needing the whole signal in advance. That is not a causal system, which is the point.

Type	Symmetry of $h [n]$	Length N	Hr has a forced zero at	Cannot build
I	Symmetric	Odd	Nothing	Any filter
II	Symmetric	Even	$ω = π$	High pass, band stop
III	Antisymmetric	Odd	$ω = 0$ and $ω = π$	Low pass, high pass
IV	Antisymmetric	Even	$ω = 0$	Low pass, band stop

Zeros come in quadruples. The symmetry $h [n] = h [M - n]$ forces $H (z) = z^{- M}$ H(1/z), so if z0 is a zero then 1/z0 is too, and for real coefficients so are their conjugates. A zero not on the unit circle therefore arrives with its mirror image outside it, which is why a linear phase FIR filter is never minimum phase unless every zero sits on the circle.

The delay that buys causality here is the $τ$ in every ideal impulse response in chapter 5, and the reason an FIR filter is chosen when phase matters.

See it drawnSymmetry buys linear phase, and costs a delay

A symmetric impulse response has exactly linear phase. Making it causal shifts it right by

(N - 1) / 2

, and that shift IS the delay.

Stability and causality, three ways of saying the same thing PIN 2/19

76 Ash · 72 Ch4+34

Before the formulas

Stability is one idea said three different ways, and the paper will ask for any of them. The impulse response must be absolutely summable. The poles must lie inside the unit circle. The region of convergence must contain the unit circle.

Those are not three facts to memorise, they are the same statement in three languages, and each one is the convenient one for a different kind of question. Knowing they are the same is what lets you answer whichever way it is phrased, and lets you check an answer in a second language for free.

	In terms of $h [n]$	In terms of the ROC	In terms of the poles
Causal	$h [n] = 0$ for n < 0	ROC is the outside of the outermost pole, and includes z = infinity	Number of finite zeros does not exceed the number of finite poles
Stable	sum of $\| h [n] \|$ over all n is finite	ROC contains the unit circle	No condition on its own: it depends which ROC is chosen
Causal and stable	Both of the above	ROC is $\| z \|$ > rmax with rmax < 1	Every pole lies strictly inside the unit circle

Why the ROC statement is the same as the

h [n]

statement Stability means sum of

| h [n] |

is finite. Evaluate the z-transform on the unit circle:

| H (e^{j ω}) |

≤ sum of |

h [n] e^{- j ω n}

| = sum of

| h [n] |

. So the sum converges on

| z | = 1

exactly when the unit circle is in the ROC.

Worked examples, the ones the question asks for

$h [n] = a^{n} u [n]$ . Causal, since $h [n] = 0$ for n < 0. Its ROC is $| z | > | a |$ , which contains the unit circle only if $| a | < 1$ . So it is stable exactly when the pole is inside the unit circle.
$h [n] = - a^{n} u [- n - 1]$ . Anticausal. Same $H (z)$ , ROC $| z | < | a |$ , which contains the unit circle only if $| a | > 1$ . The same algebra, the opposite conclusion, which is the point of always quoting the ROC.
$H (z) = 1 / (1 - 2 z^{- 1})$ with ROC $| z | > 2$ . Causal and unstable: the pole at 2 is outside the unit circle and the ROC does not contain it.
The same $H (z)$ with ROC $| z | < 2$ . Stable and not causal. A stable, anticausal system exists; it just cannot run in real time.

The pole positions this card talks about are exactly what the bilinear transformation guarantees, by mapping the whole left half plane inside the unit circle.

See it drawnStability, said three ways about the same picture

Absolutely summable, the ROC contains the unit circle, every pole inside it: three sentences, one fact.

The BIBO stability test PIN 1/19

81 Ba5

Before the formulas

The definition underneath all three of those: if every bounded input produces a bounded output, the system is stable. Not one input, every input. The test that follows is a single sum, and it is quick. The part worth practising is the other direction.

To prove a system is unstable you have to produce the one bounded input that breaks it, and there is a standard choice that always works. That construction is the answer the paper wants, not an assertion.

BIBO stable A system is bounded input bounded output stable if every input bounded by some finite Mx produces an output bounded by some finite My. For an LTI system this is equivalent to $h [n]$ being absolutely summable.

\begin{aligned} | y [n] | & = | \sum_{k} h [k] x [n - k] | \\ \leq \sum_{k} | h [k] | | x [n - k] | \\ \leq M_{x} \sum_{k} | h [k] | \\ ⟹ y bounded for every bounded x ⟺ \sum_{k} | h [k] | < \infty \end{aligned}

Worked, 2081 Baishakh

y (n) = x (n) + e^{a} y (n - 1)

. This is a first order recursive system with

H (z) = 1

/(1

- e^{a} z^{- 1}

), so the pole is at z =

e^{a}

.
For a causal system, stability needs |

e^{a}

| < 1, that is a < 0.
The impulse response is

h [n]

= (

e^{a}

)^n

u [n]

, and sum

| h [n] | = 1

/(1 - |

e^{a}

|), finite only when |

e^{a}

| < 1. Same conclusion, both ways.

See it drawnBounded in, bounded out, and the one input that breaks it

To prove instability you do not need a general argument: one bounded input whose output grows without bound is enough.

Practise thisEvery question the paper has set on this topic, worked in full1

Practise this2081 Baishakh · Q15 marksA discrete-time LTI system is given by difference equation

y (n) = x (n) + e^{a} y (n - 1)

. Check this system for BIBO stability.

What is different in this one

The coefficient is a symbol, not a number, so the answer is a condition on it rather than a yes or a no.
Every other stability question on the paper hands you numbers. The test is identical; only the last line changes.

Find $h [n]$ first. Put $x [n] = δ [n]$ and run the recursion: $h [0] = 1$ , and each later sample is $e^{a}$ times the one before it.

\begin{aligned} h [0] & = 1 \\ h [1] & = e^{a} \\ h [2] & = {(e^{a})}^{2} \\ h [n] & = {(e^{a})}^{n} u [n] \end{aligned}

Now the BIBO test, which is the same one every time: the system is stable when the impulse response is absolutely summable.

\begin{aligned} S & = \sum | h [n] | for n = 0 to  infinity \\ = \sum | e^{a} |^n \\ a geometric  series  with  ratio | e^{a} | \end{aligned}

A geometric series converges when, and only when, its ratio is below one in magnitude, so the whole question is that one comparison.

a	\| $e^{a}$ \|	the sum
$a = - 1$	0.3679	stable, $S = 1.582$
$a = - 0.1$	0.9048	stable, $S = 10.5083$
$a = 0$	1	not stable, the terms never shrink
$a = 1$	2.7183	not stable, the terms grow

The same answer in the other two languages, which is worth one line in the exam: the pole sits at z = $e^{a}$ , so it is inside the unit circle under exactly the same condition, and the region of convergence $| z | > | e^{a} |$ then contains the unit circle.

The system is BIBO stable if and only if $| e^{a} | < 1$ , that is a < 0, and the sum is then $1 / (1 - e^{a})$ . At $a = 0$ the impulse response is a constant 1 and the system is an accumulator, which is not stable; for a > 0 it grows.

The same questions grouped by how often each topic is set: checking a system for bibo stability

FIR against IIR PIN 1/19

80 Ba4+6

Before the formulas

Before the two design chapters start, the comparison you will use constantly, because chapters 5 and 6 are these two with one each. A filter with no feedback always stops, is always stable, and can have exactly linear phase.

A filter with feedback does far more with far fewer coefficients, and pays for it: it can go unstable, and it cannot have exactly linear phase. Everything in the last two chapters is a consequence of that one difference, so it is worth reading this card before you need it.

	FIR	IIR
Impulse response	Finite length, M+1 samples	Infinite length
Difference equation	Non recursive: output from inputs only	Recursive: output depends on past outputs
Poles	All at the origin, so always stable	Poles anywhere; stable only if all are inside the unit circle
Linear phase	Exactly achievable by making $h [n]$ symmetric	Not achievable exactly; phase is non linear
Order needed for a given specification	High, often 5 to 10 times higher	Low
Computation and memory	More	Less
Quantisation effects	Mild, no feedback to accumulate error	Serious: limit cycles and instability from coefficient rounding
Design from	Direct approximation of the desired response: windows, Remez	An analog prototype: Butterworth, Chebyshev, then a transform
Analog counterpart	None	Yes, which is why the design route goes through one

Which to choose

Choose FIR when linear phase matters: audio, data transmission, image processing, anything where waveform shape must survive. Also when the filter must be guaranteed stable, and when coefficients will be heavily quantised.
Choose IIR when the specification is sharp and the computation budget is small, and when phase distortion is acceptable: control loops, simple audio tone shaping, real time work on a slow processor.

The choice between them, for a given job, is the first card of chapter 5.

See it drawnThe one thing FIR can do that IIR cannot

FIR has no feedback, so its response ends and its phase can be exactly linear. IIR gets the sharper cut for fewer coefficients, and pays in phase.

Practise thisEvery question the paper has set on this topic, worked in full1

What is different in this one

A pole lies on or outside the unit circle, at magnitude 1.6621, so this system is unstable. Say so: it is a mark, and it also tells you the magnitude response will be dominated by that pole.
The poles are a conjugate pair at an angle of 1.297 rad, so the response has a peak near that frequency and the sketch is symmetric about $ω = 0$ .

\begin{aligned} H (z) & = \frac{1 - 1.16 z^{- 1} + 4.58 z^{- 2}}{1 - 0.9 z^{- 1} + 2.7625 z^{- 2}} & from the difference equation \\ = \frac{z^{2} - 1.16 z + 4.58}{z^{2} - 0.9 z + 2.7625} & \times z^{2} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	0.58 + j2.06	2.1401	74.28
zero, o	0.58 - j2.06	2.1401	-74.28
pole, x	0.45 + j1.6	1.6621	74.29
pole, x	0.45 - j1.6	1.6621	-74.29

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 1.6621, so for a causal system the ROC is $| z | > 1.6621$ , which does not contain the unit circle, so the system is unstable.

\begin{aligned} N (e^{j ω}) & = e^{j 2 ω} - 1.16 e^ jw + 4.58 \\ = \cos (2 ω) + jsin (2 ω) - 1.16 cos (ω) - j 1.16 sin (ω) + 4.58 \\ D (e^{j ω}) & = e^{j 2 ω} - 0.9 e^ jw + 2.7625 \\ = \cos (2 ω) + jsin (2 ω) - 0.9 cos (ω) - j 0.9 sin (ω) + 2.7625 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	4.42	2.8625	1.5441	1.5441
$\frac{π}{4}$	3.7598 + j0.1798	2.1261 + j0.3636	1.7322 - j0.2117	1.7451
$\frac{π}{2}$	3.58 - j1.16	1.7625 - j0.9	1.8777 + j0.3007	1.9016
$\frac{3 π}{4}$	5.4002 - j1.8202	3.3989 - j1.6364	1.4992 + j0.1862	1.5107
$π$	6.74	4.6625	1.4456	1.4456
$\frac{5 π}{4}$	5.4002 + j1.8202	3.3989 + j1.6364	1.4992 - j0.1862	1.5107
$\frac{3 π}{2}$	3.58 + j1.16	1.7625 + j0.9	1.8777 - j0.3007	1.9016
$\frac{7 π}{4}$	3.7598 - j0.1798	2.1261 - j0.3636	1.7322 + j0.2117	1.7451
$2 π$	4.42	2.8625	1.5441	1.5441

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	1.5441
$ω = 1.297$ , the angle of the poles	1.2966	2.0367
$ω = \frac{π}{2}$ , the point z = j	1.5708	1.9016
$ω = π$ , the point $z = - 1$	3.1416	1.4456

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response rises to a peak of 2.0368 near $ω = 1.291$ , with $| H |$ = 1.5441 at $ω = 0$ and 1.4456 at $ω = π$ . That makes it a low pass shape.

3.5Last minute recall

Last minute recall, chapter 3

Must memorise

$| H |$ is the product of the distances to the zeros over the product of the distances to the poles, measured from the point $e^{j ω}$ on the unit circle.
Near a pole it peaks, at the angle of the pole. Near a zero it dips, to zero if the zero is on the circle. Anything at the origin affects only the phase.
Multiply above and below by $z^{N}$ before factoring, or the zeros and poles at the origin are lost.
Check both endpoints numerically: H at $z = 1$ and at $z = - 1$ .
Eigenfunction: A $e^{j ω_{0} n}$ in gives A $H (e^{j ω_{0}}) e^{j ω_{0} n}$ out. A sinusoid is scaled by $| H |$ and shifted by the angle of H; the frequency never changes.
The characteristic equation comes from setting the input to zero and trying $y [n] = λ$ ^n; a root outside the unit circle means an unstable system.
$H (z) = B (z) / A (z)$ straight from the difference equation, with the feedback coefficients changing sign as they cross the equals sign.
An FIR filter is always stable. An IIR filter is stable only if every pole is inside the unit circle.

Most repeated in this chapter, in order

Pole zero map and magnitude sketch TOP 17/19, the single most set question on the paper
Output for a complex exponential HOT 5/19
Stability and causality in terms of $h [n]$ and the ROC PIN 2/19, and the difference equation and system function PIN 2/19
BIBO test PIN 1/19, FIR against IIR PIN 1/19, zero input response PIN 1/19

Question 4 is the pole zero sketch in seventeen of the nineteen sittings. It is a sketch, not a computation: the marks are in the map, the ROC statement and a curve whose shape you can justify.

Chapter 4 · 8 hours · 10 marks · questions 5 and 6, every paper

Discrete filter structures

The same $H (z)$ can be wired up in half a dozen ways. They compute the same thing in exact arithmetic and behave very differently in real arithmetic, which is why the paper asks you to draw them. Two questions every paper: a direct or cascade form, and a lattice.

What this chapter is about

Drawing a difference equation as a signal flow graph: adders, multipliers and unit delays.
The four IIR forms: direct form I, direct form II, cascade and parallel.
The lattice: for FIR, for IIR, and the recursions that go both ways between the lattice coefficients and $H (z)$ .
Finite word length: quantisation, rounding against truncation, limit cycles and the dead band.

Where it fits

Chapters 2 and 3 produced $H (z)$ . This chapter builds it.
Chapters 5 and 6 design $H (z)$ ; whatever they produce is realised with one of these structures.
The lattice is the structure of choice for a designed IIR filter, because its stability test is one glance.

What you will learn

4.1 The three elements, and why the structure matters, FIR structures, The frequency sampling structure, The FIR lattice, From lattice coefficients back to the system function
4.2 Direct form I and direct form II, Cascade and parallel forms, The IIR lattice ladder
4.3 Quantisation, limit cycles and the dead band, Scaling, and keeping the arithmetic inside the register, Fixed point number formats
4.4 Last minute recall, chapter 4

How it is examined

One question on a direct or cascade form, 4 to 5 marks, and one on a lattice, 5 to 10 marks.
The lattice question is the harder of the two and it is worth the most: the recursion has to be carried out correctly and the structure drawn.
Marks are for the drawing. Coefficients with no diagram score very little.

4.1FIR filters, and the structures that realize them

The three elements, and why the structure matters

Before the formulas

Chapters 1 to 3 told you what a filter is. This chapter asks how you would actually build one, and the answer is always the same three parts: a delay, a multiplier and an adder. Every structure in this chapter is those three wired in a different pattern.

The obvious question is why bother with more than one pattern for the same filter. Two reasons, and both are real: the patterns differ in how much memory they need, and in how badly they suffer when the coefficients have to be rounded to a finite number of bits.

The exam asks you to draw a structure, so this chapter is drawing. Practise until yours comes out the same every time.

\begin{aligned} ADDER        two arrows in, one out, marked with a circle and a plus \\ MULTIPLIER   an arrow with the constant written beside it \\ UNIT  DELAY a box  marked z^{- 1}, holding  one  sample \\ Every structure in this chapter is built from these three and nothing else. \end{aligned}

Why draw more than one structure for the same filter

In exact arithmetic they are identical. Every structure realises the same $H (z)$ , so on paper the choice is free.
In finite precision they are not. Coefficients get rounded, so the poles move, and different structures move them by different amounts. A direct form with a high order denominator can go unstable from rounding alone; a cascade of second order sections is far less sensitive, because each section's poles depend on only two coefficients.
They cost different amounts. Direct form II uses the fewest delays, which is why it is called the canonic form.
The lattice is the most robust of all to coefficient rounding and gives a stability test you can do by eye.

Transposition. Reverse every arrow, swap adders with branch points, and exchange input with output. The result computes the same $H (z)$ . This is where the transposed direct form II comes from, and it is worth one line in any structures answer.

Everything drawn in this chapter realizes the $H (z)$ that the convolution property defined and that chapter 3 sketched.

See it drawnEvery structure is three elements, wired differently

A delay, a multiplier and an adder. Nothing else appears in this chapter, so a structure question is only ever asking how you connected them.

FIR structures TOP 9/19

82 Bh · 79 Bh · 79 Ba · 78 Bh · 74 Ch · 74 Ash · 73 Shr · 72 Ch · 72 Ka3+72+24

Before the formulas

The simplest structure there is, and the place to start. Feed the signal down a chain of delays, multiply each tap by its coefficient, and add the taps up. Because there is no feedback anywhere, the coefficients on the taps are the impulse response, read straight off the drawing, which is a check you can use in the exam.

Nearly every FIR structure question is this picture or a small variation on it, so get this one right and the rest are rearrangements.

Direct form, or transversal, or tapped delay line. M delays in a row with the $h [k]$ multipliers tapping off them into one adder. There is no feedback, so there is nothing else to draw.
Cascade form. Factor $H (z)$ into second order sections, each ( $b_{0} + b_{1} z^{- 1} + b_{2} z^{- 2}$ ), and chain them.
Linear phase form. If $h [n] = h [M - 1 - n]$ , pair the taps that share a coefficient and add them before the multiply. That halves the number of multipliers, from M to about M/2, at no cost in accuracy. Draw this whenever the question says linear phase.
Frequency sampling form and the lattice are the other two, and the lattice is the one examined.

The symmetry that halves the multipliers is the linear phase condition, and the coefficients themselves come out of the window design in the next chapter.

See it drawnThe direct form is the delay line with the coefficients hanging off it

There is no feedback anywhere, so the numbers on the taps ARE the impulse response. That is the whole reason an FIR filter is easy to reason about.

Practise thisEvery question the paper has set on this topic, worked in full9

Practise this +5 like it2082 Bhadra · Q52+2 marksObtain the direct form I and direct form II realization of the following system:

y [n]

- 0.

75 y [n - 1]

- 0.

25 y [n - 2] = x [n]

+ 0.

5 x [n - 1]

What is different in this one

A feed forward part of order 1 and a feedback part of order 2, so direct form I needs 3 delays and direct form II needs 2.
The two halves are of different order, 1 and 2, so direct form I needs 3 delays and direct form II needs only 2. Saying both numbers is a mark.

Rearrange so that $y [n]$ stands alone. The feedback coefficients change sign as they cross the equals sign, and that is where the marks are lost.

H (z) = \frac{1 + 0.5 z^{- 1}}{1 - 0.75 z^{- 1} - 0.25 z^{- 2}}

\begin{aligned} y [n] & = x [n] + 0.5 x [n - 1] + 0.75 y [n - 1] + 0.25 y [n - 2] \\ feed  forward b_{0} & = 1, b_{1} = 0.5 \\ feedback a_{1} & = - 0.75, so  the  multiplier  is 0.75 \\ a_{2} & = - 0.25, so  the  multiplier  is 0.25 \end{aligned}

Direct form I: the all zero half first, then the all pole half, 1 + 2 = 3 delays.

Direct form II: one delay chain shared by both halves, 2 delays, which is the fewest possible.

Direct form I needs 1 + 2 = 3 delays; direct form II needs max(1, 2) = 2, and is therefore the canonic form. Both realize the same $H (z)$ .

Same as 2082 Bhadra Q52079 Bhadra · Q52+2 marksDraw direct form I and Direct form II realization of the following system.

y [n]

- 0.

25 y [n - 2] + x [n]

+ 0.

4 x [n - 1]

+ 0.

5 x [n - 2]

How this is readThe printed equation has a plus where the equals sign belongs. It reads $y [n]$ - 0. $25 y [n - 2] = x [n]$ + 0. $4 x [n - 1]$ + 0. $5 x [n - 2]$ .

What is different in this one

A feed forward part of order 2 and a feedback part of order 2, so direct form I needs 4 delays and direct form II needs 2.
One of the coefficients is zero. That delay still exists in the chain, it simply has no multiplier hanging off it: leaving it out shortens the structure and changes the filter.

Rearrange so that $y [n]$ stands alone. The feedback coefficients change sign as they cross the equals sign, and that is where the marks are lost.

H (z) = \frac{1 + 0.4 z^{- 1} + 0.5 z^{- 2}}{1 - 0.25 z^{- 2}}

\begin{aligned} y [n] & = x [n] + 0.4 x [n - 1] + 0.5 x [n - 2] + 0.25 y [n - 2] \\ feed  forward b_{0} & = 1, b_{1} = 0.4, b_{2} = 0.5 \\ feedback a_{1} & = 0, so  the  multiplier  is 0 \\ a_{2} & = - 0.25, so  the  multiplier  is 0.25 \end{aligned}

Direct form I: the all zero half first, then the all pole half, 2 + 2 = 4 delays.

Direct form II: one delay chain shared by both halves, 2 delays, which is the fewest possible.

Direct form I needs 2 + 2 = 4 delays; direct form II needs max(2, 2) = 2, and is therefore the canonic form. Both realize the same $H (z)$ .

Same as 2082 Bhadra Q52079 Baishakh · Q64 marksObtain the Direct Form I and Direct Form II realization of the following system:

y [n]

- 0.

75 y [n - 1]

- 0.

25 y [n - 2] = x [n]

+ 0.

5 x [n - 1]

What is different in this one

A feed forward part of order 1 and a feedback part of order 2, so direct form I needs 3 delays and direct form II needs 2.
The two halves are of different order, 1 and 2, so direct form I needs 3 delays and direct form II needs only 2. Saying both numbers is a mark.

Rearrange so that $y [n]$ stands alone. The feedback coefficients change sign as they cross the equals sign, and that is where the marks are lost.

H (z) = \frac{1 + 0.5 z^{- 1}}{1 - 0.75 z^{- 1} - 0.25 z^{- 2}}

\begin{aligned} y [n] & = x [n] + 0.5 x [n - 1] + 0.75 y [n - 1] + 0.25 y [n - 2] \\ feed  forward b_{0} & = 1, b_{1} = 0.5 \\ feedback a_{1} & = - 0.75, so  the  multiplier  is 0.75 \\ a_{2} & = - 0.25, so  the  multiplier  is 0.25 \end{aligned}

Direct form I: the all zero half first, then the all pole half, 1 + 2 = 3 delays.

Direct form II: one delay chain shared by both halves, 2 delays, which is the fewest possible.

Direct form I needs 1 + 2 = 3 delays; direct form II needs max(1, 2) = 2, and is therefore the canonic form. Both realize the same $H (z)$ .

Practise this +2 like it2078 Bhadra · Q63+7 marksThe system function of a filter is

H (z) = 1 + (13 / 24) z^{- 1} + (5 / 8) z^{- 2} + (1 / 3) z^{- 3}

. Draw the Direct Form and Lattice Structure implementation of the above filter.

The direct form

What is different in this one

A feed forward part of order 3 and a feedback part of order 0, so direct form I needs 3 delays and direct form II needs 3.
There is no feedback: this is an FIR filter, so direct form I and direct form II are the same drawing and there is no all pole section.

Rearrange so that $y [n]$ stands alone. The feedback coefficients change sign as they cross the equals sign, and that is where the marks are lost.

H (z) = \frac{1 + 0.5417 z^{- 1} + 0.625 z^{- 2} + 0.3333 z^{- 3}}{1}

\begin{aligned} y [n] & = x [n] + 0.5417 x [n - 1] + 0.625 x [n - 2] + 0.3333 x [n - 3] \\ feed  forward b_{0} & = 1, b_{1} = 0.5417, b_{2} = 0.625, b_{3} = 0.3333 \\ feedback \end{aligned}

Direct form II: one delay chain shared by both halves, 3 delays, which is the fewest possible.

Direct form I needs 3 + 0 = 3 delays; direct form II needs max(3, 0) = 3, and is therefore the canonic form. Both realize the same $H (z)$ .

The rest of this questionThis topic works the part that belongs to it. The other part is worked under The FIR lattice (The lattice structure), where it belongs.

Same as 2082 Bhadra Q52074 Chaitra · Q55 marksObtain the Direct Form I and Direct Form II realization of the following system.

3 y [n] + y [n - 1] + 2 y [n - 4] = 2 x [n] + x [n - 3]

What is different in this one

A feed forward part of order 3 and a feedback part of order 4, so direct form I needs 7 delays and direct form II needs 4.
The leading coefficient is 3, not 1, so the whole equation is divided through by it before anything is drawn. Every structure in this chapter assumes $a_{0} = 1$ .
The two halves are of different order, 3 and 4, so direct form I needs 7 delays and direct form II needs only 4. Saying both numbers is a mark.
One of the coefficients is zero. That delay still exists in the chain, it simply has no multiplier hanging off it: leaving it out shortens the structure and changes the filter.

The leading coefficient is 3, so divide the whole equation through by it first: $a_{0}$ must be 1 before any structure is drawn.

Rearrange so that $y [n]$ stands alone. The feedback coefficients change sign as they cross the equals sign, and that is where the marks are lost.

H (z) = \frac{0.6667 + 0.3333 z^{- 3}}{1 + 0.3333 z^{- 1} + 0.6667 z^{- 4}}

\begin{aligned} y [n] & = 0.6667 x [n] + 0.3333 x [n - 3] - 0.3333 y [n - 1] - 0.6667 y [n - 4] \\ feed  forward b_{0} & = 0.6667, b_{1} = 0, b_{2} = 0, b_{3} = 0.3333 \\ feedback a_{1} & = 0.3333, so  the  multiplier  is - 0.3333 \\ a_{2} & = 0, so  the  multiplier  is 0 \\ a_{3} & = 0, so  the  multiplier  is 0 \\ a_{4} & = 0.6667, so  the  multiplier  is - 0.6667 \end{aligned}

Direct form I: the all zero half first, then the all pole half, 3 + 4 = 7 delays.

Direct form II: one delay chain shared by both halves, 4 delays, which is the fewest possible.

Direct form I needs 3 + 4 = 7 delays; direct form II needs max(3, 4) = 4, and is therefore the canonic form. Both realize the same $H (z)$ .

Same as 2078 Bhadra Q62074 Ashwin · Q53+7 marksDraw the direct form and Lattice structure of a filter with system function

H (z) = 1 + 0.7 z^{- 1} + 1.2 z^{- 2} - z^{- 3}

The direct form

What is different in this one

A feed forward part of order 3 and a feedback part of order 0, so direct form I needs 3 delays and direct form II needs 3.
There is no feedback: this is an FIR filter, so direct form I and direct form II are the same drawing and there is no all pole section.

Rearrange so that $y [n]$ stands alone. The feedback coefficients change sign as they cross the equals sign, and that is where the marks are lost.

H (z) = \frac{1 + 0.7 z^{- 1} + 1.2 z^{- 2} - z^{- 3}}{1}

\begin{aligned} y [n] & = x [n] + 0.7 x [n - 1] + 1.2 x [n - 2] - x [n - 3] \\ feed  forward b_{0} & = 1, b_{1} = 0.7, b_{2} = 1.2, b_{3} = - 1 \\ feedback \end{aligned}

Direct form II: one delay chain shared by both halves, 3 delays, which is the fewest possible.

Direct form I needs 3 + 0 = 3 delays; direct form II needs max(3, 0) = 3, and is therefore the canonic form. Both realize the same $H (z)$ .

The rest of this questionThis topic works the part that belongs to it. The other part is worked under The FIR lattice (The lattice structure), where it belongs.

Same as 2078 Bhadra Q62073 Shrawan · Q53+7 marksThe system function of a filter is

H (z) = 2 + 1.8 z^{- 1} - 1.6 z^{- 2} + z^{- 3}

. Draw the Direct Form and Lattice Structure implementation of the above filter.

The direct form

What is different in this one

A feed forward part of order 3 and a feedback part of order 0, so direct form I needs 3 delays and direct form II needs 3.
There is no feedback: this is an FIR filter, so direct form I and direct form II are the same drawing and there is no all pole section.

Rearrange so that $y [n]$ stands alone. The feedback coefficients change sign as they cross the equals sign, and that is where the marks are lost.

H (z) = \frac{2 + 1.8 z^{- 1} - 1.6 z^{- 2} + z^{- 3}}{1}

\begin{aligned} y [n] & = 2 x [n] + 1.8 x [n - 1] - 1.6 x [n - 2] + x [n - 3] \\ feed  forward b_{0} & = 2, b_{1} = 1.8, b_{2} = - 1.6, b_{3} = 1 \\ feedback \end{aligned}

Direct form II: one delay chain shared by both halves, 3 delays, which is the fewest possible.

Direct form I needs 3 + 0 = 3 delays; direct form II needs max(3, 0) = 3, and is therefore the canonic form. Both realize the same $H (z)$ .

The rest of this questionThis topic works the part that belongs to it. The other part is worked under The FIR lattice (The lattice structure), where it belongs.

Same as 2082 Bhadra Q52072 Chaitra · Q65 marksDetermine the Direct Form I and Direct Form II realization of the following system.

y (n) = - 0.1 y (n - 1) + 0.2 y (n - 2) + 3 x (n) + 3.6 x (n - 2) + 0.6 x (n - 2)

How this is readThe paper prints $x (n - 2)$ twice. The second is $x (n - 1)$ , giving $y (n) = - 0.1 y (n - 1) + 0.2 y (n - 2) + 3 x (n) + 3.6 x (n - 1) + 0.6 x (n - 2)$ .

What is different in this one

A feed forward part of order 2 and a feedback part of order 2, so direct form I needs 4 delays and direct form II needs 2.

Rearrange so that $y [n]$ stands alone. The feedback coefficients change sign as they cross the equals sign, and that is where the marks are lost.

H (z) = \frac{3 + 3.6 z^{- 1} + 0.6 z^{- 2}}{1 + 0.1 z^{- 1} - 0.2 z^{- 2}}

\begin{aligned} y [n] & = 3 x [n] + 3.6 x [n - 1] + 0.6 x [n - 2] - 0.1 y [n - 1] + 0.2 y [n - 2] \\ feed  forward b_{0} & = 3, b_{1} = 3.6, b_{2} = 0.6 \\ feedback a_{1} & = 0.1, so  the  multiplier  is - 0.1 \\ a_{2} & = - 0.2, so  the  multiplier  is 0.2 \end{aligned}

Direct form I: the all zero half first, then the all pole half, 2 + 2 = 4 delays.

Direct form II: one delay chain shared by both halves, 2 delays, which is the fewest possible.

Direct form I needs 2 + 2 = 4 delays; direct form II needs max(2, 2) = 2, and is therefore the canonic form. Both realize the same $H (z)$ .

Same as 2082 Bhadra Q52072 Kartik · Q64 marksDetermine the Direct Form II realization of the following system

y (n) = - 0.1 y (n - 1) + 0.72 y (n - 2) + 0.7 x (n) - 0.252 x (n - 2)

What is different in this one

A feed forward part of order 2 and a feedback part of order 2, so direct form I needs 4 delays and direct form II needs 2.
One of the coefficients is zero. That delay still exists in the chain, it simply has no multiplier hanging off it: leaving it out shortens the structure and changes the filter.

Rearrange so that $y [n]$ stands alone. The feedback coefficients change sign as they cross the equals sign, and that is where the marks are lost.

H (z) = \frac{0.7 - 0.252 z^{- 2}}{1 + 0.1 z^{- 1} - 0.72 z^{- 2}}

\begin{aligned} y [n] & = 0.7 x [n] - 0.252 x [n - 2] - 0.1 y [n - 1] + 0.72 y [n - 2] \\ feed  forward b_{0} & = 0.7, b_{1} = 0, b_{2} = - 0.252 \\ feedback a_{1} & = 0.1, so  the  multiplier  is - 0.1 \\ a_{2} & = - 0.72, so  the  multiplier  is 0.72 \end{aligned}

Direct form I: the all zero half first, then the all pole half, 2 + 2 = 4 delays.

Direct form II: one delay chain shared by both halves, 2 delays, which is the fewest possible.

Direct form I needs 2 + 2 = 4 delays; direct form II needs max(2, 2) = 2, and is therefore the canonic form. Both realize the same $H (z)$ .

The same questions grouped by how often each topic is set: direct form realization

The frequency sampling structure

Before the formulas

The direct form stores the impulse response. This structure stores samples of the frequency response instead, which is what you have if the filter was specified as a set of gains rather than as a set of coefficients.

What falls out is a comb filter, which places zeros evenly around the unit circle, followed by one resonator for each sample, each one cancelling a single zero with a pole and leaving the gain you asked for at that point. It looks exotic and it comes from one line of algebra, which the card sets out.

Instead of storing the impulse response, this structure stores samples of the frequency response, H(k) for $k = 0$ to N-1, and builds the filter from a comb filter followed by a bank of resonators, one per sample.

Start from the interpolation of $H (z)$ through its N frequency samples:

H (z) = \underset{a comb filter, N zeros}{\underset{⏟}{\frac{1 - z^{- N}}{N}}} \times \underset{N resonators in parallel}{\underset{⏟}{\sum_{k = 0}^{N - 1} \frac{H (k)}{1 - W_{N}^{- k} z^{- 1}}}}

The comb puts a zero at every point $ω = 2 π$ k / N; each resonator puts a pole at one of those same points, cancelling that zero and leaving a gain of H(k) there.

How it works. The comb filter 1 $- z^{- N}$ puts a zero at every one of the N points $ω = 2 π$ k/N. Each resonator puts a pole at one of those same points, cancelling that zero and leaving a gain of H(k) there. The response therefore passes exactly through the N samples you specified.
Why it is attractive: if most of the H(k) are zero, as in a narrow band filter, those resonators are simply left out. A filter with 4 non zero samples out of 64 costs 4 resonators instead of 64 taps.
The practical catch. The poles sit on the unit circle, so the structure is only marginally stable: with quantized coefficients a pole can drift outside and the filter can run away. The cure is to move both the zeros and the poles to a radius r slightly less than 1, typically 0.99, which costs a little accuracy and buys stability.
Real coefficients: combine each conjugate pair of resonators, k and N-k, into one second order section, so the arithmetic stays real.
It is a realization of an FIR filter, even though it contains feedback. The poles are cancelled by zeros, so the overall impulse response is still finite.

In the exam

The syllabus lists this among the FIR structures and the design method that goes with it is in chapter 5. No sitting in the nineteen has asked you to draw it, so know what it is, why the comb and the resonators cancel, and the stability caveat, and spend the drawing time on the lattice.

The design method that produces the H(k) this structure stores is section 5.3.

See it drawnA comb puts the zeros, a resonator takes one back

The comb filter places N zeros evenly on the unit circle. Each resonator cancels one of them with a pole, and what is left at that point is the frequency sample you asked for.

The FIR lattice HOT 7/19

81 Ba · 78 Bh · 74 Ch · 74 Ash · 73 Shr · 72 Ch · 72 Ka3+756

Before the formulas

The same FIR filter, wired a different way: built stage by stage instead of tap by tap. Each stage raises the order by one, and the whole stage is described by a single number. It is on the syllabus for two reasons.

It behaves better than the direct form when coefficients are rounded, and the recursion that converts between the two descriptions makes a clean exam question. The one to learn properly is the backward recursion: coefficients in, the single numbers out, one order at a time.

What it is A chain of M identical two input, two output stages, each carrying one reflection coefficient k[m]. Stage m produces a forward output

f [m]

[n] and a backward output

g [m]

[n]. The forward output of the last stage is

y [n]

. The structure realises an all zero, that is FIR, system.

\begin{aligned} f_{0} [n] = g_{0} [n] & = x [n] \\ f_{m} [n] & = f_{m - 1} [n] + k_{m} g_{m - 1} [n - 1] \\ g_{m} [n] & = k_{m} f_{m - 1} [n] + g_{m - 1} [n - 1] \\ y [n] & = f_{M} [n] \end{aligned}

Coefficients to lattice, the backward recursion, which is the exam case. Given A_M(z) = 1 + a_M(1) $z^{- 1}$ + ... + a_M(M) $z^{- M}$ :

\begin{aligned} k_{M} & = a_{M} (M) & the last coefficient, always \\ a_{m - 1} (k) & = \frac{a_{m} (k) - k_{m} a_{m} (m - k)}{1 - k_{m}^{2}} & k = 1 \dots m - 1 \end{aligned}

The procedure for the exam question

Write $H (z)$ with a leading 1. If it is $H (z) = 1$ + (13/24) $z^{- 1}$ + (5/8) $z^{- 2}$ + (1/3) $z^{- 3}$ , then $M = 3$ and $a_{3}$ = (13/24, 5/8, 1/3).
Take the last coefficient as k[M]. Here k3 = 1/3.
Step down one order at a time with the recursion, writing each a[m-1] row in full before taking its last entry as the next k.
Stop at k[1] = a[1](1). Draw the M stages with the k values marked.
Check stability: the system is minimum phase, meaning all its zeros are inside the unit circle, if and only if $| k [m] | < 1$ for every m. When the question says also check whether the system is stable, this one line is the answer.

Worked,

H (z) = 1 + 2 z^{- 1} + z^{- 2}

M = 2

a_{2}

= (2, 1). So k2 = $a_{2}$ (2) = 1.

a_{1}

(1) = [

a_{2}

(1) - k2

a_{2}

(

1)] / (1 - k_{2}^{2}

) = (2 - 2)/(1 - 1) = 0/0, which is undefined: the recursion breaks down exactly because |k2| = 1. That is the signal that the filter has a zero on the unit circle, here a double zero at

z = - 1

, so it is not minimum phase and the lattice does not exist in the usual form. Saying that is worth more than forcing a number.

Worked, three stages from the k's k1 = 1/4, k2 =

\frac{1}{2}

, k3 = 1/3, all zero. Forward recursion:
A1(z) = 1 + (1/4)

z^{- 1}

.
A2:

a_{2}

(1) =

a_{1}

(1) + k2

a_{1}

(1) = 1/4 + (

\frac{1}{2}

)(1/4) = 3/8;

a_{2}

(2) = k2 =

\frac{1}{2}

. A2(z) = 1 + (3/8)

z^{- 1}

+ (

\frac{1}{2}

)

z^{- 2}

.
A3:

a_{3}

(1) =

a_{2}

(1) + k3

a_{2}

(2) = 3/8 + (1/3)(

\frac{1}{2}

) = 13/24;

a_{3}

(2) =

a_{2}

(2) + k3

a_{2}

(1) =

\frac{1}{2}

+ (1/3)(3/8) = 5/8;

a_{3}

(3) = k3 = 1/3.
$H (z) = 1$ + (13/24) $z^{- 1}$ + (5/8) $z^{- 2}$ + (1/3) $z^{- 3}$ , and the FIR coefficients are h = {1, 13/24, 5/8, 1/3}. Notice this is the same filter as the worked example above, run the other way: the paper sets both directions.

The same recursion run on a denominator gives the IIR lattice ladder, and run forwards it gives $H (z)$ back from the K values.

See it drawnOne stage per order, and the cross coupling is K

The upper rail carries the forward error, the lower rail the backward error delayed by one. Each stage raises the order by one, and K is what couples them.

Practise thisEvery question the paper has set on this topic, worked in full7

Practise this +3 like it2081 Baishakh · Q5b5 marksDraw the lattice structure for the given FIR filter and also check whether the system is stable

. H (z) = 1 + (13 / 24) z^{- 1} + (5 / 8) z^{- 2} + (1 / 3) z^{- 3}

What is different in this one

An order 3 filter, so there are 3 reflection coefficients and 3 lattice stages to draw.
Every $| K |$ comes out below 1, so the filter is minimum phase as well as stable.

The last coefficient of the polynomial is the reflection coefficient of that order. Step the order down with the Levinson Durbin recursion and read the next one off.

\begin{aligned} K_{m} & = α_{m} (m) \end{aligned}

α_{m - 1} (k) = \frac{α_{m} (k) - K_{m} α_{m} (m - k)}{1 - K_{m}^{2}}, k = 1 \dots m - 1

Polynomial at this order	Reflection coefficient	$1 - K^{2}$
$A 3 (z) = 1 + 0.5417 z^{- 1} + 0.625 z^{- 2} + 0.3333 z^{- 3}$	K3 = 0.3333	0.8889
$A 2 (z) = 1 + 0.375 z^{- 1} + 0.5 z^{- 2}$	K2 = 0.5	0.75
$A 1 (z) = 1 + 0.25 z^{- 1}$	K1 = 0.25	0.9375

K1 = 0.25, K2 = 0.5, K3 = 0.3333. An FIR filter is stable whatever its coefficients, since its impulse response is finite; here every $| K |$ is below 1, so it is minimum phase, meaning every zero lies inside the unit circle.

Also worked under Direct form realizationPractise this +2 like it2078 Bhadra · Q63+7 marksThe system function of a filter is

H (z) = 1 + (13 / 24) z^{- 1} + (5 / 8) z^{- 2} + (1 / 3) z^{- 3}

. Draw the Direct Form and Lattice Structure implementation of the above filter.

The lattice structure

What is different in this one

An order 3 filter, so there are 3 reflection coefficients and 3 lattice stages to draw.
Every $| K |$ comes out below 1, so the filter is minimum phase as well as stable.

The last coefficient of the polynomial is the reflection coefficient of that order. Step the order down with the Levinson Durbin recursion and read the next one off.

\begin{aligned} K_{m} & = α_{m} (m) \end{aligned}

α_{m - 1} (k) = \frac{α_{m} (k) - K_{m} α_{m} (m - k)}{1 - K_{m}^{2}}, k = 1 \dots m - 1

Polynomial at this order	Reflection coefficient	$1 - K^{2}$
$A 3 (z) = 1 + 0.5417 z^{- 1} + 0.625 z^{- 2} + 0.3333 z^{- 3}$	K3 = 0.3333	0.8889
$A 2 (z) = 1 + 0.375 z^{- 1} + 0.5 z^{- 2}$	K2 = 0.5	0.75
$A 1 (z) = 1 + 0.25 z^{- 1}$	K1 = 0.25	0.9375

The rest of this questionThis topic works the part that belongs to it. The other part is worked under Direct form realization (The direct form), where it belongs.

Same as 2081 Baishakh Q5b2074 Chaitra · Q65 marksDetermine the lattice coefficients coefficients corresponding to the FIR filter with the system function:

H (z) = A 3 (z) = 1 + (52 / 96) z^{- 1} + (25 / 40) z^{- 2} + (1 / 3) z^{- 3}

How this is read52/96 is 13/24 and 25/40 is 5/8, so this is the same filter the other sittings set in lowest terms.

What is different in this one

An order 3 filter, so there are 3 reflection coefficients and 3 lattice stages to draw.
Every $| K |$ comes out below 1, so the filter is minimum phase as well as stable.

The last coefficient of the polynomial is the reflection coefficient of that order. Step the order down with the Levinson Durbin recursion and read the next one off.

\begin{aligned} K_{m} & = α_{m} (m) \end{aligned}

α_{m - 1} (k) = \frac{α_{m} (k) - K_{m} α_{m} (m - k)}{1 - K_{m}^{2}}, k = 1 \dots m - 1

Polynomial at this order	Reflection coefficient	$1 - K^{2}$
$A 3 (z) = 1 + 0.5417 z^{- 1} + 0.625 z^{- 2} + 0.3333 z^{- 3}$	K3 = 0.3333	0.8889
$A 2 (z) = 1 + 0.375 z^{- 1} + 0.5 z^{- 2}$	K2 = 0.5	0.75
$A 1 (z) = 1 + 0.25 z^{- 1}$	K1 = 0.25	0.9375

Also worked under Direct form realizationSame as 2078 Bhadra Q62074 Ashwin · Q53+7 marksDraw the direct form and Lattice structure of a filter with system function

H (z) = 1 + 0.7 z^{- 1} + 1.2 z^{- 2} - z^{- 3}

The lattice structure

What is different in this one

This one has no lattice at all. The step down divides by $1 - K^{2}$ , and here a reflection coefficient comes out with magnitude exactly 1, so that divisor is zero. Reporting the breakdown and saying why IS the answer.

Step the order down as usual and read off the reflection coefficients.

A3(z) = 1 +

0.7 z^{- 1} + 1.2 z^{- 2} - z^{- 3}

K3 =

α_{3} (3)

= -1 1

- K^{2} = 1 - {(- 1)}^{2}

= 0

The recursion stops here: the step down divides by $1 - K^{2}$ , which is zero. That is not an arithmetic accident $. | K | = 1$ means a zero of the filter sits exactly on the unit circle, and a lattice cannot represent it, because the all zero lattice realizes only minimum phase polynomials, whose zeros are strictly inside.

\begin{aligned} zeros  of H (z) : - 0.6183 + j 1.2171, - 0.6183 - j 1.2171, 0.5366 \\ magnitudes:      1.3651,   1.3651,   0.5366 \end{aligned}

The lattice does not exist for this filter. K3 = -1, so $1 - K^{2} = 0$ and the recursion breaks down; the reason is the zeros on the unit circle. The filter is still perfectly stable, being FIR, and it is still realizable in direct form. Say that, show the breakdown, and draw the direct form instead: an answer that reports a genuine breakdown scores, and one that invents a number does not.

The rest of this questionThis topic works the part that belongs to it. The other part is worked under Direct form realization (The direct form), where it belongs.

Also worked under Direct form realizationSame as 2078 Bhadra Q62073 Shrawan · Q53+7 marksThe system function of a filter is

H (z) = 2 + 1.8 z^{- 1} - 1.6 z^{- 2} + z^{- 3}

. Draw the Direct Form and Lattice Structure implementation of the above filter.

The lattice structure

The leading coefficient is 2, and the recursion needs it to be 1. Divide through by it, find the lattice of $1 + 0.9 z^{- 1} - 0.8 z^{- 2} + 0.5 z^{- 3}$ , and put the gain 2 at the output.

What is different in this one

An order 3 filter, so there are 3 reflection coefficients and 3 lattice stages to draw.
One of the reflection coefficients has magnitude 2.6, above 1, so this filter is not minimum phase: at least one zero lies outside the unit circle. It is still perfectly stable, being FIR.

The last coefficient of the polynomial is the reflection coefficient of that order. Step the order down with the Levinson Durbin recursion and read the next one off.

\begin{aligned} K_{m} & = α_{m} (m) \end{aligned}

α_{m - 1} (k) = \frac{α_{m} (k) - K_{m} α_{m} (m - k)}{1 - K_{m}^{2}}, k = 1 \dots m - 1

Polynomial at this order	Reflection coefficient	$1 - K^{2}$
$A 3 (z) = 1 + 0.9 z^{- 1} - 0.8 z^{- 2} + 0.5 z^{- 3}$	K3 = 0.5	0.75
$A 2 (z) = 1 + 1.7333 z^{- 1} - 1.6667 z^{- 2}$	K2 = -1.6667	-1.7778
$A 1 (z) = 1 - 2.6 z^{- 1}$	K1 = -2.6	-5.76

K1 = -2.6, K2 = -1.6667, K3 = 0.5. An FIR filter is stable whatever its coefficients, since its impulse response is finite; here not every $| K |$ is below 1, so it is not minimum phase, meaning every zero lies inside the unit circle.

The rest of this questionThis topic works the part that belongs to it. The other part is worked under Direct form realization (The direct form), where it belongs.

Same as 2081 Baishakh Q5b2072 Chaitra · Q75 marksCompute the lattice coefficients and draw the lattice structure of following FIR system

. H (z) = 1 + 2 z^{- 1} + z^{- 2}

What is different in this one

This one has no lattice at all. The step down divides by $1 - K^{2}$ , and here a reflection coefficient comes out with magnitude exactly 1, so that divisor is zero. Reporting the breakdown and saying why IS the answer.

Step the order down as usual and read off the reflection coefficients.

A2(z) = 1 +

2 z^{- 1} + z^{- 2}

K2 =

α_{2} (2)

= 1 1

- K^{2} = 1 - {(1)}^{2}

= 0

\begin{aligned} zeros  of H (z) : - 1, - 1 \\ magnitudes:      1,   1 \end{aligned}

The lattice does not exist for this filter. K2 = 1, so $1 - K^{2} = 0$ and the recursion breaks down; the reason is the double zero at $z = - 1$ on the unit circle. The filter is still perfectly stable, being FIR, and it is still realizable in direct form. Say that, show the breakdown, and draw the direct form instead: an answer that reports a genuine breakdown scores, and one that invents a number does not.

Same as 2081 Baishakh Q5b2072 Kartik · Q76 marksCompute the lattice coefficients and draw the lattice structure of following FIR system

H (z) = 1 + 2 z^{- 1} - 3 z^{- 2} + 4 z^{- 3}

What is different in this one

An order 3 filter, so there are 3 reflection coefficients and 3 lattice stages to draw.
One of the reflection coefficients has magnitude 4, above 1, so this filter is not minimum phase: at least one zero lies outside the unit circle. It is still perfectly stable, being FIR.

The last coefficient of the polynomial is the reflection coefficient of that order. Step the order down with the Levinson Durbin recursion and read the next one off.

\begin{aligned} K_{m} & = α_{m} (m) \end{aligned}

α_{m - 1} (k) = \frac{α_{m} (k) - K_{m} α_{m} (m - k)}{1 - K_{m}^{2}}, k = 1 \dots m - 1

Polynomial at this order	Reflection coefficient	$1 - K^{2}$
$A 3 (z) = 1 + 2 z^{- 1} - 3 z^{- 2} + 4 z^{- 3}$	K3 = 4	-15
$A 2 (z) = 1 - 0.9333 z^{- 1} + 0.7333 z^{- 2}$	K2 = 0.7333	0.4622
$A 1 (z) = 1 - 0.5385 z^{- 1}$	K1 = -0.5385	0.7101

K1 = -0.5385, K2 = 0.7333, K3 = 4. An FIR filter is stable whatever its coefficients, since its impulse response is finite; here not every $| K |$ is below 1, so it is not minimum phase, meaning every zero lies inside the unit circle.

The same questions grouped by how often each topic is set: the fir lattice

From lattice coefficients back to the system function PIN 2/19

79 Bh · 71 Shr56

Before the formulas

The same recursion, run the other way. You are given the stage numbers and asked for the filter. Start from one, apply the order update once per stage, and after three stages you have a third order polynomial.

That is the answer. There is one place to be careful and the paper knows it: the second term of the update uses the previous polynomial written backwards. Write the reversed polynomial out explicitly every time rather than doing it in your head.

Asked as a 3 stage lattice filter has coefficients k1, k2, k3; obtain the system function. It is the forward recursion, run three times, and the only thing to be careful about is which system the question means.

All zero, that is an FIR lattice: the answer is $H (z)$ = A3(z), and the FIR coefficients are the coefficients of A3.
All pole, that is an IIR lattice: the answer is $H (z) = 1$ /A3(z), with the same A3. The question says which; 2079 Bhadra says all zero and 2071 Shrawan says all pole, with the same k values, which is no accident.
The recursion is A[m](z) = A[m-1](z) + k[m] $z^{- m}$ A[m-1]( $z^{- 1}$ ), started from A0(z) = 1. Write each A[m] out in full before moving on.

Worked, k1 = 1/4, k2 =

\frac{1}{2}

, k3 = 1/3 A1(z) = 1 + 0.25

z^{- 1}

.
A2(z) = A1(z) + 0.5

z^{- 2}

A1(

z^{- 1}) = 1 + 0.25 z^{- 1} + 0.5 z^{- 2} (1 + 0.25 z

) = 1 + 0.375

z^{- 1} + 0.5 z^{- 2}

.
A3(z) = A2(z) + (1/3)

z^{- 3}

A2(

z^{- 1}

) = 1 + (13/24)

z^{- 1}

+ (5/8)

z^{- 2}

+ (1/3)

z^{- 3}

.
All zero:

H (z)

= A3(z), h = {1, 13/24, 5/8, 1/3}.
All pole:

H (z) = 1

/A3(z) = 1/(1 + (13/24)

z^{- 1}

+ (5/8)

z^{- 2}

+ (1/3)

z^{- 3}

See it drawnThe order update, run once per reflection coefficient

Start from A0(z) = 1 and apply the recursion once for each K. After three stages you have a third order polynomial, and that is the answer.

Practise thisEvery question the paper has set on this topic, worked in full2

Practise this +1 like it2079 Bhadra · Q66 marksGiven a 3-stage lattice filter for all zero polynomial with coefficients K1 = 1/4, K2 =

\frac{1}{2}

and K3 = 1/3. Obtain the system function and FIR filter coefficients of this filter.

What is different in this one

3 reflection coefficients, so the recursion runs 3 times and the answer is a polynomial of order 3.
Read the question for which it wants: all zero means $H (z) = A (z)$ and the FIR coefficients are its own; all pole means $H (z) = 1 / A (z)$ and there are no FIR coefficients to give. The arithmetic is identical either way.

Run the order update once per reflection coefficient, starting from A0(z) = 1.

\begin{aligned} Am (z) & = A (m - 1) (z) + K_{m} z^{- m} A (m - 1) (z^{- 1}) \\ where A (m - 1) (z^{- 1}) is  the  same  coefficient  list  REVERSED . \end{aligned}

Order	Polynomial
A0(z)	1
A1(z)	$1 + 0.25 z^{- 1}$
A2(z)	$1 + 0.375 z^{- 1} + 0.5 z^{- 2}$
A3(z)	$1 + 0.5417 z^{- 1} + 0.625 z^{- 2} + 0.3333 z^{- 3}$

Same as 2079 Bhadra Q62071 Shrawan · Q45 marksIf a 3 stage lattice filter for all pole polynomial has coefficients K1 = 1/4, K2 =

\frac{1}{2}

and K3 = 1/3. Obtain the system function of this filter.

What is different in this one

3 reflection coefficients, so the recursion runs 3 times and the answer is a polynomial of order 3.
Read the question for which it wants: all zero means $H (z) = A (z)$ and the FIR coefficients are its own; all pole means $H (z) = 1 / A (z)$ and there are no FIR coefficients to give. The arithmetic is identical either way.

Run the order update once per reflection coefficient, starting from A0(z) = 1.

\begin{aligned} Am (z) & = A (m - 1) (z) + K_{m} z^{- m} A (m - 1) (z^{- 1}) \\ where A (m - 1) (z^{- 1}) is  the  same  coefficient  list  REVERSED . \end{aligned}

Order	Polynomial
A0(z)	1
A1(z)	$1 + 0.25 z^{- 1}$
A2(z)	$1 + 0.375 z^{- 1} + 0.5 z^{- 2}$
A3(z)	$1 + 0.5417 z^{- 1} + 0.625 z^{- 2} + 0.3333 z^{- 3}$

The same questions grouped by how often each topic is set: from lattice coefficients back to $h (z)$

4.2IIR filters, and the structures that realize them

Direct form I and direct form II TOP 9/19

82 Bh · 79 Bh · 79 Ba · 78 Bh · 74 Ch · 74 Ash · 73 Shr · 72 Ch · 72 Ka3+72+24

Before the formulas

Now the IIR structures, which have feedback to draw as well as the forward path. Direct form I writes the difference equation exactly as it stands, which needs one column of delays for the input and another for the output.

Direct form II notices that the two halves are separate LTI blocks, so they can be swapped, and after swapping the two delay columns hold the same numbers, so one column does for both. Same filter, same answer, half the memory.

The word for a structure with the fewest possible delays is canonic, and the paper wants the word and the reason.

H (z) = \frac{b_{0} + b_{1} z^{- 1} + \dots + b_{M} z^{- M}}{1 + a_{1} z^{- 1} + \dots + a_{N} z^{- N}} = \frac{B (z)}{A (z)}

Form	Order of the two sections	Delays
Direct form I	$B (z)$ first, then 1/ $A (z)$ : all zeros, then all poles	M + N, two separate chains
Direct form II	1/ $A (z)$ first, then $B (z)$ , sharing one chain	max(M, N), the fewest possible

The procedure, from a difference equation

Put it in standard form: $y [n]$ alone on the left, everything else on the right, with the coefficient of $y [n]$ made 1. If the equation reads $3 y [n] + y [n - 1] + 2 y [n - 4] = 2 x [n] + x [n - 3]$ , divide through by 3 first.
Read off b and a. The x coefficients are $b_{0}$ , $b_{1}$ , ...; the y coefficients, moved to the right hand side with their signs flipped, are $- a_{1}$ , $- a_{2}$ , ....
Direct form I: draw the feed forward chain of delays with the b multipliers summing into a node, then from that node the feedback chain of delays with the -a multipliers summing back in.
Direct form II: draw one delay chain in the middle. The input plus the feedback taps enters at the top of the chain as $w [n]$ ; the b taps come off the same chain into the output adder.

The two equations of direct form II, worth writing under the diagram

w [n] = x [n] - a_{1} w [n - 1] - a_{2} w [n - 2]

- ... - aN

w [n - N]

y [n] = b_{0} w [n] + b_{1} w [n - 1]

+ ... + bM

w [n - M]

Writing these two lines proves you know why the delays merge, which is the part of the question that separates a 2 from a 4.

Worked, the one set three times

y [n] - 0.75 y [n - 1] - 0.25 y [n - 2] = x [n] + 0.5 x [n - 1]

.
So

b_{0} = 1

b_{1} = 0.5

a_{1} = - 0.75

a_{2} = - 0.25

, and

H (z) = (1 + 0.5 z^{- 1}) / (1 - 0.75 z^{- 1} - 0.25 z^{- 2})

.
Direct form I uses three delays: one on the input side for

x [n - 1]

, two on the output side for

y [n - 1]

and

y [n - 2]

.
Direct form II uses two:

w [n] = x [n] + 0.75 w [n - 1] + 0.25 w [n - 2]

, then

y [n] = w [n] + 0.5 w [n - 1]

. Note the signs: the a coefficients are negative in the standard form, so they enter the structure as positive multipliers here.

In the exam

Normalise $a_{0}$ to 1 first. Forgetting this on $3 y [n]$ + ... = ... makes every coefficient wrong.
Watch the signs of the a's. Moving $y [n - k]$ across the equals sign flips the sign, and that flip is the commonest error in the whole chapter.
Say how many delays each form uses and that direct form II is canonic. Free marks.

A long direct form is exactly what coefficient quantization ruins, which is why cascade sections exist at all.

See it drawnDirect form II is direct form I with one delay column

The two halves of the filter are LTI, so they commute. Swap them and the two delay columns hold the same numbers, so one column does for both.

Practise thisEvery question the paper has set on this topic, worked in full9

Practise this +5 like it2082 Bhadra · Q52+2 marksObtain the direct form I and direct form II realization of the following system:

y [n]

- 0.

75 y [n - 1]

- 0.

25 y [n - 2] = x [n]

+ 0.

5 x [n - 1]

What is different in this one

A feed forward part of order 1 and a feedback part of order 2, so direct form I needs 3 delays and direct form II needs 2.
The two halves are of different order, 1 and 2, so direct form I needs 3 delays and direct form II needs only 2. Saying both numbers is a mark.

Rearrange so that $y [n]$ stands alone. The feedback coefficients change sign as they cross the equals sign, and that is where the marks are lost.

H (z) = \frac{1 + 0.5 z^{- 1}}{1 - 0.75 z^{- 1} - 0.25 z^{- 2}}

\begin{aligned} y [n] & = x [n] + 0.5 x [n - 1] + 0.75 y [n - 1] + 0.25 y [n - 2] \\ feed  forward b_{0} & = 1, b_{1} = 0.5 \\ feedback a_{1} & = - 0.75, so  the  multiplier  is 0.75 \\ a_{2} & = - 0.25, so  the  multiplier  is 0.25 \end{aligned}

Direct form I: the all zero half first, then the all pole half, 1 + 2 = 3 delays.

Direct form II: one delay chain shared by both halves, 2 delays, which is the fewest possible.

Direct form I needs 1 + 2 = 3 delays; direct form II needs max(1, 2) = 2, and is therefore the canonic form. Both realize the same $H (z)$ .

Same as 2082 Bhadra Q52079 Bhadra · Q52+2 marksDraw direct form I and Direct form II realization of the following system.

y [n]

- 0.

25 y [n - 2] + x [n]

+ 0.

4 x [n - 1]

+ 0.

5 x [n - 2]

How this is readThe printed equation has a plus where the equals sign belongs. It reads $y [n]$ - 0. $25 y [n - 2] = x [n]$ + 0. $4 x [n - 1]$ + 0. $5 x [n - 2]$ .

What is different in this one

A feed forward part of order 2 and a feedback part of order 2, so direct form I needs 4 delays and direct form II needs 2.
One of the coefficients is zero. That delay still exists in the chain, it simply has no multiplier hanging off it: leaving it out shortens the structure and changes the filter.

Rearrange so that $y [n]$ stands alone. The feedback coefficients change sign as they cross the equals sign, and that is where the marks are lost.

H (z) = \frac{1 + 0.4 z^{- 1} + 0.5 z^{- 2}}{1 - 0.25 z^{- 2}}

\begin{aligned} y [n] & = x [n] + 0.4 x [n - 1] + 0.5 x [n - 2] + 0.25 y [n - 2] \\ feed  forward b_{0} & = 1, b_{1} = 0.4, b_{2} = 0.5 \\ feedback a_{1} & = 0, so  the  multiplier  is 0 \\ a_{2} & = - 0.25, so  the  multiplier  is 0.25 \end{aligned}

Direct form I: the all zero half first, then the all pole half, 2 + 2 = 4 delays.

Direct form II: one delay chain shared by both halves, 2 delays, which is the fewest possible.

Direct form I needs 2 + 2 = 4 delays; direct form II needs max(2, 2) = 2, and is therefore the canonic form. Both realize the same $H (z)$ .

Same as 2082 Bhadra Q52079 Baishakh · Q64 marksObtain the Direct Form I and Direct Form II realization of the following system:

y [n]

- 0.

75 y [n - 1]

- 0.

25 y [n - 2] = x [n]

+ 0.

5 x [n - 1]

What is different in this one

A feed forward part of order 1 and a feedback part of order 2, so direct form I needs 3 delays and direct form II needs 2.
The two halves are of different order, 1 and 2, so direct form I needs 3 delays and direct form II needs only 2. Saying both numbers is a mark.

Rearrange so that $y [n]$ stands alone. The feedback coefficients change sign as they cross the equals sign, and that is where the marks are lost.

H (z) = \frac{1 + 0.5 z^{- 1}}{1 - 0.75 z^{- 1} - 0.25 z^{- 2}}

\begin{aligned} y [n] & = x [n] + 0.5 x [n - 1] + 0.75 y [n - 1] + 0.25 y [n - 2] \\ feed  forward b_{0} & = 1, b_{1} = 0.5 \\ feedback a_{1} & = - 0.75, so  the  multiplier  is 0.75 \\ a_{2} & = - 0.25, so  the  multiplier  is 0.25 \end{aligned}

Direct form I: the all zero half first, then the all pole half, 1 + 2 = 3 delays.

Direct form II: one delay chain shared by both halves, 2 delays, which is the fewest possible.

Direct form I needs 1 + 2 = 3 delays; direct form II needs max(1, 2) = 2, and is therefore the canonic form. Both realize the same $H (z)$ .

Practise this +2 like it2078 Bhadra · Q63+7 marksThe system function of a filter is

H (z) = 1 + (13 / 24) z^{- 1} + (5 / 8) z^{- 2} + (1 / 3) z^{- 3}

. Draw the Direct Form and Lattice Structure implementation of the above filter.

The direct form

What is different in this one

A feed forward part of order 3 and a feedback part of order 0, so direct form I needs 3 delays and direct form II needs 3.
There is no feedback: this is an FIR filter, so direct form I and direct form II are the same drawing and there is no all pole section.

Rearrange so that $y [n]$ stands alone. The feedback coefficients change sign as they cross the equals sign, and that is where the marks are lost.

H (z) = \frac{1 + 0.5417 z^{- 1} + 0.625 z^{- 2} + 0.3333 z^{- 3}}{1}

\begin{aligned} y [n] & = x [n] + 0.5417 x [n - 1] + 0.625 x [n - 2] + 0.3333 x [n - 3] \\ feed  forward b_{0} & = 1, b_{1} = 0.5417, b_{2} = 0.625, b_{3} = 0.3333 \\ feedback \end{aligned}

Direct form II: one delay chain shared by both halves, 3 delays, which is the fewest possible.

Direct form I needs 3 + 0 = 3 delays; direct form II needs max(3, 0) = 3, and is therefore the canonic form. Both realize the same $H (z)$ .

The rest of this questionThis topic works the part that belongs to it. The other part is worked under The FIR lattice (The lattice structure), where it belongs.

Same as 2082 Bhadra Q52074 Chaitra · Q55 marksObtain the Direct Form I and Direct Form II realization of the following system.

3 y [n] + y [n - 1] + 2 y [n - 4] = 2 x [n] + x [n - 3]

What is different in this one

A feed forward part of order 3 and a feedback part of order 4, so direct form I needs 7 delays and direct form II needs 4.
The leading coefficient is 3, not 1, so the whole equation is divided through by it before anything is drawn. Every structure in this chapter assumes $a_{0} = 1$ .
The two halves are of different order, 3 and 4, so direct form I needs 7 delays and direct form II needs only 4. Saying both numbers is a mark.
One of the coefficients is zero. That delay still exists in the chain, it simply has no multiplier hanging off it: leaving it out shortens the structure and changes the filter.

The leading coefficient is 3, so divide the whole equation through by it first: $a_{0}$ must be 1 before any structure is drawn.

Rearrange so that $y [n]$ stands alone. The feedback coefficients change sign as they cross the equals sign, and that is where the marks are lost.

H (z) = \frac{0.6667 + 0.3333 z^{- 3}}{1 + 0.3333 z^{- 1} + 0.6667 z^{- 4}}

\begin{aligned} y [n] & = 0.6667 x [n] + 0.3333 x [n - 3] - 0.3333 y [n - 1] - 0.6667 y [n - 4] \\ feed  forward b_{0} & = 0.6667, b_{1} = 0, b_{2} = 0, b_{3} = 0.3333 \\ feedback a_{1} & = 0.3333, so  the  multiplier  is - 0.3333 \\ a_{2} & = 0, so  the  multiplier  is 0 \\ a_{3} & = 0, so  the  multiplier  is 0 \\ a_{4} & = 0.6667, so  the  multiplier  is - 0.6667 \end{aligned}

Direct form I: the all zero half first, then the all pole half, 3 + 4 = 7 delays.

Direct form II: one delay chain shared by both halves, 4 delays, which is the fewest possible.

Direct form I needs 3 + 4 = 7 delays; direct form II needs max(3, 4) = 4, and is therefore the canonic form. Both realize the same $H (z)$ .

Same as 2078 Bhadra Q62074 Ashwin · Q53+7 marksDraw the direct form and Lattice structure of a filter with system function

H (z) = 1 + 0.7 z^{- 1} + 1.2 z^{- 2} - z^{- 3}

The direct form

What is different in this one

A feed forward part of order 3 and a feedback part of order 0, so direct form I needs 3 delays and direct form II needs 3.
There is no feedback: this is an FIR filter, so direct form I and direct form II are the same drawing and there is no all pole section.

Rearrange so that $y [n]$ stands alone. The feedback coefficients change sign as they cross the equals sign, and that is where the marks are lost.

H (z) = \frac{1 + 0.7 z^{- 1} + 1.2 z^{- 2} - z^{- 3}}{1}

\begin{aligned} y [n] & = x [n] + 0.7 x [n - 1] + 1.2 x [n - 2] - x [n - 3] \\ feed  forward b_{0} & = 1, b_{1} = 0.7, b_{2} = 1.2, b_{3} = - 1 \\ feedback \end{aligned}

Direct form II: one delay chain shared by both halves, 3 delays, which is the fewest possible.

Direct form I needs 3 + 0 = 3 delays; direct form II needs max(3, 0) = 3, and is therefore the canonic form. Both realize the same $H (z)$ .

The rest of this questionThis topic works the part that belongs to it. The other part is worked under The FIR lattice (The lattice structure), where it belongs.

Same as 2078 Bhadra Q62073 Shrawan · Q53+7 marksThe system function of a filter is

H (z) = 2 + 1.8 z^{- 1} - 1.6 z^{- 2} + z^{- 3}

. Draw the Direct Form and Lattice Structure implementation of the above filter.

The direct form

What is different in this one

A feed forward part of order 3 and a feedback part of order 0, so direct form I needs 3 delays and direct form II needs 3.
There is no feedback: this is an FIR filter, so direct form I and direct form II are the same drawing and there is no all pole section.

Rearrange so that $y [n]$ stands alone. The feedback coefficients change sign as they cross the equals sign, and that is where the marks are lost.

H (z) = \frac{2 + 1.8 z^{- 1} - 1.6 z^{- 2} + z^{- 3}}{1}

\begin{aligned} y [n] & = 2 x [n] + 1.8 x [n - 1] - 1.6 x [n - 2] + x [n - 3] \\ feed  forward b_{0} & = 2, b_{1} = 1.8, b_{2} = - 1.6, b_{3} = 1 \\ feedback \end{aligned}

Direct form II: one delay chain shared by both halves, 3 delays, which is the fewest possible.

Direct form I needs 3 + 0 = 3 delays; direct form II needs max(3, 0) = 3, and is therefore the canonic form. Both realize the same $H (z)$ .

The rest of this questionThis topic works the part that belongs to it. The other part is worked under The FIR lattice (The lattice structure), where it belongs.

Same as 2082 Bhadra Q52072 Chaitra · Q65 marksDetermine the Direct Form I and Direct Form II realization of the following system.

y (n) = - 0.1 y (n - 1) + 0.2 y (n - 2) + 3 x (n) + 3.6 x (n - 2) + 0.6 x (n - 2)

How this is readThe paper prints $x (n - 2)$ twice. The second is $x (n - 1)$ , giving $y (n) = - 0.1 y (n - 1) + 0.2 y (n - 2) + 3 x (n) + 3.6 x (n - 1) + 0.6 x (n - 2)$ .

What is different in this one

A feed forward part of order 2 and a feedback part of order 2, so direct form I needs 4 delays and direct form II needs 2.

Rearrange so that $y [n]$ stands alone. The feedback coefficients change sign as they cross the equals sign, and that is where the marks are lost.

H (z) = \frac{3 + 3.6 z^{- 1} + 0.6 z^{- 2}}{1 + 0.1 z^{- 1} - 0.2 z^{- 2}}

\begin{aligned} y [n] & = 3 x [n] + 3.6 x [n - 1] + 0.6 x [n - 2] - 0.1 y [n - 1] + 0.2 y [n - 2] \\ feed  forward b_{0} & = 3, b_{1} = 3.6, b_{2} = 0.6 \\ feedback a_{1} & = 0.1, so  the  multiplier  is - 0.1 \\ a_{2} & = - 0.2, so  the  multiplier  is 0.2 \end{aligned}

Direct form I: the all zero half first, then the all pole half, 2 + 2 = 4 delays.

Direct form II: one delay chain shared by both halves, 2 delays, which is the fewest possible.

Direct form I needs 2 + 2 = 4 delays; direct form II needs max(2, 2) = 2, and is therefore the canonic form. Both realize the same $H (z)$ .

Same as 2082 Bhadra Q52072 Kartik · Q64 marksDetermine the Direct Form II realization of the following system

y (n) = - 0.1 y (n - 1) + 0.72 y (n - 2) + 0.7 x (n) - 0.252 x (n - 2)

What is different in this one

A feed forward part of order 2 and a feedback part of order 2, so direct form I needs 4 delays and direct form II needs 2.
One of the coefficients is zero. That delay still exists in the chain, it simply has no multiplier hanging off it: leaving it out shortens the structure and changes the filter.

Rearrange so that $y [n]$ stands alone. The feedback coefficients change sign as they cross the equals sign, and that is where the marks are lost.

H (z) = \frac{0.7 - 0.252 z^{- 2}}{1 + 0.1 z^{- 1} - 0.72 z^{- 2}}

\begin{aligned} y [n] & = 0.7 x [n] - 0.252 x [n - 2] - 0.1 y [n - 1] + 0.72 y [n - 2] \\ feed  forward b_{0} & = 0.7, b_{1} = 0, b_{2} = - 0.252 \\ feedback a_{1} & = 0.1, so  the  multiplier  is - 0.1 \\ a_{2} & = - 0.72, so  the  multiplier  is 0.72 \end{aligned}

Direct form I: the all zero half first, then the all pole half, 2 + 2 = 4 delays.

Direct form II: one delay chain shared by both halves, 2 delays, which is the fewest possible.

Direct form I needs 2 + 2 = 4 delays; direct form II needs max(2, 2) = 2, and is therefore the canonic form. Both realize the same $H (z)$ .

The same questions grouped by how often each topic is set: direct form realization

Cascade and parallel forms PIN 3/19

81 Ba · 80 Ba · 76 Ch45

Before the formulas

A large filter is never built as one block. A single long list of coefficients is extremely sensitive to rounding, and one poorly placed bit can move a pole outside the unit circle and make the whole thing unstable.

The cure is to break it into small second order sections. Factorise the transfer function and the sections sit in a chain, which is the cascade form. Expand it by partial fractions instead and the same size sections sit side by side, which is the parallel form.

Both are on the syllabus, and the question is usually a given $H (z)$ and the instruction to draw one of them.

\begin{aligned} cascade H (z) & = b_{0} \prod_{k} H_{k} (z), H_{k} (z) = \frac{1 + b_{1 k} z^{- 1} + b_{2 k} z^{- 2}}{1 + a_{1 k} z^{- 1} + a_{2 k} z^{- 2}} \\ parallel H (z) & = C + \sum_{k} H_{k} (z) & from a partial fraction expansion \end{aligned}

Building the cascade form

Factor both polynomials into first and second order factors.
Pair every complex root with its conjugate in the same section, so that all the coefficients come out real. A section containing only one of a conjugate pair would need complex multipliers and cannot be built.
Pair a numerator factor with a denominator factor to make each biquad, then draw each as a direct form II and put them in series.
Which pole goes with which zero matters in practice, because it decides the internal signal levels and therefore the overflow behaviour. The usual rule is to pair each pole with its nearest zero. The exam accepts any consistent pairing.

Worked pairing, the 2080 Baishakh and 2076 Chaitra pattern The question gives

H (z)

already factored, with conjugate pairs such as (1 - 0.3

e^{j π / 6} z^{- 1}

) and (1 - 0.3

e^{- j π / 6} z^{- 1}

).
Multiply each pair out: (1 - r

e^{j θ} z^{- 1}

)(1 - r

e^{- j θ} z^{- 1}

) = 1 - 2 r cos( $θ$ ) $z^{- 1} + r^{2} z^{- 2}$ . That single identity turns every one of these questions into arithmetic.
For

r = 0.3

θ = \frac{π}{6}

: 1 - 0.6

\cos (30 d e g) z^{- 1} + 0.09 z^{- 2} = 1 - 0.5196 z^{- 1} + 0.09 z^{- 2}

Parallel form. Expand $H (z)$ by partial fractions into a constant plus a sum of first and second order terms, then realise each term as a direct form II and add the outputs. It is the least sensitive to coefficient rounding in the poles, but each section's zeros are implicit, so it is used less.

See it drawnCascade multiplies the sections, parallel adds them

Factorise

H (z)

and you get the cascade. Expand it by partial fractions and you get the parallel form. Both are built from the same second order sections.

Practise thisEvery question the paper has set on this topic, worked in full3

Practise this +2 like it2081 Baishakh · Q5a5 marksDraw the cascaded form structure of

H (z)

= 10(1

- 0.25 z^{- 1}

)(1

- 0.667 z^{- 1}

)(1 +

2 z^{- 1}

)/((1

- 0.75 z^{- 1}

)(1

- 0.125 z^{- 1}

){

1 - (0.5 + j 0.5) z^{- 1}

}{

1 - (0.5 - j 0.5) z^{- 1}

})

What is different in this one

3 zeros and 4 poles, so 2 second order sections in the chain.
2 of the roots are complex, so each must stay with its conjugate: that pairing is the one hard rule, and a section with a complex coefficient left in it cannot be built.
There is an overall gain of 10 to place, and it is usually split across the sections rather than all put at the front, so that no section overflows.

Factor $H (z)$ and group the roots into real second order sections. The one hard rule: a complex root must sit in the same section as its conjugate, or the section has complex coefficients and cannot be built.

\begin{aligned} (1 - r e^{j θ} z^{- 1}) (1 - r e^{- j θ} z^{- 1}) & = 1 - 2 r \cos (θ) z^{- 1} + r^{2} z^{- 2} \end{aligned}

Section	Numerator	Denominator
H1(z)	$1 - 0.917 z^{- 1} + 0.1668 z^{- 2}$	$1 - 0.875 z^{- 1} + 0.0938 z^{- 2}$
H2(z)	$1 + 2 z^{- 1}$	$1 - z^{- 1} + 0.5 z^{- 2}$

Each section is realized in direct form II and the sections are connected in a chain, with the gain 10 in front. In practice, pair each pole with its nearest zero and put the poles closest to the unit circle last, which keeps the internal signal levels down.

Section 1 of 2, in direct form II

Direct form II: one delay chain shared by both halves, 2 delays, which is the fewest possible.

Section 2 of 2, in direct form II

Direct form II: one delay chain shared by both halves, 2 delays, which is the fewest possible.

The chain. The input enters the gain, the gain feeds section 1, and each section feeds the next. Nothing is summed between them: that is the parallel form, not this one.

\begin{aligned} x [n] - \to [10] - \to [section 1] - \to [section 2] - \to y [n] \end{aligned}

2 second order sections in cascade, with an overall gain of 10: $H 1 = (1 - 0.917 z^{- 1} + 0.1668 z^{- 2}) / (1 - 0.875 z^{- 1} + 0.0938 z^{- 2})$ ; $H 2 = (1 + 2 z^{- 1}) / (1 - z^{- 1} + 0.5 z^{- 2})$ . Each one is drawn above with its own coefficients, and they are connected output to input.

Same as 2081 Baishakh Q5a2080 Baishakh · Q64 marksRealize the given system in Cascade Form of 2nd order section flow graph representation.

H (z) = \frac{(1 - 0.4 z^{- 1}) (1 + 0.2 z^{- 1}) (1 - 0.3 e^{j π / 6} z^{- 1}) (1 - 0.3 e^{- j π / 6} z^{- 1})}{(1 - 0.5 e^{j π / 3} z^{- 1}) (1 - 0.5 e^{- j π / 3} z^{- 1}) (1 + 0.7 e^{j π / 4} z^{- 1}) (1 + 0.7 e^{- j π / 4} z^{- 1})}

What is different in this one

4 zeros and 4 poles, so 2 second order sections in the chain.
6 of the roots are complex, so each must stay with its conjugate: that pairing is the one hard rule, and a section with a complex coefficient left in it cannot be built.

\begin{aligned} (1 - r e^{j θ} z^{- 1}) (1 - r e^{- j θ} z^{- 1}) & = 1 - 2 r \cos (θ) z^{- 1} + r^{2} z^{- 2} \end{aligned}

Section	Numerator	Denominator
H1(z)	$1 - 0.2 z^{- 1} - 0.08 z^{- 2}$	$1 - 0.5 z^{- 1} + 0.25 z^{- 2}$
H2(z)	$1 - 0.5196 z^{- 1} + 0.09 z^{- 2}$	$1 + 0.9899 z^{- 1} + 0.49 z^{- 2}$

Each section is realized in direct form II and the sections are connected in a chain, with the gain 1 in front. In practice, pair each pole with its nearest zero and put the poles closest to the unit circle last, which keeps the internal signal levels down.

Section 1 of 2, in direct form II

Direct form II: one delay chain shared by both halves, 2 delays, which is the fewest possible.

Section 2 of 2, in direct form II

Direct form II: one delay chain shared by both halves, 2 delays, which is the fewest possible.

The chain. The input enters the gain, the gain feeds section 1, and each section feeds the next. Nothing is summed between them: that is the parallel form, not this one.

\begin{aligned} x [n] - \to [1] - \to [section 1] - \to [section 2] - \to y [n] \end{aligned}

2 second order sections in cascade, with an overall gain of 1: $H 1 = (1 - 0.2 z^{- 1} - 0.08 z^{- 2}) / (1 - 0.5 z^{- 1} + 0.25 z^{- 2})$ ; $H 2 = (1 - 0.5196 z^{- 1} + 0.09 z^{- 2}) / (1 + 0.9899 z^{- 1} + 0.49 z^{- 2})$ . Each one is drawn above with its own coefficients, and they are connected output to input.

Same as 2081 Baishakh Q5a2076 Chaitra · Q74 marksRealize the given system in Cascade form of 2nd order section in signal flow graph representation.

H (z) = \frac{(1 - 0.5 z^{- 1}) (1 + 0.35 z^{- 1}) (1 - 0.3 e^{j 2 n π / 5} z^{- 1}) (1 - 0.3 e^{- j 2 n π / 5} z^{- 1})}{(1 - 0.6 e^{j n π / 3} z^{- 1}) (1 - 0.6 e^{- j n π / 3} z^{- 1}) (1 + 0.5 e^{j 2 n π / 7} z^{- 1}) (1 + 0.5 e^{- j 2 n π / 7} z^{- 1})}

How this is readThe n in the printed exponents $e^{j 2 n π / 5}$ is a scanning artifact: these are fixed angles, $\frac{2 π}{5}$ , $\frac{π}{3}$ and $\frac{2 π}{7}$ .

What is different in this one

4 zeros and 4 poles, so 2 second order sections in the chain.
6 of the roots are complex, so each must stay with its conjugate: that pairing is the one hard rule, and a section with a complex coefficient left in it cannot be built.

\begin{aligned} (1 - r e^{j θ} z^{- 1}) (1 - r e^{- j θ} z^{- 1}) & = 1 - 2 r \cos (θ) z^{- 1} + r^{2} z^{- 2} \end{aligned}

Section	Numerator	Denominator
H1(z)	$1 - 0.15 z^{- 1} - 0.175 z^{- 2}$	$1 - 0.6 z^{- 1} + 0.36 z^{- 2}$
H2(z)	$1 - 0.1854 z^{- 1} + 0.09 z^{- 2}$	$1 + 0.6235 z^{- 1} + 0.25 z^{- 2}$

Section 1 of 2, in direct form II

Direct form II: one delay chain shared by both halves, 2 delays, which is the fewest possible.

Section 2 of 2, in direct form II

Direct form II: one delay chain shared by both halves, 2 delays, which is the fewest possible.

The chain. The input enters the gain, the gain feeds section 1, and each section feeds the next. Nothing is summed between them: that is the parallel form, not this one.

\begin{aligned} x [n] - \to [1] - \to [section 1] - \to [section 2] - \to y [n] \end{aligned}

2 second order sections in cascade, with an overall gain of 1: $H 1 = (1 - 0.15 z^{- 1} - 0.175 z^{- 2}) / (1 - 0.6 z^{- 1} + 0.36 z^{- 2})$ ; $H 2 = (1 - 0.1854 z^{- 1} + 0.09 z^{- 2}) / (1 + 0.6235 z^{- 1} + 0.25 z^{- 2})$ . Each one is drawn above with its own coefficients, and they are connected output to input.

The same questions grouped by how often each topic is set: cascade form

The IIR lattice ladder TOP 10/19

82 Bh · 82 Ba · 81 Bh · 80 Bh · 80 Ba · 79 Ba · 76 Ch · 76 Ash · 75 Ch · 75 Ash6+4106

Before the formulas

The lattice from the FIR card, extended to a filter that has zeros as well as poles. The denominator builds the lattice exactly as before, so that half is work you have already done. The numerator is then added on top by tapping the backward rail at each stage and weighting the taps, which is the ladder.

It is the most asked structure in the chapter, and it is asked as a drawing with a procedure, so the procedure is the thing to rehearse.

The idea An all pole system, 1/

A (z)

, is realised by running the FIR lattice backwards: the same stages with the forward path reversed. A general pole zero system

B (z) / A (z)

adds a ladder: the backward outputs

g [m]

[n] of every stage are tapped, weighted by ladder coefficients C[m], and summed to form the output.

The denominator $A (z)$ gives the lattice coefficients by exactly the same backward recursion as the FIR case. The numerator $B (z)$ gives the ladder coefficients:

\begin{aligned} C_{M} & = b_{M} \\ C_{m} & = b_{m} - \sum_{i = m + 1}^{M} C_{i} a_{i} (i - m), m = M - 1 \dots 0 \\ y [n] & = \sum_{m = 0}^{M} C_{m} g_{m} [n] \end{aligned}

The procedure

Separate $H (z)$ into $B (z)$ over $A (z)$ , both with a leading 1 in the denominator.
Run the lattice recursion on $A (z)$ to get k[M] down to k[1], keeping every intermediate row a[m](i), because the ladder needs them.
Run the ladder recursion on $B (z)$ from C[M] downwards, using those rows.
Draw: the lattice stages across the middle, the g outputs tapped downwards into the C multipliers, and one summing node producing $y [n]$ .
Check stability: the system is stable if and only if $| k [m] | < 1$ for every m, which is the whole reason this structure is used. No root finding required.

Worked, the one set three times

H (z) = (2 - 0.7 z^{- 1} + 0.5 z^{- 2}) / (1 - 0.3 z^{- 1} + 0.25 z^{- 2})

.
Lattice, from $A (z) = 1 - 0.3 z^{- 1} + 0.25 z^{- 2}$ :

M = 2

a_{2}

= (-0.3, 0.25), so k2 = 0.25.

a_{1}

(1) = [

a_{2}

(1) - k2

a_{2}

(

1)] / (1 - k_{2}^{2}) = (- 0.3 - 0.25 (- 0.3)) / (1 - 0.0625

) = -0.225/0.9375 = -0.24, so k1 = -0.24.
Both

| k | < 1

, so the system is stable.
Ladder, from $B (z) = 2 - 0.7 z^{- 1} + 0.5 z^{- 2}$ , so

b_{0} = 2

b_{1} = - 0.7

b_{2} = 0.5

C_{2} = b_{2}

= 0.5.

C_{1} = b_{1} - C_{2} a_{2}

(1) = -0.7 - 0.5(-0.3) = -0.7 + 0.15 = -0.55.

C_{0} = b_{0}

- [

C_{1} a_{1}

(1) +

C_{2} a_{2}

(2)] = 2 - [(-0.55)(-0.24) + (0.5)(0.25)] = 2 - [0.132 + 0.125] = 1.743.

In the exam

Keep the intermediate rows. Throwing away a[1] and a[2] after finding the k's makes the ladder step impossible and costs half the question.
State the stability conclusion from the k's. It is one line and it is asked explicitly in several papers.
If the question gives $H (z)$ as a product of factors, multiply it out into a single ratio of polynomials first. The recursion needs coefficients, not roots.

The $| K m | < 1$ test here is the stability condition of chapter 3, read off without ever finding a pole.

See it drawnThe lattice makes the poles, the ladder taps make the zeros

The top rail runs forwards and the bottom rail runs backwards through the delays. The denominator

A (z)

fixes the lattice exactly as in the FIR case, and the numerator is then built by tapping the backward rail and adding.

Practise thisEvery question the paper has set on this topic, worked in full10

Practise this +1 like it2082 Bhadra · Q66 marksCompute Lattice ladder coefficients and draw lattice structure for the given system

H (z) = (2 - 0.7 z^{- 1} + 0.5 z^{- 2}) / (1 - 0.3 z^{- 1} + 0.25 z^{- 2})

What is different herea second order denominator, the usual case

What is different in this one

A denominator of order 2, so 2 reflection coefficients and 2 lattice stages.
A numerator of order 2 as well, so the ladder taps $C_{0}$ to $C_{2}$ have to be computed too, from the intermediate polynomials the recursion throws off.

The lattice half carries the poles and comes from the denominator, exactly as in the FIR case. The ladder half carries the zeros and comes from the numerator, using the intermediate polynomials the recursion threw off.

\begin{aligned} A (z) & = 1 - 0.3 z^{- 1} + 0.25 z^{- 2} \\ B (z) & = 2 - 0.7 z^{- 1} + 0.5 z^{- 2} \end{aligned}

Denominator at this order	Reflection coefficient
$A 2 (z) = 1 - 0.3 z^{- 1} + 0.25 z^{- 2}$	K2 = 0.25
$A 1 (z) = 1 - 0.24 z^{- 1}$	K1 = -0.24

\begin{aligned} M \\ CM & = bM  Cm = bm - \sum Ci α_{i} (i - m) m = M - 1 \dots 0 \\ i & = m + 1 \end{aligned}

Ladder tap	b value	Value
$C_{2}$	0.5	0.5
$C_{1}$	-0.7	-0.55
$C_{0}$	2	1.743

K1 = -0.24, K2 = 0.25, and $C_{0} = 1.743$ , $C_{1} = - 0.55$ , $C_{2} = 0.5$ . Every $| K |$ is below 1, so the filter is stable, and that test replaces finding the poles.

Practise this +3 like it2082 Baishakh · Q510 marksA third order low pass filter has a system function

H (z) = (0.2759 + 0.5121 z^{- 1} + 0.5121 z^{- 2} + 0.2759 z^{- 3}) / (1 - 0.0010 z^{- 1} + 0.6546 z^{- 2} - 0.0775 z^{- 3})

. Implement the filter using lattice-ladder structure.

What is different in this one

A denominator of order 3, so 3 reflection coefficients and 3 lattice stages.
A numerator of order 3 as well, so the ladder taps $C_{0}$ to $C_{3}$ have to be computed too, from the intermediate polynomials the recursion throws off.

\begin{aligned} A (z) & = 1 - 0.001 z^{- 1} + 0.6546 z^{- 2} - 0.0775 z^{- 3} \\ B (z) & = 0.2759 + 0.5121 z^{- 1} + 0.5121 z^{- 2} + 0.2759 z^{- 3} \end{aligned}

Denominator at this order	Reflection coefficient
$A 3 (z) = 1 - 0.001 z^{- 1} + 0.6546 z^{- 2} - 0.0775 z^{- 3}$	K3 = -0.0775
$A 2 (z) = 1 + 0.05 z^{- 1} + 0.6585 z^{- 2}$	K2 = 0.6585
$A 1 (z) = 1 + 0.0302 z^{- 1}$	K1 = 0.0302

\begin{aligned} M \\ CM & = bM  Cm = bm - \sum Ci α_{i} (i - m) m = M - 1 \dots 0 \\ i & = m + 1 \end{aligned}

Ladder tap	b value	Value
$C_{3}$	0.2759	0.2759
$C_{2}$	0.5121	0.5124
$C_{1}$	0.5121	0.3059
$C_{0}$	0.2759	-0.0493

K1 = 0.0302, K2 = 0.6585, K3 = -0.0775, and $C_{0} = - 0.0493$ , $C_{1} = 0.3059$ , $C_{2} = 0.5124$ , $C_{3} = 0.2759$ . Every $| K |$ is below 1, so the filter is stable, and that test replaces finding the poles.

Same as 2082 Baishakh Q52081 Bhadra · Q56+4 marksCompute the lattice and ladder coefficients and draw lattice-ladder structure for the given IIR system

H (z) = (1 + z^{- 1} + z^{- 2}) / ((1 + 0.5 z^{- 1}) (1 + 0.3 z^{- 1}) (1 + 0.4 z^{- 1}))

How this is read $(1 + 0.5 z^{- 1}) (1 + 0.3 z^{- 1}) (1 + 0.4 z^{- 1})$ multiplies out to $1 + 1.2 z^{- 1} + 0.47 z^{- 2} + 0.06 z^{- 3}$ .

What is different in this one

A denominator of order 3, so 3 reflection coefficients and 3 lattice stages.
A numerator of order 2 as well, so the ladder taps $C_{0}$ to $C_{2}$ have to be computed too, from the intermediate polynomials the recursion throws off.

\begin{aligned} A (z) & = 1 + 1.2 z^{- 1} + 0.47 z^{- 2} + 0.06 z^{- 3} \\ B (z) & = 1 + z^{- 1} + z^{- 2} \end{aligned}

Denominator at this order	Reflection coefficient
$A 3 (z) = 1 + 1.2 z^{- 1} + 0.47 z^{- 2} + 0.06 z^{- 3}$	K3 = 0.06
$A 2 (z) = 1 + 1.176 z^{- 1} + 0.3994 z^{- 2}$	K2 = 0.3994
$A 1 (z) = 1 + 0.8404 z^{- 1}$	K1 = 0.8404

\begin{aligned} M \\ CM & = bM  Cm = bm - \sum Ci α_{i} (i - m) m = M - 1 \dots 0 \\ i & = m + 1 \end{aligned}

Ladder tap	b value	Value
$C_{2}$	1	1
$C_{1}$	1	-0.176
$C_{0}$	1	0.7485

K1 = 0.8404, K2 = 0.3994, K3 = 0.06, and $C_{0} = 0.7485$ , $C_{1} = - 0.176$ , $C_{2} = 1$ . Every $| K |$ is below 1, so the filter is stable, and that test replaces finding the poles.

Practise this2080 Bhadra · Q510 marksDraw the Lattice structure from the following system function

. H (z) = 1 / (1 - 0.2 z^{- 1} + 0.4 z^{- 2} + 0.6 z^{- 3})

What is different herean all pole system, so the lattice has no ladder taps

What is different in this one

A denominator of order 3, so 3 reflection coefficients and 3 lattice stages.
The numerator is a constant, so this is an all pole filter: the only ladder tap is $C_{0}$ and the answer is the bare lattice.

\begin{aligned} A (z) & = 1 - 0.2 z^{- 1} + 0.4 z^{- 2} + 0.6 z^{- 3} \\ B (z) & = 1 \end{aligned}

Denominator at this order	Reflection coefficient
$A 3 (z) = 1 - 0.2 z^{- 1} + 0.4 z^{- 2} + 0.6 z^{- 3}$	K3 = 0.6
$A 2 (z) = 1 - 0.6875 z^{- 1} + 0.8125 z^{- 2}$	K2 = 0.8125
$A 1 (z) = 1 - 0.3793 z^{- 1}$	K1 = -0.3793

\begin{aligned} M \\ CM & = bM  Cm = bm - \sum Ci α_{i} (i - m) m = M - 1 \dots 0 \\ i & = m + 1 \end{aligned}

Ladder tap	b value	Value
$C_{0}$	1	1

K1 = -0.3793, K2 = 0.8125, K3 = 0.6, and $C_{0} = 1$ . Every $| K |$ is below 1, so the filter is stable, and that test replaces finding the poles.

Same as 2082 Bhadra Q62080 Baishakh · Q56 marksCompute Lattice-ladder coefficients and draw lattice structure for given system

H (z) = (2 - 0.7 z^{- 1} + 0.5 z^{- 2}) / (1 - 0.3 z^{- 1} + 0.25 z^{- 2})

What is different in this one

A denominator of order 2, so 2 reflection coefficients and 2 lattice stages.
A numerator of order 2 as well, so the ladder taps $C_{0}$ to $C_{2}$ have to be computed too, from the intermediate polynomials the recursion throws off.

\begin{aligned} A (z) & = 1 - 0.3 z^{- 1} + 0.25 z^{- 2} \\ B (z) & = 2 - 0.7 z^{- 1} + 0.5 z^{- 2} \end{aligned}

Denominator at this order	Reflection coefficient
$A 2 (z) = 1 - 0.3 z^{- 1} + 0.25 z^{- 2}$	K2 = 0.25
$A 1 (z) = 1 - 0.24 z^{- 1}$	K1 = -0.24

\begin{aligned} M \\ CM & = bM  Cm = bm - \sum Ci α_{i} (i - m) m = M - 1 \dots 0 \\ i & = m + 1 \end{aligned}

Ladder tap	b value	Value
$C_{2}$	0.5	0.5
$C_{1}$	-0.7	-0.55
$C_{0}$	2	1.743

K1 = -0.24, K2 = 0.25, and $C_{0} = 1.743$ , $C_{1} = - 0.55$ , $C_{2} = 0.5$ . Every $| K |$ is below 1, so the filter is stable, and that test replaces finding the poles.

Practise this2079 Baishakh · Q56+1 marksCompute Lattice-ladder coefficients and draw lattice structure for given system

H (z) = (1 - 0.4 z^{- 1} + 0.25 z^{- 2}) / (1 - 0.3 z^{- 1} + 0.5 z^{- 2})

. Also check the stability of given system.

The lattice ladder

What is different in this one

A denominator of order 2, so 2 reflection coefficients and 2 lattice stages.
A numerator of order 2 as well, so the ladder taps $C_{0}$ to $C_{2}$ have to be computed too, from the intermediate polynomials the recursion throws off.

\begin{aligned} A (z) & = 1 - 0.3 z^{- 1} + 0.5 z^{- 2} \\ B (z) & = 1 - 0.4 z^{- 1} + 0.25 z^{- 2} \end{aligned}

Denominator at this order	Reflection coefficient
$A 2 (z) = 1 - 0.3 z^{- 1} + 0.5 z^{- 2}$	K2 = 0.5
$A 1 (z) = 1 - 0.2 z^{- 1}$	K1 = -0.2

\begin{aligned} M \\ CM & = bM  Cm = bm - \sum Ci α_{i} (i - m) m = M - 1 \dots 0 \\ i & = m + 1 \end{aligned}

Ladder tap	b value	Value
$C_{2}$	0.25	0.25
$C_{1}$	-0.4	-0.325
$C_{0}$	1	0.81

K1 = -0.2, K2 = 0.5, and $C_{0} = 0.81$ , $C_{1} = - 0.325$ , $C_{2} = 0.25$ . Every $| K |$ is below 1, so the filter is stable, and that test replaces finding the poles.

The rest of this questionThis topic works the part that belongs to it. The other part is worked under And the stability check, the last mark, where it belongs.

Same as 2082 Baishakh Q52076 Chaitra · Q66+4 marksCompute Lattice and Ladder coefficients and Draw lattice-ladder structure for given IIR system

H (z) = (0.5 - 2 z^{- 1} + 3 z^{- 2}) / (1 - 0.5 z^{- 1} - 0.7 z^{- 2} + 0.3 z^{- 3})

What is different in this one

A denominator of order 3, so 3 reflection coefficients and 3 lattice stages.
A numerator of order 2 as well, so the ladder taps $C_{0}$ to $C_{2}$ have to be computed too, from the intermediate polynomials the recursion throws off.

\begin{aligned} A (z) & = 1 - 0.5 z^{- 1} - 0.7 z^{- 2} + 0.3 z^{- 3} \\ B (z) & = 0.5 - 2 z^{- 1} + 3 z^{- 2} \end{aligned}

Denominator at this order	Reflection coefficient
$A 3 (z) = 1 - 0.5 z^{- 1} - 0.7 z^{- 2} + 0.3 z^{- 3}$	K3 = 0.3
$A 2 (z) = 1 - 0.3187 z^{- 1} - 0.6044 z^{- 2}$	K2 = -0.6044
$A 1 (z) = 1 - 0.8056 z^{- 1}$	K1 = -0.8056

\begin{aligned} M \\ CM & = bM  Cm = bm - \sum Ci α_{i} (i - m) m = M - 1 \dots 0 \\ i & = m + 1 \end{aligned}

Ladder tap	b value	Value
$C_{2}$	3	3
$C_{1}$	-2	-1.044
$C_{0}$	0.5	1.4722

K1 = -0.8056, K2 = -0.6044, K3 = 0.3, and $C_{0} = 1.4722$ , $C_{1} = - 1.044$ , $C_{2} = 3$ . Every $| K |$ is below 1, so the filter is stable, and that test replaces finding the poles.

Same as 2082 Baishakh Q52076 Ashwin · Q56+3 marksCompute Lattice and Ladder coefficients and Draw lattice-ladder structure for given IIR system

H (z) = (0.7 - 1.5 z^{- 1} + 0.5 z^{- 2}) / (1 - 0.5 z^{- 1} - 0.7 z^{- 2} + 0.3 z^{- 3})

What is different in this one

A denominator of order 3, so 3 reflection coefficients and 3 lattice stages.
A numerator of order 2 as well, so the ladder taps $C_{0}$ to $C_{2}$ have to be computed too, from the intermediate polynomials the recursion throws off.

\begin{aligned} A (z) & = 1 - 0.5 z^{- 1} - 0.7 z^{- 2} + 0.3 z^{- 3} \\ B (z) & = 0.7 - 1.5 z^{- 1} + 0.5 z^{- 2} \end{aligned}

Denominator at this order	Reflection coefficient
$A 3 (z) = 1 - 0.5 z^{- 1} - 0.7 z^{- 2} + 0.3 z^{- 3}$	K3 = 0.3
$A 2 (z) = 1 - 0.3187 z^{- 1} - 0.6044 z^{- 2}$	K2 = -0.6044
$A 1 (z) = 1 - 0.8056 z^{- 1}$	K1 = -0.8056

\begin{aligned} M \\ CM & = bM  Cm = bm - \sum Ci α_{i} (i - m) m = M - 1 \dots 0 \\ i & = m + 1 \end{aligned}

Ladder tap	b value	Value
$C_{2}$	0.5	0.5
$C_{1}$	-1.5	-1.3407
$C_{0}$	0.7	-0.0778

K1 = -0.8056, K2 = -0.6044, K3 = 0.3, and $C_{0} = - 0.0778$ , $C_{1} = - 1.3407$ , $C_{2} = 0.5$ . Every $| K |$ is below 1, so the filter is stable, and that test replaces finding the poles.

Practise this2075 Chaitra · Q49 marksDraw the lattice structure from the following system function

H (t) = 1 / (1 + (2 / 3) z^{- 1} + (5 / 8) z^{- 2} + (2 / 3) z^{- 3} + z^{- 4})

What is different herefourth order, so the recursion runs twice more; an all pole system, so the lattice has no ladder taps

How this is readThe paper prints H(t) for $H (z)$ , and the denominator it gives is a palindrome: its last coefficient is exactly 1, so K4 = 1 and the recursion divides by zero. The answer below is right for the printed numbers, and the answer is that the lattice does not exist. A nine mark question built to be unanswerable is worth a second look at your own copy of the paper.

What is different in this one

This one has no lattice at all. A reflection coefficient comes out with magnitude exactly 1, so the step down divides by zero, and the reason is a pole sitting on the unit circle.

Step the order down on the denominator as usual.

A4(z) = 1 +

0.6667 z^{- 1} + 0.625 z^{- 2} + 0.6667 z^{- 3} + z^{- 4}

K4 = 1, so 1

- K^{2}

= 0

The recursion stops $. | K | = 1$ means a pole sits exactly on the unit circle, and the all pole lattice is defined only for | $K_{m}$ | below 1, which is the same statement as the filter being stable.

\begin{aligned} pole                       magnitude \\ 0.4429 + j 0.89661 \\ 0.4429 - j 0.89661 \\ - 0.7762 + j 0.63051 \\ - 0.7762 - j 0.63051 \end{aligned}

The lattice does not exist for this filter. K4 = 1 gives $1 - K^{2} = 0$ , and the reason is a pole of magnitude 1, on the unit circle, so the system is marginally stable at best. Report the breakdown and say what it means: that is the answer, and inventing a number is not.

Practise this2075 Ashwin · Q57+3 marksDraw the Lattice structure from the following system function:

1 / (3 + (39 / 24) Z^{- 1} + (15 / 8) Z^{- 2} + (3 / 9) Z^{- 3})

. And represent 5/8 and -5/8 in sign magnitude, 1's complement and 2's complement format.

How this is readWorked on the coefficients exactly as printed. Be aware of one thing before the exam: the other two are exactly three times the filter every other sitting sets, since 39/24 = 3 x 13/24 and 15/8 = 3 x 5/8, so the last was probably printed 3/3 and read as 3/9 by the scan. On the printed numbers K3 comes out 0.1111; on that reading it would be 0.3333 and the whole answer becomes the familiar one. Work what is on your paper, and say which you are working.

The lattice structure

The denominator does not start at 1, so divide numerator and denominator through by 3 first. Every recursion in this chapter assumes $a_{0} = 1$ .

What is different in this one

A denominator of order 3, so 3 reflection coefficients and 3 lattice stages.
The numerator is a constant, so this is an all pole filter: the only ladder tap is $C_{0}$ and the answer is the bare lattice.

\begin{aligned} A (z) & = 1 + 0.5417 z^{- 1} + 0.625 z^{- 2} + 0.1111 z^{- 3} \\ B (z) & = 0.3333 \end{aligned}

Denominator at this order	Reflection coefficient
$A 3 (z) = 1 + 0.5417 z^{- 1} + 0.625 z^{- 2} + 0.1111 z^{- 3}$	K3 = 0.1111
$A 2 (z) = 1 + 0.4781 z^{- 1} + 0.5719 z^{- 2}$	K2 = 0.5719
$A 1 (z) = 1 + 0.3042 z^{- 1}$	K1 = 0.3042

\begin{aligned} M \\ CM & = bM  Cm = bm - \sum Ci α_{i} (i - m) m = M - 1 \dots 0 \\ i & = m + 1 \end{aligned}

Ladder tap	b value	Value
$C_{0}$	0.3333	0.3333

K1 = 0.3042, K2 = 0.5719, K3 = 0.1111, and $C_{0} = 0.3333$ . Every $| K |$ is below 1, so the filter is stable, and that test replaces finding the poles.

The rest of this questionThis topic works the part that belongs to it. The other part is worked under Fixed point number formats (5/8 and -5/8 in the three formats), where it belongs.

The same questions grouped by how often each topic is set: the iir lattice ladder

4.3Quantization effects, limit cycles and scaling

Quantisation, limit cycles and the dead band PIN 1/19

71 Shr1+1+2+1

Before the formulas

Everything so far has quietly assumed exact arithmetic. Real hardware has a fixed number of bits, so every coefficient and every product gets rounded, and rounding is not the small harmless error it sounds like.

In a filter with feedback the rounding can put back exactly as much as the filter takes away. The output then stops decaying and oscillates for ever, at an amplitude that has nothing to do with the input. That is a limit cycle, and the band of amplitudes where it can happen is the dead band.

This is the chapter's one piece of real engineering, and the paper asks for it in words as often as in numbers.

Why quantisation matters A digital filter runs on a machine with a fixed number of bits, so three things get rounded: the input samples, the coefficients, and the products inside the filter. Each one introduces an error that the ideal analysis does not predict, and in a recursive filter those errors are fed back.

Effect	What it is	What it does
Input quantisation	Rounding the sampled value to b bits	Adds quantisation noise; signal to noise ratio improves by about 6 dB per bit
Coefficient quantisation	Rounding $a_{k}$ and $b_{k}$ to b bits	Moves the poles and zeros, so the response is not the designed one, and the filter can even become unstable
Product roundoff	A b bit times b bit product is 2b bits and must be cut back	Adds noise; in a recursive filter it circulates
Overflow	A sum exceeding the register range	Wraps around in two's complement, which is a large error; prevented by scaling or by saturation arithmetic

Rounding against truncation

	Rounding	Truncation
What it does	Goes to the nearest quantisation level	Drops the extra bits
Error range	-q/2 to +q/2, where q is the step	0 to -q for a positive number in sign magnitude
Mean error	Zero, so it is unbiased	Non zero, a systematic bias
Verdict	Better. Same variance and no bias	Cheaper to implement, and biased

Limit cycles and the dead band A limit cycle is a small, self sustaining oscillation at the output of a recursive filter after the input has gone to zero. It happens because the rounded product feeds back and reproduces itself instead of decaying: the filter behaves as though its pole sat on the unit circle.
The dead band is the range of output amplitudes inside which the filter is trapped during a limit cycle. For a first order filter

y [n] = a y [n - 1] + x [n]

quantised with step q, the dead band is

| y | \leq q / (2 (1 - | a |))

.
FIR filters cannot have limit cycles, because there is no feedback path for the error to circulate in. That is one of the practical arguments for FIR.

Limit cycles need feedback, so they are an IIR problem only: an FIR filter has none, and that is one of the reasons to choose one.

See it drawnRounding is why a filter can hum for ever

Inside the dead band the rounding puts back exactly as much as the pole takes away, so the output stops decaying and oscillates instead. That is a limit cycle, and only truncation or more bits removes it.

Scaling, and keeping the arithmetic inside the register

Before the formulas

The second thing finite hardware does to you, and the more violent one. A fixed point register does not clip when a sum gets too large, it wraps round: a large positive number becomes a large negative one in a single sample.

Inside a recursive filter that one sample can start an oscillation that never stops. Scaling is the cure. Shrink the signal at the input by enough that no adder anywhere inside can ever leave the range, and the question is only how much shrinking is enough.

The condition is one sum over the impulse response to the node you are protecting, and the card gives it.

In fixed point arithmetic every node has a finite range, usually -1 to just under +1. If a sum leaves that range it wraps around, turning a large positive number into a large negative one, and in a recursive filter that single error can start an oscillation that never stops. Scaling is the deliberate reduction of signal level that prevents it.

Let $f [n]$ be the impulse response from the input to the node being protected, with the input bounded by $| x [n] | \leq 1$ .

\begin{aligned} L_{1}, the safe bound s & = \frac{1}{\sum_{n} | f [n] |} & overflow becomes impossible \\ L_{2}, the energy bound s & = \frac{1}{\sqrt{\sum_{n} f^{2} [n]}} & the usual engineering choice \\ L_{\infty}, the peak bound s & = \frac{1}{max_{ω} | F (e^{j ω}) |} & for a narrowband input only \end{aligned}

Where the scaling factor goes. Split it across the sections of a cascade rather than putting it all at the input, so that no section is starved of signal and none overflows. This is why a cascade realization is usually drawn with a gain in front of each biquad.
Saturation arithmetic is the second defence: on overflow, clip to the largest representable value instead of wrapping. The error is then small and does not change sign, which kills overflow limit cycles.
Ordering and pairing. In a cascade, pair each pole with its nearest zero, so the section's gain stays flat, and put the sections with poles closest to the unit circle last, so their large internal signals pass through as little of the filter as possible.
The trade. Scaling down avoids overflow and raises the relative round off noise, since the noise floor does not scale with the signal. Too much scaling is as bad as too little; both show up as a poorer signal to noise ratio at the output.

See it drawnScaling is what keeps a sum inside the register

A fixed point register wraps instead of clipping, so one overflow turns a large positive number into a large negative one. Scale the input down until the worst case sum cannot leave the range.

Fixed point number formats PIN 1/19

75 Ash7+3

Before the formulas

The arithmetic underneath the whole chapter, and the easiest marks in it. Fixed point means the bits are read as a sign followed by a binary fraction: the first bit after the point is a half, then a quarter, then an eighth.

Once you see the number that way the question is mechanical. A positive number looks identical in all three formats on the syllabus. They differ only in how each one writes a negative number, so read the card for the three negative rules and do all three carefully rather than quickly.

Asked as represent 5/8 and -5/8 in sign magnitude, 1's complement and 2's complement format. Work in binary fractions: the bit after the point is $\frac{1}{2}$ , then 1/4, then 1/8.

\begin{aligned} 5 / 8 & = 0.101 in  binary (\frac{1}{2} + 0 + 1 / 8) \\ + 5 / 8 - 5 / 8 \\ SIGN MAGNITUDE       0.101           1.101      flip the sign bit only \\ 1' s COMPLEMENT 0.1011 .010 invert  every  bit \\ 2' s COMPLEMENT 0.1011 .011 invert  every  bit, then  add 1 LSB \end{aligned}

Positive numbers are identical in all three formats. Only the negative representation differs.
Two's complement is what hardware uses, because addition and subtraction use the same adder and there is only one representation of zero. Sign magnitude and one's complement both have a +0 and a -0.
Two's complement overflow wraps, turning a large positive into a large negative, which is why overflow is so damaging in a recursive filter and why saturation arithmetic is preferred.

See it drawnThree formats, and only the negative number differs

A positive fraction is written the same way in all three. The question is always really about how each one writes the minus sign.

Practise thisEvery question the paper has set on this topic, worked in full1

Also worked under The IIR lattice ladderPractise this2075 Ashwin · Q57+3 marksDraw the Lattice structure from the following system function:

1 / (3 + (39 / 24) Z^{- 1} + (15 / 8) Z^{- 2} + (3 / 9) Z^{- 3})

. And represent 5/8 and -5/8 in sign magnitude, 1's complement and 2's complement format.

5/8 and -5/8 in the three formats

With the binary point just after the sign bit the weights are 0.5, 0.25, 0.125 and so on, so 0.625 = 0.101 in binary.

Format	+0.625	-0.625
Sign magnitude	0.101	1.101, only the sign bit changes
One's complement	0.101	1.010, every bit flipped
Two's complement	0.101	1.011, flipped then plus one in the last place

Check the two's complement by reading the sign bit with weight -1: 1.011 = -1 + 0.375 = -0.625, as required.

The rest of this questionThis topic works the part that belongs to it. The other part is worked under The IIR lattice ladder (The lattice structure), where it belongs.

The same questions grouped by how often each topic is set: fixed point number formats

4.4Last minute recall

Last minute recall, chapter 4

Must memorise

Direct form I uses M + N delays; direct form II uses max(M, N) and is therefore canonic. Say the numbers.
The feedback multipliers are $- a_{1}$ , $- a_{2}$ , not $a_{1}$ and $a_{2}$ .
$K_{m}$ = the last coefficient of the polynomial at that order, and the step down divides by 1 $- K_{m}$ ^2.
Am(z) = A(m-1)(z) + $K_{m} z^{- m}$ A(m-1)( $z^{- 1}$ ), the reverse of the coefficient list, is the way back up.
| $K_{m}$ | < 1 for every m if and only if the filter is stable for an all pole lattice, and minimum phase for an all zero one. That test replaces finding the poles.
The ladder taps come from the numerator, top down, using the intermediate polynomials the recursion throws off.
A conjugate pair must stay in one section in a cascade, or the section has complex coefficients.
Rounding beats truncation because its error has zero mean; both have variance $q^{2}$ /12.
Dead band $| y | \leq q / (2 (1 - | a |))$ , and an FIR filter has no limit cycles at all.

Most repeated in this chapter, in order

Lattice ladder coefficients for an IIR system TOP 10/19
Direct form I and II realization TOP 9/19
The FIR lattice HOT 7/19
Cascade form with second order sections PIN 3/19
Lattice coefficients back to $H (z)$ PIN 2/19, quantization and limit cycles PIN 1/19, number formats PIN 1/19

Questions 5 and 6 are both from this chapter in most sittings, and they are drawings. Label every multiplier and every delay: an unlabelled signal flow graph scores nothing however correct its shape.

Chapter 5 · 6 hours · 15 marks · questions 7 and 8, every paper

FIR filter design

Two questions every paper, worth about fourteen marks between them: a design by window or by Kaiser window, and the Remez exchange algorithm. The second of those is pure bookwork, it has been asked in fifteen of nineteen sittings, and it never changes.

What this chapter is about

Why FIR: exactly linear phase, always stable, and what it costs.
The window method: truncate the ideal impulse response, and pay for the truncation with Gibbs' oscillation unless the truncation is tapered.
The Kaiser window: the one window with a knob, and the two formulas that turn a specification into a filter length.
The optimum filter: the alternation theorem and the Remez exchange algorithm.

Where it fits

Chapter 1's multiplication property is the whole theory of the window method: multiplying in time convolves in frequency.
Chapter 3 said only FIR can have exactly linear phase; here is how.
Chapter 4 realises whatever comes out, usually as a linear phase direct form.
Chapter 6 is the other half of filter design, and the paper always asks for a comparison somewhere.

What you will learn

5.1 Why FIR, and when to choose it, The linear phase condition, The ideal impulse responses, The window method, step by step, Gibbs' phenomenon, The fixed windows
5.2 The Kaiser window
5.3 Design by frequency sampling
5.4 The optimum filter and the Remez exchange algorithm
5.5 Last minute recall, chapter 5

How it is examined

One design question, 6 to 10 marks, using either a named window or the Kaiser window from a ripple specification.
One bookwork question, 4 to 9 marks: the Remez exchange algorithm with its flow chart, or Gibbs' phenomenon, or why Kaiser beats the fixed windows.
The bookwork is the easiest fourteen marks on the paper. Learn the flow chart cold.

5.1Design by the window method, and the common windows

Why FIR, and when to choose it PIN 3/19

82 Ba · 80 Ba · 76 Ash2+4+42+62+8

Before the formulas

Chapters 5 and 6 are the two design chapters, and this is the FIR one. Design here means something specific: you are handed a specification, a band to pass and a band to stop and how much error is allowed in each, and you have to produce the actual numbers $h [n]$ .

Before any method, this card is the argument for choosing FIR at all: always stable, exactly linear phase if you want it, easy arithmetic, and no feedback to go wrong. It is paid for with a much higher order than an IIR filter would need for the same sharpness.

The paper asks for that list of advantages and disadvantages directly, so it is marks as well as orientation.

Advantages of the FIR digital filter

Always stable. All the poles sit at the origin, so stability is not something you have to check or protect.
Exactly linear phase is achievable, by making the impulse response symmetric. Every frequency is then delayed by the same amount and the waveform shape survives, so there is no phase distortion.
No feedback, so no accumulation of quantisation error, and no limit cycles.
Relatively easy to design, and efficient to implement: a linear phase filter of length N needs only about N/2 multiplications.
Realisable in hardware or software, and well suited to multirate and adaptive work.

Disadvantages, against IIR

More memory and more computation for the same response: a sharp specification can need a very large number of coefficients, often five to ten times the IIR order.
Some responses are impractical to build as FIR at all.
No analog counterpart, so the mature analog design tables cannot be reused.

In which case do we choose FIR and which IIR

Choose FIR when	Choose IIR when
Linear phase is required: audio, data transmission, biomedical, image work	Phase distortion is acceptable
Guaranteed stability matters	A sharp cutoff is needed at low order
The coefficients will be heavily quantised	Computation and memory are tight
An arbitrary, non standard response is wanted	An analog design already exists to convert

Analog against digital filter, asked as its own comparison

	Analog filter	Digital filter
Operates on	Continuous time signals	Sampled, quantised sequences
Built from	R, L, C and op amps	Adders, multipliers, delays, in software or hardware
Accuracy	Limited by component tolerance, and drifts with temperature and age	Set by word length; perfectly repeatable
Changing the response	Change components	Change coefficients, at run time if wanted
Linear phase	Very hard	Exact, with FIR
Frequency range	Very high frequencies possible	Limited by the sampling rate and the processor
Cost of a sharp filter	Grows fast	Grows with order, but only in arithmetic

The comparison itself is in chapter 3; what this chapter adds is how to actually build the FIR filter the comparison recommends.

See it drawnAn FIR filter has no poles to go wrong, and pays for it in length

Every pole of an FIR filter sits at the origin, so it is stable whatever the coefficients are. An IIR filter buys its sharpness with poles near the circle, and those are what can go unstable.

Practise thisEvery question the paper has set on this topic, worked in full3

Practise this +1 like it2082 Baishakh · Q62+6 marksDifferentiate between analog and digital filter. Design an FIR lowpass digital filter that will have a -3dB cut-off at

30 π

rad/s and an attenuation of 50 dB at

45 π

rad/s. The filter is required to have a linear phase and the system uses a sampling rate of 100 Hz.

How this is readA -3 dB cut off at $30 π$ rad/s with a 100 Hz sampling rate is $ω = \frac{30 π}{100} = 0.3 π$ , and $45 π$ rad/s is $0.45 π$ .

What is different in this one

The stopband asks for 50 dB, and Hamming is the smallest window that reaches it, at 53 dB. Choosing a bigger one is not wrong, it is wasteful: the order roughly doubles.

The window is chosen by the stopband attenuation, never by the transition width: take the smallest window that reaches 50 dB.

Window	Stopband attenuation	Transition width
Rectangular	21 dB	$1.8 π$ / N	too weak
Bartlett	25 dB	$6.1 π$ / N	too weak
Hanning	44 dB	$6.2 π$ / N	too weak
Hamming	53 dB	$6.6 π$ / N	chosen
Blackman	74 dB	$11 π$ / N

Hamming gives 53 dB, which meets the 50 dB asked for, and is the smallest that does.

\begin{aligned} transition  width  dw & = ω_{s} - ω_{p} = 1.4137 - 0.9425 = 0.4712 rad \\ N \geq 6.6 π / dw & = \frac{6.6 π}{0.4712} = 44 \\ round  up, and  take  the  next  ODD  value : N & = 45 \end{aligned}

\begin{aligned} cut  off ω_{c} & = (ω_{p} + ω_{s}) / 2 = (0.9425 + 1.4137) / 2 = 1.1781 rad \\ delay τ & = (N - 1) / 2 = 22 samples \\ ideal  response h_{d} [n] & = \sin (ω_{c} (n - t a u)) / (π (n - τ)) \\ h_{d} [τ] & = ω_{c} / π = 0.375 \\ window ω [n] & = 0.54 - 0.46 \cos (2 π n / (N - 1)) \end{aligned}

n	$h_{d} [n]$	$w [n]$	$h [n] = h_{d} [n] w [n]$
0	0.01023	0.08	0.00082
1	-0.0058	0.08468	-0.00049
2	-0.01592	0.09863	-0.00157
3	-0.00641	0.12157	-0.00078
4	0.0125	0.15302	0.00191
5	0.0173	0.19236	0.00333
6	0	0.23876	0
7	-0.01961	0.29131	-0.00571
8	-0.01608	0.34891	-0.00561
9	0.00937	0.4104	0.00385
10	0.02653	0.47454	0.01259
11	0.01107	0.54	0.00598
12	-0.02251	0.60546	-0.01363

The remaining coefficients follow from the symmetry $h [n] = h [44 - n]$ , which every linear phase filter has.

A Hamming window, length $N = 45$ , cut off $ω_{c} = 1.1781$ rad and group delay 22 samples. The first coefficients are ${0.00082, - 0.00049, - 0.00157, - 0.00078, 0.00191, 0.00333}$ , and $H (z)$ = sum of $h [n] z^{- n}$ over $n = 0$ to 44.

Same as 2080 Bhadra Q72080 Baishakh · Q72+8 marksIn which case do we choose FIR filter and IIR filter? Design a linear phase FIR filter using Kaiser Window to meet the following specifications:

0.99 \leq | H (e^{j ω}) | \leq 1.01

for

0 \leq w \geq 0.016 π

;

| H (e^{j ω}) | \leq 0.01

for

0.08 π \leq ω \leq 2 π

How this is readThe paper prints "0 <= w >= $0.016 π$ " and a stopband running to $2 π$ . Read as a passband up to $0.016 π$ and a stopband from $0.08 π$ , since beyond $π$ the response simply repeats.

What is different in this one

Both ripples are 0.01, so there is no choosing to do: delta is that number.

Take the smaller of the two ripples, delta = 0.01, and turn it into decibels: $A = - 20 \log_{10} (0.01)$ = 40 dB.

\begin{aligned} 21 \leq A \leq 50, so β & = 0.5842 {(A - 21)}^{0.4} + 0.07886 (A - 21) \\ = 0.5842 {(19)}^{0.4} + 0.07886 (19) = 3.3953 \\ transition  dw & = ω_{s} - ω_{p} = 0.2513 - 0.0503 = 0.2011 rad \\ order M \geq (A - 8) / (2.285 dw) \\ = (40 - 8) / (2.285 x 0.2011) = 69.65 \\ round  up : M & = 70, length N = M + 1 = 71 \\ α & = τ = M / 2 = 35, ω_{c} = (ω_{p} + ω_{s}) / 2 = 0.1508 rad \end{aligned}

w [n] = \frac{I_{0} (β \sqrt{1 - {[\frac{2 n - (N - 1)}{N - 1}]}^{2}})}{I_{0} (β)}

I_0 is the modified Bessel function, $I_{0} (x) = \sum_{k}$ ( $x^{2} / 4)^{k} / (k$ !)^2. It converges slowly enough that stopping at four terms is not close enough: at $β = 3.3953$ the four term sum is 6.6236 while the true value is 6.7583, about 2 percent out. The standard text stops there and prints 6.6236. Carry terms until they stop changing the fourth decimal, which here takes about eight.

n	$h_{d} [n]$	$w [n]$	$h [n]$
0	-0.00768	0.14797	-0.00114
1	-0.00857	0.17298	-0.00148
2	-0.00931	0.19923	-0.00186
3	-0.00988	0.22666	-0.00224
4	-0.01026	0.25516	-0.00262
5	-0.01042	0.28466	-0.00297
6	-0.01035	0.31504	-0.00326
7	-0.01003	0.34621	-0.00347
8	-0.00945	0.37803	-0.00357
9	-0.0086	0.4104	-0.00353
10	-0.00748	0.44319	-0.00332
11	-0.0061	0.47625	-0.0029
12	-0.00444	0.50946	-0.00226

The rest follow from $h [n] = h [70 - n]$ .

$A = 40$ dB, $β = 3.3953$ , $N = 71$ , cut off $ω_{c} = 0.1508$ rad, group delay 35 samples. The first coefficients are ${- 0.00114, - 0.00148, - 0.00186, - 0.00224, - 0.00262, - 0.00297}$ .

Practise this2076 Ashwin · Q92+4+4 marksIn which case do we choose FIR filter and IIR filter? Design a Kaiser Window to meet the following specifications:

0.99 \leq | H (e^{j ω}) | \leq 1.01

for

0 \leq w \leq 0.16 π

;

| H (e^{j ω}) | \leq 0.01

for

0.18 π \leq ω \leq 2 π

. Draw the flow chart for Remez-Exchange algorithmWhat is different hereThe transition band is only

0.02 π

wide, which is what makes the filter long. A number in the hundreds is the right answer here, not a mistake.

How this is readThe stopband is printed as $0.18 π$ to $2 π$ , which runs past $π$ . A digital spectrum stops at $π$ , so the stopband is $0.18 π$ to $π$ and it is worked that way.

What is different in this one

Both ripples are 0.01, so there is no choosing to do: delta is that number.
The transition band is only $0.02 π$ wide, which is very narrow. A filter length in the hundreds is the right answer here, not a mistake.

Take the smaller of the two ripples, delta = 0.01, and turn it into decibels: $A = - 20 \log_{10} (0.01)$ = 40 dB.

\begin{aligned} 21 \leq A \leq 50, so β & = 0.5842 {(A - 21)}^{0.4} + 0.07886 (A - 21) \\ = 0.5842 {(19)}^{0.4} + 0.07886 (19) = 3.3953 \\ transition  dw & = ω_{s} - ω_{p} = 0.5655 - 0.5027 = 0.0628 rad \\ order M \geq (A - 8) / (2.285 dw) \\ = (40 - 8) / (2.285 x 0.0628) = 222.89 \\ round  up : M & = 223, length N = M + 1 = 224 \\ α & = τ = M / 2 = 111.5, ω_{c} = (ω_{p} + ω_{s}) / 2 = 0.5341 rad \end{aligned}

M is odd, so $α = 111.5$ is half a sample. That is not a mistake and it does not need fixing: the filter is still exactly linear phase, it just delays by half a sample, and there is no middle tap to treat as a special case. If an integer delay is wanted instead, take the length up to the next odd number, 225, which costs one extra tap and changes nothing else.

w [n] = \frac{I_{0} (β \sqrt{1 - {[\frac{2 n - (N - 1)}{N - 1}]}^{2}})}{I_{0} (β)}

n	$h_{d} [n]$	$w [n]$	$h [n]$
0	0.0004	0.14797	0.00006
1	0.0018	0.15568	0.00028
2	0.00272	0.16352	0.00044
3	0.00289	0.17149	0.0005
4	0.00225	0.17957	0.0004
5	0.00097	0.18778	0.00018
6	-0.00061	0.19611	-0.00012
7	-0.00205	0.20455	-0.00042
8	-0.00294	0.21311	-0.00063
9	-0.00302	0.22178	-0.00067
10	-0.00225	0.23056	-0.00052
11	-0.00084	0.23944	-0.0002
12	0.00084	0.24842	0.00021

The rest follow from $h [n] = h [223 - n]$ .

$A = 40$ dB, $β = 3.3953$ , $N = 224$ , cut off $ω_{c} = 0.5341$ rad, group delay 111.5 samples. The first coefficients are ${0.00006, 0.00028, 0.00044, 0.0005, 0.0004, 0.00018}$ .

The linear phase condition PIN 1/19

80 Bh10

Before the formulas

The main reason to choose FIR, stated precisely enough to use. If the coefficients are symmetric about the middle of the filter, the phase comes out as a straight line, which means every frequency is delayed by the same amount and the shape of the signal survives.

A good half of the design work in this chapter exists to preserve that symmetry, which is why it comes so early. The condition is one line, the delay is half the filter length, and both are quotable results.

Linear phase The phase response is a straight line in

ω

, so the group delay

τ

= -d

θ

/dw is constant: every frequency component is delayed by the same number of samples. The filter delays the signal without changing its shape.

An FIR filter of length M has linear phase if

\begin{aligned} h [n] & = h [M - 1 - n] & symmetric \\ h [n] & = - h [M - 1 - n] & antisymmetric \end{aligned}

and the constant group delay is then $τ$ = (M-1)/2 samples.

Why symmetry gives linear phase. Group the terms of $H (e^{j ω})$ = sum $h [n] e^{- j ω n}$ in pairs, n with M-1-n. Each pair contributes $h [n]$ ( $e^{- j ω n} + e^{- j ω (M - 1 - n)}$ ), and factoring out $e^{- j ω (M - 1) / 2}$ turns the bracket into 2 $\cos (ω (M - 1) / 2 - w n)$ , which is real. So the whole response is a real function times $e^{- j ω (M - 1) / 2}$ , and the phase is exactly $- ω$ (M-1)/2: a straight line.

Type	Symmetry	Length M	Can realise	Cannot
I	Symmetric	Odd	Low pass, high pass, band pass, band stop
II	Symmetric	Even	Low pass, band pass	High pass and band stop: H must be zero at $ω = π$
III	Antisymmetric	Odd	Band pass, differentiator, Hilbert transformer	Low pass and high pass: H is zero at $ω = 0$ and $ω = π$
IV	Antisymmetric	Even	High pass, differentiator, Hilbert	Low pass and band stop: H is zero at $ω = 0$

The practical consequence: for an ordinary low pass design use type I, an odd length with symmetric coefficients. That is what the window method produces, and it is why the length is nearly always rounded up to the next odd number.

Why the symmetry is forced on you, rather than chosen, is the causality argument in chapter 3.

See it drawnSymmetry in the coefficients IS the straight line phase

Fold the coefficients about the middle of the filter and they land on themselves. That symmetry is what makes the phase a straight line, and the slope of that line is the delay every frequency gets.

Practise thisEvery question the paper has set on this topic, worked in full1

Practise this2080 Bhadra · Q610 marksDesign the symmetric FIR Low Pass Filter (LPF) for which the desired frequency response is expressed as Hd(W) =

e^{- j W t a u}

for

| W | \leq W c

and 0 elsewhere. The length of the filter should be 7 and

Ω_{c} = 1

rad/sample. Make use of the Hanning window.What is different herethe window and the length are given, so the choosing step is skipped

How this is readHd(W) = $e^{- j W t a u}$ for $| W | \leq W c$ is the ideal low pass with the delay already in it, so $h_{d} [n]$ is the shifted sinc. The question fixes $N = 7$ , $Ω_{c} = 1$ rad per sample and the Hanning window.

What is different in this one

The stopband asks for 44 dB, and Hanning is the smallest window that reaches it, at 44 dB. Choosing a bigger one is not wrong, it is wasteful: the order roughly doubles.
The question fixes the length at $N = 7$ , so there is no order calculation: go straight to the cut off and the coefficients.
The question names the window, so the choosing step is skipped. Most of these do not, and picking the smallest window that reaches the attenuation is the first mark.
The cut off is given as 1 rad per sample rather than derived from the band edges.

The question names the Hanning window, so there is no choosing to do.

The question fixes the length at $N = 7$ .

\begin{aligned} cut  off ω_{c} & = 1 rad, given  by  the  question \\ delay τ & = (N - 1) / 2 = 3 samples \\ ideal  response h_{d} [n] & = \sin (ω_{c} (n - t a u)) / (π (n - τ)) \\ h_{d} [τ] & = ω_{c} / π = 0.3183 \\ window ω [n] & = 0.5 - 0.5 \cos (2 π n / (N - 1)) \end{aligned}

n	$h_{d} [n]$	$w [n]$	$h [n] = h_{d} [n] w [n]$
0	0.01497	0	0
1	0.14472	0.25	0.03618
2	0.26785	0.75	0.20089
3	0.31831	1	0.31831
4	0.26785	0.75	0.20089
5	0.14472	0.25	0.03618
6	0.01497	0	0

A Hanning window, length $N = 7$ , cut off $ω_{c} = 1$ rad and group delay 3 samples. The first coefficients are ${0, 0.03618, 0.20089, 0.31831, 0.20089, 0.03618}$ , and $H (z)$ = sum of $h [n] z^{- n}$ over $n = 0$ to 6.

The same questions grouped by how often each topic is set: fir design by windowing

The ideal impulse responses

Before the formulas

Every design in this chapter starts from the filter you wish you had: a brick wall that passes one band perfectly, stops the rest dead, and does it with no transition at all. Transform that wish back into time and you get a sinc: infinitely long, and starting before time zero.

It cannot be built, and that is not a failure of the method, it is the starting point of it. Everything that follows in the chapter is a different way of cutting that ideal response down to something you can actually build, so derive the low pass once and the other three follow the same way.

Every window design starts by writing down the impulse response of the ideal filter you want, which comes from the inverse DTFT of a rectangle. Derive the low pass once and the others follow.

\begin{aligned} h_{d} [n] & = \frac{1}{2 π} \int_{- π}^{π} H_{d} (e^{j ω}) e^{j ω n} d ω \\ = \frac{1}{2 π} \int_{- ω_{c}}^{ω_{c}} e^{- j ω α} e^{j ω n} d ω & ideal low pass, α = \frac{M - 1}{2} \\ = \frac{\sin (ω_{c} (n - α))}{π (n - α)}, h_{d} [α] = \frac{ω_{c}}{π} \end{aligned}

Ideal filter	$h_{d} [n]$ for n not equal to $α$	hd[ $α$ ]
Low pass, cutoff $ω_{c}$	sin( $ω_{c}$ (n- $α$ )) / ( $π$ (n- $α$ ))	$ω_{c} / π$
High pass, cutoff $ω_{c}$	-sin( $ω_{c}$ (n- $α$ )) / ( $π$ (n- $α$ ))	1 $- ω_{c} / π$
Band pass, wc1 to wc2	[ $\sin (w c 2 (n - α)) - \sin (w c 1 (n - α))$ ] / ( $π$ (n- $α$ ))	(wc2 - wc1)/ $π$
Band stop, wc1 to wc2	[ $\sin (w c 1 (n - α)) - \sin (w c 2 (n - α))$ ] / ( $π$ (n- $α$ ))	1 - (wc2 - wc1)/ $π$

Every one of these is infinitely long and non causal, which is precisely the problem the window solves.
Notice the pattern: high pass is all pass minus low pass, band stop is all pass minus band pass. If you remember the low pass you can rebuild the rest.
hd is symmetric about $α$ , so the design automatically has linear phase.

These come from the inverse DTFT, and the $τ$ in every one of them is the delay that makes the filter causal.

See it drawnThe filter you wish you had, and why you cannot have it

A perfect brick wall in frequency is a sinc in time: infinitely long, and starting before time zero. Every method in this chapter is a way of cutting it down to something buildable.

The window method, step by step TOP 11/19

82 Bh · 82 Ba · 81 Bh · 81 Ba · 80 Bh · 79 Ba · 76 Ch · 74 Ash · 73 Shr · 72 Ch · 71 Shr1062+6

Before the formulas

The first method and the most asked. The idea is exactly what it sounds like: take the ideal response from the previous card, keep a finite piece of it, and throw the rest away. Keeping a piece is multiplying by a window.

Chopping it off square is itself a choice, and a bad one, so the window is shaped to taper instead. The one thing worth understanding rather than memorising: multiplying in time is smearing in frequency, and that smearing is what turns your brick wall into a transition band. Once that clicks, every line of the five step procedure has a reason behind it.

The idea in one line Take the ideal impulse response, which is infinite, and multiply it by a finite length window

w [n]

. Multiplication in time is convolution in frequency, so the sharp ideal response gets smeared by the window's spectrum: the transition band comes from the width of the window's main lobe and the stopband ripple comes from its side lobes.

\begin{aligned} h [n] & = h_{d} [n] w [n] \\ H (e^{j ω}) & = \frac{1}{2 π} \int_{- π}^{π} H_{d} (e^{j θ}) W (e^{j (ω - θ)}) d θ \end{aligned}

A wider main lobe widens the transition band; lower side lobes improve the stopband, and for a given length the two always trade against each other.

The procedure

Read the specification and get $ω_{p}$ , $ω_{s}$ , and the attenuations. If the frequencies are in Hz, convert: $ω = 2 π$ f / Fs.
Cutoff: put $ω_{c}$ in the middle of the transition band, $ω_{c}$ = ( $ω_{p} + ω_{s}$ )/2.
Transition width: delta $ω = ω_{s} - ω_{p}$ .
Choose the window from the required stopband attenuation, using the table below. Pick the narrowest main lobe that meets the attenuation, because a wider window than necessary means a longer filter.
Get the length M from the window's transition width entry, for example M = $6.6 π$ / delta $ω$ for Hamming. Round up, and to an odd number for a type I filter.
Set $α$ = (M-1)/2 and evaluate $h_{d} [n]$ for $n = 0$ to M-1.
Evaluate $w [n]$ over the same range and multiply: $h [n] = h_{d} [n] w [n]$ .
State the answer as the list of $h [n]$ , note its symmetry, and draw the linear phase structure if asked.

Worked, 2082 Bhadra Passband edge 2 kHz, stopband edge 5 kHz, stopband attenuation 42 dB, sampling 20 kHz.

ω_{p} = 2 π

(2000)/20000 = 0.

2 π

ω_{s} = 2 π

(5000)/20000 =

0.5 π

.
delta

ω = 0

π

, and

ω_{c}

= (0.

2 π + 0.5 π

)/2 =

0.35 π

.
42 dB of stopband attenuation rules out rectangular (21 dB) and Bartlett (25 dB); the Hanning window gives 44 dB, which is the narrowest window that clears it.
For Hanning, M = 6.

2 π

/ delta

ω = 6

2 π / 0.3 π = 20.7

, so take $M = 21$ , odd, and

α = 10

.
Then

h [n]

= [

\sin (0.35 π (n - 10))

π

(n-10))] . [0.5 - 0.5 cos(

2 π

n/20)] for

n = 0

..20, with

h [10] = 0.35

In the exam

Convert to digital frequency first. Every Hz in the question must become radians per sample before anything else happens.
Say why you chose that window, naming its attenuation. That sentence is a mark on its own and it is the step the question is really testing.
Round M up and make it odd. Rounding down silently fails the specification.
Use the symmetry. You only need to compute $h [0]$ to $h [a l p h a]$ ; the rest are mirrored, and saying so saves half the arithmetic.

The whole method is the multiplication property of the DTFT: multiplying by a window in time convolves the ideal response with the window spectrum in frequency. Gibbs' phenomenon and every row of the window table follow from that one line.

See it drawnKeep a piece of the ideal response, and shape how you cut it

Multiplying in time is smearing in frequency. That smearing is exactly what turns the brick wall into a transition band, which is why the shape of the window decides the shape of the response.

Practise thisEvery question the paper has set on this topic, worked in full8

Practise this2082 Bhadra · Q710 marksDesign a linear phase FIR system to meet the following specifications: Passband edge = 2 kHz, Stopband edge = 5 kHz, Stopband attenuation = 42 dB, Sampling frequency = 20 kHzWhat is different herethe band edges arrive in hertz, so they are converted to radians per sample first

How this is read $ω = 2 π$ f / fs turns the edges into $2 π$ (2000)/20000 = $0.2 π$ and $2 π$ (5000)/20000 = $0.5 π$ radians per sample.

What is different in this one

The stopband asks for 42 dB, and Hanning is the smallest window that reaches it, at 44 dB. Choosing a bigger one is not wrong, it is wasteful: the order roughly doubles.

Convert the edges to radians per sample with $ω = 2 π$ f / fs, with fs = 20000 Hz.

The window is chosen by the stopband attenuation, never by the transition width: take the smallest window that reaches 42 dB.

Window	Stopband attenuation	Transition width
Rectangular	21 dB	$1.8 π$ / N	too weak
Bartlett	25 dB	$6.1 π$ / N	too weak
Hanning	44 dB	$6.2 π$ / N	chosen
Hamming	53 dB	$6.6 π$ / N
Blackman	74 dB	$11 π$ / N

Hanning gives 44 dB, which meets the 42 dB asked for, and is the smallest that does.

\begin{aligned} transition  width  dw & = ω_{s} - ω_{p} = 1.5708 - 0.6283 = 0.9425 rad \\ N \geq 6.2 π / dw & = \frac{6.2 π}{0.9425} = 20.67 \\ round  up, and  take  the  next  ODD  value : N & = 21 \end{aligned}

\begin{aligned} cut  off ω_{c} & = (ω_{p} + ω_{s}) / 2 = (0.6283 + 1.5708) / 2 = 1.0996 rad \\ delay τ & = (N - 1) / 2 = 10 samples \\ ideal  response h_{d} [n] & = \sin (ω_{c} (n - t a u)) / (π (n - τ)) \\ h_{d} [τ] & = ω_{c} / π = 0.35 \\ window ω [n] & = 0.5 - 0.5 \cos (2 π n / (N - 1)) \end{aligned}

n	$h_{d} [n]$	$w [n]$	$h [n] = h_{d} [n] w [n]$
0	-0.03183	0	0
1	-0.01606	0.02447	-0.00039
2	0.02339	0.09549	0.00223
3	0.04491	0.20611	0.00926
4	0.01639	0.34549	0.00566
5	-0.04502	0.5	-0.02251
6	-0.07568	0.65451	-0.04953
7	-0.0166	0.79389	-0.01318
8	0.12876	0.90451	0.11646
9	0.28362	0.97553	0.27668
10	0.35	1	0.35
11	0.28362	0.97553	0.27668
12	0.12876	0.90451	0.11646

The remaining coefficients follow from the symmetry $h [n] = h [20 - n]$ , which every linear phase filter has.

A Hanning window, length $N = 21$ , cut off $ω_{c} = 1.0996$ rad and group delay 10 samples. The first coefficients are ${0, - 0.00039, 0.00223, 0.00926, 0.00566, - 0.02251}$ , and $H (z)$ = sum of $h [n] z^{- n}$ over $n = 0$ to 20.

Practise this +1 like it2082 Baishakh · Q62+6 marksDifferentiate between analog and digital filter. Design an FIR lowpass digital filter that will have a -3dB cut-off at

30 π

rad/s and an attenuation of 50 dB at

45 π

rad/s. The filter is required to have a linear phase and the system uses a sampling rate of 100 Hz.

How this is readA -3 dB cut off at $30 π$ rad/s with a 100 Hz sampling rate is $ω = \frac{30 π}{100} = 0.3 π$ , and $45 π$ rad/s is $0.45 π$ .

What is different in this one

The stopband asks for 50 dB, and Hamming is the smallest window that reaches it, at 53 dB. Choosing a bigger one is not wrong, it is wasteful: the order roughly doubles.

The window is chosen by the stopband attenuation, never by the transition width: take the smallest window that reaches 50 dB.

Window	Stopband attenuation	Transition width
Rectangular	21 dB	$1.8 π$ / N	too weak
Bartlett	25 dB	$6.1 π$ / N	too weak
Hanning	44 dB	$6.2 π$ / N	too weak
Hamming	53 dB	$6.6 π$ / N	chosen
Blackman	74 dB	$11 π$ / N

Hamming gives 53 dB, which meets the 50 dB asked for, and is the smallest that does.

\begin{aligned} transition  width  dw & = ω_{s} - ω_{p} = 1.4137 - 0.9425 = 0.4712 rad \\ N \geq 6.6 π / dw & = \frac{6.6 π}{0.4712} = 44 \\ round  up, and  take  the  next  ODD  value : N & = 45 \end{aligned}

\begin{aligned} cut  off ω_{c} & = (ω_{p} + ω_{s}) / 2 = (0.9425 + 1.4137) / 2 = 1.1781 rad \\ delay τ & = (N - 1) / 2 = 22 samples \\ ideal  response h_{d} [n] & = \sin (ω_{c} (n - t a u)) / (π (n - τ)) \\ h_{d} [τ] & = ω_{c} / π = 0.375 \\ window ω [n] & = 0.54 - 0.46 \cos (2 π n / (N - 1)) \end{aligned}

n	$h_{d} [n]$	$w [n]$	$h [n] = h_{d} [n] w [n]$
0	0.01023	0.08	0.00082
1	-0.0058	0.08468	-0.00049
2	-0.01592	0.09863	-0.00157
3	-0.00641	0.12157	-0.00078
4	0.0125	0.15302	0.00191
5	0.0173	0.19236	0.00333
6	0	0.23876	0
7	-0.01961	0.29131	-0.00571
8	-0.01608	0.34891	-0.00561
9	0.00937	0.4104	0.00385
10	0.02653	0.47454	0.01259
11	0.01107	0.54	0.00598
12	-0.02251	0.60546	-0.01363

The remaining coefficients follow from the symmetry $h [n] = h [44 - n]$ , which every linear phase filter has.

Practise this +2 like it2081 Bhadra · Q68 marksDesign a low pass FIR filter using suitable window to meet following specifications:

0.99 \leq | H (e^{j ω}) | \leq 1.01

for

0 \leq | w | \leq 0.3 π

;

| H (e^{j ω}) | \leq 0.01

for

0.35 π

≤ |

ω

| ≤

π

What is different herethe ripple is given as a number rather than in dB, so it is converted first

What is different in this one

The stopband asks for 40 dB, and Hanning is the smallest window that reaches it, at 44 dB. Choosing a bigger one is not wrong, it is wasteful: the order roughly doubles.

Turn the ripple into decibels first: passband ripple 0.01 means Ap = -20 $\log_{10} (1 - 0.01) = 0.09$ dB; stopband ripple 0.01 means As = -20 $\log_{10} (0.01)$ = 40 dB.

The window is chosen by the stopband attenuation, never by the transition width: take the smallest window that reaches 40 dB.

Window	Stopband attenuation	Transition width
Rectangular	21 dB	$1.8 π$ / N	too weak
Bartlett	25 dB	$6.1 π$ / N	too weak
Hanning	44 dB	$6.2 π$ / N	chosen
Hamming	53 dB	$6.6 π$ / N
Blackman	74 dB	$11 π$ / N

Hanning gives 44 dB, which meets the 40 dB asked for, and is the smallest that does.

\begin{aligned} transition  width  dw & = ω_{s} - ω_{p} = 1.0996 - 0.9425 = 0.1571 rad \\ N \geq 6.2 π / dw & = \frac{6.2 π}{0.1571} = 124 \\ round  up, and  take  the  next  ODD  value : N & = 125 \end{aligned}

\begin{aligned} cut  off ω_{c} & = (ω_{p} + ω_{s}) / 2 = (0.9425 + 1.0996) / 2 = 1.021 rad \\ delay τ & = (N - 1) / 2 = 62 samples \\ ideal  response h_{d} [n] & = \sin (ω_{c} (n - t a u)) / (π (n - τ)) \\ h_{d} [τ] & = ω_{c} / π = 0.325 \\ window ω [n] & = 0.5 - 0.5 \cos (2 π n / (N - 1)) \end{aligned}

n	$h_{d} [n]$	$w [n]$	$h [n] = h_{d} [n] w [n]$
0	0.00233	0	0
1	-0.00273	0.00064	0
2	-0.00531	0.00257	-0.00001
3	-0.00282	0.00577	-0.00002
4	0.00249	0.01024	0.00003
5	0.00557	0.01596	0.00009
6	0.00334	0.02293	0.00008
7	-0.00221	0.03112	-0.00007
8	-0.00582	0.04052	-0.00024
9	-0.0039	0.0511	-0.0002
10	0.00189	0.06283	0.00012
11	0.00607	0.07568	0.00046
12	0.0045	0.08962	0.0004

The remaining coefficients follow from the symmetry $h [n] = h [124 - n]$ , which every linear phase filter has.

A Hanning window, length $N = 125$ , cut off $ω_{c} = 1.021$ rad and group delay 62 samples. The first coefficients are ${0, 0, - 0.00001, - 0.00002, 0.00003, 0.00009}$ , and $H (z)$ = sum of $h [n] z^{- n}$ over $n = 0$ to 124.

Same as 2081 Bhadra Q62081 Baishakh · Q610 marksDesign a linear phase FIR filter using suitable window to meet following specifications:

0.99 \leq | H (e^{j ω}) | \leq 1.01

for

0 \leq | w | \leq 0.3 π

;

| H (e^{j ω}) | \leq 0.01

for

0.35 π

≤ |

ω

| ≤

π

What is different in this one

The stopband asks for 40 dB, and Hanning is the smallest window that reaches it, at 44 dB. Choosing a bigger one is not wrong, it is wasteful: the order roughly doubles.

Turn the ripple into decibels first: passband ripple 0.01 means Ap = -20 $\log_{10} (1 - 0.01) = 0.09$ dB; stopband ripple 0.01 means As = -20 $\log_{10} (0.01)$ = 40 dB.

The window is chosen by the stopband attenuation, never by the transition width: take the smallest window that reaches 40 dB.

Window	Stopband attenuation	Transition width
Rectangular	21 dB	$1.8 π$ / N	too weak
Bartlett	25 dB	$6.1 π$ / N	too weak
Hanning	44 dB	$6.2 π$ / N	chosen
Hamming	53 dB	$6.6 π$ / N
Blackman	74 dB	$11 π$ / N

Hanning gives 44 dB, which meets the 40 dB asked for, and is the smallest that does.

\begin{aligned} transition  width  dw & = ω_{s} - ω_{p} = 1.0996 - 0.9425 = 0.1571 rad \\ N \geq 6.2 π / dw & = \frac{6.2 π}{0.1571} = 124 \\ round  up, and  take  the  next  ODD  value : N & = 125 \end{aligned}

\begin{aligned} cut  off ω_{c} & = (ω_{p} + ω_{s}) / 2 = (0.9425 + 1.0996) / 2 = 1.021 rad \\ delay τ & = (N - 1) / 2 = 62 samples \\ ideal  response h_{d} [n] & = \sin (ω_{c} (n - t a u)) / (π (n - τ)) \\ h_{d} [τ] & = ω_{c} / π = 0.325 \\ window ω [n] & = 0.5 - 0.5 \cos (2 π n / (N - 1)) \end{aligned}

n	$h_{d} [n]$	$w [n]$	$h [n] = h_{d} [n] w [n]$
0	0.00233	0	0
1	-0.00273	0.00064	0
2	-0.00531	0.00257	-0.00001
3	-0.00282	0.00577	-0.00002
4	0.00249	0.01024	0.00003
5	0.00557	0.01596	0.00009
6	0.00334	0.02293	0.00008
7	-0.00221	0.03112	-0.00007
8	-0.00582	0.04052	-0.00024
9	-0.0039	0.0511	-0.0002
10	0.00189	0.06283	0.00012
11	0.00607	0.07568	0.00046
12	0.0045	0.08962	0.0004

The remaining coefficients follow from the symmetry $h [n] = h [124 - n]$ , which every linear phase filter has.

Practise this2080 Bhadra · Q610 marksDesign the symmetric FIR Low Pass Filter (LPF) for which the desired frequency response is expressed as Hd(W) =

e^{- j W t a u}

for

| W | \leq W c

and 0 elsewhere. The length of the filter should be 7 and

Ω_{c} = 1

rad/sample. Make use of the Hanning window.What is different herethe window and the length are given, so the choosing step is skipped

What is different in this one

The stopband asks for 44 dB, and Hanning is the smallest window that reaches it, at 44 dB. Choosing a bigger one is not wrong, it is wasteful: the order roughly doubles.
The question fixes the length at $N = 7$ , so there is no order calculation: go straight to the cut off and the coefficients.
The question names the window, so the choosing step is skipped. Most of these do not, and picking the smallest window that reaches the attenuation is the first mark.
The cut off is given as 1 rad per sample rather than derived from the band edges.

The question names the Hanning window, so there is no choosing to do.

The question fixes the length at $N = 7$ .

\begin{aligned} cut  off ω_{c} & = 1 rad, given  by  the  question \\ delay τ & = (N - 1) / 2 = 3 samples \\ ideal  response h_{d} [n] & = \sin (ω_{c} (n - t a u)) / (π (n - τ)) \\ h_{d} [τ] & = ω_{c} / π = 0.3183 \\ window ω [n] & = 0.5 - 0.5 \cos (2 π n / (N - 1)) \end{aligned}

n	$h_{d} [n]$	$w [n]$	$h [n] = h_{d} [n] w [n]$
0	0.01497	0	0
1	0.14472	0.25	0.03618
2	0.26785	0.75	0.20089
3	0.31831	1	0.31831
4	0.26785	0.75	0.20089
5	0.14472	0.25	0.03618
6	0.01497	0	0

Same as 2082 Baishakh Q62079 Baishakh · Q75+3 marksDesign a low pass digital FIR filter having Pass band edge frequency w_p =

0.2 π

, Stop band edge frequency w_s =

0.45 π

and Stop band attenuation a_s = 51 dB using any appropriate window function.

What is different in this one

The stopband asks for 51 dB, and Hamming is the smallest window that reaches it, at 53 dB. Choosing a bigger one is not wrong, it is wasteful: the order roughly doubles.

The window is chosen by the stopband attenuation, never by the transition width: take the smallest window that reaches 51 dB.

Window	Stopband attenuation	Transition width
Rectangular	21 dB	$1.8 π$ / N	too weak
Bartlett	25 dB	$6.1 π$ / N	too weak
Hanning	44 dB	$6.2 π$ / N	too weak
Hamming	53 dB	$6.6 π$ / N	chosen
Blackman	74 dB	$11 π$ / N

Hamming gives 53 dB, which meets the 51 dB asked for, and is the smallest that does.

\begin{aligned} transition  width  dw & = ω_{s} - ω_{p} = 1.4137 - 0.6283 = 0.7854 rad \\ N \geq 6.6 π / dw & = \frac{6.6 π}{0.7854} = 26.4 \\ round  up, and  take  the  next  ODD  value : N & = 27 \end{aligned}

\begin{aligned} cut  off ω_{c} & = (ω_{p} + ω_{s}) / 2 = (0.6283 + 1.4137) / 2 = 1.021 rad \\ delay τ & = (N - 1) / 2 = 13 samples \\ ideal  response h_{d} [n] & = \sin (ω_{c} (n - t a u)) / (π (n - τ)) \\ h_{d} [τ] & = ω_{c} / π = 0.325 \\ window ω [n] & = 0.54 - 0.46 \cos (2 π n / (N - 1)) \end{aligned}

n	$h_{d} [n]$	$w [n]$	$h [n] = h_{d} [n] w [n]$
0	0.0159	0.08	0.00127
1	-0.0082	0.09337	-0.00077
2	-0.02814	0.13269	-0.00373
3	-0.02251	0.19569	-0.0044
4	0.00826	0.27869	0.0023
5	0.03784	0.37688	0.01426
6	0.03458	0.48455	0.01675
7	-0.0083	0.59545	-0.00494
8	-0.05882	0.70312	-0.04135
9	-0.06438	0.80131	-0.05159
10	0.00832	0.88431	0.00736
11	0.14181	0.94731	0.13434
12	0.2714	0.98663	0.26778

The remaining coefficients follow from the symmetry $h [n] = h [26 - n]$ , which every linear phase filter has.

A Hamming window, length $N = 27$ , cut off $ω_{c} = 1.021$ rad and group delay 13 samples. The first coefficients are ${0.00127, - 0.00077, - 0.00373, - 0.0044, 0.0023, 0.01426}$ , and $H (z)$ = sum of $h [n] z^{- n}$ over $n = 0$ to 26.

Same as 2081 Bhadra Q62076 Chaitra · Q86 marksDesign the FIR filter using suitable window for the specifications:

0.899 \leq | H (e^{j ω}) | \leq 1

for

| w | \leq 0.2 π

;

| H (e^{j ω}) | \leq 0.01

for

0.4 π \leq ω \leq π

How this is read $0.899 \leq | H | \leq 1$ gives a passband ripple of 0.101 and $| H | \leq 0.01$ gives As = 40 dB, which is what chooses the window.

What is different in this one

The stopband asks for 40 dB, and Hanning is the smallest window that reaches it, at 44 dB. Choosing a bigger one is not wrong, it is wasteful: the order roughly doubles.
The two ripples differ, 0.101 in the passband and 0.01 in the stopband, so it is the stopband figure that chooses the window.

Turn the ripple into decibels first: passband ripple 0.101 means Ap = -20 $\log_{10} (1 - 0.101) = 0.92$ dB; stopband ripple 0.01 means As = -20 $\log_{10} (0.01)$ = 40 dB.

The window is chosen by the stopband attenuation, never by the transition width: take the smallest window that reaches 40 dB.

Window	Stopband attenuation	Transition width
Rectangular	21 dB	$1.8 π$ / N	too weak
Bartlett	25 dB	$6.1 π$ / N	too weak
Hanning	44 dB	$6.2 π$ / N	chosen
Hamming	53 dB	$6.6 π$ / N
Blackman	74 dB	$11 π$ / N

Hanning gives 44 dB, which meets the 40 dB asked for, and is the smallest that does.

\begin{aligned} transition  width  dw & = ω_{s} - ω_{p} = 1.2566 - 0.6283 = 0.6283 rad \\ N \geq 6.2 π / dw & = \frac{6.2 π}{0.6283} = 31 \\ round  up, and  take  the  next  ODD  value : N & = 31 \end{aligned}

\begin{aligned} cut  off ω_{c} & = (ω_{p} + ω_{s}) / 2 = (0.6283 + 1.2566) / 2 = 0.9425 rad \\ delay τ & = (N - 1) / 2 = 15 samples \\ ideal  response h_{d} [n] & = \sin (ω_{c} (n - t a u)) / (π (n - τ)) \\ h_{d} [τ] & = ω_{c} / π = 0.3 \\ window ω [n] & = 0.5 - 0.5 \cos (2 π n / (N - 1)) \end{aligned}

n	$h_{d} [n]$	$w [n]$	$h [n] = h_{d} [n] w [n]$
0	0.02122	0	0
1	0.01336	0.01093	0.00015
2	-0.00757	0.04323	-0.00033
3	-0.02523	0.09549	-0.00241
4	-0.02341	0.16543	-0.00387
5	0	0.25	0
6	0.02861	0.34549	0.00989
7	0.03784	0.44774	0.01694
8	0.01405	0.55226	0.00776
9	-0.03118	0.65451	-0.02041
10	-0.06366	0.75	-0.04775
11	-0.04677	0.83457	-0.03904
12	0.03279	0.90451	0.02966

The remaining coefficients follow from the symmetry $h [n] = h [30 - n]$ , which every linear phase filter has.

A Hanning window, length $N = 31$ , cut off $ω_{c} = 0.9425$ rad and group delay 15 samples. The first coefficients are ${0, 0.00015, - 0.00033, - 0.00241, - 0.00387, 0}$ , and $H (z)$ = sum of $h [n] z^{- n}$ over $n = 0$ to 30.

Practise this2074 Ashwin · Q62+6 marksWhy Kaiser window is better than other fixed windows in FIR filter design? Find out first six coefficients of impulse response of a low pass FIR filter having Pass band edge frequency w_p =

0.2 π

, Stop band edge frequency w_s =

0.5 π

and Stop band attenuation a_s = 41dB using any appropriate window function.What is different here41 dB is just past what Hanning gives at 44 dB, so Hanning is the smallest window that does it. The question wants only the first six coefficients.

What is different in this one

The stopband asks for 41 dB, and Hanning is the smallest window that reaches it, at 44 dB. Choosing a bigger one is not wrong, it is wasteful: the order roughly doubles.

The window is chosen by the stopband attenuation, never by the transition width: take the smallest window that reaches 41 dB.

Window	Stopband attenuation	Transition width
Rectangular	21 dB	$1.8 π$ / N	too weak
Bartlett	25 dB	$6.1 π$ / N	too weak
Hanning	44 dB	$6.2 π$ / N	chosen
Hamming	53 dB	$6.6 π$ / N
Blackman	74 dB	$11 π$ / N

Hanning gives 44 dB, which meets the 41 dB asked for, and is the smallest that does.

\begin{aligned} transition  width  dw & = ω_{s} - ω_{p} = 1.5708 - 0.6283 = 0.9425 rad \\ N \geq 6.2 π / dw & = \frac{6.2 π}{0.9425} = 20.67 \\ round  up, and  take  the  next  ODD  value : N & = 21 \end{aligned}

\begin{aligned} cut  off ω_{c} & = (ω_{p} + ω_{s}) / 2 = (0.6283 + 1.5708) / 2 = 1.0996 rad \\ delay τ & = (N - 1) / 2 = 10 samples \\ ideal  response h_{d} [n] & = \sin (ω_{c} (n - t a u)) / (π (n - τ)) \\ h_{d} [τ] & = ω_{c} / π = 0.35 \\ window ω [n] & = 0.5 - 0.5 \cos (2 π n / (N - 1)) \end{aligned}

n	$h_{d} [n]$	$w [n]$	$h [n] = h_{d} [n] w [n]$
0	-0.03183	0	0
1	-0.01606	0.02447	-0.00039
2	0.02339	0.09549	0.00223
3	0.04491	0.20611	0.00926
4	0.01639	0.34549	0.00566
5	-0.04502	0.5	-0.02251
6	-0.07568	0.65451	-0.04953
7	-0.0166	0.79389	-0.01318
8	0.12876	0.90451	0.11646
9	0.28362	0.97553	0.27668
10	0.35	1	0.35
11	0.28362	0.97553	0.27668
12	0.12876	0.90451	0.11646

The remaining coefficients follow from the symmetry $h [n] = h [20 - n]$ , which every linear phase filter has.

The same questions grouped by how often each topic is set: fir design by windowing

Gibbs' phenomenon HOT 6/19

82 Ba · 81 Ba · 80 Ba · 79 Bh · 73 Shr · 71 Shr62+52+8

Before the formulas

What goes wrong if you cut the ideal response off square, and it is worth seeing clearly because the fix is the whole of the next card. An overshoot appears on each side of the edge, about nine percent of the jump.

Here is the part that surprises people: making the filter longer does not shrink it. The ripples get narrower and crowd towards the edge, and stay just as tall. So length is not the cure. A window that tapers instead of chopping is, and that is why the fixed windows exist at all.

Gibbs' phenomenon The oscillatory behaviour and the fixed overshoot that appear near a discontinuity when a Fourier series or an ideal frequency response is truncated. In FIR design it shows as ripples on both sides of the cutoff, with an overshoot of about 9 percent of the jump that does not shrink as the filter gets longer.

How it arises with the rectangular window, which is the question as asked

Truncation is multiplication by a rectangle. Keeping only M samples of the infinite $h_{d} [n]$ is $h [n] = h_{d} [n] w [n]$ with $w [n] = 1$ over the window and 0 outside.
So the spectrum is convolved with the rectangle's spectrum, which is the Dirichlet kernel, $\sin (w M / 2) / \sin (ω / 2)$ . That kernel has a narrow main lobe and large side lobes, the first at about -13 dB.
Convolving a step with that kernel smears the edge into a transition band and wraps ripples round it. The ripples are the side lobes sliding past the discontinuity.
Increasing M narrows the main lobe, so the transition gets sharper and the ripples get faster, but the peak overshoot stays at about 9 percent. That is the surprising part, and it is what makes it a named phenomenon rather than just an error.

How it is minimised

Taper the window. A window that goes smoothly to zero at its ends, such as Hanning, Hamming or Blackman, has much smaller side lobes, so the ripples shrink. The price is a wider main lobe and therefore a wider transition band.
Use the Kaiser window, which lets you trade the two continuously instead of picking from a fixed menu.
Use an equiripple design. The Remez exchange algorithm does not truncate at all; it minimises the maximum error directly, so the ripple is spread evenly and is as small as the length allows.
It can never be removed by lengthening the filter alone. Saying this is the point of the question.

The overshoot is the same one that a truncated Fourier series shows at a discontinuity, and the cure, a tapered window, is what the Kaiser window makes adjustable.

See it drawnMaking the filter longer does not make the overshoot smaller

The ripples crowd towards the band edge as the length grows, and the first one settles at about nine percent of the jump and stays there. Only a tapered window brings it down.

Practise thisEvery question the paper has set on this topic, worked in full1

Practise this2079 Bhadra · Q72+8 marksDefine Gibb's phenomenon. Design the FIR filter using Kaiser window technique for the specifications:

0.899 \leq | H (e^{j ω}) | \leq 1

for

| w | \leq 0.2 π

;

| H (e^{j ω}) | \leq 0.01

for

0.4 π \leq ω \leq π

What is different hereThe passband here is 0.899 to 1, not 0.99 to 1.01, so the passband ripple is 0.101 and the STOPBAND ripple 0.01 is the smaller one. Taking the wrong one gives too short a filter and the stopband then fails.

What is different in this one

The ripples are different, 0.101 and 0.01, and the design takes the smaller one. Taking the wrong one gives too short a filter and the stopband then fails.

Take the smaller of the two ripples, delta = 0.01, and turn it into decibels: $A = - 20 \log_{10} (0.01)$ = 40 dB.

\begin{aligned} 21 \leq A \leq 50, so β & = 0.5842 {(A - 21)}^{0.4} + 0.07886 (A - 21) \\ = 0.5842 {(19)}^{0.4} + 0.07886 (19) = 3.3953 \\ transition  dw & = ω_{s} - ω_{p} = 1.2566 - 0.6283 = 0.6283 rad \\ order M \geq (A - 8) / (2.285 dw) \\ = (40 - 8) / (2.285 x 0.6283) = 22.29 \\ round  up : M & = 23, length N = M + 1 = 24 \\ α & = τ = M / 2 = 11.5, ω_{c} = (ω_{p} + ω_{s}) / 2 = 0.9425 rad \end{aligned}

M is odd, so $α = 11.5$ is half a sample. That is not a mistake and it does not need fixing: the filter is still exactly linear phase, it just delays by half a sample, and there is no middle tap to treat as a special case. If an integer delay is wanted instead, take the length up to the next odd number, 25, which costs one extra tap and changes nothing else.

w [n] = \frac{I_{0} (β \sqrt{1 - {[\frac{2 n - (N - 1)}{N - 1}]}^{2}})}{I_{0} (β)}

n	$h_{d} [n]$	$w [n]$	$h [n]$
0	-0.02734	0.14797	-0.00405
1	-0.01376	0.22787	-0.00314
2	0.01521	0.31772	0.00483
3	0.03699	0.41466	0.01534
4	0.03001	0.51524	0.01546
5	-0.00766	0.6156	-0.00472
6	-0.05157	0.71165	-0.0367
7	-0.06303	0.79931	-0.05038
8	-0.01423	0.87475	-0.01245
9	0.09003	0.93459	0.08414
10	0.20959	0.97608	0.20458
11	0.28902	0.99732	0.28825
12	0.28902	0.99732	0.28825

The rest follow from $h [n] = h [23 - n]$ .

$A = 40$ dB, $β = 3.3953$ , $N = 24$ , cut off $ω_{c} = 0.9425$ rad, group delay 11.5 samples. The first coefficients are ${- 0.00405, - 0.00314, 0.00483, 0.01534, 0.01546, - 0.00472}$ .

The fixed windows TOP 11/19

82 Bh · 82 Ba · 81 Bh · 81 Ba · 80 Bh · 79 Ba · 76 Ch · 74 Ash · 73 Shr · 72 Ch · 71 Shr1062+6

Before the formulas

The four windows the paper names, with the numbers that go with them. They all do the same job, and they sit at different points on one trade. A window that tapers gently has low side lobes, which gives a clean stopband, and a wide main lobe, which gives a wide transition band.

A window that tapers abruptly gives you the opposite. There is no window that wins both, and no amount of cleverness produces one. So the question is never which window is best. It is which window the required stopband attenuation forces on you, and then how long it has to be.

\begin{aligned} rectangular w [n] & = 1 \\ Bartlett w [n] & = 1 - \frac{| 2 n - (M - 1) |}{M - 1} \\ Hanning w [n] & = 0.5 - 0.5 \cos (\frac{2 π n}{M - 1}) \\ Hamming w [n] & = 0.54 - 0.46 \cos (\frac{2 π n}{M - 1}) \\ Blackman w [n] & = 0.42 - 0.5 \cos (\frac{2 π n}{M - 1}) + 0.08 \cos (\frac{4 π n}{M - 1}) \end{aligned}

All over $0 \leq n \leq M - 1$ , and zero outside.

Window	Main lobe width	Approximate transition width	Peak side lobe	Minimum stopband attenuation	Length from delta $ω$
Rectangular	$4 π$ / M	$0.9 π$ / M	-13 dB	21 dB	M = $1.8 π$ / delta $ω$
Bartlett	$8 π$ / M	$3.1 π$ / M	-25 dB	25 dB	M = 6. $2 π$ / delta $ω$
Hanning	$8 π$ / M	$3.1 π$ / M	-31 dB	44 dB	M = 6. $2 π$ / delta $ω$
Hamming	$8 π$ / M	$3.3 π$ / M	-41 dB	53 dB	M = $6.6 π$ / delta $ω$
Blackman	$12 π$ / M	$5.5 π$ / M	-57 dB	74 dB	M = $11 π$ / delta $ω$

Read the table from the right. The specification gives a stopband attenuation, so pick the first window in the list that meets it, then use its length formula.
The trade is visible in the table: going down the list the attenuation improves and the main lobe widens, so the filter gets longer for the same transition band.
The attenuation column is a property of the window, not of M. Lengthening the filter narrows the transition; it does not improve the stopband.

Each window is a different trade between main lobe width and side lobe height, which by the multiplication property is a trade between transition width and stopband ripple.

See it drawnEvery window is a point on one trade, and none of them wins both

A gentle taper gives low side lobes, so a clean stopband, and a wide main lobe, so a wide transition. The question is never which window is best, it is which one the required attenuation forces on you.

Practise thisEvery question the paper has set on this topic, worked in full8

How this is read $ω = 2 π$ f / fs turns the edges into $2 π$ (2000)/20000 = $0.2 π$ and $2 π$ (5000)/20000 = $0.5 π$ radians per sample.

What is different in this one

The stopband asks for 42 dB, and Hanning is the smallest window that reaches it, at 44 dB. Choosing a bigger one is not wrong, it is wasteful: the order roughly doubles.

Convert the edges to radians per sample with $ω = 2 π$ f / fs, with fs = 20000 Hz.

The window is chosen by the stopband attenuation, never by the transition width: take the smallest window that reaches 42 dB.

Window	Stopband attenuation	Transition width
Rectangular	21 dB	$1.8 π$ / N	too weak
Bartlett	25 dB	$6.1 π$ / N	too weak
Hanning	44 dB	$6.2 π$ / N	chosen
Hamming	53 dB	$6.6 π$ / N
Blackman	74 dB	$11 π$ / N

Hanning gives 44 dB, which meets the 42 dB asked for, and is the smallest that does.

\begin{aligned} transition  width  dw & = ω_{s} - ω_{p} = 1.5708 - 0.6283 = 0.9425 rad \\ N \geq 6.2 π / dw & = \frac{6.2 π}{0.9425} = 20.67 \\ round  up, and  take  the  next  ODD  value : N & = 21 \end{aligned}

\begin{aligned} cut  off ω_{c} & = (ω_{p} + ω_{s}) / 2 = (0.6283 + 1.5708) / 2 = 1.0996 rad \\ delay τ & = (N - 1) / 2 = 10 samples \\ ideal  response h_{d} [n] & = \sin (ω_{c} (n - t a u)) / (π (n - τ)) \\ h_{d} [τ] & = ω_{c} / π = 0.35 \\ window ω [n] & = 0.5 - 0.5 \cos (2 π n / (N - 1)) \end{aligned}

n	$h_{d} [n]$	$w [n]$	$h [n] = h_{d} [n] w [n]$
0	-0.03183	0	0
1	-0.01606	0.02447	-0.00039
2	0.02339	0.09549	0.00223
3	0.04491	0.20611	0.00926
4	0.01639	0.34549	0.00566
5	-0.04502	0.5	-0.02251
6	-0.07568	0.65451	-0.04953
7	-0.0166	0.79389	-0.01318
8	0.12876	0.90451	0.11646
9	0.28362	0.97553	0.27668
10	0.35	1	0.35
11	0.28362	0.97553	0.27668
12	0.12876	0.90451	0.11646

The remaining coefficients follow from the symmetry $h [n] = h [20 - n]$ , which every linear phase filter has.

Practise this +1 like it2082 Baishakh · Q62+6 marksDifferentiate between analog and digital filter. Design an FIR lowpass digital filter that will have a -3dB cut-off at

30 π

rad/s and an attenuation of 50 dB at

45 π

rad/s. The filter is required to have a linear phase and the system uses a sampling rate of 100 Hz.

How this is readA -3 dB cut off at $30 π$ rad/s with a 100 Hz sampling rate is $ω = \frac{30 π}{100} = 0.3 π$ , and $45 π$ rad/s is $0.45 π$ .

What is different in this one

The stopband asks for 50 dB, and Hamming is the smallest window that reaches it, at 53 dB. Choosing a bigger one is not wrong, it is wasteful: the order roughly doubles.

The window is chosen by the stopband attenuation, never by the transition width: take the smallest window that reaches 50 dB.

Window	Stopband attenuation	Transition width
Rectangular	21 dB	$1.8 π$ / N	too weak
Bartlett	25 dB	$6.1 π$ / N	too weak
Hanning	44 dB	$6.2 π$ / N	too weak
Hamming	53 dB	$6.6 π$ / N	chosen
Blackman	74 dB	$11 π$ / N

Hamming gives 53 dB, which meets the 50 dB asked for, and is the smallest that does.

\begin{aligned} transition  width  dw & = ω_{s} - ω_{p} = 1.4137 - 0.9425 = 0.4712 rad \\ N \geq 6.6 π / dw & = \frac{6.6 π}{0.4712} = 44 \\ round  up, and  take  the  next  ODD  value : N & = 45 \end{aligned}

\begin{aligned} cut  off ω_{c} & = (ω_{p} + ω_{s}) / 2 = (0.9425 + 1.4137) / 2 = 1.1781 rad \\ delay τ & = (N - 1) / 2 = 22 samples \\ ideal  response h_{d} [n] & = \sin (ω_{c} (n - t a u)) / (π (n - τ)) \\ h_{d} [τ] & = ω_{c} / π = 0.375 \\ window ω [n] & = 0.54 - 0.46 \cos (2 π n / (N - 1)) \end{aligned}

n	$h_{d} [n]$	$w [n]$	$h [n] = h_{d} [n] w [n]$
0	0.01023	0.08	0.00082
1	-0.0058	0.08468	-0.00049
2	-0.01592	0.09863	-0.00157
3	-0.00641	0.12157	-0.00078
4	0.0125	0.15302	0.00191
5	0.0173	0.19236	0.00333
6	0	0.23876	0
7	-0.01961	0.29131	-0.00571
8	-0.01608	0.34891	-0.00561
9	0.00937	0.4104	0.00385
10	0.02653	0.47454	0.01259
11	0.01107	0.54	0.00598
12	-0.02251	0.60546	-0.01363

The remaining coefficients follow from the symmetry $h [n] = h [44 - n]$ , which every linear phase filter has.

Practise this +2 like it2081 Bhadra · Q68 marksDesign a low pass FIR filter using suitable window to meet following specifications:

0.99 \leq | H (e^{j ω}) | \leq 1.01

for

0 \leq | w | \leq 0.3 π

;

| H (e^{j ω}) | \leq 0.01

for

0.35 π

≤ |

ω

| ≤

π

What is different herethe ripple is given as a number rather than in dB, so it is converted first

What is different in this one

The stopband asks for 40 dB, and Hanning is the smallest window that reaches it, at 44 dB. Choosing a bigger one is not wrong, it is wasteful: the order roughly doubles.

Turn the ripple into decibels first: passband ripple 0.01 means Ap = -20 $\log_{10} (1 - 0.01) = 0.09$ dB; stopband ripple 0.01 means As = -20 $\log_{10} (0.01)$ = 40 dB.

The window is chosen by the stopband attenuation, never by the transition width: take the smallest window that reaches 40 dB.

Window	Stopband attenuation	Transition width
Rectangular	21 dB	$1.8 π$ / N	too weak
Bartlett	25 dB	$6.1 π$ / N	too weak
Hanning	44 dB	$6.2 π$ / N	chosen
Hamming	53 dB	$6.6 π$ / N
Blackman	74 dB	$11 π$ / N

Hanning gives 44 dB, which meets the 40 dB asked for, and is the smallest that does.

\begin{aligned} transition  width  dw & = ω_{s} - ω_{p} = 1.0996 - 0.9425 = 0.1571 rad \\ N \geq 6.2 π / dw & = \frac{6.2 π}{0.1571} = 124 \\ round  up, and  take  the  next  ODD  value : N & = 125 \end{aligned}

\begin{aligned} cut  off ω_{c} & = (ω_{p} + ω_{s}) / 2 = (0.9425 + 1.0996) / 2 = 1.021 rad \\ delay τ & = (N - 1) / 2 = 62 samples \\ ideal  response h_{d} [n] & = \sin (ω_{c} (n - t a u)) / (π (n - τ)) \\ h_{d} [τ] & = ω_{c} / π = 0.325 \\ window ω [n] & = 0.5 - 0.5 \cos (2 π n / (N - 1)) \end{aligned}

n	$h_{d} [n]$	$w [n]$	$h [n] = h_{d} [n] w [n]$
0	0.00233	0	0
1	-0.00273	0.00064	0
2	-0.00531	0.00257	-0.00001
3	-0.00282	0.00577	-0.00002
4	0.00249	0.01024	0.00003
5	0.00557	0.01596	0.00009
6	0.00334	0.02293	0.00008
7	-0.00221	0.03112	-0.00007
8	-0.00582	0.04052	-0.00024
9	-0.0039	0.0511	-0.0002
10	0.00189	0.06283	0.00012
11	0.00607	0.07568	0.00046
12	0.0045	0.08962	0.0004

The remaining coefficients follow from the symmetry $h [n] = h [124 - n]$ , which every linear phase filter has.

Same as 2081 Bhadra Q62081 Baishakh · Q610 marksDesign a linear phase FIR filter using suitable window to meet following specifications:

0.99 \leq | H (e^{j ω}) | \leq 1.01

for

0 \leq | w | \leq 0.3 π

;

| H (e^{j ω}) | \leq 0.01

for

0.35 π

≤ |

ω

| ≤

π

What is different in this one

The stopband asks for 40 dB, and Hanning is the smallest window that reaches it, at 44 dB. Choosing a bigger one is not wrong, it is wasteful: the order roughly doubles.

Turn the ripple into decibels first: passband ripple 0.01 means Ap = -20 $\log_{10} (1 - 0.01) = 0.09$ dB; stopband ripple 0.01 means As = -20 $\log_{10} (0.01)$ = 40 dB.

The window is chosen by the stopband attenuation, never by the transition width: take the smallest window that reaches 40 dB.

Window	Stopband attenuation	Transition width
Rectangular	21 dB	$1.8 π$ / N	too weak
Bartlett	25 dB	$6.1 π$ / N	too weak
Hanning	44 dB	$6.2 π$ / N	chosen
Hamming	53 dB	$6.6 π$ / N
Blackman	74 dB	$11 π$ / N

Hanning gives 44 dB, which meets the 40 dB asked for, and is the smallest that does.

\begin{aligned} transition  width  dw & = ω_{s} - ω_{p} = 1.0996 - 0.9425 = 0.1571 rad \\ N \geq 6.2 π / dw & = \frac{6.2 π}{0.1571} = 124 \\ round  up, and  take  the  next  ODD  value : N & = 125 \end{aligned}

\begin{aligned} cut  off ω_{c} & = (ω_{p} + ω_{s}) / 2 = (0.9425 + 1.0996) / 2 = 1.021 rad \\ delay τ & = (N - 1) / 2 = 62 samples \\ ideal  response h_{d} [n] & = \sin (ω_{c} (n - t a u)) / (π (n - τ)) \\ h_{d} [τ] & = ω_{c} / π = 0.325 \\ window ω [n] & = 0.5 - 0.5 \cos (2 π n / (N - 1)) \end{aligned}

n	$h_{d} [n]$	$w [n]$	$h [n] = h_{d} [n] w [n]$
0	0.00233	0	0
1	-0.00273	0.00064	0
2	-0.00531	0.00257	-0.00001
3	-0.00282	0.00577	-0.00002
4	0.00249	0.01024	0.00003
5	0.00557	0.01596	0.00009
6	0.00334	0.02293	0.00008
7	-0.00221	0.03112	-0.00007
8	-0.00582	0.04052	-0.00024
9	-0.0039	0.0511	-0.0002
10	0.00189	0.06283	0.00012
11	0.00607	0.07568	0.00046
12	0.0045	0.08962	0.0004

The remaining coefficients follow from the symmetry $h [n] = h [124 - n]$ , which every linear phase filter has.

Practise this2080 Bhadra · Q610 marksDesign the symmetric FIR Low Pass Filter (LPF) for which the desired frequency response is expressed as Hd(W) =

e^{- j W t a u}

for

| W | \leq W c

and 0 elsewhere. The length of the filter should be 7 and

Ω_{c} = 1

rad/sample. Make use of the Hanning window.What is different herethe window and the length are given, so the choosing step is skipped

What is different in this one

The stopband asks for 44 dB, and Hanning is the smallest window that reaches it, at 44 dB. Choosing a bigger one is not wrong, it is wasteful: the order roughly doubles.
The question fixes the length at $N = 7$ , so there is no order calculation: go straight to the cut off and the coefficients.
The question names the window, so the choosing step is skipped. Most of these do not, and picking the smallest window that reaches the attenuation is the first mark.
The cut off is given as 1 rad per sample rather than derived from the band edges.

The question names the Hanning window, so there is no choosing to do.

The question fixes the length at $N = 7$ .

\begin{aligned} cut  off ω_{c} & = 1 rad, given  by  the  question \\ delay τ & = (N - 1) / 2 = 3 samples \\ ideal  response h_{d} [n] & = \sin (ω_{c} (n - t a u)) / (π (n - τ)) \\ h_{d} [τ] & = ω_{c} / π = 0.3183 \\ window ω [n] & = 0.5 - 0.5 \cos (2 π n / (N - 1)) \end{aligned}

n	$h_{d} [n]$	$w [n]$	$h [n] = h_{d} [n] w [n]$
0	0.01497	0	0
1	0.14472	0.25	0.03618
2	0.26785	0.75	0.20089
3	0.31831	1	0.31831
4	0.26785	0.75	0.20089
5	0.14472	0.25	0.03618
6	0.01497	0	0

Same as 2082 Baishakh Q62079 Baishakh · Q75+3 marksDesign a low pass digital FIR filter having Pass band edge frequency w_p =

0.2 π

, Stop band edge frequency w_s =

0.45 π

and Stop band attenuation a_s = 51 dB using any appropriate window function.

What is different in this one

The stopband asks for 51 dB, and Hamming is the smallest window that reaches it, at 53 dB. Choosing a bigger one is not wrong, it is wasteful: the order roughly doubles.

The window is chosen by the stopband attenuation, never by the transition width: take the smallest window that reaches 51 dB.

Window	Stopband attenuation	Transition width
Rectangular	21 dB	$1.8 π$ / N	too weak
Bartlett	25 dB	$6.1 π$ / N	too weak
Hanning	44 dB	$6.2 π$ / N	too weak
Hamming	53 dB	$6.6 π$ / N	chosen
Blackman	74 dB	$11 π$ / N

Hamming gives 53 dB, which meets the 51 dB asked for, and is the smallest that does.

\begin{aligned} transition  width  dw & = ω_{s} - ω_{p} = 1.4137 - 0.6283 = 0.7854 rad \\ N \geq 6.6 π / dw & = \frac{6.6 π}{0.7854} = 26.4 \\ round  up, and  take  the  next  ODD  value : N & = 27 \end{aligned}

\begin{aligned} cut  off ω_{c} & = (ω_{p} + ω_{s}) / 2 = (0.6283 + 1.4137) / 2 = 1.021 rad \\ delay τ & = (N - 1) / 2 = 13 samples \\ ideal  response h_{d} [n] & = \sin (ω_{c} (n - t a u)) / (π (n - τ)) \\ h_{d} [τ] & = ω_{c} / π = 0.325 \\ window ω [n] & = 0.54 - 0.46 \cos (2 π n / (N - 1)) \end{aligned}

n	$h_{d} [n]$	$w [n]$	$h [n] = h_{d} [n] w [n]$
0	0.0159	0.08	0.00127
1	-0.0082	0.09337	-0.00077
2	-0.02814	0.13269	-0.00373
3	-0.02251	0.19569	-0.0044
4	0.00826	0.27869	0.0023
5	0.03784	0.37688	0.01426
6	0.03458	0.48455	0.01675
7	-0.0083	0.59545	-0.00494
8	-0.05882	0.70312	-0.04135
9	-0.06438	0.80131	-0.05159
10	0.00832	0.88431	0.00736
11	0.14181	0.94731	0.13434
12	0.2714	0.98663	0.26778

The remaining coefficients follow from the symmetry $h [n] = h [26 - n]$ , which every linear phase filter has.

Same as 2081 Bhadra Q62076 Chaitra · Q86 marksDesign the FIR filter using suitable window for the specifications:

0.899 \leq | H (e^{j ω}) | \leq 1

for

| w | \leq 0.2 π

;

| H (e^{j ω}) | \leq 0.01

for

0.4 π \leq ω \leq π

How this is read $0.899 \leq | H | \leq 1$ gives a passband ripple of 0.101 and $| H | \leq 0.01$ gives As = 40 dB, which is what chooses the window.

What is different in this one

The stopband asks for 40 dB, and Hanning is the smallest window that reaches it, at 44 dB. Choosing a bigger one is not wrong, it is wasteful: the order roughly doubles.
The two ripples differ, 0.101 in the passband and 0.01 in the stopband, so it is the stopband figure that chooses the window.

Turn the ripple into decibels first: passband ripple 0.101 means Ap = -20 $\log_{10} (1 - 0.101) = 0.92$ dB; stopband ripple 0.01 means As = -20 $\log_{10} (0.01)$ = 40 dB.

The window is chosen by the stopband attenuation, never by the transition width: take the smallest window that reaches 40 dB.

Window	Stopband attenuation	Transition width
Rectangular	21 dB	$1.8 π$ / N	too weak
Bartlett	25 dB	$6.1 π$ / N	too weak
Hanning	44 dB	$6.2 π$ / N	chosen
Hamming	53 dB	$6.6 π$ / N
Blackman	74 dB	$11 π$ / N

Hanning gives 44 dB, which meets the 40 dB asked for, and is the smallest that does.

\begin{aligned} transition  width  dw & = ω_{s} - ω_{p} = 1.2566 - 0.6283 = 0.6283 rad \\ N \geq 6.2 π / dw & = \frac{6.2 π}{0.6283} = 31 \\ round  up, and  take  the  next  ODD  value : N & = 31 \end{aligned}

\begin{aligned} cut  off ω_{c} & = (ω_{p} + ω_{s}) / 2 = (0.6283 + 1.2566) / 2 = 0.9425 rad \\ delay τ & = (N - 1) / 2 = 15 samples \\ ideal  response h_{d} [n] & = \sin (ω_{c} (n - t a u)) / (π (n - τ)) \\ h_{d} [τ] & = ω_{c} / π = 0.3 \\ window ω [n] & = 0.5 - 0.5 \cos (2 π n / (N - 1)) \end{aligned}

n	$h_{d} [n]$	$w [n]$	$h [n] = h_{d} [n] w [n]$
0	0.02122	0	0
1	0.01336	0.01093	0.00015
2	-0.00757	0.04323	-0.00033
3	-0.02523	0.09549	-0.00241
4	-0.02341	0.16543	-0.00387
5	0	0.25	0
6	0.02861	0.34549	0.00989
7	0.03784	0.44774	0.01694
8	0.01405	0.55226	0.00776
9	-0.03118	0.65451	-0.02041
10	-0.06366	0.75	-0.04775
11	-0.04677	0.83457	-0.03904
12	0.03279	0.90451	0.02966

The remaining coefficients follow from the symmetry $h [n] = h [30 - n]$ , which every linear phase filter has.

0.2 π

, Stop band edge frequency w_s =

0.5 π

What is different in this one

The stopband asks for 41 dB, and Hanning is the smallest window that reaches it, at 44 dB. Choosing a bigger one is not wrong, it is wasteful: the order roughly doubles.

The window is chosen by the stopband attenuation, never by the transition width: take the smallest window that reaches 41 dB.

Window	Stopband attenuation	Transition width
Rectangular	21 dB	$1.8 π$ / N	too weak
Bartlett	25 dB	$6.1 π$ / N	too weak
Hanning	44 dB	$6.2 π$ / N	chosen
Hamming	53 dB	$6.6 π$ / N
Blackman	74 dB	$11 π$ / N

Hanning gives 44 dB, which meets the 41 dB asked for, and is the smallest that does.

\begin{aligned} transition  width  dw & = ω_{s} - ω_{p} = 1.5708 - 0.6283 = 0.9425 rad \\ N \geq 6.2 π / dw & = \frac{6.2 π}{0.9425} = 20.67 \\ round  up, and  take  the  next  ODD  value : N & = 21 \end{aligned}

\begin{aligned} cut  off ω_{c} & = (ω_{p} + ω_{s}) / 2 = (0.6283 + 1.5708) / 2 = 1.0996 rad \\ delay τ & = (N - 1) / 2 = 10 samples \\ ideal  response h_{d} [n] & = \sin (ω_{c} (n - t a u)) / (π (n - τ)) \\ h_{d} [τ] & = ω_{c} / π = 0.35 \\ window ω [n] & = 0.5 - 0.5 \cos (2 π n / (N - 1)) \end{aligned}

n	$h_{d} [n]$	$w [n]$	$h [n] = h_{d} [n] w [n]$
0	-0.03183	0	0
1	-0.01606	0.02447	-0.00039
2	0.02339	0.09549	0.00223
3	0.04491	0.20611	0.00926
4	0.01639	0.34549	0.00566
5	-0.04502	0.5	-0.02251
6	-0.07568	0.65451	-0.04953
7	-0.0166	0.79389	-0.01318
8	0.12876	0.90451	0.11646
9	0.28362	0.97553	0.27668
10	0.35	1	0.35
11	0.28362	0.97553	0.27668
12	0.12876	0.90451	0.11646

The remaining coefficients follow from the symmetry $h [n] = h [20 - n]$ , which every linear phase filter has.

The same questions grouped by how often each topic is set: fir design by windowing

5.2Design by the Kaiser window

The Kaiser window TOP 11/19

81 Ba · 80 Bh · 80 Ba · 79 Bh · 78 Bh · 76 Ash · 75 Ash · 74 Ch · 74 Ash · 72 Ch · 72 Ka2+82+2+22+4+4

Before the formulas

The four fixed windows give you four fixed points on that trade. If the specification lands between two of them you have to take the better one and overshoot, which wastes filter length. Kaiser's window gives you the whole curve instead of four points.

One parameter slides it continuously from rectangular at one end to gentler than Blackman at the other. That is why it is the one used in practice: there are formulas that take the attenuation and transition width you were asked for and hand back the parameter and the length directly, so the design stops being a search.

Why Kaiser is better than the fixed windows The fixed windows offer a fixed trade between transition width and stopband attenuation: choosing Hamming commits you to 53 dB whether you needed 45 or 60. The Kaiser window has a shape parameter $β$ that slides continuously between them, so you can meet the specification exactly rather than exceed it, and the filter comes out as short as the specification allows. It is very close to the optimal window in the sense of concentrating energy in the main lobe.

\begin{aligned} w [n] & = \frac{I_{0} (β \sqrt{1 - {(\frac{2 n}{M - 1} - 1)}^{2}})}{I_{0} (β)}, 0 \leq n \leq M - 1 \\ δ & = min (δ_{p}, δ_{s}), A = - 20 \log_{10} δ \\ M & \geq \frac{A - 8}{2.285 Δ ω} + 1 \end{aligned}

I_0 is the zeroth order modified Bessel function of the first kind.

The procedure

Turn the magnitude limits into ripples. A specification such as $0.99 \leq | H | \leq 1.01$ means $δ_{p} = 0.01$ ; $| H | \leq 0.01$ in the stopband means $δ_{s} = 0.01$ .
Take the smaller of the two as delta, because the window method produces approximately equal passband and stopband ripple, so the tighter one governs.
Compute A in dB, then $β$ from the piecewise formula, then M from the length formula, rounding up to the next odd integer.
Set $α$ = (M-1)/2, compute $h_{d} [n]$ as usual with $ω_{c}$ = ( $ω_{p} + ω_{s}$ )/2, compute $w [n]$ , and multiply.

Worked, the 2075 Ashwin and 2072 Kartik specification

0.99 \leq | H | \leq 1.01

for

0 \leq ω \leq 0.19 π

, and

| H | \leq 0.01

for

0.21 π \leq ω \leq π

δ_{p} = 0.01

and

δ_{s} = 0.01

, so delta = 0.01 and $A = - 20 \log_{10} (0.01) = 40$ dB.
A is between 21 and 50, so

β = 0.5842 {(40 - 21)}^{0}

.4 + 0.07886(40-21) = 0.5842(3.255) + 1.498 = 1.902 + 1.498 = 3.40.
delta

ω = 0

.21

π - 0.19 π = 0.02 π = 0.0628

M \geq (40 - 8) / (2.285 x 0.0628) + 1 = 32 / 0.1435 + 1 = 223 + 1 = 224

, so take $M = 225$ , odd, and

α = 112

In the exam

Use the smaller ripple. Using $δ_{p}$ when $δ_{s}$ is smaller under designs the filter.
Check which branch of the $β$ formula you are in before substituting. $A = 40$ is the middle branch; $A = 60$ is the top one.
A very narrow transition gives a very long filter, and that is the correct answer, not a mistake. Say so rather than quietly rounding M down.
The question sometimes asks only for ripple, attenuation and length, as in 2080 Bhadra. Then delta, A, $β$ and M are the entire answer.

It beats the fixed windows for the same reason the Remez design beats it: the less the error is allowed to exceed the specification, the lower the order.

See it drawnOne parameter, and the whole trade curve instead of four points

Beta slides the window continuously from rectangular at one end to gentler than Blackman at the other, so you design to the specification instead of picking the nearest fixed window and overshooting.

Practise thisEvery question the paper has set on this topic, worked in full8

Practise this +3 like it2080 Bhadra · Q72+2+2 marksKaiser window is to be used to design a linear phase FIR filter that meets following specification:

| H (e^{j ω}) | \leq 0.01

for

0.21 π

≤ |

ω

| ≤

π

;

0.95 \leq | H (e^{j ω}) | \leq 1.05

for

0 \leq | w | \leq 0.19 π

. Calculate the optimum value of ripple, attenuation and window length.

What is different in this one

The ripples are different, 0.05 and 0.01, and the design takes the smaller one. Taking the wrong one gives too short a filter and the stopband then fails.
The transition band is only $0.02 π$ wide, which is very narrow. A filter length in the hundreds is the right answer here, not a mistake.

Take the smaller of the two ripples, delta = 0.01, and turn it into decibels: $A = - 20 \log_{10} (0.01)$ = 40 dB.

\begin{aligned} 21 \leq A \leq 50, so β & = 0.5842 {(A - 21)}^{0.4} + 0.07886 (A - 21) \\ = 0.5842 {(19)}^{0.4} + 0.07886 (19) = 3.3953 \\ transition  dw & = ω_{s} - ω_{p} = 0.6597 - 0.5969 = 0.0628 rad \\ order M \geq (A - 8) / (2.285 dw) \\ = (40 - 8) / (2.285 x 0.0628) = 222.89 \\ round  up : M & = 223, length N = M + 1 = 224 \\ α & = τ = M / 2 = 111.5, ω_{c} = (ω_{p} + ω_{s}) / 2 = 0.6283 rad \end{aligned}

w [n] = \frac{I_{0} (β \sqrt{1 - {[\frac{2 n - (N - 1)}{N - 1}]}^{2}})}{I_{0} (β)}

n	$h_{d} [n]$	$w [n]$	$h [n]$
0	0.00231	0.14797	0.00034
1	0.00089	0.15568	0.00014
2	-0.0009	0.16352	-0.00015
3	-0.00237	0.17149	-0.00041
4	-0.00296	0.17957	-0.00053
5	-0.00242	0.18778	-0.00045
6	-0.00093	0.19611	-0.00018
7	0.00094	0.20455	0.00019
8	0.00249	0.21311	0.00053
9	0.00311	0.22178	0.00069
10	0.00254	0.23056	0.00058
11	0.00098	0.23944	0.00023
12	-0.00099	0.24842	-0.00025

The rest follow from $h [n] = h [223 - n]$ .

$A = 40$ dB, $β = 3.3953$ , $N = 224$ , cut off $ω_{c} = 0.6283$ rad, group delay 111.5 samples. The first coefficients are ${0.00034, 0.00014, - 0.00015, - 0.00041, - 0.00053, - 0.00045}$ .

Same as 2080 Bhadra Q72080 Baishakh · Q72+8 marksIn which case do we choose FIR filter and IIR filter? Design a linear phase FIR filter using Kaiser Window to meet the following specifications:

0.99 \leq | H (e^{j ω}) | \leq 1.01

for

0 \leq w \geq 0.016 π

;

| H (e^{j ω}) | \leq 0.01

for

0.08 π \leq ω \leq 2 π

What is different in this one

Both ripples are 0.01, so there is no choosing to do: delta is that number.

Take the smaller of the two ripples, delta = 0.01, and turn it into decibels: $A = - 20 \log_{10} (0.01)$ = 40 dB.

\begin{aligned} 21 \leq A \leq 50, so β & = 0.5842 {(A - 21)}^{0.4} + 0.07886 (A - 21) \\ = 0.5842 {(19)}^{0.4} + 0.07886 (19) = 3.3953 \\ transition  dw & = ω_{s} - ω_{p} = 0.2513 - 0.0503 = 0.2011 rad \\ order M \geq (A - 8) / (2.285 dw) \\ = (40 - 8) / (2.285 x 0.2011) = 69.65 \\ round  up : M & = 70, length N = M + 1 = 71 \\ α & = τ = M / 2 = 35, ω_{c} = (ω_{p} + ω_{s}) / 2 = 0.1508 rad \end{aligned}

w [n] = \frac{I_{0} (β \sqrt{1 - {[\frac{2 n - (N - 1)}{N - 1}]}^{2}})}{I_{0} (β)}

n	$h_{d} [n]$	$w [n]$	$h [n]$
0	-0.00768	0.14797	-0.00114
1	-0.00857	0.17298	-0.00148
2	-0.00931	0.19923	-0.00186
3	-0.00988	0.22666	-0.00224
4	-0.01026	0.25516	-0.00262
5	-0.01042	0.28466	-0.00297
6	-0.01035	0.31504	-0.00326
7	-0.01003	0.34621	-0.00347
8	-0.00945	0.37803	-0.00357
9	-0.0086	0.4104	-0.00353
10	-0.00748	0.44319	-0.00332
11	-0.0061	0.47625	-0.0029
12	-0.00444	0.50946	-0.00226

The rest follow from $h [n] = h [70 - n]$ .

Practise this2079 Bhadra · Q72+8 marksDefine Gibb's phenomenon. Design the FIR filter using Kaiser window technique for the specifications:

0.899 \leq | H (e^{j ω}) | \leq 1

for

| w | \leq 0.2 π

;

| H (e^{j ω}) | \leq 0.01

for

0.4 π \leq ω \leq π

What is different in this one

The ripples are different, 0.101 and 0.01, and the design takes the smaller one. Taking the wrong one gives too short a filter and the stopband then fails.

Take the smaller of the two ripples, delta = 0.01, and turn it into decibels: $A = - 20 \log_{10} (0.01)$ = 40 dB.

\begin{aligned} 21 \leq A \leq 50, so β & = 0.5842 {(A - 21)}^{0.4} + 0.07886 (A - 21) \\ = 0.5842 {(19)}^{0.4} + 0.07886 (19) = 3.3953 \\ transition  dw & = ω_{s} - ω_{p} = 1.2566 - 0.6283 = 0.6283 rad \\ order M \geq (A - 8) / (2.285 dw) \\ = (40 - 8) / (2.285 x 0.6283) = 22.29 \\ round  up : M & = 23, length N = M + 1 = 24 \\ α & = τ = M / 2 = 11.5, ω_{c} = (ω_{p} + ω_{s}) / 2 = 0.9425 rad \end{aligned}

w [n] = \frac{I_{0} (β \sqrt{1 - {[\frac{2 n - (N - 1)}{N - 1}]}^{2}})}{I_{0} (β)}

n	$h_{d} [n]$	$w [n]$	$h [n]$
0	-0.02734	0.14797	-0.00405
1	-0.01376	0.22787	-0.00314
2	0.01521	0.31772	0.00483
3	0.03699	0.41466	0.01534
4	0.03001	0.51524	0.01546
5	-0.00766	0.6156	-0.00472
6	-0.05157	0.71165	-0.0367
7	-0.06303	0.79931	-0.05038
8	-0.01423	0.87475	-0.01245
9	0.09003	0.93459	0.08414
10	0.20959	0.97608	0.20458
11	0.28902	0.99732	0.28825
12	0.28902	0.99732	0.28825

The rest follow from $h [n] = h [23 - n]$ .

Practise this +1 like it2078 Bhadra · Q78 marksDesign a linear phase FIR filter using KAISER window to meet the following specifications:

| H (e^{j ω}) | \leq 0.01

for

0 \leq | w | \leq 0.25 π

;

0.95 \leq | H (e^{j ω}) | \leq 1.05

for

0.35 π \leq | w | \leq 0.6 π

;

| H (e^{j ω}) | \leq 0.01

for

0.65 π

≤ |

ω

| ≤

π

What is different in this one

This is a band pass specification, not low pass: a stopband, a passband, then a second stopband. The ideal response is therefore the difference of two low pass responses.
There are two transition bands, $0.1 π$ and $0.05 π$ wide, and the narrower one sets the length, because one filter has to meet both.

The narrower transition is what the order formula is fed.

\begin{aligned} lower  transition 1.0996 - 0.7854 & = 0.3142 rad \\ upper  transition 2.042 - 1.885 & = 0.1571 rad \\ take  the  narrower  dw & = 0.1571 rad \\ delta & = min (0.05, 0.01) = 0.01 \\ A & = - 20 \log_{10} (0.01) = 40 dB \\ β & = 3.3953 \\ N \geq (A - 8) / (2.285 dw) + 1 & = 90.15, so N = 91, τ = 45 \\ cut  offsω_{c} 1 & = (0.7854 + 1.0996) / 2 = 0.9425 rad \\ ω_{c} 2 & = (1.885 + 2.042) / 2 = 1.9635 rad \end{aligned}

\begin{aligned} ideal  band  pass h_{d} [n] & = [\sin (ω c 2 (n - t a u)) - \sin (ω c 1 (n - t a u))] / (π (n - τ)) \\ h_{d} [τ] & = (ω_{c} 2 - ω_{c} 1) / π = 0.325 \end{aligned}

n	$h_{d} [n]$	$w [n]$	$h [n]$
0	0.00978	0.14797	0.00145
1	-0.00298	0.16731	-0.0005
2	0.00055	0.18742	0.0001
3	-0.00185	0.20825	-0.00039
4	-0.01345	0.22977	-0.00309
5	0	0.25194	0
6	0.01414	0.27472	0.00389
7	0.00204	0.29806	0.00061
8	-0.00063	0.3219	-0.0002
9	0.00364	0.34621	0.00126
10	-0.01257	0.37091	-0.00466
11	-0.01212	0.39596	-0.0048
12	0.01189	0.42129	0.00501

The rest follow from $h [n] = h [90 - n]$ .

delta = 0.01, $A = 40$ dB, $β = 3.3953$ , $N = 91$ , with cut offs wc1 = 0.9425 and wc2 = 1.9635 rad and a group delay of 45 samples.

Practise this2076 Ashwin · Q92+4+4 marksIn which case do we choose FIR filter and IIR filter? Design a Kaiser Window to meet the following specifications:

0.99 \leq | H (e^{j ω}) | \leq 1.01

for

0 \leq w \leq 0.16 π

;

| H (e^{j ω}) | \leq 0.01

for

0.18 π \leq ω \leq 2 π

. Draw the flow chart for Remez-Exchange algorithmWhat is different hereThe transition band is only

0.02 π

wide, which is what makes the filter long. A number in the hundreds is the right answer here, not a mistake.

How this is readThe stopband is printed as $0.18 π$ to $2 π$ , which runs past $π$ . A digital spectrum stops at $π$ , so the stopband is $0.18 π$ to $π$ and it is worked that way.

What is different in this one

Both ripples are 0.01, so there is no choosing to do: delta is that number.
The transition band is only $0.02 π$ wide, which is very narrow. A filter length in the hundreds is the right answer here, not a mistake.

Take the smaller of the two ripples, delta = 0.01, and turn it into decibels: $A = - 20 \log_{10} (0.01)$ = 40 dB.

\begin{aligned} 21 \leq A \leq 50, so β & = 0.5842 {(A - 21)}^{0.4} + 0.07886 (A - 21) \\ = 0.5842 {(19)}^{0.4} + 0.07886 (19) = 3.3953 \\ transition  dw & = ω_{s} - ω_{p} = 0.5655 - 0.5027 = 0.0628 rad \\ order M \geq (A - 8) / (2.285 dw) \\ = (40 - 8) / (2.285 x 0.0628) = 222.89 \\ round  up : M & = 223, length N = M + 1 = 224 \\ α & = τ = M / 2 = 111.5, ω_{c} = (ω_{p} + ω_{s}) / 2 = 0.5341 rad \end{aligned}

w [n] = \frac{I_{0} (β \sqrt{1 - {[\frac{2 n - (N - 1)}{N - 1}]}^{2}})}{I_{0} (β)}

n	$h_{d} [n]$	$w [n]$	$h [n]$
0	0.0004	0.14797	0.00006
1	0.0018	0.15568	0.00028
2	0.00272	0.16352	0.00044
3	0.00289	0.17149	0.0005
4	0.00225	0.17957	0.0004
5	0.00097	0.18778	0.00018
6	-0.00061	0.19611	-0.00012
7	-0.00205	0.20455	-0.00042
8	-0.00294	0.21311	-0.00063
9	-0.00302	0.22178	-0.00067
10	-0.00225	0.23056	-0.00052
11	-0.00084	0.23944	-0.0002
12	0.00084	0.24842	0.00021

The rest follow from $h [n] = h [223 - n]$ .

$A = 40$ dB, $β = 3.3953$ , $N = 224$ , cut off $ω_{c} = 0.5341$ rad, group delay 111.5 samples. The first coefficients are ${0.00006, 0.00028, 0.00044, 0.0005, 0.0004, 0.00018}$ .

Same as 2080 Bhadra Q72075 Ashwin · Q78+4 marksDesign a linear phase FIR filter using Kaiser Window to meet the following specifications:

0.99 \leq | H (e^{j ω}) | \leq 1.01

for

0 \leq w \leq 0.19 π

;

| H (e^{j ω}) | \leq 0.01

for

0.21 π \leq ω \leq π

. Draw the flow chart for Optimum filter design.

What is different in this one

Both ripples are 0.01, so there is no choosing to do: delta is that number.
The transition band is only $0.02 π$ wide, which is very narrow. A filter length in the hundreds is the right answer here, not a mistake.

Take the smaller of the two ripples, delta = 0.01, and turn it into decibels: $A = - 20 \log_{10} (0.01)$ = 40 dB.

\begin{aligned} 21 \leq A \leq 50, so β & = 0.5842 {(A - 21)}^{0.4} + 0.07886 (A - 21) \\ = 0.5842 {(19)}^{0.4} + 0.07886 (19) = 3.3953 \\ transition  dw & = ω_{s} - ω_{p} = 0.6597 - 0.5969 = 0.0628 rad \\ order M \geq (A - 8) / (2.285 dw) \\ = (40 - 8) / (2.285 x 0.0628) = 222.89 \\ round  up : M & = 223, length N = M + 1 = 224 \\ α & = τ = M / 2 = 111.5, ω_{c} = (ω_{p} + ω_{s}) / 2 = 0.6283 rad \end{aligned}

w [n] = \frac{I_{0} (β \sqrt{1 - {[\frac{2 n - (N - 1)}{N - 1}]}^{2}})}{I_{0} (β)}

n	$h_{d} [n]$	$w [n]$	$h [n]$
0	0.00231	0.14797	0.00034
1	0.00089	0.15568	0.00014
2	-0.0009	0.16352	-0.00015
3	-0.00237	0.17149	-0.00041
4	-0.00296	0.17957	-0.00053
5	-0.00242	0.18778	-0.00045
6	-0.00093	0.19611	-0.00018
7	0.00094	0.20455	0.00019
8	0.00249	0.21311	0.00053
9	0.00311	0.22178	0.00069
10	0.00254	0.23056	0.00058
11	0.00098	0.23944	0.00023
12	-0.00099	0.24842	-0.00025

The rest follow from $h [n] = h [223 - n]$ .

Same as 2078 Bhadra Q72074 Chaitra · Q89+3 marksDesign a FIR linear phase filter using Kaiser window that meets the following specifications:

| H (e^{j ω}) | \leq 0.01

for

0 \leq | w | \leq 0.25 π

;

0.95 \leq | H (e^{j ω}) | \leq 1.05

for

0.35 π \leq | w | \leq 0.6 π

;

| H (e^{j ω}) | \leq 0.01

for

0.65 π

≤ |

ω

| ≤

π

. Also determine the minimum length (M+1) of the impulse response and Kaiser window parameter

β

What is different in this one

This is a band pass specification, not low pass: a stopband, a passband, then a second stopband. The ideal response is therefore the difference of two low pass responses.
There are two transition bands, $0.1 π$ and $0.05 π$ wide, and the narrower one sets the length, because one filter has to meet both.

The narrower transition is what the order formula is fed.

\begin{aligned} lower  transition 1.0996 - 0.7854 & = 0.3142 rad \\ upper  transition 2.042 - 1.885 & = 0.1571 rad \\ take  the  narrower  dw & = 0.1571 rad \\ delta & = min (0.05, 0.01) = 0.01 \\ A & = - 20 \log_{10} (0.01) = 40 dB \\ β & = 3.3953 \\ N \geq (A - 8) / (2.285 dw) + 1 & = 90.15, so N = 91, τ = 45 \\ cut  offsω_{c} 1 & = (0.7854 + 1.0996) / 2 = 0.9425 rad \\ ω_{c} 2 & = (1.885 + 2.042) / 2 = 1.9635 rad \end{aligned}

\begin{aligned} ideal  band  pass h_{d} [n] & = [\sin (ω c 2 (n - t a u)) - \sin (ω c 1 (n - t a u))] / (π (n - τ)) \\ h_{d} [τ] & = (ω_{c} 2 - ω_{c} 1) / π = 0.325 \end{aligned}

n	$h_{d} [n]$	$w [n]$	$h [n]$
0	0.00978	0.14797	0.00145
1	-0.00298	0.16731	-0.0005
2	0.00055	0.18742	0.0001
3	-0.00185	0.20825	-0.00039
4	-0.01345	0.22977	-0.00309
5	0	0.25194	0
6	0.01414	0.27472	0.00389
7	0.00204	0.29806	0.00061
8	-0.00063	0.3219	-0.0002
9	0.00364	0.34621	0.00126
10	-0.01257	0.37091	-0.00466
11	-0.01212	0.39596	-0.0048
12	0.01189	0.42129	0.00501

The rest follow from $h [n] = h [90 - n]$ .

delta = 0.01, $A = 40$ dB, $β = 3.3953$ , $N = 91$ , with cut offs wc1 = 0.9425 and wc2 = 1.9635 rad and a group delay of 45 samples.

Same as 2080 Bhadra Q72072 Kartik · Q86+8 marksDraw the flowchart of Remez-Exchange theorem and explain it. Design an FIR linear phase filter using Kaiser window to meet the following specifications:

0.99 \leq | H (e^{j ω}) | \leq 1.01

for

0 \geq w \geq 0.19 π

;

| H (e^{j ω}) | \leq 0.01

for

0.21 π \leq ω \leq π

How this is readThe paper prints the passband as $0 \geq w \geq 0.19 π$ , its inequalities reversed by the scan. It means 0 to $0.19 π$ , and it is worked that way.

What is different in this one

Both ripples are 0.01, so there is no choosing to do: delta is that number.
The transition band is only $0.02 π$ wide, which is very narrow. A filter length in the hundreds is the right answer here, not a mistake.

Take the smaller of the two ripples, delta = 0.01, and turn it into decibels: $A = - 20 \log_{10} (0.01)$ = 40 dB.

\begin{aligned} 21 \leq A \leq 50, so β & = 0.5842 {(A - 21)}^{0.4} + 0.07886 (A - 21) \\ = 0.5842 {(19)}^{0.4} + 0.07886 (19) = 3.3953 \\ transition  dw & = ω_{s} - ω_{p} = 0.6597 - 0.5969 = 0.0628 rad \\ order M \geq (A - 8) / (2.285 dw) \\ = (40 - 8) / (2.285 x 0.0628) = 222.89 \\ round  up : M & = 223, length N = M + 1 = 224 \\ α & = τ = M / 2 = 111.5, ω_{c} = (ω_{p} + ω_{s}) / 2 = 0.6283 rad \end{aligned}

w [n] = \frac{I_{0} (β \sqrt{1 - {[\frac{2 n - (N - 1)}{N - 1}]}^{2}})}{I_{0} (β)}

n	$h_{d} [n]$	$w [n]$	$h [n]$
0	0.00231	0.14797	0.00034
1	0.00089	0.15568	0.00014
2	-0.0009	0.16352	-0.00015
3	-0.00237	0.17149	-0.00041
4	-0.00296	0.17957	-0.00053
5	-0.00242	0.18778	-0.00045
6	-0.00093	0.19611	-0.00018
7	0.00094	0.20455	0.00019
8	0.00249	0.21311	0.00053
9	0.00311	0.22178	0.00069
10	0.00254	0.23056	0.00058
11	0.00098	0.23944	0.00023
12	-0.00099	0.24842	-0.00025

The rest follow from $h [n] = h [223 - n]$ .

The same questions grouped by how often each topic is set: fir design with the kaiser window

5.3Design by frequency sampling

Design by frequency sampling

Before the formulas

The third method, and a genuinely different idea from the first two. Instead of truncating an ideal response, you state what you want the response to be at N evenly spaced frequencies and transform that back to get $h [n]$ .

The filter then passes exactly through every point you specified, and does as it pleases between them. So the design work is not the transform, which is mechanical. It is choosing the few samples that sit in the transition region, because those are what decide how much ripple you get everywhere else.

The third FIR design method on the syllabus. Instead of truncating an ideal impulse response, you specify the frequency response at N equally spaced points and take the inverse DFT to get $h [n]$ . The designed filter then passes exactly through those points, and does whatever the interpolation does in between.

\begin{aligned} 1. choose N, the filter length \\ 2. sample the desired response at ω_{k} = \frac{2 π k}{N} : \\ H (k) = \underset{the magnitude wanted}{\underset{⏟}{H_{d} (e^{j 2 π k / N})}} \cdot \underset{the linear phase term}{\underset{⏟}{e^{- j π k (N - 1) / N}}} \\ 3. impose H (N - k) = H^{*} (k), so that h [n] is real \\ 4. h [n] = \frac{1}{N} \sum_{k} H (k) e^{j 2 π k n / N} the inverse DFT \\ 5. check between the samples, and free the transition samples if needed \end{aligned}

The naive version is poor. Jumping straight from 1 to 0 between two adjacent samples is the same discontinuity that causes Gibbs' phenomenon, and the stopband comes out at about -20 dB, which is useless.
The fix is transition samples. Leave one, two or three samples in the transition band unspecified and choose their values to minimize the peak stopband ripple. One free sample buys roughly -44 dB, two about -65 dB, three about -85 dB. Tables of the optimum values are standard, and this optimisation is what makes the method usable.
Where it wins: a response given as a set of numbers rather than as a formula, and narrow band filters, where most H(k) are zero and the frequency sampling structure from chapter 4 then realizes it cheaply.
Where it loses: control of the response between the samples, which the window method gives smoothly and Remez gives optimally.

	Window	Frequency sampling	Optimum, Remez
You specify	Band edges and ripple	N samples of H	Band edges and a ripple ratio
Error shape	Largest at the band edge	Zero at the samples, free between	Equiripple everywhere
Order needed	Highest	Middling	Lowest
Design effort	A formula	An IDFT, plus an optimisation for the transition samples	An iterative program

The structure that realizes a filter designed this way, cheaply when most samples are zero, is the frequency sampling structure.

See it drawnYou fix the response at N points and it does as it likes between them

The design passes exactly through every sample you specify. The ripple in between is decided almost entirely by the one or two samples you place in the transition.

5.4Design by optimum approximation: the Remez exchange algorithm

The optimum filter and the Remez exchange algorithm TOP 15/19

82 Bh · 82 Ba · 81 Bh · 79 Bh · 79 Ba · 78 Bh · 76 Ch · 76 Ash · 75 Ch · 75 Ash · 74 Ash · 73 Shr · 72 Ch · 72 Ka · 71 Shr1+692+4+4

Before the formulas

The last method and, in a precise sense, the best one. The window method spreads its error unevenly: large near the band edges, tiny in the middle of a band. That is wasted filter length, because the length is set by the worst error and most of the filter is doing better than it needs to.

The optimum design spreads the error evenly so that the worst case is as small as it can possibly be for that length. That is the minimax idea. The alternation theorem tells you how to recognise when you have reached it, and the Remez algorithm is the search that gets there. Asked in fifteen of nineteen papers, always as the same five part answer.

Asked in fifteen of nineteen papers, worth 4 to 9 marks, and always the same answer: define the optimum filter, state the minimax criterion, state the alternation theorem, write the matrix equation, and draw the flow chart.

What an optimum filter is A filter is optimum, or equiripple, when the maximum weighted error between the designed response and the desired response is as small as possible. The window method minimises nothing in particular; it just truncates, and its error is largest near the band edges. The optimum filter spreads the error evenly, so for a given length the worst case error is the least it can be. It is also called the minimax or weighted Chebyshev design, and the algorithm that finds it is the Remez exchange algorithm, applied to filters by Parks and McClellan.

For a type I linear phase FIR of even order M, write

H (e^{j ω}) = e^{- j ω M / 2} \underset{H_{r} (ω), real}{\underset{⏟}{\sum_{k = 0}^{L} a_{k} \cos (ω k)}}, L = \frac{M}{2}

The alternation theorem, which is why the algorithm works A(

ω

) is the unique best approximation to

H_{d} (ω)

if and only if the error

E (ω)

exhibits at least L + 2 extremal frequencies: there are frequencies

w 1 < w 2 < . .

. <

ω

(L+2) in the bands such that
E( $w [k]$ ) = -E( $w [k + 1]$ ), the error alternates in sign between successive extrema, and
|E( $w [k]$ )| = delta = max | $E (ω)$ |, every one of them reaches the maximum error.
So the optimum filter is recognised by its equiripple error, and the design problem becomes: find the extremal frequencies.

\begin{aligned} THE REMEZ EXCHANGE ALGORITHM \\ 1 . GUESS  an  initial  set  of L + 2 extremal  frequencies ω 1 \dots ω (L + 2), \\ usually spaced evenly across the passband and the stopband. \\ 2 . SOLVE  for  the a_{k} and  for  delta  that  make  the  error  alternate  exactly, \\ that is, force \\ {(- 1)}^{k} . delta & = W (ω [k]) [Hd (ω [k]) - A (ω [k])] \\ which  rearranges  to A (ω [k]) + {(- 1)}^{k} delta / W (ω [k]) & = Hd (ω [k]) \\ and is the linear system \\ | 1 cos ω 1 cos 2 ω 1 \dots \cos (L ω 1) - 1 / W (ω 1) | | a [0] | | Hd (ω 1) | \\ | 1 cos ω 2 cos 2 ω 2 \dots \cos (L ω 2) + 1 / W (ω 2) | | a [1] | | Hd (ω 2) | \\ | . | | \dots | & = | \dots | \\ | 1 cos ω (L + 2) \dots \cos (L ω (L + 2)) {(- 1)}^{L + 2} / W | | delta | | Hd (wL + 2) | \\ In  practice  delta  is  obtained  in  closed  form  and A (ω) by  Lagrange \\ interpolation, rather than by inverting the matrix. \\ 3 . COMPUTE  the  error E (ω) on a dense  grid  of  frequencies . \\ 4 . FIND  the  new  extremal  frequencies : the L + 2 largest  local  peaks  of | E (ω) | . \\ 5 . If  the  new  set  equals  the  old  set, STOP : the  filter  is  optimal . \\ Otherwise EXCHANGE, replacing the old set by the new one, and go to 2. \\ 6 . On  convergence, compute h [n] from  the a_{k} by  the  inverse  relation \\ and  by  the  symmetry h [n] & = h [M - n] . \end{aligned}

The flow chart, which the question nearly always asks for. Draw it as a single loop:

Start with the specification: band edges, desired response, weights, and the order M.
Box: initial guess of L+2 extremal frequencies.
Box: compute delta and the interpolated A( $ω$ ) on the extremal set.
Box: evaluate $E (ω)$ on a dense grid.
Box: find the new extremal frequencies.
Decision diamond: have the extremal frequencies changed? Yes loops back to the delta and A( $ω$ ) box; No falls through.
Box: compute the impulse response $h [n]$ from $a_{k}$ .
Stop.

Estimating the order before you start. The paper sometimes wants the length first, and the standard estimate for an equiripple design is

M = \frac{- 10 \log_{10} (δ_{p} δ_{s}) - 15}{14 Δ f}, Δ f = \frac{ω_{s} - ω_{p}}{2 π}

In the exam

Answer in five parts: what optimum means, the minimax statement, the alternation theorem, the algorithm steps, the flow chart. Papers that ask for 9 marks want all five; papers that ask for 4 want the first three and the chart.
The phrase that earns the definition mark is that the maximum error is minimised, so the error is equiripple across the bands.
L + 2, not L. Getting the number of extremal frequencies wrong is the one detail markers look for.
Draw the chart even when the question does not say draw. It takes a minute and it is the clearest evidence you know the loop.

Compare the error shapes: a window design is worst at the band edge and tiny elsewhere, which is the waste that equiripple removes.

See it drawnSpread the error evenly and the worst error gets as small as it can be

The window method wastes length by being far too good in the middle of a band. The optimum filter makes the error ripple between the same two bounds everywhere, and the alternation theorem is how you know you have got there.

Practise thisEvery question the paper has set on this topic, worked in full3

Practise this2076 Ashwin · Q92+4+4 marksIn which case do we choose FIR filter and IIR filter? Design a Kaiser Window to meet the following specifications:

0.99 \leq | H (e^{j ω}) | \leq 1.01

for

0 \leq w \leq 0.16 π

;

| H (e^{j ω}) | \leq 0.01

for

0.18 π \leq ω \leq 2 π

. Draw the flow chart for Remez-Exchange algorithmWhat is different hereThe transition band is only

0.02 π

wide, which is what makes the filter long. A number in the hundreds is the right answer here, not a mistake.

How this is readThe stopband is printed as $0.18 π$ to $2 π$ , which runs past $π$ . A digital spectrum stops at $π$ , so the stopband is $0.18 π$ to $π$ and it is worked that way.

What is different in this one

Both ripples are 0.01, so there is no choosing to do: delta is that number.
The transition band is only $0.02 π$ wide, which is very narrow. A filter length in the hundreds is the right answer here, not a mistake.

Take the smaller of the two ripples, delta = 0.01, and turn it into decibels: $A = - 20 \log_{10} (0.01)$ = 40 dB.

\begin{aligned} 21 \leq A \leq 50, so β & = 0.5842 {(A - 21)}^{0.4} + 0.07886 (A - 21) \\ = 0.5842 {(19)}^{0.4} + 0.07886 (19) = 3.3953 \\ transition  dw & = ω_{s} - ω_{p} = 0.5655 - 0.5027 = 0.0628 rad \\ order M \geq (A - 8) / (2.285 dw) \\ = (40 - 8) / (2.285 x 0.0628) = 222.89 \\ round  up : M & = 223, length N = M + 1 = 224 \\ α & = τ = M / 2 = 111.5, ω_{c} = (ω_{p} + ω_{s}) / 2 = 0.5341 rad \end{aligned}

w [n] = \frac{I_{0} (β \sqrt{1 - {[\frac{2 n - (N - 1)}{N - 1}]}^{2}})}{I_{0} (β)}

n	$h_{d} [n]$	$w [n]$	$h [n]$
0	0.0004	0.14797	0.00006
1	0.0018	0.15568	0.00028
2	0.00272	0.16352	0.00044
3	0.00289	0.17149	0.0005
4	0.00225	0.17957	0.0004
5	0.00097	0.18778	0.00018
6	-0.00061	0.19611	-0.00012
7	-0.00205	0.20455	-0.00042
8	-0.00294	0.21311	-0.00063
9	-0.00302	0.22178	-0.00067
10	-0.00225	0.23056	-0.00052
11	-0.00084	0.23944	-0.0002
12	0.00084	0.24842	0.00021

The rest follow from $h [n] = h [223 - n]$ .

$A = 40$ dB, $β = 3.3953$ , $N = 224$ , cut off $ω_{c} = 0.5341$ rad, group delay 111.5 samples. The first coefficients are ${0.00006, 0.00028, 0.00044, 0.0005, 0.0004, 0.00018}$ .

Same as 2080 Bhadra Q72075 Ashwin · Q78+4 marksDesign a linear phase FIR filter using Kaiser Window to meet the following specifications:

0.99 \leq | H (e^{j ω}) | \leq 1.01

for

0 \leq w \leq 0.19 π

;

| H (e^{j ω}) | \leq 0.01

for

0.21 π \leq ω \leq π

. Draw the flow chart for Optimum filter design.

What is different in this one

Both ripples are 0.01, so there is no choosing to do: delta is that number.
The transition band is only $0.02 π$ wide, which is very narrow. A filter length in the hundreds is the right answer here, not a mistake.

Take the smaller of the two ripples, delta = 0.01, and turn it into decibels: $A = - 20 \log_{10} (0.01)$ = 40 dB.

\begin{aligned} 21 \leq A \leq 50, so β & = 0.5842 {(A - 21)}^{0.4} + 0.07886 (A - 21) \\ = 0.5842 {(19)}^{0.4} + 0.07886 (19) = 3.3953 \\ transition  dw & = ω_{s} - ω_{p} = 0.6597 - 0.5969 = 0.0628 rad \\ order M \geq (A - 8) / (2.285 dw) \\ = (40 - 8) / (2.285 x 0.0628) = 222.89 \\ round  up : M & = 223, length N = M + 1 = 224 \\ α & = τ = M / 2 = 111.5, ω_{c} = (ω_{p} + ω_{s}) / 2 = 0.6283 rad \end{aligned}

w [n] = \frac{I_{0} (β \sqrt{1 - {[\frac{2 n - (N - 1)}{N - 1}]}^{2}})}{I_{0} (β)}

n	$h_{d} [n]$	$w [n]$	$h [n]$
0	0.00231	0.14797	0.00034
1	0.00089	0.15568	0.00014
2	-0.0009	0.16352	-0.00015
3	-0.00237	0.17149	-0.00041
4	-0.00296	0.17957	-0.00053
5	-0.00242	0.18778	-0.00045
6	-0.00093	0.19611	-0.00018
7	0.00094	0.20455	0.00019
8	0.00249	0.21311	0.00053
9	0.00311	0.22178	0.00069
10	0.00254	0.23056	0.00058
11	0.00098	0.23944	0.00023
12	-0.00099	0.24842	-0.00025

The rest follow from $h [n] = h [223 - n]$ .

0.99 \leq | H (e^{j ω}) | \leq 1.01

for

0 \geq w \geq 0.19 π

;

| H (e^{j ω}) | \leq 0.01

for

0.21 π \leq ω \leq π

How this is readThe paper prints the passband as $0 \geq w \geq 0.19 π$ , its inequalities reversed by the scan. It means 0 to $0.19 π$ , and it is worked that way.

What is different in this one

Both ripples are 0.01, so there is no choosing to do: delta is that number.
The transition band is only $0.02 π$ wide, which is very narrow. A filter length in the hundreds is the right answer here, not a mistake.

Take the smaller of the two ripples, delta = 0.01, and turn it into decibels: $A = - 20 \log_{10} (0.01)$ = 40 dB.

\begin{aligned} 21 \leq A \leq 50, so β & = 0.5842 {(A - 21)}^{0.4} + 0.07886 (A - 21) \\ = 0.5842 {(19)}^{0.4} + 0.07886 (19) = 3.3953 \\ transition  dw & = ω_{s} - ω_{p} = 0.6597 - 0.5969 = 0.0628 rad \\ order M \geq (A - 8) / (2.285 dw) \\ = (40 - 8) / (2.285 x 0.0628) = 222.89 \\ round  up : M & = 223, length N = M + 1 = 224 \\ α & = τ = M / 2 = 111.5, ω_{c} = (ω_{p} + ω_{s}) / 2 = 0.6283 rad \end{aligned}

w [n] = \frac{I_{0} (β \sqrt{1 - {[\frac{2 n - (N - 1)}{N - 1}]}^{2}})}{I_{0} (β)}

n	$h_{d} [n]$	$w [n]$	$h [n]$
0	0.00231	0.14797	0.00034
1	0.00089	0.15568	0.00014
2	-0.0009	0.16352	-0.00015
3	-0.00237	0.17149	-0.00041
4	-0.00296	0.17957	-0.00053
5	-0.00242	0.18778	-0.00045
6	-0.00093	0.19611	-0.00018
7	0.00094	0.20455	0.00019
8	0.00249	0.21311	0.00053
9	0.00311	0.22178	0.00069
10	0.00254	0.23056	0.00058
11	0.00098	0.23944	0.00023
12	-0.00099	0.24842	-0.00025

The rest follow from $h [n] = h [223 - n]$ .

5.5Last minute recall

Last minute recall, chapter 5

Must memorise

$h_{d} [n]$ = sin( $ω_{c}$ (n $- τ$ ))/( $π$ (n $- τ$ )), with hd[ $τ$ ] = $ω_{c} / π$ and $τ = (N - 1) / 2$ .
The window is chosen by the attenuation, the length by the transition width. Rectangular 21 dB, Bartlett 25, Hanning 44, Hamming 53, Blackman 74.
Transition widths: 1.8, 6.1, 6.2, 6.6 and 11 times $π$ /N in the same order.
As = -20 log10( $δ_{s}$ ), so a stopband ripple of 0.01 is 40 dB.
$ω_{c}$ = ( $ω_{p} + ω_{s}$ )/2, and N is rounded up to the next odd value.
Kaiser: $A = - 20 \log_{10} (δ)$ , $β$ from A, N ≥ ( $A - 8) / (2.285$ dw) + 1. Take the smaller ripple, and the narrower transition band.
Gibbs overshoot is about 9 percent whatever N is. Lengthening the filter narrows the ripples, it does not shrink them; only a tapered window does.
Linear phase needs $h [n] = h [N - 1 - n]$ , and the group delay is then $(N - 1) / 2$ samples at every frequency.
The alternation theorem: the best Chebyshev approximation has at least L + 2 extrema of alternating sign and equal size.

Most repeated in this chapter, in order

The optimum filter and the Remez exchange algorithm TOP 15/19, almost always with the flowchart
Design by a window TOP 11/19 and design with a Kaiser window TOP 11/19
Gibbs phenomenon HOT 6/19
When to choose FIR against IIR PIN 2/19

The Remez question is worth more marks than any other single theory question on the paper and is almost entirely bookwork: the definition of an optimum filter, the weighted error, the alternation theorem, and the flowchart. Learn the flowchart.

Chapter 6 · 6 hours · 15 marks · question 9, every paper

IIR filter design

One question, twelve to fifteen marks, in every single paper: design a digital Butterworth filter, almost always by the bilinear transformation. It is the longest single question on the paper and it is entirely procedural, so it is also the most reliable block of marks available.

What this chapter is about

The route: a digital IIR filter is designed by borrowing an analog prototype and mapping it into the z-plane.
The Butterworth approximation and the two formulas that turn a specification into an order and a cutoff.
The two mappings: impulse invariance, which aliases, and the bilinear transformation, which warps.
Spectral transformation, which turns the low pass you designed into a high pass without starting again.

Where it fits

Chapter 3 gave the magnitude response this chapter is trying to hit.
Chapter 4 realises the $H (z)$ that comes out, usually as a cascade of biquads or as a lattice ladder.
Chapter 5 is the other half of filter design, and the comparison between them is asked repeatedly.

What you will learn

6.1 The design route, Impulse invariance
6.2 The bilinear transformation, Bilinear transformation against impulse invariance
6.3 The Butterworth approximation
6.4 Chebyshev, elliptic and Bessel: the other approximations, Digital domain spectral transformation
6.5 Last minute recall, chapter 6

How it is examined

Question 9 is the design, 10 to 15 marks, with a Butterworth prototype and usually the bilinear transformation.
A 2 to 4 mark rider is attached: compare the two methods, explain warping, explain prewarping, or convert the result to a high pass.
The specification arrives in one of three dresses: as magnitudes $(0.8 \leq | H | \leq 1)$ , as decibels (-3 dB and -10 dB), or as ripples ( $δ_{p} = 0.11$ ). All three reduce to the same two numbers.

6.1Design by the impulse invariance method

The design route TOP 18/19

82 Bh · 82 Ba · 81 Bh · 81 Ba · 80 Bh · 80 Ba · 79 Bh · 79 Ba · 78 Bh · 76 Ch · 76 Ash · 75 Ch · 75 Ash · 74 Ch · 74 Ash · 73 Shr · 72 Ka · 71 Shr1212+315

Before the formulas

This is the IIR design chapter, and its method is a detour that is worth understanding before any single technique. Nobody designs an IIR digital filter from nothing. Analog approximation theory is a century old, complete and tabulated: Butterworth, Chebyshev and elliptic responses are known, optimal in a stated sense, and easy to compute.

There is no comparable direct theory in the digital domain. So you convert the digital specification into an analog one, design the analog filter, and map it back. The five steps on this card are the skeleton of almost every answer in the chapter, so learn the route first and the methods second.

Why go through an analog filter at all. Analog approximation theory is a century old and complete: Butterworth, Chebyshev and elliptic responses are tabulated, optimal in a known sense, and easy to compute. There is no equivalent direct theory for IIR digital filters, so the practical route is to design the analog filter and map it.

\begin{aligned} DIGITAL SPECω_{p},ω_{s}in radians per sample, with attenuations \\ | \\ | 1 . PREWARP (bilinear  only) Ω & = (2 / T) \tan (ω / 2) \\ v \\ ANALOG  SPEC Ω_{p}, Ω_{s} with  the  same  attenuations \\ | \\ | 2 . ORDER N and  CUTOFF Ω_{c} & from the Butterworth formulas \\ v \\ ANALOG  PROTOTYPE H_{a} (s), from  the  normalised  Butterworth  polynomial \\ | \\ | 3 . TRANSFORM s \to z & by bilinear or by impulse invariance \\ v \\ DIGITAL  FILTER H (z), then  realise  it  with a chapter 4 structure \end{aligned}

The five steps written out, which is the skeleton of every answer

Convert the specification to $ω_{p}$ and $ω_{s}$ in radians per sample, and the attenuations to dB. If Hz are given, $ω = 2 π$ f / Fs.
Prewarp both band edges, if using the bilinear transformation.
Find N and round up to an integer, then find $Ω_{c}$ .
Write $H_{a} (s)$ from the normalised polynomial of order N with s replaced by s/ $Ω_{c}$ .
Substitute the mapping and simplify to $H (z)$ in powers of $z^{- 1}$ .

Every step below produces the $H (z)$ that chapter 4 then realizes, and whose poles chapter 3 would sketch.

See it drawnNobody designs an IIR filter directly: everyone goes round

Analog approximation theory is finished and tabulated, and there is no comparable direct theory in the digital domain. So the route is always the same five steps, and the paper marks the route as much as the arithmetic.

Practise thisEvery question the paper has set on this topic, worked in full18

Practise this +1 like it2082 Bhadra · Q93+12 marksExplain frequency warping effect in detail. The passband and stopband frequencies are 350Hz and 1,000Hz respectively. The attenuation at passband and stopband are -3dB and -10dB respectively. The sampling frequency is 5,000Hz. Design a digital low-pass Butterworth filter using bilinear transformation technique.What is different herethe specification arrives in hertz, so it has to be turned into radians per sample first

How this is readThe attenuations are printed as -3 dB and -10 dB; the design formulas take them as positive numbers, so Ap = 3 and As = 10.

What is different in this one

The edges are given in hertz, not radians, so the first step is $ω = 2 π$ f / fs. A question that already gives radians skips it.
The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.
$T = 0.0002$ here, not 1, so the factor 2/T in the prewarping is 10000. It cancels in the end, but only if it is carried consistently.

Digital edges first: $ω = 2 π$ f / fs gives $ω_{p} = 2 π$ (350)/5000 = 0.4398 rad and $ω_{s} = 2 π$ (1000)/5000 = 1.2566 rad.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 0.0002) \tan (0.4398 / 2) = 2235.2648 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 0.0002) \tan (1.2566 / 2) = 7265.4253 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1} - 1) / (10^{0.3} - 1)] / (2 \log_{10} (3.2504)) \\ = \log_{10} (9 / 0.9953) / (2 log 103.2504) \\ = 0.934 so  take N = 1 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 2235.2648 / {(0.9953)}^{1 / 2} \\ = 2240.5787 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 7265.4253 / {(9)}^{1 / 2} \\ = 2421.8084 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 2240.5787$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 1 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 0 \\ s 0 & = - 2240.5787 \end{aligned}

Multiply the factors back together in conjugate pairs, which is what keeps every coefficient real. A pair at s = $- σ$ +/- j $ω$ gives $s^{2} + 2 σ s + W c^{2}$ , since the product of a pole and its conjugate is always $Ω_{c}^{2}$ .

H_{a} (s) = \frac{{(2240.5787)}^{1}}{(s + 2240.5787)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 10000 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.183 + 0.183 z^{- 1}}{1 - 0.6339 z^{- 1}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.4398$	0.7079	3 dB, against the 3 dB allowed
$ω = ω_{s} = 1.2566$	0.2947	10.61 dB, against the 10 dB required
$ω = π$	0	at Nyquist

$N = 1$ and $Ω_{c} = 2240.5787$ rad/s. The specification is met at both edges.

H (z) = \frac{0.183 + 0.183 z^{- 1}}{1 - 0.6339 z^{- 1}}

Practise this +12 like it2082 Baishakh · Q812+3 marksUsing bilinear transformation technique, design a Butterworth filter which satisfies the following conditions:

0.8 \leq | H (e^{j ω}) | \leq 1

for

0 \leq w \leq 0.2 π

;

| H (e^{j ω}) | \leq 0.2

for

0.6 π \leq ω \leq π

. Consider sampling frequency 1 Hz. Compare Impulse Invariance Method and Bilinear Transformation technique for analog filter to digital filter conversion.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 1) \tan (0.6283 / 2) = 0.6498 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 1) \tan (1.885 / 2) = 2.7528 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.398} - 1) / (10^{0.194} - 1)] / (2 \log_{10} (4.2361)) \\ = \log_{10} (24 / 0.5625) / (2 log 104.2361) \\ = 1.3 so  take N = 2 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.6498 / {(0.5625)}^{1 / 4} \\ = 0.7504 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 2.7528 / {(24)}^{1 / 4} \\ = 1.2437 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.7504$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 2 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 1 \\ s 0 & = - 0.5306 + j 0.5306 \\ s 1 & = - 0.5306 - j 0.5306 \end{aligned}

H_{a} (s) = \frac{{(0.7504)}^{2}}{(s^{2} + 1.0612 s + 0.5631)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 2 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.0842 + 0.1684 z^{- 1} + 0.0842 z^{- 2}}{1 - 1.0282 z^{- 1} + 0.3651 z^{- 2}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.6283$	0.8	1.94 dB, against the 1.94 dB allowed
$ω = ω_{s} = 1.885$	0.0741	22.6 dB, against the 13.98 dB required
$ω = π$	0	at Nyquist

$N = 2$ and $Ω_{c} = 0.7504$ rad/s. The specification is met at both edges.

H (z) = \frac{0.0842 + 0.1684 z^{- 1} + 0.0842 z^{- 2}}{1 - 1.0282 z^{- 1} + 0.3651 z^{- 2}}

Same as 2082 Baishakh Q82081 Bhadra · Q812+3 marksDesign a digital low pass IIR filter using Bilinear transformation to meet following specifications:

0.9 \leq | H (e^{j ω}) | \leq 1

for

0 \leq | w | \leq \frac{π}{2}

;

| H (e^{j ω}) | \leq 0.2

for

3 π / 4 \leq | w | \leq π

. Use Butterworth approximation for design with sampling frequency of 1Hz. Compare Impulse Invariance Method and Bilinear Transformation method.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 1) \tan (1.5708 / 2) = 2 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 1) \tan (2.3562 / 2) = 4.8284 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.398} - 1) / (10^{0.092} - 1)] / (2 \log_{10} (2.4142)) \\ = \log_{10} (24 / 0.2346) / (2 log 102.4142) \\ = 2.6255 so  take N = 3 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 2 / {(0.2346)}^{1 / 6} \\ = 2.5467 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 4.8284 / {(24)}^{1 / 6} \\ = 2.843 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 2.5467$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 3 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 2 \\ s 0 & = - 1.2734 + j 2.2055 \\ s 1 & = - 2.5467 \\ s 2 & = - 1.2734 - j 2.2055 \end{aligned}

H_{a} (s) = \frac{{(2.5467)}^{3}}{(s^{2} + 2.5467 s + 6.4859) (s + 2.5467)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 2 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.2332 + 0.6996 z^{- 1} + 0.6996 z^{- 2} + 0.2332 z^{- 3}}{1 + 0.4394 z^{- 1} + 0.3845 z^{- 2} + 0.0416 z^{- 3}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 1.5708$	0.9	0.92 dB, against the 0.92 dB allowed
$ω = ω_{s} = 2.3562$	0.1452	16.76 dB, against the 13.98 dB required
$ω = π$	0	at Nyquist

$N = 3$ and $Ω_{c} = 2.5467$ rad/s. The specification is met at both edges.

H (z) = \frac{0.2332 + 0.6996 z^{- 1} + 0.6996 z^{- 2} + 0.2332 z^{- 3}}{1 + 0.4394 z^{- 1} + 0.3845 z^{- 2} + 0.0416 z^{- 3}}

Same as 2082 Baishakh Q82081 Baishakh · Q82+10 marksDifferentiate between bilinear transformation and impulse invariance. Design a Butterworth digital IIR lowpass filter using bilinear transformation by taking

T = 0.1

second, to satisfy the following specifications:

0.6 \leq | H (e^{j ω}) | \leq 1

for

0 \leq w \leq 0.35 π

;

| H (e^{j ω}) | \leq 0.1

for

0.7 π \leq ω \leq π

How this is readThe question fixes $T = 0.1$ s rather than giving a sampling frequency, so the prewarping uses 2/T = 20.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.
$T = 0.1$ here, not 1, so the factor 2/T in the prewarping is 20. It cancels in the end, but only if it is carried consistently.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 0.1) \tan (1.0996 / 2) = 12.256 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 0.1) \tan (2.1991 / 2) = 39.2522 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{2} - 1) / (10^{0.444} - 1)] / (2 \log_{10} (3.2027)) \\ = \log_{10} (99 / 1.7778) / (2 log 103.2027) \\ = 1.7267 so  take N = 2 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 12.256 / {(1.7778)}^{1 / 4} \\ = 10.614 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 39.2522 / {(99)}^{1 / 4} \\ = 12.4439 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 10.614$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 2 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 1 \\ s 0 & = - 7.5052 + j 7.5052 \\ s 1 & = - 7.5052 - j 7.5052 \end{aligned}

H_{a} (s) = \frac{{(10.614)}^{2}}{(s^{2} + 15.0105 s + 112.6574)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 20 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.1386 + 0.2772 z^{- 1} + 0.1386 z^{- 2}}{1 - 0.707 z^{- 1} + 0.2614 z^{- 2}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 1.0996$	0.6	4.44 dB, against the 4.44 dB allowed
$ω = ω_{s} = 2.1991$	0.0729	22.74 dB, against the 20 dB required
$ω = π$	0	at Nyquist

$N = 2$ and $Ω_{c} = 10.614$ rad/s. The specification is met at both edges.

H (z) = \frac{0.1386 + 0.2772 z^{- 1} + 0.1386 z^{- 2}}{1 - 0.707 z^{- 1} + 0.2614 z^{- 2}}

Same as 2082 Baishakh Q82080 Bhadra · Q812 marksUsing Bilinear transformation, design a Butterworth low pass filter which satisfies following conditions:

0.9 \leq | H (e^{j ω}) | \leq 1

for

0 \leq ω \leq \frac{π}{2}

;

| H (e^{j ω}) | \leq 0.2

for

\frac{3 π}{4} \leq ω \leq π

. Consider sampling frequency of 1 Hz.

How this is read $0.9 \leq | H | \leq 1$ in the passband means Ap = -20 $\log_{10} (0.9) = 0.9151$ dB, and $| H | \leq 0.2$ in the stopband means As = -20 $\log_{10} (0.2) = 13.9794$ dB. A sampling frequency of 1 Hz gives $T = 1$ s.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 1) \tan (1.5708 / 2) = 2 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 1) \tan (2.3562 / 2) = 4.8284 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.398} - 1) / (10^{0.092} - 1)] / (2 \log_{10} (2.4142)) \\ = \log_{10} (24 / 0.2346) / (2 log 102.4142) \\ = 2.6255 so  take N = 3 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 2 / {(0.2346)}^{1 / 6} \\ = 2.5467 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 4.8284 / {(24)}^{1 / 6} \\ = 2.843 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 2.5467$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 3 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 2 \\ s 0 & = - 1.2734 + j 2.2055 \\ s 1 & = - 2.5467 \\ s 2 & = - 1.2734 - j 2.2055 \end{aligned}

H_{a} (s) = \frac{{(2.5467)}^{3}}{(s^{2} + 2.5467 s + 6.4859) (s + 2.5467)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 2 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.2332 + 0.6996 z^{- 1} + 0.6996 z^{- 2} + 0.2332 z^{- 3}}{1 + 0.4394 z^{- 1} + 0.3845 z^{- 2} + 0.0416 z^{- 3}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 1.5708$	0.9	0.92 dB, against the 0.92 dB allowed
$ω = ω_{s} = 2.3562$	0.1452	16.76 dB, against the 13.98 dB required
$ω = π$	0	at Nyquist

$N = 3$ and $Ω_{c} = 2.5467$ rad/s. The specification is met at both edges.

H (z) = \frac{0.2332 + 0.6996 z^{- 1} + 0.6996 z^{- 2} + 0.2332 z^{- 3}}{1 + 0.4394 z^{- 1} + 0.3845 z^{- 2} + 0.0416 z^{- 3}}

Same as 2082 Baishakh Q82080 Baishakh · Q911 marksDesign a low pass digital IIR filter by Bilinear Transformation method to an approximate Butterworth low pass filter, if passband edge frequency is

0.26 π

radians and maximum deviation of 0.99 dB below 0 dB gain in the passband. The maximum gain of -14.99 dB and frequency is

0.58 π

radians in stopband, Consider sampling frequency 0.5 Hz.

How this is readA sampling frequency of 0.5 Hz gives $T = 2$ s.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.
$T = 2$ here, not 1, so the factor 2/T in the prewarping is 1. It cancels in the end, but only if it is carried consistently.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 2) \tan (0.8168 / 2) = 0.4327 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 2) \tan (1.8221 / 2) = 1.2892 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.499} - 1) / (10^{0.099} - 1)] / (2 \log_{10} (2.9791)) \\ = \log_{10} (30.55 / 0.256) / (2 log 102.9791) \\ = 2.1902 so  take N = 3 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.4327 / {(0.256)}^{1 / 6} \\ = 0.5431 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 1.2892 / {(30.55)}^{1 / 6} \\ = 0.7291 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.5431$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 3 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 2 \\ s 0 & = - 0.2715 + j 0.4703 \\ s 1 & = - 0.5431 \\ s 2 & = - 0.2715 - j 0.4703 \end{aligned}

H_{a} (s) = \frac{{(0.5431)}^{3}}{(s^{2} + 0.5431 s + 0.2949) (s + 0.5431)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 1 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.0565 + 0.1694 z^{- 1} + 0.1694 z^{- 2} + 0.0565 z^{- 3}}{1 - 1.0634 z^{- 1} + 0.6363 z^{- 2} - 0.1211 z^{- 3}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.8168$	0.8923	0.99 dB, against the 0.99 dB allowed
$ω = ω_{s} = 1.8221$	0.0745	22.55 dB, against the 14.99 dB required
$ω = π$	0	at Nyquist

$N = 3$ and $Ω_{c} = 0.5431$ rad/s. The specification is met at both edges.

H (z) = \frac{0.0565 + 0.1694 z^{- 1} + 0.1694 z^{- 2} + 0.0565 z^{- 3}}{1 - 1.0634 z^{- 1} + 0.6363 z^{- 2} - 0.1211 z^{- 3}}

Practise this2079 Bhadra · Q911+4 marksDesign a low pass discrete time Butterworth filter using bilinear transformation having following specifications: Passband frequency (

Ω_{p}

) =

0.25 π

radians, Stopband frequency (

Ω_{s}

) =

0.55 π

radians, Passband ripple (deltaP) = 0.11, Stopband ripple (deltaS) = 0.21. Consider sampling frequency of 0.5 Hz. Also, convert the obtained digital low pass filter to high pass filter with new pass band frequency, W'p =

0.45 π

using digital domain transformation.

The design

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.
$T = 2$ here, not 1, so the factor 2/T in the prewarping is 1. It cancels in the end, but only if it is carried consistently.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 2) \tan (0.7854 / 2) = 0.4142 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 2) \tan (1.7279 / 2) = 1.1708 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.356} - 1) / (10^{0.101} - 1)] / (2 \log_{10} (2.8267)) \\ = \log_{10} (21.676 / 0.2625) / (2 log 102.8267) \\ = 2.1239 so  take N = 3 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.4142 / {(0.2625)}^{1 / 6} \\ = 0.5177 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 1.1708 / {(21.676)}^{1 / 6} \\ = 0.7012 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.5177$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 3 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 2 \\ s 0 & = - 0.2588 + j 0.4483 \\ s 1 & = - 0.5177 \\ s 2 & = - 0.2588 - j 0.4483 \end{aligned}

H_{a} (s) = \frac{{(0.5177)}^{3}}{(s^{2} + 0.5177 s + 0.268) (s + 0.5177)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 1 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.0512 + 0.1536 z^{- 1} + 0.1536 z^{- 2} + 0.0512 z^{- 3}}{1 - 1.1377 z^{- 1} + 0.6808 z^{- 2} - 0.1335 z^{- 3}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.7854$	0.89	1.01 dB, against the 1.01 dB allowed
$ω = ω_{s} = 1.7279$	0.0861	21.3 dB, against the 13.56 dB required
$ω = π$	0	at Nyquist

$N = 3$ and $Ω_{c} = 0.5177$ rad/s. The specification is met at both edges.

H (z) = \frac{0.0512 + 0.1536 z^{- 1} + 0.1536 z^{- 2} + 0.0512 z^{- 3}}{1 - 1.1377 z^{- 1} + 0.6808 z^{- 2} - 0.1335 z^{- 3}}

The rest of this questionThis topic works the part that belongs to it. The other part is worked under Converting it to a high pass, the last four marks, where it belongs.

Same as 2082 Baishakh Q82079 Baishakh · Q911+3 marksDesign a low pass digital IIR filter by Bilinear Transformation method to an approximate Butterworth low pass filter, if passband edge frequency is

0.24 π

radians and maximum deviation of 0.98 dB below 0 dB gain in the passband. The maximum gain of -14.95 dB and frequency is

0.57 π

radians in stopband, consider sampling frequency 0.5 Hz. Compare impulse invariance method with bilinear transformation method.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.
$T = 2$ here, not 1, so the factor 2/T in the prewarping is 1. It cancels in the end, but only if it is carried consistently.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 2) \tan (0.754 / 2) = 0.3959 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 2) \tan (1.7907 / 2) = 1.2482 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.495} - 1) / (10^{0.098} - 1)] / (2 \log_{10} (3.1526)) \\ = \log_{10} (30.261 / 0.2531) / (2 log 103.1526) \\ = 2.0831 so  take N = 3 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.3959 / {(0.2531)}^{1 / 6} \\ = 0.4978 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 1.2482 / {(30.261)}^{1 / 6} \\ = 0.7071 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.4978$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 3 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 2 \\ s 0 & = - 0.2489 + j 0.4311 \\ s 1 & = - 0.4978 \\ s 2 & = - 0.2489 - j 0.4311 \end{aligned}

H_{a} (s) = \frac{{(0.4978)}^{3}}{(s^{2} + 0.4978 s + 0.2478) (s + 0.4978)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 1 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.0472 + 0.1415 z^{- 1} + 0.1415 z^{- 2} + 0.0472 z^{- 3}}{1 - 1.1971 z^{- 1} + 0.7186 z^{- 2} - 0.1441 z^{- 3}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.754$	0.8933	0.98 dB, against the 0.98 dB allowed
$ω = ω_{s} = 1.7907$	0.0633	23.97 dB, against the 14.95 dB required
$ω = π$	0	at Nyquist

$N = 3$ and $Ω_{c} = 0.4978$ rad/s. The specification is met at both edges.

H (z) = \frac{0.0472 + 0.1415 z^{- 1} + 0.1415 z^{- 2} + 0.0472 z^{- 3}}{1 - 1.1971 z^{- 1} + 0.7186 z^{- 2} - 0.1441 z^{- 3}}

Practise this2078 Bhadra · Q912+3 marksDesign a LPF Butterworth filter using Impulse Invariance Method (IIM) method with passband and stopband frequencies 200Hz and 500Hz respectively. The passband and stopband attenuations are 5dB and 12dB respectively. The sampling frequency is 5000Hz. What is pre-warping and why it is necessary? Explain.What is different hereimpulse invariance, not the bilinear transformation, so there is no prewarping and the mapping is a scaling; the specification arrives in hertz, so it has to be turned into radians per sample first. Impulse invariance, not the bilinear transformation, so there is no prewarping: the edges scale linearly,

Ω = ω / T

. The second half of the question asks what prewarping is and why it is necessary, which is exactly the step this design does not take. Watch the standard text here: its worked impulse invariance example turns

2 / ((s + 1) (s + 2))

T = 0.2

into 0.2774z over

z^{2} - 1.4887 z + 0.54853

, but the numerator is wrong. It is

2 (e^{- 0} .2 - e^{- 0} .4) = 2 (0.8187 - 0.6703) = < b > 0.2968 < / b

>; the book doubled 0.67 as 1.36 instead of 1.34. The denominator is right.

What is different in this one

The edges are given in hertz, not radians, so the first step is $ω = 2 π$ f / fs. A question that already gives radians skips it.
Impulse invariance, not the bilinear transformation. There is therefore no prewarping: the edges scale linearly, $Ω = ω / T$ , and the mapping is applied to each partial fraction term separately rather than to $H (s)$ as a whole.
$T = 0.0002$ here, not 1, so the factor 2/T in the prewarping is 10000. It cancels in the end, but only if it is carried consistently.
Impulse invariance, not the bilinear transformation, so there is no prewarping: the edges scale linearly, $Ω = ω / T$ . The second half of the question asks what prewarping is and why it is necessary, which is exactly the step this design does not take. Watch the standard text here: its worked impulse invariance example turns $2 / ((s + 1) (s + 2))$ at $T = 0.2$ into 0.2774z over $z^{2} - 1.4887 z + 0.54853$ , but the numerator is wrong. It is $2 (e^{- 0} .2 - e^{- 0} .4) = 2 (0.8187 - 0.6703)$ = 0.2968; the book doubled 0.67 as 1.36 instead of 1.34. The denominator is right.

Digital edges first: $ω = 2 π$ f / fs gives $ω_{p} = 2 π$ (200)/5000 = 0.2513 rad and $ω_{s} = 2 π$ (500)/5000 = 0.6283 rad.

Impulse invariance maps frequency linearly, $Ω = ω / T$ , so there is no prewarping to do.

\begin{aligned} Ω_{p} & = ω_{p} / T = 1256.6371 rad / s \\ Ω_{s} & = ω_{s} / T = 3141.5927 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.2} - 1) / (10^{0.5} - 1)] / (2 \log_{10} (2.5)) \\ = \log_{10} (14.849 / 2.1623) / (2 log 102.5) \\ = 1.0514 so  take N = 2 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 1256.6371 / {(2.1623)}^{1 / 4} \\ = 1036.2916 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 3141.5927 / {(14.849)}^{1 / 4} \\ = 1600.3909 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 1036.2916$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 2 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 1 \\ s 0 & = - 732.7689 + j 732.7689 \\ s 1 & = - 732.7689 - j 732.7689 \end{aligned}

H_{a} (s) = \frac{{(1036.2916)}^{2}}{(s^{2} + 1465.5377 s + 1073900.3832)}

Expand $H_{a} (s)$ in partial fractions and map each pole on its own with s = pk going to z = $e^{p k T}$ , which is what impulse invariance does.

\begin{aligned} residue - j 732.7689 at  pole - 732.7689 + j 732.7689 \\ \to - j 0.1466 / (1 - (0.8544 + j 0.1261) z^{- 1}) \\ residue j 732.7689 at  pole - 732.7689 - j 732.7689 \\ \to j 0.1466 / (1 - (0.8544 - j 0.1261) z^{- 1}) \end{aligned}

H (z) = \frac{0.037 z^{- 1}}{1 - 1.7088 z^{- 1} + 0.7459 z^{- 2}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	0.9964	the DC gain
$ω = ω_{p} = 0.2513$	0.5633	4.99 dB, against the 5 dB allowed
$ω = ω_{s} = 0.6283$	0.1114	19.06 dB, against the 12 dB required
$ω = π$	0.0107	at Nyquist

$N = 2$ and $Ω_{c} = 1036.2916$ rad/s. The specification is met at both edges.

H (z) = \frac{0.037 z^{- 1}}{1 - 1.7088 z^{- 1} + 0.7459 z^{- 2}}

Same as 2082 Bhadra Q92076 Chaitra · Q1010 marksDesign a digital low pass Butterworth filter by applying bilinear transformation techniques for the given specifications: Passband peak to peak ripple ≤ 1dB, Passband edge frequency = 1.2KHz, Stopband Attenuation ≥ 40dB, Stopband edge frequency = 2.5 KHz, Sample rate = 8KHz

What is different in this one

The edges are given in hertz, not radians, so the first step is $ω = 2 π$ f / fs. A question that already gives radians skips it.
The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.
$T = 0.0001$ here, not 1, so the factor 2/T in the prewarping is 16000. It cancels in the end, but only if it is carried consistently.

Digital edges first: $ω = 2 π$ f / fs gives $ω_{p} = 2 π$ (1200)/8000 = 0.9425 rad and $ω_{s} = 2 π$ (2500)/8000 = 1.9635 rad.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 0.0001) \tan (0.9425 / 2) = 8152.4072 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 0.0001) \tan (1.9635 / 2) = 23945.6922 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{4} - 1) / (10^{0.1} - 1)] / (2 \log_{10} (2.9373)) \\ = \log_{10} (9999 / 0.2589) / (2 log 102.9373) \\ = 4.901 so  take N = 5 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 8152.4072 / {(0.2589)}^{1 / 10} \\ = 9331.8639 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 23945.6922 / {(9999)}^{1 / 10} \\ = 9533.0471 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 9331.8639$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 5 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 4 \\ s 0 & = - 2883.7045 + j 8875.13 \\ s 1 & = - 7549.6365 + j 5485.132 \\ s 2 & = - 9331.8639 \\ s 3 & = - 7549.6365 - j 5485.132 \\ s 4 & = - 2883.7045 - j 8875.13 \end{aligned}

\begin{aligned} H_{a} (s) & = \frac{{(9331.8639)}^{5}}{D (s)} \\ D (s) & = (s^{2} + 5767.4091 s + 87083683.7083) \\ \times (s^{2} + 15099.273 s + 87083683.7083) \\ \times (s + 9331.8639) \end{aligned}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 16000 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.011 + 0.0549 z^{- 1} + 0.1098 z^{- 2} + 0.1098 z^{- 3} + 0.0549 z^{- 4} + 0.011 z^{- 5}}{1 - 1.617 z^{- 1} + 1.5544 z^{- 2} - 0.7829 z^{- 3} + 0.2231 z^{- 4} - 0.0263 z^{- 5}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.9425$	0.8913	1 dB, against the 1 dB allowed
$ω = ω_{s} = 1.9635$	0.009	40.93 dB, against the 40 dB required
$ω = π$	0	at Nyquist

$N = 5$ and $Ω_{c} = 9331.8639$ rad/s. The specification is met at both edges.

H (z) = \frac{0.011 + 0.0549 z^{- 1} + 0.1098 z^{- 2} + 0.1098 z^{- 3} + 0.0549 z^{- 4} + 0.011 z^{- 5}}{1 - 1.617 z^{- 1} + 1.5544 z^{- 2} - 0.7829 z^{- 3} + 0.2231 z^{- 4} - 0.0263 z^{- 5}}

Same as 2082 Baishakh Q82076 Ashwin · Q712 marksDesign a low pass discrete IIR filter by Bilinear Transformation method to an approximate Butterworth filter having specifications as below: Pass bandedge frequency (

ω_{p}

) =

0.22 π

radians, Stop bandedge frequency (

ω_{s}

) =

0.54 π

radians, Passband ripple (deltap) = 0.11, Stopband ripple (deltas) = 0.22, Consider sampling frequency 0.5 Hz.

How this is readA passband ripple of 0.11 means Ap = -20 $\log_{10} (0.89) = 1.0122$ dB, and a stopband ripple of 0.22 means As = -20 $\log_{10} (0.22) = 13.1515$ dB. fs = 0.5 Hz gives $T = 2$ s.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.
$T = 2$ here, not 1, so the factor 2/T in the prewarping is 1. It cancels in the end, but only if it is carried consistently.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 2) \tan (0.6912 / 2) = 0.36 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 2) \tan (1.6965 / 2) = 1.1343 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.315} - 1) / (10^{0.101} - 1)] / (2 \log_{10} (3.1506)) \\ = \log_{10} (19.661 / 0.2625) / (2 log 103.1506) \\ = 1.8806 so  take N = 2 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.36 / {(0.2625)}^{1 / 4} \\ = 0.503 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 1.1343 / {(19.661)}^{1 / 4} \\ = 0.5387 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.503$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 2 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 1 \\ s 0 & = - 0.3557 + j 0.3557 \\ s 1 & = - 0.3557 - j 0.3557 \end{aligned}

H_{a} (s) = \frac{{(0.503)}^{2}}{(s^{2} + 0.7113 s + 0.253)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 1 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.1288 + 0.2576 z^{- 1} + 0.1288 z^{- 2}}{1 - 0.7606 z^{- 1} + 0.2757 z^{- 2}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.6912$	0.89	1.01 dB, against the 1.01 dB allowed
$ω = ω_{s} = 1.6965$	0.1929	14.29 dB, against the 13.15 dB required
$ω = π$	0	at Nyquist

$N = 2$ and $Ω_{c} = 0.503$ rad/s. The specification is met at both edges.

H (z) = \frac{0.1288 + 0.2576 z^{- 1} + 0.1288 z^{- 2}}{1 - 0.7606 z^{- 1} + 0.2757 z^{- 2}}

Same as 2082 Baishakh Q82075 Chaitra · Q710 marksUsing bilinear transformation, design a digital filter using Butterworth approximation which satisfies the following conditions:

0.8 \leq | H (e^{j ω}) | \leq 1

for

0 \leq W \leq 0.2 π

;

| H (e^{j ω}) | \leq 0.2

for

0.6 π \leq W \leq π

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 1) \tan (0.6283 / 2) = 0.6498 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 1) \tan (1.885 / 2) = 2.7528 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.398} - 1) / (10^{0.194} - 1)] / (2 \log_{10} (4.2361)) \\ = \log_{10} (24 / 0.5625) / (2 log 104.2361) \\ = 1.3 so  take N = 2 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.6498 / {(0.5625)}^{1 / 4} \\ = 0.7504 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 2.7528 / {(24)}^{1 / 4} \\ = 1.2437 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.7504$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 2 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 1 \\ s 0 & = - 0.5306 + j 0.5306 \\ s 1 & = - 0.5306 - j 0.5306 \end{aligned}

H_{a} (s) = \frac{{(0.7504)}^{2}}{(s^{2} + 1.0612 s + 0.5631)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 2 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.0842 + 0.1684 z^{- 1} + 0.0842 z^{- 2}}{1 - 1.0282 z^{- 1} + 0.3651 z^{- 2}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.6283$	0.8	1.94 dB, against the 1.94 dB allowed
$ω = ω_{s} = 1.885$	0.0741	22.6 dB, against the 13.98 dB required
$ω = π$	0	at Nyquist

$N = 2$ and $Ω_{c} = 0.7504$ rad/s. The specification is met at both edges.

H (z) = \frac{0.0842 + 0.1684 z^{- 1} + 0.0842 z^{- 2}}{1 - 1.0282 z^{- 1} + 0.3651 z^{- 2}}

Same as 2082 Baishakh Q82075 Ashwin · Q612 marksDesign a digital low-pass filter with the following specification: i) Pass-band magnitude constant to 0.7 dB below the frequency of

0.15 π

ii) Stop-band attenuation at least 14 dB for the frequencies between

0.6 π

π

. Use Butterworth approximation as a prototype and use bilinear transformation method to obtain the digital filter.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 1) \tan (0.4712 / 2) = 0.4802 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 1) \tan (1.885 / 2) = 2.7528 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.4} - 1) / (10^{0.07} - 1)] / (2 \log_{10} (5.733)) \\ = \log_{10} (24.119 / 0.1749) / (2 log 105.733) \\ = 1.4106 so  take N = 2 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.4802 / {(0.1749)}^{1 / 4} \\ = 0.7425 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 2.7528 / {(24.119)}^{1 / 4} \\ = 1.2422 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.7425$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 2 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 1 \\ s 0 & = - 0.525 + j 0.525 \\ s 1 & = - 0.525 - j 0.525 \end{aligned}

H_{a} (s) = \frac{{(0.7425)}^{2}}{(s^{2} + 1.05 s + 0.5513)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 2 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.0829 + 0.1658 z^{- 1} + 0.0829 z^{- 2}}{1 - 1.037 z^{- 1} + 0.3685 z^{- 2}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.4712$	0.9226	0.7 dB, against the 0.7 dB allowed
$ω = ω_{s} = 1.885$	0.0726	22.79 dB, against the 14 dB required
$ω = π$	0	at Nyquist

$N = 2$ and $Ω_{c} = 0.7425$ rad/s. The specification is met at both edges.

H (z) = \frac{0.0829 + 0.1658 z^{- 1} + 0.0829 z^{- 2}}{1 - 1.037 z^{- 1} + 0.3685 z^{- 2}}

Practise this2074 Chaitra · Q712 marksDesign a digital low-pass filter with the following specification: i) Pass-band magnitude constant to 0.7 dB below the frequency of

0.15 π

ii) Stop-band attenuation at least 14 dB for the frequencies between

0.6 π

π

. Use Butterworth approximation as a prototype and use impulse invariance method to obtain the digital filter.What is different hereimpulse invariance, not the bilinear transformation, so there is no prewarping and the mapping is a scaling

What is different in this one

Impulse invariance, not the bilinear transformation. There is therefore no prewarping: the edges scale linearly, $Ω = ω / T$ , and the mapping is applied to each partial fraction term separately rather than to $H (s)$ as a whole.

Impulse invariance maps frequency linearly, $Ω = ω / T$ , so there is no prewarping to do.

\begin{aligned} Ω_{p} & = ω_{p} / T = 0.4712 rad / s \\ Ω_{s} & = ω_{s} / T = 1.885 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.4} - 1) / (10^{0.07} - 1)] / (2 \log_{10} (4)) \\ = \log_{10} (24.119 / 0.1749) / (2 log 104) \\ = 1.7769 so  take N = 2 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.4712 / {(0.1749)}^{1 / 4} \\ = 0.7287 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 1.885 / {(24.119)}^{1 / 4} \\ = 0.8506 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.7287$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 2 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 1 \\ s 0 & = - 0.5153 + j 0.5153 \\ s 1 & = - 0.5153 - j 0.5153 \end{aligned}

H_{a} (s) = \frac{{(0.7287)}^{2}}{(s^{2} + 1.0305 s + 0.531)}

Expand $H_{a} (s)$ in partial fractions and map each pole on its own with s = pk going to z = $e^{p k T}$ , which is what impulse invariance does.

\begin{aligned} residue - j 0.5153 at  pole - 0.5153 + j 0.5153 \\ \to - j 0.5153 / (1 - (0.5198 + j 0.2943) z^{- 1}) \\ residue j 0.5153 at  pole - 0.5153 - j 0.5153 \\ \to j 0.5153 / (1 - (0.5198 - j 0.2943) z^{- 1}) \end{aligned}

H (z) = \frac{0.3033 z^{- 1}}{1 - 1.0396 z^{- 1} + 0.3568 z^{- 2}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	0.9561	the DC gain
$ω = ω_{p} = 0.4712$	0.8986	0.93 dB, against the 0.7 dB allowed
$ω = ω_{s} = 1.885$	0.1918	14.35 dB, against the 14 dB required
$ω = π$	0.1266	at Nyquist

$N = 2$ and $Ω_{c} = 0.7287$ rad/s. The specification is met at the passband edge and exceeded at the stopband edge, as rounding the order up guarantees.

H (z) = \frac{0.3033 z^{- 1}}{1 - 1.0396 z^{- 1} + 0.3568 z^{- 2}}

Same as 2082 Baishakh Q82074 Ashwin · Q815 marksDesign a low pass discrete IIR filter by Bilinear Transformation method to an approximate Butterworth filter having specifications as below: Pass bandedge frequency (

ω_{p}

) =

0.27 π

radians, Stop bandedge frequency (

ω_{s}

) =

0.58 π

radians, Passband ripple (deltap) = 0.11, Stopband ripple (deltas) = 0.21, Consider sampling frequency 0.5 Hz.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.
$T = 2$ here, not 1, so the factor 2/T in the prewarping is 1. It cancels in the end, but only if it is carried consistently.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 2) \tan (0.8482 / 2) = 0.4515 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 2) \tan (1.8221 / 2) = 1.2892 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.356} - 1) / (10^{0.101} - 1)] / (2 \log_{10} (2.8552)) \\ = \log_{10} (21.676 / 0.2625) / (2 log 102.8552) \\ = 2.1035 so  take N = 3 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.4515 / {(0.2625)}^{1 / 6} \\ = 0.5643 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 1.2892 / {(21.676)}^{1 / 6} \\ = 0.7721 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.5643$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 3 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 2 \\ s 0 & = - 0.2821 + j 0.4887 \\ s 1 & = - 0.5643 \\ s 2 & = - 0.2821 - j 0.4887 \end{aligned}

H_{a} (s) = \frac{{(0.5643)}^{3}}{(s^{2} + 0.5643 s + 0.3184) (s + 0.5643)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 1 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.061 + 0.183 z^{- 1} + 0.183 z^{- 2} + 0.061 z^{- 3}}{1 - 1.0026 z^{- 1} + 0.6022 z^{- 2} - 0.1116 z^{- 3}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.8482$	0.89	1.01 dB, against the 1.01 dB allowed
$ω = ω_{s} = 1.8221$	0.0836	21.56 dB, against the 13.56 dB required
$ω = π$	0	at Nyquist

$N = 3$ and $Ω_{c} = 0.5643$ rad/s. The specification is met at both edges.

H (z) = \frac{0.061 + 0.183 z^{- 1} + 0.183 z^{- 2} + 0.061 z^{- 3}}{1 - 1.0026 z^{- 1} + 0.6022 z^{- 2} - 0.1116 z^{- 3}}

Same as 2082 Baishakh Q82073 Shrawan · Q812 marksUsing bilinear transformation, design a butterworth low pass filter which satisfies the following Magnitude Response:

0.89125 \leq | H (e^{j ω}) | \leq 1

for

0 \leq w \leq 0.2 π

;

| H (e^{j ω}) | \leq 0.17783

for

0.3 π \leq ω \leq π

How this is read0.89125 and 0.17783 are the familiar 1 dB and 15 dB written as ratios: -20 $\log_{10} (0.89125) = 1$ dB and -20 $\log_{10} (0.17783) = 15$ dB.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 1) \tan (0.6283 / 2) = 0.6498 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 1) \tan (0.9425 / 2) = 1.0191 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.5} - 1) / (10^{0.1} - 1)] / (2 \log_{10} (1.5682)) \\ = \log_{10} (30.622 / 0.2589) / (2 log 101.5682) \\ = 5.3044 so  take N = 6 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.6498 / {(0.2589)}^{1 / 12} \\ = 0.7273 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 1.0191 / {(30.622)}^{1 / 12} \\ = 0.7662 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.7273$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 6 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 5 \\ s 0 & = - 0.1882 + j 0.7025 \\ s 1 & = - 0.5143 + j 0.5143 \\ s 2 & = - 0.7025 + j 0.1882 \\ s 3 & = - 0.7025 - j 0.1882 \\ s 4 & = - 0.5143 - j 0.5143 \\ s 5 & = - 0.1882 - j 0.7025 \end{aligned}

\begin{aligned} H_{a} (s) & = \frac{{(0.7273)}^{6}}{D (s)} \\ D (s) & = (s^{2} + 0.3765 s + 0.529) \\ \times (s^{2} + 1.0285 s + 0.529) \\ \times (s^{2} + 1.405 s + 0.529) \end{aligned}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 2 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.0006 + 0.0035 z^{- 1} + 0.0087 z^{- 2} + 0.0116 z^{- 3} + 0.0087 z^{- 4} + 0.0035 z^{- 5} + 0.0006 z^{- 6}}{1 - 3.3143 z^{- 1} + 4.9501 z^{- 2} - 4.1433 z^{- 3} + 2.0275 z^{- 4} - 0.5458 z^{- 5} + 0.0628 z^{- 6}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.6283$	0.8913	1 dB, against the 1 dB allowed
$ω = ω_{s} = 0.9425$	0.131	17.65 dB, against the 15 dB required
$ω = π$	0	at Nyquist

$N = 6$ and $Ω_{c} = 0.7273$ rad/s. The specification is met at both edges.

H (z) = \frac{0.0006 + 0.0035 z^{- 1} + 0.0087 z^{- 2} + 0.0116 z^{- 3} + 0.0087 z^{- 4} + 0.0035 z^{- 5} + 0.0006 z^{- 6}}{1 - 3.3143 z^{- 1} + 4.9501 z^{- 2} - 4.1433 z^{- 3} + 2.0275 z^{- 4} - 0.5458 z^{- 5} + 0.0628 z^{- 6}}

Same as 2082 Baishakh Q82072 Kartik · Q915 marksDesign a low pass digital filter by Bilinear Transformation method to an approximate Butterworth filter, if passband edge frequency is

0.25 π

radians and maximum deviation of 1 dB below 0 dB gain in the passband. The maximum gain of -15 dB and frequency is

0.45 π

radians in stopband, Consider sampling frequency 1Hz.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 1) \tan (0.7854 / 2) = 0.8284 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 1) \tan (1.4137 / 2) = 1.7082 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.5} - 1) / (10^{0.1} - 1)] / (2 \log_{10} (2.0619)) \\ = \log_{10} (30.623 / 0.2589) / (2 log 102.0619) \\ = 3.2979 so  take N = 4 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.8284 / {(0.2589)}^{1 / 8} \\ = 0.9809 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 1.7082 / {(30.623)}^{1 / 8} \\ = 1.1137 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.9809$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 4 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 3 \\ s 0 & = - 0.3754 + j 0.9062 \\ s 1 & = - 0.9062 + j 0.3754 \\ s 2 & = - 0.9062 - j 0.3754 \\ s 3 & = - 0.3754 - j 0.9062 \end{aligned}

H_{a} (s) = \frac{{(0.9809)}^{4}}{(s^{2} + 0.7507 s + 0.9621) (s^{2} + 1.8124 s + 0.9621)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 2 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.0167 + 0.0667 z^{- 1} + 0.1001 z^{- 2} + 0.0667 z^{- 3} + 0.0167 z^{- 4}}{1 - 1.6476 z^{- 1} + 1.3563 z^{- 2} - 0.5252 z^{- 3} + 0.0834 z^{- 4}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.7854$	0.8913	1 dB, against the 1 dB allowed
$ω = ω_{s} = 1.4137$	0.1081	19.32 dB, against the 15 dB required
$ω = π$	0	at Nyquist

$N = 4$ and $Ω_{c} = 0.9809$ rad/s. The specification is met at both edges.

H (z) = \frac{0.0167 + 0.0667 z^{- 1} + 0.1001 z^{- 2} + 0.0667 z^{- 3} + 0.0167 z^{- 4}}{1 - 1.6476 z^{- 1} + 1.3563 z^{- 2} - 0.5252 z^{- 3} + 0.0834 z^{- 4}}

Same as 2082 Baishakh Q82071 Shrawan · Q815 marksDesign a low pass digital filter by Bilinear Transformation method to an approximate Butterworth filter it passband frequency is

0.2 π

radians and maximum deviation of 1 db below 0 dB gain in the pass band. The maximum gain of -15 db and frequency is

0.4 π

radians in stop band, consider sampling frequency 1 Hz.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 1) \tan (0.6283 / 2) = 0.6498 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 1) \tan (1.2566 / 2) = 1.4531 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.5} - 1) / (10^{0.1} - 1)] / (2 \log_{10} (2.2361)) \\ = \log_{10} (30.623 / 0.2589) / (2 log 102.2361) \\ = 2.9656 so  take N = 3 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.6498 / {(0.2589)}^{1 / 6} \\ = 0.814 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 1.4531 / {(30.623)}^{1 / 6} \\ = 0.8215 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.814$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 3 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 2 \\ s 0 & = - 0.407 + j 0.7049 \\ s 1 & = - 0.814 \\ s 2 & = - 0.407 - j 0.7049 \end{aligned}

H_{a} (s) = \frac{{(0.814)}^{3}}{(s^{2} + 0.814 s + 0.6626) (s + 0.814)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 2 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.0305 + 0.0914 z^{- 1} + 0.0914 z^{- 2} + 0.0305 z^{- 3}}{1 - 1.4826 z^{- 1} + 0.9296 z^{- 2} - 0.2033 z^{- 3}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.6283$	0.8913	1 dB, against the 1 dB allowed
$ω = ω_{s} = 1.2566$	0.1731	15.23 dB, against the 15 dB required
$ω = π$	0	at Nyquist

$N = 3$ and $Ω_{c} = 0.814$ rad/s. The specification is met at both edges.

H (z) = \frac{0.0305 + 0.0914 z^{- 1} + 0.0914 z^{- 2} + 0.0305 z^{- 3}}{1 - 1.4826 z^{- 1} + 0.9296 z^{- 2} - 0.2033 z^{- 3}}

The same questions grouped by how often each topic is set: butterworth iir design

Impulse invariance PIN 3/19

78 Bh · 74 Ch · 72 Ch12+33+1212

Before the formulas

The first of the two mappings, and the obvious one: sample the analog filter's impulse response and use the samples as the digital filter's impulse response. Because it preserves the response in time, it is the right choice when the time behaviour is what matters to you.

It has one serious flaw, and it is the same flaw as section 1.10 of chapter 1. Sampling copies the spectrum, so anything the analog filter still passes above half the sampling rate folds back on top of the band you wanted. That makes it safe only for filters that are already nearly dead at high frequencies.

The idea Make the digital filter's impulse response equal to samples of the analog filter's impulse response:

h [n] = T

ha(nT). The digital filter then behaves like the analog one in the time domain, sample for sample.

The derivation. Expand $H_{a} (s)$ in partial fractions, sample, and transform each term.

\begin{aligned} H_{a} (s) & = \sum_{k = 1}^{N} \frac{A_{k}}{s - p_{k}} ⟹ h_{a} (t) = \sum_{k} A_{k} e^{p_{k} t} u (t) \\ h [n] & = T h_{a} (n T) = T \sum_{k} A_{k} {(e^{p_{k} T})}^{n} u [n] \\ H (z) & = \sum_{k} \frac{T A_{k}}{1 - e^{p_{k} T} z^{- 1}} & term by term \end{aligned}

The consequences

The frequency axis maps linearly, $ω = Ω$ T, so there is no warping: the shape of the passband is preserved exactly.
But sampling aliases. The analog response does not stop at $Ω = π$ /T, so everything above that folds back:
$H (e^{j ω})$ = ( $\frac{1}{T}$ ) $\sum_{k}$ Ha(j( $ω / T - 2 π$ k/T)).
The digital response is the sum of shifted copies of the analog one, and the tails overlap. The stopband is filled in and the specification can fail.
So it is only suitable for band limited filters: low pass and band pass, where the analog response is already small above $π$ /T. A high pass or band stop filter cannot be designed this way, because its response does not decay.
Zeros do not map by z = $e^{s T}$ . Only the poles do; the zeros come out of the partial fraction algebra and land somewhere else, which is why the two responses are not identical even before aliasing.
The T in $h [n] = T$ ha(nT) is a scaling so the gain does not depend on the sampling rate. Some books leave it out; say which you use.

The procedure: find N and $Ω_{c}$ as usual, but with no prewarping, using $Ω = ω$ /T directly; write $H_{a} (s)$ ; expand it in partial fractions; replace each term $A / (s - p)$ by TA/(1 $- e^{p T} z^{- 1}$ ); combine into one rational $H (z)$ .

The aliasing is the same overlap as in sampling a continuous time signal, for the same reason, and the partial fraction step is the one from chapter 2.

See it drawnSampling the impulse response folds every high frequency back

Each horizontal strip of the s plane maps onto the whole z plane, so every strip lands on top of the first one. Anything the analog filter still passes above half the sampling rate comes back inside the band.

Practise thisEvery question the paper has set on this topic, worked in full3

Ω = ω / T

2 / ((s + 1) (s + 2))

T = 0.2

into 0.2774z over

z^{2} - 1.4887 z + 0.54853

, but the numerator is wrong. It is

2 (e^{- 0} .2 - e^{- 0} .4) = 2 (0.8187 - 0.6703) = < b > 0.2968 < / b

>; the book doubled 0.67 as 1.36 instead of 1.34. The denominator is right.

What is different in this one

The edges are given in hertz, not radians, so the first step is $ω = 2 π$ f / fs. A question that already gives radians skips it.
Impulse invariance, not the bilinear transformation. There is therefore no prewarping: the edges scale linearly, $Ω = ω / T$ , and the mapping is applied to each partial fraction term separately rather than to $H (s)$ as a whole.
$T = 0.0002$ here, not 1, so the factor 2/T in the prewarping is 10000. It cancels in the end, but only if it is carried consistently.
Impulse invariance, not the bilinear transformation, so there is no prewarping: the edges scale linearly, $Ω = ω / T$ . The second half of the question asks what prewarping is and why it is necessary, which is exactly the step this design does not take. Watch the standard text here: its worked impulse invariance example turns $2 / ((s + 1) (s + 2))$ at $T = 0.2$ into 0.2774z over $z^{2} - 1.4887 z + 0.54853$ , but the numerator is wrong. It is $2 (e^{- 0} .2 - e^{- 0} .4) = 2 (0.8187 - 0.6703)$ = 0.2968; the book doubled 0.67 as 1.36 instead of 1.34. The denominator is right.

Digital edges first: $ω = 2 π$ f / fs gives $ω_{p} = 2 π$ (200)/5000 = 0.2513 rad and $ω_{s} = 2 π$ (500)/5000 = 0.6283 rad.

Impulse invariance maps frequency linearly, $Ω = ω / T$ , so there is no prewarping to do.

\begin{aligned} Ω_{p} & = ω_{p} / T = 1256.6371 rad / s \\ Ω_{s} & = ω_{s} / T = 3141.5927 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.2} - 1) / (10^{0.5} - 1)] / (2 \log_{10} (2.5)) \\ = \log_{10} (14.849 / 2.1623) / (2 log 102.5) \\ = 1.0514 so  take N = 2 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 1256.6371 / {(2.1623)}^{1 / 4} \\ = 1036.2916 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 3141.5927 / {(14.849)}^{1 / 4} \\ = 1600.3909 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 1036.2916$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 2 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 1 \\ s 0 & = - 732.7689 + j 732.7689 \\ s 1 & = - 732.7689 - j 732.7689 \end{aligned}

H_{a} (s) = \frac{{(1036.2916)}^{2}}{(s^{2} + 1465.5377 s + 1073900.3832)}

Expand $H_{a} (s)$ in partial fractions and map each pole on its own with s = pk going to z = $e^{p k T}$ , which is what impulse invariance does.

\begin{aligned} residue - j 732.7689 at  pole - 732.7689 + j 732.7689 \\ \to - j 0.1466 / (1 - (0.8544 + j 0.1261) z^{- 1}) \\ residue j 732.7689 at  pole - 732.7689 - j 732.7689 \\ \to j 0.1466 / (1 - (0.8544 - j 0.1261) z^{- 1}) \end{aligned}

H (z) = \frac{0.037 z^{- 1}}{1 - 1.7088 z^{- 1} + 0.7459 z^{- 2}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	0.9964	the DC gain
$ω = ω_{p} = 0.2513$	0.5633	4.99 dB, against the 5 dB allowed
$ω = ω_{s} = 0.6283$	0.1114	19.06 dB, against the 12 dB required
$ω = π$	0.0107	at Nyquist

$N = 2$ and $Ω_{c} = 1036.2916$ rad/s. The specification is met at both edges.

H (z) = \frac{0.037 z^{- 1}}{1 - 1.7088 z^{- 1} + 0.7459 z^{- 2}}

Practise this2074 Chaitra · Q712 marksDesign a digital low-pass filter with the following specification: i) Pass-band magnitude constant to 0.7 dB below the frequency of

0.15 π

ii) Stop-band attenuation at least 14 dB for the frequencies between

0.6 π

π

What is different in this one

Impulse invariance, not the bilinear transformation. There is therefore no prewarping: the edges scale linearly, $Ω = ω / T$ , and the mapping is applied to each partial fraction term separately rather than to $H (s)$ as a whole.

Impulse invariance maps frequency linearly, $Ω = ω / T$ , so there is no prewarping to do.

\begin{aligned} Ω_{p} & = ω_{p} / T = 0.4712 rad / s \\ Ω_{s} & = ω_{s} / T = 1.885 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.4} - 1) / (10^{0.07} - 1)] / (2 \log_{10} (4)) \\ = \log_{10} (24.119 / 0.1749) / (2 log 104) \\ = 1.7769 so  take N = 2 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.4712 / {(0.1749)}^{1 / 4} \\ = 0.7287 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 1.885 / {(24.119)}^{1 / 4} \\ = 0.8506 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.7287$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 2 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 1 \\ s 0 & = - 0.5153 + j 0.5153 \\ s 1 & = - 0.5153 - j 0.5153 \end{aligned}

H_{a} (s) = \frac{{(0.7287)}^{2}}{(s^{2} + 1.0305 s + 0.531)}

Expand $H_{a} (s)$ in partial fractions and map each pole on its own with s = pk going to z = $e^{p k T}$ , which is what impulse invariance does.

\begin{aligned} residue - j 0.5153 at  pole - 0.5153 + j 0.5153 \\ \to - j 0.5153 / (1 - (0.5198 + j 0.2943) z^{- 1}) \\ residue j 0.5153 at  pole - 0.5153 - j 0.5153 \\ \to j 0.5153 / (1 - (0.5198 - j 0.2943) z^{- 1}) \end{aligned}

H (z) = \frac{0.3033 z^{- 1}}{1 - 1.0396 z^{- 1} + 0.3568 z^{- 2}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	0.9561	the DC gain
$ω = ω_{p} = 0.4712$	0.8986	0.93 dB, against the 0.7 dB allowed
$ω = ω_{s} = 1.885$	0.1918	14.35 dB, against the 14 dB required
$ω = π$	0.1266	at Nyquist

$N = 2$ and $Ω_{c} = 0.7287$ rad/s. The specification is met at the passband edge and exceeded at the stopband edge, as rounding the order up guarantees.

H (z) = \frac{0.3033 z^{- 1}}{1 - 1.0396 z^{- 1} + 0.3568 z^{- 2}}

Practise this2072 Chaitra · Q103+12 marksExplain about the advantages of selecting bilinear transformation method over impulse invariance method (I I M). Design a digital low pass Butterworth filter using impulse invariant transformation with pass band and stop band frequencies 200Hz and 500Hz respectively. The pass band and stop band attenuation are -5dB and -12dB respectively. The sampling frequency is 5kHz. Use IIM method.What is different hereimpulse invariance, not the bilinear transformation, so there is no prewarping and the mapping is a scaling; the specification arrives in hertz, so it has to be turned into radians per sample first

How this is readThe attenuations are printed as -5 dB and -12 dB; the formulas take them positive.

What is different in this one

The edges are given in hertz, not radians, so the first step is $ω = 2 π$ f / fs. A question that already gives radians skips it.
Impulse invariance, not the bilinear transformation. There is therefore no prewarping: the edges scale linearly, $Ω = ω / T$ , and the mapping is applied to each partial fraction term separately rather than to $H (s)$ as a whole.
$T = 0.0002$ here, not 1, so the factor 2/T in the prewarping is 10000. It cancels in the end, but only if it is carried consistently.

Digital edges first: $ω = 2 π$ f / fs gives $ω_{p} = 2 π$ (200)/5000 = 0.2513 rad and $ω_{s} = 2 π$ (500)/5000 = 0.6283 rad.

Impulse invariance maps frequency linearly, $Ω = ω / T$ , so there is no prewarping to do.

\begin{aligned} Ω_{p} & = ω_{p} / T = 1256.6371 rad / s \\ Ω_{s} & = ω_{s} / T = 3141.5927 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.2} - 1) / (10^{0.5} - 1)] / (2 \log_{10} (2.5)) \\ = \log_{10} (14.849 / 2.1623) / (2 log 102.5) \\ = 1.0514 so  take N = 2 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 1256.6371 / {(2.1623)}^{1 / 4} \\ = 1036.2916 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 3141.5927 / {(14.849)}^{1 / 4} \\ = 1600.3909 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 1036.2916$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 2 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 1 \\ s 0 & = - 732.7689 + j 732.7689 \\ s 1 & = - 732.7689 - j 732.7689 \end{aligned}

H_{a} (s) = \frac{{(1036.2916)}^{2}}{(s^{2} + 1465.5377 s + 1073900.3832)}

Expand $H_{a} (s)$ in partial fractions and map each pole on its own with s = pk going to z = $e^{p k T}$ , which is what impulse invariance does.

\begin{aligned} residue - j 732.7689 at  pole - 732.7689 + j 732.7689 \\ \to - j 0.1466 / (1 - (0.8544 + j 0.1261) z^{- 1}) \\ residue j 732.7689 at  pole - 732.7689 - j 732.7689 \\ \to j 0.1466 / (1 - (0.8544 - j 0.1261) z^{- 1}) \end{aligned}

H (z) = \frac{0.037 z^{- 1}}{1 - 1.7088 z^{- 1} + 0.7459 z^{- 2}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	0.9964	the DC gain
$ω = ω_{p} = 0.2513$	0.5633	4.99 dB, against the 5 dB allowed
$ω = ω_{s} = 0.6283$	0.1114	19.06 dB, against the 12 dB required
$ω = π$	0.0107	at Nyquist

$N = 2$ and $Ω_{c} = 1036.2916$ rad/s. The specification is met at both edges.

H (z) = \frac{0.037 z^{- 1}}{1 - 1.7088 z^{- 1} + 0.7459 z^{- 2}}

The same questions grouped by how often each topic is set: butterworth iir design

6.2Design using the bilinear transformation

The bilinear transformation TOP 13/19

82 Bh · 80 Bh · 80 Ba · 79 Bh · 78 Bh · 76 Ch · 76 Ash · 75 Ch · 75 Ash · 74 Ash · 73 Shr · 72 Ka · 71 Shr121510

Before the formulas

The second mapping, and the one that avoids that flaw completely. Instead of sampling, it squeezes the entire infinite analog frequency axis into the finite digital one, so there is nothing left over to fold back and no aliasing at all.

It also maps the stable half of the analog plane exactly onto the inside of the unit circle, so a stable analog filter always gives a stable digital one. Nothing is free. The squeeze is not uniform, so frequencies get moved, which is called warping.

The fix is to bend the specification the other way before you design, which is prewarping, and the edges then land exactly where you asked.

The mapping A one to one, algebraic substitution that maps the entire left half of the s-plane into the inside of the unit circle, with no aliasing at all, at the cost of compressing the infinite analog frequency axis into the finite digital one.

s = \frac{2}{T} \frac{1 - z^{- 1}}{1 + z^{- 1}} and z = \frac{1 + s T / 2}{1 - s T / 2}

Where it comes from. It is the trapezoidal rule for integration, written as a transfer function.

An integrator has $H_{a} (s) = 1$ /s, that is y'(t) = $x (t)$ . Integrate from (n-1)T to nT by the trapezoidal rule:

\begin{aligned} y [n] & = y [n - 1] + \frac{T}{2} (x [n] + x [n - 1]) \\ Y (z) (1 - z^{- 1}) & = \frac{T}{2} X (z) (1 + z^{- 1}) \\ \frac{Y (z)}{X (z)} & = \frac{T}{2} \frac{1 + z^{- 1}}{1 - z^{- 1}} = \frac{1}{s} with s = \frac{2}{T} \frac{1 - z^{- 1}}{1 + z^{- 1}} \end{aligned}

The frequency relation, and warping. Put s = j $Ω$ and z = $e^{j ω}$ :

\begin{aligned} j Ω & = \frac{2}{T} \frac{1 - e^{- j ω}}{1 + e^{- j ω}} \\ = \frac{2}{T} \frac{e^{- j ω / 2} (e^{j ω / 2} - e^{- j ω / 2})}{e^{- j ω / 2} (e^{j ω / 2} + e^{- j ω / 2})} \\ = \frac{2}{T} \frac{2 j \sin (ω / 2)}{2 \cos (ω / 2)} = j \frac{2}{T} \tan (\frac{ω}{2}) \\ ⟹ Ω & = \frac{2}{T} \tan (\frac{ω}{2}) \end{aligned}

Frequency warping, and prewarping The relation between

Ω

and

ω

is a tangent, not a straight line. It is nearly linear for small

ω

and bends sharply as

ω

approaches

π

, squeezing the whole infinite range

0 \leq Ω

< infinity into

0 \leq ω < π

. This non linear compression is frequency warping.
Warping does not distort the magnitude values, only the frequencies at which they occur, so a piecewise constant specification survives it. A phase response or a differentiator does not.
Prewarping is the cure: before designing, push every critical digital frequency through

Ω

= (2/T)

\tan (ω / 2)

, design the analog filter at those warped analog frequencies, and the bilinear transformation then bends them back to exactly where the specification wanted them.

Why the mapping is safe

No aliasing. The mapping is one to one: each point of the s-plane goes to exactly one point of the z-plane, so no two analog frequencies land on the same digital one.
Stability is preserved. The left half plane, Re $(s) < 0$ , maps to the inside of the unit circle, $| z | < 1 . A$ stable analog filter always gives a stable digital filter.
The imaginary axis maps to the unit circle, so the analog frequency response becomes the digital frequency response, rearranged in frequency but not in value.
It works for every filter type, high pass and band stop included, which impulse invariance cannot do.

Worked, the 2075 Chaitra and 2082 Baishakh specification

0.8 \leq | H | \leq 1

for

0 \leq ω \leq 0

2 π

, and

| H | \leq 0.2

for

0.6 π \leq ω \leq π

, with Fs = 1 Hz so T = 1.
Attenuations:

α_{p} = - 20 \log_{10} (0.8) = 1.9382

dB;

α_{s} = - 20 \log_{10} (0.2) = 13.9794

dB.
Prewarp:

Ω_{p} = 2 \tan (0.1 π) = 2

(0.3249) = 0.6498;

Ω_{s} = 2 \tan (0.3 π) = 2

(1.3764) = 2.7528.
Order:

10^{0.1 x 1.9382}

- 1 = 0.5625;

10^{0.1 x 13.9794}

- 1 = 24.0.

N \geq l o g 10 (24.0 / 0.5625) / (2 l o g 10 (2.7528 / 0.6498)) = l o g 10 (42.67) / (2 l o g 10 (4.236)) = 1.630

/(2 x 0.6270) = 1.30, so $N = 2$ .
Cutoff:

Ω_{c} = 0.6498 / {(0.5625)}^{1 / 4} = \frac{0.6498}{0}

.8660 = 0.7504.
Prototype:

H_{a} (s) = Ω_{c}^{2} /

(

s^{2} + 1.4142 Ω_{c} s + Ω_{c}^{2}

) =

0.5631 /

(

s^{2}

+ 1.0613 s + 0.5631).
Substitute

s = 2

- z^{- 1}

)/(1 +

z^{- 1}

) and clear the fractions to get

H (z)

in powers of

z^{- 1}

In the exam

Prewarp before you compute N, not after. Computing the order from the unwarped edges is the single commonest error in this question.
T cancels out of the order calculation, since only the ratio $Ω_{s} / Ω_{p}$ appears, but it does not cancel out of the substitution. Keep it.
Show the substitution line. Several marks live in the algebra of clearing (1 + $z^{- 1}$ ) out of the denominator.
Check the gain at DC. $H (z)$ at $z = 1$ should be 1 for a low pass. It is a one line check on a long calculation.

It maps the whole left half plane inside the unit circle, which is the stability condition made automatic. The price, warping, is the price of avoiding the aliasing of chapter 1.

See it drawnThe whole left half plane is squeezed inside the circle, once

Nothing is left over to fold back, so there is no aliasing at all, and a stable analog filter always gives a stable digital one. The price is that the frequency axis is bent.

Practise thisEvery question the paper has set on this topic, worked in full13

How this is readThe attenuations are printed as -3 dB and -10 dB; the design formulas take them as positive numbers, so Ap = 3 and As = 10.

What is different in this one

The edges are given in hertz, not radians, so the first step is $ω = 2 π$ f / fs. A question that already gives radians skips it.
The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.
$T = 0.0002$ here, not 1, so the factor 2/T in the prewarping is 10000. It cancels in the end, but only if it is carried consistently.

Digital edges first: $ω = 2 π$ f / fs gives $ω_{p} = 2 π$ (350)/5000 = 0.4398 rad and $ω_{s} = 2 π$ (1000)/5000 = 1.2566 rad.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 0.0002) \tan (0.4398 / 2) = 2235.2648 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 0.0002) \tan (1.2566 / 2) = 7265.4253 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1} - 1) / (10^{0.3} - 1)] / (2 \log_{10} (3.2504)) \\ = \log_{10} (9 / 0.9953) / (2 log 103.2504) \\ = 0.934 so  take N = 1 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 2235.2648 / {(0.9953)}^{1 / 2} \\ = 2240.5787 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 7265.4253 / {(9)}^{1 / 2} \\ = 2421.8084 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 2240.5787$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 1 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 0 \\ s 0 & = - 2240.5787 \end{aligned}

H_{a} (s) = \frac{{(2240.5787)}^{1}}{(s + 2240.5787)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 10000 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.183 + 0.183 z^{- 1}}{1 - 0.6339 z^{- 1}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.4398$	0.7079	3 dB, against the 3 dB allowed
$ω = ω_{s} = 1.2566$	0.2947	10.61 dB, against the 10 dB required
$ω = π$	0	at Nyquist

$N = 1$ and $Ω_{c} = 2240.5787$ rad/s. The specification is met at both edges.

H (z) = \frac{0.183 + 0.183 z^{- 1}}{1 - 0.6339 z^{- 1}}

Same as 2082 Baishakh Q82080 Bhadra · Q812 marksUsing Bilinear transformation, design a Butterworth low pass filter which satisfies following conditions:

0.9 \leq | H (e^{j ω}) | \leq 1

for

0 \leq ω \leq \frac{π}{2}

;

| H (e^{j ω}) | \leq 0.2

for

\frac{3 π}{4} \leq ω \leq π

. Consider sampling frequency of 1 Hz.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 1) \tan (1.5708 / 2) = 2 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 1) \tan (2.3562 / 2) = 4.8284 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.398} - 1) / (10^{0.092} - 1)] / (2 \log_{10} (2.4142)) \\ = \log_{10} (24 / 0.2346) / (2 log 102.4142) \\ = 2.6255 so  take N = 3 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 2 / {(0.2346)}^{1 / 6} \\ = 2.5467 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 4.8284 / {(24)}^{1 / 6} \\ = 2.843 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 2.5467$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 3 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 2 \\ s 0 & = - 1.2734 + j 2.2055 \\ s 1 & = - 2.5467 \\ s 2 & = - 1.2734 - j 2.2055 \end{aligned}

H_{a} (s) = \frac{{(2.5467)}^{3}}{(s^{2} + 2.5467 s + 6.4859) (s + 2.5467)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 2 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.2332 + 0.6996 z^{- 1} + 0.6996 z^{- 2} + 0.2332 z^{- 3}}{1 + 0.4394 z^{- 1} + 0.3845 z^{- 2} + 0.0416 z^{- 3}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 1.5708$	0.9	0.92 dB, against the 0.92 dB allowed
$ω = ω_{s} = 2.3562$	0.1452	16.76 dB, against the 13.98 dB required
$ω = π$	0	at Nyquist

$N = 3$ and $Ω_{c} = 2.5467$ rad/s. The specification is met at both edges.

H (z) = \frac{0.2332 + 0.6996 z^{- 1} + 0.6996 z^{- 2} + 0.2332 z^{- 3}}{1 + 0.4394 z^{- 1} + 0.3845 z^{- 2} + 0.0416 z^{- 3}}

Same as 2082 Baishakh Q82080 Baishakh · Q911 marksDesign a low pass digital IIR filter by Bilinear Transformation method to an approximate Butterworth low pass filter, if passband edge frequency is

0.26 π

radians and maximum deviation of 0.99 dB below 0 dB gain in the passband. The maximum gain of -14.99 dB and frequency is

0.58 π

radians in stopband, Consider sampling frequency 0.5 Hz.

How this is readA sampling frequency of 0.5 Hz gives $T = 2$ s.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.
$T = 2$ here, not 1, so the factor 2/T in the prewarping is 1. It cancels in the end, but only if it is carried consistently.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 2) \tan (0.8168 / 2) = 0.4327 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 2) \tan (1.8221 / 2) = 1.2892 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.499} - 1) / (10^{0.099} - 1)] / (2 \log_{10} (2.9791)) \\ = \log_{10} (30.55 / 0.256) / (2 log 102.9791) \\ = 2.1902 so  take N = 3 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.4327 / {(0.256)}^{1 / 6} \\ = 0.5431 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 1.2892 / {(30.55)}^{1 / 6} \\ = 0.7291 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.5431$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 3 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 2 \\ s 0 & = - 0.2715 + j 0.4703 \\ s 1 & = - 0.5431 \\ s 2 & = - 0.2715 - j 0.4703 \end{aligned}

H_{a} (s) = \frac{{(0.5431)}^{3}}{(s^{2} + 0.5431 s + 0.2949) (s + 0.5431)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 1 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.0565 + 0.1694 z^{- 1} + 0.1694 z^{- 2} + 0.0565 z^{- 3}}{1 - 1.0634 z^{- 1} + 0.6363 z^{- 2} - 0.1211 z^{- 3}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.8168$	0.8923	0.99 dB, against the 0.99 dB allowed
$ω = ω_{s} = 1.8221$	0.0745	22.55 dB, against the 14.99 dB required
$ω = π$	0	at Nyquist

$N = 3$ and $Ω_{c} = 0.5431$ rad/s. The specification is met at both edges.

H (z) = \frac{0.0565 + 0.1694 z^{- 1} + 0.1694 z^{- 2} + 0.0565 z^{- 3}}{1 - 1.0634 z^{- 1} + 0.6363 z^{- 2} - 0.1211 z^{- 3}}

Practise this2079 Bhadra · Q911+4 marksDesign a low pass discrete time Butterworth filter using bilinear transformation having following specifications: Passband frequency (

Ω_{p}

) =

0.25 π

radians, Stopband frequency (

Ω_{s}

) =

0.55 π

0.45 π

using digital domain transformation.

The design

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.
$T = 2$ here, not 1, so the factor 2/T in the prewarping is 1. It cancels in the end, but only if it is carried consistently.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 2) \tan (0.7854 / 2) = 0.4142 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 2) \tan (1.7279 / 2) = 1.1708 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.356} - 1) / (10^{0.101} - 1)] / (2 \log_{10} (2.8267)) \\ = \log_{10} (21.676 / 0.2625) / (2 log 102.8267) \\ = 2.1239 so  take N = 3 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.4142 / {(0.2625)}^{1 / 6} \\ = 0.5177 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 1.1708 / {(21.676)}^{1 / 6} \\ = 0.7012 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.5177$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 3 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 2 \\ s 0 & = - 0.2588 + j 0.4483 \\ s 1 & = - 0.5177 \\ s 2 & = - 0.2588 - j 0.4483 \end{aligned}

H_{a} (s) = \frac{{(0.5177)}^{3}}{(s^{2} + 0.5177 s + 0.268) (s + 0.5177)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 1 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.0512 + 0.1536 z^{- 1} + 0.1536 z^{- 2} + 0.0512 z^{- 3}}{1 - 1.1377 z^{- 1} + 0.6808 z^{- 2} - 0.1335 z^{- 3}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.7854$	0.89	1.01 dB, against the 1.01 dB allowed
$ω = ω_{s} = 1.7279$	0.0861	21.3 dB, against the 13.56 dB required
$ω = π$	0	at Nyquist

$N = 3$ and $Ω_{c} = 0.5177$ rad/s. The specification is met at both edges.

H (z) = \frac{0.0512 + 0.1536 z^{- 1} + 0.1536 z^{- 2} + 0.0512 z^{- 3}}{1 - 1.1377 z^{- 1} + 0.6808 z^{- 2} - 0.1335 z^{- 3}}

The rest of this questionThis topic works the part that belongs to it. The other part is worked under Converting it to a high pass, the last four marks, where it belongs.

Ω = ω / T

2 / ((s + 1) (s + 2))

T = 0.2

into 0.2774z over

z^{2} - 1.4887 z + 0.54853

, but the numerator is wrong. It is

2 (e^{- 0} .2 - e^{- 0} .4) = 2 (0.8187 - 0.6703) = < b > 0.2968 < / b

>; the book doubled 0.67 as 1.36 instead of 1.34. The denominator is right.

What is different in this one

The edges are given in hertz, not radians, so the first step is $ω = 2 π$ f / fs. A question that already gives radians skips it.
Impulse invariance, not the bilinear transformation. There is therefore no prewarping: the edges scale linearly, $Ω = ω / T$ , and the mapping is applied to each partial fraction term separately rather than to $H (s)$ as a whole.
$T = 0.0002$ here, not 1, so the factor 2/T in the prewarping is 10000. It cancels in the end, but only if it is carried consistently.
Impulse invariance, not the bilinear transformation, so there is no prewarping: the edges scale linearly, $Ω = ω / T$ . The second half of the question asks what prewarping is and why it is necessary, which is exactly the step this design does not take. Watch the standard text here: its worked impulse invariance example turns $2 / ((s + 1) (s + 2))$ at $T = 0.2$ into 0.2774z over $z^{2} - 1.4887 z + 0.54853$ , but the numerator is wrong. It is $2 (e^{- 0} .2 - e^{- 0} .4) = 2 (0.8187 - 0.6703)$ = 0.2968; the book doubled 0.67 as 1.36 instead of 1.34. The denominator is right.

Digital edges first: $ω = 2 π$ f / fs gives $ω_{p} = 2 π$ (200)/5000 = 0.2513 rad and $ω_{s} = 2 π$ (500)/5000 = 0.6283 rad.

Impulse invariance maps frequency linearly, $Ω = ω / T$ , so there is no prewarping to do.

\begin{aligned} Ω_{p} & = ω_{p} / T = 1256.6371 rad / s \\ Ω_{s} & = ω_{s} / T = 3141.5927 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.2} - 1) / (10^{0.5} - 1)] / (2 \log_{10} (2.5)) \\ = \log_{10} (14.849 / 2.1623) / (2 log 102.5) \\ = 1.0514 so  take N = 2 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 1256.6371 / {(2.1623)}^{1 / 4} \\ = 1036.2916 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 3141.5927 / {(14.849)}^{1 / 4} \\ = 1600.3909 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 1036.2916$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 2 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 1 \\ s 0 & = - 732.7689 + j 732.7689 \\ s 1 & = - 732.7689 - j 732.7689 \end{aligned}

H_{a} (s) = \frac{{(1036.2916)}^{2}}{(s^{2} + 1465.5377 s + 1073900.3832)}

Expand $H_{a} (s)$ in partial fractions and map each pole on its own with s = pk going to z = $e^{p k T}$ , which is what impulse invariance does.

\begin{aligned} residue - j 732.7689 at  pole - 732.7689 + j 732.7689 \\ \to - j 0.1466 / (1 - (0.8544 + j 0.1261) z^{- 1}) \\ residue j 732.7689 at  pole - 732.7689 - j 732.7689 \\ \to j 0.1466 / (1 - (0.8544 - j 0.1261) z^{- 1}) \end{aligned}

H (z) = \frac{0.037 z^{- 1}}{1 - 1.7088 z^{- 1} + 0.7459 z^{- 2}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	0.9964	the DC gain
$ω = ω_{p} = 0.2513$	0.5633	4.99 dB, against the 5 dB allowed
$ω = ω_{s} = 0.6283$	0.1114	19.06 dB, against the 12 dB required
$ω = π$	0.0107	at Nyquist

$N = 2$ and $Ω_{c} = 1036.2916$ rad/s. The specification is met at both edges.

H (z) = \frac{0.037 z^{- 1}}{1 - 1.7088 z^{- 1} + 0.7459 z^{- 2}}

What is different in this one

The edges are given in hertz, not radians, so the first step is $ω = 2 π$ f / fs. A question that already gives radians skips it.
The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.
$T = 0.0001$ here, not 1, so the factor 2/T in the prewarping is 16000. It cancels in the end, but only if it is carried consistently.

Digital edges first: $ω = 2 π$ f / fs gives $ω_{p} = 2 π$ (1200)/8000 = 0.9425 rad and $ω_{s} = 2 π$ (2500)/8000 = 1.9635 rad.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 0.0001) \tan (0.9425 / 2) = 8152.4072 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 0.0001) \tan (1.9635 / 2) = 23945.6922 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{4} - 1) / (10^{0.1} - 1)] / (2 \log_{10} (2.9373)) \\ = \log_{10} (9999 / 0.2589) / (2 log 102.9373) \\ = 4.901 so  take N = 5 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 8152.4072 / {(0.2589)}^{1 / 10} \\ = 9331.8639 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 23945.6922 / {(9999)}^{1 / 10} \\ = 9533.0471 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 9331.8639$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 5 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 4 \\ s 0 & = - 2883.7045 + j 8875.13 \\ s 1 & = - 7549.6365 + j 5485.132 \\ s 2 & = - 9331.8639 \\ s 3 & = - 7549.6365 - j 5485.132 \\ s 4 & = - 2883.7045 - j 8875.13 \end{aligned}

\begin{aligned} H_{a} (s) & = \frac{{(9331.8639)}^{5}}{D (s)} \\ D (s) & = (s^{2} + 5767.4091 s + 87083683.7083) \\ \times (s^{2} + 15099.273 s + 87083683.7083) \\ \times (s + 9331.8639) \end{aligned}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 16000 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.011 + 0.0549 z^{- 1} + 0.1098 z^{- 2} + 0.1098 z^{- 3} + 0.0549 z^{- 4} + 0.011 z^{- 5}}{1 - 1.617 z^{- 1} + 1.5544 z^{- 2} - 0.7829 z^{- 3} + 0.2231 z^{- 4} - 0.0263 z^{- 5}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.9425$	0.8913	1 dB, against the 1 dB allowed
$ω = ω_{s} = 1.9635$	0.009	40.93 dB, against the 40 dB required
$ω = π$	0	at Nyquist

$N = 5$ and $Ω_{c} = 9331.8639$ rad/s. The specification is met at both edges.

H (z) = \frac{0.011 + 0.0549 z^{- 1} + 0.1098 z^{- 2} + 0.1098 z^{- 3} + 0.0549 z^{- 4} + 0.011 z^{- 5}}{1 - 1.617 z^{- 1} + 1.5544 z^{- 2} - 0.7829 z^{- 3} + 0.2231 z^{- 4} - 0.0263 z^{- 5}}

ω_{p}

) =

0.22 π

radians, Stop bandedge frequency (

ω_{s}

) =

0.54 π

radians, Passband ripple (deltap) = 0.11, Stopband ripple (deltas) = 0.22, Consider sampling frequency 0.5 Hz.

How this is readA passband ripple of 0.11 means Ap = -20 $\log_{10} (0.89) = 1.0122$ dB, and a stopband ripple of 0.22 means As = -20 $\log_{10} (0.22) = 13.1515$ dB. fs = 0.5 Hz gives $T = 2$ s.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.
$T = 2$ here, not 1, so the factor 2/T in the prewarping is 1. It cancels in the end, but only if it is carried consistently.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 2) \tan (0.6912 / 2) = 0.36 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 2) \tan (1.6965 / 2) = 1.1343 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.315} - 1) / (10^{0.101} - 1)] / (2 \log_{10} (3.1506)) \\ = \log_{10} (19.661 / 0.2625) / (2 log 103.1506) \\ = 1.8806 so  take N = 2 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.36 / {(0.2625)}^{1 / 4} \\ = 0.503 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 1.1343 / {(19.661)}^{1 / 4} \\ = 0.5387 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.503$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 2 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 1 \\ s 0 & = - 0.3557 + j 0.3557 \\ s 1 & = - 0.3557 - j 0.3557 \end{aligned}

H_{a} (s) = \frac{{(0.503)}^{2}}{(s^{2} + 0.7113 s + 0.253)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 1 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.1288 + 0.2576 z^{- 1} + 0.1288 z^{- 2}}{1 - 0.7606 z^{- 1} + 0.2757 z^{- 2}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.6912$	0.89	1.01 dB, against the 1.01 dB allowed
$ω = ω_{s} = 1.6965$	0.1929	14.29 dB, against the 13.15 dB required
$ω = π$	0	at Nyquist

$N = 2$ and $Ω_{c} = 0.503$ rad/s. The specification is met at both edges.

H (z) = \frac{0.1288 + 0.2576 z^{- 1} + 0.1288 z^{- 2}}{1 - 0.7606 z^{- 1} + 0.2757 z^{- 2}}

Same as 2082 Baishakh Q82075 Chaitra · Q710 marksUsing bilinear transformation, design a digital filter using Butterworth approximation which satisfies the following conditions:

0.8 \leq | H (e^{j ω}) | \leq 1

for

0 \leq W \leq 0.2 π

;

| H (e^{j ω}) | \leq 0.2

for

0.6 π \leq W \leq π

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 1) \tan (0.6283 / 2) = 0.6498 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 1) \tan (1.885 / 2) = 2.7528 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.398} - 1) / (10^{0.194} - 1)] / (2 \log_{10} (4.2361)) \\ = \log_{10} (24 / 0.5625) / (2 log 104.2361) \\ = 1.3 so  take N = 2 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.6498 / {(0.5625)}^{1 / 4} \\ = 0.7504 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 2.7528 / {(24)}^{1 / 4} \\ = 1.2437 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.7504$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 2 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 1 \\ s 0 & = - 0.5306 + j 0.5306 \\ s 1 & = - 0.5306 - j 0.5306 \end{aligned}

H_{a} (s) = \frac{{(0.7504)}^{2}}{(s^{2} + 1.0612 s + 0.5631)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 2 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.0842 + 0.1684 z^{- 1} + 0.0842 z^{- 2}}{1 - 1.0282 z^{- 1} + 0.3651 z^{- 2}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.6283$	0.8	1.94 dB, against the 1.94 dB allowed
$ω = ω_{s} = 1.885$	0.0741	22.6 dB, against the 13.98 dB required
$ω = π$	0	at Nyquist

$N = 2$ and $Ω_{c} = 0.7504$ rad/s. The specification is met at both edges.

H (z) = \frac{0.0842 + 0.1684 z^{- 1} + 0.0842 z^{- 2}}{1 - 1.0282 z^{- 1} + 0.3651 z^{- 2}}

Same as 2082 Baishakh Q82075 Ashwin · Q612 marksDesign a digital low-pass filter with the following specification: i) Pass-band magnitude constant to 0.7 dB below the frequency of

0.15 π

ii) Stop-band attenuation at least 14 dB for the frequencies between

0.6 π

π

. Use Butterworth approximation as a prototype and use bilinear transformation method to obtain the digital filter.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 1) \tan (0.4712 / 2) = 0.4802 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 1) \tan (1.885 / 2) = 2.7528 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.4} - 1) / (10^{0.07} - 1)] / (2 \log_{10} (5.733)) \\ = \log_{10} (24.119 / 0.1749) / (2 log 105.733) \\ = 1.4106 so  take N = 2 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.4802 / {(0.1749)}^{1 / 4} \\ = 0.7425 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 2.7528 / {(24.119)}^{1 / 4} \\ = 1.2422 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.7425$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 2 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 1 \\ s 0 & = - 0.525 + j 0.525 \\ s 1 & = - 0.525 - j 0.525 \end{aligned}

H_{a} (s) = \frac{{(0.7425)}^{2}}{(s^{2} + 1.05 s + 0.5513)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 2 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.0829 + 0.1658 z^{- 1} + 0.0829 z^{- 2}}{1 - 1.037 z^{- 1} + 0.3685 z^{- 2}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.4712$	0.9226	0.7 dB, against the 0.7 dB allowed
$ω = ω_{s} = 1.885$	0.0726	22.79 dB, against the 14 dB required
$ω = π$	0	at Nyquist

$N = 2$ and $Ω_{c} = 0.7425$ rad/s. The specification is met at both edges.

H (z) = \frac{0.0829 + 0.1658 z^{- 1} + 0.0829 z^{- 2}}{1 - 1.037 z^{- 1} + 0.3685 z^{- 2}}

ω_{p}

) =

0.27 π

radians, Stop bandedge frequency (

ω_{s}

) =

0.58 π

radians, Passband ripple (deltap) = 0.11, Stopband ripple (deltas) = 0.21, Consider sampling frequency 0.5 Hz.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.
$T = 2$ here, not 1, so the factor 2/T in the prewarping is 1. It cancels in the end, but only if it is carried consistently.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 2) \tan (0.8482 / 2) = 0.4515 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 2) \tan (1.8221 / 2) = 1.2892 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.356} - 1) / (10^{0.101} - 1)] / (2 \log_{10} (2.8552)) \\ = \log_{10} (21.676 / 0.2625) / (2 log 102.8552) \\ = 2.1035 so  take N = 3 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.4515 / {(0.2625)}^{1 / 6} \\ = 0.5643 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 1.2892 / {(21.676)}^{1 / 6} \\ = 0.7721 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.5643$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 3 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 2 \\ s 0 & = - 0.2821 + j 0.4887 \\ s 1 & = - 0.5643 \\ s 2 & = - 0.2821 - j 0.4887 \end{aligned}

H_{a} (s) = \frac{{(0.5643)}^{3}}{(s^{2} + 0.5643 s + 0.3184) (s + 0.5643)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 1 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.061 + 0.183 z^{- 1} + 0.183 z^{- 2} + 0.061 z^{- 3}}{1 - 1.0026 z^{- 1} + 0.6022 z^{- 2} - 0.1116 z^{- 3}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.8482$	0.89	1.01 dB, against the 1.01 dB allowed
$ω = ω_{s} = 1.8221$	0.0836	21.56 dB, against the 13.56 dB required
$ω = π$	0	at Nyquist

$N = 3$ and $Ω_{c} = 0.5643$ rad/s. The specification is met at both edges.

H (z) = \frac{0.061 + 0.183 z^{- 1} + 0.183 z^{- 2} + 0.061 z^{- 3}}{1 - 1.0026 z^{- 1} + 0.6022 z^{- 2} - 0.1116 z^{- 3}}

Same as 2082 Baishakh Q82073 Shrawan · Q812 marksUsing bilinear transformation, design a butterworth low pass filter which satisfies the following Magnitude Response:

0.89125 \leq | H (e^{j ω}) | \leq 1

for

0 \leq w \leq 0.2 π

;

| H (e^{j ω}) | \leq 0.17783

for

0.3 π \leq ω \leq π

How this is read0.89125 and 0.17783 are the familiar 1 dB and 15 dB written as ratios: -20 $\log_{10} (0.89125) = 1$ dB and -20 $\log_{10} (0.17783) = 15$ dB.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 1) \tan (0.6283 / 2) = 0.6498 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 1) \tan (0.9425 / 2) = 1.0191 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.5} - 1) / (10^{0.1} - 1)] / (2 \log_{10} (1.5682)) \\ = \log_{10} (30.622 / 0.2589) / (2 log 101.5682) \\ = 5.3044 so  take N = 6 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.6498 / {(0.2589)}^{1 / 12} \\ = 0.7273 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 1.0191 / {(30.622)}^{1 / 12} \\ = 0.7662 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.7273$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 6 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 5 \\ s 0 & = - 0.1882 + j 0.7025 \\ s 1 & = - 0.5143 + j 0.5143 \\ s 2 & = - 0.7025 + j 0.1882 \\ s 3 & = - 0.7025 - j 0.1882 \\ s 4 & = - 0.5143 - j 0.5143 \\ s 5 & = - 0.1882 - j 0.7025 \end{aligned}

\begin{aligned} H_{a} (s) & = \frac{{(0.7273)}^{6}}{D (s)} \\ D (s) & = (s^{2} + 0.3765 s + 0.529) \\ \times (s^{2} + 1.0285 s + 0.529) \\ \times (s^{2} + 1.405 s + 0.529) \end{aligned}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 2 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.0006 + 0.0035 z^{- 1} + 0.0087 z^{- 2} + 0.0116 z^{- 3} + 0.0087 z^{- 4} + 0.0035 z^{- 5} + 0.0006 z^{- 6}}{1 - 3.3143 z^{- 1} + 4.9501 z^{- 2} - 4.1433 z^{- 3} + 2.0275 z^{- 4} - 0.5458 z^{- 5} + 0.0628 z^{- 6}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.6283$	0.8913	1 dB, against the 1 dB allowed
$ω = ω_{s} = 0.9425$	0.131	17.65 dB, against the 15 dB required
$ω = π$	0	at Nyquist

$N = 6$ and $Ω_{c} = 0.7273$ rad/s. The specification is met at both edges.

H (z) = \frac{0.0006 + 0.0035 z^{- 1} + 0.0087 z^{- 2} + 0.0116 z^{- 3} + 0.0087 z^{- 4} + 0.0035 z^{- 5} + 0.0006 z^{- 6}}{1 - 3.3143 z^{- 1} + 4.9501 z^{- 2} - 4.1433 z^{- 3} + 2.0275 z^{- 4} - 0.5458 z^{- 5} + 0.0628 z^{- 6}}

Same as 2082 Baishakh Q82072 Kartik · Q915 marksDesign a low pass digital filter by Bilinear Transformation method to an approximate Butterworth filter, if passband edge frequency is

0.25 π

radians and maximum deviation of 1 dB below 0 dB gain in the passband. The maximum gain of -15 dB and frequency is

0.45 π

radians in stopband, Consider sampling frequency 1Hz.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 1) \tan (0.7854 / 2) = 0.8284 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 1) \tan (1.4137 / 2) = 1.7082 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.5} - 1) / (10^{0.1} - 1)] / (2 \log_{10} (2.0619)) \\ = \log_{10} (30.623 / 0.2589) / (2 log 102.0619) \\ = 3.2979 so  take N = 4 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.8284 / {(0.2589)}^{1 / 8} \\ = 0.9809 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 1.7082 / {(30.623)}^{1 / 8} \\ = 1.1137 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.9809$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 4 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 3 \\ s 0 & = - 0.3754 + j 0.9062 \\ s 1 & = - 0.9062 + j 0.3754 \\ s 2 & = - 0.9062 - j 0.3754 \\ s 3 & = - 0.3754 - j 0.9062 \end{aligned}

H_{a} (s) = \frac{{(0.9809)}^{4}}{(s^{2} + 0.7507 s + 0.9621) (s^{2} + 1.8124 s + 0.9621)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 2 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.0167 + 0.0667 z^{- 1} + 0.1001 z^{- 2} + 0.0667 z^{- 3} + 0.0167 z^{- 4}}{1 - 1.6476 z^{- 1} + 1.3563 z^{- 2} - 0.5252 z^{- 3} + 0.0834 z^{- 4}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.7854$	0.8913	1 dB, against the 1 dB allowed
$ω = ω_{s} = 1.4137$	0.1081	19.32 dB, against the 15 dB required
$ω = π$	0	at Nyquist

$N = 4$ and $Ω_{c} = 0.9809$ rad/s. The specification is met at both edges.

H (z) = \frac{0.0167 + 0.0667 z^{- 1} + 0.1001 z^{- 2} + 0.0667 z^{- 3} + 0.0167 z^{- 4}}{1 - 1.6476 z^{- 1} + 1.3563 z^{- 2} - 0.5252 z^{- 3} + 0.0834 z^{- 4}}

Same as 2082 Baishakh Q82071 Shrawan · Q815 marksDesign a low pass digital filter by Bilinear Transformation method to an approximate Butterworth filter it passband frequency is

0.2 π

radians and maximum deviation of 1 db below 0 dB gain in the pass band. The maximum gain of -15 db and frequency is

0.4 π

radians in stop band, consider sampling frequency 1 Hz.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 1) \tan (0.6283 / 2) = 0.6498 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 1) \tan (1.2566 / 2) = 1.4531 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.5} - 1) / (10^{0.1} - 1)] / (2 \log_{10} (2.2361)) \\ = \log_{10} (30.623 / 0.2589) / (2 log 102.2361) \\ = 2.9656 so  take N = 3 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.6498 / {(0.2589)}^{1 / 6} \\ = 0.814 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 1.4531 / {(30.623)}^{1 / 6} \\ = 0.8215 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.814$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 3 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 2 \\ s 0 & = - 0.407 + j 0.7049 \\ s 1 & = - 0.814 \\ s 2 & = - 0.407 - j 0.7049 \end{aligned}

H_{a} (s) = \frac{{(0.814)}^{3}}{(s^{2} + 0.814 s + 0.6626) (s + 0.814)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 2 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.0305 + 0.0914 z^{- 1} + 0.0914 z^{- 2} + 0.0305 z^{- 3}}{1 - 1.4826 z^{- 1} + 0.9296 z^{- 2} - 0.2033 z^{- 3}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.6283$	0.8913	1 dB, against the 1 dB allowed
$ω = ω_{s} = 1.2566$	0.1731	15.23 dB, against the 15 dB required
$ω = π$	0	at Nyquist

$N = 3$ and $Ω_{c} = 0.814$ rad/s. The specification is met at both edges.

H (z) = \frac{0.0305 + 0.0914 z^{- 1} + 0.0914 z^{- 2} + 0.0305 z^{- 3}}{1 - 1.4826 z^{- 1} + 0.9296 z^{- 2} - 0.2033 z^{- 3}}

The same questions grouped by how often each topic is set: butterworth iir design

Bilinear transformation against impulse invariance HOT 5/19

82 Ba · 81 Bh · 81 Ba · 79 Ba · 72 Ch12+311+32+10

Before the formulas

The comparison the paper asks for in words, so it is worth having the sentence ready. Impulse invariance keeps the shape of the response in time and suffers aliasing. The bilinear transformation keeps stability and every band edge you prewarp for, and suffers warping.

The line that earns the marks is the difference between those two faults: warping can be corrected in advance and aliasing cannot be corrected at all, once it has happened.

	Impulse invariance	Bilinear transformation
Based on	Sampling the analog impulse response, $h [n] = T$ ha(nT)	An algebraic substitution, from trapezoidal integration
Mapping	z = $e^{s T}$ , many to one	s = (2/T)(1- $z^{- 1}$ )/(1+ $z^{- 1}$ ), one to one
Frequency relation	$ω = Ω$ T, linear	$Ω$ = (2/T) $\tan (ω / 2)$ , non linear
Aliasing	Yes, and it fills in the stopband	None
Warping	None	Yes, corrected by prewarping
Usable for	Low pass and band pass only	All types, high pass and band stop included
Preserves	The shape of the impulse response	The magnitude values, and stability
Stability	Preserved, provided sampling is fast enough	Always preserved
Poles and zeros	Poles map by $e^{p T}$ ; the zeros do not map	Both map by the same substitution
Verdict	Use when the time domain shape matters and the filter is band limited	The default choice, and the one the paper asks for

The advantages of the bilinear method over impulse invariance, which is the wording the paper uses: no aliasing, so the designed stopband is actually achieved; a one to one mapping, so stability is guaranteed; it works for every filter type; and the warping it introduces is completely correctable by prewarping, whereas aliasing is not correctable at all.

See it drawnOne fault can be corrected in advance and the other cannot

Warping moves every frequency by a known amount, so you bend the specification the other way first and the edges land where you asked. Aliasing destroys information, and nothing afterwards brings it back.

Practise thisEvery question the paper has set on this topic, worked in full5

Practise this +12 like it2082 Baishakh · Q812+3 marksUsing bilinear transformation technique, design a Butterworth filter which satisfies the following conditions:

0.8 \leq | H (e^{j ω}) | \leq 1

for

0 \leq w \leq 0.2 π

;

| H (e^{j ω}) | \leq 0.2

for

0.6 π \leq ω \leq π

. Consider sampling frequency 1 Hz. Compare Impulse Invariance Method and Bilinear Transformation technique for analog filter to digital filter conversion.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 1) \tan (0.6283 / 2) = 0.6498 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 1) \tan (1.885 / 2) = 2.7528 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.398} - 1) / (10^{0.194} - 1)] / (2 \log_{10} (4.2361)) \\ = \log_{10} (24 / 0.5625) / (2 log 104.2361) \\ = 1.3 so  take N = 2 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.6498 / {(0.5625)}^{1 / 4} \\ = 0.7504 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 2.7528 / {(24)}^{1 / 4} \\ = 1.2437 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.7504$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 2 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 1 \\ s 0 & = - 0.5306 + j 0.5306 \\ s 1 & = - 0.5306 - j 0.5306 \end{aligned}

H_{a} (s) = \frac{{(0.7504)}^{2}}{(s^{2} + 1.0612 s + 0.5631)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 2 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.0842 + 0.1684 z^{- 1} + 0.0842 z^{- 2}}{1 - 1.0282 z^{- 1} + 0.3651 z^{- 2}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.6283$	0.8	1.94 dB, against the 1.94 dB allowed
$ω = ω_{s} = 1.885$	0.0741	22.6 dB, against the 13.98 dB required
$ω = π$	0	at Nyquist

$N = 2$ and $Ω_{c} = 0.7504$ rad/s. The specification is met at both edges.

H (z) = \frac{0.0842 + 0.1684 z^{- 1} + 0.0842 z^{- 2}}{1 - 1.0282 z^{- 1} + 0.3651 z^{- 2}}

Same as 2082 Baishakh Q82081 Bhadra · Q812+3 marksDesign a digital low pass IIR filter using Bilinear transformation to meet following specifications:

0.9 \leq | H (e^{j ω}) | \leq 1

for

0 \leq | w | \leq \frac{π}{2}

;

| H (e^{j ω}) | \leq 0.2

for

3 π / 4 \leq | w | \leq π

. Use Butterworth approximation for design with sampling frequency of 1Hz. Compare Impulse Invariance Method and Bilinear Transformation method.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 1) \tan (1.5708 / 2) = 2 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 1) \tan (2.3562 / 2) = 4.8284 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.398} - 1) / (10^{0.092} - 1)] / (2 \log_{10} (2.4142)) \\ = \log_{10} (24 / 0.2346) / (2 log 102.4142) \\ = 2.6255 so  take N = 3 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 2 / {(0.2346)}^{1 / 6} \\ = 2.5467 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 4.8284 / {(24)}^{1 / 6} \\ = 2.843 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 2.5467$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 3 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 2 \\ s 0 & = - 1.2734 + j 2.2055 \\ s 1 & = - 2.5467 \\ s 2 & = - 1.2734 - j 2.2055 \end{aligned}

H_{a} (s) = \frac{{(2.5467)}^{3}}{(s^{2} + 2.5467 s + 6.4859) (s + 2.5467)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 2 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.2332 + 0.6996 z^{- 1} + 0.6996 z^{- 2} + 0.2332 z^{- 3}}{1 + 0.4394 z^{- 1} + 0.3845 z^{- 2} + 0.0416 z^{- 3}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 1.5708$	0.9	0.92 dB, against the 0.92 dB allowed
$ω = ω_{s} = 2.3562$	0.1452	16.76 dB, against the 13.98 dB required
$ω = π$	0	at Nyquist

$N = 3$ and $Ω_{c} = 2.5467$ rad/s. The specification is met at both edges.

H (z) = \frac{0.2332 + 0.6996 z^{- 1} + 0.6996 z^{- 2} + 0.2332 z^{- 3}}{1 + 0.4394 z^{- 1} + 0.3845 z^{- 2} + 0.0416 z^{- 3}}

T = 0.1

second, to satisfy the following specifications:

0.6 \leq | H (e^{j ω}) | \leq 1

for

0 \leq w \leq 0.35 π

;

| H (e^{j ω}) | \leq 0.1

for

0.7 π \leq ω \leq π

How this is readThe question fixes $T = 0.1$ s rather than giving a sampling frequency, so the prewarping uses 2/T = 20.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.
$T = 0.1$ here, not 1, so the factor 2/T in the prewarping is 20. It cancels in the end, but only if it is carried consistently.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 0.1) \tan (1.0996 / 2) = 12.256 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 0.1) \tan (2.1991 / 2) = 39.2522 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{2} - 1) / (10^{0.444} - 1)] / (2 \log_{10} (3.2027)) \\ = \log_{10} (99 / 1.7778) / (2 log 103.2027) \\ = 1.7267 so  take N = 2 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 12.256 / {(1.7778)}^{1 / 4} \\ = 10.614 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 39.2522 / {(99)}^{1 / 4} \\ = 12.4439 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 10.614$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 2 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 1 \\ s 0 & = - 7.5052 + j 7.5052 \\ s 1 & = - 7.5052 - j 7.5052 \end{aligned}

H_{a} (s) = \frac{{(10.614)}^{2}}{(s^{2} + 15.0105 s + 112.6574)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 20 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.1386 + 0.2772 z^{- 1} + 0.1386 z^{- 2}}{1 - 0.707 z^{- 1} + 0.2614 z^{- 2}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 1.0996$	0.6	4.44 dB, against the 4.44 dB allowed
$ω = ω_{s} = 2.1991$	0.0729	22.74 dB, against the 20 dB required
$ω = π$	0	at Nyquist

$N = 2$ and $Ω_{c} = 10.614$ rad/s. The specification is met at both edges.

H (z) = \frac{0.1386 + 0.2772 z^{- 1} + 0.1386 z^{- 2}}{1 - 0.707 z^{- 1} + 0.2614 z^{- 2}}

0.24 π

radians and maximum deviation of 0.98 dB below 0 dB gain in the passband. The maximum gain of -14.95 dB and frequency is

0.57 π

radians in stopband, consider sampling frequency 0.5 Hz. Compare impulse invariance method with bilinear transformation method.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.
$T = 2$ here, not 1, so the factor 2/T in the prewarping is 1. It cancels in the end, but only if it is carried consistently.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 2) \tan (0.754 / 2) = 0.3959 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 2) \tan (1.7907 / 2) = 1.2482 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.495} - 1) / (10^{0.098} - 1)] / (2 \log_{10} (3.1526)) \\ = \log_{10} (30.261 / 0.2531) / (2 log 103.1526) \\ = 2.0831 so  take N = 3 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.3959 / {(0.2531)}^{1 / 6} \\ = 0.4978 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 1.2482 / {(30.261)}^{1 / 6} \\ = 0.7071 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.4978$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 3 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 2 \\ s 0 & = - 0.2489 + j 0.4311 \\ s 1 & = - 0.4978 \\ s 2 & = - 0.2489 - j 0.4311 \end{aligned}

H_{a} (s) = \frac{{(0.4978)}^{3}}{(s^{2} + 0.4978 s + 0.2478) (s + 0.4978)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 1 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.0472 + 0.1415 z^{- 1} + 0.1415 z^{- 2} + 0.0472 z^{- 3}}{1 - 1.1971 z^{- 1} + 0.7186 z^{- 2} - 0.1441 z^{- 3}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.754$	0.8933	0.98 dB, against the 0.98 dB allowed
$ω = ω_{s} = 1.7907$	0.0633	23.97 dB, against the 14.95 dB required
$ω = π$	0	at Nyquist

$N = 3$ and $Ω_{c} = 0.4978$ rad/s. The specification is met at both edges.

H (z) = \frac{0.0472 + 0.1415 z^{- 1} + 0.1415 z^{- 2} + 0.0472 z^{- 3}}{1 - 1.1971 z^{- 1} + 0.7186 z^{- 2} - 0.1441 z^{- 3}}

How this is readThe attenuations are printed as -5 dB and -12 dB; the formulas take them positive.

What is different in this one

The edges are given in hertz, not radians, so the first step is $ω = 2 π$ f / fs. A question that already gives radians skips it.
Impulse invariance, not the bilinear transformation. There is therefore no prewarping: the edges scale linearly, $Ω = ω / T$ , and the mapping is applied to each partial fraction term separately rather than to $H (s)$ as a whole.
$T = 0.0002$ here, not 1, so the factor 2/T in the prewarping is 10000. It cancels in the end, but only if it is carried consistently.

Digital edges first: $ω = 2 π$ f / fs gives $ω_{p} = 2 π$ (200)/5000 = 0.2513 rad and $ω_{s} = 2 π$ (500)/5000 = 0.6283 rad.

Impulse invariance maps frequency linearly, $Ω = ω / T$ , so there is no prewarping to do.

\begin{aligned} Ω_{p} & = ω_{p} / T = 1256.6371 rad / s \\ Ω_{s} & = ω_{s} / T = 3141.5927 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.2} - 1) / (10^{0.5} - 1)] / (2 \log_{10} (2.5)) \\ = \log_{10} (14.849 / 2.1623) / (2 log 102.5) \\ = 1.0514 so  take N = 2 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 1256.6371 / {(2.1623)}^{1 / 4} \\ = 1036.2916 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 3141.5927 / {(14.849)}^{1 / 4} \\ = 1600.3909 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 1036.2916$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 2 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 1 \\ s 0 & = - 732.7689 + j 732.7689 \\ s 1 & = - 732.7689 - j 732.7689 \end{aligned}

H_{a} (s) = \frac{{(1036.2916)}^{2}}{(s^{2} + 1465.5377 s + 1073900.3832)}

Expand $H_{a} (s)$ in partial fractions and map each pole on its own with s = pk going to z = $e^{p k T}$ , which is what impulse invariance does.

\begin{aligned} residue - j 732.7689 at  pole - 732.7689 + j 732.7689 \\ \to - j 0.1466 / (1 - (0.8544 + j 0.1261) z^{- 1}) \\ residue j 732.7689 at  pole - 732.7689 - j 732.7689 \\ \to j 0.1466 / (1 - (0.8544 - j 0.1261) z^{- 1}) \end{aligned}

H (z) = \frac{0.037 z^{- 1}}{1 - 1.7088 z^{- 1} + 0.7459 z^{- 2}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	0.9964	the DC gain
$ω = ω_{p} = 0.2513$	0.5633	4.99 dB, against the 5 dB allowed
$ω = ω_{s} = 0.6283$	0.1114	19.06 dB, against the 12 dB required
$ω = π$	0.0107	at Nyquist

$N = 2$ and $Ω_{c} = 1036.2916$ rad/s. The specification is met at both edges.

H (z) = \frac{0.037 z^{- 1}}{1 - 1.7088 z^{- 1} + 0.7459 z^{- 2}}

6.3Design of a digital low pass Butterworth filter

The Butterworth approximation TOP 18/19

Before the formulas

The analog filter the paper always sets, so this card is the one to know cold. Its defining property is flatness: it is as flat as a response can be made in the passband, with no ripple anywhere in either band, and it pays for that with the slowest rolloff of the standard approximations.

Every Butterworth question is the same four steps: turn the specification into an order, find the cut off frequency, write down the poles, and map to the digital domain. The order formula is four lines of derivation, and show how the order is obtained is a question the paper actually asks, so learn it as a derivation.

The Butterworth low pass The magnitude squared response is maximally flat at

Ω = 0

: the first 2N-1 derivatives vanish there. There is no ripple in either band, and the response falls monotonically, at 20N dB per decade far out.

| H_{a} (j Ω) |^{2} = \frac{1}{1 + {(Ω / Ω_{c})}^{2 N}}

At $Ω = Ω_{c}$ the magnitude squared is $\frac{1}{2}$ , that is -3 dB, for any N, so $Ω_{c}$ is always the 3 dB cut off.

Deriving the order formula. This is the derivation the paper means by show how the order is obtained, and it is four lines.

\begin{aligned} | H_{a} (j Ω_{p}) |^{2} & \geq \frac{1}{1 + ε^{2}} & the passband requirement \\ | H_{a} (j Ω_{s}) |^{2} & \leq \frac{1}{λ^{2}} & the stopband requirement \\ ε^{2} & = 10^{0.1 α_{p}} - 1, λ^{2} = 10^{0.1 α_{s}} - 1 \\ N & \geq \frac{\log_{10} (λ^{2} / ε^{2})}{2 \log_{10} (Ω_{s} / Ω_{p})} & divide one by the other: Ω_{c} cancels \end{aligned}

Which $Ω_{c}$ to use. Rounding N up means the filter over performs, so the two formulas give slightly different cutoffs. Using the passband one meets the passband exactly and beats the stopband; using the stopband one does the reverse. Either is accepted; say which you used.
If the specification is given as ripples rather than dB, convert first: $α_{p} = - 20$ log10(1 $- δ_{p}$ ) and $α_{s} = - 20$ log10( $δ_{s}$ ). A specification such as $0.8 \leq | H | \leq 1$ with $| H | \leq 0.2$ gives $α_{p} = - 20 \log_{10} (0.8) = 1.94$ dB and $α_{s} = - 20 \log_{10} (0.2) = 13.98$ dB.

The poles, and the prototype The poles of |

H_{a} (s)

|^2 lie on a circle of radius

Ω_{c}

, equally spaced, and the stable filter takes the left half plane ones:
$s [k] = Ω_{c} e^{j π (2 k + N + 1) / (2 N)}$ , for

k = 0

, 1, ..., N-1.
They are symmetric about the real axis, never on the imaginary axis, and separated by

π

/N radians.

N	Normalised Butterworth denominator, with $Ω_{c} = 1$
1	s + 1
2	$s^{2}$ + 1.4142 s + 1
3	(s + 1)( $s^{2}$ + s + 1) = $s^{3} + 2 s^{2} + 2 s + 1$
4	( $s^{2}$ + 0.7654s + 1)( $s^{2}$ + 1.8478s + 1)
5	(s + 1)( $s^{2}$ + 0.6180s + 1)( $s^{2}$ + 1.6180s + 1)

To denormalise, replace s by s/ $Ω_{c}$ throughout, so that for $N = 2 H_{a} (s) = Ω_{c}^{2} /$ ( $s^{2} + 1.4142 Ω_{c} s + Ω_{c}^{2}$ ). The numerator is whatever makes Ha(0) = 1.

In the exam

Round N up, always, and say what N you are using before computing $Ω_{c}$ .
Keep four decimal places through the prewarping and the cutoff. These questions carry the error forward, and a two decimal $Ω_{c}$ gives a visibly wrong $H (z)$ .
N is usually 2 or 3 in these papers. If you get 7, re check the attenuations.

The poles this places on a circle are what a pole zero sketch would show, and a cascade of biquads is how the finished filter is actually built.

See it drawnFlat everywhere, and N only decides how fast it falls

There is no ripple in either band at any order, every curve passes through the same point at the cut off, and raising N only steepens the descent. That is what maximally flat means.

Practise thisEvery question the paper has set on this topic, worked in full18

How this is readThe attenuations are printed as -3 dB and -10 dB; the design formulas take them as positive numbers, so Ap = 3 and As = 10.

What is different in this one

The edges are given in hertz, not radians, so the first step is $ω = 2 π$ f / fs. A question that already gives radians skips it.
The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.
$T = 0.0002$ here, not 1, so the factor 2/T in the prewarping is 10000. It cancels in the end, but only if it is carried consistently.

Digital edges first: $ω = 2 π$ f / fs gives $ω_{p} = 2 π$ (350)/5000 = 0.4398 rad and $ω_{s} = 2 π$ (1000)/5000 = 1.2566 rad.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 0.0002) \tan (0.4398 / 2) = 2235.2648 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 0.0002) \tan (1.2566 / 2) = 7265.4253 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1} - 1) / (10^{0.3} - 1)] / (2 \log_{10} (3.2504)) \\ = \log_{10} (9 / 0.9953) / (2 log 103.2504) \\ = 0.934 so  take N = 1 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 2235.2648 / {(0.9953)}^{1 / 2} \\ = 2240.5787 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 7265.4253 / {(9)}^{1 / 2} \\ = 2421.8084 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 2240.5787$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 1 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 0 \\ s 0 & = - 2240.5787 \end{aligned}

H_{a} (s) = \frac{{(2240.5787)}^{1}}{(s + 2240.5787)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 10000 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.183 + 0.183 z^{- 1}}{1 - 0.6339 z^{- 1}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.4398$	0.7079	3 dB, against the 3 dB allowed
$ω = ω_{s} = 1.2566$	0.2947	10.61 dB, against the 10 dB required
$ω = π$	0	at Nyquist

$N = 1$ and $Ω_{c} = 2240.5787$ rad/s. The specification is met at both edges.

H (z) = \frac{0.183 + 0.183 z^{- 1}}{1 - 0.6339 z^{- 1}}

Practise this +12 like it2082 Baishakh · Q812+3 marksUsing bilinear transformation technique, design a Butterworth filter which satisfies the following conditions:

0.8 \leq | H (e^{j ω}) | \leq 1

for

0 \leq w \leq 0.2 π

;

| H (e^{j ω}) | \leq 0.2

for

0.6 π \leq ω \leq π

. Consider sampling frequency 1 Hz. Compare Impulse Invariance Method and Bilinear Transformation technique for analog filter to digital filter conversion.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 1) \tan (0.6283 / 2) = 0.6498 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 1) \tan (1.885 / 2) = 2.7528 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.398} - 1) / (10^{0.194} - 1)] / (2 \log_{10} (4.2361)) \\ = \log_{10} (24 / 0.5625) / (2 log 104.2361) \\ = 1.3 so  take N = 2 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.6498 / {(0.5625)}^{1 / 4} \\ = 0.7504 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 2.7528 / {(24)}^{1 / 4} \\ = 1.2437 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.7504$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 2 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 1 \\ s 0 & = - 0.5306 + j 0.5306 \\ s 1 & = - 0.5306 - j 0.5306 \end{aligned}

H_{a} (s) = \frac{{(0.7504)}^{2}}{(s^{2} + 1.0612 s + 0.5631)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 2 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.0842 + 0.1684 z^{- 1} + 0.0842 z^{- 2}}{1 - 1.0282 z^{- 1} + 0.3651 z^{- 2}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.6283$	0.8	1.94 dB, against the 1.94 dB allowed
$ω = ω_{s} = 1.885$	0.0741	22.6 dB, against the 13.98 dB required
$ω = π$	0	at Nyquist

$N = 2$ and $Ω_{c} = 0.7504$ rad/s. The specification is met at both edges.

H (z) = \frac{0.0842 + 0.1684 z^{- 1} + 0.0842 z^{- 2}}{1 - 1.0282 z^{- 1} + 0.3651 z^{- 2}}

Same as 2082 Baishakh Q82081 Bhadra · Q812+3 marksDesign a digital low pass IIR filter using Bilinear transformation to meet following specifications:

0.9 \leq | H (e^{j ω}) | \leq 1

for

0 \leq | w | \leq \frac{π}{2}

;

| H (e^{j ω}) | \leq 0.2

for

3 π / 4 \leq | w | \leq π

. Use Butterworth approximation for design with sampling frequency of 1Hz. Compare Impulse Invariance Method and Bilinear Transformation method.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 1) \tan (1.5708 / 2) = 2 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 1) \tan (2.3562 / 2) = 4.8284 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.398} - 1) / (10^{0.092} - 1)] / (2 \log_{10} (2.4142)) \\ = \log_{10} (24 / 0.2346) / (2 log 102.4142) \\ = 2.6255 so  take N = 3 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 2 / {(0.2346)}^{1 / 6} \\ = 2.5467 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 4.8284 / {(24)}^{1 / 6} \\ = 2.843 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 2.5467$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 3 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 2 \\ s 0 & = - 1.2734 + j 2.2055 \\ s 1 & = - 2.5467 \\ s 2 & = - 1.2734 - j 2.2055 \end{aligned}

H_{a} (s) = \frac{{(2.5467)}^{3}}{(s^{2} + 2.5467 s + 6.4859) (s + 2.5467)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 2 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.2332 + 0.6996 z^{- 1} + 0.6996 z^{- 2} + 0.2332 z^{- 3}}{1 + 0.4394 z^{- 1} + 0.3845 z^{- 2} + 0.0416 z^{- 3}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 1.5708$	0.9	0.92 dB, against the 0.92 dB allowed
$ω = ω_{s} = 2.3562$	0.1452	16.76 dB, against the 13.98 dB required
$ω = π$	0	at Nyquist

$N = 3$ and $Ω_{c} = 2.5467$ rad/s. The specification is met at both edges.

H (z) = \frac{0.2332 + 0.6996 z^{- 1} + 0.6996 z^{- 2} + 0.2332 z^{- 3}}{1 + 0.4394 z^{- 1} + 0.3845 z^{- 2} + 0.0416 z^{- 3}}

T = 0.1

second, to satisfy the following specifications:

0.6 \leq | H (e^{j ω}) | \leq 1

for

0 \leq w \leq 0.35 π

;

| H (e^{j ω}) | \leq 0.1

for

0.7 π \leq ω \leq π

How this is readThe question fixes $T = 0.1$ s rather than giving a sampling frequency, so the prewarping uses 2/T = 20.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.
$T = 0.1$ here, not 1, so the factor 2/T in the prewarping is 20. It cancels in the end, but only if it is carried consistently.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 0.1) \tan (1.0996 / 2) = 12.256 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 0.1) \tan (2.1991 / 2) = 39.2522 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{2} - 1) / (10^{0.444} - 1)] / (2 \log_{10} (3.2027)) \\ = \log_{10} (99 / 1.7778) / (2 log 103.2027) \\ = 1.7267 so  take N = 2 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 12.256 / {(1.7778)}^{1 / 4} \\ = 10.614 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 39.2522 / {(99)}^{1 / 4} \\ = 12.4439 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 10.614$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 2 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 1 \\ s 0 & = - 7.5052 + j 7.5052 \\ s 1 & = - 7.5052 - j 7.5052 \end{aligned}

H_{a} (s) = \frac{{(10.614)}^{2}}{(s^{2} + 15.0105 s + 112.6574)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 20 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.1386 + 0.2772 z^{- 1} + 0.1386 z^{- 2}}{1 - 0.707 z^{- 1} + 0.2614 z^{- 2}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 1.0996$	0.6	4.44 dB, against the 4.44 dB allowed
$ω = ω_{s} = 2.1991$	0.0729	22.74 dB, against the 20 dB required
$ω = π$	0	at Nyquist

$N = 2$ and $Ω_{c} = 10.614$ rad/s. The specification is met at both edges.

H (z) = \frac{0.1386 + 0.2772 z^{- 1} + 0.1386 z^{- 2}}{1 - 0.707 z^{- 1} + 0.2614 z^{- 2}}

Same as 2082 Baishakh Q82080 Bhadra · Q812 marksUsing Bilinear transformation, design a Butterworth low pass filter which satisfies following conditions:

0.9 \leq | H (e^{j ω}) | \leq 1

for

0 \leq ω \leq \frac{π}{2}

;

| H (e^{j ω}) | \leq 0.2

for

\frac{3 π}{4} \leq ω \leq π

. Consider sampling frequency of 1 Hz.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 1) \tan (1.5708 / 2) = 2 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 1) \tan (2.3562 / 2) = 4.8284 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.398} - 1) / (10^{0.092} - 1)] / (2 \log_{10} (2.4142)) \\ = \log_{10} (24 / 0.2346) / (2 log 102.4142) \\ = 2.6255 so  take N = 3 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 2 / {(0.2346)}^{1 / 6} \\ = 2.5467 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 4.8284 / {(24)}^{1 / 6} \\ = 2.843 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 2.5467$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 3 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 2 \\ s 0 & = - 1.2734 + j 2.2055 \\ s 1 & = - 2.5467 \\ s 2 & = - 1.2734 - j 2.2055 \end{aligned}

H_{a} (s) = \frac{{(2.5467)}^{3}}{(s^{2} + 2.5467 s + 6.4859) (s + 2.5467)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 2 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.2332 + 0.6996 z^{- 1} + 0.6996 z^{- 2} + 0.2332 z^{- 3}}{1 + 0.4394 z^{- 1} + 0.3845 z^{- 2} + 0.0416 z^{- 3}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 1.5708$	0.9	0.92 dB, against the 0.92 dB allowed
$ω = ω_{s} = 2.3562$	0.1452	16.76 dB, against the 13.98 dB required
$ω = π$	0	at Nyquist

$N = 3$ and $Ω_{c} = 2.5467$ rad/s. The specification is met at both edges.

H (z) = \frac{0.2332 + 0.6996 z^{- 1} + 0.6996 z^{- 2} + 0.2332 z^{- 3}}{1 + 0.4394 z^{- 1} + 0.3845 z^{- 2} + 0.0416 z^{- 3}}

Same as 2082 Baishakh Q82080 Baishakh · Q911 marksDesign a low pass digital IIR filter by Bilinear Transformation method to an approximate Butterworth low pass filter, if passband edge frequency is

0.26 π

radians and maximum deviation of 0.99 dB below 0 dB gain in the passband. The maximum gain of -14.99 dB and frequency is

0.58 π

radians in stopband, Consider sampling frequency 0.5 Hz.

How this is readA sampling frequency of 0.5 Hz gives $T = 2$ s.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.
$T = 2$ here, not 1, so the factor 2/T in the prewarping is 1. It cancels in the end, but only if it is carried consistently.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 2) \tan (0.8168 / 2) = 0.4327 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 2) \tan (1.8221 / 2) = 1.2892 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.499} - 1) / (10^{0.099} - 1)] / (2 \log_{10} (2.9791)) \\ = \log_{10} (30.55 / 0.256) / (2 log 102.9791) \\ = 2.1902 so  take N = 3 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.4327 / {(0.256)}^{1 / 6} \\ = 0.5431 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 1.2892 / {(30.55)}^{1 / 6} \\ = 0.7291 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.5431$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 3 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 2 \\ s 0 & = - 0.2715 + j 0.4703 \\ s 1 & = - 0.5431 \\ s 2 & = - 0.2715 - j 0.4703 \end{aligned}

H_{a} (s) = \frac{{(0.5431)}^{3}}{(s^{2} + 0.5431 s + 0.2949) (s + 0.5431)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 1 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.0565 + 0.1694 z^{- 1} + 0.1694 z^{- 2} + 0.0565 z^{- 3}}{1 - 1.0634 z^{- 1} + 0.6363 z^{- 2} - 0.1211 z^{- 3}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.8168$	0.8923	0.99 dB, against the 0.99 dB allowed
$ω = ω_{s} = 1.8221$	0.0745	22.55 dB, against the 14.99 dB required
$ω = π$	0	at Nyquist

$N = 3$ and $Ω_{c} = 0.5431$ rad/s. The specification is met at both edges.

H (z) = \frac{0.0565 + 0.1694 z^{- 1} + 0.1694 z^{- 2} + 0.0565 z^{- 3}}{1 - 1.0634 z^{- 1} + 0.6363 z^{- 2} - 0.1211 z^{- 3}}

Practise this2079 Bhadra · Q911+4 marksDesign a low pass discrete time Butterworth filter using bilinear transformation having following specifications: Passband frequency (

Ω_{p}

) =

0.25 π

radians, Stopband frequency (

Ω_{s}

) =

0.55 π

0.45 π

using digital domain transformation.

The design

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.
$T = 2$ here, not 1, so the factor 2/T in the prewarping is 1. It cancels in the end, but only if it is carried consistently.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 2) \tan (0.7854 / 2) = 0.4142 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 2) \tan (1.7279 / 2) = 1.1708 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.356} - 1) / (10^{0.101} - 1)] / (2 \log_{10} (2.8267)) \\ = \log_{10} (21.676 / 0.2625) / (2 log 102.8267) \\ = 2.1239 so  take N = 3 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.4142 / {(0.2625)}^{1 / 6} \\ = 0.5177 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 1.1708 / {(21.676)}^{1 / 6} \\ = 0.7012 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.5177$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 3 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 2 \\ s 0 & = - 0.2588 + j 0.4483 \\ s 1 & = - 0.5177 \\ s 2 & = - 0.2588 - j 0.4483 \end{aligned}

H_{a} (s) = \frac{{(0.5177)}^{3}}{(s^{2} + 0.5177 s + 0.268) (s + 0.5177)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 1 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.0512 + 0.1536 z^{- 1} + 0.1536 z^{- 2} + 0.0512 z^{- 3}}{1 - 1.1377 z^{- 1} + 0.6808 z^{- 2} - 0.1335 z^{- 3}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.7854$	0.89	1.01 dB, against the 1.01 dB allowed
$ω = ω_{s} = 1.7279$	0.0861	21.3 dB, against the 13.56 dB required
$ω = π$	0	at Nyquist

$N = 3$ and $Ω_{c} = 0.5177$ rad/s. The specification is met at both edges.

H (z) = \frac{0.0512 + 0.1536 z^{- 1} + 0.1536 z^{- 2} + 0.0512 z^{- 3}}{1 - 1.1377 z^{- 1} + 0.6808 z^{- 2} - 0.1335 z^{- 3}}

The rest of this questionThis topic works the part that belongs to it. The other part is worked under Converting it to a high pass, the last four marks, where it belongs.

0.24 π

radians and maximum deviation of 0.98 dB below 0 dB gain in the passband. The maximum gain of -14.95 dB and frequency is

0.57 π

radians in stopband, consider sampling frequency 0.5 Hz. Compare impulse invariance method with bilinear transformation method.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.
$T = 2$ here, not 1, so the factor 2/T in the prewarping is 1. It cancels in the end, but only if it is carried consistently.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 2) \tan (0.754 / 2) = 0.3959 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 2) \tan (1.7907 / 2) = 1.2482 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.495} - 1) / (10^{0.098} - 1)] / (2 \log_{10} (3.1526)) \\ = \log_{10} (30.261 / 0.2531) / (2 log 103.1526) \\ = 2.0831 so  take N = 3 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.3959 / {(0.2531)}^{1 / 6} \\ = 0.4978 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 1.2482 / {(30.261)}^{1 / 6} \\ = 0.7071 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.4978$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 3 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 2 \\ s 0 & = - 0.2489 + j 0.4311 \\ s 1 & = - 0.4978 \\ s 2 & = - 0.2489 - j 0.4311 \end{aligned}

H_{a} (s) = \frac{{(0.4978)}^{3}}{(s^{2} + 0.4978 s + 0.2478) (s + 0.4978)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 1 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.0472 + 0.1415 z^{- 1} + 0.1415 z^{- 2} + 0.0472 z^{- 3}}{1 - 1.1971 z^{- 1} + 0.7186 z^{- 2} - 0.1441 z^{- 3}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.754$	0.8933	0.98 dB, against the 0.98 dB allowed
$ω = ω_{s} = 1.7907$	0.0633	23.97 dB, against the 14.95 dB required
$ω = π$	0	at Nyquist

$N = 3$ and $Ω_{c} = 0.4978$ rad/s. The specification is met at both edges.

H (z) = \frac{0.0472 + 0.1415 z^{- 1} + 0.1415 z^{- 2} + 0.0472 z^{- 3}}{1 - 1.1971 z^{- 1} + 0.7186 z^{- 2} - 0.1441 z^{- 3}}

Ω = ω / T

2 / ((s + 1) (s + 2))

T = 0.2

into 0.2774z over

z^{2} - 1.4887 z + 0.54853

, but the numerator is wrong. It is

2 (e^{- 0} .2 - e^{- 0} .4) = 2 (0.8187 - 0.6703) = < b > 0.2968 < / b

>; the book doubled 0.67 as 1.36 instead of 1.34. The denominator is right.

What is different in this one

The edges are given in hertz, not radians, so the first step is $ω = 2 π$ f / fs. A question that already gives radians skips it.
Impulse invariance, not the bilinear transformation. There is therefore no prewarping: the edges scale linearly, $Ω = ω / T$ , and the mapping is applied to each partial fraction term separately rather than to $H (s)$ as a whole.
$T = 0.0002$ here, not 1, so the factor 2/T in the prewarping is 10000. It cancels in the end, but only if it is carried consistently.
Impulse invariance, not the bilinear transformation, so there is no prewarping: the edges scale linearly, $Ω = ω / T$ . The second half of the question asks what prewarping is and why it is necessary, which is exactly the step this design does not take. Watch the standard text here: its worked impulse invariance example turns $2 / ((s + 1) (s + 2))$ at $T = 0.2$ into 0.2774z over $z^{2} - 1.4887 z + 0.54853$ , but the numerator is wrong. It is $2 (e^{- 0} .2 - e^{- 0} .4) = 2 (0.8187 - 0.6703)$ = 0.2968; the book doubled 0.67 as 1.36 instead of 1.34. The denominator is right.

Digital edges first: $ω = 2 π$ f / fs gives $ω_{p} = 2 π$ (200)/5000 = 0.2513 rad and $ω_{s} = 2 π$ (500)/5000 = 0.6283 rad.

Impulse invariance maps frequency linearly, $Ω = ω / T$ , so there is no prewarping to do.

\begin{aligned} Ω_{p} & = ω_{p} / T = 1256.6371 rad / s \\ Ω_{s} & = ω_{s} / T = 3141.5927 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.2} - 1) / (10^{0.5} - 1)] / (2 \log_{10} (2.5)) \\ = \log_{10} (14.849 / 2.1623) / (2 log 102.5) \\ = 1.0514 so  take N = 2 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 1256.6371 / {(2.1623)}^{1 / 4} \\ = 1036.2916 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 3141.5927 / {(14.849)}^{1 / 4} \\ = 1600.3909 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 1036.2916$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 2 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 1 \\ s 0 & = - 732.7689 + j 732.7689 \\ s 1 & = - 732.7689 - j 732.7689 \end{aligned}

H_{a} (s) = \frac{{(1036.2916)}^{2}}{(s^{2} + 1465.5377 s + 1073900.3832)}

Expand $H_{a} (s)$ in partial fractions and map each pole on its own with s = pk going to z = $e^{p k T}$ , which is what impulse invariance does.

\begin{aligned} residue - j 732.7689 at  pole - 732.7689 + j 732.7689 \\ \to - j 0.1466 / (1 - (0.8544 + j 0.1261) z^{- 1}) \\ residue j 732.7689 at  pole - 732.7689 - j 732.7689 \\ \to j 0.1466 / (1 - (0.8544 - j 0.1261) z^{- 1}) \end{aligned}

H (z) = \frac{0.037 z^{- 1}}{1 - 1.7088 z^{- 1} + 0.7459 z^{- 2}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	0.9964	the DC gain
$ω = ω_{p} = 0.2513$	0.5633	4.99 dB, against the 5 dB allowed
$ω = ω_{s} = 0.6283$	0.1114	19.06 dB, against the 12 dB required
$ω = π$	0.0107	at Nyquist

$N = 2$ and $Ω_{c} = 1036.2916$ rad/s. The specification is met at both edges.

H (z) = \frac{0.037 z^{- 1}}{1 - 1.7088 z^{- 1} + 0.7459 z^{- 2}}

What is different in this one

The edges are given in hertz, not radians, so the first step is $ω = 2 π$ f / fs. A question that already gives radians skips it.
The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.
$T = 0.0001$ here, not 1, so the factor 2/T in the prewarping is 16000. It cancels in the end, but only if it is carried consistently.

Digital edges first: $ω = 2 π$ f / fs gives $ω_{p} = 2 π$ (1200)/8000 = 0.9425 rad and $ω_{s} = 2 π$ (2500)/8000 = 1.9635 rad.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 0.0001) \tan (0.9425 / 2) = 8152.4072 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 0.0001) \tan (1.9635 / 2) = 23945.6922 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{4} - 1) / (10^{0.1} - 1)] / (2 \log_{10} (2.9373)) \\ = \log_{10} (9999 / 0.2589) / (2 log 102.9373) \\ = 4.901 so  take N = 5 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 8152.4072 / {(0.2589)}^{1 / 10} \\ = 9331.8639 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 23945.6922 / {(9999)}^{1 / 10} \\ = 9533.0471 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 9331.8639$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 5 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 4 \\ s 0 & = - 2883.7045 + j 8875.13 \\ s 1 & = - 7549.6365 + j 5485.132 \\ s 2 & = - 9331.8639 \\ s 3 & = - 7549.6365 - j 5485.132 \\ s 4 & = - 2883.7045 - j 8875.13 \end{aligned}

\begin{aligned} H_{a} (s) & = \frac{{(9331.8639)}^{5}}{D (s)} \\ D (s) & = (s^{2} + 5767.4091 s + 87083683.7083) \\ \times (s^{2} + 15099.273 s + 87083683.7083) \\ \times (s + 9331.8639) \end{aligned}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 16000 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.011 + 0.0549 z^{- 1} + 0.1098 z^{- 2} + 0.1098 z^{- 3} + 0.0549 z^{- 4} + 0.011 z^{- 5}}{1 - 1.617 z^{- 1} + 1.5544 z^{- 2} - 0.7829 z^{- 3} + 0.2231 z^{- 4} - 0.0263 z^{- 5}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.9425$	0.8913	1 dB, against the 1 dB allowed
$ω = ω_{s} = 1.9635$	0.009	40.93 dB, against the 40 dB required
$ω = π$	0	at Nyquist

$N = 5$ and $Ω_{c} = 9331.8639$ rad/s. The specification is met at both edges.

H (z) = \frac{0.011 + 0.0549 z^{- 1} + 0.1098 z^{- 2} + 0.1098 z^{- 3} + 0.0549 z^{- 4} + 0.011 z^{- 5}}{1 - 1.617 z^{- 1} + 1.5544 z^{- 2} - 0.7829 z^{- 3} + 0.2231 z^{- 4} - 0.0263 z^{- 5}}

ω_{p}

) =

0.22 π

radians, Stop bandedge frequency (

ω_{s}

) =

0.54 π

radians, Passband ripple (deltap) = 0.11, Stopband ripple (deltas) = 0.22, Consider sampling frequency 0.5 Hz.

How this is readA passband ripple of 0.11 means Ap = -20 $\log_{10} (0.89) = 1.0122$ dB, and a stopband ripple of 0.22 means As = -20 $\log_{10} (0.22) = 13.1515$ dB. fs = 0.5 Hz gives $T = 2$ s.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.
$T = 2$ here, not 1, so the factor 2/T in the prewarping is 1. It cancels in the end, but only if it is carried consistently.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 2) \tan (0.6912 / 2) = 0.36 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 2) \tan (1.6965 / 2) = 1.1343 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.315} - 1) / (10^{0.101} - 1)] / (2 \log_{10} (3.1506)) \\ = \log_{10} (19.661 / 0.2625) / (2 log 103.1506) \\ = 1.8806 so  take N = 2 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.36 / {(0.2625)}^{1 / 4} \\ = 0.503 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 1.1343 / {(19.661)}^{1 / 4} \\ = 0.5387 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.503$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 2 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 1 \\ s 0 & = - 0.3557 + j 0.3557 \\ s 1 & = - 0.3557 - j 0.3557 \end{aligned}

H_{a} (s) = \frac{{(0.503)}^{2}}{(s^{2} + 0.7113 s + 0.253)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 1 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.1288 + 0.2576 z^{- 1} + 0.1288 z^{- 2}}{1 - 0.7606 z^{- 1} + 0.2757 z^{- 2}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.6912$	0.89	1.01 dB, against the 1.01 dB allowed
$ω = ω_{s} = 1.6965$	0.1929	14.29 dB, against the 13.15 dB required
$ω = π$	0	at Nyquist

$N = 2$ and $Ω_{c} = 0.503$ rad/s. The specification is met at both edges.

H (z) = \frac{0.1288 + 0.2576 z^{- 1} + 0.1288 z^{- 2}}{1 - 0.7606 z^{- 1} + 0.2757 z^{- 2}}

Same as 2082 Baishakh Q82075 Chaitra · Q710 marksUsing bilinear transformation, design a digital filter using Butterworth approximation which satisfies the following conditions:

0.8 \leq | H (e^{j ω}) | \leq 1

for

0 \leq W \leq 0.2 π

;

| H (e^{j ω}) | \leq 0.2

for

0.6 π \leq W \leq π

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 1) \tan (0.6283 / 2) = 0.6498 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 1) \tan (1.885 / 2) = 2.7528 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.398} - 1) / (10^{0.194} - 1)] / (2 \log_{10} (4.2361)) \\ = \log_{10} (24 / 0.5625) / (2 log 104.2361) \\ = 1.3 so  take N = 2 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.6498 / {(0.5625)}^{1 / 4} \\ = 0.7504 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 2.7528 / {(24)}^{1 / 4} \\ = 1.2437 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.7504$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 2 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 1 \\ s 0 & = - 0.5306 + j 0.5306 \\ s 1 & = - 0.5306 - j 0.5306 \end{aligned}

H_{a} (s) = \frac{{(0.7504)}^{2}}{(s^{2} + 1.0612 s + 0.5631)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 2 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.0842 + 0.1684 z^{- 1} + 0.0842 z^{- 2}}{1 - 1.0282 z^{- 1} + 0.3651 z^{- 2}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.6283$	0.8	1.94 dB, against the 1.94 dB allowed
$ω = ω_{s} = 1.885$	0.0741	22.6 dB, against the 13.98 dB required
$ω = π$	0	at Nyquist

$N = 2$ and $Ω_{c} = 0.7504$ rad/s. The specification is met at both edges.

H (z) = \frac{0.0842 + 0.1684 z^{- 1} + 0.0842 z^{- 2}}{1 - 1.0282 z^{- 1} + 0.3651 z^{- 2}}

Same as 2082 Baishakh Q82075 Ashwin · Q612 marksDesign a digital low-pass filter with the following specification: i) Pass-band magnitude constant to 0.7 dB below the frequency of

0.15 π

ii) Stop-band attenuation at least 14 dB for the frequencies between

0.6 π

π

. Use Butterworth approximation as a prototype and use bilinear transformation method to obtain the digital filter.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 1) \tan (0.4712 / 2) = 0.4802 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 1) \tan (1.885 / 2) = 2.7528 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.4} - 1) / (10^{0.07} - 1)] / (2 \log_{10} (5.733)) \\ = \log_{10} (24.119 / 0.1749) / (2 log 105.733) \\ = 1.4106 so  take N = 2 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.4802 / {(0.1749)}^{1 / 4} \\ = 0.7425 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 2.7528 / {(24.119)}^{1 / 4} \\ = 1.2422 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.7425$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 2 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 1 \\ s 0 & = - 0.525 + j 0.525 \\ s 1 & = - 0.525 - j 0.525 \end{aligned}

H_{a} (s) = \frac{{(0.7425)}^{2}}{(s^{2} + 1.05 s + 0.5513)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 2 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.0829 + 0.1658 z^{- 1} + 0.0829 z^{- 2}}{1 - 1.037 z^{- 1} + 0.3685 z^{- 2}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.4712$	0.9226	0.7 dB, against the 0.7 dB allowed
$ω = ω_{s} = 1.885$	0.0726	22.79 dB, against the 14 dB required
$ω = π$	0	at Nyquist

$N = 2$ and $Ω_{c} = 0.7425$ rad/s. The specification is met at both edges.

H (z) = \frac{0.0829 + 0.1658 z^{- 1} + 0.0829 z^{- 2}}{1 - 1.037 z^{- 1} + 0.3685 z^{- 2}}

Practise this2074 Chaitra · Q712 marksDesign a digital low-pass filter with the following specification: i) Pass-band magnitude constant to 0.7 dB below the frequency of

0.15 π

ii) Stop-band attenuation at least 14 dB for the frequencies between

0.6 π

π

What is different in this one

Impulse invariance, not the bilinear transformation. There is therefore no prewarping: the edges scale linearly, $Ω = ω / T$ , and the mapping is applied to each partial fraction term separately rather than to $H (s)$ as a whole.

Impulse invariance maps frequency linearly, $Ω = ω / T$ , so there is no prewarping to do.

\begin{aligned} Ω_{p} & = ω_{p} / T = 0.4712 rad / s \\ Ω_{s} & = ω_{s} / T = 1.885 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.4} - 1) / (10^{0.07} - 1)] / (2 \log_{10} (4)) \\ = \log_{10} (24.119 / 0.1749) / (2 log 104) \\ = 1.7769 so  take N = 2 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.4712 / {(0.1749)}^{1 / 4} \\ = 0.7287 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 1.885 / {(24.119)}^{1 / 4} \\ = 0.8506 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.7287$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 2 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 1 \\ s 0 & = - 0.5153 + j 0.5153 \\ s 1 & = - 0.5153 - j 0.5153 \end{aligned}

H_{a} (s) = \frac{{(0.7287)}^{2}}{(s^{2} + 1.0305 s + 0.531)}

Expand $H_{a} (s)$ in partial fractions and map each pole on its own with s = pk going to z = $e^{p k T}$ , which is what impulse invariance does.

\begin{aligned} residue - j 0.5153 at  pole - 0.5153 + j 0.5153 \\ \to - j 0.5153 / (1 - (0.5198 + j 0.2943) z^{- 1}) \\ residue j 0.5153 at  pole - 0.5153 - j 0.5153 \\ \to j 0.5153 / (1 - (0.5198 - j 0.2943) z^{- 1}) \end{aligned}

H (z) = \frac{0.3033 z^{- 1}}{1 - 1.0396 z^{- 1} + 0.3568 z^{- 2}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	0.9561	the DC gain
$ω = ω_{p} = 0.4712$	0.8986	0.93 dB, against the 0.7 dB allowed
$ω = ω_{s} = 1.885$	0.1918	14.35 dB, against the 14 dB required
$ω = π$	0.1266	at Nyquist

$N = 2$ and $Ω_{c} = 0.7287$ rad/s. The specification is met at the passband edge and exceeded at the stopband edge, as rounding the order up guarantees.

H (z) = \frac{0.3033 z^{- 1}}{1 - 1.0396 z^{- 1} + 0.3568 z^{- 2}}

ω_{p}

) =

0.27 π

radians, Stop bandedge frequency (

ω_{s}

) =

0.58 π

radians, Passband ripple (deltap) = 0.11, Stopband ripple (deltas) = 0.21, Consider sampling frequency 0.5 Hz.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.
$T = 2$ here, not 1, so the factor 2/T in the prewarping is 1. It cancels in the end, but only if it is carried consistently.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 2) \tan (0.8482 / 2) = 0.4515 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 2) \tan (1.8221 / 2) = 1.2892 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.356} - 1) / (10^{0.101} - 1)] / (2 \log_{10} (2.8552)) \\ = \log_{10} (21.676 / 0.2625) / (2 log 102.8552) \\ = 2.1035 so  take N = 3 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.4515 / {(0.2625)}^{1 / 6} \\ = 0.5643 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 1.2892 / {(21.676)}^{1 / 6} \\ = 0.7721 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.5643$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 3 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 2 \\ s 0 & = - 0.2821 + j 0.4887 \\ s 1 & = - 0.5643 \\ s 2 & = - 0.2821 - j 0.4887 \end{aligned}

H_{a} (s) = \frac{{(0.5643)}^{3}}{(s^{2} + 0.5643 s + 0.3184) (s + 0.5643)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 1 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.061 + 0.183 z^{- 1} + 0.183 z^{- 2} + 0.061 z^{- 3}}{1 - 1.0026 z^{- 1} + 0.6022 z^{- 2} - 0.1116 z^{- 3}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.8482$	0.89	1.01 dB, against the 1.01 dB allowed
$ω = ω_{s} = 1.8221$	0.0836	21.56 dB, against the 13.56 dB required
$ω = π$	0	at Nyquist

$N = 3$ and $Ω_{c} = 0.5643$ rad/s. The specification is met at both edges.

H (z) = \frac{0.061 + 0.183 z^{- 1} + 0.183 z^{- 2} + 0.061 z^{- 3}}{1 - 1.0026 z^{- 1} + 0.6022 z^{- 2} - 0.1116 z^{- 3}}

Same as 2082 Baishakh Q82073 Shrawan · Q812 marksUsing bilinear transformation, design a butterworth low pass filter which satisfies the following Magnitude Response:

0.89125 \leq | H (e^{j ω}) | \leq 1

for

0 \leq w \leq 0.2 π

;

| H (e^{j ω}) | \leq 0.17783

for

0.3 π \leq ω \leq π

How this is read0.89125 and 0.17783 are the familiar 1 dB and 15 dB written as ratios: -20 $\log_{10} (0.89125) = 1$ dB and -20 $\log_{10} (0.17783) = 15$ dB.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 1) \tan (0.6283 / 2) = 0.6498 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 1) \tan (0.9425 / 2) = 1.0191 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.5} - 1) / (10^{0.1} - 1)] / (2 \log_{10} (1.5682)) \\ = \log_{10} (30.622 / 0.2589) / (2 log 101.5682) \\ = 5.3044 so  take N = 6 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.6498 / {(0.2589)}^{1 / 12} \\ = 0.7273 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 1.0191 / {(30.622)}^{1 / 12} \\ = 0.7662 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.7273$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 6 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 5 \\ s 0 & = - 0.1882 + j 0.7025 \\ s 1 & = - 0.5143 + j 0.5143 \\ s 2 & = - 0.7025 + j 0.1882 \\ s 3 & = - 0.7025 - j 0.1882 \\ s 4 & = - 0.5143 - j 0.5143 \\ s 5 & = - 0.1882 - j 0.7025 \end{aligned}

\begin{aligned} H_{a} (s) & = \frac{{(0.7273)}^{6}}{D (s)} \\ D (s) & = (s^{2} + 0.3765 s + 0.529) \\ \times (s^{2} + 1.0285 s + 0.529) \\ \times (s^{2} + 1.405 s + 0.529) \end{aligned}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 2 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.0006 + 0.0035 z^{- 1} + 0.0087 z^{- 2} + 0.0116 z^{- 3} + 0.0087 z^{- 4} + 0.0035 z^{- 5} + 0.0006 z^{- 6}}{1 - 3.3143 z^{- 1} + 4.9501 z^{- 2} - 4.1433 z^{- 3} + 2.0275 z^{- 4} - 0.5458 z^{- 5} + 0.0628 z^{- 6}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.6283$	0.8913	1 dB, against the 1 dB allowed
$ω = ω_{s} = 0.9425$	0.131	17.65 dB, against the 15 dB required
$ω = π$	0	at Nyquist

$N = 6$ and $Ω_{c} = 0.7273$ rad/s. The specification is met at both edges.

H (z) = \frac{0.0006 + 0.0035 z^{- 1} + 0.0087 z^{- 2} + 0.0116 z^{- 3} + 0.0087 z^{- 4} + 0.0035 z^{- 5} + 0.0006 z^{- 6}}{1 - 3.3143 z^{- 1} + 4.9501 z^{- 2} - 4.1433 z^{- 3} + 2.0275 z^{- 4} - 0.5458 z^{- 5} + 0.0628 z^{- 6}}

Same as 2082 Baishakh Q82072 Kartik · Q915 marksDesign a low pass digital filter by Bilinear Transformation method to an approximate Butterworth filter, if passband edge frequency is

0.25 π

radians and maximum deviation of 1 dB below 0 dB gain in the passband. The maximum gain of -15 dB and frequency is

0.45 π

radians in stopband, Consider sampling frequency 1Hz.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 1) \tan (0.7854 / 2) = 0.8284 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 1) \tan (1.4137 / 2) = 1.7082 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.5} - 1) / (10^{0.1} - 1)] / (2 \log_{10} (2.0619)) \\ = \log_{10} (30.623 / 0.2589) / (2 log 102.0619) \\ = 3.2979 so  take N = 4 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.8284 / {(0.2589)}^{1 / 8} \\ = 0.9809 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 1.7082 / {(30.623)}^{1 / 8} \\ = 1.1137 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.9809$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 4 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 3 \\ s 0 & = - 0.3754 + j 0.9062 \\ s 1 & = - 0.9062 + j 0.3754 \\ s 2 & = - 0.9062 - j 0.3754 \\ s 3 & = - 0.3754 - j 0.9062 \end{aligned}

H_{a} (s) = \frac{{(0.9809)}^{4}}{(s^{2} + 0.7507 s + 0.9621) (s^{2} + 1.8124 s + 0.9621)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 2 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.0167 + 0.0667 z^{- 1} + 0.1001 z^{- 2} + 0.0667 z^{- 3} + 0.0167 z^{- 4}}{1 - 1.6476 z^{- 1} + 1.3563 z^{- 2} - 0.5252 z^{- 3} + 0.0834 z^{- 4}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.7854$	0.8913	1 dB, against the 1 dB allowed
$ω = ω_{s} = 1.4137$	0.1081	19.32 dB, against the 15 dB required
$ω = π$	0	at Nyquist

$N = 4$ and $Ω_{c} = 0.9809$ rad/s. The specification is met at both edges.

H (z) = \frac{0.0167 + 0.0667 z^{- 1} + 0.1001 z^{- 2} + 0.0667 z^{- 3} + 0.0167 z^{- 4}}{1 - 1.6476 z^{- 1} + 1.3563 z^{- 2} - 0.5252 z^{- 3} + 0.0834 z^{- 4}}

Same as 2082 Baishakh Q82071 Shrawan · Q815 marksDesign a low pass digital filter by Bilinear Transformation method to an approximate Butterworth filter it passband frequency is

0.2 π

radians and maximum deviation of 1 db below 0 dB gain in the pass band. The maximum gain of -15 db and frequency is

0.4 π

radians in stop band, consider sampling frequency 1 Hz.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 1) \tan (0.6283 / 2) = 0.6498 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 1) \tan (1.2566 / 2) = 1.4531 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.5} - 1) / (10^{0.1} - 1)] / (2 \log_{10} (2.2361)) \\ = \log_{10} (30.623 / 0.2589) / (2 log 102.2361) \\ = 2.9656 so  take N = 3 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.6498 / {(0.2589)}^{1 / 6} \\ = 0.814 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 1.4531 / {(30.623)}^{1 / 6} \\ = 0.8215 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.814$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 3 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 2 \\ s 0 & = - 0.407 + j 0.7049 \\ s 1 & = - 0.814 \\ s 2 & = - 0.407 - j 0.7049 \end{aligned}

H_{a} (s) = \frac{{(0.814)}^{3}}{(s^{2} + 0.814 s + 0.6626) (s + 0.814)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 2 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.0305 + 0.0914 z^{- 1} + 0.0914 z^{- 2} + 0.0305 z^{- 3}}{1 - 1.4826 z^{- 1} + 0.9296 z^{- 2} - 0.2033 z^{- 3}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.6283$	0.8913	1 dB, against the 1 dB allowed
$ω = ω_{s} = 1.2566$	0.1731	15.23 dB, against the 15 dB required
$ω = π$	0	at Nyquist

$N = 3$ and $Ω_{c} = 0.814$ rad/s. The specification is met at both edges.

H (z) = \frac{0.0305 + 0.0914 z^{- 1} + 0.0914 z^{- 2} + 0.0305 z^{- 3}}{1 - 1.4826 z^{- 1} + 0.9296 z^{- 2} - 0.2033 z^{- 3}}

The same questions grouped by how often each topic is set: butterworth iir design

6.4Chebyshev, elliptic and Bessel filters, and spectral transformation

Chebyshev, elliptic and Bessel: the other approximations

Before the formulas

Butterworth is not the only choice, and the syllabus wants the comparison even though it sets the numbers on Butterworth. There is one pattern here said four ways: allow ripple somewhere and you need fewer poles for the same sharpness.

Chebyshev puts the ripple in the passband. The inverse type puts it in the stopband. The elliptic allows both and is the sharpest of the lot. Bessel gives up sharpness altogether to keep the phase linear. Knowing the trade is enough. The paper asks for properties and a comparison, not for elliptic design.

Butterworth is the one the paper sets, but the syllabus asks for the properties of the other three, and the comparison is worth knowing because it is the same trade every time: allow ripple, and you need fewer poles.

	Passband	Stopband	Roll off	Phase and delay	Order for the same job
Butterworth	Maximally flat, monotonic	Monotonic	Slowest, 20N dB per decade	Best behaved of the three ripple types	Highest
Chebyshev I	Equiripple	Monotonic	Faster	Worse near the edge	Lower
Chebyshev II	Monotonic, flat	Equiripple	Faster	Better than type I	Lower
Elliptic, Cauer	Equiripple	Equiripple	Fastest	Worst, very non linear near the edge	Lowest
Bessel, Thomson	Monotonic	Monotonic	Slowest of all	Maximally flat GROUP DELAY, nearly linear phase	Highest

\begin{aligned} | H_{a} (j Ω) |^{2} & = \frac{1}{1 + ε^{2} T_{N}^{2} (Ω / Ω_{p})} & Chebyshev type I \\ T_{N} (x) & = \cos (N \arccos x) & | x | \leq 1, ripples between \pm 1 \\ T_{N} (x) & = \cosh (N arccosh x) & | x | > 1, grows fast \\ N & \geq \frac{arccosh \sqrt{(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)}}{arccosh (Ω_{s} / Ω_{p})} \end{aligned}

Where the ripple count goes. A Chebyshev type I filter of order N has N ripples in the passband, and the response starts at 1 for odd N and at 1/sqrt(1 + $e p s^{2}$ ) for even N. That detail is a favourite short question.
The elliptic filter uses Jacobian elliptic functions in place of the Chebyshev polynomial and ripples in both bands. For a given specification it always has the lowest order of any of them, which is why it is used wherever phase does not matter.
The Bessel filter approximates a constant group delay rather than a flat magnitude, so a pulse passes through with its shape intact. It has the poorest selectivity, and note that the bilinear transformation destroys its one virtue, because warping bends the delay, so a digital Bessel filter is designed by matching the delay directly rather than by transforming.
The choice, in one line: flat passband and gentle phase, Butterworth; sharper for the same order and ripple allowed in one band, Chebyshev; the sharpest possible and phase irrelevant, elliptic; pulse shape must survive, Bessel.

The trade here, ripple against order, is the same one the optimum FIR design makes: allow the error to oscillate and you need fewer coefficients.

See it drawnAllow ripple somewhere and you need fewer poles

That is the whole comparison, said four ways. Where you let the ripple sit is the only real choice, and the elliptic filter, which allows it in both bands, is the sharpest of the four.

Digital domain spectral transformation PIN 2/19

80 Ba · 79 Bh11+44

Before the formulas

Every method in the chapter so far designs a low pass filter. You are rarely asked for a low pass filter. The spectral transformation is the last step that converts the low pass prototype into the high pass, band pass or band stop filter you actually wanted, by substituting for z.

The prototype does the work; the substitution moves it where it is needed. The low pass to high pass case is the one the paper sets, and it is almost embarrassingly short once you have the substitution written down.

What it is for Once a digital low pass prototype exists with cutoff

θ_{p}

, another filter of a different type or a different cutoff can be obtained by substituting an all pass function for

z^{- 1}

. There is no need to go back to the analog prototype, and because the substitution is all pass it maps the unit circle onto itself, so stability and the magnitude values are preserved.

To get	Replace $z^{- 1}$ by	Parameter
Low pass, new cutoff $ω_{p}$	( $z^{- 1} - a) / (1 - a z^{- 1}$ )	a = sin(( $θ_{p} - ω_{p}$ )/2) / sin(( $θ_{p} + ω_{p}$ )/2)
High pass, cutoff $ω_{p}$	-( $z^{- 1} + a) / (1 + a z^{- 1}$ )	a = -cos(( $θ_{p} + ω_{p}$ )/2) / cos(( $θ_{p} - ω_{p}$ )/2)
Band pass, wl to wu	-( $z^{- 2} - a_{1} z^{- 1} + a_{2}$ )/( $a_{2} z^{- 2} - a_{1} z^{- 1}$ + 1)	From the two band edges and $θ_{p}$
Band stop, wl to wu	( $z^{- 2} - a_{1} z^{- 1} + a_{2}$ )/( $a_{2} z^{- 2} - a_{1} z^{- 1}$ + 1)	From the two band edges and $θ_{p}$

Features and parameters, for the low pass to high pass case, which is the one asked

One parameter, a, computed from the prototype cutoff $θ_{p}$ and the required high pass cutoff $ω_{p}$ .
The order does not change. A second order low pass gives a second order high pass.
The minus sign is what flips the band. It rotates the unit circle by $π$ , taking $ω = 0$ to $ω = π$ , so the passband moves from around DC to around Nyquist.
Stability survives because the substitution is all pass and maps the inside of the unit circle to itself.
The magnitude values survive exactly; only the frequencies at which they occur move, in the same spirit as warping.

Worked, the 2079 Bhadra rider: a digital low pass has been designed with $θ_{p} = 0.25 π$ and a high pass with $ω$ 'p = $0.45 π$ is wanted. a = $- \cos ((0.25 π + 0.45 π) / 2) / \cos ((0.25 π - 0.45 π) / 2) = - \cos (0.35 π) / \cos (- 0.1 π) = \frac{- 0.4540}{0}$ .9511 = -0.4774. Substitute $z^{- 1}$ by -( $z^{- 1} - 0.4774) / (1 - 0.4774 z^{- 1}$ ) in the low pass $H (z)$ .

Because the substitution is all pass, it moves the band edges without touching the ripple, so one Butterworth prototype serves every band type.

See it drawnOne substitution turns the low pass prototype into what you wanted

Every method in the chapter designs a low pass filter, and you are rarely asked for one. The transformation folds the response about the middle of the band, and it is one line of algebra.

Practise thisEvery question the paper has set on this topic, worked in full1

Practise this2079 Bhadra · Q911+4 marksDesign a low pass discrete time Butterworth filter using bilinear transformation having following specifications: Passband frequency (

Ω_{p}

) =

0.25 π

radians, Stopband frequency (

Ω_{s}

) =

0.55 π

0.45 π

using digital domain transformation.

The design

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.
$T = 2$ here, not 1, so the factor 2/T in the prewarping is 1. It cancels in the end, but only if it is carried consistently.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 2) \tan (0.7854 / 2) = 0.4142 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 2) \tan (1.7279 / 2) = 1.1708 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.356} - 1) / (10^{0.101} - 1)] / (2 \log_{10} (2.8267)) \\ = \log_{10} (21.676 / 0.2625) / (2 log 102.8267) \\ = 2.1239 so  take N = 3 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.4142 / {(0.2625)}^{1 / 6} \\ = 0.5177 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 1.1708 / {(21.676)}^{1 / 6} \\ = 0.7012 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.5177$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 3 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 2 \\ s 0 & = - 0.2588 + j 0.4483 \\ s 1 & = - 0.5177 \\ s 2 & = - 0.2588 - j 0.4483 \end{aligned}

H_{a} (s) = \frac{{(0.5177)}^{3}}{(s^{2} + 0.5177 s + 0.268) (s + 0.5177)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 1 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.0512 + 0.1536 z^{- 1} + 0.1536 z^{- 2} + 0.0512 z^{- 3}}{1 - 1.1377 z^{- 1} + 0.6808 z^{- 2} - 0.1335 z^{- 3}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.7854$	0.89	1.01 dB, against the 1.01 dB allowed
$ω = ω_{s} = 1.7279$	0.0861	21.3 dB, against the 13.56 dB required
$ω = π$	0	at Nyquist

$N = 3$ and $Ω_{c} = 0.5177$ rad/s. The specification is met at both edges.

H (z) = \frac{0.0512 + 0.1536 z^{- 1} + 0.1536 z^{- 2} + 0.0512 z^{- 3}}{1 - 1.1377 z^{- 1} + 0.6808 z^{- 2} - 0.1335 z^{- 3}}

The rest of this questionThis topic works the part that belongs to it. The other part is worked under Converting it to a high pass, the last four marks, where it belongs.

6.5Last minute recall

Last minute recall, chapter 6

Must memorise

Prewarp: $Ω_{p}$ = (2/T) tan( $ω_{p}$ /2). Impulse invariance instead scales linearly, $Ω = ω / T$ .
$N \geq l o g 10 [$ ( $10^{0.1 A s} - 1) / (10^{0.1 A p}$ - 1)] / (2 log10( $Ω_{s} / Ω_{p}$ )), rounded up.
$Ω_{c} = Ω_{p}$ /( $10^{0.1 A p}$ - 1)^(1/2N) is exact at the passband edge; the stopband version is exact at the other.
Butterworth poles lie on a circle of radius $Ω_{c}$ , evenly spaced, left half plane only: $s_{k} = Ω_{c}$ exp(j $π$ (2k + $N + 1$ )/2N).
s = (2/T)(1 $- z^{- 1}$ )/(1 + $z^{- 1}$ ), and it maps the whole left half plane inside the unit circle, so stability is automatic.
W = (2/T) $\tan (ω / 2)$ is the warping; prewarping fixes the band edges exactly and cannot straighten the curve between them.
Impulse invariance aliases and suits low pass and band pass only; the bilinear transformation does not alias and suits everything.
Ripple to decibels: $| H | \geq 0.9$ is Ap = 0.9151 dB; $| H | \leq 0.2$ is As = 13.98 dB; 0.89125 is 1 dB and 0.17783 is 15 dB.
Chebyshev ripples in one band, elliptic in both, Bessel keeps the group delay flat, and the order falls in that order.

Most repeated in this chapter, in order

Design a Butterworth digital filter TOP 18/19, the most set question in the whole subject
By the bilinear transformation TOP 13/19, which is the route in almost every one of them
Bilinear against impulse invariance HOT 5/19
Impulse invariance itself PIN 3/19 and spectral transformation PIN 2/19

Eighteen of nineteen sittings set this design, usually for fifteen marks or close to it. The steps never change, only the numbers, so the whole chapter is one procedure carried out carefully.

Chapter 7 · 7 hours · 15 marks · the last two or three questions, every paper

The discrete Fourier transform

The only Fourier transform a computer can actually compute, the fast algorithm that makes it practical, and the circular convolution that the DFT quietly performs instead of the linear one you wanted. Two or three questions in every paper, worth fifteen marks between them.

What this chapter is about

The DFT: what it is, how it relates to the DTFT, and why it has to exist.
Circular convolution: what the DFT gives you when you multiply two spectra, how it differs from linear convolution, and how zero padding reconciles them.
The FFT: where the saving comes from, and the two ways of splitting the sum.
The butterfly diagrams for DIT and DIF, which is what the exam actually wants drawn.

Where it fits

Chapter 3's frequency response is continuous in $ω$ ; the DFT samples it at N points.
Chapter 1's convolution reappears here in circular form.
Chapter 5's window method is the other side of the same sampling question.

What you will learn

7.1 The DFT and the IDFT, DFT against DTFT, The properties, Circular convolution, and how it differs from linear, Zero padding
7.2 Why the FFT is fast, Decimation in time, Decimation in frequency
7.3 The computational complexity of the FFT
7.4 Last minute recall, chapter 7

How it is examined

An 8 point FFT, DIT or DIF, 6 to 8 marks, with the butterfly diagram drawn.
A circular convolution, 5 to 7 marks, sometimes disguised as a product of two DFTs.
A short bookwork rider: why we need the DFT, how fast the FFT is, DFT against DTFT, or zero padding.

7.1The DFT, its properties, and circular convolution

The DFT and the IDFT HOT 5/19

79 Ba · 76 Ash · 74 Ch · 73 Shr · 72 Ka2+62+83+7

Before the formulas

Chapter 1 gave you a transform whose answer is a continuous curve. No computer can hold a continuous curve. The DFT is the version a machine can actually compute: N numbers in, N numbers out, both finite, nothing continuous anywhere.

That is the entire reason it exists, and it is the transform every piece of real software uses. The formulas will look familiar from the Fourier series in 1.5, and that is not a coincidence: it is the same finite sum, which is also the first hint of the one assumption that makes the DFT behave oddly later.

The N point DFT Takes a finite sequence of N samples and produces N frequency samples. Both the sequence and its transform are treated as one period of a periodic signal, which is where every peculiarity of the DFT comes from.

\begin{aligned} X (k) & = \sum_{n = 0}^{N - 1} x [n] W_{N}^{k n}, k = 0 \dots N - 1, W_{N} = e^{- j 2 π / N} \\ x [n] & = \frac{1}{N} \sum_{k = 0}^{N - 1} X (k) W_{N}^{- k n}, n = 0 \dots N - 1 \\ W_{N}^{N} & = 1, W_{N}^{k + N / 2} = - W_{N}^{k}, W_{N}^{2} = W_{N / 2} \end{aligned}

Why we need the DFT, which is asked as a two mark rider in eight papers

The DTFT is not computable. $X (e^{j ω})$ is a function of a continuous variable, so it has infinitely many values. A computer cannot store or compute it.
The DFT is finite in both domains: N samples in, N samples out, all of them numbers a machine can hold.
It is the sampled DTFT. X(k) = $X (e^{j ω})$ evaluated at $ω = 2 π$ k / N, so it is the DTFT read at N equally spaced points around the unit circle.
It is invertible, exactly, from those N samples, provided the sequence really has only N non zero samples. No information is lost.
It has a fast algorithm. The FFT makes spectrum analysis, fast convolution and filtering practical, and that is the reason the DFT is the transform used in practice.

Computing one by hand For a small N, write the twiddle factors first. For

N = 4

: W = -j, so

W^{0} = 1

W^{1}

= -j,

W^{2} = - 1

W^{3} = j

. For

N = 8

W^{0} = 1

W^{1} = 0.707

- j0.707,

W^{2}

= -j,

W^{3} = - 0.707

- j0.707, and

W^{k + 4} = - W^{k}

.
Then X(k) = sum

x [n] W^{n k}

. The whole thing is a matrix multiplication by a symmetric matrix of twiddle factors.

This is the DTFT sampled at N points, which is why the time sequence it implies is periodic, and therefore why its convolution is circular.

See it drawnN numbers in, N numbers out, and nothing continuous anywhere

This is the transform a machine can actually carry out. The price of making both sides finite is that the sequence behaves as though it repeats, which is where every surprise later in the chapter comes from.

Practise thisEvery question the paper has set on this topic, worked in full5

Same as 2082 Bhadra Q112079 Baishakh · Q102+6 marksWhy we need DFT? Find 8-point DFT of sequence

x [n] = {1, 2, 4, 3, 5, - 1, 3}

using Decimation in Frequency Fast Fourier Transform (DIFFFT) algorithm.

What is different in this one

Decimation in frequency: 8 points, 3 stages of 4 butterflies, with the output in bit reversed order.
The sequence has 7 samples, not a power of two, so it is zero padded to $N = 8$ before the algorithm starts. Say so in the answer: a radix 2 FFT cannot run on 7 points.

The sequence has 7 samples, and a radix 2 FFT needs a power of two, so zero pad to $N = 8$ and say so: $x [n] = {1, 2, 4, 3, 5, - 1, 3, 0}$ .

Decimation in frequency: the input stays in natural order and the output comes out bit reversed, so the last step is to relabel it.

Stage 1, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ applied to the differences:

\begin{aligned} 6 \\ 1 \\ 7 \\ 3 \\ - 4 \\ 2.1213 - j 2.1213 \\ - j 1 \\ - 2.1213 - j 2.1213 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ applied to the differences:

\begin{aligned} 13 \\ 4 \\ - 1 \\ j2 \\ - 4 - j 1 \\ - j 4.2426 \\ - 4 + j 1 \\ - j 4.2426 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{2}^{0}$ applied to the differences:

\begin{aligned} 17 \\ 9 \\ - 1 + j 2 \\ - 1 - j 2 \\ - 4 - j 5.2426 \\ - 4 + j 3.2426 \\ - 4 - j 3.2426 \\ - 4 + j 5.2426 \end{aligned}

Read the output in bit reversed order: $X [0]$ , $X [4]$ , $X [2]$ , $X [6]$ , $X [1]$ , $X [5]$ , $X [3]$ , $X [7]$ .

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	17	17	0
$X [1]$	-4 - j5.2426	6.5943	-127.34
$X [2]$	-1 + j2	2.2361	116.57
$X [3]$	-4 - j3.2426	5.1492	-140.97
$X [4]$	9	9	0
$X [5]$	-4 + j3.2426	5.1492	140.97
$X [6]$	-1 - j2	2.2361	-116.57
$X [7]$	-4 + j5.2426	6.5943	127.34

$X [k]$ = {17, -4 - j5.2426, -1 + j2, -4 - j3.2426, 9, -4 + j3.2426, -1 - j2, -4 + j5.2426}. Two free checks: $X [0] = 17$ , which is the sum of the samples; $X [7] = X^{*} [1]$ and $X [6] = X^{*} [2]$ , the conjugate symmetry a real input forces.

Same as 2082 Bhadra Q112076 Ashwin · Q82+8 marksWhy we need DFT? Find 8-point DFT of sequence

x [n] = {1, 2, 3, 3, 5, 1, 4, 2}

using Decimation in frequency Fast Fourier Transform (DIFFFT) algorithm.

What is different in this one

Decimation in frequency: 8 points, 3 stages of 4 butterflies, with the output in bit reversed order.

$N = 8$ , which is $2^{3}$ , so the transform takes 3 stages of 4 butterflies.

Decimation in frequency: the input stays in natural order and the output comes out bit reversed, so the last step is to relabel it.

Stage 1, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ applied to the differences:

\begin{aligned} 6 \\ 3 \\ 7 \\ 5 \\ - 4 \\ 0.7071 - j 0.7071 \\ j1 \\ - 0.7071 - j 0.7071 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ applied to the differences:

\begin{aligned} 13 \\ 8 \\ - 1 \\ j2 \\ - 4 + j 1 \\ - j 1.4142 \\ - 4 - j 1 \\ - j 1.4142 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{2}^{0}$ applied to the differences:

\begin{aligned} 21 \\ 5 \\ - 1 + j 2 \\ - 1 - j 2 \\ - 4 - j 0.4142 \\ - 4 + j 2.4142 \\ - 4 - j 2.4142 \\ - 4 + j 0.4142 \end{aligned}

Read the output in bit reversed order: $X [0]$ , $X [4]$ , $X [2]$ , $X [6]$ , $X [1]$ , $X [5]$ , $X [3]$ , $X [7]$ .

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	21	21	0
$X [1]$	-4 - j0.4142	4.0214	-174.09
$X [2]$	-1 + j2	2.2361	116.57
$X [3]$	-4 - j2.4142	4.6721	-148.89
$X [4]$	5	5	0
$X [5]$	-4 + j2.4142	4.6721	148.89
$X [6]$	-1 - j2	2.2361	-116.57
$X [7]$	-4 + j0.4142	4.0214	174.09

$X [k]$ = {21, -4 - j0.4142, -1 + j2, -4 - j2.4142, 5, -4 + j2.4142, -1 - j2, -4 + j0.4142}. Two free checks: $X [0] = 21$ , which is the sum of the samples; $X [7] = X^{*} [1]$ and $X [6] = X^{*} [2]$ , the conjugate symmetry a real input forces.

Practise this2074 Chaitra · Q93+7 marksWhy do we need DFT? Draw the butterfly structure to compute the DFT of the following signal using Radix-2 DIFFFT algorithm, and compute X(2) and X(1) only

x [n] = {1.5, - 1, 1.8, 0.6, 3, 1.7}

.What is different heredecimation in frequency, where the input stays in order and the output comes out bit reversed. The question asks only for X(2) and X(1), but the butterfly diagram has to be drawn in full to reach them, so the whole transform is here.

What is different in this one

Decimation in frequency: 8 points, 3 stages of 4 butterflies, with the output in bit reversed order.
The sequence has 6 samples, not a power of two, so it is zero padded to $N = 8$ before the algorithm starts. Say so in the answer: a radix 2 FFT cannot run on 6 points.
The question asks only for X(2) and X(1), but the butterfly diagram has to be drawn in full to reach them, so the whole transform is here.

The sequence has 6 samples, and a radix 2 FFT needs a power of two, so zero pad to $N = 8$ and say so: $x [n] = {1.5, - 1, 1.8, 0.6, 3, 1.7, 0, 0}$ .

Decimation in frequency: the input stays in natural order and the output comes out bit reversed, so the last step is to relabel it.

Stage 1, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ applied to the differences:

\begin{aligned} 4.5 \\ 0.7 \\ 1.8 \\ 0.6 \\ - 1.5 \\ - 1.9092 + j 1.9092 \\ - j 1.8 \\ - 0.4243 - j 0.4243 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ applied to the differences:

\begin{aligned} 6.3 \\ 1.3 \\ 2.7 \\ - j 0.1 \\ - 1.5 - j 1.8 \\ - 2.3335 + j 1.4849 \\ - 1.5 + j 1.8 \\ 2.3335 + j 1.4849 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{2}^{0}$ applied to the differences:

\begin{aligned} 7.6 \\ 5 \\ 2.7 - j 0.1 \\ 2.7 + j 0.1 \\ - 3.8335 - j 0.3151 \\ 0.8335 - j 3.2849 \\ 0.8335 + j 3.2849 \\ - 3.8335 + j 0.3151 \end{aligned}

Read the output in bit reversed order: $X [0]$ , $X [4]$ , $X [2]$ , $X [6]$ , $X [1]$ , $X [5]$ , $X [3]$ , $X [7]$ .

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	7.6	7.6	0
$X [1]$	-3.8335 - j0.3151	3.8464	-175.3
$X [2]$	2.7 - j0.1	2.7019	-2.12
$X [3]$	0.8335 + j3.2849	3.389	75.76
$X [4]$	5	5	0
$X [5]$	0.8335 - j3.2849	3.389	-75.76
$X [6]$	2.7 + j0.1	2.7019	2.12
$X [7]$	-3.8335 + j0.3151	3.8464	175.3

$X [k]$ = {7.6, -3.8335 - j0.3151, 2.7 - j0.1, 0.8335 + j3.2849, 5, 0.8335 - j3.2849, 2.7 + j0.1, -3.8335 + j0.3151}. Two free checks: $X [0] = 7.6$ , which is the sum of the samples; $X [7] = X^{*} [1]$ and $X [6] = X^{*} [2]$ , the conjugate symmetry a real input forces.

Same as 2081 Bhadra Q92073 Shrawan · Q102+6 marksWhy do we need DFT? Find 8-point DFT of sequence

x [n] = {1, - 1, 2, 2, 1, 1, 2, 2}

using Fast Fourier Transform algorithm.

What is different in this one

Decimation in time: 8 points, 3 stages of 4 butterflies, with the input in bit reversed order.

$N = 8$ , which is $2^{3}$ , so the transform takes 3 stages of 4 butterflies.

Decimation in time: the input goes in bit reversed order and the output comes out in natural order. For $N = 8$ that input order is $x [0]$ , $x [4]$ , $x [2]$ , $x [6]$ , $x [1]$ , $x [5]$ , $x [3]$ , $x [7]$ .

Stage 1, 4 butterflies with twiddles $W_{2}^{0}$ :

\begin{aligned} 2 \\ 0 \\ 4 \\ 0 \\ 0 \\ - 2 \\ 4 \\ 0 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ :

\begin{aligned} 6 \\ 0 \\ - 2 \\ 0 \\ 4 \\ - 2 \\ - 4 \\ - 2 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ :

\begin{aligned} 10 \\ - 1.4142 + j 1.4142 \\ - 2 + j 4 \\ 1.4142 + j 1.4142 \\ 2 \\ 1.4142 - j 1.4142 \\ - 2 - j 4 \\ - 1.4142 - j 1.4142 \end{aligned}

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	10	10	0
$X [1]$	-1.4142 + j1.4142	2	135
$X [2]$	-2 + j4	4.4721	116.57
$X [3]$	1.4142 + j1.4142	2	45
$X [4]$	2	2	0
$X [5]$	1.4142 - j1.4142	2	-45
$X [6]$	-2 - j4	4.4721	-116.57
$X [7]$	-1.4142 - j1.4142	2	-135

$X [k]$ = {10, -1.4142 + j1.4142, -2 + j4, 1.4142 + j1.4142, 2, 1.4142 - j1.4142, -2 - j4, -1.4142 - j1.4142}. Two free checks: $X [0] = 10$ , which is the sum of the samples; $X [7] = X^{*} [1]$ and $X [6] = X^{*} [2]$ , the conjugate symmetry a real input forces.

Same as 2082 Bhadra Q102072 Kartik · Q112+6 marksWhy we need DFT? If X1(k) and X2(k) are DFT of sequence

x_{1} [n] = {1, 2, 4}

and

x_{2} [n] = {- 1, 2, 3, 1}

respectively, then find the sequence

x_{3} [n]

, if DFT of

x_{3} [n]

is given by X3(k) = X1(k) X2(k).

What is different in this one

The two sequences are different lengths, 3 and 4, so the shorter is zero padded to $N = 4$ before anything else happens. Getting N wrong makes every sample wrong.
$N = 4$ is smaller than $N_{1} + N_{2} - 1 = 6$ , so this circular answer is not the linear convolution: the last 2 samples have wrapped round and added to the front. That wrap is the whole point of the question.

The DFT length is $N = 4$ , the longer of the two. Zero pad to that length, since $x_{1} [n]$ has 3 samples. The circular convolution is $y [n] = \sum_{m} x_{1} [m] x_{2}$ [(n - m) mod N], and the wrap is what makes it circular rather than linear.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=1	$x_{1}$ [1]=2	$x_{1}$ [2]=4	$x_{1}$ [3]=0	sum
$y [0]$	-1	1	3	2	13
$y [1]$	2	-1	1	3	4
$y [2]$	3	2	-1	1	3
$y [3]$	1	3	2	-1	15

y [0]

= 1(-1) + 2(1) + 4(3) = 13

y [1]

= 1(2) + 2(-1) + 4(1) = 4

y [2]

= 1(3) + 2(2) + 4(-1) = 3

y [3]

= 1(1) + 2(3) + 4(2) = 15

$y [n] = {13, 4, 3, 15}$ . The check holds: 7 × 5 = 35, the sum of y.

DFT against DTFT PIN 1/19

78 Bh2+6

Before the formulas

Worth being precise about, because the paper asks for it directly and because everything else in the chapter follows from it. The DFT is not a different transform. It is the transform of chapter 1 evaluated at N evenly spaced frequencies and nowhere else.

Sampling in frequency has a price, and it is the exact mirror of sampling in time from 1.10: the sequence behaves as though it repeats every N samples. Every strange thing the DFT does, circular shifts, circular convolution, leakage, is that one assumption showing itself.

	DTFT	DFT
Input	An infinite sequence	A finite sequence of N samples
Output	A continuous function of $ω$	N discrete samples
Formula	$X (e^{j ω}) = \sum_{n = - \infty}^{\infty} x [n] e^{- j ω n}$	X(k) = sum over $n = 0$ ..N-1 of $x [n] W^{n k}$
Periodic in frequency	Yes, period $2 π$	Yes, period N
Computable	No, infinitely many values	Yes, and fast with the FFT
Convolution property gives	Linear convolution	Circular convolution
Relation	X(k) = $X (e^{j ω})$ sampled at $ω = 2 π$ k / N. The DFT is N samples of one period of the DTFT.

The consequence worth stating: because the DFT is a sampled DTFT, it implicitly assumes the sequence repeats every N samples. Every surprising property of the DFT, circular shifts, circular convolution and spectral leakage, is that assumption showing through.

The one consequence worth carrying: sampling in frequency makes the signal periodic in time, which is the whole reason for zero padding.

See it drawnThe DFT is the DTFT, read at N points and nowhere else

Between the points the DFT says nothing at all. Sampling the spectrum this way is exactly what forces the sequence to behave as though it repeats every N samples.

Practise thisEvery question the paper has set on this topic, worked in full1

Same as 2081 Baishakh Q102078 Bhadra · Q102+6 marksDifferentiate between DFT and DTFT. Find the circular convolution of

x_{1} [n] = {2, 1, 2, 1}

and

x_{2} [n] = {1, 2, 3, 4}

What is different in this one

$N = 4$ is smaller than $N_{1} + N_{2} - 1 = 7$ , so this circular answer is not the linear convolution: the last 3 samples have wrapped round and added to the front. That wrap is the whole point of the question.

The DFT length is $N = 4$ . The circular convolution is $y [n] = \sum_{m} x_{1} [m] x_{2}$ [(n - m) mod N], and the wrap is what makes it circular rather than linear.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=2	$x_{1}$ [1]=1	$x_{1}$ [2]=2	$x_{1}$ [3]=1	sum
$y [0]$	1	4	3	2	14
$y [1]$	2	1	4	3	16
$y [2]$	3	2	1	4	14
$y [3]$	4	3	2	1	16

y [0]

= 2(1) + 1(4) + 2(3) + 1(2) = 14

y [1]

= 2(2) + 1(1) + 2(4) + 1(3) = 16

y [2]

= 2(3) + 1(2) + 2(1) + 1(4) = 14

y [3]

= 2(4) + 1(3) + 2(2) + 1(1) = 16

$y [n] = {14, 16, 14, 16}$ . The check holds: 6 × 10 = 60, the sum of y.

The properties PIN 1/19

75 Ash1+5

Before the formulas

The DFT's properties are the ones you already know from chapters 1 and 2, each with the word circular inserted. A shift does not fall off the end, it wraps round to the front. A fold is about zero measured round a circle.

Conjugate symmetry means the second half of the output is a mirror of the first, so half of a real sequence's spectrum is redundant and computing it is wasted work. Read the table with the wrapping in mind and none of it is surprising. Read it without, and all of it is.

Property	$x [n]$	X(k)
Linearity	a $x_{1} [n] + b x_{2} [n]$	a X1(k) + b X2(k)
Periodicity	$x [n + N] = x [n]$	X(k + N) = X(k)
Circular time shift	x[(n - m) mod N]	X(k) $W^{k m}$
Circular frequency shift	$x [n] W^{- l n}$	X[(k - l) mod N]
Circular convolution	$x_{1} [n]$ (N) $x_{2} [n]$	X1(k) X2(k)
Multiplication	$x_{1} [n] x_{2} [n]$	$\frac{1}{N}$ X1(k) (N) X2(k)
Time reversal	x[(-n) mod N]	X[(-k) mod N]
Conjugate symmetry, real x	$x [n]$ real	X(N-k) = X*(k), so the magnitude is even and the phase is odd
Parseval	sum $\| x [n] \|^{2}$	$\frac{1}{N}$ sum \|X(k)\|^2

The circular convolution property, stated The DFT of the circular convolution of two N point sequences is the product of their DFTs:
DFT{ $x_{1} [n]$ (N) $x_{2} [n]}$ = X1(k) . X2(k)
and therefore

x_{3} [n]

= IDFT{ X1(k) X2(k) } is the circular convolution, not the linear one. Whenever a question says find $x_{3} [n]$ if X3(k) = X1(k) X2(k), it is asking for a circular convolution and nothing else.

The convolution property here is the convolution sum with the shift taken modulo N, and nothing else.

See it drawnEvery property is the one you know, with the word circular in it

A shift wraps round instead of falling off the end, and for a real sequence the second half of the spectrum is a mirror of the first, so half the computation is already done.

Practise thisEvery question the paper has set on this topic, worked in full1

Same as 2081 Baishakh Q102075 Ashwin · Q91+5 marksState the circular convolution property of DFT. Find the circular convolution of: (a)

{1, 2, - 1, 1}

and

x_{2}

(n) =

{1, 3, 5, 7}

What is different in this one

$N = 4$ is smaller than $N_{1} + N_{2} - 1 = 7$ , so this circular answer is not the linear convolution: the last 3 samples have wrapped round and added to the front. That wrap is the whole point of the question.

The DFT length is $N = 4$ . The circular convolution is $y [n] = \sum_{m} x_{1} [m] x_{2}$ [(n - m) mod N], and the wrap is what makes it circular rather than linear.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=1	$x_{1}$ [1]=2	$x_{1}$ [2]=-1	$x_{1}$ [3]=1	sum
$y [0]$	1	7	5	3	13
$y [1]$	3	1	7	5	3
$y [2]$	5	3	1	7	17
$y [3]$	7	5	3	1	15

y [0]

= 1(1) + 2(7) + -1(5) + 1(3) = 13

y [1]

= 1(3) + 2(1) + -1(7) + 1(5) = 3

y [2]

= 1(5) + 2(3) + -1(1) + 1(7) = 17

y [3]

= 1(7) + 2(5) + -1(3) + 1(1) = 15

$y [n] = {13, 3, 17, 15}$ . The check holds: 3 × 16 = 48, the sum of y.

Circular convolution, and how it differs from linear TOP 17/19

82 Bh · 82 Ba · 81 Bh · 81 Ba · 80 Bh · 80 Ba · 79 Bh · 79 Ba · 78 Bh · 76 Ch · 75 Ch · 75 Ash · 74 Ash · 73 Shr · 72 Ch · 72 Ka · 71 Shr762+6

Before the formulas

The most asked idea in the chapter, and the one place the DFT will quietly give you a wrong answer if you are not careful. Multiplying two DFTs together does not give ordinary convolution. It gives the circular kind, where the tail of the answer wraps round and adds onto the front.

That is not a quirk, it is exactly what the repeat assumption in the previous card forces. You need to be able to compute it three ways, and to say precisely how the result differs from the linear answer, because that difference is the next card.

Circular convolution For two sequences of the same length N,
$y [m]$ = sum over $n = 0$ to N-1 of $x_{1} [n] x_{2}$ [(m - n) mod N].
The index wraps around instead of running off the end, which is exactly what the DFT's assumption of periodicity implies.

	Linear convolution	Circular convolution
Index arithmetic	Ordinary; the shifted sequence slides off the end	Modulo N; what falls off one end comes back at the other
Length of the result	$N_{1} + N_{2} - 1$	N, the same as the inputs
What implements it	The convolution sum, or the z-transform	The DFT: multiply the spectra and invert
Models	A real filter processing a real signal	A filter acting on a periodic extension of the signal
Relation	They are equal when both sequences are zero padded to at least $N_{1} + N_{2} - 1$ points.

Three ways to compute it, and when each is quickest

The concentric circle method. Write $x_{1}$ anticlockwise on an outer circle. Write $x_{2}$ clockwise on an inner circle, starting from the same point. Multiply facing pairs and add: that is $y [0]$ . Rotate the inner circle one step anticlockwise for $y [1]$ , and so on. Fast for $N = 4$ or 5, which is what the exam sets.
The matrix method. Build the N by N circulant matrix whose first column is $x_{2}$ and whose every next column is the previous one rotated down by one. Multiply by the column vector $x_{1}$ . Reliable, and easy to check.
The DFT method. Take both DFTs, multiply point by point, take the IDFT. This is what the question means when it says using DFT, and it is the route that matters in practice because the FFT makes it cheap.

If the two sequences have different lengths Circular convolution is only defined for equal lengths, so pad the shorter one with zeros up to the length of the longer. A question giving h =

{1, 2, 1, - 1, 1}

and x =

{1, 2, 3, 1}

is a 5 point circular convolution with x padded to

{1, 2, 3, 1, 0}

In the exam

Say which N you are using before you start, and pad to it.
Check with the sum rule: sum of y = (sum of $x_{1}$ )(sum of $x_{2}$ ), exactly as for linear convolution.
When the question says find $x_{3}$ given X3(k) = X1(k)X2(k), do not compute any DFTs. Just circularly convolve $x_{1}$ and $x_{2}$ . It is a quarter of the work and it is the same answer.

It differs from linear convolution only in the wrap, and zero padding removes even that.

See it drawnThe tail of the answer wraps round and adds onto the front

Multiplying two DFTs gives the circular result, not the linear one. The difference is exactly the samples that ran off the end of the frame and came back at the beginning.

Practise thisEvery question the paper has set on this topic, worked in full17

Practise this +8 like it2082 Bhadra · Q107 marksCompute the circular convolution of the following sequences:

h (n) = {1, 2, 1, - 1, 1}

and

x (n) = {1, 2, 3, 1}

What is different hereone sequence is shorter than N, so it is zero padded first

What is different in this one

The two sequences are different lengths, 5 and 4, so the shorter is zero padded to $N = 5$ before anything else happens. Getting N wrong makes every sample wrong.
$N = 5$ is smaller than $N_{1} + N_{2} - 1 = 8$ , so this circular answer is not the linear convolution: the last 3 samples have wrapped round and added to the front. That wrap is the whole point of the question.

The DFT length is $N = 5$ , the longer of the two. Zero pad to that length, since $x_{2} [n]$ has 4 samples. The circular convolution is $y [n] = \sum_{m} x_{1} [m] x_{2}$ [(n - m) mod N], and the wrap is what makes it circular rather than linear.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=1	$x_{1}$ [1]=2	$x_{1}$ [2]=1	$x_{1}$ [3]=-1	$x_{1}$ [4]=1	sum
$y [0]$	1	0	1	3	2	1
$y [1]$	2	1	0	1	3	6
$y [2]$	3	2	1	0	1	9
$y [3]$	1	3	2	1	0	8
$y [4]$	0	1	3	2	1	4

y [0]

= 1(1) + 1(1) + -1(3) + 1(2) = 1

y [1]

= 1(2) + 2(1) + -1(1) + 1(3) = 6

y [2]

= 1(3) + 2(2) + 1(1) + 1(1) = 9

y [3]

= 1(1) + 2(3) + 1(2) + -1(1) = 8

y [4]

= 2(1) + 1(3) + -1(2) + 1(1) = 4

$y [n] = {1, 6, 9, 8, 4}$ . The check holds: 4 × 7 = 28, the sum of y.

Same as 2082 Bhadra Q102082 Baishakh · Q106 marksFind the circular convolution of the sequences

x_{1} [n] = {1, 1, - 1, - 1, 2}

and

x_{2} [n] = {1, - 1, - 2}

What is different in this one

The two sequences are different lengths, 5 and 3, so the shorter is zero padded to $N = 5$ before anything else happens. Getting N wrong makes every sample wrong.
$N = 5$ is smaller than $N_{1} + N_{2} - 1 = 7$ , so this circular answer is not the linear convolution: the last 2 samples have wrapped round and added to the front. That wrap is the whole point of the question.

The DFT length is $N = 5$ , the longer of the two. Zero pad to that length, since $x_{2} [n]$ has 3 samples. The circular convolution is $y [n] = \sum_{m} x_{1} [m] x_{2}$ [(n - m) mod N], and the wrap is what makes it circular rather than linear.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=1	$x_{1}$ [1]=1	$x_{1}$ [2]=-1	$x_{1}$ [3]=-1	$x_{1}$ [4]=2	sum
$y [0]$	1	0	0	-2	-1	1
$y [1]$	-1	1	0	0	-2	-4
$y [2]$	-2	-1	1	0	0	-4
$y [3]$	0	-2	-1	1	0	-2
$y [4]$	0	0	-2	-1	1	5

y [0]

= 1(1) + -1(-2) + 2(-1) = 1

y [1]

= 1(-1) + 1(1) + 2(-2) = -4

y [2]

= 1(-2) + 1(-1) + -1(1) = -4

y [3]

= 1(-2) + -1(-1) + -1(1) = -2

y [4]

= -1(-2) + -1(-1) + 2(1) = 5

$y [n] = {1, - 4, - 4, - 2, 5}$ . The check holds: 2 × -2 = -4, the sum of y.

Same as 2082 Bhadra Q102081 Bhadra · Q106 marksIf X1(k) and X2(k) are the 5-point DFT of

x_{1}

(n) =

3^{n}

for

0 \leq n \leq 3

and

x_{2}

(n) =

2^{n}

for

0 \leq n \leq 4

. Find

x_{3}

(n) if X3(k) = X1(k)X2(k)

How this is read $x_{1}$ (n) = $3^{n}$ over 0 <= n <= 3 is {1, 3, 9, 27} and x2(n) = 2^n over 0 <= n <= 4 is {1, 2, 4, 8, 16}. X3 = X1 X2 means CIRCULAR convolution, at the 5 point length the question sets.

What is different in this one

The two sequences are different lengths, 4 and 5, so the shorter is zero padded to $N = 5$ before anything else happens. Getting N wrong makes every sample wrong.
$N = 5$ is smaller than $N_{1} + N_{2} - 1 = 8$ , so this circular answer is not the linear convolution: the last 3 samples have wrapped round and added to the front. That wrap is the whole point of the question.

The DFT length is $N = 5$ , the longer of the two. Zero pad to that length, since $x_{1} [n]$ has 4 samples. The circular convolution is $y [n] = \sum_{m} x_{1} [m] x_{2}$ [(n - m) mod N], and the wrap is what makes it circular rather than linear.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=1	$x_{1}$ [1]=3	$x_{1}$ [2]=9	$x_{1}$ [3]=27	$x_{1}$ [4]=0	sum
$y [0]$	1	16	8	4	2	229
$y [1]$	2	1	16	8	4	365
$y [2]$	4	2	1	16	8	451
$y [3]$	8	4	2	1	16	65
$y [4]$	16	8	4	2	1	130

y [0]

= 1(1) + 3(16) + 9(8) + 27(4) = 229

y [1]

= 1(2) + 3(1) + 9(16) + 27(8) = 365

y [2]

= 1(4) + 3(2) + 9(1) + 27(16) = 451

y [3]

= 1(8) + 3(4) + 9(2) + 27(1) = 65

y [4]

= 1(16) + 3(8) + 9(4) + 27(2) = 130

$y [n] = {229, 365, 451, 65, 130}$ . The check holds: 40 × 31 = 1240, the sum of y.

Practise this +6 like it2081 Baishakh · Q106 marksCompute circular convolution of the following two sequences using DFT.

x (n) = {1, 2, 4, 5}

and

h (n) = {2, 1, 6, 8}

What is different in this one

$N = 4$ is smaller than $N_{1} + N_{2} - 1 = 7$ , so this circular answer is not the linear convolution: the last 3 samples have wrapped round and added to the front. That wrap is the whole point of the question.

The DFT length is $N = 4$ . The circular convolution is $y [n] = \sum_{m} x_{1} [m] x_{2}$ [(n - m) mod N], and the wrap is what makes it circular rather than linear.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=1	$x_{1}$ [1]=2	$x_{1}$ [2]=4	$x_{1}$ [3]=5	sum
$y [0]$	2	8	6	1	47
$y [1]$	1	2	8	6	67
$y [2]$	6	1	2	8	56
$y [3]$	8	6	1	2	34

y [0]

= 1(2) + 2(8) + 4(6) + 5(1) = 47

y [1]

= 1(1) + 2(2) + 4(8) + 5(6) = 67

y [2]

= 1(6) + 2(1) + 4(2) + 5(8) = 56

y [3]

= 1(8) + 2(6) + 4(1) + 5(2) = 34

$y [n] = {47, 67, 56, 34}$ . The check holds: 12 × 17 = 204, the sum of y.

Same as 2081 Baishakh Q102080 Bhadra · Q106 marksObtain the circular convolution of the following sequences:

x_{1}

(n) =

{1, 2, 3, 1}

and

x_{2}

(n) =

{4, 3, 2, 2}

What is different in this one

$N = 4$ is smaller than $N_{1} + N_{2} - 1 = 7$ , so this circular answer is not the linear convolution: the last 3 samples have wrapped round and added to the front. That wrap is the whole point of the question.

The DFT length is $N = 4$ . The circular convolution is $y [n] = \sum_{m} x_{1} [m] x_{2}$ [(n - m) mod N], and the wrap is what makes it circular rather than linear.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=1	$x_{1}$ [1]=2	$x_{1}$ [2]=3	$x_{1}$ [3]=1	sum
$y [0]$	4	2	2	3	17
$y [1]$	3	4	2	2	19
$y [2]$	2	3	4	2	22
$y [3]$	2	2	3	4	19

y [0]

= 1(4) + 2(2) + 3(2) + 1(3) = 17

y [1]

= 1(3) + 2(4) + 3(2) + 1(2) = 19

y [2]

= 1(2) + 2(3) + 3(4) + 1(2) = 22

y [3]

= 1(2) + 2(2) + 3(3) + 1(4) = 19

$y [n] = {17, 19, 22, 19}$ . The check holds: 7 × 11 = 77, the sum of y.

Same as 2081 Baishakh Q102080 Baishakh · Q122+5 marksWrite the complexity of DFT and FFT? Obtain the circular convolution of the following sequences:

X_{1} [n] = {1, 2, 3, 1}

and

X_{2} [n] = {4, 3, 2, 2}

What is different in this one

$N = 4$ is smaller than $N_{1} + N_{2} - 1 = 7$ , so this circular answer is not the linear convolution: the last 3 samples have wrapped round and added to the front. That wrap is the whole point of the question.

The DFT length is $N = 4$ . The circular convolution is $y [n] = \sum_{m} x_{1} [m] x_{2}$ [(n - m) mod N], and the wrap is what makes it circular rather than linear.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=1	$x_{1}$ [1]=2	$x_{1}$ [2]=3	$x_{1}$ [3]=1	sum
$y [0]$	4	2	2	3	17
$y [1]$	3	4	2	2	19
$y [2]$	2	3	4	2	22
$y [3]$	2	2	3	4	19

y [0]

= 1(4) + 2(2) + 3(2) + 1(3) = 17

y [1]

= 1(3) + 2(4) + 3(2) + 1(2) = 19

y [2]

= 1(2) + 2(3) + 3(4) + 1(2) = 22

y [3]

= 1(2) + 2(2) + 3(3) + 1(4) = 19

$y [n] = {17, 19, 22, 19}$ . The check holds: 7 × 11 = 77, the sum of y.

Same as 2082 Bhadra Q102079 Bhadra · Q117 marksIf X1(k) and X2(k) are DFT of sequence

x_{1} [n] = {1, 0, 0, 1}

and

x_{2} [n] = {2, 0, 2}

respectively then find the sequence

x_{3} [n]

; if DFT of

x_{3} [n]

is given by X3(k) = X1(k) X2(k).

What is different in this one

The two sequences are different lengths, 4 and 3, so the shorter is zero padded to $N = 4$ before anything else happens. Getting N wrong makes every sample wrong.
$N = 4$ is smaller than $N_{1} + N_{2} - 1 = 6$ , so this circular answer is not the linear convolution: the last 2 samples have wrapped round and added to the front. That wrap is the whole point of the question.

The DFT length is $N = 4$ , the longer of the two. Zero pad to that length, since $x_{2} [n]$ has 3 samples. The circular convolution is $y [n] = \sum_{m} x_{1} [m] x_{2}$ [(n - m) mod N], and the wrap is what makes it circular rather than linear.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=1	$x_{1}$ [1]=0	$x_{1}$ [2]=0	$x_{1}$ [3]=1	sum
$y [0]$	2	0	2	0	2
$y [1]$	0	2	0	2	2
$y [2]$	2	0	2	0	2
$y [3]$	0	2	0	2	2

y [0]

= 1(2) = 2

y [1]

= 1(2) = 2

y [2]

= 1(2) = 2

y [3]

= 1(2) = 2

$y [n] = {2, 2, 2, 2}$ . The check holds: 2 × 4 = 8, the sum of y.

Same as 2082 Bhadra Q102079 Baishakh · Q117 marksFind the circular convolution of the sequences

x_{1} [n] = {1, - 1, - 2, 3, - 1}

and

x_{2} [n] = {1, 2, 3}

What is different in this one

The two sequences are different lengths, 5 and 3, so the shorter is zero padded to $N = 5$ before anything else happens. Getting N wrong makes every sample wrong.
$N = 5$ is smaller than $N_{1} + N_{2} - 1 = 7$ , so this circular answer is not the linear convolution: the last 2 samples have wrapped round and added to the front. That wrap is the whole point of the question.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=1	$x_{1}$ [1]=-1	$x_{1}$ [2]=-2	$x_{1}$ [3]=3	$x_{1}$ [4]=-1	sum
$y [0]$	1	0	0	3	2	8
$y [1]$	2	1	0	0	3	-2
$y [2]$	3	2	1	0	0	-1
$y [3]$	0	3	2	1	0	-4
$y [4]$	0	0	3	2	1	-1

y [0]

= 1(1) + 3(3) + -1(2) = 8

y [1]

= 1(2) + -1(1) + -1(3) = -2

y [2]

= 1(3) + -1(2) + -2(1) = -1

y [3]

= -1(3) + -2(2) + 3(1) = -4

y [4]

= -2(3) + 3(2) + -1(1) = -1

$y [n] = {8, - 2, - 1, - 4, - 1}$ . The check holds: 0 × 6 = 0, the sum of y.

Same as 2081 Baishakh Q102078 Bhadra · Q102+6 marksDifferentiate between DFT and DTFT. Find the circular convolution of

x_{1} [n] = {2, 1, 2, 1}

and

x_{2} [n] = {1, 2, 3, 4}

What is different in this one

$N = 4$ is smaller than $N_{1} + N_{2} - 1 = 7$ , so this circular answer is not the linear convolution: the last 3 samples have wrapped round and added to the front. That wrap is the whole point of the question.

The DFT length is $N = 4$ . The circular convolution is $y [n] = \sum_{m} x_{1} [m] x_{2}$ [(n - m) mod N], and the wrap is what makes it circular rather than linear.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=2	$x_{1}$ [1]=1	$x_{1}$ [2]=2	$x_{1}$ [3]=1	sum
$y [0]$	1	4	3	2	14
$y [1]$	2	1	4	3	16
$y [2]$	3	2	1	4	14
$y [3]$	4	3	2	1	16

y [0]

= 2(1) + 1(4) + 2(3) + 1(2) = 14

y [1]

= 2(2) + 1(1) + 2(4) + 1(3) = 16

y [2]

= 2(3) + 1(2) + 2(1) + 1(4) = 14

y [3]

= 2(4) + 1(3) + 2(2) + 1(1) = 16

$y [n] = {14, 16, 14, 16}$ . The check holds: 6 × 10 = 60, the sum of y.

Same as 2082 Bhadra Q102076 Chaitra · Q125 marksFind

x_{3} [n]

if DFT of

x_{3} [n]

is given by X3(k) = X1(k) * X2(k) where X1(k) and X2(k) are 4-point DFT of

x_{1} [n] = {1, 2, - 2}

and

x_{2} [n] = {1, 2, 3, - 1}

respectively.

What is different in this one

The two sequences are different lengths, 3 and 4, so the shorter is zero padded to $N = 4$ before anything else happens. Getting N wrong makes every sample wrong.
$N = 4$ is smaller than $N_{1} + N_{2} - 1 = 6$ , so this circular answer is not the linear convolution: the last 2 samples have wrapped round and added to the front. That wrap is the whole point of the question.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=1	$x_{1}$ [1]=2	$x_{1}$ [2]=-2	$x_{1}$ [3]=0	sum
$y [0]$	1	-1	3	2	-7
$y [1]$	2	1	-1	3	6
$y [2]$	3	2	1	-1	5
$y [3]$	-1	3	2	1	1

y [0]

= 1(1) + 2(-1) + -2(3) = -7

y [1]

= 1(2) + 2(1) + -2(-1) = 6

y [2]

= 1(3) + 2(2) + -2(1) = 5

y [3]

= 1(-1) + 2(3) + -2(2) = 1

$y [n] = {- 7, 6, 5, 1}$ . The check holds: 1 × 5 = 5, the sum of y.

Practise this2075 Chaitra · Q93+7 marksDifferentiate between linear convolution and circular convolution compute circular convolution of signals

X_{1} [n] = {0, 0, 1, 1}

and

X_{2} [n] = {1, 1, 1, 1}

What is different hereOne sequence is a constant, so every output sample is the same number: the sum of the other sequence. A five second answer once seen.

What is different in this one

$N = 4$ is smaller than $N_{1} + N_{2} - 1 = 7$ , so this circular answer is not the linear convolution: the last 3 samples have wrapped round and added to the front. That wrap is the whole point of the question.

The DFT length is $N = 4$ . The circular convolution is $y [n] = \sum_{m} x_{1} [m] x_{2}$ [(n - m) mod N], and the wrap is what makes it circular rather than linear.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=0	$x_{1}$ [1]=0	$x_{1}$ [2]=1	$x_{1}$ [3]=1	sum
$y [0]$	1	1	1	1	2
$y [1]$	1	1	1	1	2
$y [2]$	1	1	1	1	2
$y [3]$	1	1	1	1	2

y [0]

= 1(1) + 1(1) = 2

y [1]

= 1(1) + 1(1) = 2

y [2]

= 1(1) + 1(1) = 2

y [3]

= 1(1) + 1(1) = 2

$y [n] = {2, 2, 2, 2}$ . The check holds: 2 × 4 = 8, the sum of y.

Same as 2081 Baishakh Q102075 Ashwin · Q91+5 marksState the circular convolution property of DFT. Find the circular convolution of: (a)

{1, 2, - 1, 1}

and

x_{2}

(n) =

{1, 3, 5, 7}

What is different in this one

$N = 4$ is smaller than $N_{1} + N_{2} - 1 = 7$ , so this circular answer is not the linear convolution: the last 3 samples have wrapped round and added to the front. That wrap is the whole point of the question.

The DFT length is $N = 4$ . The circular convolution is $y [n] = \sum_{m} x_{1} [m] x_{2}$ [(n - m) mod N], and the wrap is what makes it circular rather than linear.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=1	$x_{1}$ [1]=2	$x_{1}$ [2]=-1	$x_{1}$ [3]=1	sum
$y [0]$	1	7	5	3	13
$y [1]$	3	1	7	5	3
$y [2]$	5	3	1	7	17
$y [3]$	7	5	3	1	15

y [0]

= 1(1) + 2(7) + -1(5) + 1(3) = 13

y [1]

= 1(3) + 2(1) + -1(7) + 1(5) = 3

y [2]

= 1(5) + 2(3) + -1(1) + 1(7) = 17

y [3]

= 1(7) + 2(5) + -1(3) + 1(1) = 15

$y [n] = {13, 3, 17, 15}$ . The check holds: 3 × 16 = 48, the sum of y.

Same as 2081 Baishakh Q102074 Ashwin · Q101+7 marksWhat is a zero padding? If X1(k) and X2(k) are DFT of sequence

x_{1} [n] = {1, 2, 0, 1, - 2}

and

x_{2} [n] = {1, 0, 1, 1, 2}

respectively then find the sequence

x_{3} [n]

; If DFT of

x_{3} [n]

is given by X3(k) = X1(k) X2(k).

What is different in this one

$N = 5$ is smaller than $N_{1} + N_{2} - 1 = 9$ , so this circular answer is not the linear convolution: the last 4 samples have wrapped round and added to the front. That wrap is the whole point of the question.

The DFT length is $N = 5$ . The circular convolution is $y [n] = \sum_{m} x_{1} [m] x_{2}$ [(n - m) mod N], and the wrap is what makes it circular rather than linear.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=1	$x_{1}$ [1]=2	$x_{1}$ [2]=0	$x_{1}$ [3]=1	$x_{1}$ [4]=-2	sum
$y [0]$	1	2	1	1	0	6
$y [1]$	0	1	2	1	1	1
$y [2]$	1	0	1	2	1	1
$y [3]$	1	1	0	1	2	0
$y [4]$	2	1	1	0	1	2

y [0]

= 1(1) + 2(2) + 1(1) = 6

y [1]

= 2(1) + 1(1) + -2(1) = 1

y [2]

= 1(1) + 1(2) + -2(1) = 1

y [3]

= 1(1) + 2(1) + 1(1) + -2(2) = 0

y [4]

= 1(2) + 2(1) + -2(1) = 2

$y [n] = {6, 1, 1, 0, 2}$ . The check holds: 2 × 5 = 10, the sum of y.

Same as 2082 Bhadra Q102073 Shrawan · Q117 marksFind

x_{3} [n]

if DFT of

x_{3} [n]

is given by X3(k) = X1(k) X2(k) where X1(k) and X2(k) are 5-point DFT of

x_{1} [n] = {1, - 2, 2, 1, 4}

and

x_{2} [n] = {2, 1, - 3, - 1}

respectively.

What is different in this one

The two sequences are different lengths, 5 and 4, so the shorter is zero padded to $N = 5$ before anything else happens. Getting N wrong makes every sample wrong.
$N = 5$ is smaller than $N_{1} + N_{2} - 1 = 8$ , so this circular answer is not the linear convolution: the last 3 samples have wrapped round and added to the front. That wrap is the whole point of the question.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=1	$x_{1}$ [1]=-2	$x_{1}$ [2]=2	$x_{1}$ [3]=1	$x_{1}$ [4]=4	sum
$y [0]$	2	0	-1	-3	1	1
$y [1]$	1	2	0	-1	-3	-16
$y [2]$	-3	1	2	0	-1	-5
$y [3]$	-1	-3	1	2	0	9
$y [4]$	0	-1	-3	1	2	5

y [0]

= 1(2) + 2(-1) + 1(-3) + 4(1) = 1

y [1]

= 1(1) + -2(2) + 1(-1) + 4(-3) = -16

y [2]

= 1(-3) + -2(1) + 2(2) + 4(-1) = -5

y [3]

= 1(-1) + -2(-3) + 2(1) + 1(2) = 9

y [4]

= -2(-1) + 2(-3) + 1(1) + 4(2) = 5

$y [n] = {1, - 16, - 5, 9, 5}$ . The check holds: 6 × -1 = -6, the sum of y.

Same as 2082 Bhadra Q102072 Chaitra · Q127 marksCompute Circular Convolution of

h (n) = {1, 2, 1, - 1, 1}

and

x [n] = {1, 2, 3, 1}

What is different in this one

The two sequences are different lengths, 5 and 4, so the shorter is zero padded to $N = 5$ before anything else happens. Getting N wrong makes every sample wrong.
$N = 5$ is smaller than $N_{1} + N_{2} - 1 = 8$ , so this circular answer is not the linear convolution: the last 3 samples have wrapped round and added to the front. That wrap is the whole point of the question.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=1	$x_{1}$ [1]=2	$x_{1}$ [2]=1	$x_{1}$ [3]=-1	$x_{1}$ [4]=1	sum
$y [0]$	1	0	1	3	2	1
$y [1]$	2	1	0	1	3	6
$y [2]$	3	2	1	0	1	9
$y [3]$	1	3	2	1	0	8
$y [4]$	0	1	3	2	1	4

y [0]

= 1(1) + 1(1) + -1(3) + 1(2) = 1

y [1]

= 1(2) + 2(1) + -1(1) + 1(3) = 6

y [2]

= 1(3) + 2(2) + 1(1) + 1(1) = 9

y [3]

= 1(1) + 2(3) + 1(2) + -1(1) = 8

y [4]

= 2(1) + 1(3) + -1(2) + 1(1) = 4

$y [n] = {1, 6, 9, 8, 4}$ . The check holds: 4 × 7 = 28, the sum of y.

Same as 2082 Bhadra Q102072 Kartik · Q112+6 marksWhy we need DFT? If X1(k) and X2(k) are DFT of sequence

x_{1} [n] = {1, 2, 4}

and

x_{2} [n] = {- 1, 2, 3, 1}

respectively, then find the sequence

x_{3} [n]

, if DFT of

x_{3} [n]

is given by X3(k) = X1(k) X2(k).

What is different in this one

The two sequences are different lengths, 3 and 4, so the shorter is zero padded to $N = 4$ before anything else happens. Getting N wrong makes every sample wrong.
$N = 4$ is smaller than $N_{1} + N_{2} - 1 = 6$ , so this circular answer is not the linear convolution: the last 2 samples have wrapped round and added to the front. That wrap is the whole point of the question.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=1	$x_{1}$ [1]=2	$x_{1}$ [2]=4	$x_{1}$ [3]=0	sum
$y [0]$	-1	1	3	2	13
$y [1]$	2	-1	1	3	4
$y [2]$	3	2	-1	1	3
$y [3]$	1	3	2	-1	15

y [0]

= 1(-1) + 2(1) + 4(3) = 13

y [1]

= 1(2) + 2(-1) + 4(1) = 4

y [2]

= 1(3) + 2(2) + 4(-1) = 3

y [3]

= 1(1) + 2(3) + 4(2) = 15

$y [n] = {13, 4, 3, 15}$ . The check holds: 7 × 5 = 35, the sum of y.

Same as 2081 Baishakh Q102071 Shrawan · Q97 marksA system has input signal

x [n] = {1, 2, 3, 4}

and impulse response

h [n] = {1, 3, 5, 7}

and the DFT of

x [n]

X [k]

and the DFT of

h [n]

H [k]

. Find the output of the system

y [n]

G [k] = X [k]

H [k]

What is different in this one

$N = 4$ is smaller than $N_{1} + N_{2} - 1 = 7$ , so this circular answer is not the linear convolution: the last 3 samples have wrapped round and added to the front. That wrap is the whole point of the question.

The DFT length is $N = 4$ . The circular convolution is $y [n] = \sum_{m} x_{1} [m] x_{2}$ [(n - m) mod N], and the wrap is what makes it circular rather than linear.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=1	$x_{1}$ [1]=2	$x_{1}$ [2]=3	$x_{1}$ [3]=4	sum
$y [0]$	1	7	5	3	42
$y [1]$	3	1	7	5	46
$y [2]$	5	3	1	7	42
$y [3]$	7	5	3	1	30

y [0]

= 1(1) + 2(7) + 3(5) + 4(3) = 42

y [1]

= 1(3) + 2(1) + 3(7) + 4(5) = 46

y [2]

= 1(5) + 2(3) + 3(1) + 4(7) = 42

y [3]

= 1(7) + 2(5) + 3(3) + 4(1) = 30

$y [n] = {42, 46, 42, 30}$ . The check holds: 10 × 16 = 160, the sum of y.

The same questions grouped by how often each topic is set: circular convolution

Zero padding PIN 2/19

74 Ch · 74 Ash1+51+7

Before the formulas

And here is the fix, which is simpler than the problem. If the wrapped tail is what spoils the answer, make the frame long enough that there is nothing to wrap: pad both sequences with zeros until the length is at least the length of the linear result.

The circular answer and the linear answer are then identical, which is what makes fast convolution possible at all. Zero padding also gives a denser set of frequency points, which looks like more resolution and is not, and the paper does like asking about that difference.

Zero padding Appending zeros to a sequence before taking its DFT. It does not add information and does not improve the true resolution; it interpolates the spectrum, giving a finer grid of samples of the same underlying DTFT. Its other and more important use is to make circular convolution produce the linear one.

To get linear convolution from the DFT:

\begin{aligned} x_{1} has N_{1} samples, x_{2} has N_{2} \\ the linear convolution has N_{1} + N_{2} - 1 samples \\ pad BOTH to N \geq N_{1} + N_{2} - 1 \\ y = IDFT {DFT (x_{1}) \cdot DFT (x_{2})} is then exactly the linear convolution \end{aligned}

If N is smaller, the tail wraps round and adds to the front: time domain aliasing.

Worked, 2074 Chaitra

x [n] = {1, 1, 1, 1}

with

N_{1} = 4

h [n] = {2, 3}

with

N_{2} = 2

. Linear convolution has 4 + 2 - 1 = 5 samples, so pad both to

N = 5

: x =

{1, 1, 1, 1, 0}

and h =

{2, 3, 0, 0, 0}

.
The 5 point circular convolution gives

{2, 5, 5, 5, 3}

, which is exactly the linear convolution of

{1, 1, 1, 1}

with

{2, 3}

.
With

N = 4

instead, the last sample would have wrapped onto the first, giving

{5, 5, 5, 5}

: wrong, and wrong in the characteristic way.

Two uses worth naming

Fast convolution. Filtering a long signal is done in blocks, using the overlap add or overlap save method, each block zero padded so that the circular convolution of the block equals its linear convolution.
A smoother spectrum plot. Padding from 8 to 64 points gives 64 samples of the same DTFT, which looks like more detail but is only interpolation.

Padding to $N_{1} + N_{2} - 1$ is what makes circular convolution equal the linear convolution a real filter performs.

See it drawnMake the frame long enough and nothing has to wrap

Pad both sequences to at least the length of the linear answer and the circular result becomes the linear result exactly. That is what makes fast convolution possible at all.

Practise thisEvery question the paper has set on this topic, worked in full1

Practise this2074 Chaitra · Q101+5 marksDefine zero padding. Find the linear convolution through circular convolution with padding of zeros for the following sequences:

x [n] = {1, 1, 1, 1}

and

h [n] {2, 3}

What is different in this one

x has 4 samples and h has 2, so the linear convolution has 5 and the DFT length must be at least that. Pad both to N = 5.
At the unpadded length of 4 the answer would be wrong: the last 1 samples wrap round and add to the front. Showing that is the point of the question.

x has 4 samples and h has 2, so the linear convolution has 4 + 2 - 1 = 5 samples and the DFT length must be at least that. Pad both sequences to N = 5.

Now do the circular convolution at that padded length. This is where the numbers come from, and at $N = 5$ it is the linear answer exactly.

The DFT length is $N = 5$ . The circular convolution is $y [n] = \sum_{m} x_{1} [m] x_{2}$ [(n - m) mod N], and the wrap is what makes it circular rather than linear.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=1	$x_{1}$ [1]=1	$x_{1}$ [2]=1	$x_{1}$ [3]=1	$x_{1}$ [4]=0	sum
$y [0]$	2	0	0	0	3	2
$y [1]$	3	2	0	0	0	5
$y [2]$	0	3	2	0	0	5
$y [3]$	0	0	3	2	0	5
$y [4]$	0	0	0	3	2	3

y [0]

= 1(2) = 2

y [1]

= 1(3) + 1(2) = 5

y [2]

= 1(3) + 1(2) = 5

y [3]

= 1(3) + 1(2) = 5

y [4]

= 1(3) = 3

$y [n] = {2, 5, 5, 5, 3}$ . The check holds: 4 × 5 = 20, the sum of y.

What the padding bought. Do the same convolution at the unpadded length of 4 and compare, sample by sample: the tail has nowhere to go and lands on the front.

Sample	Linear answer	At $N = 4$	What happened
$n = 0$	2	5	wrapped: $y [0] + y [4] = 2$ + 3
$n = 1$	5	5	unchanged
$n = 2$	5	5	unchanged
$n = 3$	5	5	unchanged
$n = 4$	3	not there	lost off the end

$y [n] = {2, 5, 5, 5, 3}$ , which is the linear convolution exactly.

The same questions grouped by how often each topic is set: linear convolution through the dft

7.2The FFT: decimation in time and decimation in frequency

Why the FFT is fast HOT 7/19

82 Ba · 81 Bh · 81 Ba · 80 Ba · 79 Bh · 75 Ch · 75 Ash2+61+2+62+5

Before the formulas

The DFT as written costs about N squared multiplications. For a thousand points that is a million, which is why the transform sat mostly unused for decades. The FFT is not a different transform. It is the same numbers, computed in a smarter order.

Split the sum into two half length sums, notice that each of those splits again, and keep going until nothing is left to split. The cost falls to roughly N log N. For a thousand points that is about ten thousand operations instead of a million, for an answer that is identical to the last digit.

\begin{aligned} direct DFT & N^{2} complex multiplications, N (N - 1) additions \\ each X (k) needs N multiplications, and there are N \\ radix 2 FFT & \frac{N}{2} \log_{2} N multiplications, N \log_{2} N additions \\ \log_{2} N stages, each of \frac{N}{2} butterflies \end{aligned}

N	DFT multiplications	FFT multiplications	Speed up
8	64	12	about 5 times
64	4,096	192	about 21 times
1,024	1,048,576	5,120	about 200 times

Where the saving comes from

Divide and conquer. An N point DFT is split into two N/2 point DFTs, then each of those into two more, and so on for log2(N) stages. Two half length transforms cost $2 (N / 2)^{2} = N^{2}$ /2, which is half of $N^{2}$ , and recursing keeps halving it.
The twiddle factor identities make the recombination almost free: $W_{N}^{k + N / 2} = - W_{N}^{k}$ means the two outputs of a butterfly differ only in a sign, and $W_{N}^{2} = W_{N / 2}$ means the sub transforms really are ordinary DFTs of half the length.
Symmetry and periodicity of W remove the repeated work that the direct sum does over and over.
It needs N to be a power of 2 for radix 2. Sequences of other lengths are zero padded to the next power of two.

The counts, the in place computation and the bit reversal are the last section of this chapter.

See it drawnSplit the sum in half, then keep splitting

Each split halves the work and costs only a handful of extra multiplications, and there are log two of N splits to make. That is the whole idea, and the answer is identical to the direct sum.

Practise thisEvery question the paper has set on this topic, worked in full8

Practise this2082 Baishakh · Q91+2+6 marksWhat is Fast Fourier Transform algorithm? How Decimation in Frequency Fast Fourier Transform algorithm reduces the calculation complexity of DFT? Find 4 point DFT of the sequence

x [n] = {1, 4, - 1}

using Decimation in Frequency Fast Fourier Transform algorithm.What is different heredecimation in frequency, where the input stays in order and the output comes out bit reversed; only four points, so two stages, and the only twiddles are 1 and -j. Only 4 points, so there are two stages of two butterflies and the only twiddles are

W 4^{0} = 1

and

W 4^{1} = - j

. Most sittings set the 8 point version.

What is different in this one

Decimation in frequency: 4 points, 2 stages of 2 butterflies, with the output in bit reversed order.
The sequence has 3 samples, not a power of two, so it is zero padded to $N = 4$ before the algorithm starts. Say so in the answer: a radix 2 FFT cannot run on 3 points.
This is a 4 point transform, so there are only 2 stages of 2 butterflies and the only twiddles are $W 4^{0} = 1$ and $W 4^{1} = - j$ . Much shorter than the usual 8 point question.
Only 4 points, so there are two stages of two butterflies and the only twiddles are $W 4^{0} = 1$ and $W 4^{1} = - j$ . Most sittings set the 8 point version.

The sequence has 3 samples, and a radix 2 FFT needs a power of two, so zero pad to $N = 4$ and say so: $x [n] = {1, 4, - 1, 0}$ .

Decimation in frequency: the input stays in natural order and the output comes out bit reversed, so the last step is to relabel it.

Stage 1, 2 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ applied to the differences:

\begin{aligned} 0 \\ 4 \\ 2 \\ - j 4 \end{aligned}

Stage 2, 2 butterflies with twiddles $W_{2}^{0}$ applied to the differences:

\begin{aligned} 4 \\ - 4 \\ 2 - j 4 \\ 2 + j 4 \end{aligned}

Read the output in bit reversed order: $X [0]$ , $X [2]$ , $X [1]$ , $X [3]$ .

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	4	4	0
$X [1]$	2 - j4	4.4721	-63.43
$X [2]$	-4	4	180
$X [3]$	2 + j4	4.4721	63.43

$X [k]$ = {4, 2 - j4, -4, 2 + j4}. Two free checks: $X [0] = 4$ , which is the sum of the samples; $X [3] = X^{*} [1]$ and $X [2] = X^{*} [2]$ , the conjugate symmetry a real input forces.

Practise this +8 like it2081 Bhadra · Q92+7 marksHow does FFT reduce computational complexity compared to direct calculation of DFT? Compute 8-point DFT of

x [n] = {1, - 2, - 4, 3, 0, 1}

using DIT-FFT algorithm.

What is different in this one

Decimation in time: 8 points, 3 stages of 4 butterflies, with the input in bit reversed order.
The sequence has 6 samples, not a power of two, so it is zero padded to $N = 8$ before the algorithm starts. Say so in the answer: a radix 2 FFT cannot run on 6 points.

The sequence has 6 samples, and a radix 2 FFT needs a power of two, so zero pad to $N = 8$ and say so: $x [n] = {1, - 2, - 4, 3, 0, 1, 0, 0}$ .

Stage 1, 4 butterflies with twiddles $W_{2}^{0}$ :

\begin{aligned} 1 \\ 1 \\ - 4 \\ - 4 \\ - 1 \\ - 3 \\ 3 \\ 3 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ :

\begin{aligned} - 3 \\ 1 + j 4 \\ 5 \\ 1 - j 4 \\ 2 \\ - 3 - j 3 \\ - 4 \\ - 3 + j 3 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ :

\begin{aligned} - 1 \\ - 3.2426 + j 4 \\ 5 + j 4 \\ 5.2426 - j 4 \\ - 5 \\ 5.2426 + j 4 \\ 5 - j 4 \\ - 3.2426 - j 4 \end{aligned}

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	-1	1	180
$X [1]$	-3.2426 + j4	5.1492	129.03
$X [2]$	5 + j4	6.4031	38.66
$X [3]$	5.2426 - j4	6.5943	-37.34
$X [4]$	-5	5	180
$X [5]$	5.2426 + j4	6.5943	37.34
$X [6]$	5 - j4	6.4031	-38.66
$X [7]$	-3.2426 - j4	5.1492	-129.03

$X [k]$ = {-1, -3.2426 + j4, 5 + j4, 5.2426 - j4, -5, 5.2426 + j4, 5 - j4, -3.2426 - j4}. Two free checks: $X [0] = - 1$ , which is the sum of the samples; $X [7] = X^{*} [1]$ and $X [6] = X^{*} [2]$ , the conjugate symmetry a real input forces.

Practise this2081 Baishakh · Q92+6 marksWhy Decimation in Time Fast Fourier Transform (DITFFT) Algorithm is better than direct computation of DFT? Find 4 point DFT of the sequence

x (n) = {2, 2, 4}

using DITFFT algorithm.What is different hereonly four points, so two stages, and the only twiddles are 1 and -j

What is different in this one

Decimation in time: 4 points, 2 stages of 2 butterflies, with the input in bit reversed order.
The sequence has 3 samples, not a power of two, so it is zero padded to $N = 4$ before the algorithm starts. Say so in the answer: a radix 2 FFT cannot run on 3 points.
This is a 4 point transform, so there are only 2 stages of 2 butterflies and the only twiddles are $W 4^{0} = 1$ and $W 4^{1} = - j$ . Much shorter than the usual 8 point question.

The sequence has 3 samples, and a radix 2 FFT needs a power of two, so zero pad to $N = 4$ and say so: $x [n] = {2, 2, 4, 0}$ .

Decimation in time: the input goes in bit reversed order and the output comes out in natural order. For $N = 4$ that input order is $x [0]$ , $x [2]$ , $x [1]$ , $x [3]$ .

Stage 1, 2 butterflies with twiddles $W_{2}^{0}$ :

\begin{aligned} 6 \\ - 2 \\ 2 \\ 2 \end{aligned}

Stage 2, 2 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ :

\begin{aligned} 8 \\ - 2 - j 2 \\ 4 \\ - 2 + j 2 \end{aligned}

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	8	8	0
$X [1]$	-2 - j2	2.8284	-135
$X [2]$	4	4	0
$X [3]$	-2 + j2	2.8284	135

$X [k]$ = {8, -2 - j2, 4, -2 + j2}. Two free checks: $X [0] = 8$ , which is the sum of the samples; $X [3] = X^{*} [1]$ and $X [2] = X^{*} [2]$ , the conjugate symmetry a real input forces.

Same as 2081 Bhadra Q92080 Baishakh · Q112+6 marksHow fast is FFT? Find 8-point DFT of sequence

x [n] = {1, 1, 0, 0, 1, 1, 2}

using Decimation in Time Fast Fourier Transform (DITFFT) algorithm.

What is different in this one

Decimation in time: 8 points, 3 stages of 4 butterflies, with the input in bit reversed order.
The sequence has 7 samples, not a power of two, so it is zero padded to $N = 8$ before the algorithm starts. Say so in the answer: a radix 2 FFT cannot run on 7 points.

The sequence has 7 samples, and a radix 2 FFT needs a power of two, so zero pad to $N = 8$ and say so: $x [n] = {1, 1, 0, 0, 1, 1, 2, 0}$ .

Stage 1, 4 butterflies with twiddles $W_{2}^{0}$ :

\begin{aligned} 2 \\ 0 \\ 2 \\ - 2 \\ 2 \\ 0 \\ 0 \\ 0 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ :

\begin{aligned} 4 \\ j2 \\ 0 \\ - j 2 \\ 2 \\ 0 \\ 2 \\ 0 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ :

\begin{aligned} 6 \\ j2 \\ - j 2 \\ - j 2 \\ 2 \\ j2 \\ j2 \\ - j 2 \end{aligned}

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	6	6	0
$X [1]$	j2	2	90
$X [2]$	-j2	2	-90
$X [3]$	-j2	2	-90
$X [4]$	2	2	0
$X [5]$	j2	2	90
$X [6]$	j2	2	90
$X [7]$	-j2	2	-90

$X [k]$ = {6, j2, -j2, -j2, 2, j2, j2, -j2}. Two free checks: $X [0] = 6$ , which is the sum of the samples; $X [7] = X^{*} [1]$ and $X [6] = X^{*} [2]$ , the conjugate symmetry a real input forces.

Same as 2081 Baishakh Q102080 Baishakh · Q122+5 marksWrite the complexity of DFT and FFT? Obtain the circular convolution of the following sequences:

X_{1} [n] = {1, 2, 3, 1}

and

X_{2} [n] = {4, 3, 2, 2}

What is different in this one

$N = 4$ is smaller than $N_{1} + N_{2} - 1 = 7$ , so this circular answer is not the linear convolution: the last 3 samples have wrapped round and added to the front. That wrap is the whole point of the question.

The DFT length is $N = 4$ . The circular convolution is $y [n] = \sum_{m} x_{1} [m] x_{2}$ [(n - m) mod N], and the wrap is what makes it circular rather than linear.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=1	$x_{1}$ [1]=2	$x_{1}$ [2]=3	$x_{1}$ [3]=1	sum
$y [0]$	4	2	2	3	17
$y [1]$	3	4	2	2	19
$y [2]$	2	3	4	2	22
$y [3]$	2	2	3	4	19

y [0]

= 1(4) + 2(2) + 3(2) + 1(3) = 17

y [1]

= 1(3) + 2(4) + 3(2) + 1(2) = 19

y [2]

= 1(2) + 2(3) + 3(4) + 1(2) = 22

y [3]

= 1(2) + 2(2) + 3(3) + 1(4) = 19

$y [n] = {17, 19, 22, 19}$ . The check holds: 7 × 11 = 77, the sum of y.

Same as 2081 Bhadra Q92079 Bhadra · Q102+6 marksWhy we need FFT? Find the 8-point DFT of the following sequence using radix-2 DITFFT algorithm.

How this is readThe transcription of this paper lost the sequence, which the paper gives as a plot. It is recovered from the standard text, which works this very paper as its Example 7.18: $x [n]$ = {1, 2j, 3, 4j, 0, 0, -j, 0}. Note that it is complex, which is unusual for this question and is the whole point of it: nothing in the butterflies cares, because every step is already a complex multiply and add.

What is different in this one

Decimation in time: 8 points, 3 stages of 4 butterflies, with the input in bit reversed order.

$N = 8$ , which is $2^{3}$ , so the transform takes 3 stages of 4 butterflies.

Stage 1, 4 butterflies with twiddles $W_{2}^{0}$ :

\begin{aligned} 1 \\ 1 \\ 3 - j 1 \\ 3 + j 1 \\ j2 \\ j2 \\ j4 \\ j4 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ :

\begin{aligned} 4 - j 1 \\ 2 - j 3 \\ - 2 + j 1 \\ j3 \\ j6 \\ 4 + j 2 \\ - j 2 \\ - 4 + j 2 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ :

\begin{aligned} 4 + j 5 \\ 6.2426 - j 4.4142 \\ - 4 + j 1 \\ 4.2426 + j 4.4142 \\ 4 - j 7 \\ - 2.2426 - j 1.5858 \\ j1 \\ - 4.2426 + j 1.5858 \end{aligned}

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	4 + j5	6.4031	51.34
$X [1]$	6.2426 - j4.4142	7.6456	-35.26
$X [2]$	-4 + j1	4.1231	165.96
$X [3]$	4.2426 + j4.4142	6.1225	46.14
$X [4]$	4 - j7	8.0623	-60.26
$X [5]$	-2.2426 - j1.5858	2.7467	-144.74
$X [6]$	j1	1	90
$X [7]$	-4.2426 + j1.5858	4.5293	159.51

$X [k]$ = {4 + j5, 6.2426 - j4.4142, -4 + j1, 4.2426 + j4.4142, 4 - j7, -2.2426 - j1.5858, j1, -4.2426 + j1.5858}. Two free checks: $X [0] = 4$ + j5, which is the sum of the samples.

Practise this2075 Chaitra · Q82+8 marksHow fast is FFT? Find X(3) and X(5) for given sequence

x [n] = {1, - 2, 3, 2}

using DITFFT algorithm.What is different hereonly four points, so two stages, and the only twiddles are 1 and -j. The question asks for X(3) and X(5) of a FOUR point sequence, and a 4 point DFT only has

k = 0

to 3. X(5) is read by periodicity:

X [k + N] = X [k]

, so X(5) = X(1). Saying that is the question.

What is different in this one

Decimation in time: 4 points, 2 stages of 2 butterflies, with the input in bit reversed order.
This is a 4 point transform, so there are only 2 stages of 2 butterflies and the only twiddles are $W 4^{0} = 1$ and $W 4^{1} = - j$ . Much shorter than the usual 8 point question.
The question asks for X(3) and X(5) of a FOUR point sequence, and a 4 point DFT only has $k = 0$ to 3. X(5) is read by periodicity: $X [k + N] = X [k]$ , so X(5) = X(1). Saying that is the question.

$N = 4$ , which is $2^{2}$ , so the transform takes 2 stages of 2 butterflies.

Decimation in time: the input goes in bit reversed order and the output comes out in natural order. For $N = 4$ that input order is $x [0]$ , $x [2]$ , $x [1]$ , $x [3]$ .

Stage 1, 2 butterflies with twiddles $W_{2}^{0}$ :

\begin{aligned} 4 \\ - 2 \\ 0 \\ - 4 \end{aligned}

Stage 2, 2 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ :

\begin{aligned} 4 \\ - 2 + j 4 \\ 4 \\ - 2 - j 4 \end{aligned}

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	4	4	0
$X [1]$	-2 + j4	4.4721	116.57
$X [2]$	4	4	0
$X [3]$	-2 - j4	4.4721	-116.57

$X [k]$ = {4, -2 + j4, 4, -2 - j4}. Two free checks: $X [0] = 4$ , which is the sum of the samples; $X [3] = X^{*} [1]$ and $X [2] = X^{*} [2]$ , the conjugate symmetry a real input forces.

Same as 2081 Bhadra Q92075 Ashwin · Q83+7 marksHow fast is FFT compare to DFT? Draw the butterfly diagram of 8-point DFT of a sequence as

x [n] = n

+ 1 using Decimation in Time FFT algorithm.

How this is read $x [n] = n$ + 1 over $n = 0$ to 7 is ${1, 2, 3, 4, 5, 6, 7, 8}$ .

What is different in this one

Decimation in time: 8 points, 3 stages of 4 butterflies, with the input in bit reversed order.

$N = 8$ , which is $2^{3}$ , so the transform takes 3 stages of 4 butterflies.

Stage 1, 4 butterflies with twiddles $W_{2}^{0}$ :

\begin{aligned} 6 \\ - 4 \\ 10 \\ - 4 \\ 8 \\ - 4 \\ 12 \\ - 4 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ :

\begin{aligned} 16 \\ - 4 + j 4 \\ - 4 \\ - 4 - j 4 \\ 20 \\ - 4 + j 4 \\ - 4 \\ - 4 - j 4 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ :

\begin{aligned} 36 \\ - 4 + j 9.6569 \\ - 4 + j 4 \\ - 4 + j 1.6569 \\ - 4 \\ - 4 - j 1.6569 \\ - 4 - j 4 \\ - 4 - j 9.6569 \end{aligned}

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	36	36	0
$X [1]$	-4 + j9.6569	10.4525	112.5
$X [2]$	-4 + j4	5.6569	135
$X [3]$	-4 + j1.6569	4.3296	157.5
$X [4]$	-4	4	180
$X [5]$	-4 - j1.6569	4.3296	-157.5
$X [6]$	-4 - j4	5.6569	-135
$X [7]$	-4 - j9.6569	10.4525	-112.5

$X [k]$ = {36, -4 + j9.6569, -4 + j4, -4 + j1.6569, -4, -4 - j1.6569, -4 - j4, -4 - j9.6569}. Two free checks: $X [0] = 36$ , which is the sum of the samples; $X [7] = X^{*} [1]$ and $X [6] = X^{*} [2]$ , the conjugate symmetry a real input forces.

Decimation in time TOP 11/19

81 Bh · 81 Ba · 80 Ba · 79 Bh · 78 Bh · 75 Ch · 75 Ash · 73 Shr · 72 Ch · 72 Ka · 71 Shr2+672+7

Before the formulas

The first of the two ways to do that splitting, and the one asked most often. Decimation in time splits the input sequence into its even numbered and odd numbered samples, then does the same to each half, and so on.

The bookkeeping consequence is the thing to remember: the input has to be fed in bit reversed order and the output comes out in natural order. After that it is one butterfly repeated across three stages for eight points, and the exam question is that flow graph, drawn and then evaluated with real numbers.

The split Divide the input sequence into its even indexed and odd indexed samples. Each half is an N/2 point DFT, and the two are recombined with one twiddle factor per output.

\begin{aligned} X (k) & = \sum_{n = 0}^{N - 1} x [n] W_{N}^{n k} \\ = \sum_{r = 0}^{N / 2 - 1} x [2 r] W_{N}^{2 r k} + \sum_{r = 0}^{N / 2 - 1} x [2 r + 1] W_{N}^{(2 r + 1) k} & split n by parity \\ = \underset{G (k)}{\underset{⏟}{\sum_{r} x [2 r] W_{N / 2}^{r k}}} + W_{N}^{k} \underset{H (k)}{\underset{⏟}{\sum_{r} x [2 r + 1] W_{N / 2}^{r k}}} & using W_{N}^{2} = W_{N / 2} \\ X (k) & = G (k) + W_{N}^{k} H (k) \\ X (k + \frac{N}{2}) & = G (k) - W_{N}^{k} H (k) & since W_{N}^{k + N / 2} = - W_{N}^{k} \end{aligned}

The procedure for an 8 point DIT FFT

Bit reverse the input order. For $N = 8$ the order is $x [0]$ , $x [4]$ , $x [2]$ , $x [6]$ , $x [1]$ , $x [5]$ , $x [3]$ , $x [7]$ , which is the indices 0 to 7 with their three bits reversed. The output comes out in natural order.
Stage 1: four butterflies on adjacent pairs, twiddle $W^{0} = 1$ only.
Stage 2: four butterflies spanning two, with twiddles $W^{0}$ and $W^{2}$ .
Stage 3: four butterflies spanning four, with twiddles $W^{0}$ , $W^{1}$ , $W^{2}$ , $W^{3}$ .
Write the numbers on the diagram as you go, and read X(0) to X(7) off the right hand edge in order.

The twiddle factors for

N = 8

, worth writing out first

W^{0} = 1

W^{1} = 0.7071

- j0.7071

W^{2}

= -j

W^{3} = - 0.7071

- j0.7071
and

W^{k + 4} = - W^{k}

for the rest.

Reverse every arrow of this flow graph and the decimation in frequency algorithm appears: the two are transposes, and cost exactly the same.

See it drawnSplit the input into its even and odd samples, three times over

The input then has to be fed in bit reversed order and the output comes out naturally ordered. Everything between is one butterfly, repeated: three stages of four for eight points.

Practise thisEvery question the paper has set on this topic, worked in full11

Practise this +8 like it2081 Bhadra · Q92+7 marksHow does FFT reduce computational complexity compared to direct calculation of DFT? Compute 8-point DFT of

x [n] = {1, - 2, - 4, 3, 0, 1}

using DIT-FFT algorithm.

What is different in this one

Decimation in time: 8 points, 3 stages of 4 butterflies, with the input in bit reversed order.
The sequence has 6 samples, not a power of two, so it is zero padded to $N = 8$ before the algorithm starts. Say so in the answer: a radix 2 FFT cannot run on 6 points.

The sequence has 6 samples, and a radix 2 FFT needs a power of two, so zero pad to $N = 8$ and say so: $x [n] = {1, - 2, - 4, 3, 0, 1, 0, 0}$ .

Stage 1, 4 butterflies with twiddles $W_{2}^{0}$ :

\begin{aligned} 1 \\ 1 \\ - 4 \\ - 4 \\ - 1 \\ - 3 \\ 3 \\ 3 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ :

\begin{aligned} - 3 \\ 1 + j 4 \\ 5 \\ 1 - j 4 \\ 2 \\ - 3 - j 3 \\ - 4 \\ - 3 + j 3 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ :

\begin{aligned} - 1 \\ - 3.2426 + j 4 \\ 5 + j 4 \\ 5.2426 - j 4 \\ - 5 \\ 5.2426 + j 4 \\ 5 - j 4 \\ - 3.2426 - j 4 \end{aligned}

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	-1	1	180
$X [1]$	-3.2426 + j4	5.1492	129.03
$X [2]$	5 + j4	6.4031	38.66
$X [3]$	5.2426 - j4	6.5943	-37.34
$X [4]$	-5	5	180
$X [5]$	5.2426 + j4	6.5943	37.34
$X [6]$	5 - j4	6.4031	-38.66
$X [7]$	-3.2426 - j4	5.1492	-129.03

Practise this2081 Baishakh · Q92+6 marksWhy Decimation in Time Fast Fourier Transform (DITFFT) Algorithm is better than direct computation of DFT? Find 4 point DFT of the sequence

x (n) = {2, 2, 4}

using DITFFT algorithm.What is different hereonly four points, so two stages, and the only twiddles are 1 and -j

What is different in this one

Decimation in time: 4 points, 2 stages of 2 butterflies, with the input in bit reversed order.
The sequence has 3 samples, not a power of two, so it is zero padded to $N = 4$ before the algorithm starts. Say so in the answer: a radix 2 FFT cannot run on 3 points.
This is a 4 point transform, so there are only 2 stages of 2 butterflies and the only twiddles are $W 4^{0} = 1$ and $W 4^{1} = - j$ . Much shorter than the usual 8 point question.

The sequence has 3 samples, and a radix 2 FFT needs a power of two, so zero pad to $N = 4$ and say so: $x [n] = {2, 2, 4, 0}$ .

Decimation in time: the input goes in bit reversed order and the output comes out in natural order. For $N = 4$ that input order is $x [0]$ , $x [2]$ , $x [1]$ , $x [3]$ .

Stage 1, 2 butterflies with twiddles $W_{2}^{0}$ :

\begin{aligned} 6 \\ - 2 \\ 2 \\ 2 \end{aligned}

Stage 2, 2 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ :

\begin{aligned} 8 \\ - 2 - j 2 \\ 4 \\ - 2 + j 2 \end{aligned}

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	8	8	0
$X [1]$	-2 - j2	2.8284	-135
$X [2]$	4	4	0
$X [3]$	-2 + j2	2.8284	135

$X [k]$ = {8, -2 - j2, 4, -2 + j2}. Two free checks: $X [0] = 8$ , which is the sum of the samples; $X [3] = X^{*} [1]$ and $X [2] = X^{*} [2]$ , the conjugate symmetry a real input forces.

Same as 2081 Bhadra Q92080 Baishakh · Q112+6 marksHow fast is FFT? Find 8-point DFT of sequence

x [n] = {1, 1, 0, 0, 1, 1, 2}

using Decimation in Time Fast Fourier Transform (DITFFT) algorithm.

What is different in this one

Decimation in time: 8 points, 3 stages of 4 butterflies, with the input in bit reversed order.
The sequence has 7 samples, not a power of two, so it is zero padded to $N = 8$ before the algorithm starts. Say so in the answer: a radix 2 FFT cannot run on 7 points.

The sequence has 7 samples, and a radix 2 FFT needs a power of two, so zero pad to $N = 8$ and say so: $x [n] = {1, 1, 0, 0, 1, 1, 2, 0}$ .

Stage 1, 4 butterflies with twiddles $W_{2}^{0}$ :

\begin{aligned} 2 \\ 0 \\ 2 \\ - 2 \\ 2 \\ 0 \\ 0 \\ 0 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ :

\begin{aligned} 4 \\ j2 \\ 0 \\ - j 2 \\ 2 \\ 0 \\ 2 \\ 0 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ :

\begin{aligned} 6 \\ j2 \\ - j 2 \\ - j 2 \\ 2 \\ j2 \\ j2 \\ - j 2 \end{aligned}

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	6	6	0
$X [1]$	j2	2	90
$X [2]$	-j2	2	-90
$X [3]$	-j2	2	-90
$X [4]$	2	2	0
$X [5]$	j2	2	90
$X [6]$	j2	2	90
$X [7]$	-j2	2	-90

Same as 2081 Bhadra Q92079 Bhadra · Q102+6 marksWhy we need FFT? Find the 8-point DFT of the following sequence using radix-2 DITFFT algorithm.

What is different in this one

Decimation in time: 8 points, 3 stages of 4 butterflies, with the input in bit reversed order.

$N = 8$ , which is $2^{3}$ , so the transform takes 3 stages of 4 butterflies.

Stage 1, 4 butterflies with twiddles $W_{2}^{0}$ :

\begin{aligned} 1 \\ 1 \\ 3 - j 1 \\ 3 + j 1 \\ j2 \\ j2 \\ j4 \\ j4 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ :

\begin{aligned} 4 - j 1 \\ 2 - j 3 \\ - 2 + j 1 \\ j3 \\ j6 \\ 4 + j 2 \\ - j 2 \\ - 4 + j 2 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ :

\begin{aligned} 4 + j 5 \\ 6.2426 - j 4.4142 \\ - 4 + j 1 \\ 4.2426 + j 4.4142 \\ 4 - j 7 \\ - 2.2426 - j 1.5858 \\ j1 \\ - 4.2426 + j 1.5858 \end{aligned}

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	4 + j5	6.4031	51.34
$X [1]$	6.2426 - j4.4142	7.6456	-35.26
$X [2]$	-4 + j1	4.1231	165.96
$X [3]$	4.2426 + j4.4142	6.1225	46.14
$X [4]$	4 - j7	8.0623	-60.26
$X [5]$	-2.2426 - j1.5858	2.7467	-144.74
$X [6]$	j1	1	90
$X [7]$	-4.2426 + j1.5858	4.5293	159.51

$X [k]$ = {4 + j5, 6.2426 - j4.4142, -4 + j1, 4.2426 + j4.4142, 4 - j7, -2.2426 - j1.5858, j1, -4.2426 + j1.5858}. Two free checks: $X [0] = 4$ + j5, which is the sum of the samples.

Same as 2081 Bhadra Q92078 Bhadra · Q117 marksFind the 8-point DFT of

x [n] = u [n] - u [n - 4]

using FFT DIT algorithm.

How this is read $x [n] = u [n] - u [n - 4]$ is ${1, 1, 1, 1}$ for $n = 0$ to 3, and zero to n = 7.

What is different in this one

Decimation in time: 8 points, 3 stages of 4 butterflies, with the input in bit reversed order.

$N = 8$ , which is $2^{3}$ , so the transform takes 3 stages of 4 butterflies.

Stage 1, 4 butterflies with twiddles $W_{2}^{0}$ :

\begin{aligned} 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ :

\begin{aligned} 2 \\ 1 - j 1 \\ 0 \\ 1 + j 1 \\ 2 \\ 1 - j 1 \\ 0 \\ 1 + j 1 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ :

\begin{aligned} 4 \\ 1 - j 2.4142 \\ 0 \\ 1 - j 0.4142 \\ 0 \\ 1 + j 0.4142 \\ 0 \\ 1 + j 2.4142 \end{aligned}

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	4	4	0
$X [1]$	1 - j2.4142	2.6131	-67.5
$X [2]$	0	0	0
$X [3]$	1 - j0.4142	1.0824	-22.5
$X [4]$	0	0	0
$X [5]$	1 + j0.4142	1.0824	22.5
$X [6]$	0	0	0
$X [7]$	1 + j2.4142	2.6131	67.5

$X [k]$ = {4, 1 - j2.4142, 0, 1 - j0.4142, 0, 1 + j0.4142, 0, 1 + j2.4142}. Two free checks: $X [0] = 4$ , which is the sum of the samples; $X [7] = X^{*} [1]$ and $X [6] = X^{*} [2]$ , the conjugate symmetry a real input forces.

Practise this2075 Chaitra · Q82+8 marksHow fast is FFT? Find X(3) and X(5) for given sequence

x [n] = {1, - 2, 3, 2}

k = 0

to 3. X(5) is read by periodicity:

X [k + N] = X [k]

, so X(5) = X(1). Saying that is the question.

What is different in this one

Decimation in time: 4 points, 2 stages of 2 butterflies, with the input in bit reversed order.
This is a 4 point transform, so there are only 2 stages of 2 butterflies and the only twiddles are $W 4^{0} = 1$ and $W 4^{1} = - j$ . Much shorter than the usual 8 point question.
The question asks for X(3) and X(5) of a FOUR point sequence, and a 4 point DFT only has $k = 0$ to 3. X(5) is read by periodicity: $X [k + N] = X [k]$ , so X(5) = X(1). Saying that is the question.

$N = 4$ , which is $2^{2}$ , so the transform takes 2 stages of 2 butterflies.

Decimation in time: the input goes in bit reversed order and the output comes out in natural order. For $N = 4$ that input order is $x [0]$ , $x [2]$ , $x [1]$ , $x [3]$ .

Stage 1, 2 butterflies with twiddles $W_{2}^{0}$ :

\begin{aligned} 4 \\ - 2 \\ 0 \\ - 4 \end{aligned}

Stage 2, 2 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ :

\begin{aligned} 4 \\ - 2 + j 4 \\ 4 \\ - 2 - j 4 \end{aligned}

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	4	4	0
$X [1]$	-2 + j4	4.4721	116.57
$X [2]$	4	4	0
$X [3]$	-2 - j4	4.4721	-116.57

$X [k]$ = {4, -2 + j4, 4, -2 - j4}. Two free checks: $X [0] = 4$ , which is the sum of the samples; $X [3] = X^{*} [1]$ and $X [2] = X^{*} [2]$ , the conjugate symmetry a real input forces.

Same as 2081 Bhadra Q92075 Ashwin · Q83+7 marksHow fast is FFT compare to DFT? Draw the butterfly diagram of 8-point DFT of a sequence as

x [n] = n

+ 1 using Decimation in Time FFT algorithm.

How this is read $x [n] = n$ + 1 over $n = 0$ to 7 is ${1, 2, 3, 4, 5, 6, 7, 8}$ .

What is different in this one

Decimation in time: 8 points, 3 stages of 4 butterflies, with the input in bit reversed order.

$N = 8$ , which is $2^{3}$ , so the transform takes 3 stages of 4 butterflies.

Stage 1, 4 butterflies with twiddles $W_{2}^{0}$ :

\begin{aligned} 6 \\ - 4 \\ 10 \\ - 4 \\ 8 \\ - 4 \\ 12 \\ - 4 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ :

\begin{aligned} 16 \\ - 4 + j 4 \\ - 4 \\ - 4 - j 4 \\ 20 \\ - 4 + j 4 \\ - 4 \\ - 4 - j 4 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ :

\begin{aligned} 36 \\ - 4 + j 9.6569 \\ - 4 + j 4 \\ - 4 + j 1.6569 \\ - 4 \\ - 4 - j 1.6569 \\ - 4 - j 4 \\ - 4 - j 9.6569 \end{aligned}

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	36	36	0
$X [1]$	-4 + j9.6569	10.4525	112.5
$X [2]$	-4 + j4	5.6569	135
$X [3]$	-4 + j1.6569	4.3296	157.5
$X [4]$	-4	4	180
$X [5]$	-4 - j1.6569	4.3296	-157.5
$X [6]$	-4 - j4	5.6569	-135
$X [7]$	-4 - j9.6569	10.4525	-112.5

Same as 2081 Bhadra Q92073 Shrawan · Q102+6 marksWhy do we need DFT? Find 8-point DFT of sequence

x [n] = {1, - 1, 2, 2, 1, 1, 2, 2}

using Fast Fourier Transform algorithm.

What is different in this one

Decimation in time: 8 points, 3 stages of 4 butterflies, with the input in bit reversed order.

$N = 8$ , which is $2^{3}$ , so the transform takes 3 stages of 4 butterflies.

Stage 1, 4 butterflies with twiddles $W_{2}^{0}$ :

\begin{aligned} 2 \\ 0 \\ 4 \\ 0 \\ 0 \\ - 2 \\ 4 \\ 0 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ :

\begin{aligned} 6 \\ 0 \\ - 2 \\ 0 \\ 4 \\ - 2 \\ - 4 \\ - 2 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ :

\begin{aligned} 10 \\ - 1.4142 + j 1.4142 \\ - 2 + j 4 \\ 1.4142 + j 1.4142 \\ 2 \\ 1.4142 - j 1.4142 \\ - 2 - j 4 \\ - 1.4142 - j 1.4142 \end{aligned}

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	10	10	0
$X [1]$	-1.4142 + j1.4142	2	135
$X [2]$	-2 + j4	4.4721	116.57
$X [3]$	1.4142 + j1.4142	2	45
$X [4]$	2	2	0
$X [5]$	1.4142 - j1.4142	2	-45
$X [6]$	-2 - j4	4.4721	-116.57
$X [7]$	-1.4142 - j1.4142	2	-135

Same as 2081 Bhadra Q92072 Chaitra · Q118 marksFind the FFT of the signal

x [n] {1, 1, 2, 4, 3, 1, 2, 1}

using DIT-FFT algorithm.

What is different in this one

Decimation in time: 8 points, 3 stages of 4 butterflies, with the input in bit reversed order.

$N = 8$ , which is $2^{3}$ , so the transform takes 3 stages of 4 butterflies.

Stage 1, 4 butterflies with twiddles $W_{2}^{0}$ :

\begin{aligned} 4 \\ - 2 \\ 4 \\ 0 \\ 2 \\ 0 \\ 5 \\ 3 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ :

\begin{aligned} 8 \\ - 2 \\ 0 \\ - 2 \\ 7 \\ - j 3 \\ - 3 \\ j3 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ :

\begin{aligned} 15 \\ - 4.1213 - j 2.1213 \\ j3 \\ 0.1213 - j 2.1213 \\ 1 \\ 0.1213 + j 2.1213 \\ - j 3 \\ - 4.1213 + j 2.1213 \end{aligned}

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	15	15	0
$X [1]$	-4.1213 - j2.1213	4.6352	-152.76
$X [2]$	j3	3	90
$X [3]$	0.1213 - j2.1213	2.1248	-86.73
$X [4]$	1	1	0
$X [5]$	0.1213 + j2.1213	2.1248	86.73
$X [6]$	-j3	3	-90
$X [7]$	-4.1213 + j2.1213	4.6352	152.76

$X [k]$ = {15, -4.1213 - j2.1213, j3, 0.1213 - j2.1213, 1, 0.1213 + j2.1213, -j3, -4.1213 + j2.1213}. Two free checks: $X [0] = 15$ , which is the sum of the samples; $X [7] = X^{*} [1]$ and $X [6] = X^{*} [2]$ , the conjugate symmetry a real input forces.

Same as 2081 Bhadra Q92072 Kartik · Q107 marksFind 8-point DFT of sequence

x [n] = {1, 1, 0, 1, 0, 1, 2}

using Decimation in Time Fast Fourier Transform (DITFFT) algorithm.

What is different in this one

Decimation in time: 8 points, 3 stages of 4 butterflies, with the input in bit reversed order.
The sequence has 7 samples, not a power of two, so it is zero padded to $N = 8$ before the algorithm starts. Say so in the answer: a radix 2 FFT cannot run on 7 points.

The sequence has 7 samples, and a radix 2 FFT needs a power of two, so zero pad to $N = 8$ and say so: $x [n] = {1, 1, 0, 1, 0, 1, 2, 0}$ .

Stage 1, 4 butterflies with twiddles $W_{2}^{0}$ :

\begin{aligned} 1 \\ 1 \\ 2 \\ - 2 \\ 2 \\ 0 \\ 1 \\ 1 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ :

\begin{aligned} 3 \\ 1 + j 2 \\ - 1 \\ 1 - j 2 \\ 3 \\ - j 1 \\ 1 \\ j1 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ :

\begin{aligned} 6 \\ 0.2929 + j 1.2929 \\ - 1 - j 1 \\ 1.7071 - j 2.7071 \\ 0 \\ 1.7071 + j 2.7071 \\ - 1 + j 1 \\ 0.2929 - j 1.2929 \end{aligned}

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	6	6	0
$X [1]$	0.2929 + j1.2929	1.3257	77.24
$X [2]$	-1 - j1	1.4142	-135
$X [3]$	1.7071 - j2.7071	3.2004	-57.76
$X [4]$	0	0	0
$X [5]$	1.7071 + j2.7071	3.2004	57.76
$X [6]$	-1 + j1	1.4142	135
$X [7]$	0.2929 - j1.2929	1.3257	-77.24

$X [k]$ = {6, 0.2929 + j1.2929, -1 - j1, 1.7071 - j2.7071, 0, 1.7071 + j2.7071, -1 + j1, 0.2929 - j1.2929}. Two free checks: $X [0] = 6$ , which is the sum of the samples; $X [7] = X^{*} [1]$ and $X [6] = X^{*} [2]$ , the conjugate symmetry a real input forces.

Same as 2081 Bhadra Q92071 Shrawan · Q106+2 marksFind DFT for

{1, 1, 2, 0, 1, 2, 0, 1}

using FFT DIT butterfly algorithm and plot the spectrum.

What is different in this one

Decimation in time: 8 points, 3 stages of 4 butterflies, with the input in bit reversed order.

$N = 8$ , which is $2^{3}$ , so the transform takes 3 stages of 4 butterflies.

Stage 1, 4 butterflies with twiddles $W_{2}^{0}$ :

\begin{aligned} 2 \\ 0 \\ 2 \\ 2 \\ 3 \\ - 1 \\ 1 \\ - 1 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ :

\begin{aligned} 4 \\ - j 2 \\ 0 \\ j2 \\ 4 \\ - 1 + j 1 \\ 2 \\ - 1 - j 1 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ :

\begin{aligned} 8 \\ - j 0.5858 \\ - j 2 \\ j3.4142 \\ 0 \\ - j 3.4142 \\ j2 \\ j0.5858 \end{aligned}

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	8	8	0
$X [1]$	-j0.5858	0.5858	-90
$X [2]$	-j2	2	-90
$X [3]$	j3.4142	3.4142	90
$X [4]$	0	0	0
$X [5]$	-j3.4142	3.4142	-90
$X [6]$	j2	2	90
$X [7]$	j0.5858	0.5858	90

The spectrum. The question asks for it plotted, so draw the magnitude against k. It is symmetric about the middle for a real input, which is the check.

X [k]

|, drawn from the computed samples.

$X [k]$ = {8, -j0.5858, -j2, j3.4142, 0, -j3.4142, j2, j0.5858}. Two free checks: $X [0] = 8$ , which is the sum of the samples; $X [7] = X^{*} [1]$ and $X [6] = X^{*} [2]$ , the conjugate symmetry a real input forces.

The same questions grouped by how often each topic is set: the fft, decimation in time

Decimation in frequency TOP 8/19

82 Bh · 82 Ba · 80 Bh · 79 Ba · 76 Ch · 76 Ash · 74 Ch · 74 Ash781+2+6

Before the formulas

The other way to split: by output rather than by input. Decimation in frequency splits the output points into the even numbered and odd numbered ones, and everything ends up mirrored against the previous card.

The input goes in in natural order and the output comes out bit reversed. The butterfly itself differs from the decimation in time one only in where the twiddle factor sits, before the subtraction rather than after it. Knowing that one difference is usually enough to answer either question.

The split Divide the output, that is the frequency index k, into even and odd. The input is split into its first half and its second half instead, and the twiddle factor is applied after the subtraction.

\begin{aligned} X (k) & = \sum_{n = 0}^{N / 2 - 1} x [n] W_{N}^{n k} + \sum_{n = N / 2}^{N - 1} x [n] W_{N}^{n k} \\ = \sum_{n = 0}^{N / 2 - 1} (x [n] + (- 1)^{k} x [n + \frac{N}{2}]) W_{N}^{n k} & put n = m + \frac{N}{2}, W_{N}^{N k / 2} = (- 1)^{k} \\ X (2 r) & = \sum_{n} (x [n] + x [n + \frac{N}{2}]) W_{N / 2}^{r n} & even k \\ X (2 r + 1) & = \sum_{n} ((x [n] - x [n + \frac{N}{2}]) W_{N}^{n}) W_{N / 2}^{r n} & odd k \end{aligned}

The procedure for an 8 point DIF FFT

Input in natural order, $x [0]$ to $x [7]$ .
Stage 1: butterflies between $x [n]$ and $x [n + 4]$ , twiddles $W^{0}$ to $W^{3}$ on the differences.
Stage 2: butterflies spanning two, twiddles $W^{0}$ and $W^{2}$ .
Stage 3: butterflies on adjacent pairs, twiddle $W^{0}$ .
The output comes out bit reversed: X(0), X(4), X(2), X(6), X(1), X(5), X(3), X(7). Re order it before writing the answer.

	DIT	DIF
Splits	The input sequence, into even and odd n	The output, into even and odd k
Input order	Bit reversed	Natural
Output order	Natural	Bit reversed
Butterfly	Twiddle before the add and subtract	Twiddle after the subtraction
Operation count	Identical: (N/2)log2(N) multiplications, N log2(N) additions
In place	Both: each butterfly overwrites its own two values

In the exam

Read which algorithm is asked for and get the ordering right. Handing in a DIT diagram for a DIF question loses most of the marks even if the numbers are correct.
Pad to 8 points. A sequence of 5 or 6 samples in an 8 point question is padded with zeros; say so.
Draw the diagram. The numbers alone are worth little; the butterfly structure with its twiddles is the answer.
Check X(0). It is always the sum of all the samples, and it costs nothing to verify.
Check Parseval if there is time: sum $| x [n] |^{2} = \frac{1}{N}$ sum |X(k)|^2.

It is the decimation in time algorithm with the twiddle moved to the other side of the subtraction and the bit reversal at the other end.

See it drawnThe same saving, split by output instead of by input

Everything ends up mirrored: the input goes in naturally ordered and the output comes out bit reversed. The butterfly differs only in where the twiddle factor sits.

Practise thisEvery question the paper has set on this topic, worked in full8

Practise this +5 like it2082 Bhadra · Q118 marksFind 8-point DFT using DIF-FFT of the sequence

x [n]

= {1,

\frac{1}{2}

, -1,

- \frac{1}{2}

, 2, -3/2}.What is different heredecimation in frequency, where the input stays in order and the output comes out bit reversed

What is different in this one

Decimation in frequency: 8 points, 3 stages of 4 butterflies, with the output in bit reversed order.
The sequence has 6 samples, not a power of two, so it is zero padded to $N = 8$ before the algorithm starts. Say so in the answer: a radix 2 FFT cannot run on 6 points.

The sequence has 6 samples, and a radix 2 FFT needs a power of two, so zero pad to $N = 8$ and say so: $x [n] = {1, 0.5, - 1, - 0.5, 2, - 1.5, 0, 0}$ .

Decimation in frequency: the input stays in natural order and the output comes out bit reversed, so the last step is to relabel it.

Stage 1, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ applied to the differences:

\begin{aligned} 3 \\ - 1 \\ - 1 \\ - 0.5 \\ - 1 \\ 1.4142 - j 1.4142 \\ j1 \\ 0.3536 + j 0.3536 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ applied to the differences:

\begin{aligned} 2 \\ - 1.5 \\ 4 \\ j0.5 \\ - 1 + j 1 \\ 1.7678 - j 1.0607 \\ - 1 - j 1 \\ - 1.7678 - j 1.0607 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{2}^{0}$ applied to the differences:

\begin{aligned} 0.5 \\ 3.5 \\ 4 + j 0.5 \\ 4 - j 0.5 \\ 0.7678 - j 0.0607 \\ - 2.7678 + j 2.0607 \\ - 2.7678 - j 2.0607 \\ 0.7678 + j 0.0607 \end{aligned}

Read the output in bit reversed order: $X [0]$ , $X [4]$ , $X [2]$ , $X [6]$ , $X [1]$ , $X [5]$ , $X [3]$ , $X [7]$ .

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	0.5	0.5	0
$X [1]$	0.7678 - j0.0607	0.7702	-4.52
$X [2]$	4 + j0.5	4.0311	7.13
$X [3]$	-2.7678 - j2.0607	3.4506	-143.33
$X [4]$	3.5	3.5	0
$X [5]$	-2.7678 + j2.0607	3.4506	143.33
$X [6]$	4 - j0.5	4.0311	-7.13
$X [7]$	0.7678 + j0.0607	0.7702	4.52

$X [k]$ = {0.5, 0.7678 - j0.0607, 4 + j0.5, -2.7678 - j2.0607, 3.5, -2.7678 + j2.0607, 4 - j0.5, 0.7678 + j0.0607}. Two free checks: $X [0] = 0.5$ , which is the sum of the samples; $X [7] = X^{*} [1]$ and $X [6] = X^{*} [2]$ , the conjugate symmetry a real input forces.

x [n] = {1, 4, - 1}

W 4^{0} = 1

and

W 4^{1} = - j

. Most sittings set the 8 point version.

What is different in this one

Decimation in frequency: 4 points, 2 stages of 2 butterflies, with the output in bit reversed order.
The sequence has 3 samples, not a power of two, so it is zero padded to $N = 4$ before the algorithm starts. Say so in the answer: a radix 2 FFT cannot run on 3 points.
This is a 4 point transform, so there are only 2 stages of 2 butterflies and the only twiddles are $W 4^{0} = 1$ and $W 4^{1} = - j$ . Much shorter than the usual 8 point question.
Only 4 points, so there are two stages of two butterflies and the only twiddles are $W 4^{0} = 1$ and $W 4^{1} = - j$ . Most sittings set the 8 point version.

The sequence has 3 samples, and a radix 2 FFT needs a power of two, so zero pad to $N = 4$ and say so: $x [n] = {1, 4, - 1, 0}$ .

Decimation in frequency: the input stays in natural order and the output comes out bit reversed, so the last step is to relabel it.

Stage 1, 2 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ applied to the differences:

\begin{aligned} 0 \\ 4 \\ 2 \\ - j 4 \end{aligned}

Stage 2, 2 butterflies with twiddles $W_{2}^{0}$ applied to the differences:

\begin{aligned} 4 \\ - 4 \\ 2 - j 4 \\ 2 + j 4 \end{aligned}

Read the output in bit reversed order: $X [0]$ , $X [2]$ , $X [1]$ , $X [3]$ .

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	4	4	0
$X [1]$	2 - j4	4.4721	-63.43
$X [2]$	-4	4	180
$X [3]$	2 + j4	4.4721	63.43

$X [k]$ = {4, 2 - j4, -4, 2 + j4}. Two free checks: $X [0] = 4$ , which is the sum of the samples; $X [3] = X^{*} [1]$ and $X [2] = X^{*} [2]$ , the conjugate symmetry a real input forces.

Same as 2082 Bhadra Q112080 Bhadra · Q98 marksCompute 8-point DIF-FFT of sequence

x (n) = {2, 1, 2, 1, 1, 2, 1, 2}

What is different in this one

Decimation in frequency: 8 points, 3 stages of 4 butterflies, with the output in bit reversed order.

$N = 8$ , which is $2^{3}$ , so the transform takes 3 stages of 4 butterflies.

Decimation in frequency: the input stays in natural order and the output comes out bit reversed, so the last step is to relabel it.

Stage 1, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ applied to the differences:

\begin{aligned} 3 \\ 3 \\ 3 \\ 3 \\ 1 \\ - 0.7071 + j 0.7071 \\ - j 1 \\ 0.7071 + j 0.7071 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ applied to the differences:

\begin{aligned} 6 \\ 6 \\ 0 \\ 0 \\ 1 - j 1 \\ j1.4142 \\ 1 + j 1 \\ j1.4142 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{2}^{0}$ applied to the differences:

\begin{aligned} 12 \\ 0 \\ 0 \\ 0 \\ 1 + j 0.4142 \\ 1 - j 2.4142 \\ 1 + j 2.4142 \\ 1 - j 0.4142 \end{aligned}

Read the output in bit reversed order: $X [0]$ , $X [4]$ , $X [2]$ , $X [6]$ , $X [1]$ , $X [5]$ , $X [3]$ , $X [7]$ .

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	12	12	0
$X [1]$	1 + j0.4142	1.0824	22.5
$X [2]$	0	0	0
$X [3]$	1 + j2.4142	2.6131	67.5
$X [4]$	0	0	0
$X [5]$	1 - j2.4142	2.6131	-67.5
$X [6]$	0	0	0
$X [7]$	1 - j0.4142	1.0824	-22.5

$X [k]$ = {12, 1 + j0.4142, 0, 1 + j2.4142, 0, 1 - j2.4142, 0, 1 - j0.4142}. Two free checks: $X [0] = 12$ , which is the sum of the samples; $X [7] = X^{*} [1]$ and $X [6] = X^{*} [2]$ , the conjugate symmetry a real input forces.

Same as 2082 Bhadra Q112079 Baishakh · Q102+6 marksWhy we need DFT? Find 8-point DFT of sequence

x [n] = {1, 2, 4, 3, 5, - 1, 3}

using Decimation in Frequency Fast Fourier Transform (DIFFFT) algorithm.

What is different in this one

Decimation in frequency: 8 points, 3 stages of 4 butterflies, with the output in bit reversed order.
The sequence has 7 samples, not a power of two, so it is zero padded to $N = 8$ before the algorithm starts. Say so in the answer: a radix 2 FFT cannot run on 7 points.

The sequence has 7 samples, and a radix 2 FFT needs a power of two, so zero pad to $N = 8$ and say so: $x [n] = {1, 2, 4, 3, 5, - 1, 3, 0}$ .

Decimation in frequency: the input stays in natural order and the output comes out bit reversed, so the last step is to relabel it.

Stage 1, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ applied to the differences:

\begin{aligned} 6 \\ 1 \\ 7 \\ 3 \\ - 4 \\ 2.1213 - j 2.1213 \\ - j 1 \\ - 2.1213 - j 2.1213 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ applied to the differences:

\begin{aligned} 13 \\ 4 \\ - 1 \\ j2 \\ - 4 - j 1 \\ - j 4.2426 \\ - 4 + j 1 \\ - j 4.2426 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{2}^{0}$ applied to the differences:

\begin{aligned} 17 \\ 9 \\ - 1 + j 2 \\ - 1 - j 2 \\ - 4 - j 5.2426 \\ - 4 + j 3.2426 \\ - 4 - j 3.2426 \\ - 4 + j 5.2426 \end{aligned}

Read the output in bit reversed order: $X [0]$ , $X [4]$ , $X [2]$ , $X [6]$ , $X [1]$ , $X [5]$ , $X [3]$ , $X [7]$ .

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	17	17	0
$X [1]$	-4 - j5.2426	6.5943	-127.34
$X [2]$	-1 + j2	2.2361	116.57
$X [3]$	-4 - j3.2426	5.1492	-140.97
$X [4]$	9	9	0
$X [5]$	-4 + j3.2426	5.1492	140.97
$X [6]$	-1 - j2	2.2361	-116.57
$X [7]$	-4 + j5.2426	6.5943	127.34

Same as 2082 Bhadra Q112076 Chaitra · Q117 marksFind 8-point DFT of sequence

x [n] = {1, 2, 3, 3, 5, 0, 4, 6}

using Decimation in frequency Fast Fourier Transform (DIFFFT) algorithm.

What is different in this one

Decimation in frequency: 8 points, 3 stages of 4 butterflies, with the output in bit reversed order.

$N = 8$ , which is $2^{3}$ , so the transform takes 3 stages of 4 butterflies.

Decimation in frequency: the input stays in natural order and the output comes out bit reversed, so the last step is to relabel it.

Stage 1, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ applied to the differences:

\begin{aligned} 6 \\ 2 \\ 7 \\ 9 \\ - 4 \\ 1.4142 - j 1.4142 \\ j1 \\ 2.1213 + j 2.1213 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ applied to the differences:

\begin{aligned} 13 \\ 11 \\ - 1 \\ j7 \\ - 4 + j 1 \\ 3.5355 + j 0.7071 \\ - 4 - j 1 \\ - 3.5355 + j 0.7071 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{2}^{0}$ applied to the differences:

\begin{aligned} 24 \\ 2 \\ - 1 + j 7 \\ - 1 - j 7 \\ - 0.4645 + j 1.7071 \\ - 7.5355 + j 0.2929 \\ - 7.5355 - j 0.2929 \\ - 0.4645 - j 1.7071 \end{aligned}

Read the output in bit reversed order: $X [0]$ , $X [4]$ , $X [2]$ , $X [6]$ , $X [1]$ , $X [5]$ , $X [3]$ , $X [7]$ .

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	24	24	0
$X [1]$	-0.4645 + j1.7071	1.7692	105.22
$X [2]$	-1 + j7	7.0711	98.13
$X [3]$	-7.5355 - j0.2929	7.5412	-177.77
$X [4]$	2	2	0
$X [5]$	-7.5355 + j0.2929	7.5412	177.77
$X [6]$	-1 - j7	7.0711	-98.13
$X [7]$	-0.4645 - j1.7071	1.7692	-105.22

$X [k]$ = {24, -0.4645 + j1.7071, -1 + j7, -7.5355 - j0.2929, 2, -7.5355 + j0.2929, -1 - j7, -0.4645 - j1.7071}. Two free checks: $X [0] = 24$ , which is the sum of the samples; $X [7] = X^{*} [1]$ and $X [6] = X^{*} [2]$ , the conjugate symmetry a real input forces.

Same as 2082 Bhadra Q112076 Ashwin · Q82+8 marksWhy we need DFT? Find 8-point DFT of sequence

x [n] = {1, 2, 3, 3, 5, 1, 4, 2}

using Decimation in frequency Fast Fourier Transform (DIFFFT) algorithm.

What is different in this one

Decimation in frequency: 8 points, 3 stages of 4 butterflies, with the output in bit reversed order.

$N = 8$ , which is $2^{3}$ , so the transform takes 3 stages of 4 butterflies.

Decimation in frequency: the input stays in natural order and the output comes out bit reversed, so the last step is to relabel it.

Stage 1, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ applied to the differences:

\begin{aligned} 6 \\ 3 \\ 7 \\ 5 \\ - 4 \\ 0.7071 - j 0.7071 \\ j1 \\ - 0.7071 - j 0.7071 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ applied to the differences:

\begin{aligned} 13 \\ 8 \\ - 1 \\ j2 \\ - 4 + j 1 \\ - j 1.4142 \\ - 4 - j 1 \\ - j 1.4142 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{2}^{0}$ applied to the differences:

\begin{aligned} 21 \\ 5 \\ - 1 + j 2 \\ - 1 - j 2 \\ - 4 - j 0.4142 \\ - 4 + j 2.4142 \\ - 4 - j 2.4142 \\ - 4 + j 0.4142 \end{aligned}

Read the output in bit reversed order: $X [0]$ , $X [4]$ , $X [2]$ , $X [6]$ , $X [1]$ , $X [5]$ , $X [3]$ , $X [7]$ .

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	21	21	0
$X [1]$	-4 - j0.4142	4.0214	-174.09
$X [2]$	-1 + j2	2.2361	116.57
$X [3]$	-4 - j2.4142	4.6721	-148.89
$X [4]$	5	5	0
$X [5]$	-4 + j2.4142	4.6721	148.89
$X [6]$	-1 - j2	2.2361	-116.57
$X [7]$	-4 + j0.4142	4.0214	174.09

Practise this2074 Chaitra · Q93+7 marksWhy do we need DFT? Draw the butterfly structure to compute the DFT of the following signal using Radix-2 DIFFFT algorithm, and compute X(2) and X(1) only

x [n] = {1.5, - 1, 1.8, 0.6, 3, 1.7}

What is different in this one

Decimation in frequency: 8 points, 3 stages of 4 butterflies, with the output in bit reversed order.
The sequence has 6 samples, not a power of two, so it is zero padded to $N = 8$ before the algorithm starts. Say so in the answer: a radix 2 FFT cannot run on 6 points.
The question asks only for X(2) and X(1), but the butterfly diagram has to be drawn in full to reach them, so the whole transform is here.

The sequence has 6 samples, and a radix 2 FFT needs a power of two, so zero pad to $N = 8$ and say so: $x [n] = {1.5, - 1, 1.8, 0.6, 3, 1.7, 0, 0}$ .

Decimation in frequency: the input stays in natural order and the output comes out bit reversed, so the last step is to relabel it.

Stage 1, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ applied to the differences:

\begin{aligned} 4.5 \\ 0.7 \\ 1.8 \\ 0.6 \\ - 1.5 \\ - 1.9092 + j 1.9092 \\ - j 1.8 \\ - 0.4243 - j 0.4243 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ applied to the differences:

\begin{aligned} 6.3 \\ 1.3 \\ 2.7 \\ - j 0.1 \\ - 1.5 - j 1.8 \\ - 2.3335 + j 1.4849 \\ - 1.5 + j 1.8 \\ 2.3335 + j 1.4849 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{2}^{0}$ applied to the differences:

\begin{aligned} 7.6 \\ 5 \\ 2.7 - j 0.1 \\ 2.7 + j 0.1 \\ - 3.8335 - j 0.3151 \\ 0.8335 - j 3.2849 \\ 0.8335 + j 3.2849 \\ - 3.8335 + j 0.3151 \end{aligned}

Read the output in bit reversed order: $X [0]$ , $X [4]$ , $X [2]$ , $X [6]$ , $X [1]$ , $X [5]$ , $X [3]$ , $X [7]$ .

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	7.6	7.6	0
$X [1]$	-3.8335 - j0.3151	3.8464	-175.3
$X [2]$	2.7 - j0.1	2.7019	-2.12
$X [3]$	0.8335 + j3.2849	3.389	75.76
$X [4]$	5	5	0
$X [5]$	0.8335 - j3.2849	3.389	-75.76
$X [6]$	2.7 + j0.1	2.7019	2.12
$X [7]$	-3.8335 + j0.3151	3.8464	175.3

Same as 2082 Bhadra Q112074 Ashwin · Q97 marksCompute the 8-point DFT of the sequence

x [n]

= {

\frac{1}{2}

\frac{1}{2}

\frac{1}{2}

\frac{1}{2}

, 0, 0, 0, 0} using Decimation in Frequency Fast Fourier Transform (DIF-FFT) algorithm.

What is different in this one

Decimation in frequency: 8 points, 3 stages of 4 butterflies, with the output in bit reversed order.

$N = 8$ , which is $2^{3}$ , so the transform takes 3 stages of 4 butterflies.

Decimation in frequency: the input stays in natural order and the output comes out bit reversed, so the last step is to relabel it.

Stage 1, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ applied to the differences:

\begin{aligned} 0.5 \\ 0.5 \\ 0.5 \\ 0.5 \\ 0.5 \\ 0.3536 - j 0.3536 \\ - j 0.5 \\ - 0.3536 - j 0.3536 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ applied to the differences:

\begin{aligned} 1 \\ 1 \\ 0 \\ 0 \\ 0.5 - j 0.5 \\ - j 0.7071 \\ 0.5 + j 0.5 \\ - j 0.7071 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{2}^{0}$ applied to the differences:

\begin{aligned} 2 \\ 0 \\ 0 \\ 0 \\ 0.5 - j 1.2071 \\ 0.5 + j 0.2071 \\ 0.5 - j 0.2071 \\ 0.5 + j 1.2071 \end{aligned}

Read the output in bit reversed order: $X [0]$ , $X [4]$ , $X [2]$ , $X [6]$ , $X [1]$ , $X [5]$ , $X [3]$ , $X [7]$ .

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	2	2	0
$X [1]$	0.5 - j1.2071	1.3066	-67.5
$X [2]$	0	0	0
$X [3]$	0.5 - j0.2071	0.5412	-22.5
$X [4]$	0	0	0
$X [5]$	0.5 + j0.2071	0.5412	22.5
$X [6]$	0	0	0
$X [7]$	0.5 + j1.2071	1.3066	67.5

$X [k]$ = {2, 0.5 - j1.2071, 0, 0.5 - j0.2071, 0, 0.5 + j0.2071, 0, 0.5 + j1.2071}. Two free checks: $X [0] = 2$ , which is the sum of the samples; $X [7] = X^{*} [1]$ and $X [6] = X^{*} [2]$ , the conjugate symmetry a real input forces.

The same questions grouped by how often each topic is set: the fft, decimation in frequency

7.3Computational complexity of the FFT

The computational complexity of the FFT HOT 7/19

82 Ba · 81 Bh · 81 Ba · 80 Ba · 79 Bh · 75 Ch · 75 Ash2+61+2+62+5

Before the formulas

The last card is the arithmetic that justifies the whole chapter, and it is pure counting. There are log to base two of N stages. Each stage holds N over two butterflies. Each butterfly is one complex multiplication and two complex additions.

Multiply those together and compare with N squared. Put real numbers in and the point makes itself: the saving is not a fixed factor, it grows as N grows. Two or three marks, and no cleverness required.

\begin{aligned} direct DFT: & N^{2} multiplications, & N (N - 1) additions \\ radix 2 FFT: & \frac{N}{2} \log_{2} N, & N \log_{2} N \\ speed up & = \frac{N^{2}}{\frac{N}{2} \log_{2} N} = \frac{2 N}{\log_{2} N} \end{aligned}

N	Stages, log2 N	Butterflies, (N/2)log2 N	DFT multiplies	Speed up
8	3	12	64	5.3
16	4	32	256	8
64	6	192	4,096	21.3
256	8	1,024	65,536	64
1,024	10	5,120	1,048,576	204.8
4,096	12	24,576	16,777,216	682.7

Where the count comes from. Each stage has N/2 butterflies, each butterfly costs one complex multiplication and two complex additions, and there are log2 N stages because the sequence can be halved that many times.
In place computation. A butterfly reads two values and writes two values to the same two locations, so the whole transform runs in one array of N complex numbers. No extra memory is needed, which mattered enormously when the algorithm was published and still matters on a small processor.
Bit reversal is the price of being in place. One end of the transform must be in bit reversed order: the input for DIT, the output for DIF. Reversing the index bits is a cheap shuffle, and it is not counted in the arithmetic above.
Real arithmetic, if the question asks. One complex multiplication is 4 real multiplications and 2 real additions, so an N point radix 2 FFT costs about 2N log2 N real multiplications.
The count can be reduced further. The twiddle factors 1, -1, j and -j need no multiplication at all, which removes a good fraction of the count in practice; radix 4 and split radix algorithms cut it further; and a real input halves the work by transforming two real sequences at once.
Fast convolution. Convolving two N point sequences directly costs $N^{2}$ multiplications; through the FFT it costs three transforms and N multiplications, that is about 3(N/2)log2 N + N, which is cheaper for N above roughly 64. This is the reason long filters are run in the frequency domain.

The saving is what makes fast convolution worth doing, and therefore what makes a long FIR filter practical at all.

See it drawnThe saving is not a fixed factor: it grows with N

Count the butterflies. Log two of N stages, N over two in each, one complex multiplication apiece. Put numbers in and the gap widens with every doubling.

Practise thisEvery question the paper has set on this topic, worked in full8

x [n] = {1, 4, - 1}

W 4^{0} = 1

and

W 4^{1} = - j

. Most sittings set the 8 point version.

What is different in this one

Decimation in frequency: 4 points, 2 stages of 2 butterflies, with the output in bit reversed order.
The sequence has 3 samples, not a power of two, so it is zero padded to $N = 4$ before the algorithm starts. Say so in the answer: a radix 2 FFT cannot run on 3 points.
This is a 4 point transform, so there are only 2 stages of 2 butterflies and the only twiddles are $W 4^{0} = 1$ and $W 4^{1} = - j$ . Much shorter than the usual 8 point question.
Only 4 points, so there are two stages of two butterflies and the only twiddles are $W 4^{0} = 1$ and $W 4^{1} = - j$ . Most sittings set the 8 point version.

The sequence has 3 samples, and a radix 2 FFT needs a power of two, so zero pad to $N = 4$ and say so: $x [n] = {1, 4, - 1, 0}$ .

Decimation in frequency: the input stays in natural order and the output comes out bit reversed, so the last step is to relabel it.

Stage 1, 2 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ applied to the differences:

\begin{aligned} 0 \\ 4 \\ 2 \\ - j 4 \end{aligned}

Stage 2, 2 butterflies with twiddles $W_{2}^{0}$ applied to the differences:

\begin{aligned} 4 \\ - 4 \\ 2 - j 4 \\ 2 + j 4 \end{aligned}

Read the output in bit reversed order: $X [0]$ , $X [2]$ , $X [1]$ , $X [3]$ .

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	4	4	0
$X [1]$	2 - j4	4.4721	-63.43
$X [2]$	-4	4	180
$X [3]$	2 + j4	4.4721	63.43

$X [k]$ = {4, 2 - j4, -4, 2 + j4}. Two free checks: $X [0] = 4$ , which is the sum of the samples; $X [3] = X^{*} [1]$ and $X [2] = X^{*} [2]$ , the conjugate symmetry a real input forces.

Practise this +8 like it2081 Bhadra · Q92+7 marksHow does FFT reduce computational complexity compared to direct calculation of DFT? Compute 8-point DFT of

x [n] = {1, - 2, - 4, 3, 0, 1}

using DIT-FFT algorithm.

What is different in this one

Decimation in time: 8 points, 3 stages of 4 butterflies, with the input in bit reversed order.
The sequence has 6 samples, not a power of two, so it is zero padded to $N = 8$ before the algorithm starts. Say so in the answer: a radix 2 FFT cannot run on 6 points.

The sequence has 6 samples, and a radix 2 FFT needs a power of two, so zero pad to $N = 8$ and say so: $x [n] = {1, - 2, - 4, 3, 0, 1, 0, 0}$ .

Stage 1, 4 butterflies with twiddles $W_{2}^{0}$ :

\begin{aligned} 1 \\ 1 \\ - 4 \\ - 4 \\ - 1 \\ - 3 \\ 3 \\ 3 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ :

\begin{aligned} - 3 \\ 1 + j 4 \\ 5 \\ 1 - j 4 \\ 2 \\ - 3 - j 3 \\ - 4 \\ - 3 + j 3 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ :

\begin{aligned} - 1 \\ - 3.2426 + j 4 \\ 5 + j 4 \\ 5.2426 - j 4 \\ - 5 \\ 5.2426 + j 4 \\ 5 - j 4 \\ - 3.2426 - j 4 \end{aligned}

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	-1	1	180
$X [1]$	-3.2426 + j4	5.1492	129.03
$X [2]$	5 + j4	6.4031	38.66
$X [3]$	5.2426 - j4	6.5943	-37.34
$X [4]$	-5	5	180
$X [5]$	5.2426 + j4	6.5943	37.34
$X [6]$	5 - j4	6.4031	-38.66
$X [7]$	-3.2426 - j4	5.1492	-129.03

Practise this2081 Baishakh · Q92+6 marksWhy Decimation in Time Fast Fourier Transform (DITFFT) Algorithm is better than direct computation of DFT? Find 4 point DFT of the sequence

x (n) = {2, 2, 4}

using DITFFT algorithm.What is different hereonly four points, so two stages, and the only twiddles are 1 and -j

What is different in this one

Decimation in time: 4 points, 2 stages of 2 butterflies, with the input in bit reversed order.
The sequence has 3 samples, not a power of two, so it is zero padded to $N = 4$ before the algorithm starts. Say so in the answer: a radix 2 FFT cannot run on 3 points.
This is a 4 point transform, so there are only 2 stages of 2 butterflies and the only twiddles are $W 4^{0} = 1$ and $W 4^{1} = - j$ . Much shorter than the usual 8 point question.

The sequence has 3 samples, and a radix 2 FFT needs a power of two, so zero pad to $N = 4$ and say so: $x [n] = {2, 2, 4, 0}$ .

Decimation in time: the input goes in bit reversed order and the output comes out in natural order. For $N = 4$ that input order is $x [0]$ , $x [2]$ , $x [1]$ , $x [3]$ .

Stage 1, 2 butterflies with twiddles $W_{2}^{0}$ :

\begin{aligned} 6 \\ - 2 \\ 2 \\ 2 \end{aligned}

Stage 2, 2 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ :

\begin{aligned} 8 \\ - 2 - j 2 \\ 4 \\ - 2 + j 2 \end{aligned}

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	8	8	0
$X [1]$	-2 - j2	2.8284	-135
$X [2]$	4	4	0
$X [3]$	-2 + j2	2.8284	135

$X [k]$ = {8, -2 - j2, 4, -2 + j2}. Two free checks: $X [0] = 8$ , which is the sum of the samples; $X [3] = X^{*} [1]$ and $X [2] = X^{*} [2]$ , the conjugate symmetry a real input forces.

Same as 2081 Bhadra Q92080 Baishakh · Q112+6 marksHow fast is FFT? Find 8-point DFT of sequence

x [n] = {1, 1, 0, 0, 1, 1, 2}

using Decimation in Time Fast Fourier Transform (DITFFT) algorithm.

What is different in this one

Decimation in time: 8 points, 3 stages of 4 butterflies, with the input in bit reversed order.
The sequence has 7 samples, not a power of two, so it is zero padded to $N = 8$ before the algorithm starts. Say so in the answer: a radix 2 FFT cannot run on 7 points.

The sequence has 7 samples, and a radix 2 FFT needs a power of two, so zero pad to $N = 8$ and say so: $x [n] = {1, 1, 0, 0, 1, 1, 2, 0}$ .

Stage 1, 4 butterflies with twiddles $W_{2}^{0}$ :

\begin{aligned} 2 \\ 0 \\ 2 \\ - 2 \\ 2 \\ 0 \\ 0 \\ 0 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ :

\begin{aligned} 4 \\ j2 \\ 0 \\ - j 2 \\ 2 \\ 0 \\ 2 \\ 0 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ :

\begin{aligned} 6 \\ j2 \\ - j 2 \\ - j 2 \\ 2 \\ j2 \\ j2 \\ - j 2 \end{aligned}

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	6	6	0
$X [1]$	j2	2	90
$X [2]$	-j2	2	-90
$X [3]$	-j2	2	-90
$X [4]$	2	2	0
$X [5]$	j2	2	90
$X [6]$	j2	2	90
$X [7]$	-j2	2	-90

Same as 2081 Baishakh Q102080 Baishakh · Q122+5 marksWrite the complexity of DFT and FFT? Obtain the circular convolution of the following sequences:

X_{1} [n] = {1, 2, 3, 1}

and

X_{2} [n] = {4, 3, 2, 2}

What is different in this one

$N = 4$ is smaller than $N_{1} + N_{2} - 1 = 7$ , so this circular answer is not the linear convolution: the last 3 samples have wrapped round and added to the front. That wrap is the whole point of the question.

The DFT length is $N = 4$ . The circular convolution is $y [n] = \sum_{m} x_{1} [m] x_{2}$ [(n - m) mod N], and the wrap is what makes it circular rather than linear.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=1	$x_{1}$ [1]=2	$x_{1}$ [2]=3	$x_{1}$ [3]=1	sum
$y [0]$	4	2	2	3	17
$y [1]$	3	4	2	2	19
$y [2]$	2	3	4	2	22
$y [3]$	2	2	3	4	19

y [0]

= 1(4) + 2(2) + 3(2) + 1(3) = 17

y [1]

= 1(3) + 2(4) + 3(2) + 1(2) = 19

y [2]

= 1(2) + 2(3) + 3(4) + 1(2) = 22

y [3]

= 1(2) + 2(2) + 3(3) + 1(4) = 19

$y [n] = {17, 19, 22, 19}$ . The check holds: 7 × 11 = 77, the sum of y.

Same as 2081 Bhadra Q92079 Bhadra · Q102+6 marksWhy we need FFT? Find the 8-point DFT of the following sequence using radix-2 DITFFT algorithm.

What is different in this one

Decimation in time: 8 points, 3 stages of 4 butterflies, with the input in bit reversed order.

$N = 8$ , which is $2^{3}$ , so the transform takes 3 stages of 4 butterflies.

Stage 1, 4 butterflies with twiddles $W_{2}^{0}$ :

\begin{aligned} 1 \\ 1 \\ 3 - j 1 \\ 3 + j 1 \\ j2 \\ j2 \\ j4 \\ j4 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ :

\begin{aligned} 4 - j 1 \\ 2 - j 3 \\ - 2 + j 1 \\ j3 \\ j6 \\ 4 + j 2 \\ - j 2 \\ - 4 + j 2 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ :

\begin{aligned} 4 + j 5 \\ 6.2426 - j 4.4142 \\ - 4 + j 1 \\ 4.2426 + j 4.4142 \\ 4 - j 7 \\ - 2.2426 - j 1.5858 \\ j1 \\ - 4.2426 + j 1.5858 \end{aligned}

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	4 + j5	6.4031	51.34
$X [1]$	6.2426 - j4.4142	7.6456	-35.26
$X [2]$	-4 + j1	4.1231	165.96
$X [3]$	4.2426 + j4.4142	6.1225	46.14
$X [4]$	4 - j7	8.0623	-60.26
$X [5]$	-2.2426 - j1.5858	2.7467	-144.74
$X [6]$	j1	1	90
$X [7]$	-4.2426 + j1.5858	4.5293	159.51

$X [k]$ = {4 + j5, 6.2426 - j4.4142, -4 + j1, 4.2426 + j4.4142, 4 - j7, -2.2426 - j1.5858, j1, -4.2426 + j1.5858}. Two free checks: $X [0] = 4$ + j5, which is the sum of the samples.

Practise this2075 Chaitra · Q82+8 marksHow fast is FFT? Find X(3) and X(5) for given sequence

x [n] = {1, - 2, 3, 2}

k = 0

to 3. X(5) is read by periodicity:

X [k + N] = X [k]

, so X(5) = X(1). Saying that is the question.

What is different in this one

Decimation in time: 4 points, 2 stages of 2 butterflies, with the input in bit reversed order.
This is a 4 point transform, so there are only 2 stages of 2 butterflies and the only twiddles are $W 4^{0} = 1$ and $W 4^{1} = - j$ . Much shorter than the usual 8 point question.
The question asks for X(3) and X(5) of a FOUR point sequence, and a 4 point DFT only has $k = 0$ to 3. X(5) is read by periodicity: $X [k + N] = X [k]$ , so X(5) = X(1). Saying that is the question.

$N = 4$ , which is $2^{2}$ , so the transform takes 2 stages of 2 butterflies.

Decimation in time: the input goes in bit reversed order and the output comes out in natural order. For $N = 4$ that input order is $x [0]$ , $x [2]$ , $x [1]$ , $x [3]$ .

Stage 1, 2 butterflies with twiddles $W_{2}^{0}$ :

\begin{aligned} 4 \\ - 2 \\ 0 \\ - 4 \end{aligned}

Stage 2, 2 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ :

\begin{aligned} 4 \\ - 2 + j 4 \\ 4 \\ - 2 - j 4 \end{aligned}

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	4	4	0
$X [1]$	-2 + j4	4.4721	116.57
$X [2]$	4	4	0
$X [3]$	-2 - j4	4.4721	-116.57

$X [k]$ = {4, -2 + j4, 4, -2 - j4}. Two free checks: $X [0] = 4$ , which is the sum of the samples; $X [3] = X^{*} [1]$ and $X [2] = X^{*} [2]$ , the conjugate symmetry a real input forces.

Same as 2081 Bhadra Q92075 Ashwin · Q83+7 marksHow fast is FFT compare to DFT? Draw the butterfly diagram of 8-point DFT of a sequence as

x [n] = n

+ 1 using Decimation in Time FFT algorithm.

How this is read $x [n] = n$ + 1 over $n = 0$ to 7 is ${1, 2, 3, 4, 5, 6, 7, 8}$ .

What is different in this one

Decimation in time: 8 points, 3 stages of 4 butterflies, with the input in bit reversed order.

$N = 8$ , which is $2^{3}$ , so the transform takes 3 stages of 4 butterflies.

Stage 1, 4 butterflies with twiddles $W_{2}^{0}$ :

\begin{aligned} 6 \\ - 4 \\ 10 \\ - 4 \\ 8 \\ - 4 \\ 12 \\ - 4 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ :

\begin{aligned} 16 \\ - 4 + j 4 \\ - 4 \\ - 4 - j 4 \\ 20 \\ - 4 + j 4 \\ - 4 \\ - 4 - j 4 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ :

\begin{aligned} 36 \\ - 4 + j 9.6569 \\ - 4 + j 4 \\ - 4 + j 1.6569 \\ - 4 \\ - 4 - j 1.6569 \\ - 4 - j 4 \\ - 4 - j 9.6569 \end{aligned}

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	36	36	0
$X [1]$	-4 + j9.6569	10.4525	112.5
$X [2]$	-4 + j4	5.6569	135
$X [3]$	-4 + j1.6569	4.3296	157.5
$X [4]$	-4	4	180
$X [5]$	-4 - j1.6569	4.3296	-157.5
$X [6]$	-4 - j4	5.6569	-135
$X [7]$	-4 - j9.6569	10.4525	-112.5

7.4Last minute recall

Last minute recall, chapter 7

Must memorise

$X [k]$ = sum of $x [n] W_{N}$ ^(kn), with $W_{N} = e^{- j 2 p i / N}$ , and the DFT is the sampled DTFT at N points round the unit circle.
$W_{N}$ ^(k + N/2) = $- W_{N}^{k}$ is the symmetry that makes the butterfly work, and $W_{N}$ ^2 = W(N/2) is the halving that makes the recursion work.
A product of DFTs is a CIRCULAR convolution, never a linear one. "Find $x_{3}$ if X3 = X1 X2" is a circular convolution question.
Zero pad both sequences to N ≥ $N_{1} + N_{2} - 1$ and circular becomes linear exactly.
Zero padding adds points, not resolution. Resolution comes only from more data.
$N^{2}$ against (N/2) log2 N, a speed up of 2N/log2 N. For $N = 1024$ that is about 205 times.
DIT: input bit reversed, twiddle before the add. DIF: output bit reversed, twiddle after the subtraction.
For $N = 8$ the bit reversed order is 0, 4, 2, 6, 1, 5, 3, 7.
$X [0]$ is the sum of the samples, and for a real input $X [N - k] = X^{*} [k]$ .

Most repeated in this chapter, in order

Circular convolution TOP 17/19, often disguised as a product of DFTs
The DIT FFT TOP 11/19 and the DIF FFT TOP 8/19
Why the FFT is fast HOT 6/19 and why we need the DFT HOT 5/19
Zero padding PIN 2/19, DFT against DTFT PIN 1/19, the properties PIN 1/19

This chapter carries fifteen marks and its two questions are both mechanical: a circular convolution and an eight point butterfly. Both have a free check, so there is no reason to lose a mark on either.

39 theory questions · asked 105 times in 19 papers · exam answers only

Theory answers

Only the theory part of each past question, with the answer as it is written in the exam. Where a question also has a numerical part, only its theory part is here; the numerical is in Numerical solutions. Read them by chapter, each question once with every paper that set it, or by paper, question by question.

1Signals and systems

Energy and power signals HOT 6/19

Asked 6 times

2082 Bhadra · Q12 marksCompare between energy signal and power signal.

2079 Bhadra · Q12 marksDefine energy and power signal.

2079 Baishakh · Q12 marksCompare between energy signal and power signal.

2075 Chaitra · Q13 marksDefine Power and Energy type discrete time signal with suitable example.

2074 Ashwin · Q12 marksDefine Energy and Power type discrete time signal.

2072 Kartik · Q12 marksDefine energy and power signal.

Energy signal. A discrete time signal $x [n]$ is an energy signal if its total energy is finite and non zero, $0 < E < \infty$ . Its average power is then zero.

E = \sum_{n = - \infty}^{\infty} | x [n] |^{2}

Example: $x [n] = (1 / 2)^{n} u [n]$ , for which $E = \sum_{n = 0}^{\infty} (1 / 4)^{n} = 4 / 3$ , finite.

Power signal. $x [n]$ is a power signal if its average power is finite and non zero, $0 < P < \infty$ . Its energy is then infinite.

P = lim_{N \to \infty} \frac{1}{2 N + 1} \sum_{n = - N}^{N} | x [n] |^{2}

Example: $x [n] = u [n]$ , for which $P = {lim}_{N \to \infty} \frac{N + 1}{2 N + 1} = \frac{1}{2}$ , finite.

Energy signal	Power signal
Energy finite, $0 < E < \infty$	Power finite, $0 < P < \infty$
Average power $P = 0$	Energy $E = \infty$
Usually finite duration or decaying, aperiodic	Usually periodic or of infinite duration
Examples: $δ [n]$ , $(0.5)^{n} u [n]$	Examples: $u [n]$ , $\cos ω_{0} n$

A signal cannot be both. A signal can be neither: $x [n] = n u [n]$ has $E = \infty$ and $P = \infty$ .

Calculating the Fourier series coefficients PIN 2/19

Asked 2 times

2073 Shrawan · Q13 marksExplain the process of calculating fourier series coefficients.

2072 Chaitra · Q14 marksHow fourier series coefficients are calculated? Explain.

A periodic discrete time signal $x [n]$ with fundamental period N is represented by its Fourier series as a sum of N harmonically related complex exponentials:

x [n] = \sum_{k = 0}^{N - 1} c_{k} e^{j 2 π k n / N}

Derivation of $c_{k}$ . Multiply both sides by $e^{- j 2 π m n / N}$ and sum over one period, $n = 0$ to $N - 1$ :

\sum_{n = 0}^{N - 1} x [n] e^{- j 2 π m n / N} = \sum_{k = 0}^{N - 1} c_{k} \sum_{n = 0}^{N - 1} e^{j 2 π (k - m) n / N}

The inner sum is the orthogonality of the harmonics. It is a geometric series:

\sum_{n = 0}^{N - 1} e^{j 2 π (k - m) n / N} = {\begin{matrix} N, & k = m \\ 0, & k \neq m \end{matrix}

So only the term $k = m$ remains, and

\sum_{n = 0}^{N - 1} x [n] e^{- j 2 π m n / N} = N c_{m}

c_{k} = \frac{1}{N} \sum_{n = 0}^{N - 1} x [n] e^{- j 2 π k n / N}, k = 0, 1, \dots, N - 1

Process.

Find the fundamental period N of $x [n]$ .
Take one period of $x [n]$ , $n = 0$ to $N - 1$ .
For each $k = 0, 1, \dots, N - 1$ , evaluate $c_{k} = \frac{1}{N} \sum_{n = 0}^{N - 1} x [n] e^{- j 2 π k n / N}$ .
Only N coefficients are distinct, since $c_{k + N} = c_{k}$ .

Even and odd signals PIN 1/19

Asked once

2076 Chaitra · Q12 marksDefine even and odd type discrete time signals with suitable example.

Even signal. A discrete time signal is even (symmetric) if

x [- n] = x [n] for all n

Example: $x [n] = \cos ω_{0} n$ , or $x [n] = {1, 2, \underset{―}{3}, 2, 1}$ .

Odd signal. A discrete time signal is odd (antisymmetric) if

x [- n] = - x [n] for all n

An odd signal always has $x [0] = 0$ . Example: $x [n] = \sin ω_{0} n$ , or $x [n] = {- 2, - 1, \underset{―}{0}, 1, 2}$ .

Every signal is the sum of an even part and an odd part:

x_{e} [n] = \frac{1}{2} [x [n] + x [- n]], x_{o} [n] = \frac{1}{2} [x [n] - x [- n]], x [n] = x_{e} [n] + x_{o} [n]

Fourier series against Fourier transform PIN 1/19

Asked once

2075 Chaitra · Q14 marksDifferentiate between Fourier Series and Fourier Transform.

Fourier series	Fourier transform
Applies to periodic signals	Applies to aperiodic signals
Signal is a sum of harmonics of the fundamental frequency	Signal is an integral over a continuum of frequencies
Spectrum is discrete, a line spectrum at $ω = 2 π k / N$	Spectrum is continuous in $ω$
$x [n] = \sum_{k = 0}^{N - 1} c_{k} e^{j 2 π k n / N}$	$x [n] = \frac{1}{2 π} \int_{- π}^{π} X (e^{j ω}) e^{j ω n} d ω$
$c_{k} = \frac{1}{N} \sum_{n = 0}^{N - 1} x [n] e^{- j 2 π k n / N}$	$X (e^{j ω}) = \sum_{n = - \infty}^{\infty} x [n] e^{- j ω n}$
Finite number (N) of coefficients, periodic in k with period N	$X (e^{j ω})$ is continuous and periodic in $ω$ with period $2 π$
Always exists (a finite sum over one period)	Exists if $x [n]$ is absolutely summable, $\sum \| x [n] \| < \infty$

Dirichlet's conditions for Fourier series PIN 1/19

Asked once

2076 Ashwin · Q13 marksWrite Drichlet's conditions for Fourier series.

A periodic signal $x (t)$ with period T has a convergent Fourier series if it satisfies the Dirichlet conditions:

Absolutely integrable over one period: $\int_{T} | x (t) | d t < \infty$ .
Finite number of maxima and minima within one period.
Finite number of discontinuities within one period, each of finite size.

When these hold, the series converges to $x (t)$ at every point where $x (t)$ is continuous, and to the average value $\frac{1}{2} [x (t^{-}) + x (t^{+})]$ at a discontinuity.

Fourier transform multiplication property PIN 1/19

Asked once

2076 Ashwin · Q14 marksExplain Fourier transform multiplication property for two sequences.

Statement. If $x_{1} [n] \leftrightarrow X_{1} (e^{j ω})$ and $x_{2} [n] \leftrightarrow X_{2} (e^{j ω})$ , then the product of the two sequences has the transform

x_{1} [n] x_{2} [n] ⟷ \frac{1}{2 π} \int_{- π}^{π} X_{1} (e^{j θ}) X_{2} (e^{j (ω - θ)}) d θ

that is, multiplication in the time domain is periodic convolution in the frequency domain, scaled by $1 / 2 π$ .

Proof. Let $y [n] = x_{1} [n] x_{2} [n]$ . Then

\begin{aligned} Y (e^{j ω}) & = \sum_{n = - \infty}^{\infty} x_{1} [n] x_{2} [n] e^{- j ω n} \\ = \sum_{n} [\frac{1}{2 π} \int_{- π}^{π} X_{1} (e^{j θ}) e^{j θ n} d θ] x_{2} [n] e^{- j ω n} \\ = \frac{1}{2 π} \int_{- π}^{π} X_{1} (e^{j θ}) [\sum_{n} x_{2} [n] e^{- j (ω - θ) n}] d θ \\ = \frac{1}{2 π} \int_{- π}^{π} X_{1} (e^{j θ}) X_{2} (e^{j (ω - θ)}) d θ \end{aligned}

Application. Truncating an impulse response by a window, $h [n] = h_{d} [n] w [n]$ , convolves the desired response with the spectrum of the window. This is why windowing produces ripples in FIR filter design.

Properties of an LTI system PIN 1/19

Asked once

2074 Chaitra · Q12 marksList out the properties of LTI system.

For an LTI system with impulse response $h [n]$ , the output is $y [n] = x [n] * h [n]$ .

Commutative: $x [n] * h [n] = h [n] * x [n]$ .
Associative: $[x [n] * h_{1} [n]] * h_{2} [n] = x [n] * [h_{1} [n] * h_{2} [n]]$ ; two systems in cascade equal one system with impulse response $h_{1} [n] * h_{2} [n]$ .
Distributive: $x [n] * [h_{1} [n] + h_{2} [n]] = x [n] * h_{1} [n] + x [n] * h_{2} [n]$ ; two systems in parallel equal one system with impulse response $h_{1} [n] + h_{2} [n]$ .
Identity: $x [n] * δ [n] = x [n]$ .
Causality: causal if and only if $h [n] = 0$ for $n < 0$ .
Stability: BIBO stable if and only if $\sum_{n = - \infty}^{\infty} | h [n] | < \infty$ .
Memoryless: if and only if $h [n] = K δ [n]$ .
Invertibility: invertible if there exists $h_{i} [n]$ with $h [n] * h_{i} [n] = δ [n]$ .

2The z-transform

Region of convergence TOP 8/19

Asked 8 times

2082 Baishakh · Q31 markDefine ROC of a Z-transform.

2081 Bhadra · Q31 markDefine ROC of z-transform.

2080 Bhadra · Q31 markDefine the Region of Convergence (ROC).

2080 Baishakh · Q31 markDefine ROC.

2078 Bhadra · Q31 markDefine ROC.

2076 Chaitra · Q42 marksDefine ROC of z-transform.

2074 Chaitra · Q31 markDefine a ROC.

2073 Shrawan · Q31 markDefine a ROC.

The region of convergence (ROC) of the z-transform $X (z) = \sum_{n = - \infty}^{\infty} x [n] z^{- n}$ is the set of all values of z in the complex plane for which this sum converges, that is, for which

\sum_{n = - \infty}^{\infty} | x [n] | | z |^{- n} < \infty

Since convergence depends only on $| z |$ , the ROC is a ring centred at the origin, $r_{1} < | z | < r_{2}$ .

Example: $x [n] = a^{n} u [n]$ gives $X (z) = \frac{1}{1 - a z^{- 1}}$ with ROC $| z | > | a |$ .

Properties of the region of convergence HOT 6/19

Asked 6 times

2075 Chaitra · Q64 marksList out the properties of Region of convergence

2075 Ashwin · Q33 marksList out the properties of Region of convergence

2074 Chaitra · Q33 marksWhat are the properties of ROC of z-transform?

2074 Ashwin · Q33 marksState and explain the properties of a Region of Convergence (ROC).

2072 Chaitra · Q36 marksExplain the properties of Region of Convergence with examples.

2072 Kartik · Q33 marksState the properties of region of convergence (ROC).

The ROC is a ring or a disc centred at the origin, $0 \leq r_{1} < | z | < r_{2} \leq \infty$ .
The ROC contains no poles, since $X (z)$ is infinite at a pole.
The ROC is a single connected region. It cannot be two separate rings.
Finite duration sequence: the ROC is the entire z-plane, except possibly $z = 0$ or $z = \infty$ .
- Causal, e.g. $δ [n - 1] \to z^{- 1}$ : all z except $z = 0$ .
- Anticausal, e.g. $δ [n + 1] \to z$ : all z except $z = \infty$ .
- Two sided, e.g. $δ [n + 1] + δ [n - 1]$ : all z except $z = 0$ and $z = \infty$ .
Right sided (causal) sequence: the ROC is the outside of the circle through the outermost pole, $| z | > r_{max}$ . Example: $a^{n} u [n] \leftrightarrow \frac{1}{1 - a z^{- 1}}$ , ROC $| z | > | a |$ .
Left sided (anticausal) sequence: the ROC is the inside of the circle through the innermost pole, $| z | < r_{min}$ . Example: $- b^{n} u [- n - 1] \leftrightarrow \frac{1}{1 - b z^{- 1}}$ , ROC $| z | < | b |$ .
Two sided sequence: the ROC is a ring between two poles, $r_{1} < | z | < r_{2}$ , if one exists. Example: $a^{n} u [n] - b^{n} u [- n - 1]$ has ROC $| a | < | z | < | b |$ , which exists only when $| a | < | b |$ .
Stability: an LTI system is stable if and only if the ROC of $H (z)$ contains the unit circle $| z | = 1$ .

The z-transform PIN 2/19

Asked 2 times

2082 Bhadra · Q31 markDefine z-transform for a discrete time signal.

2079 Baishakh · Q31 markDefine z-transform for a discrete time signal.

The z-transform of a discrete time signal $x [n]$ is the power series

X (z) = \sum_{n = - \infty}^{\infty} x [n] z^{- n}

where z is a complex variable, $z = r e^{j ω}$ . It exists only for those values of z for which the series converges, called the region of convergence (ROC). On the unit circle, $z = e^{j ω}$ , it reduces to the discrete time Fourier transform $X (e^{j ω})$ .

Convolution property of the z-transform PIN 2/19

Asked 2 times

2076 Ashwin · Q33 marksState Convolution property of Z-transform.

2072 Kartik · Q33 marksDrive the convolution property of Z-transform.

Statement. If $x_{1} [n] \leftrightarrow X_{1} (z)$ with ROC $R_{1}$ and $x_{2} [n] \leftrightarrow X_{2} (z)$ with ROC $R_{2}$ , then

x_{1} [n] * x_{2} [n] ⟷ X_{1} (z) X_{2} (z), ROC contains R_{1} \cap R_{2}

Convolution in the time domain is multiplication in the z-domain.

Derivation. Let $y [n] = x_{1} [n] * x_{2} [n] = \sum_{k = - \infty}^{\infty} x_{1} [k] x_{2} [n - k]$ .

\begin{aligned} Y (z) & = \sum_{n = - \infty}^{\infty} \sum_{k = - \infty}^{\infty} x_{1} [k] x_{2} [n - k] z^{- n} \\ = \sum_{k} x_{1} [k] \sum_{n} x_{2} [n - k] z^{- n} (interchanging the sums) \\ = \sum_{k} x_{1} [k] \sum_{m} x_{2} [m] z^{- (m + k)} (putting m = n - k) \\ = \sum_{k} x_{1} [k] z^{- k} \sum_{m} x_{2} [m] z^{- m} \\ = X_{1} (z) X_{2} (z) \end{aligned}

The ROC is at least $R_{1} \cap R_{2}$ , and larger if a pole is cancelled by a zero. For an LTI system this gives $Y (z) = H (z) X (z)$ .

3Frequency domain

Stability and causality in terms of impulse response and ROC PIN 2/19

Asked 2 times

2076 Ashwin · Q44+3 marksDescribe stability and causality characteristics of LTI system in terms of Impulse Response and ROC of its transfer function with suitable examples.

2072 Chaitra · Q44 marksDescribe stability and causality characteristics of LTI system in terms of Impulse Response and ROC of its transfer function with suitable examples.

Causality.

Impulse response: an LTI system is causal if and only if $h [n] = 0$ for $n < 0$ .
ROC: the ROC of $H (z)$ is the exterior of a circle, $| z | > r_{max}$ , including $z = \infty$ .
Example: $h [n] = (0.5)^{n} u [n]$ is causal; $H (z) = \frac{1}{1 - 0.5 z^{- 1}}$ , ROC $| z | > 0.5$ . The system $h [n] = - (0.5)^{n} u [- n - 1]$ is not causal; its ROC is $| z | < 0.5$ .

Stability (BIBO).

Impulse response: an LTI system is stable if and only if $h [n]$ is absolutely summable, $\sum_{n = - \infty}^{\infty} | h [n] | < \infty$ .
ROC: the ROC of $H (z)$ contains the unit circle $| z | = 1$ .
Example: $h [n] = (0.5)^{n} u [n]$ gives $\sum | h [n] | = \frac{1}{1 - 0.5} = 2$ , finite, so it is stable; its ROC $| z | > 0.5$ contains $| z | = 1$ . The system $h [n] = 2^{n} u [n]$ gives $\sum 2^{n} = \infty$ , so it is unstable; its ROC $| z | > 2$ does not contain the unit circle.

Causal and stable together: the ROC is $| z | > r_{max}$ with $r_{max} < 1$ , that is, all poles of $H (z)$ lie inside the unit circle.

Linear constant coefficient difference equation and its system function PIN 2/19

Asked 2 times

2073 Shrawan · Q43 marksState linear constant coefficient difference equation and corresponding system function.

2071 Shrawan · Q33 marksState linear constant coefficient difference equation and corresponding system function.

A linear constant coefficient difference equation (LCCDE) of order N relates the output $y [n]$ and input $x [n]$ of an LTI system as

\sum_{k = 0}^{N} a_{k} y [n - k] = \sum_{k = 0}^{M} b_{k} x [n - k]

where $a_{k}$ and $b_{k}$ are constants. With $a_{0} = 1$ ,

y [n] = - \sum_{k = 1}^{N} a_{k} y [n - k] + \sum_{k = 0}^{M} b_{k} x [n - k]

System function. Taking the z-transform with zero initial conditions and using $x [n - k] \leftrightarrow z^{- k} X (z)$ :

\sum_{k = 0}^{N} a_{k} z^{- k} Y (z) = \sum_{k = 0}^{M} b_{k} z^{- k} X (z)

H (z) = \frac{Y (z)}{X (z)} = \frac{\sum_{k = 0}^{M} b_{k} z^{- k}}{\sum_{k = 0}^{N} a_{k} z^{- k}} = \frac{b_{0} + b_{1} z^{- 1} + \dots + b_{M} z^{- M}}{1 + a_{1} z^{- 1} + \dots + a_{N} z^{- N}}

The roots of the numerator are the zeros and the roots of the denominator are the poles. If $a_{k} = 0$ for all $k \geq 1$ , the system is FIR; otherwise it is IIR.

Why a difference equation is needed PIN 1/19

Asked once

2071 Shrawan · Q32 marksWhy do we need difference equation?

It describes the input output relation of a discrete time LTI system in the time domain, using present and past samples of the input and output.
It is directly implementable: the output is computed recursively, sample by sample, using only adders, multipliers and delays.
It represents a system with an infinitely long impulse response (IIR) using a finite number of coefficients.
Its z-transform gives the system function $H (z)$ directly, from which the poles, zeros, stability and frequency response are found.
It gives the complete response, including the effect of initial conditions (the zero input response).

FIR system against IIR system PIN 1/19

Asked once

2080 Baishakh · Q44 marksDifferentiate between FIR system and IIR System.

FIR system	IIR system
Impulse response has a finite number of non zero samples	Impulse response is of infinite duration
Non recursive: $y [n] = \sum_{k = 0}^{M} b_{k} x [n - k]$	Recursive: $y [n] = - \sum_{k = 1}^{N} a_{k} y [n - k] + \sum_{k = 0}^{M} b_{k} x [n - k]$
$H (z) = \sum b_{k} z^{- k}$ : zeros only, poles at the origin	$H (z) = B (z) / A (z)$ : poles and zeros
Always stable	May be unstable; all poles must lie inside the unit circle
Exact linear phase is possible, with symmetric $h [n]$	Exact linear phase is not possible
Needs a higher order for a sharp cutoff	Lower order for the same specification
More memory and computation	Less memory and computation
No feedback, so no limit cycles	Feedback, so limit cycles can occur
Designed by window, frequency sampling and optimum (Remez) methods	Designed from analog filters by impulse invariance or bilinear transformation

4Filter structures

Importance of quantization in DSP PIN 1/19

Asked once

2071 Shrawan · Q51 markWhat is the importance of quantization in Digital Signal Processing?

Quantization converts each sample of a continuous amplitude signal into one of a finite set of levels, so that it can be represented with a finite number of bits. It is essential because a digital processor stores and computes only finite word length numbers. It introduces quantization error (noise) and limits the accuracy of filter coefficients and arithmetic, so its effects (quantization noise, coefficient sensitivity, overflow and limit cycles) decide the word length needed in an implementation.

Rounding against truncation PIN 1/19

Asked once

2071 Shrawan · Q51 markWhich one is better rounding or truncation?

Rounding is better. With quantization step $Δ$ :

Rounding: error $- \frac{Δ}{2} < e \leq \frac{Δ}{2}$ , mean 0, variance $Δ^{2} / 12$ .
Truncation (two's complement): error $- Δ < e \leq 0$ , mean $- Δ / 2$ , variance $Δ^{2} / 12$ .

Rounding gives a zero mean error and half the maximum error, so its mean square error ( $Δ^{2} / 12$ ) is smaller than that of truncation ( $Δ^{2} / 3$ ). Truncation is simpler to implement.

Limit cycles in a recursive system PIN 1/19

Asked once

2071 Shrawan · Q52 marksExplain about limit cycles in recursive system?

In a recursive (IIR) digital filter implemented with finite word length, the products in the feedback path are quantized (rounded or truncated). This non linearity can make the output oscillate for ever, or settle at a non zero constant, even when the input is zero and the ideal output should decay to zero. Such zero input oscillations are called limit cycles.

They are caused by quantization of the feedback products; the amplitude range in which they occur is the dead band.
They occur only in recursive filters, never in FIR (non recursive) filters.
Overflow limit cycles are large amplitude oscillations caused by overflow in the adders.

Example: $y [n] = Q [\frac{1}{2} y [n - 1]] + x [n]$ with rounding to $Δ = 1 / 8$ and $x [0] = 7 / 8$ gives $7 / 8, 1 / 2, 1 / 4, 1 / 8, 1 / 8, 1 / 8, \dots$ : the output never reaches zero.

Dead band PIN 1/19

Asked once

2071 Shrawan · Q51 markDefine dead band.

The dead band is the range of output amplitudes in which a recursive filter with zero input gets trapped in a limit cycle, because rounding stops the output from decaying further. For the first order filter $y [n] = a y [n - 1] + x [n]$ with rounding step $Δ$ , the dead band is

| y [n - 1] | \leq \frac{Δ}{2 (1 - | a |)}

5FIR design

Remez exchange algorithm TOP 15/19

Asked 15 times

2082 Bhadra · Q84 marksExplain Remez exchange algorithm for FIR filter design with the help of a flow chart.

2082 Baishakh · Q75 marksDescribe Remez exchange algorithm for FIR filter design along with flowchart.

2081 Bhadra · Q76 marksDescribe the Remez exchange algorithm for FIR filter design along with the flowchart.

2079 Bhadra · Q85 marksDiscuss the Remez exchange algorithm for FIR filter design.

2079 Baishakh · Q86 marksDescribe Remez exchange algorithm for FIR filter design along with the flowchart.

2078 Bhadra · Q86 marksShow mathematical expression of Remez exchange algorithm for FIR filter design.

2076 Chaitra · Q95 marksShow mathematical expression of Remez exchange algorithm for FIR filter design.

2076 Ashwin · Q94 marksDraw the flow chart for Remez-Exchange algorithm

2075 Chaitra · Q56 marksShow mathematical expression of Remez exchange algorithm for FIR filter design.

2075 Ashwin · Q74 marksDraw the flow chart for Optimum filter design.

2074 Ashwin · Q76 marksShow mathematical expression of the Remez exchange algorithm for FIR filter design with flow chart.

2073 Shrawan · Q79 marksExplain about Remaz exchange algorithm with suitable derivation and flow chart.

2072 Chaitra · Q96 marksShow mathematical expression of Remez exchange algorithm for FIR filter design.

2072 Kartik · Q86 marksDraw the flowchart of Remez-Exchange theorem and explain it.

2071 Shrawan · Q79 marksWhat is a Remez exchange algorithm? Derive its equation and draw its flow chart.

The Remez exchange algorithm (Parks McClellan algorithm) designs an optimum linear phase FIR filter: it minimises the maximum weighted error between the desired and the actual frequency response, giving equal ripples in the passband and the stopband.

1. Mathematical expression. For a linear phase FIR filter of length M (M odd, symmetric $h [n]$ ), the frequency response is $H (e^{j ω}) = e^{- j ω (M - 1) / 2} H_{r} (ω)$ with the real amplitude response

H_{r} (ω) = \sum_{k = 0}^{L} α (k) \cos ω k, L = \frac{M - 1}{2}

where $α (0) = h [\frac{M - 1}{2}]$ and $α (k) = 2 h [\frac{M - 1}{2} - k]$ for $k = 1, \dots, L$ .

With the desired response $H_{d r} (ω)$ (1 in the passband, 0 in the stopband) and the weighting function

W (ω) = {\begin{matrix} δ_{2} / δ_{1}, & passband \\ 1, & stopband \end{matrix}

the weighted approximation error is

E (ω) = W (ω) [H_{d r} (ω) - H_{r} (ω)]

The design problem (Chebyshev or minimax criterion) is to find the coefficients $α (k)$ that minimise

δ = max_{ω \in S} | E (ω) |

over the set S of passband and stopband frequencies.

2. Alternation theorem. $H_{r} (ω)$ is the unique best approximation if and only if $E (ω)$ has at least $L + 2$ extremal frequencies $ω_{0} < ω_{1} < \dots < ω_{L + 1}$ in S at which the error alternates in sign with the maximum magnitude:

E (ω_{i}) = - E (ω_{i + 1}), | E (ω_{i}) | = δ

so that

W (ω_{i}) [H_{d r} (ω_{i}) - H_{r} (ω_{i})] = (- 1)^{i} δ, i = 0, 1, \dots, L + 1

3. Derivation of the working equations. Substituting $H_{r} (ω)$ ,

\sum_{k = 0}^{L} α (k) \cos ω_{i} k + \frac{(- 1)^{i} δ}{W (ω_{i})} = H_{d r} (ω_{i}), i = 0, 1, \dots, L + 1

These are $L + 2$ equations in the $L + 2$ unknowns $α (0), \dots, α (L)$ and $δ$ . Solving them for $δ$ :

δ = \frac{\sum_{k = 0}^{L + 1} γ_{k} H_{d r} (ω_{k})}{\sum_{k = 0}^{L + 1} (- 1)^{k} γ_{k} / W (ω_{k})}, γ_{k} = \prod_{i = 0, i \neq k}^{L + 1} \frac{1}{\cos ω_{k} - \cos ω_{i}}

Then $H_{r} (ω)$ is obtained by Lagrange interpolation through $L + 1$ of the extremal points, with $x = \cos ω$ and $x_{k} = \cos ω_{k}$ :

H_{r} (ω) = \frac{\sum_{k = 0}^{L} \frac{β_{k}}{x - x_{k}} C_{k}}{\sum_{k = 0}^{L} \frac{β_{k}}{x - x_{k}}}

C_{k} = H_{d r} (ω_{k}) - \frac{(- 1)^{k} δ}{W (ω_{k})}, β_{k} = \prod_{i = 0, i \neq k}^{L} \frac{1}{x_{k} - x_{i}}

4. Algorithm.

Specify the band edges $ω_{p}$ , $ω_{s}$ , the ripple ratio $δ_{1} / δ_{2}$ and the filter length M, so $L = (M - 1) / 2$ .
Choose an initial set of $L + 2$ extremal frequencies spread over the passband and the stopband, including $ω_{p}$ and $ω_{s}$ .
Compute $δ$ from the formula above.
Compute $H_{r} (ω)$ by Lagrange interpolation and evaluate the error $E (ω)$ on a dense grid of frequencies in S.
Find the $L + 2$ frequencies where $| E (ω) |$ has its largest local maxima. If any $| E (ω) | > δ$ , exchange the old extremal frequencies for these new ones.
Repeat steps 3 to 5 until the extremal frequencies no longer change. The error is then equiripple, with $| E (ω) | \leq δ$ everywhere in S.
Compute the impulse response $h [n]$ from $α (k)$ , using $h [n] = h [M - 1 - n]$ .

5. Flowchart.

Optimum filter TOP 8/19

Asked 8 times

2082 Bhadra · Q81 markWhat is an optimal filter?

2081 Bhadra · Q71 markWhat do you mean by optimum filter?

2079 Baishakh · Q81 markWhat do you understand by optimum filter?

2078 Bhadra · Q81 markWhat is optimum filter?

2076 Chaitra · Q91 markWhat is optimum filter?

2075 Chaitra · Q52 marksWhat is optimum filter?

2074 Ashwin · Q71 markWhat is an optimum filter?

2072 Chaitra · Q91 markWhat is an optimum filter?

An optimum filter is a linear phase FIR filter designed so that the maximum weighted error between the desired frequency response and the actual frequency response is as small as possible over the passband and the stopband (the minimax or Chebyshev criterion). The error is spread evenly, so the passband and stopband ripples are equal in size; it is therefore also called an equiripple filter. It is designed by the Parks McClellan (Remez exchange) algorithm and, for a given specification, it has the smallest possible length.

Gibbs phenomenon PIN 3/19

Asked 3 times

2082 Baishakh · Q72 marksExplain Gibbs phenomenon in FIR filter design.

2081 Baishakh · Q73 marksWhat is Gibbs phenomenon and how can it be minimized?

2079 Bhadra · Q72 marksDefine Gibb's phenomenon.

Gibbs phenomenon. When an FIR filter is designed by truncating the infinitely long ideal impulse response $h_{d} [n]$ (multiplying it by a rectangular window), the frequency response of the resulting filter shows oscillations (ripples) in the passband and the stopband, with the largest overshoot close to the band edge. This is called the Gibbs phenomenon.

The peak overshoot is about 9% of the jump and does not decrease when the filter length M is increased; a longer filter only makes the ripples narrower and moves them closer to the band edge.
Cause: $H (e^{j ω})$ is the convolution of the ideal response with the spectrum of the window. The side lobes of the rectangular window, produced by its abrupt truncation, cause the ripples.

Minimisation.

Use a tapered window (Hanning, Hamming, Blackman) instead of the rectangular window. Its lower side lobes give smaller ripples, at the cost of a wider transition band.
Use the Kaiser window and choose its parameter $β$ to set the ripple required.
Specify a transition band instead of an abrupt edge, as in the optimum (equiripple) and frequency sampling designs.

Rectangular window in FIR design and how Gibbs oscillations arise PIN 3/19

Asked 3 times

2080 Baishakh · Q85 marksExplain in detail about how Gibb's oscillation arise while using the rectangular window in FIR filter design.

2073 Shrawan · Q66 marksExplain in detail about how rectangular window is used in FIR filter design. How Gibb's oscillations arise in this process.

2071 Shrawan · Q66 marksExplain in detail about how rectangular window is used in FIR filter design. How Gibb's oscillations arise in this process.

Use of the rectangular window.

The desired frequency response of an ideal low pass filter is
$H_{d} (e^{j ω}) = {\begin{matrix} e^{- j ω α}, & | ω | \leq ω_{c} \\ 0, & ω_{c} < | ω | \leq π \end{matrix}$
Its impulse response, by the inverse transform, is
$h_{d} [n] = \frac{1}{2 π} \int_{- ω_{c}}^{ω_{c}} e^{- j ω α} e^{j ω n} d ω = \frac{\sin ω_{c} (n - α)}{π (n - α)}, n \neq α$
with $h_{d} [α] = ω_{c} / π$ . It is infinitely long and non causal, so it cannot be implemented.
It is truncated to M samples by multiplying with the rectangular window
$w_{R} [n] = {\begin{matrix} 1, & 0 \leq n \leq M - 1 \\ 0, & otherwise \end{matrix}$

$h [n] = h_{d} [n] w_{R} [n], α = \frac{M - 1}{2}$
The choice $α = (M - 1) / 2$ makes $h [n]$ symmetric, so the filter has linear phase.

How Gibbs oscillations arise. Multiplication in the time domain is convolution in the frequency domain:

H (e^{j ω}) = \frac{1}{2 π} \int_{- π}^{π} H_{d} (e^{j θ}) W_{R} (e^{j (ω - θ)}) d θ

The spectrum of the rectangular window is

W_{R} (e^{j ω}) = \sum_{n = 0}^{M - 1} e^{- j ω n} = e^{- j ω (M - 1) / 2} \frac{\sin (ω M / 2)}{\sin (ω / 2)}

which has a main lobe of width $4 π / M$ and side lobes on either side.

The main lobe smears the sharp edge of the ideal response into a transition band.
As the window spectrum slides across the discontinuity at $ω_{c}$ , its side lobes produce ripples in the passband and the stopband. These are the Gibbs oscillations.
The largest ripple occurs next to $ω_{c}$ : an overshoot of about 9%, giving a minimum stopband attenuation of only about 21 dB.
Increasing M narrows the main lobe and the transition band, but the side lobe area, and so the peak overshoot, stays the same. The oscillations can be reduced only by a tapered window.

Why the Kaiser window is better than the fixed windows PIN 3/19

Asked 3 times

2081 Baishakh · Q73 marksWhy Kaiser window is better than other fixed windows in FIR filter design?

2074 Ashwin · Q62 marksWhy Kaiser window is better than other fixed windows in FIR filter design?

2072 Chaitra · Q83 marksWhy Kaiser window is better than other fixed windows in FIR filter design?

The fixed windows (rectangular, Hanning, Hamming, Blackman) have no adjustable parameter. Each gives a fixed minimum stopband attenuation (about 21, 44, 53 and 74 dB) whatever its length; only the transition width changes with M. The designer must pick a window that exceeds the required attenuation, which usually gives a longer filter than necessary.

The Kaiser window is better because:

It has an adjustable shape parameter $β$ besides the length:
$w [n] = \frac{I_{0} [β \sqrt{1 - {(\frac{n - α}{α})}^{2}}]}{I_{0} (β)}, 0 \leq n \leq M, α = \frac{M}{2}$
$β$ sets the stopband attenuation and M sets the transition width, independently, so any required ripple is met exactly.
Direct design formulas give $β$ and M from the specification:
$A = - 20 \log_{10} δ, M = \frac{A - 8}{2.285 Δ ω}$

$β = {\begin{matrix} 0.1102 (A - 8.7), & A > 50 \\ 0.5842 (A - 21)^{0.4} + 0.07886 (A - 21), & 21 \leq A \leq 50 \\ 0, & A < 21 \end{matrix}$
It gives a near optimum trade off between main lobe width and side lobe level, so the specification is met with close to the minimum filter length.
It covers the fixed windows as special cases: $β = 0$ is the rectangular window, and $β \approx 5$ , 6 and 8.6 behave like the Hamming, Hanning and Blackman windows.

When to choose an FIR filter and when an IIR filter PIN 2/19

Asked 2 times

2080 Baishakh · Q72 marksIn which case do we choose FIR filter and IIR filter?

2076 Ashwin · Q92 marksIn which case do we choose FIR filter and IIR filter?

An FIR filter is chosen when:

Exact linear phase (no phase distortion) is required, as in data transmission, image processing and audio.
Guaranteed stability is required.
The filter must be free of limit cycles and robust to finite word length, as in adaptive filtering.
A higher order and more computation can be afforded.

An IIR filter is chosen when:

A sharp cutoff (narrow transition band) is needed with a low filter order.
Low computation, low memory and a small delay are required.
Phase linearity is not important, as in speech processing.
A digital version of a standard analog filter (Butterworth, Chebyshev, elliptic) is wanted.

FIR filter design by the window method PIN 1/19

Asked once

2072 Chaitra · Q85 marksDescribe how digital FIR filter can be design by window method.

Steps of FIR filter design by the window method.

Specify the desired frequency response $H_{d} (e^{j ω})$ , for example an ideal low pass filter with cutoff $ω_{c}$ and linear phase:
$H_{d} (e^{j ω}) = {\begin{matrix} e^{- j ω α}, & | ω | \leq ω_{c} \\ 0, & ω_{c} < | ω | \leq π \end{matrix}$
Find the desired impulse response by the inverse Fourier transform:
$h_{d} [n] = \frac{1}{2 π} \int_{- π}^{π} H_{d} (e^{j ω}) e^{j ω n} d ω = \frac{\sin ω_{c} (n - α)}{π (n - α)}, n \neq α$
with $h_{d} [α] = ω_{c} / π$ . It is infinite and non causal.

Choose the window from the required stopband attenuation, and the length M from the transition width

Δ ω = ω_{s} - ω_{p}

Window	$w [n]$ , $0 \leq n \leq M - 1$	Transition width	Stopband attenuation
Rectangular	$1$	$1.8 π / M$	21 dB
Hanning	$0.5 - 0.5 \cos \frac{2 π n}{M - 1}$	$6.2 π / M$	44 dB
Hamming	$0.54 - 0.46 \cos \frac{2 π n}{M - 1}$	$6.6 π / M$	53 dB
Blackman	$0.42 - 0.5 \cos \frac{2 π n}{M - 1} + 0.08 \cos \frac{4 π n}{M - 1}$	$11 π / M$	74 dB

Make the filter causal and linear phase by choosing $α = (M - 1) / 2$ , and multiply by the window:
$h [n] = h_{d} [n] w [n], 0 \leq n \leq M - 1$
The FIR filter is $H (z) = \sum_{n = 0}^{M - 1} h [n] z^{- n}$ , with $h [n] = h [M - 1 - n]$ . Its frequency response is checked against the specification.

Analog filter against digital filter PIN 1/19

Asked once

2082 Baishakh · Q62 marksDifferentiate between analog and digital filter.

Analog filter	Digital filter
Processes continuous time signals $x (t)$	Processes discrete time (sampled) signals $x [n]$
Built from R, L, C and op amps	Built from adders, multipliers and delays, in hardware or software
Described by a differential equation and $H (s)$	Described by a difference equation and $H (z)$
Frequency response $H (j Ω)$ is not periodic	Frequency response $H (e^{j ω})$ is periodic with period $2 π$
Component values drift with temperature and ageing	No drift; exactly reproducible
Characteristics are hard to change	Changed simply by changing the coefficients
No sampling, so no aliasing or quantization noise	Limited by the sampling rate (aliasing) and by quantization
Can process very high frequencies	Highest frequency limited to half the sampling rate
Exact linear phase is difficult	Exact linear phase is possible (FIR)

6IIR design

Impulse invariance against bilinear transformation HOT 5/19

Asked 5 times

2082 Baishakh · Q83 marksCompare Impulse Invariance Method and Bilinear Transformation technique for analog filter to digital filter conversion.

2081 Bhadra · Q83 marksCompare Impulse Invariance Method and Bilinear Transformation method.

2081 Baishakh · Q82 marksDifferentiate between bilinear transformation and impulse invariance.

2079 Baishakh · Q93 marksCompare impulse invariance method with bilinear transformation method.

2072 Chaitra · Q103 marksExplain about the advantages of selecting bilinear transformation method over impulse invariance method (I I M).

Impulse invariance method (IIM)	Bilinear transformation (BLT)
Digital impulse response is the sampled analog one, $h [n] = h_{a} (n T)$	Obtained by the substitution $s = \frac{2}{T} \frac{1 - z^{- 1}}{1 + z^{- 1}}$
Pole mapping $\frac{1}{s - p_{k}} \to \frac{1}{1 - e^{p_{k} T} z^{- 1}}$	Every point is mapped by $z = \frac{1 + s T / 2}{1 - s T / 2}$
Frequency relation $ω = Ω T$ , linear	Frequency relation $Ω = \frac{2}{T} \tan \frac{ω}{2}$ , non linear
Many to one mapping: strips of width $2 π / T$ in the s-plane map onto the same z-plane	One to one mapping: the whole $j Ω$ axis maps once onto the unit circle
Suffers from aliasing	No aliasing
No frequency warping	Frequency warping, corrected by pre-warping
Only the poles are mapped by the formula	Both poles and zeros are mapped
Suitable only for band limited filters: low pass and band pass	Suitable for low pass, high pass, band pass and band stop filters
Preserves the shape of the impulse response	Preserves the shape of the magnitude response, with warping

Advantages of the bilinear transformation over impulse invariance.

No aliasing, since the mapping is one to one.
High pass and band stop filters can be designed, which impulse invariance cannot do.
Both poles and zeros are mapped directly by one algebraic substitution.
The frequency warping is exactly known and is removed by pre-warping the band edges.
A stable analog filter always gives a stable digital filter.

Frequency warping PIN 1/19

Asked once

2082 Bhadra · Q93 marksExplain frequency warping effect in detail.

In the bilinear transformation $s = \frac{2}{T} \frac{z - 1}{z + 1}$ , put $s = j Ω$ and $z = e^{j ω}$ :

j Ω = \frac{2}{T} \frac{e^{j ω} - 1}{e^{j ω} + 1} = \frac{2}{T} \frac{e^{j ω / 2} - e^{- j ω / 2}}{e^{j ω / 2} + e^{- j ω / 2}} = j \frac{2}{T} \tan \frac{ω}{2}

Ω = \frac{2}{T} \tan \frac{ω}{2} or ω = 2 \tan^{- 1} \frac{Ω T}{2}

The relation between the analog frequency $Ω$ and the digital frequency $ω$ is non linear:

For small $ω$ , $\tan (ω / 2) \approx ω / 2$ , so $Ω \approx ω / T$ : nearly linear.
As $Ω \to \infty$ , $ω \to π$ : the entire infinite analog axis $0 \leq Ω < \infty$ is compressed into the finite range $0 \leq ω < π$ .

This non linear compression of the frequency axis is called frequency warping.

Effects.

The band edges of the digital filter are shifted from $ω = Ω T$ , the shift being larger at higher frequencies.
The amplitude shape of a piecewise constant response is preserved, but the frequency locations are distorted, and a linear phase analog response does not stay linear.
Because the mapping is one to one, there is no aliasing.

Remedy: pre-warping. The specified digital edge frequencies are converted to analog ones by $Ω = \frac{2}{T} \tan \frac{ω}{2}$ before the analog filter is designed, so that the edges land exactly where they are wanted after the transformation.

Pre-warping PIN 1/19

Asked once

2078 Bhadra · Q93 marksWhat is pre-warping and why it is necessary? Explain.

Pre-warping. In the bilinear transformation the analog frequency $Ω$ and the digital frequency $ω$ are related non linearly by

Ω = \frac{2}{T} \tan \frac{ω}{2}

Pre-warping is the conversion of the specified digital critical frequencies into the corresponding analog frequencies using this relation, before the analog prototype filter is designed:

Ω_{p} = \frac{2}{T} \tan \frac{ω_{p}}{2}, Ω_{s} = \frac{2}{T} \tan \frac{ω_{s}}{2}

Why it is necessary.

The bilinear transformation compresses the whole analog axis $0 \leq Ω < \infty$ into $0 \leq ω < π$ , warping the frequencies non linearly, most strongly at high frequencies.
If the analog filter were designed at $Ω = ω / T$ , its band edges would map to the wrong digital frequencies and the digital filter would not meet the specification.
Pre-warping compensates in advance, so that after the transformation the band edges fall exactly at the required $ω_{p}$ and $ω_{s}$ .
It is not needed in the impulse invariance method, where $ω = Ω T$ is linear.

Bilinear transformation method of IIR filter design PIN 1/19

Asked once

2073 Shrawan · Q93 marksExplain briefly about bilinear transformation method of IIR filter design.

The bilinear transformation converts an analog filter $H_{a} (s)$ into a digital filter $H (z)$ by the substitution

s = \frac{2}{T} \frac{1 - z^{- 1}}{1 + z^{- 1}}, H (z) = H_{a} (s) |_{s = \frac{2}{T} \frac{1 - z^{- 1}}{1 + z^{- 1}}}

Properties.

Frequency relation $Ω = \frac{2}{T} \tan \frac{ω}{2}$ , which is non linear (frequency warping).
The left half s-plane maps inside the unit circle, the $j Ω$ axis onto the unit circle, and the right half plane outside it, so a stable analog filter gives a stable digital filter.
The mapping is one to one, so there is no aliasing.

Design steps.

Pre-warp the digital band edges: $Ω_{p} = \frac{2}{T} \tan \frac{ω_{p}}{2}$ , $Ω_{s} = \frac{2}{T} \tan \frac{ω_{s}}{2}$ .
Design the analog prototype $H_{a} (s)$ (for example Butterworth: find the order N and the cutoff $Ω_{c}$ ) for the pre-warped specification.
Substitute $s = \frac{2}{T} \frac{1 - z^{- 1}}{1 + z^{- 1}}$ to obtain $H (z)$ .

Digital spectral transformation, low pass to high pass PIN 1/19

Asked once

2080 Baishakh · Q104 marksDescribe digital domain Spectral Transformation features and parameters for low pass to high pass filter in IIR Filter design.

Digital domain spectral transformation converts a digital low pass prototype filter $H_{L P} (z)$ into another type of digital filter by replacing $z^{- 1}$ with a rational function $g (z^{- 1})$ .

Low pass to high pass.

z^{- 1} \to - \frac{z^{- 1} + α}{1 + α z^{- 1}}, H_{H P} (z) = H_{L P} (z) |_{z^{- 1} \to - \frac{z^{- 1} + α}{1 + α z^{- 1}}}

α = - \frac{\cos [(ω_{p} + ω_{p}^{'}) / 2]}{\cos [(ω_{p} - ω_{p}^{'}) / 2]}

Parameters. $ω_{p}$ is the passband edge of the low pass prototype, $ω_{p}^{'}$ is the required passband edge of the high pass filter, and $α$ is computed from them.

Features.

The unit circle maps onto the unit circle, so the frequency response is only moved along the frequency axis, not distorted in amplitude.
The inside of the unit circle maps to the inside, so a stable and causal prototype gives a stable and causal filter.
The mapping is an all pass rational function of $z^{- 1}$ , so the new filter is still rational and realisable.
The passband around $ω = 0$ of the low pass filter moves to around $ω = π$ , giving a high pass response.
It is done entirely in the digital domain, so no aliasing is introduced and the analog design is not repeated.

7DFT and FFT

How fast the FFT is: complexity of the DFT and the FFT HOT 6/19

Asked 7 times, in 6 papers

2082 Baishakh · Q92 marksHow Decimation in Frequency Fast Fourier Transform algorithm reduces the calculation complexity of DFT?

2081 Bhadra · Q92 marksHow does FFT reduce computational complexity compared to direct calculation of DFT?

2081 Baishakh · Q92 marksWhy Decimation in Time Fast Fourier Transform (DITFFT) Algorithm is better than direct computation of DFT?

2080 Baishakh · Q112 marksHow fast is FFT?

2080 Baishakh · Q122 marksWrite the complexity of DFT and FFT?

2075 Chaitra · Q82 marksHow fast is FFT?

2075 Ashwin · Q83 marksHow fast is FFT compare to DFT?

Direct computation of the DFT.

X (k) = \sum_{n = 0}^{N - 1} x [n] W_{N}^{k n}, W_{N} = e^{- j 2 π / N}, k = 0, 1, \dots, N - 1

For each of the N values of k, N products are formed and added. This needs $N^{2}$ complex multiplications and $N (N - 1)$ complex additions, a complexity of $O (N^{2})$ .

Computation by the FFT (radix 2, $N = 2^{v}$ ).

The N point DFT is split into two N/2 point DFTs, each of those into two again, and so on, giving $\log_{2} N$ stages. DIT-FFT splits the input samples into even and odd indexed ones; DIF-FFT splits the output (frequency) samples in the same way.
Each stage has $N / 2$ butterflies. Each butterfly needs one complex multiplication and two complex additions:
$A = a + W_{N}^{r} b, B = a - W_{N}^{r} b$

So the FFT needs $\frac{N}{2} \log_{2} N$ complex multiplications and $N \log_{2} N$ complex additions, a complexity of $O (N \log_{2} N)$ .

The saving comes from two properties of the twiddle factor:

W_{N}^{k + N} = W_{N}^{k} (periodicity), W_{N}^{k + N / 2} = - W_{N}^{k} (symmetry)

which let products be shared instead of recomputed.

Speed up = \frac{N^{2}}{\frac{N}{2} \log_{2} N} = \frac{2 N}{\log_{2} N}

N	DFT multiplications $N^{2}$	DFT additions $N (N - 1)$	FFT multiplications $\frac{N}{2} \log_{2} N$	FFT additions $N \log_{2} N$
4	16	12	4	8
8	64	56	12	24
16	256	240	32	64
32	1024	992	80	160
64	4096	4032	192	384
128	16384	16256	448	896

Both DIT-FFT and DIF-FFT have the same complexity.

Why the DFT is needed HOT 5/19

Asked 5 times

2079 Baishakh · Q102 marksWhy we need DFT?

2076 Ashwin · Q82 marksWhy we need DFT?

2074 Chaitra · Q93 marksWhy do we need DFT?

2073 Shrawan · Q102 marksWhy do we need DFT?

2072 Kartik · Q112 marksWhy we need DFT?

The DTFT $X (e^{j ω}) = \sum_{n = - \infty}^{\infty} x [n] e^{- j ω n}$ is a continuous function of $ω$ , so it cannot be computed or stored by a digital computer.
The DFT samples the DTFT at N equally spaced frequencies $ω_{k} = 2 π k / N$ , giving a finite set of numbers computed from a finite length sequence:
$X (k) = \sum_{n = 0}^{N - 1} x [n] e^{- j 2 π k n / N}, k = 0, 1, \dots, N - 1$
It is exactly invertible, so no information of the finite sequence is lost:
$x [n] = \frac{1}{N} \sum_{k = 0}^{N - 1} X (k) e^{j 2 π k n / N}$
It allows frequency (spectrum) analysis of signals on digital hardware and software.
It can be computed efficiently by the FFT, with $\frac{N}{2} \log_{2} N$ multiplications instead of $N^{2}$ .
It is used for filtering, convolution through the circular convolution property, correlation and spectral estimation.

What the FFT is, and why it is needed PIN 2/19

Asked 2 times

2082 Baishakh · Q91 markWhat is Fast Fourier Transform algorithm?

2079 Bhadra · Q102 marksWhy we need FFT?

The Fast Fourier Transform (FFT) is a set of efficient algorithms for computing the DFT, and its inverse, of $N = 2^{v}$ samples. It repeatedly splits the N point transform into two N/2 point transforms (radix 2), using the periodicity and symmetry of the twiddle factor $W_{N} = e^{- j 2 π / N}$ . It gives exactly the same result as the DFT. Its two forms are the decimation in time FFT (DIT-FFT) and the decimation in frequency FFT (DIF-FFT).

It is needed because:

Direct computation of an N point DFT needs $N^{2}$ complex multiplications, which is too slow for large N.
The FFT needs only $\frac{N}{2} \log_{2} N$ multiplications. For $N = 1024$ : 1,048,576 by the direct DFT against 5,120 by the FFT, about 205 times faster.
This makes real time spectrum analysis, filtering and other signal processing practical.

Zero padding PIN 2/19

Asked 2 times

2074 Chaitra · Q101 markDefine zero padding.

2074 Ashwin · Q101 markWhat is a zero padding?

Zero padding is the process of appending zeros to the end of a finite length sequence to increase its length, without changing its original samples.

Example: $x [n] = {1, 2, 3}$ zero padded to length 5 is ${1, 2, 3, 0, 0}$ .

It is used to make the length a power of 2 for the radix 2 FFT, to make two sequences the same length for circular convolution, to make circular convolution equal to linear convolution (padding to $N \geq N_{1} + N_{2} - 1$ ), and to sample the spectrum at more closely spaced frequencies.

DFT against DTFT PIN 1/19

Asked once

2078 Bhadra · Q102 marksDifferentiate between DFT and DTFT.

DTFT	DFT
Input is a sequence of any length, $n \in (- \infty, \infty)$	Input is a finite length sequence, $n = 0, 1, \dots, N - 1$
$X (e^{j ω}) = \sum_{n = - \infty}^{\infty} x [n] e^{- j ω n}$	$X (k) = \sum_{n = 0}^{N - 1} x [n] e^{- j 2 π k n / N}$
Output is continuous in frequency $ω$	Output is N discrete samples at $ω_{k} = 2 π k / N$
Periodic in $ω$ with period $2 π$	Periodic in k with period N
Inverse: $x [n] = \frac{1}{2 π} \int_{- π}^{π} X (e^{j ω}) e^{j ω n} d ω$	Inverse: $x [n] = \frac{1}{N} \sum_{k = 0}^{N - 1} X (k) e^{j 2 π k n / N}$
Cannot be computed directly; used for theoretical analysis	Computable, efficiently by the FFT; used in practical DSP

The DFT is the DTFT of a finite length sequence sampled at N equally spaced frequencies.

Linear convolution against circular convolution PIN 1/19

Asked once

2075 Chaitra · Q93 marksDifferentiate between linear convolution and circular convolution

Linear convolution	Circular convolution
$y [n] = \sum_{k = - \infty}^{\infty} x [k] h [n - k]$	$y [n] = \sum_{k = 0}^{N - 1} x [k] h [(n - k) mod N]$
The shifted sequence slides off the ends; samples outside are zero	The shift is circular: samples moved past $N - 1$ wrap around to the start
Output length $N_{1} + N_{2} - 1$	Output length N
The two sequences may have different lengths	Both sequences have the same length N; the shorter is zero padded
Gives the output of an LTI system for any input	Gives the convolution of periodic sequences, as used with the DFT
Frequency domain: $Y (e^{j ω}) = X (e^{j ω}) H (e^{j ω})$	Frequency domain: $Y (k) = X (k) H (k)$
Computed on a linear time axis	Computed on a circular scale

The two are equal when both sequences are zero padded to a length $N \geq N_{1} + N_{2} - 1$ .

Circular convolution property of the DFT PIN 1/19

Asked once

2075 Ashwin · Q91 markState the circular convolution property of DFT.

If $x_{1} [n] \leftrightarrow X_{1} (k)$ and $x_{2} [n] \leftrightarrow X_{2} (k)$ are N point DFT pairs, then the circular convolution of the two sequences has the DFT equal to the product of their DFTs:

x_{1} [n] (N) x_{2} [n] = \sum_{m = 0}^{N - 1} x_{1} [m] x_{2} [(n - m) mod N] ⟷ X_{1} (k) X_{2} (k)

2082 Bhadra Regular · 5 theory questions

Q12 marksCompare between energy signal and power signal.Ch 1

Q31 markDefine z-transform for a discrete time signal.Ch 2

Q81 markWhat is an optimal filter?Ch 5

Q84 marksExplain Remez exchange algorithm for FIR filter design with the help of a flow chart.Ch 5

Q93 marksExplain frequency warping effect in detail.Ch 6

2082 Baishakh Back · 7 theory questions

Q31 markDefine ROC of a Z-transform.Ch 2

Q62 marksDifferentiate between analog and digital filter.Ch 5

Q72 marksExplain Gibbs phenomenon in FIR filter design.Ch 5

Q75 marksDescribe Remez exchange algorithm for FIR filter design along with flowchart.Ch 5

Q83 marksCompare Impulse Invariance Method and Bilinear Transformation technique for analog filter to digital filter conversion.Ch 6

Q91 markWhat is Fast Fourier Transform algorithm?Ch 7

Q92 marksHow Decimation in Frequency Fast Fourier Transform algorithm reduces the calculation complexity of DFT?Ch 7

2081 Bhadra Regular · 5 theory questions

Q31 markDefine ROC of z-transform.Ch 2

Q71 markWhat do you mean by optimum filter?Ch 5

Q76 marksDescribe the Remez exchange algorithm for FIR filter design along with the flowchart.Ch 5

Q83 marksCompare Impulse Invariance Method and Bilinear Transformation method.Ch 6

Q92 marksHow does FFT reduce computational complexity compared to direct calculation of DFT?Ch 7

2081 Baishakh Back · 4 theory questions

Q73 marksWhat is Gibbs phenomenon and how can it be minimized?Ch 5

Q73 marksWhy Kaiser window is better than other fixed windows in FIR filter design?Ch 5

Q82 marksDifferentiate between bilinear transformation and impulse invariance.Ch 6

Q92 marksWhy Decimation in Time Fast Fourier Transform (DITFFT) Algorithm is better than direct computation of DFT?Ch 7

2080 Bhadra Regular · 1 theory question

Q31 markDefine the Region of Convergence (ROC).Ch 2

2080 Baishakh Back · 7 theory questions

Q31 markDefine ROC.Ch 2

Q44 marksDifferentiate between FIR system and IIR System.Ch 3

Q72 marksIn which case do we choose FIR filter and IIR filter?Ch 5

Q85 marksExplain in detail about how Gibb's oscillation arise while using the rectangular window in FIR filter design.Ch 5

Q104 marksDescribe digital domain Spectral Transformation features and parameters for low pass to high pass filter in IIR Filter design.Ch 6

Q112 marksHow fast is FFT?Ch 7

Q122 marksWrite the complexity of DFT and FFT?Ch 7

2079 Bhadra Regular · 4 theory questions

Q12 marksDefine energy and power signal.Ch 1

Q72 marksDefine Gibb's phenomenon.Ch 5

Q85 marksDiscuss the Remez exchange algorithm for FIR filter design.Ch 5

Q102 marksWhy we need FFT?Ch 7

2079 Baishakh Back · 6 theory questions

Q12 marksCompare between energy signal and power signal.Ch 1

Q31 markDefine z-transform for a discrete time signal.Ch 2

Q81 markWhat do you understand by optimum filter?Ch 5

Q86 marksDescribe Remez exchange algorithm for FIR filter design along with the flowchart.Ch 5

Q93 marksCompare impulse invariance method with bilinear transformation method.Ch 6

Q102 marksWhy we need DFT?Ch 7

2078 Bhadra Regular · 5 theory questions

Q31 markDefine ROC.Ch 2

Q81 markWhat is optimum filter?Ch 5

Q86 marksShow mathematical expression of Remez exchange algorithm for FIR filter design.Ch 5

Q93 marksWhat is pre-warping and why it is necessary? Explain.Ch 6

Q102 marksDifferentiate between DFT and DTFT.Ch 7

2076 Chaitra Regular · 4 theory questions

Q12 marksDefine even and odd type discrete time signals with suitable example.Ch 1

Q42 marksDefine ROC of z-transform.Ch 2

Q91 markWhat is optimum filter?Ch 5

Q95 marksShow mathematical expression of Remez exchange algorithm for FIR filter design.Ch 5

2076 Ashwin Back · 7 theory questions

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Q13 marksWrite Drichlet's conditions for Fourier series.Ch 1

Q33 marksState Convolution property of Z-transform.Ch 2

Q44+3 marksDescribe stability and causality characteristics of LTI system in terms of Impulse Response and ROC of its transfer function with suitable examples.Ch 3

Q82 marksWhy we need DFT?Ch 7

Q92 marksIn which case do we choose FIR filter and IIR filter?Ch 5

Q94 marksDraw the flow chart for Remez-Exchange algorithmCh 5

2075 Chaitra Regular and Back · 7 theory questions

Q13 marksDefine Power and Energy type discrete time signal with suitable example.Ch 1

Q14 marksDifferentiate between Fourier Series and Fourier Transform.Ch 1

Q52 marksWhat is optimum filter?Ch 5

Q56 marksShow mathematical expression of Remez exchange algorithm for FIR filter design.Ch 5

Q64 marksList out the properties of Region of convergenceCh 2

Q82 marksHow fast is FFT?Ch 7

Q93 marksDifferentiate between linear convolution and circular convolutionCh 7

2075 Ashwin Back · 4 theory questions

Q33 marksList out the properties of Region of convergenceCh 2

Q74 marksDraw the flow chart for Optimum filter design.Ch 5

Q83 marksHow fast is FFT compare to DFT?Ch 7

Q91 markState the circular convolution property of DFT.Ch 7

2074 Chaitra Regular · 5 theory questions

Q12 marksList out the properties of LTI system.Ch 1

Q31 markDefine a ROC.Ch 2

Q33 marksWhat are the properties of ROC of z-transform?Ch 2

Q93 marksWhy do we need DFT?Ch 7

Q101 markDefine zero padding.Ch 7

2074 Ashwin Back · 6 theory questions

Q12 marksDefine Energy and Power type discrete time signal.Ch 1

Q33 marksState and explain the properties of a Region of Convergence (ROC).Ch 2

Q62 marksWhy Kaiser window is better than other fixed windows in FIR filter design?Ch 5

Q71 markWhat is an optimum filter?Ch 5

Q76 marksShow mathematical expression of the Remez exchange algorithm for FIR filter design with flow chart.Ch 5

Q101 markWhat is a zero padding?Ch 7

2073 Shrawan New Back · 7 theory questions

Q13 marksExplain the process of calculating fourier series coefficients.Ch 1

Q31 markDefine a ROC.Ch 2

Q43 marksState linear constant coefficient difference equation and corresponding system function.Ch 3

Q66 marksExplain in detail about how rectangular window is used in FIR filter design. How Gibb's oscillations arise in this process.Ch 5

Q79 marksExplain about Remaz exchange algorithm with suitable derivation and flow chart.Ch 5

Q93 marksExplain briefly about bilinear transformation method of IIR filter design.Ch 6

Q102 marksWhy do we need DFT?Ch 7

2072 Chaitra Regular · 8 theory questions

Q14 marksHow fourier series coefficients are calculated? Explain.Ch 1

Q36 marksExplain the properties of Region of Convergence with examples.Ch 2

Q44 marksDescribe stability and causality characteristics of LTI system in terms of Impulse Response and ROC of its transfer function with suitable examples.Ch 3

Q85 marksDescribe how digital FIR filter can be design by window method.Ch 5

Q83 marksWhy Kaiser window is better than other fixed windows in FIR filter design?Ch 5

Q91 markWhat is an optimum filter?Ch 5

Q96 marksShow mathematical expression of Remez exchange algorithm for FIR filter design.Ch 5

Q103 marksExplain about the advantages of selecting bilinear transformation method over impulse invariance method (I I M).Ch 6

2072 Kartik New Back · 5 theory questions

Q12 marksDefine energy and power signal.Ch 1

Q33 marksState the properties of region of convergence (ROC).Ch 2

Q33 marksDrive the convolution property of Z-transform.Ch 2

Q86 marksDraw the flowchart of Remez-Exchange theorem and explain it.Ch 5

Q112 marksWhy we need DFT?Ch 7

2071 Shrawan New Back · 8 theory questions

Q32 marksWhy do we need difference equation?Ch 3

Q33 marksState linear constant coefficient difference equation and corresponding system function.Ch 3

Q51 markWhat is the importance of quantization in Digital Signal Processing?Ch 4

Q51 markWhich one is better rounding or truncation?Ch 4

Q52 marksExplain about limit cycles in recursive system?Ch 4

Q51 markDefine dead band.Ch 4

Q66 marksExplain in detail about how rectangular window is used in FIR filter design. How Gibb's oscillations arise in this process.Ch 5

Q79 marksWhat is a Remez exchange algorithm? Derive its equation and draw its flow chart.Ch 5

182 worked numericals · grouped by topic · ordered by how often each is set

Every numerical, worked

The 182 computational questions the paper has set, grouped into 25 topics and ordered by how many sittings set each one. Open a topic, then open a question. Each one is worked in full, in the order the examiner wants it, and most carry a check at the end that costs one line.

How to use this page

By topic, not by paper. The same computation comes back every year with different numbers, so practise the method once and the year does not matter. The topic that has been set most often is at the top.
Every solution shows the working, in the order the examiner wants it: the setup, the step that carries the marks, the arithmetic, then the answer in bold.
Do the check. Most of these have a check that costs one line and catches almost every slip: the sums multiply in a convolution, $X [0]$ is the sum of the samples in a DFT, the two parts add back to the signal in an even and odd split. Each solution shows its own.
Where the scan is ambiguous, or the printed equation is plainly mistyped, the reading used is stated above the solution rather than assumed.
The theory behind each method is in its chapter and in the theory answers; this page is the arithmetic.

Butterworth IIR design TOP 19/19

Ch 6 · IIR design19 questions, from 19 of the 19 sittings6 to practise

The procedure, the same for every question in this topic

Digital edges. $ω = 2 π f / f_{s}$ , if the question gives the edges in hertz. Skip this when it already gives them in radians.
Prewarp to analog. The bilinear map is not linear in frequency, so the analog edges are not the digital ones.
$Ω_{p} = \frac{2}{T} \tan \frac{ω_{p}}{2}, Ω_{s} = \frac{2}{T} \tan \frac{ω_{s}}{2}$
Impulse invariance instead: $Ω = ω / T$ , a straight scaling with no warping to undo.
The order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.
$N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})} round UP, never down$
The cut off. Either edge may be made exact; the passband one is the usual choice, and it meets the stopband with room to spare.
$\begin{aligned} Ω_{c} & = \frac{Ω_{p}}{{(10^{0.1 A_{p}} - 1)}^{1 / 2 N}} & exact at ω_{p} \\ Ω_{c} & = \frac{Ω_{s}}{{(10^{0.1 A_{s}} - 1)}^{1 / 2 N}} & exact at ω_{s} \end{aligned}$
The prototype. The poles sit evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.
$s_{k} = Ω_{c} \exp (\frac{j π (2 k + N + 1)}{2 N}), k = 0 \dots N - 1$
Transform back.
$s = \frac{2}{T} \frac{1 - z^{- 1}}{1 + z^{- 1}}$
Simplify, and check the DC gain is $H (e^{j 0}) = 1$ . That check costs one substitution and catches an algebra slip in the whole transform.

Turn ripples into decibels first. $| H | \geq 0.9$ means Ap = -20 $\log_{10} (0.9) = 0.9151$ dB; $| H | \leq 0.2$ means As = 13.9794 dB; "1 dB below 0 dB gain" already is Ap.
T comes from the sampling frequency. fs = 1 Hz gives $T = 1$ , fs = 0.5 Hz gives T = 2. If every critical frequency is prewarped with the same T it cancels, but say so rather than letting it vanish.

The checks that cost nothing

N lands between 2 and 6 in every one of these papers. A large N means the prewarping went wrong.
$Ω_{c}$ lies between $Ω_{p}$ and $Ω_{s}$ . If it does not, the order or the cut off formula was misapplied.
Substitute $z = 1$ into the finished $H (z)$ : a low pass filter must give 1.
Check both edges of the finished digital filter against the specification it was built for. Rounding the order up means the stopband is always beaten, never just met.

How this is readThe attenuations are printed as -3 dB and -10 dB; the design formulas take them as positive numbers, so Ap = 3 and As = 10.

What is different in this one

The edges are given in hertz, not radians, so the first step is $ω = 2 π$ f / fs. A question that already gives radians skips it.
The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.
$T = 0.0002$ here, not 1, so the factor 2/T in the prewarping is 10000. It cancels in the end, but only if it is carried consistently.

Digital edges first: $ω = 2 π$ f / fs gives $ω_{p} = 2 π$ (350)/5000 = 0.4398 rad and $ω_{s} = 2 π$ (1000)/5000 = 1.2566 rad.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 0.0002) \tan (0.4398 / 2) = 2235.2648 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 0.0002) \tan (1.2566 / 2) = 7265.4253 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1} - 1) / (10^{0.3} - 1)] / (2 \log_{10} (3.2504)) \\ = \log_{10} (9 / 0.9953) / (2 log 103.2504) \\ = 0.934 so  take N = 1 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 2235.2648 / {(0.9953)}^{1 / 2} \\ = 2240.5787 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 7265.4253 / {(9)}^{1 / 2} \\ = 2421.8084 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 2240.5787$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 1 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 0 \\ s 0 & = - 2240.5787 \end{aligned}

H_{a} (s) = \frac{{(2240.5787)}^{1}}{(s + 2240.5787)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 10000 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.183 + 0.183 z^{- 1}}{1 - 0.6339 z^{- 1}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.4398$	0.7079	3 dB, against the 3 dB allowed
$ω = ω_{s} = 1.2566$	0.2947	10.61 dB, against the 10 dB required
$ω = π$	0	at Nyquist

$N = 1$ and $Ω_{c} = 2240.5787$ rad/s. The specification is met at both edges.

H (z) = \frac{0.183 + 0.183 z^{- 1}}{1 - 0.6339 z^{- 1}}

Practise this +12 like it2082 Baishakh · Q812+3 marksUsing bilinear transformation technique, design a Butterworth filter which satisfies the following conditions:

0.8 \leq | H (e^{j ω}) | \leq 1

for

0 \leq w \leq 0.2 π

;

| H (e^{j ω}) | \leq 0.2

for

0.6 π \leq ω \leq π

. Consider sampling frequency 1 Hz. Compare Impulse Invariance Method and Bilinear Transformation technique for analog filter to digital filter conversion.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 1) \tan (0.6283 / 2) = 0.6498 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 1) \tan (1.885 / 2) = 2.7528 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.398} - 1) / (10^{0.194} - 1)] / (2 \log_{10} (4.2361)) \\ = \log_{10} (24 / 0.5625) / (2 log 104.2361) \\ = 1.3 so  take N = 2 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.6498 / {(0.5625)}^{1 / 4} \\ = 0.7504 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 2.7528 / {(24)}^{1 / 4} \\ = 1.2437 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.7504$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 2 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 1 \\ s 0 & = - 0.5306 + j 0.5306 \\ s 1 & = - 0.5306 - j 0.5306 \end{aligned}

H_{a} (s) = \frac{{(0.7504)}^{2}}{(s^{2} + 1.0612 s + 0.5631)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 2 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.0842 + 0.1684 z^{- 1} + 0.0842 z^{- 2}}{1 - 1.0282 z^{- 1} + 0.3651 z^{- 2}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.6283$	0.8	1.94 dB, against the 1.94 dB allowed
$ω = ω_{s} = 1.885$	0.0741	22.6 dB, against the 13.98 dB required
$ω = π$	0	at Nyquist

$N = 2$ and $Ω_{c} = 0.7504$ rad/s. The specification is met at both edges.

H (z) = \frac{0.0842 + 0.1684 z^{- 1} + 0.0842 z^{- 2}}{1 - 1.0282 z^{- 1} + 0.3651 z^{- 2}}

Same as 2082 Baishakh Q82081 Bhadra · Q812+3 marksDesign a digital low pass IIR filter using Bilinear transformation to meet following specifications:

0.9 \leq | H (e^{j ω}) | \leq 1

for

0 \leq | w | \leq \frac{π}{2}

;

| H (e^{j ω}) | \leq 0.2

for

3 π / 4 \leq | w | \leq π

. Use Butterworth approximation for design with sampling frequency of 1Hz. Compare Impulse Invariance Method and Bilinear Transformation method.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 1) \tan (1.5708 / 2) = 2 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 1) \tan (2.3562 / 2) = 4.8284 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.398} - 1) / (10^{0.092} - 1)] / (2 \log_{10} (2.4142)) \\ = \log_{10} (24 / 0.2346) / (2 log 102.4142) \\ = 2.6255 so  take N = 3 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 2 / {(0.2346)}^{1 / 6} \\ = 2.5467 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 4.8284 / {(24)}^{1 / 6} \\ = 2.843 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 2.5467$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 3 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 2 \\ s 0 & = - 1.2734 + j 2.2055 \\ s 1 & = - 2.5467 \\ s 2 & = - 1.2734 - j 2.2055 \end{aligned}

H_{a} (s) = \frac{{(2.5467)}^{3}}{(s^{2} + 2.5467 s + 6.4859) (s + 2.5467)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 2 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.2332 + 0.6996 z^{- 1} + 0.6996 z^{- 2} + 0.2332 z^{- 3}}{1 + 0.4394 z^{- 1} + 0.3845 z^{- 2} + 0.0416 z^{- 3}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 1.5708$	0.9	0.92 dB, against the 0.92 dB allowed
$ω = ω_{s} = 2.3562$	0.1452	16.76 dB, against the 13.98 dB required
$ω = π$	0	at Nyquist

$N = 3$ and $Ω_{c} = 2.5467$ rad/s. The specification is met at both edges.

H (z) = \frac{0.2332 + 0.6996 z^{- 1} + 0.6996 z^{- 2} + 0.2332 z^{- 3}}{1 + 0.4394 z^{- 1} + 0.3845 z^{- 2} + 0.0416 z^{- 3}}

T = 0.1

second, to satisfy the following specifications:

0.6 \leq | H (e^{j ω}) | \leq 1

for

0 \leq w \leq 0.35 π

;

| H (e^{j ω}) | \leq 0.1

for

0.7 π \leq ω \leq π

How this is readThe question fixes $T = 0.1$ s rather than giving a sampling frequency, so the prewarping uses 2/T = 20.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.
$T = 0.1$ here, not 1, so the factor 2/T in the prewarping is 20. It cancels in the end, but only if it is carried consistently.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 0.1) \tan (1.0996 / 2) = 12.256 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 0.1) \tan (2.1991 / 2) = 39.2522 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{2} - 1) / (10^{0.444} - 1)] / (2 \log_{10} (3.2027)) \\ = \log_{10} (99 / 1.7778) / (2 log 103.2027) \\ = 1.7267 so  take N = 2 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 12.256 / {(1.7778)}^{1 / 4} \\ = 10.614 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 39.2522 / {(99)}^{1 / 4} \\ = 12.4439 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 10.614$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 2 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 1 \\ s 0 & = - 7.5052 + j 7.5052 \\ s 1 & = - 7.5052 - j 7.5052 \end{aligned}

H_{a} (s) = \frac{{(10.614)}^{2}}{(s^{2} + 15.0105 s + 112.6574)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 20 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.1386 + 0.2772 z^{- 1} + 0.1386 z^{- 2}}{1 - 0.707 z^{- 1} + 0.2614 z^{- 2}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 1.0996$	0.6	4.44 dB, against the 4.44 dB allowed
$ω = ω_{s} = 2.1991$	0.0729	22.74 dB, against the 20 dB required
$ω = π$	0	at Nyquist

$N = 2$ and $Ω_{c} = 10.614$ rad/s. The specification is met at both edges.

H (z) = \frac{0.1386 + 0.2772 z^{- 1} + 0.1386 z^{- 2}}{1 - 0.707 z^{- 1} + 0.2614 z^{- 2}}

Same as 2082 Baishakh Q82080 Bhadra · Q812 marksUsing Bilinear transformation, design a Butterworth low pass filter which satisfies following conditions:

0.9 \leq | H (e^{j ω}) | \leq 1

for

0 \leq ω \leq \frac{π}{2}

;

| H (e^{j ω}) | \leq 0.2

for

\frac{3 π}{4} \leq ω \leq π

. Consider sampling frequency of 1 Hz.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 1) \tan (1.5708 / 2) = 2 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 1) \tan (2.3562 / 2) = 4.8284 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.398} - 1) / (10^{0.092} - 1)] / (2 \log_{10} (2.4142)) \\ = \log_{10} (24 / 0.2346) / (2 log 102.4142) \\ = 2.6255 so  take N = 3 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 2 / {(0.2346)}^{1 / 6} \\ = 2.5467 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 4.8284 / {(24)}^{1 / 6} \\ = 2.843 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 2.5467$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 3 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 2 \\ s 0 & = - 1.2734 + j 2.2055 \\ s 1 & = - 2.5467 \\ s 2 & = - 1.2734 - j 2.2055 \end{aligned}

H_{a} (s) = \frac{{(2.5467)}^{3}}{(s^{2} + 2.5467 s + 6.4859) (s + 2.5467)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 2 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.2332 + 0.6996 z^{- 1} + 0.6996 z^{- 2} + 0.2332 z^{- 3}}{1 + 0.4394 z^{- 1} + 0.3845 z^{- 2} + 0.0416 z^{- 3}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 1.5708$	0.9	0.92 dB, against the 0.92 dB allowed
$ω = ω_{s} = 2.3562$	0.1452	16.76 dB, against the 13.98 dB required
$ω = π$	0	at Nyquist

$N = 3$ and $Ω_{c} = 2.5467$ rad/s. The specification is met at both edges.

H (z) = \frac{0.2332 + 0.6996 z^{- 1} + 0.6996 z^{- 2} + 0.2332 z^{- 3}}{1 + 0.4394 z^{- 1} + 0.3845 z^{- 2} + 0.0416 z^{- 3}}

Same as 2082 Baishakh Q82080 Baishakh · Q911 marksDesign a low pass digital IIR filter by Bilinear Transformation method to an approximate Butterworth low pass filter, if passband edge frequency is

0.26 π

radians and maximum deviation of 0.99 dB below 0 dB gain in the passband. The maximum gain of -14.99 dB and frequency is

0.58 π

radians in stopband, Consider sampling frequency 0.5 Hz.

How this is readA sampling frequency of 0.5 Hz gives $T = 2$ s.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.
$T = 2$ here, not 1, so the factor 2/T in the prewarping is 1. It cancels in the end, but only if it is carried consistently.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 2) \tan (0.8168 / 2) = 0.4327 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 2) \tan (1.8221 / 2) = 1.2892 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.499} - 1) / (10^{0.099} - 1)] / (2 \log_{10} (2.9791)) \\ = \log_{10} (30.55 / 0.256) / (2 log 102.9791) \\ = 2.1902 so  take N = 3 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.4327 / {(0.256)}^{1 / 6} \\ = 0.5431 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 1.2892 / {(30.55)}^{1 / 6} \\ = 0.7291 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.5431$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 3 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 2 \\ s 0 & = - 0.2715 + j 0.4703 \\ s 1 & = - 0.5431 \\ s 2 & = - 0.2715 - j 0.4703 \end{aligned}

H_{a} (s) = \frac{{(0.5431)}^{3}}{(s^{2} + 0.5431 s + 0.2949) (s + 0.5431)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 1 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.0565 + 0.1694 z^{- 1} + 0.1694 z^{- 2} + 0.0565 z^{- 3}}{1 - 1.0634 z^{- 1} + 0.6363 z^{- 2} - 0.1211 z^{- 3}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.8168$	0.8923	0.99 dB, against the 0.99 dB allowed
$ω = ω_{s} = 1.8221$	0.0745	22.55 dB, against the 14.99 dB required
$ω = π$	0	at Nyquist

$N = 3$ and $Ω_{c} = 0.5431$ rad/s. The specification is met at both edges.

H (z) = \frac{0.0565 + 0.1694 z^{- 1} + 0.1694 z^{- 2} + 0.0565 z^{- 3}}{1 - 1.0634 z^{- 1} + 0.6363 z^{- 2} - 0.1211 z^{- 3}}

Practise this2079 Bhadra · Q911+4 marksDesign a low pass discrete time Butterworth filter using bilinear transformation having following specifications: Passband frequency (

Ω_{p}

) =

0.25 π

radians, Stopband frequency (

Ω_{s}

) =

0.55 π

0.45 π

using digital domain transformation.

The design

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.
$T = 2$ here, not 1, so the factor 2/T in the prewarping is 1. It cancels in the end, but only if it is carried consistently.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 2) \tan (0.7854 / 2) = 0.4142 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 2) \tan (1.7279 / 2) = 1.1708 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.356} - 1) / (10^{0.101} - 1)] / (2 \log_{10} (2.8267)) \\ = \log_{10} (21.676 / 0.2625) / (2 log 102.8267) \\ = 2.1239 so  take N = 3 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.4142 / {(0.2625)}^{1 / 6} \\ = 0.5177 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 1.1708 / {(21.676)}^{1 / 6} \\ = 0.7012 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.5177$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 3 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 2 \\ s 0 & = - 0.2588 + j 0.4483 \\ s 1 & = - 0.5177 \\ s 2 & = - 0.2588 - j 0.4483 \end{aligned}

H_{a} (s) = \frac{{(0.5177)}^{3}}{(s^{2} + 0.5177 s + 0.268) (s + 0.5177)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 1 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.0512 + 0.1536 z^{- 1} + 0.1536 z^{- 2} + 0.0512 z^{- 3}}{1 - 1.1377 z^{- 1} + 0.6808 z^{- 2} - 0.1335 z^{- 3}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.7854$	0.89	1.01 dB, against the 1.01 dB allowed
$ω = ω_{s} = 1.7279$	0.0861	21.3 dB, against the 13.56 dB required
$ω = π$	0	at Nyquist

$N = 3$ and $Ω_{c} = 0.5177$ rad/s. The specification is met at both edges.

H (z) = \frac{0.0512 + 0.1536 z^{- 1} + 0.1536 z^{- 2} + 0.0512 z^{- 3}}{1 - 1.1377 z^{- 1} + 0.6808 z^{- 2} - 0.1335 z^{- 3}}

The rest of this questionThis topic works the part that belongs to it. The other part is worked under Converting it to a high pass, the last four marks, where it belongs.

0.24 π

radians and maximum deviation of 0.98 dB below 0 dB gain in the passband. The maximum gain of -14.95 dB and frequency is

0.57 π

radians in stopband, consider sampling frequency 0.5 Hz. Compare impulse invariance method with bilinear transformation method.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.
$T = 2$ here, not 1, so the factor 2/T in the prewarping is 1. It cancels in the end, but only if it is carried consistently.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 2) \tan (0.754 / 2) = 0.3959 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 2) \tan (1.7907 / 2) = 1.2482 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.495} - 1) / (10^{0.098} - 1)] / (2 \log_{10} (3.1526)) \\ = \log_{10} (30.261 / 0.2531) / (2 log 103.1526) \\ = 2.0831 so  take N = 3 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.3959 / {(0.2531)}^{1 / 6} \\ = 0.4978 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 1.2482 / {(30.261)}^{1 / 6} \\ = 0.7071 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.4978$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 3 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 2 \\ s 0 & = - 0.2489 + j 0.4311 \\ s 1 & = - 0.4978 \\ s 2 & = - 0.2489 - j 0.4311 \end{aligned}

H_{a} (s) = \frac{{(0.4978)}^{3}}{(s^{2} + 0.4978 s + 0.2478) (s + 0.4978)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 1 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.0472 + 0.1415 z^{- 1} + 0.1415 z^{- 2} + 0.0472 z^{- 3}}{1 - 1.1971 z^{- 1} + 0.7186 z^{- 2} - 0.1441 z^{- 3}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.754$	0.8933	0.98 dB, against the 0.98 dB allowed
$ω = ω_{s} = 1.7907$	0.0633	23.97 dB, against the 14.95 dB required
$ω = π$	0	at Nyquist

$N = 3$ and $Ω_{c} = 0.4978$ rad/s. The specification is met at both edges.

H (z) = \frac{0.0472 + 0.1415 z^{- 1} + 0.1415 z^{- 2} + 0.0472 z^{- 3}}{1 - 1.1971 z^{- 1} + 0.7186 z^{- 2} - 0.1441 z^{- 3}}

Ω = ω / T

2 / ((s + 1) (s + 2))

T = 0.2

into 0.2774z over

z^{2} - 1.4887 z + 0.54853

, but the numerator is wrong. It is

2 (e^{- 0} .2 - e^{- 0} .4) = 2 (0.8187 - 0.6703) = < b > 0.2968 < / b

>; the book doubled 0.67 as 1.36 instead of 1.34. The denominator is right.

What is different in this one

The edges are given in hertz, not radians, so the first step is $ω = 2 π$ f / fs. A question that already gives radians skips it.
Impulse invariance, not the bilinear transformation. There is therefore no prewarping: the edges scale linearly, $Ω = ω / T$ , and the mapping is applied to each partial fraction term separately rather than to $H (s)$ as a whole.
$T = 0.0002$ here, not 1, so the factor 2/T in the prewarping is 10000. It cancels in the end, but only if it is carried consistently.
Impulse invariance, not the bilinear transformation, so there is no prewarping: the edges scale linearly, $Ω = ω / T$ . The second half of the question asks what prewarping is and why it is necessary, which is exactly the step this design does not take. Watch the standard text here: its worked impulse invariance example turns $2 / ((s + 1) (s + 2))$ at $T = 0.2$ into 0.2774z over $z^{2} - 1.4887 z + 0.54853$ , but the numerator is wrong. It is $2 (e^{- 0} .2 - e^{- 0} .4) = 2 (0.8187 - 0.6703)$ = 0.2968; the book doubled 0.67 as 1.36 instead of 1.34. The denominator is right.

Digital edges first: $ω = 2 π$ f / fs gives $ω_{p} = 2 π$ (200)/5000 = 0.2513 rad and $ω_{s} = 2 π$ (500)/5000 = 0.6283 rad.

Impulse invariance maps frequency linearly, $Ω = ω / T$ , so there is no prewarping to do.

\begin{aligned} Ω_{p} & = ω_{p} / T = 1256.6371 rad / s \\ Ω_{s} & = ω_{s} / T = 3141.5927 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.2} - 1) / (10^{0.5} - 1)] / (2 \log_{10} (2.5)) \\ = \log_{10} (14.849 / 2.1623) / (2 log 102.5) \\ = 1.0514 so  take N = 2 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 1256.6371 / {(2.1623)}^{1 / 4} \\ = 1036.2916 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 3141.5927 / {(14.849)}^{1 / 4} \\ = 1600.3909 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 1036.2916$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 2 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 1 \\ s 0 & = - 732.7689 + j 732.7689 \\ s 1 & = - 732.7689 - j 732.7689 \end{aligned}

H_{a} (s) = \frac{{(1036.2916)}^{2}}{(s^{2} + 1465.5377 s + 1073900.3832)}

Expand $H_{a} (s)$ in partial fractions and map each pole on its own with s = pk going to z = $e^{p k T}$ , which is what impulse invariance does.

\begin{aligned} residue - j 732.7689 at  pole - 732.7689 + j 732.7689 \\ \to - j 0.1466 / (1 - (0.8544 + j 0.1261) z^{- 1}) \\ residue j 732.7689 at  pole - 732.7689 - j 732.7689 \\ \to j 0.1466 / (1 - (0.8544 - j 0.1261) z^{- 1}) \end{aligned}

H (z) = \frac{0.037 z^{- 1}}{1 - 1.7088 z^{- 1} + 0.7459 z^{- 2}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	0.9964	the DC gain
$ω = ω_{p} = 0.2513$	0.5633	4.99 dB, against the 5 dB allowed
$ω = ω_{s} = 0.6283$	0.1114	19.06 dB, against the 12 dB required
$ω = π$	0.0107	at Nyquist

$N = 2$ and $Ω_{c} = 1036.2916$ rad/s. The specification is met at both edges.

H (z) = \frac{0.037 z^{- 1}}{1 - 1.7088 z^{- 1} + 0.7459 z^{- 2}}

What is different in this one

The edges are given in hertz, not radians, so the first step is $ω = 2 π$ f / fs. A question that already gives radians skips it.
The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.
$T = 0.0001$ here, not 1, so the factor 2/T in the prewarping is 16000. It cancels in the end, but only if it is carried consistently.

Digital edges first: $ω = 2 π$ f / fs gives $ω_{p} = 2 π$ (1200)/8000 = 0.9425 rad and $ω_{s} = 2 π$ (2500)/8000 = 1.9635 rad.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 0.0001) \tan (0.9425 / 2) = 8152.4072 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 0.0001) \tan (1.9635 / 2) = 23945.6922 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{4} - 1) / (10^{0.1} - 1)] / (2 \log_{10} (2.9373)) \\ = \log_{10} (9999 / 0.2589) / (2 log 102.9373) \\ = 4.901 so  take N = 5 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 8152.4072 / {(0.2589)}^{1 / 10} \\ = 9331.8639 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 23945.6922 / {(9999)}^{1 / 10} \\ = 9533.0471 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 9331.8639$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 5 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 4 \\ s 0 & = - 2883.7045 + j 8875.13 \\ s 1 & = - 7549.6365 + j 5485.132 \\ s 2 & = - 9331.8639 \\ s 3 & = - 7549.6365 - j 5485.132 \\ s 4 & = - 2883.7045 - j 8875.13 \end{aligned}

\begin{aligned} H_{a} (s) & = \frac{{(9331.8639)}^{5}}{D (s)} \\ D (s) & = (s^{2} + 5767.4091 s + 87083683.7083) \\ \times (s^{2} + 15099.273 s + 87083683.7083) \\ \times (s + 9331.8639) \end{aligned}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 16000 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.011 + 0.0549 z^{- 1} + 0.1098 z^{- 2} + 0.1098 z^{- 3} + 0.0549 z^{- 4} + 0.011 z^{- 5}}{1 - 1.617 z^{- 1} + 1.5544 z^{- 2} - 0.7829 z^{- 3} + 0.2231 z^{- 4} - 0.0263 z^{- 5}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.9425$	0.8913	1 dB, against the 1 dB allowed
$ω = ω_{s} = 1.9635$	0.009	40.93 dB, against the 40 dB required
$ω = π$	0	at Nyquist

$N = 5$ and $Ω_{c} = 9331.8639$ rad/s. The specification is met at both edges.

H (z) = \frac{0.011 + 0.0549 z^{- 1} + 0.1098 z^{- 2} + 0.1098 z^{- 3} + 0.0549 z^{- 4} + 0.011 z^{- 5}}{1 - 1.617 z^{- 1} + 1.5544 z^{- 2} - 0.7829 z^{- 3} + 0.2231 z^{- 4} - 0.0263 z^{- 5}}

ω_{p}

) =

0.22 π

radians, Stop bandedge frequency (

ω_{s}

) =

0.54 π

radians, Passband ripple (deltap) = 0.11, Stopband ripple (deltas) = 0.22, Consider sampling frequency 0.5 Hz.

How this is readA passband ripple of 0.11 means Ap = -20 $\log_{10} (0.89) = 1.0122$ dB, and a stopband ripple of 0.22 means As = -20 $\log_{10} (0.22) = 13.1515$ dB. fs = 0.5 Hz gives $T = 2$ s.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.
$T = 2$ here, not 1, so the factor 2/T in the prewarping is 1. It cancels in the end, but only if it is carried consistently.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 2) \tan (0.6912 / 2) = 0.36 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 2) \tan (1.6965 / 2) = 1.1343 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.315} - 1) / (10^{0.101} - 1)] / (2 \log_{10} (3.1506)) \\ = \log_{10} (19.661 / 0.2625) / (2 log 103.1506) \\ = 1.8806 so  take N = 2 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.36 / {(0.2625)}^{1 / 4} \\ = 0.503 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 1.1343 / {(19.661)}^{1 / 4} \\ = 0.5387 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.503$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 2 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 1 \\ s 0 & = - 0.3557 + j 0.3557 \\ s 1 & = - 0.3557 - j 0.3557 \end{aligned}

H_{a} (s) = \frac{{(0.503)}^{2}}{(s^{2} + 0.7113 s + 0.253)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 1 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.1288 + 0.2576 z^{- 1} + 0.1288 z^{- 2}}{1 - 0.7606 z^{- 1} + 0.2757 z^{- 2}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.6912$	0.89	1.01 dB, against the 1.01 dB allowed
$ω = ω_{s} = 1.6965$	0.1929	14.29 dB, against the 13.15 dB required
$ω = π$	0	at Nyquist

$N = 2$ and $Ω_{c} = 0.503$ rad/s. The specification is met at both edges.

H (z) = \frac{0.1288 + 0.2576 z^{- 1} + 0.1288 z^{- 2}}{1 - 0.7606 z^{- 1} + 0.2757 z^{- 2}}

Same as 2082 Baishakh Q82075 Chaitra · Q710 marksUsing bilinear transformation, design a digital filter using Butterworth approximation which satisfies the following conditions:

0.8 \leq | H (e^{j ω}) | \leq 1

for

0 \leq W \leq 0.2 π

;

| H (e^{j ω}) | \leq 0.2

for

0.6 π \leq W \leq π

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 1) \tan (0.6283 / 2) = 0.6498 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 1) \tan (1.885 / 2) = 2.7528 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.398} - 1) / (10^{0.194} - 1)] / (2 \log_{10} (4.2361)) \\ = \log_{10} (24 / 0.5625) / (2 log 104.2361) \\ = 1.3 so  take N = 2 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.6498 / {(0.5625)}^{1 / 4} \\ = 0.7504 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 2.7528 / {(24)}^{1 / 4} \\ = 1.2437 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.7504$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 2 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 1 \\ s 0 & = - 0.5306 + j 0.5306 \\ s 1 & = - 0.5306 - j 0.5306 \end{aligned}

H_{a} (s) = \frac{{(0.7504)}^{2}}{(s^{2} + 1.0612 s + 0.5631)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 2 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.0842 + 0.1684 z^{- 1} + 0.0842 z^{- 2}}{1 - 1.0282 z^{- 1} + 0.3651 z^{- 2}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.6283$	0.8	1.94 dB, against the 1.94 dB allowed
$ω = ω_{s} = 1.885$	0.0741	22.6 dB, against the 13.98 dB required
$ω = π$	0	at Nyquist

$N = 2$ and $Ω_{c} = 0.7504$ rad/s. The specification is met at both edges.

H (z) = \frac{0.0842 + 0.1684 z^{- 1} + 0.0842 z^{- 2}}{1 - 1.0282 z^{- 1} + 0.3651 z^{- 2}}

Same as 2082 Baishakh Q82075 Ashwin · Q612 marksDesign a digital low-pass filter with the following specification: i) Pass-band magnitude constant to 0.7 dB below the frequency of

0.15 π

ii) Stop-band attenuation at least 14 dB for the frequencies between

0.6 π

π

. Use Butterworth approximation as a prototype and use bilinear transformation method to obtain the digital filter.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 1) \tan (0.4712 / 2) = 0.4802 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 1) \tan (1.885 / 2) = 2.7528 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.4} - 1) / (10^{0.07} - 1)] / (2 \log_{10} (5.733)) \\ = \log_{10} (24.119 / 0.1749) / (2 log 105.733) \\ = 1.4106 so  take N = 2 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.4802 / {(0.1749)}^{1 / 4} \\ = 0.7425 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 2.7528 / {(24.119)}^{1 / 4} \\ = 1.2422 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.7425$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 2 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 1 \\ s 0 & = - 0.525 + j 0.525 \\ s 1 & = - 0.525 - j 0.525 \end{aligned}

H_{a} (s) = \frac{{(0.7425)}^{2}}{(s^{2} + 1.05 s + 0.5513)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 2 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.0829 + 0.1658 z^{- 1} + 0.0829 z^{- 2}}{1 - 1.037 z^{- 1} + 0.3685 z^{- 2}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.4712$	0.9226	0.7 dB, against the 0.7 dB allowed
$ω = ω_{s} = 1.885$	0.0726	22.79 dB, against the 14 dB required
$ω = π$	0	at Nyquist

$N = 2$ and $Ω_{c} = 0.7425$ rad/s. The specification is met at both edges.

H (z) = \frac{0.0829 + 0.1658 z^{- 1} + 0.0829 z^{- 2}}{1 - 1.037 z^{- 1} + 0.3685 z^{- 2}}

Practise this2074 Chaitra · Q712 marksDesign a digital low-pass filter with the following specification: i) Pass-band magnitude constant to 0.7 dB below the frequency of

0.15 π

ii) Stop-band attenuation at least 14 dB for the frequencies between

0.6 π

π

What is different in this one

Impulse invariance, not the bilinear transformation. There is therefore no prewarping: the edges scale linearly, $Ω = ω / T$ , and the mapping is applied to each partial fraction term separately rather than to $H (s)$ as a whole.

Impulse invariance maps frequency linearly, $Ω = ω / T$ , so there is no prewarping to do.

\begin{aligned} Ω_{p} & = ω_{p} / T = 0.4712 rad / s \\ Ω_{s} & = ω_{s} / T = 1.885 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.4} - 1) / (10^{0.07} - 1)] / (2 \log_{10} (4)) \\ = \log_{10} (24.119 / 0.1749) / (2 log 104) \\ = 1.7769 so  take N = 2 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.4712 / {(0.1749)}^{1 / 4} \\ = 0.7287 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 1.885 / {(24.119)}^{1 / 4} \\ = 0.8506 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.7287$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 2 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 1 \\ s 0 & = - 0.5153 + j 0.5153 \\ s 1 & = - 0.5153 - j 0.5153 \end{aligned}

H_{a} (s) = \frac{{(0.7287)}^{2}}{(s^{2} + 1.0305 s + 0.531)}

Expand $H_{a} (s)$ in partial fractions and map each pole on its own with s = pk going to z = $e^{p k T}$ , which is what impulse invariance does.

\begin{aligned} residue - j 0.5153 at  pole - 0.5153 + j 0.5153 \\ \to - j 0.5153 / (1 - (0.5198 + j 0.2943) z^{- 1}) \\ residue j 0.5153 at  pole - 0.5153 - j 0.5153 \\ \to j 0.5153 / (1 - (0.5198 - j 0.2943) z^{- 1}) \end{aligned}

H (z) = \frac{0.3033 z^{- 1}}{1 - 1.0396 z^{- 1} + 0.3568 z^{- 2}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	0.9561	the DC gain
$ω = ω_{p} = 0.4712$	0.8986	0.93 dB, against the 0.7 dB allowed
$ω = ω_{s} = 1.885$	0.1918	14.35 dB, against the 14 dB required
$ω = π$	0.1266	at Nyquist

$N = 2$ and $Ω_{c} = 0.7287$ rad/s. The specification is met at the passband edge and exceeded at the stopband edge, as rounding the order up guarantees.

H (z) = \frac{0.3033 z^{- 1}}{1 - 1.0396 z^{- 1} + 0.3568 z^{- 2}}

ω_{p}

) =

0.27 π

radians, Stop bandedge frequency (

ω_{s}

) =

0.58 π

radians, Passband ripple (deltap) = 0.11, Stopband ripple (deltas) = 0.21, Consider sampling frequency 0.5 Hz.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.
$T = 2$ here, not 1, so the factor 2/T in the prewarping is 1. It cancels in the end, but only if it is carried consistently.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 2) \tan (0.8482 / 2) = 0.4515 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 2) \tan (1.8221 / 2) = 1.2892 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.356} - 1) / (10^{0.101} - 1)] / (2 \log_{10} (2.8552)) \\ = \log_{10} (21.676 / 0.2625) / (2 log 102.8552) \\ = 2.1035 so  take N = 3 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.4515 / {(0.2625)}^{1 / 6} \\ = 0.5643 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 1.2892 / {(21.676)}^{1 / 6} \\ = 0.7721 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.5643$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 3 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 2 \\ s 0 & = - 0.2821 + j 0.4887 \\ s 1 & = - 0.5643 \\ s 2 & = - 0.2821 - j 0.4887 \end{aligned}

H_{a} (s) = \frac{{(0.5643)}^{3}}{(s^{2} + 0.5643 s + 0.3184) (s + 0.5643)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 1 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.061 + 0.183 z^{- 1} + 0.183 z^{- 2} + 0.061 z^{- 3}}{1 - 1.0026 z^{- 1} + 0.6022 z^{- 2} - 0.1116 z^{- 3}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.8482$	0.89	1.01 dB, against the 1.01 dB allowed
$ω = ω_{s} = 1.8221$	0.0836	21.56 dB, against the 13.56 dB required
$ω = π$	0	at Nyquist

$N = 3$ and $Ω_{c} = 0.5643$ rad/s. The specification is met at both edges.

H (z) = \frac{0.061 + 0.183 z^{- 1} + 0.183 z^{- 2} + 0.061 z^{- 3}}{1 - 1.0026 z^{- 1} + 0.6022 z^{- 2} - 0.1116 z^{- 3}}

Same as 2082 Baishakh Q82073 Shrawan · Q812 marksUsing bilinear transformation, design a butterworth low pass filter which satisfies the following Magnitude Response:

0.89125 \leq | H (e^{j ω}) | \leq 1

for

0 \leq w \leq 0.2 π

;

| H (e^{j ω}) | \leq 0.17783

for

0.3 π \leq ω \leq π

How this is read0.89125 and 0.17783 are the familiar 1 dB and 15 dB written as ratios: -20 $\log_{10} (0.89125) = 1$ dB and -20 $\log_{10} (0.17783) = 15$ dB.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 1) \tan (0.6283 / 2) = 0.6498 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 1) \tan (0.9425 / 2) = 1.0191 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.5} - 1) / (10^{0.1} - 1)] / (2 \log_{10} (1.5682)) \\ = \log_{10} (30.622 / 0.2589) / (2 log 101.5682) \\ = 5.3044 so  take N = 6 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.6498 / {(0.2589)}^{1 / 12} \\ = 0.7273 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 1.0191 / {(30.622)}^{1 / 12} \\ = 0.7662 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.7273$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 6 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 5 \\ s 0 & = - 0.1882 + j 0.7025 \\ s 1 & = - 0.5143 + j 0.5143 \\ s 2 & = - 0.7025 + j 0.1882 \\ s 3 & = - 0.7025 - j 0.1882 \\ s 4 & = - 0.5143 - j 0.5143 \\ s 5 & = - 0.1882 - j 0.7025 \end{aligned}

\begin{aligned} H_{a} (s) & = \frac{{(0.7273)}^{6}}{D (s)} \\ D (s) & = (s^{2} + 0.3765 s + 0.529) \\ \times (s^{2} + 1.0285 s + 0.529) \\ \times (s^{2} + 1.405 s + 0.529) \end{aligned}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 2 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.0006 + 0.0035 z^{- 1} + 0.0087 z^{- 2} + 0.0116 z^{- 3} + 0.0087 z^{- 4} + 0.0035 z^{- 5} + 0.0006 z^{- 6}}{1 - 3.3143 z^{- 1} + 4.9501 z^{- 2} - 4.1433 z^{- 3} + 2.0275 z^{- 4} - 0.5458 z^{- 5} + 0.0628 z^{- 6}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.6283$	0.8913	1 dB, against the 1 dB allowed
$ω = ω_{s} = 0.9425$	0.131	17.65 dB, against the 15 dB required
$ω = π$	0	at Nyquist

$N = 6$ and $Ω_{c} = 0.7273$ rad/s. The specification is met at both edges.

H (z) = \frac{0.0006 + 0.0035 z^{- 1} + 0.0087 z^{- 2} + 0.0116 z^{- 3} + 0.0087 z^{- 4} + 0.0035 z^{- 5} + 0.0006 z^{- 6}}{1 - 3.3143 z^{- 1} + 4.9501 z^{- 2} - 4.1433 z^{- 3} + 2.0275 z^{- 4} - 0.5458 z^{- 5} + 0.0628 z^{- 6}}

How this is readThe attenuations are printed as -5 dB and -12 dB; the formulas take them positive.

What is different in this one

The edges are given in hertz, not radians, so the first step is $ω = 2 π$ f / fs. A question that already gives radians skips it.
Impulse invariance, not the bilinear transformation. There is therefore no prewarping: the edges scale linearly, $Ω = ω / T$ , and the mapping is applied to each partial fraction term separately rather than to $H (s)$ as a whole.
$T = 0.0002$ here, not 1, so the factor 2/T in the prewarping is 10000. It cancels in the end, but only if it is carried consistently.

Digital edges first: $ω = 2 π$ f / fs gives $ω_{p} = 2 π$ (200)/5000 = 0.2513 rad and $ω_{s} = 2 π$ (500)/5000 = 0.6283 rad.

Impulse invariance maps frequency linearly, $Ω = ω / T$ , so there is no prewarping to do.

\begin{aligned} Ω_{p} & = ω_{p} / T = 1256.6371 rad / s \\ Ω_{s} & = ω_{s} / T = 3141.5927 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.2} - 1) / (10^{0.5} - 1)] / (2 \log_{10} (2.5)) \\ = \log_{10} (14.849 / 2.1623) / (2 log 102.5) \\ = 1.0514 so  take N = 2 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 1256.6371 / {(2.1623)}^{1 / 4} \\ = 1036.2916 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 3141.5927 / {(14.849)}^{1 / 4} \\ = 1600.3909 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 1036.2916$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 2 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 1 \\ s 0 & = - 732.7689 + j 732.7689 \\ s 1 & = - 732.7689 - j 732.7689 \end{aligned}

H_{a} (s) = \frac{{(1036.2916)}^{2}}{(s^{2} + 1465.5377 s + 1073900.3832)}

Expand $H_{a} (s)$ in partial fractions and map each pole on its own with s = pk going to z = $e^{p k T}$ , which is what impulse invariance does.

\begin{aligned} residue - j 732.7689 at  pole - 732.7689 + j 732.7689 \\ \to - j 0.1466 / (1 - (0.8544 + j 0.1261) z^{- 1}) \\ residue j 732.7689 at  pole - 732.7689 - j 732.7689 \\ \to j 0.1466 / (1 - (0.8544 - j 0.1261) z^{- 1}) \end{aligned}

H (z) = \frac{0.037 z^{- 1}}{1 - 1.7088 z^{- 1} + 0.7459 z^{- 2}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	0.9964	the DC gain
$ω = ω_{p} = 0.2513$	0.5633	4.99 dB, against the 5 dB allowed
$ω = ω_{s} = 0.6283$	0.1114	19.06 dB, against the 12 dB required
$ω = π$	0.0107	at Nyquist

$N = 2$ and $Ω_{c} = 1036.2916$ rad/s. The specification is met at both edges.

H (z) = \frac{0.037 z^{- 1}}{1 - 1.7088 z^{- 1} + 0.7459 z^{- 2}}

Same as 2082 Baishakh Q82072 Kartik · Q915 marksDesign a low pass digital filter by Bilinear Transformation method to an approximate Butterworth filter, if passband edge frequency is

0.25 π

radians and maximum deviation of 1 dB below 0 dB gain in the passband. The maximum gain of -15 dB and frequency is

0.45 π

radians in stopband, Consider sampling frequency 1Hz.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 1) \tan (0.7854 / 2) = 0.8284 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 1) \tan (1.4137 / 2) = 1.7082 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.5} - 1) / (10^{0.1} - 1)] / (2 \log_{10} (2.0619)) \\ = \log_{10} (30.623 / 0.2589) / (2 log 102.0619) \\ = 3.2979 so  take N = 4 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.8284 / {(0.2589)}^{1 / 8} \\ = 0.9809 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 1.7082 / {(30.623)}^{1 / 8} \\ = 1.1137 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.9809$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 4 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 3 \\ s 0 & = - 0.3754 + j 0.9062 \\ s 1 & = - 0.9062 + j 0.3754 \\ s 2 & = - 0.9062 - j 0.3754 \\ s 3 & = - 0.3754 - j 0.9062 \end{aligned}

H_{a} (s) = \frac{{(0.9809)}^{4}}{(s^{2} + 0.7507 s + 0.9621) (s^{2} + 1.8124 s + 0.9621)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 2 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.0167 + 0.0667 z^{- 1} + 0.1001 z^{- 2} + 0.0667 z^{- 3} + 0.0167 z^{- 4}}{1 - 1.6476 z^{- 1} + 1.3563 z^{- 2} - 0.5252 z^{- 3} + 0.0834 z^{- 4}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.7854$	0.8913	1 dB, against the 1 dB allowed
$ω = ω_{s} = 1.4137$	0.1081	19.32 dB, against the 15 dB required
$ω = π$	0	at Nyquist

$N = 4$ and $Ω_{c} = 0.9809$ rad/s. The specification is met at both edges.

H (z) = \frac{0.0167 + 0.0667 z^{- 1} + 0.1001 z^{- 2} + 0.0667 z^{- 3} + 0.0167 z^{- 4}}{1 - 1.6476 z^{- 1} + 1.3563 z^{- 2} - 0.5252 z^{- 3} + 0.0834 z^{- 4}}

Same as 2082 Baishakh Q82071 Shrawan · Q815 marksDesign a low pass digital filter by Bilinear Transformation method to an approximate Butterworth filter it passband frequency is

0.2 π

radians and maximum deviation of 1 db below 0 dB gain in the pass band. The maximum gain of -15 db and frequency is

0.4 π

radians in stop band, consider sampling frequency 1 Hz.

What is different in this one

The bilinear transformation, so every critical frequency must be prewarped with W = (2/T) $\tan (ω / 2)$ before the analog filter is designed.

Prewarp, because the bilinear transformation bends the frequency axis. Design the analog filter at these frequencies and the mapping pulls the edges back to exactly where they were asked for.

\begin{aligned} Ω_{p} & = (2 / T) \tan (ω_{p} / 2) = (2 / 1) \tan (0.6283 / 2) = 0.6498 rad / s \\ Ω_{s} & = (2 / T) \tan (ω_{s} / 2) = (2 / 1) \tan (1.2566 / 2) = 1.4531 rad / s \end{aligned}

Now the order. Put the two specification points into the Butterworth magnitude and divide one by the other, so that $Ω_{c}$ cancels.

N \geq \frac{\log_{10} [(10^{0.1 A_{s}} - 1) / (10^{0.1 A_{p}} - 1)]}{2 \log_{10} (Ω_{s} / Ω_{p})}

\begin{aligned} N \geq log 10 [(10^{1.5} - 1) / (10^{0.1} - 1)] / (2 \log_{10} (2.2361)) \\ = \log_{10} (30.623 / 0.2589) / (2 log 102.2361) \\ = 2.9656 so  take N = 3 \end{aligned}

\begin{aligned} Ω_{c} & = Ω_{p} / {(10^{0.1 A p} - 1)}^{1 / 2 N} \\ = 0.6498 / {(0.2589)}^{1 / 6} \\ = 0.814 rad / s & exact at the passband edge \\ Ω_{c} & = Ω_{s} / {(10^{0.1 A s} - 1)}^{1 / 2 N} \\ = 1.4531 / {(30.623)}^{1 / 6} \\ = 0.8215 rad / s & exact at the stopband edge \end{aligned}

Either is acceptable. Take $Ω_{c} = 0.814$ rad/s, the passband value, which meets the passband exactly and the stopband with room to spare.

The prototype has its 3 poles evenly spaced on a circle of radius $Ω_{c}$ , and only the left half plane ones belong to a stable filter.

\begin{aligned} s_{k} & = Ω_{c} \exp (j π (2 k + N + 1) / 2 N), k = 0 \dots 2 \\ s 0 & = - 0.407 + j 0.7049 \\ s 1 & = - 0.814 \\ s 2 & = - 0.407 - j 0.7049 \end{aligned}

H_{a} (s) = \frac{{(0.814)}^{3}}{(s^{2} + 0.814 s + 0.6626) (s + 0.814)}

Now substitute, which is one line of algebra per power of s.

\begin{aligned} s & = (2 / T) (1 - z^{- 1}) / (1 + z^{- 1}) = 2 (1 - z^{- 1}) / (1 + z^{- 1}) \end{aligned}

H (z) = \frac{0.0305 + 0.0914 z^{- 1} + 0.0914 z^{- 2} + 0.0305 z^{- 3}}{1 - 1.4826 z^{- 1} + 0.9296 z^{- 2} - 0.2033 z^{- 3}}

Check the finished filter against the specification it was built for, which costs four substitutions:

Frequency	$\| H (e^{j ω}) \|$
$ω = 0$	1	the DC gain
$ω = ω_{p} = 0.6283$	0.8913	1 dB, against the 1 dB allowed
$ω = ω_{s} = 1.2566$	0.1731	15.23 dB, against the 15 dB required
$ω = π$	0	at Nyquist

$N = 3$ and $Ω_{c} = 0.814$ rad/s. The specification is met at both edges.

H (z) = \frac{0.0305 + 0.0914 z^{- 1} + 0.0914 z^{- 2} + 0.0305 z^{- 3}}{1 - 1.4826 z^{- 1} + 0.9296 z^{- 2} - 0.2033 z^{- 3}}

Convolution TOP 17/19

Ch 1 · Signals and systems20 questions, from 17 of the 19 sittings9 to practise

The procedure, the same for every question in this topic

y [n] = \sum_{k = - \infty}^{\infty} x [k] h [n - k]

Method	How it goes, and when to use it
Tabular	x along the top, h down the side, fill the grid with products, add along the anti diagonals. Fastest for two short finite sequences, which is most of what the paper sets.
Graphical	Fold $h [k]$ to $h [- k]$ , slide it by n, multiply the overlap, add. This is what "using the graphical method" means.
Analytical	For sequences carrying $u [n]$ : write the sum, fix the limits from where both are non zero, and sum the geometric series.

Where the answer sits, before you compute a single sample. This is the check that costs nothing.

\begin{aligned} x runs n_{1} \dots n_{2}, & h runs m_{1} \dots m_{2} \\ ⟹ y runs n_{1} + m_{1} & \dots n_{2} + m_{2}, N_{x} + N_{h} - 1 samples \end{aligned}

Read the gate before anything else. $h [n] = {(1 / 2)}^{n} {u [n + 2] - u [n - 2]}$ runs over $n = - 2$ , -1, 0, 1 and takes the values 4, 2, 1, 0.5. An off by one here loses the whole question.
A sequence of impulses is a sequence. $2 δ [n + 1] + 2 δ [n - 1]$ is just ${2, 0, 2}$ starting at n = -1.
An infinite sequence is not a table. If either side runs to infinity, do it analytically: split into the region where the overlap is still growing and the region where it is complete.

The checks that cost nothing

The sums multiply. (sum of x) times (sum of h) equals the sum of y. One line, catches almost everything.
Length and position. $N_{x} + N_{h} - 1$ samples, starting at n1 + m1. If your answer is the wrong length you have dropped a term.
Two regions must agree at the join when the answer was split into a rising part and a falling part. Evaluate both at the boundary n.

Practise this +2 like it2082 Bhadra · Q25 marksFind the output of an LTI system whose impulse response is given by

{0.5}^{n} {u [n] - u [n - 3]}

and input is given by

{2, 1, 0.5, - 1}

.What is different here

h [n]

starts at

n = 0

, so the answer starts where

x [n]

does

How this is read $h [n] = {0.5}^{n} {u [n] - u [n - 3]}$ is the exponential kept for $n = 0$ , 1, 2 only, so $h [n] = {1, 0.5, 0.25}$ .

What is different in this one

Two finite sequences, 4 samples and 3, so the answer has 4 + 3 - 1 = 6 samples and the table is 4 by 3.

$x [n]$ runs from $n = 0$ to 3, 4 samples; $h [n]$ runs from $n = 0$ to 2, 3 samples. So $y [n]$ runs from $n = 0$ to 5, which is 4 + 3 - 1 = 6 samples.

Multiply every sample of one by every sample of the other and add along the anti diagonals, which is the tabular method:

	$h [0] = 1$	$h [1] = 0.5$	$h [2] = 0.25$
$x [0] = 2$	2	1	0.5
$x [1] = 1$	1	0.5	0.25
$x [2] = 0.5$	0.5	0.25	0.125
$x [3] = - 1$	-1	-0.5	-0.25

y [0] = 2

= 2

y [1] = 1

+ 1 = 2

y [2] = 0.5

+ 0.5 + 0.5 = 1.5

y [3] = 0.25

+ 0.25 + -1 = -0.5

y [4] = 0.125

+ -0.5 = -0.375

y [5] = - 0.25

= -0.25

$y [n]$ = {2, 2, 1.5, -0.5, -0.375, -0.25} for $n = 0$ to 5. The check holds: the samples of x add to 2.5, those of h add to 1.75, and 2.5 × 1.75 = 4.375, which is the sum of y.

Also worked under The output for an exponential inputPractise this +2 like it2082 Baishakh · Q25 marksFind the output of an LTI system having impulse response

h [n] = {(1 / 2)}^{n} u [n]

excited by an input

x [n] = 5 e^{j π n / 3}

What is different in this one

1 component at one frequency, so H is evaluated once and the results added: superposition.
Do not convolve. The input runs over all n, so it is not a finite sequence and the graphical method is the wrong tool.

x [n] = e^{j ω_{0} n} ⟶ y [n] = H (e^{j ω_{0}}) e^{j ω_{0} n}, A \cos (ω_{0} n) ⟶ A | H | \cos (ω_{0} n + ∠ H)

\begin{aligned} H (e^{j ω}) & = 1 / (1 - 0.5 e^{- j ω}), the  transform  of h [n] = {(1 / 2)}^{n} u [n] \end{aligned}

Component	$ω_{0}$	$H (e^{j ω_{0}})$	in polar form	Output amplitude
$x [n] = 5 e^{j π n / 3}$	$0.3333 π$	1 - j0.5774	1.1547 angle -30 deg	5.7735

Where	By convolution	By the formula
$n = 0$	5	5
$n = 1$	5	5
$n = 2$	0	0
$n = 5$	0	0

$y [n] = 5.7735 e^{j (0.3333 π n - 30 d e g)}$ .

Practise this +4 like it2081 Bhadra · Q25 marksFind the output of an LTI System with impulse response

h [n] = 2 δ [n + 1] + 2 δ [n - 1]

when an input of

x [n] = δ [n] + 2 δ [n - 1] - δ [n - 3]

is applied to it.

How this is read $h [n] = 2 δ [n + 1] + 2 δ [n - 1]$ is ${2, 0, 2}$ starting at $n = - 1$ ; $x [n] = δ [n] + 2 δ [n - 1] - δ [n - 3]$ is ${1, 2, 0, - 1}$ from $n = 0$ .

What is different in this one

Two finite sequences, 4 samples and 3, so the answer has 4 + 3 - 1 = 6 samples and the table is 4 by 3.
One of the sequences starts at a negative n, so the output starts at $n = - 1$ , not at zero. Mark the origin on the answer or the plot is wrong even when the numbers are right.
There is a zero sample in the middle of one sequence. It still occupies an index: dropping it shortens the answer and shifts everything after it.

$x [n]$ runs from $n = 0$ to 3, 4 samples; $h [n]$ runs from $n = - 1$ to 1, 3 samples. So $y [n]$ runs from $n = - 1$ to 4, which is 4 + 3 - 1 = 6 samples.

Multiply every sample of one by every sample of the other and add along the anti diagonals, which is the tabular method:

	$h [- 1] = 2$	$h [1] = 2$
$x [0] = 1$	2	2
$x [1] = 2$	4	4
$x [2] = 0$	0	0
$x [3] = - 1$	-2	-2

y [- 1] = 2

= 2

y [0] = 4

= 4

y [1] = 2

= 2

y [2] = 4

+ -2 = 2

y [3] = 0

= 0

y [4] = - 2

= -2

$y [n]$ = {2, 4, 2, 2, 0, -2} for $n = - 1$ to 4. The check holds: the samples of x add to 2, those of h add to 4, and 2 × 4 = 8, which is the sum of y.

Practise this +3 like it2081 Baishakh · Q26 marksFind the output

y (n)

of LTI system having impulse response

h (n) = {(1 / 3)}^{n} u (n - 3)

and input

x (n) = {(1 / 6)}^{n - 6} u (n)

What is different in this one

At least one of the sequences is infinite, so this is an analytical convolution and not a table. Fix the limits of the sum from where BOTH are non zero, then sum the geometric series.
The answer comes out in regions: one while the overlap is still growing and one once it is complete. They must agree at the join, and that is the check.

Neither sequence is finite, so this is an analytical convolution, not a table.

h [k] = {(1 / 3)}^{k}

for k ≥ 3,

x [m] = {(1 / 6)}^{m - 6} u [m] = 6^{6} {(1 / 6)}^{m}

for m ≥ 0

y [n] = \sum_{k} h [k] x [n - k]

h needs k ≥ 3 and x needs n - k ≥ 0, so k runs from 3 to n, and

y [n] = 0

for n < 3. n

y [n] = 6^{6} \sum {(1 / 3)}^{k} {(1 / 6)}^{n - k}

k=3 n =

6^{6} {(1 / 6)}^{n} \sum {(6 / 3)}^{k}

k=3 n = 46656

{(1 / 6)}^{n}

[

2^{n + 1} - 2^{3}

] since

\sum 2^{k} = 2^{n + 1}

- 8 k=3

Check the closed form against the convolution sum carried out term by term, at several n. They agree, which is what makes the closed form an answer rather than a guess.

n	$\sum_{k} h [k] x [n - k]$	the closed form
3	1728	1728
4	864	864
5	336	336
6	120	120
8	14	14
10	1.57407	1.57407

Also worked under The output for an exponential inputSame as 2082 Baishakh Q22080 Baishakh · Q25 marksFind the output of LTI system having impulse response

h [n] = {(1 / 2)}^{n} u [n]

and input

x [n] = 5 e^{j π n / 3}

for -infinity < n < infinity.

What is different in this one

1 component at one frequency, so H is evaluated once and the results added: superposition.
Do not convolve. The input runs over all n, so it is not a finite sequence and the graphical method is the wrong tool.

x [n] = e^{j ω_{0} n} ⟶ y [n] = H (e^{j ω_{0}}) e^{j ω_{0} n}, A \cos (ω_{0} n) ⟶ A | H | \cos (ω_{0} n + ∠ H)

\begin{aligned} H (e^{j ω}) & = 1 / (1 - 0.5 e^{- j ω}), the  transform  of h [n] = {(1 / 2)}^{n} u [n] \end{aligned}

Component	$ω_{0}$	$H (e^{j ω_{0}})$	in polar form	Output amplitude
$x [n] = 5 e^{j π n / 3}$	$0.3333 π$	1 - j0.5774	1.1547 angle -30 deg	5.7735

Where	By convolution	By the formula
$n = 0$	5	5
$n = 1$	5	5
$n = 2$	0	0
$n = 5$	0	0

$y [n] = 5.7735 e^{j (0.3333 π n - 30 d e g)}$ .

Practise this2079 Bhadra · Q25 marksFind the output of LTI system having input signal

x [n] = δ [n] + 2 δ [n - 1] - δ [n - 3]

and

h [n] = 2 δ [n + 1] + 2 δ [n - 1]

.What is different hereTwo transcriptions of this paper disagree about the input, and the past paper itself is not on this machine to settle it. This one reads

x [n] = δ [n] + 2 δ [n - 1] - δ [n - 3]

, so x =

{1, 2, 0, - 1}

from

n = 0

. The standard text works the same paper as its Example 1.26 and reads

x [n] = δ [n + 2] + δ [n - 1] - δ [n - 3]

, so x =

{1, 0, 0, 1, 0, - 1}

from n = -2.

h [n] = 2 δ [n + 1] + 2 δ [n - 1]

either way. The method is identical and so is the shape of the answer,

y [n] = 2 x [n + 1] + 2 x [n - 1]

: only which samples are non zero moves. Work it with whichever your paper shows; practising one gives you the other.

What is different in this one

Two finite sequences, 4 samples and 3, so the answer has 4 + 3 - 1 = 6 samples and the table is 4 by 3.
One of the sequences starts at a negative n, so the output starts at $n = - 1$ , not at zero. Mark the origin on the answer or the plot is wrong even when the numbers are right.
There is a zero sample in the middle of one sequence. It still occupies an index: dropping it shortens the answer and shifts everything after it.

$x [n]$ runs from $n = 0$ to 3, 4 samples; $h [n]$ runs from $n = - 1$ to 1, 3 samples. So $y [n]$ runs from $n = - 1$ to 4, which is 4 + 3 - 1 = 6 samples.

Multiply every sample of one by every sample of the other and add along the anti diagonals, which is the tabular method:

	$h [- 1] = 2$	$h [1] = 2$
$x [0] = 1$	2	2
$x [1] = 2$	4	4
$x [2] = 0$	0	0
$x [3] = - 1$	-2	-2

y [- 1] = 2

= 2

y [0] = 4

= 4

y [1] = 2

= 2

y [2] = 4

+ -2 = 2

y [3] = 0

= 0

y [4] = - 2

= -2

$y [n]$ = {2, 4, 2, 2, 0, -2} for $n = - 1$ to 4. The check holds: the samples of x add to 2, those of h add to 4, and 2 × 4 = 8, which is the sum of y.

Same as 2081 Bhadra Q22079 Baishakh · Q25 marksFind the output of LTI system having impulse response

h [n] = {(1 / 2)}^{n} {u [n + 2] - u [n - 2]}

to the input

x [n] = {2, 1, 0, - 1, 4}

How this is read $h [n] = {(1 / 2)}^{n} {u [n + 2] - u [n - 2]}$ runs over $n = - 2$ , -1, 0, 1, taking the values 4, 2, 1 and 0.5.

What is different in this one

Two finite sequences, 5 samples and 4, so the answer has 5 + 4 - 1 = 8 samples and the table is 5 by 4.
One of the sequences starts at a negative n, so the output starts at $n = - 2$ , not at zero. Mark the origin on the answer or the plot is wrong even when the numbers are right.
There is a zero sample in the middle of one sequence. It still occupies an index: dropping it shortens the answer and shifts everything after it.

$x [n]$ runs from $n = 0$ to 4, 5 samples; $h [n]$ runs from $n = - 2$ to 1, 4 samples. So $y [n]$ runs from $n = - 2$ to 5, which is 5 + 4 - 1 = 8 samples.

Multiply every sample of one by every sample of the other and add along the anti diagonals, which is the tabular method:

	$h [- 2] = 4$	$h [- 1] = 2$	$h [0] = 1$	$h [1] = 0.5$
$x [0] = 2$	8	4	2	1
$x [1] = 1$	4	2	1	0.5
$x [2] = 0$	0	0	0	0
$x [3] = - 1$	-4	-2	-1	-0.5
$x [4] = 4$	16	8	4	2

y [- 2] = 8

= 8

y [- 1] = 4

+ 4 = 8

y [0] = 2

+ 2 = 4

y [1] = 1

+ 1 + -4 = -2

y [2] = 0.5

+ -2 + 16 = 14.5

y [3] = - 1

+ 8 = 7

y [4] = - 0.5

+ 4 = 3.5

y [5] = 2

= 2

$y [n]$ = {8, 8, 4, -2, 14.5, 7, 3.5, 2} for $n = - 2$ to 5. The check holds: the samples of x add to 6, those of h add to 7.5, and 6 × 7.5 = 45, which is the sum of y.

Same as 2081 Baishakh Q22078 Bhadra · Q25 marksFind the output of LTI system having impulse response

h [n] = u [n] - u [n - 4]

and input signal

x [n] = {(1 / 2)}^{n} u [n]

What is different in this one

At least one of the sequences is infinite, so this is an analytical convolution and not a table. Fix the limits of the sum from where BOTH are non zero, then sum the geometric series.
The answer comes out in regions: one while the overlap is still growing and one once it is complete. They must agree at the join, and that is the check.

$h [n]$ is a four sample rectangle and $x [n]$ is an infinite exponential, so the answer has two regions: while the rectangle is still sliding onto the input, and after it is fully overlapped.

\begin{aligned} h [n] & = u [n] - u [n - 4] = 1 for n = 0, 1, 2, 3 \\ x [n] & = {(1 / 2)}^{n} u [n] \\ 3 \\ y [n] & = sum h [k] x [n - k] = \sum {(1 / 2)}^{n - k}, keeping  only n - k \geq 0 \\ k & = 0 \\ PARTIAL  OVERLAP, 0 \leq n \leq 3 : k runs 0 \dots n \\ y [n] & = sum  from j = 0 to n of {(1 / 2)}^{j} = 2 - {(1 / 2)}^{n} \\ FULL  OVERLAP, n \geq 3 : k runs 0 \dots 3 \\ y [n] & = {(1 / 2)}^{n} sum  from k = 0 to 3 of 2^{k} = 15 {(1 / 2)}^{n} \end{aligned}

Check the closed form against the convolution sum carried out term by term, at several n. They agree, which is what makes the closed form an answer rather than a guess.

n	$\sum_{k} h [k] x [n - k]$	the closed form
0	1	1
1	1.5	1.5
2	1.75	1.75
3	1.875	1.875
4	0.9375	0.9375
5	0.46875	0.46875
8	0.05859	0.05859

Same as 2081 Baishakh Q22076 Chaitra · Q36 marksFind the output of LTI system having input signal

x [n] = u [n + 1] - u [n - 4]

and impulse response

h [n] = {(1 / 2)}^{n} u [n - 1]

What is different in this one

At least one of the sequences is infinite, so this is an analytical convolution and not a table. Fix the limits of the sum from where BOTH are non zero, then sum the geometric series.
The answer comes out in regions: one while the overlap is still growing and one once it is complete. They must agree at the join, and that is the check.

$x [n]$ is a five sample rectangle from $n = - 1$ and $h [n]$ is an infinite exponential that starts at $n = 1$ , so again there are two regions.

x [n] = u [n + 1] - u [n - 4] = 1

for

n = - 1

.. 3

h [n] = {(1 / 2)}^{n} u [n - 1] = {(1 / 2)}^{n}

for n ≥ 1

y [n] = \sum_{k} x [k] h [n - k]

, needing -1 ≤ k ≤ 3 and n - k ≥ 1 so k runs from -1 to min(3, n-1), and

y [n] = 0

for n < 0 K

y [n] = {(1 / 2)}^{n} \sum 2^{k}

with K = min(3, n-1) k=-1 RISING, 0 ≤ n ≤ 4: K = n-1, sum =

2^{n} - 0.5 y [n] = 1 - {(1 / 2)}^{n + 1}

FALLING, n ≥ 4:

K = 3

, sum = 16 - 0.5 = 15.5

y [n] = 15.5 {(1 / 2)}^{n}

Check the closed form against the convolution sum carried out term by term, at several n. They agree, which is what makes the closed form an answer rather than a guess.

n	$\sum_{k} h [k] x [n - k]$	the closed form
0	0.5	0.5
1	0.75	0.75
2	0.875	0.875
3	0.9375	0.9375
4	0.96875	0.96875
5	0.48438	0.48438
7	0.12109	0.12109

$y [n] = 1 - {(1 / 2)}^{n + 1}$ for $0 \leq n \leq 4$ , and 15.5 ${(1 / 2)}^{n}$ for n ≥ 4. Both give 0.96875 at $n = 4$ , which is the check.

Same as 2081 Baishakh Q22076 Ashwin · Q26 marksFind convolution between two signals

x [n] = 2^{n}

4[-n],

0 < a < 1

and

h [n]

= 4[n]

What is different in this one

At least one of the sequences is infinite, so this is an analytical convolution and not a table. Fix the limits of the sum from where BOTH are non zero, then sum the geometric series.
The answer comes out in regions: one while the overlap is still growing and one once it is complete. They must agree at the join, and that is the check.

The scan has mangled this one: the u has come out as a 4, and the base as a 2. It reads $x [n] = a^{n} u [- n]$ with $0 < a < 1$ , and $h [n] = u [n]$ .

x [k] = a^{- k}

for k ≤ 0,

h [m] = 1

for m ≥ 0

y [n] = \sum_{k} x [k] h [n - k]

, needing k ≤ 0 and k ≤ n n ≥ 0: k runs from -inf to 0

y [n]

= sum from m=0 to inf of

a^{m} = 1

/ (1 - a) n < 0: k runs from -inf to n

y [n]

= sum from m=

| n |

to inf of

a^{m} = a

| n |

/ (1 - a)

Check the closed form against the convolution sum carried out term by term, at several n. They agree, which is what makes the closed form an answer rather than a guess.

n	$\sum_{k} h [k] x [n - k]$	the closed form
-4	0.125	0.125
-2	0.5	0.5
-1	1	1
0	2	2
1	2	2
3	2	2

Practise this2075 Chaitra · Q25+2 marksFind the output of LTI system having impulse response

h [n]

with

h [- 2] = 3

h [0] = 2

h [1] = 1

and input signal

x [n] = {(2)}^{n}

, for

- 1 \leq n \leq 3

. Also check the answer.What is different here

x [n]

starts before

n = 0

How this is readh has $h [- 2] = 3$ , $h [0] = 2$ , $h [1] = 1$ and nothing else, so h = ${3, 0, 2, 1}$ from n = -2. $x [n] = 2^{n}$ over $- 1 \leq n \leq 3$ is ${0.5, 1, 2, 4, 8}$ from n = -1.

What is different in this one

Two finite sequences, 5 samples and 4, so the answer has 5 + 4 - 1 = 8 samples and the table is 5 by 4.
One of the sequences starts at a negative n, so the output starts at $n = - 3$ , not at zero. Mark the origin on the answer or the plot is wrong even when the numbers are right.
There is a zero sample in the middle of one sequence. It still occupies an index: dropping it shortens the answer and shifts everything after it.

$x [n]$ runs from $n = - 1$ to 3, 5 samples; $h [n]$ runs from $n = - 2$ to 1, 4 samples. So $y [n]$ runs from $n = - 3$ to 4, which is 5 + 4 - 1 = 8 samples.

Multiply every sample of one by every sample of the other and add along the anti diagonals, which is the tabular method:

	$h [- 2] = 3$	$h [0] = 2$	$h [1] = 1$
$x [- 1] = 0.5$	1.5	1	0.5
$x [0] = 1$	3	2	1
$x [1] = 2$	6	4	2
$x [2] = 4$	12	8	4
$x [3] = 8$	24	16	8

y [- 3] = 1.5

= 1.5

y [- 2] = 3

= 3

y [- 1] = 1

+ 6 = 7

y [0] = 0.5

+ 2 + 12 = 14.5

y [1] = 1

+ 4 + 24 = 29

y [2] = 2

+ 8 = 10

y [3] = 4

+ 16 = 20

y [4] = 8

= 8

$y [n]$ = {1.5, 3, 7, 14.5, 29, 10, 20, 8} for $n = - 3$ to 4. The check holds: the samples of x add to 15.5, those of h add to 6, and 15.5 × 6 = 93, which is the sum of y.

Same as 2082 Bhadra Q22075 Ashwin · Q25 marksFind the output of LTI system having impulse response

h [n] = 2^{n} {u [n] - u [n - 3]}

and input signal

x [n] = δ [n] + δ [n - 1] + δ [n - 2]

How this is read $h [n] = 2^{n} {u [n] - u [n - 3]}$ is ${1, 2, 4}$ over $n = 0$ , 1, 2, and $x [n] = δ [n] + δ [n - 1] + δ [n - 2]$ is ${1, 1, 1}$ .

What is different in this one

Two finite sequences, 3 samples and 3, so the answer has 3 + 3 - 1 = 5 samples and the table is 3 by 3.
Both sequences are the same length, so the table is square and the answer is symmetric in the two if the sequences are.

$x [n]$ runs from $n = 0$ to 2, 3 samples; $h [n]$ runs from $n = 0$ to 2, 3 samples. So $y [n]$ runs from $n = 0$ to 4, which is 3 + 3 - 1 = 5 samples.

Multiply every sample of one by every sample of the other and add along the anti diagonals, which is the tabular method:

	$h [0] = 1$	$h [1] = 2$	$h [2] = 4$
$x [0] = 1$	1	2	4
$x [1] = 1$	1	2	4
$x [2] = 1$	1	2	4

y [0] = 1

= 1

y [1] = 2

+ 1 = 3

y [2] = 4

+ 2 + 1 = 7

y [3] = 4

+ 2 = 6

y [4] = 4

= 4

$y [n]$ = {1, 3, 7, 6, 4} for $n = 0$ to 4. The check holds: the samples of x add to 3, those of h add to 7, and 3 × 7 = 21, which is the sum of y.

Same as 2081 Bhadra Q22074 Ashwin · Q23+2 marksFind the output of LTI system having impulse response

h [n] = {(1 / 2)}^{n} {u [n + 2] - u [n - 2]}

and input signal

x [n] = {2, 1, 0.5, - 1}

. Also check the answer.

What is different in this one

Two finite sequences, 4 samples and 4, so the answer has 4 + 4 - 1 = 7 samples and the table is 4 by 4.
One of the sequences starts at a negative n, so the output starts at $n = - 2$ , not at zero. Mark the origin on the answer or the plot is wrong even when the numbers are right.
Both sequences are the same length, so the table is square and the answer is symmetric in the two if the sequences are.

$x [n]$ runs from $n = 0$ to 3, 4 samples; $h [n]$ runs from $n = - 2$ to 1, 4 samples. So $y [n]$ runs from $n = - 2$ to 4, which is 4 + 4 - 1 = 7 samples.

Multiply every sample of one by every sample of the other and add along the anti diagonals, which is the tabular method:

	$h [- 2] = 4$	$h [- 1] = 2$	$h [0] = 1$	$h [1] = 0.5$
$x [0] = 2$	8	4	2	1
$x [1] = 1$	4	2	1	0.5
$x [2] = 0.5$	2	1	0.5	0.25
$x [3] = - 1$	-4	-2	-1	-0.5

y [- 2] = 8

= 8

y [- 1] = 4

+ 4 = 8

y [0] = 2

+ 2 + 2 = 6

y [1] = 1

+ 1 + 1 + -4 = -1

y [2] = 0.5

+ 0.5 + -2 = -1

y [3] = 0.25

+ -1 = -0.75

y [4] = - 0.5

= -0.5

$y [n]$ = {8, 8, 6, -1, -1, -0.75, -0.5} for $n = - 2$ to 4. The check holds: the samples of x add to 2.5, those of h add to 7.5, and 2.5 × 7.5 = 18.75, which is the sum of y.

Same as 2082 Bhadra Q22073 Shrawan · Q26 marksDetermine the system output

y (n)

of the following signals:

h (n) = {1, 1, 1}

and

x (n) = {1, 1, 1, 1}

What is different in this one

Two finite sequences, 4 samples and 3, so the answer has 4 + 3 - 1 = 6 samples and the table is 4 by 3.

$x [n]$ runs from $n = 0$ to 3, 4 samples; $h [n]$ runs from $n = 0$ to 2, 3 samples. So $y [n]$ runs from $n = 0$ to 5, which is 4 + 3 - 1 = 6 samples.

Multiply every sample of one by every sample of the other and add along the anti diagonals, which is the tabular method:

	$h [0] = 1$	$h [1] = 1$	$h [2] = 1$
$x [0] = 1$	1	1	1
$x [1] = 1$	1	1	1
$x [2] = 1$	1	1	1
$x [3] = 1$	1	1	1

y [0] = 1

= 1

y [1] = 1

+ 1 = 2

y [2] = 1

+ 1 + 1 = 3

y [3] = 1

+ 1 + 1 = 3

y [4] = 1

+ 1 = 2

y [5] = 1

= 1

$y [n]$ = {1, 2, 3, 3, 2, 1} for $n = 0$ to 5. The check holds: the samples of x add to 4, those of h add to 3, and 4 × 3 = 12, which is the sum of y.

h [n] = {(1 / 2)}^{n} u [n]

when the input signal is

x [n] = 10

- 5sin(

π

n/2) + 20cos(

π

What is different in this one

3 components at 3 different frequencies, so H is evaluated 3 times and the results added: superposition.
One of them is a constant, which is simply the $ω = 0$ case: the input times the DC gain.
The input is real, so the answer must be real. A sine goes in and a sine of the same frequency comes out, scaled by $| H |$ and shifted by the angle of H. Leaving the answer as complex exponentials is not finishing the question.
Do not convolve. The input runs over all n, so it is not a finite sequence and the graphical method is the wrong tool.

x [n] = e^{j ω_{0} n} ⟶ y [n] = H (e^{j ω_{0}}) e^{j ω_{0} n}, A \cos (ω_{0} n) ⟶ A | H | \cos (ω_{0} n + ∠ H)

\begin{aligned} H (e^{j ω}) & = 1 / (1 - 0.5 e^{- j ω}), the  transform  of h [n] = {(1 / 2)}^{n} u [n] \end{aligned}

Component	$ω_{0}$	$H (e^{j ω_{0}})$	in polar form	Output amplitude
the constant 10, which is the $ω = 0$ case	$0 π$	2	2 angle 0 deg	20
-5 $\sin (π n / 2)$ , taken at $ω = \frac{π}{2}$	$0.5 π$	0.8 - j0.4	0.8944 angle -26.57 deg	4.4721
20 $\cos (π n)$ , taken at $ω = π$	$1 π$	0.6667	0.6667 angle 0 deg	13.3333

Where	By convolution	By the formula
$n = 0$	35.3333	35.3333
$n = 1$	2.6667	2.6667
$n = 2$	31.3333	31.3333
$n = 5$	2.6667	2.6667

Same as 2081 Bhadra Q22072 Chaitra · Q23+2 marksFind the output of LTI system having impulse response

h [n]

with

h [- 2] = 1

h [0] = 2

h [1] = 3

and input signal

x [n]

with

x [0] = \frac{1}{2}

x [2] = 2

x [3] = 3

. Also check the answer.

How this is read $h [- 2] = 1$ , $h [0] = 2$ , $h [1] = 3$ gives h = ${1, 0, 2, 3}$ from $n = - 2$ ; $x [0] = \frac{1}{2}$ , $x [2] = 2$ , $x [3] = 3$ gives x = ${0.5, 0, 2, 3}$ from $n = 0$ .

What is different in this one

Two finite sequences, 4 samples and 4, so the answer has 4 + 4 - 1 = 7 samples and the table is 4 by 4.
One of the sequences starts at a negative n, so the output starts at $n = - 2$ , not at zero. Mark the origin on the answer or the plot is wrong even when the numbers are right.
Both sequences are the same length, so the table is square and the answer is symmetric in the two if the sequences are.
There is a zero sample in the middle of one sequence. It still occupies an index: dropping it shortens the answer and shifts everything after it.

$x [n]$ runs from $n = 0$ to 3, 4 samples; $h [n]$ runs from $n = - 2$ to 1, 4 samples. So $y [n]$ runs from $n = - 2$ to 4, which is 4 + 4 - 1 = 7 samples.

Multiply every sample of one by every sample of the other and add along the anti diagonals, which is the tabular method:

	$h [- 2] = 1$	$h [0] = 2$	$h [1] = 3$
$x [0] = 0.5$	0.5	1	1.5
$x [1] = 0$	0	0	0
$x [2] = 2$	2	4	6
$x [3] = 3$	3	6	9

y [- 2] = 0.5

= 0.5

y [- 1] = 0

= 0

y [0] = 1

+ 2 = 3

y [1] = 1.5

+ 3 = 4.5

y [2] = 4

= 4

y [3] = 6

+ 6 = 12

y [4] = 9

= 9

$y [n]$ = {0.5, 0, 3, 4.5, 4, 12, 9} for $n = - 2$ to 4. The check holds: the samples of x add to 5.5, those of h add to 6, and 5.5 × 6 = 33, which is the sum of y.

Same as 2081 Bhadra Q22072 Kartik · Q25 marksFind the output of LTI system having impulse response

h [n] = {(1 / 3)}^{n} {u [n + 1] - u [n - 2]}

and input signal

x [n] = {2, 1, 0.5, 3}

How this is read $h [n] = {(1 / 3)}^{n} {u [n + 1] - u [n - 2]}$ runs over $n = - 1$ , 0, 1 with the values 3, 1 and 1/3.

What is different in this one

Two finite sequences, 4 samples and 3, so the answer has 4 + 3 - 1 = 6 samples and the table is 4 by 3.
One of the sequences starts at a negative n, so the output starts at $n = - 1$ , not at zero. Mark the origin on the answer or the plot is wrong even when the numbers are right.

$x [n]$ runs from $n = 0$ to 3, 4 samples; $h [n]$ runs from $n = - 1$ to 1, 3 samples. So $y [n]$ runs from $n = - 1$ to 4, which is 4 + 3 - 1 = 6 samples.

Multiply every sample of one by every sample of the other and add along the anti diagonals, which is the tabular method:

	$h [- 1] = 3$	$h [0] = 1$	$h [1] = 0.3333$
$x [0] = 2$	6	2	0.6667
$x [1] = 1$	3	1	0.3333
$x [2] = 0.5$	1.5	0.5	0.1667
$x [3] = 3$	9	3	1

y [- 1] = 6

= 6

y [0] = 2

+ 3 = 5

y [1] = 0.6667

+ 1 + 1.5 = 3.1667

y [2] = 0.3333

+ 0.5 + 9 = 9.8333

y [3] = 0.1667

+ 3 = 3.1667

y [4] = 1

= 1

$y [n]$ = {6, 5, 3.1667, 9.8333, 3.1667, 1} for $n = - 1$ to 4. The check holds: the samples of x add to 6.5, those of h add to 4.3333, and 6.5 × 4.3333 = 28.1667, which is the sum of y.

Also worked under The output for an exponential inputSame as 2082 Baishakh Q22072 Kartik · Q44 marksFind the output of LTI System having impulse response

h [n] = {(1 / 2)}^{n} u [n]

and input signal

x [n] = 5 e^{j π n / 3}

for -infinity < n < infinity.

What is different in this one

1 component at one frequency, so H is evaluated once and the results added: superposition.
Do not convolve. The input runs over all n, so it is not a finite sequence and the graphical method is the wrong tool.

x [n] = e^{j ω_{0} n} ⟶ y [n] = H (e^{j ω_{0}}) e^{j ω_{0} n}, A \cos (ω_{0} n) ⟶ A | H | \cos (ω_{0} n + ∠ H)

\begin{aligned} H (e^{j ω}) & = 1 / (1 - 0.5 e^{- j ω}), the  transform  of h [n] = {(1 / 2)}^{n} u [n] \end{aligned}

Component	$ω_{0}$	$H (e^{j ω_{0}})$	in polar form	Output amplitude
$x [n] = 5 e^{j π n / 3}$	$0.3333 π$	1 - j0.5774	1.1547 angle -30 deg	5.7735

Where	By convolution	By the formula
$n = 0$	5	5
$n = 1$	5	5
$n = 2$	0	0
$n = 5$	0	0

$y [n] = 5.7735 e^{j (0.3333 π n - 30 d e g)}$ .

Also worked under Even and odd partsPractise this2071 Shrawan · Q14+5 marksFind the odd and even part of the following signal: [a stem plot of

x [n]

with values 1 at

n = - 4

, -3, -2, -1 and 0, and 2 at

n = 1

, 2, 3, 4]. A discrete time LTI system has input signal and impulse response as,

x [n] = 1

for

- 1 \leq n \leq 1

and 0 elsewhere, and

h [n] = 1

for

- 1 \leq n \leq 1

and 0 elsewhere. Find the output of the system using graphical method.

How this is readThe stem plot is $x [n] = 1$ for $n = - 4$ to 0 and 2 for $n = 1$ to 4. The second half convolves two identical three sample rectangles, each from $n = - 1$ to 1.

The output, by the graphical method

What is different in this one

Two finite sequences, 3 samples and 3, so the answer has 3 + 3 - 1 = 5 samples and the table is 3 by 3.
One of the sequences starts at a negative n, so the output starts at $n = - 2$ , not at zero. Mark the origin on the answer or the plot is wrong even when the numbers are right.
Both sequences are the same length, so the table is square and the answer is symmetric in the two if the sequences are.

$x [n]$ runs from $n = - 1$ to 1, 3 samples; $h [n]$ runs from $n = - 1$ to 1, 3 samples. So $y [n]$ runs from $n = - 2$ to 2, which is 3 + 3 - 1 = 5 samples.

Multiply every sample of one by every sample of the other and add along the anti diagonals, which is the tabular method:

	$h [- 1] = 1$	$h [0] = 1$	$h [1] = 1$
$x [- 1] = 1$	1	1	1
$x [0] = 1$	1	1	1
$x [1] = 1$	1	1	1

y [- 2] = 1

= 1

y [- 1] = 1

+ 1 = 2

y [0] = 1

+ 1 + 1 = 3

y [1] = 1

+ 1 = 2

y [2] = 1

= 1

$y [n]$ = {1, 2, 3, 2, 1} for $n = - 2$ to 2. The check holds: the samples of x add to 3, those of h add to 3, and 3 × 3 = 9, which is the sum of y.

The rest of this questionThis topic works the part that belongs to it. The other part is worked under Even and odd parts (The even and odd parts), where it belongs.

h [n] = {(1 / 2)}^{n} u [n]

. Determine

y [n]

, if the input is

x [n] = A e^{j π n}

.What is different hereAt

ω = π

the exponential

e^{- j ω}

is exactly -1, so H is

1 / (1 + 0.5) = 2 / 3

with no complex arithmetic at all. The answer is

y [n]

= (2A/3)

e^{j π n}

What is different in this one

1 component at one frequency, so H is evaluated once and the results added: superposition.
Do not convolve. The input runs over all n, so it is not a finite sequence and the graphical method is the wrong tool.

x [n] = e^{j ω_{0} n} ⟶ y [n] = H (e^{j ω_{0}}) e^{j ω_{0} n}, A \cos (ω_{0} n) ⟶ A | H | \cos (ω_{0} n + ∠ H)

\begin{aligned} H (e^{j ω}) & = 1 / (1 - 0.5 e^{- j ω}), the  transform  of h [n] = {(1 / 2)}^{n} u [n] \end{aligned}

Component	$ω_{0}$	$H (e^{j ω_{0}})$	in polar form	Output amplitude
$x [n] = A e^{j π n}$ , with A factored out	$1 π$	0.6667	0.6667 angle 0 deg	0.6667

Where	By convolution	By the formula
$n = 0$	0.6667	0.6667
$n = 1$	-0.6667	-0.6667
$n = 2$	0.6667	0.6667
$n = 5$	-0.6667	-0.6667

$y [n] = 0.6667 e^{j (π n)}$ .

The pole zero map and the magnitude sketch TOP 17/19

Ch 3 · Frequency domain17 questions, from 17 of the 19 sittings7 to practise

The procedure, the same for every question in this topic

Difference equation to $H (z)$ . Take the z-transform of both sides with $x [n - k] ⟶ z^{- k} X (z)$ , then form $H (z) = Y (z) / X (z)$ .
Multiply above and below by $z^{N}$ so both are POSITIVE powers of z. The zeros and poles AT THE ORIGIN appear at this step, and forgetting them is the usual lost mark.
Factor. The roots of the numerator are the zeros, drawn o; the roots of the denominator are the poles, drawn x.
Draw the unit circle, mark Re and Im, plot everything, and state the ROC: $| z | >$ the largest pole magnitude, for a causal system.
Sketch the magnitude by the geometric rule.
$| H (e^{j ω}) | = \frac{\prod distances from e^{j ω} to each ZERO}{\prod distances from e^{j ω} to each POLE} \times | gain |$
Walk $ω$ from 0, the point $z = 1$ , round to $π$ , the point $z = - 1$ , reading those distances off your own sketch.

Near a pole the response peaks, at $ω$ equal to the angle of the pole, and the closer the pole is to the circle the sharper the peak.
Near a zero it dips, to exactly zero if the zero is on the circle.
Anything at the origin changes only the phase, since its distance to every point of the unit circle is 1.
A pole or zero far outside the circle barely changes the shape: it scales the curve. That is why the 0.45 +/- j1.6 family gives a gentle response and not a resonance.

The checks that cost nothing

Both endpoints, numerically. |H( $e^{j_{0}}$ )| is $H (z)$ at $z = 1$ and |H( $e^{j p i}$ )| is $H (z)$ at $z = - 1$ . Two substitutions, and they anchor the whole sketch.
Conjugate pairs. Real coefficients force poles and zeros into conjugate pairs, so the magnitude is even in $ω$ : sketching 0 to $π$ is the whole answer.
Count them. The number of poles equals the number of zeros once those at the origin and at infinity are counted. If they differ, a $z^{N}$ was dropped.

What is different in this one

A pole lies on or outside the unit circle, at magnitude 1.6621, so this system is unstable. Say so: it is a mark, and it also tells you the magnitude response will be dominated by that pole.
The poles are a conjugate pair at an angle of 1.297 rad, so the response has a peak near that frequency and the sketch is symmetric about $ω = 0$ .

\begin{aligned} H (z) & = \frac{1 - 1.16 z^{- 1} + 4.58 z^{- 2}}{1 - 0.9 z^{- 1} + 2.7625 z^{- 2}} & from the difference equation \\ = \frac{z^{2} - 1.16 z + 4.58}{z^{2} - 0.9 z + 2.7625} & \times z^{2} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	0.58 + j2.06	2.1401	74.28
zero, o	0.58 - j2.06	2.1401	-74.28
pole, x	0.45 + j1.6	1.6621	74.29
pole, x	0.45 - j1.6	1.6621	-74.29

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 1.6621, so for a causal system the ROC is $| z | > 1.6621$ , which does not contain the unit circle, so the system is unstable.

\begin{aligned} N (e^{j ω}) & = e^{j 2 ω} - 1.16 e^ jw + 4.58 \\ = \cos (2 ω) + jsin (2 ω) - 1.16 cos (ω) - j 1.16 sin (ω) + 4.58 \\ D (e^{j ω}) & = e^{j 2 ω} - 0.9 e^ jw + 2.7625 \\ = \cos (2 ω) + jsin (2 ω) - 0.9 cos (ω) - j 0.9 sin (ω) + 2.7625 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	4.42	2.8625	1.5441	1.5441
$\frac{π}{4}$	3.7598 + j0.1798	2.1261 + j0.3636	1.7322 - j0.2117	1.7451
$\frac{π}{2}$	3.58 - j1.16	1.7625 - j0.9	1.8777 + j0.3007	1.9016
$\frac{3 π}{4}$	5.4002 - j1.8202	3.3989 - j1.6364	1.4992 + j0.1862	1.5107
$π$	6.74	4.6625	1.4456	1.4456
$\frac{5 π}{4}$	5.4002 + j1.8202	3.3989 + j1.6364	1.4992 - j0.1862	1.5107
$\frac{3 π}{2}$	3.58 + j1.16	1.7625 + j0.9	1.8777 - j0.3007	1.9016
$\frac{7 π}{4}$	3.7598 - j0.1798	2.1261 - j0.3636	1.7322 + j0.2117	1.7451
$2 π$	4.42	2.8625	1.5441	1.5441

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	1.5441
$ω = 1.297$ , the angle of the poles	1.2966	2.0367
$ω = \frac{π}{2}$ , the point z = j	1.5708	1.9016
$ω = π$ , the point $z = - 1$	3.1416	1.4456

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response rises to a peak of 2.0368 near $ω = 1.291$ , with $| H |$ = 1.5441 at $ω = 0$ and 1.4456 at $ω = π$ . That makes it a low pass shape.

What is different in this one

A pole lies on or outside the unit circle, at magnitude 1.6621, so this system is unstable. Say so: it is a mark, and it also tells you the magnitude response will be dominated by that pole.
The poles are a conjugate pair at an angle of 1.297 rad, so the response has a peak near that frequency and the sketch is symmetric about $ω = 0$ .

\begin{aligned} H (z) & = \frac{1 - 1.16 z^{- 1} + 4.58 z^{- 2}}{1 - 0.9 z^{- 1} + 2.7625 z^{- 2}} & from the difference equation \\ = \frac{z^{2} - 1.16 z + 4.58}{z^{2} - 0.9 z + 2.7625} & \times z^{2} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	0.58 + j2.06	2.1401	74.28
zero, o	0.58 - j2.06	2.1401	-74.28
pole, x	0.45 + j1.6	1.6621	74.29
pole, x	0.45 - j1.6	1.6621	-74.29

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 1.6621, so for a causal system the ROC is $| z | > 1.6621$ , which does not contain the unit circle, so the system is unstable.

\begin{aligned} N (e^{j ω}) & = e^{j 2 ω} - 1.16 e^ jw + 4.58 \\ = \cos (2 ω) + jsin (2 ω) - 1.16 cos (ω) - j 1.16 sin (ω) + 4.58 \\ D (e^{j ω}) & = e^{j 2 ω} - 0.9 e^ jw + 2.7625 \\ = \cos (2 ω) + jsin (2 ω) - 0.9 cos (ω) - j 0.9 sin (ω) + 2.7625 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	4.42	2.8625	1.5441	1.5441
$\frac{π}{4}$	3.7598 + j0.1798	2.1261 + j0.3636	1.7322 - j0.2117	1.7451
$\frac{π}{2}$	3.58 - j1.16	1.7625 - j0.9	1.8777 + j0.3007	1.9016
$\frac{3 π}{4}$	5.4002 - j1.8202	3.3989 - j1.6364	1.4992 + j0.1862	1.5107
$π$	6.74	4.6625	1.4456	1.4456
$\frac{5 π}{4}$	5.4002 + j1.8202	3.3989 + j1.6364	1.4992 - j0.1862	1.5107
$\frac{3 π}{2}$	3.58 + j1.16	1.7625 + j0.9	1.8777 - j0.3007	1.9016
$\frac{7 π}{4}$	3.7598 - j0.1798	2.1261 - j0.3636	1.7322 + j0.2117	1.7451
$2 π$	4.42	2.8625	1.5441	1.5441

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	1.5441
$ω = 1.297$ , the angle of the poles	1.2966	2.0367
$ω = \frac{π}{2}$ , the point z = j	1.5708	1.9016
$ω = π$ , the point $z = - 1$	3.1416	1.4456

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

arg

H (e^{j ω})

in radians, unwrapped, drawn from the computed response.

Where	$ω$	arg H, radians
$ω = 0$	0	0
$ω = \frac{π}{2}$	1.5708	0.159
$ω = π$	3.1416	0

Practise this +5 like it2081 Bhadra · Q43+7 marksPlot the pole zero in z plane and draw magnitude response (not to the scale) of the system described by difference equation

y [n] - 0.4 y [n - 1] + 0.25 y [n - 2] = x [n] - 0.4 x [n - 1]

What is different in this one

There is a zero at the origin, which appears only after multiplying by $z^{2}$ . It changes the phase and not the magnitude, since its distance to every point of the unit circle is 1, and forgetting it is the usual lost mark.
The poles are a conjugate pair at an angle of 1.159 rad, so the response has a peak near that frequency and the sketch is symmetric about $ω = 0$ .

\begin{aligned} H (z) & = \frac{1 - 0.4 z^{- 1}}{1 - 0.4 z^{- 1} + 0.25 z^{- 2}} & from the difference equation \\ = \frac{z^{2} - 0.4 z}{z^{2} - 0.4 z + 0.25} & \times z^{2} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	0.4	0.4	0
zero, o	0	0	0
pole, x	0.2 + j0.4583	0.5	66.42
pole, x	0.2 - j0.4583	0.5	-66.42

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 0.5, so for a causal system the ROC is $| z | > 0.5$ , which contains the unit circle, so the system is stable.

\begin{aligned} N (e^{j ω}) & = e^{j 2 ω} - 0.4 e^ jw \\ = \cos (2 ω) + jsin (2 ω) - 0.4 cos (ω) - j 0.4 sin (ω) \\ D (e^{j ω}) & = e^{j 2 ω} - 0.4 e^ jw + 0.25 \\ = \cos (2 ω) + jsin (2 ω) - 0.4 cos (ω) - j 0.4 sin (ω) + 0.25 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	0.6	0.85	0.7059	0.7059
$\frac{π}{4}$	-0.2828 + j0.7172	-0.0328 + j0.7172	1.0159 + j0.3479	1.0738
$\frac{π}{2}$	-1 - j0.4	-0.75 - j0.4	1.2595 - j0.1384	1.2671
$\frac{3 π}{4}$	0.2828 - j1.2828	0.5328 - j1.2828	0.931 - j0.1662	0.9457
$π$	1.4	1.65	0.8485	0.8485
$\frac{5 π}{4}$	0.2828 + j1.2828	0.5328 + j1.2828	0.931 + j0.1662	0.9457
$\frac{3 π}{2}$	-1 + j0.4	-0.75 + j0.4	1.2595 + j0.1384	1.2671
$\frac{7 π}{4}$	-0.2828 - j0.7172	-0.0328 - j0.7172	1.0159 - j0.3479	1.0738
$2 π$	0.6	0.85	0.7059	0.7059

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	0.7059
$ω = 1.159$ , the angle of the poles	1.1593	1.3194
$ω = \frac{π}{2}$ , the point z = j	1.5708	1.2671
$ω = π$ , the point $z = - 1$	3.1416	0.8485

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response rises to a peak of 1.3405 near $ω = 1.291$ , with $| H |$ = 0.7059 at $ω = 0$ and 0.8485 at $ω = π$ . That makes it a high pass shape.

Practise this +1 like it2081 Baishakh · Q43+7 marksPlot the pole-zero in z plane and draw the magnitude response (not to the scale) of the system described by difference equation

y (n) = 0.67 x (n) - 0.3 x (n - 1) + 2.75 y (n - 1)

What is different herea first order denominator; the poles are real, so the response falls away without a peak

What is different in this one

A pole lies on or outside the unit circle, at magnitude 2.75, so this system is unstable. Say so: it is a mark, and it also tells you the magnitude response will be dominated by that pole.
Every pole is real, so the response has no resonant peak: it rises or falls monotonically between the two ends.

\begin{aligned} H (z) & = \frac{0.67 - 0.3 z^{- 1}}{1 - 2.75 z^{- 1}} & from the difference equation \\ = \frac{0.67 z - 0.3}{z - 2.75} & \times z^{1} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	0.4478	0.4478	0
pole, x	2.75	2.75	0

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 2.75, so for a causal system the ROC is $| z | > 2.75$ , which does not contain the unit circle, so the system is unstable.

\begin{aligned} N (e^{j ω}) & = 0.67 e^ jw - 0.3 \\ = 0.67 cos (ω) + j 0.67 sin (ω) - 0.3 \\ D (e^{j ω}) & = e^{j ω} - 2.75 \\ = \cos (ω) + jsin (ω) - 2.75 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	0.37	-1.75	-0.2114	0.2114
$\frac{π}{4}$	0.1738 + j0.4738	-2.0429 + j0.7071	-0.0043 - j0.2334	0.2334
$\frac{π}{2}$	-0.3 + j0.67	-2.75 + j1	0.1746 - j0.1801	0.2509
$\frac{3 π}{4}$	-0.7738 + j0.4738	-3.4571 + j0.7071	0.2417 - j0.0876	0.2571
$π$	-0.97	-3.75	0.2587	0.2587
$\frac{5 π}{4}$	-0.7738 - j0.4738	-3.4571 - j0.7071	0.2417 + j0.0876	0.2571
$\frac{3 π}{2}$	-0.3 - j0.67	-2.75 - j1	0.1746 + j0.1801	0.2509
$\frac{7 π}{4}$	0.1738 - j0.4738	-2.0429 - j0.7071	-0.0043 + j0.2334	0.2334
$2 π$	0.37	-1.75	-0.2114	0.2114

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	0.2114
$ω = \frac{π}{2}$ , the point z = j	1.5708	0.2509
$ω = π$ , the point $z = - 1$	3.1416	0.2587

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response rises monotonically, with $| H |$ = 0.2114 at $ω = 0$ and 0.2587 at $ω = π$ . That makes it a high pass shape.

Practise this2080 Bhadra · Q43+7 marksPlot the pole-zero on the z-plane and draw Magnitude response (not to the scale) of an LTI system described by the equation,

y (n) = x (n) + 0.8 x (n - 1) + 0.8 x (n - 2) + 0.49 y (n - 2)

.What is different herethe poles are real, so the response falls away without a peak

What is different in this one

Every pole is real, so the response has no resonant peak: it rises or falls monotonically between the two ends.

\begin{aligned} H (z) & = \frac{1 + 0.8 z^{- 1} + 0.8 z^{- 2}}{1 - 0.49 z^{- 2}} & from the difference equation \\ = \frac{z^{2} + 0.8 z + 0.8}{z^{2} - 0.49} & \times z^{2} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	-0.4 + j0.8	0.8944	116.57
zero, o	-0.4 - j0.8	0.8944	-116.57
pole, x	-0.7	0.7	180
pole, x	0.7	0.7	0

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 0.7, so for a causal system the ROC is $| z | > 0.7$ , which contains the unit circle, so the system is stable.

\begin{aligned} N (e^{j ω}) & = e^{j 2 ω} + 0.8 e^ jw + 0.8 \\ = \cos (2 ω) + jsin (2 ω) + 0.8 cos (ω) + j 0.8 sin (ω) + 0.8 \\ D (e^{j ω}) & = e^{j 2 ω} - 0.49 \\ = \cos (2 ω) + jsin (2 ω) - 0.49 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	2.6	0.51	5.098	5.098
$\frac{π}{4}$	1.3657 + j1.5657	-0.49 + j1	0.7229 - j1.7199	1.8657
$\frac{π}{2}$	-0.2 + j0.8	-1.49	0.1342 - j0.5369	0.5534
$\frac{3 π}{4}$	0.2343 - j0.4343	-0.49 - j1	0.2576 + j0.3606	0.4431
$π$	1	0.51	1.9608	1.9608
$\frac{5 π}{4}$	0.2343 + j0.4343	-0.49 + j1	0.2576 - j0.3606	0.4431
$\frac{3 π}{2}$	-0.2 - j0.8	-1.49	0.1342 + j0.5369	0.5534
$\frac{7 π}{4}$	1.3657 - j1.5657	-0.49 - j1	0.7229 + j1.7199	1.8657
$2 π$	2.6	0.51	5.098	5.098

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	5.098
$ω = \frac{π}{2}$ , the point z = j	1.5708	0.5534
$ω = π$ , the point $z = - 1$	3.1416	1.9608

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response falls monotonically, with $| H |$ = 5.098 at $ω = 0$ and 1.9608 at $ω = π$ . That makes it a low pass shape.

What is different in this one

A pole lies on or outside the unit circle, at magnitude 1.6621, so this system is unstable. Say so: it is a mark, and it also tells you the magnitude response will be dominated by that pole.
The poles are a conjugate pair at an angle of 1.297 rad, so the response has a peak near that frequency and the sketch is symmetric about $ω = 0$ .

\begin{aligned} H (z) & = \frac{1 - 1.16 z^{- 1} + 4.58 z^{- 2}}{1 - 0.9 z^{- 1} + 2.7625 z^{- 2}} & from the difference equation \\ = \frac{z^{2} - 1.16 z + 4.58}{z^{2} - 0.9 z + 2.7625} & \times z^{2} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	0.58 + j2.06	2.1401	74.28
zero, o	0.58 - j2.06	2.1401	-74.28
pole, x	0.45 + j1.6	1.6621	74.29
pole, x	0.45 - j1.6	1.6621	-74.29

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 1.6621, so for a causal system the ROC is $| z | > 1.6621$ , which does not contain the unit circle, so the system is unstable.

\begin{aligned} N (e^{j ω}) & = e^{j 2 ω} - 1.16 e^ jw + 4.58 \\ = \cos (2 ω) + jsin (2 ω) - 1.16 cos (ω) - j 1.16 sin (ω) + 4.58 \\ D (e^{j ω}) & = e^{j 2 ω} - 0.9 e^ jw + 2.7625 \\ = \cos (2 ω) + jsin (2 ω) - 0.9 cos (ω) - j 0.9 sin (ω) + 2.7625 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	4.42	2.8625	1.5441	1.5441
$\frac{π}{4}$	3.7598 + j0.1798	2.1261 + j0.3636	1.7322 - j0.2117	1.7451
$\frac{π}{2}$	3.58 - j1.16	1.7625 - j0.9	1.8777 + j0.3007	1.9016
$\frac{3 π}{4}$	5.4002 - j1.8202	3.3989 - j1.6364	1.4992 + j0.1862	1.5107
$π$	6.74	4.6625	1.4456	1.4456
$\frac{5 π}{4}$	5.4002 + j1.8202	3.3989 + j1.6364	1.4992 - j0.1862	1.5107
$\frac{3 π}{2}$	3.58 + j1.16	1.7625 + j0.9	1.8777 - j0.3007	1.9016
$\frac{7 π}{4}$	3.7598 - j0.1798	2.1261 - j0.3636	1.7322 + j0.2117	1.7451
$2 π$	4.42	2.8625	1.5441	1.5441

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	1.5441
$ω = 1.297$ , the angle of the poles	1.2966	2.0367
$ω = \frac{π}{2}$ , the point z = j	1.5708	1.9016
$ω = π$ , the point $z = - 1$	3.1416	1.4456

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response rises to a peak of 2.0368 near $ω = 1.291$ , with $| H |$ = 1.5441 at $ω = 0$ and 1.4456 at $ω = π$ . That makes it a low pass shape.

Practise this2079 Bhadra · Q43+7 marksPlot the pole-zero in z-plane and draw the magnitude response (not to the scale) of the equation of the system describe by difference equation:

y [n]

- 0.

35 y [n - 1]

+ 0.

25 y [n - 2] = x [n]

- 0.

75 x [n - 1]

.What is different hereThe standard text works this one as its Example 3.3 and slips in the

ω = \frac{π}{4}

row: it prints

D (e^{j ω}) = - 0.025

+ j0.75, but 0.25 - 0.2475 is <

b > 0.0025 < / b

>, not -0.025, so

D = 0.0025

+ j0.7525 and

H = 0.6218

+ j0.7068. The same slip is repeated at

\frac{7 π}{4}

. The magnitude survives it almost unchanged, 0.9414 against their 0.943, which is exactly why it is worth checking the real and imaginary parts and not only the modulus.

What is different in this one

There is a zero at the origin, which appears only after multiplying by $z^{2}$ . It changes the phase and not the magnitude, since its distance to every point of the unit circle is 1, and forgetting it is the usual lost mark.
The poles are a conjugate pair at an angle of 1.213 rad, so the response has a peak near that frequency and the sketch is symmetric about $ω = 0$ .

\begin{aligned} H (z) & = \frac{1 - 0.75 z^{- 1}}{1 - 0.35 z^{- 1} + 0.25 z^{- 2}} & from the difference equation \\ = \frac{z^{2} - 0.75 z}{z^{2} - 0.35 z + 0.25} & \times z^{2} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	0.75	0.75	0
zero, o	0	0	0
pole, x	0.175 + j0.4684	0.5	69.51
pole, x	0.175 - j0.4684	0.5	-69.51

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 0.5, so for a causal system the ROC is $| z | > 0.5$ , which contains the unit circle, so the system is stable.

\begin{aligned} N (e^{j ω}) & = e^{j 2 ω} - 0.75 e^ jw \\ = \cos (2 ω) + jsin (2 ω) - 0.75 cos (ω) - j 0.75 sin (ω) \\ D (e^{j ω}) & = e^{j 2 ω} - 0.35 e^ jw + 0.25 \\ = \cos (2 ω) + jsin (2 ω) - 0.35 cos (ω) - j 0.35 sin (ω) + 0.25 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	0.25	0.9	0.2778	0.2778
$\frac{π}{4}$	-0.5303 + j0.4697	0.0025 + j0.7525	0.6218 + j0.7068	0.9414
$\frac{π}{2}$	-1 - j0.75	-0.75 - j0.35	1.4781 + j0.3102	1.5103
$\frac{3 π}{4}$	0.5303 - j1.5303	0.4975 - j1.2475	1.2047 - j0.0553	1.2059
$π$	1.75	1.6	1.0938	1.0938
$\frac{5 π}{4}$	0.5303 + j1.5303	0.4975 + j1.2475	1.2047 + j0.0553	1.2059
$\frac{3 π}{2}$	-1 + j0.75	-0.75 + j0.35	1.4781 - j0.3102	1.5103
$\frac{7 π}{4}$	-0.5303 - j0.4697	0.0025 - j0.7525	0.6218 - j0.7068	0.9414
$2 π$	0.25	0.9	0.2778	0.2778

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	0.2778
$ω = 1.213$ , the angle of the poles	1.2132	1.4387
$ω = \frac{π}{2}$ , the point z = j	1.5708	1.5103
$ω = π$ , the point $z = - 1$	3.1416	1.0938

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response rises to a peak of 1.5251 near $ω = 1.456$ , with $| H |$ = 0.2778 at $ω = 0$ and 1.0938 at $ω = π$ . That makes it a high pass shape.

Same as 2081 Bhadra Q42079 Baishakh · Q43+7 marksPlot pole-zero in z-plane and draw magnitude response (not to the scale) of the system described by difference equation

y [n] - 0.3 y [n - 1]

+ 0.

2 y [n - 2] = x [n]

- 0.

5 x [n - 1]

What is different in this one

There is a zero at the origin, which appears only after multiplying by $z^{2}$ . It changes the phase and not the magnitude, since its distance to every point of the unit circle is 1, and forgetting it is the usual lost mark.
The poles are a conjugate pair at an angle of 1.229 rad, so the response has a peak near that frequency and the sketch is symmetric about $ω = 0$ .

\begin{aligned} H (z) & = \frac{1 - 0.5 z^{- 1}}{1 - 0.3 z^{- 1} + 0.2 z^{- 2}} & from the difference equation \\ = \frac{z^{2} - 0.5 z}{z^{2} - 0.3 z + 0.2} & \times z^{2} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	0.5	0.5	0
zero, o	0	0	0
pole, x	0.15 + j0.4213	0.4472	70.4
pole, x	0.15 - j0.4213	0.4472	-70.4

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 0.4472, so for a causal system the ROC is $| z | > 0.4472$ , which contains the unit circle, so the system is stable.

\begin{aligned} N (e^{j ω}) & = e^{j 2 ω} - 0.5 e^ jw \\ = \cos (2 ω) + jsin (2 ω) - 0.5 cos (ω) - j 0.5 sin (ω) \\ D (e^{j ω}) & = e^{j 2 ω} - 0.3 e^ jw + 0.2 \\ = \cos (2 ω) + jsin (2 ω) - 0.3 cos (ω) - j 0.3 sin (ω) + 0.2 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	0.5	0.9	0.5556	0.5556
$\frac{π}{4}$	-0.3536 + j0.6464	-0.0121 + j0.7879	0.8272 + j0.436	0.9351
$\frac{π}{2}$	-1 - j0.5	-0.8 - j0.3	1.3014 + j0.137	1.3086
$\frac{3 π}{4}$	0.3536 - j1.3536	0.4121 - j1.2121	1.0899 - j0.0789	1.0927
$π$	1.5	1.5	1	1
$\frac{5 π}{4}$	0.3536 + j1.3536	0.4121 + j1.2121	1.0899 + j0.0789	1.0927
$\frac{3 π}{2}$	-1 + j0.5	-0.8 + j0.3	1.3014 - j0.137	1.3086
$\frac{7 π}{4}$	-0.3536 - j0.6464	-0.0121 - j0.7879	0.8272 - j0.436	0.9351
$2 π$	0.5	0.9	0.5556	0.5556

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	0.5556
$ω = 1.229$ , the angle of the poles	1.2288	1.2574
$ω = \frac{π}{2}$ , the point z = j	1.5708	1.3086
$ω = π$ , the point $z = - 1$	3.1416	1

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response rises to a peak of 1.3142 near $ω = 1.482$ , with $| H |$ = 0.5556 at $ω = 0$ and 1 at $ω = π$ . That makes it a high pass shape.

Same as 2081 Bhadra Q42078 Bhadra · Q52+4 marksPlot the pole-zero in z-plane and draw magnitude response (not to the scale) of the system described by difference equation.

y [n] - 0.4 y [n - 1] + 0.25 y [n - 2] = x [n]

- 0.

4 x [n - 1]

What is different in this one

There is a zero at the origin, which appears only after multiplying by $z^{2}$ . It changes the phase and not the magnitude, since its distance to every point of the unit circle is 1, and forgetting it is the usual lost mark.
The poles are a conjugate pair at an angle of 1.159 rad, so the response has a peak near that frequency and the sketch is symmetric about $ω = 0$ .

\begin{aligned} H (z) & = \frac{1 - 0.4 z^{- 1}}{1 - 0.4 z^{- 1} + 0.25 z^{- 2}} & from the difference equation \\ = \frac{z^{2} - 0.4 z}{z^{2} - 0.4 z + 0.25} & \times z^{2} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	0.4	0.4	0
zero, o	0	0	0
pole, x	0.2 + j0.4583	0.5	66.42
pole, x	0.2 - j0.4583	0.5	-66.42

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 0.5, so for a causal system the ROC is $| z | > 0.5$ , which contains the unit circle, so the system is stable.

\begin{aligned} N (e^{j ω}) & = e^{j 2 ω} - 0.4 e^ jw \\ = \cos (2 ω) + jsin (2 ω) - 0.4 cos (ω) - j 0.4 sin (ω) \\ D (e^{j ω}) & = e^{j 2 ω} - 0.4 e^ jw + 0.25 \\ = \cos (2 ω) + jsin (2 ω) - 0.4 cos (ω) - j 0.4 sin (ω) + 0.25 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	0.6	0.85	0.7059	0.7059
$\frac{π}{4}$	-0.2828 + j0.7172	-0.0328 + j0.7172	1.0159 + j0.3479	1.0738
$\frac{π}{2}$	-1 - j0.4	-0.75 - j0.4	1.2595 - j0.1384	1.2671
$\frac{3 π}{4}$	0.2828 - j1.2828	0.5328 - j1.2828	0.931 - j0.1662	0.9457
$π$	1.4	1.65	0.8485	0.8485
$\frac{5 π}{4}$	0.2828 + j1.2828	0.5328 + j1.2828	0.931 + j0.1662	0.9457
$\frac{3 π}{2}$	-1 + j0.4	-0.75 + j0.4	1.2595 + j0.1384	1.2671
$\frac{7 π}{4}$	-0.2828 - j0.7172	-0.0328 - j0.7172	1.0159 - j0.3479	1.0738
$2 π$	0.6	0.85	0.7059	0.7059

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	0.7059
$ω = 1.159$ , the angle of the poles	1.1593	1.3194
$ω = \frac{π}{2}$ , the point z = j	1.5708	1.2671
$ω = π$ , the point $z = - 1$	3.1416	0.8485

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response rises to a peak of 1.3405 near $ω = 1.291$ , with $| H |$ = 0.7059 at $ω = 0$ and 0.8485 at $ω = π$ . That makes it a high pass shape.

What is different in this one

A pole lies on or outside the unit circle, at magnitude 1.1516, so this system is unstable. Say so: it is a mark, and it also tells you the magnitude response will be dominated by that pole.
The poles are a conjugate pair at an angle of 1.169 rad, so the response has a peak near that frequency and the sketch is symmetric about $ω = 0$ .

\begin{aligned} H (z) & = \frac{1 - 1.16 z^{- 1} + 4.58 z^{- 2}}{1 - 0.9 z^{- 1} + 1.3261 z^{- 2}} & from the difference equation \\ = \frac{z^{2} - 1.16 z + 4.58}{z^{2} - 0.9 z + 1.3261} & \times z^{2} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	0.58 + j2.06	2.1401	74.28
zero, o	0.58 - j2.06	2.1401	-74.28
pole, x	0.45 + j1.06	1.1516	67
pole, x	0.45 - j1.06	1.1516	-67

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 1.1516, so for a causal system the ROC is $| z | > 1.1516$ , which does not contain the unit circle, so the system is unstable.

\begin{aligned} N (e^{j ω}) & = e^{j 2 ω} - 1.16 e^ jw + 4.58 \\ = \cos (2 ω) + jsin (2 ω) - 1.16 cos (ω) - j 1.16 sin (ω) + 4.58 \\ D (e^{j ω}) & = e^{j 2 ω} - 0.9 e^ jw + 1.3261 \\ = \cos (2 ω) + jsin (2 ω) - 0.9 cos (ω) - j 0.9 sin (ω) + 1.3261 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	4.42	1.4261	3.0994	3.0994
$\frac{π}{4}$	3.7598 + j0.1798	0.6897 + j0.3636	4.3732 - j2.0449	4.8277
$\frac{π}{2}$	3.58 - j1.16	0.3261 - j0.9	2.4133 + j3.1033	3.9313
$\frac{3 π}{4}$	5.4002 - j1.8202	1.9625 - j1.6364	2.0794 + j0.8063	2.2302
$π$	6.74	3.2261	2.0892	2.0892
$\frac{5 π}{4}$	5.4002 + j1.8202	1.9625 + j1.6364	2.0794 - j0.8063	2.2302
$\frac{3 π}{2}$	3.58 + j1.16	0.3261 + j0.9	2.4133 - j3.1033	3.9313
$\frac{7 π}{4}$	3.7598 - j0.1798	0.6897 - j0.3636	4.3732 + j2.0449	4.8277
$2 π$	4.42	1.4261	3.0994	3.0994

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	3.0994
$ω = 1.169$ , the angle of the poles	1.1693	11.4875
$ω = \frac{π}{2}$ , the point z = j	1.5708	3.9313
$ω = π$ , the point $z = - 1$	3.1416	2.0892

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response rises to a peak of 11.4954 near $ω = 1.165$ , with $| H |$ = 3.0994 at $ω = 0$ and 2.0892 at $ω = π$ . That makes it a low pass shape.

Practise this2076 Ashwin · Q62+8 marksFor the system described by the following difference equation:

y [n] = 0

67 x [n]

- 0.

3 x [n - 1]

+ 2.

75 y [n - 1]

What is different in this one

A pole lies on or outside the unit circle, at magnitude 2.75, so this system is unstable. Say so: it is a mark, and it also tells you the magnitude response will be dominated by that pole.
Every pole is real, so the response has no resonant peak: it rises or falls monotonically between the two ends.

\begin{aligned} H (z) & = \frac{0.67 - 0.3 z^{- 1}}{1 - 2.75 z^{- 1}} & from the difference equation \\ = \frac{0.67 z - 0.3}{z - 2.75} & \times z^{1} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	0.4478	0.4478	0
pole, x	2.75	2.75	0

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 2.75, so for a causal system the ROC is $| z | > 2.75$ , which does not contain the unit circle, so the system is unstable.

\begin{aligned} N (e^{j ω}) & = 0.67 e^ jw - 0.3 \\ = 0.67 cos (ω) + j 0.67 sin (ω) - 0.3 \\ D (e^{j ω}) & = e^{j ω} - 2.75 \\ = \cos (ω) + jsin (ω) - 2.75 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	0.37	-1.75	-0.2114	0.2114
$\frac{π}{4}$	0.1738 + j0.4738	-2.0429 + j0.7071	-0.0043 - j0.2334	0.2334
$\frac{π}{2}$	-0.3 + j0.67	-2.75 + j1	0.1746 - j0.1801	0.2509
$\frac{3 π}{4}$	-0.7738 + j0.4738	-3.4571 + j0.7071	0.2417 - j0.0876	0.2571
$π$	-0.97	-3.75	0.2587	0.2587
$\frac{5 π}{4}$	-0.7738 - j0.4738	-3.4571 - j0.7071	0.2417 + j0.0876	0.2571
$\frac{3 π}{2}$	-0.3 - j0.67	-2.75 - j1	0.1746 + j0.1801	0.2509
$\frac{7 π}{4}$	0.1738 - j0.4738	-2.0429 - j0.7071	-0.0043 + j0.2334	0.2334
$2 π$	0.37	-1.75	-0.2114	0.2114

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	0.2114
$ω = \frac{π}{2}$ , the point z = j	1.5708	0.2509
$ω = π$ , the point $z = - 1$	3.1416	0.2587

arg

H (e^{j ω})

in radians, unwrapped, drawn from the computed response.

Where	$ω$	arg H, radians
$ω = 0$	0	-3.142
$ω = \frac{π}{2}$	1.5708	-0.801
$ω = π$	3.1416	0

The magnitude is not asked here, only the pole zero map and the phase. The phase runs from -3.142 rad at $ω = 0$ to 0 rad at $ω = π$ , rising steadily.

Same as 2081 Baishakh Q42075 Chaitra · Q32+7 marksPlot the pole-zero in z-plane and draw magnitude response (not to scale) of the system described by differential equation

y (n) - 0.3 y (n - 1) = 2 x (n - 2) + 0.7 x (n - 1) + 4 x (n)

What is different in this one

Every pole is real, so the response has no resonant peak: it rises or falls monotonically between the two ends.

\begin{aligned} H (z) & = \frac{4 + 0.7 z^{- 1} + 2 z^{- 2}}{1 - 0.3 z^{- 1}} & from the difference equation \\ = \frac{4 z^{2} + 0.7 z + 2}{z^{2} - 0.3 z} & \times z^{2} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	-0.0875 + j0.7017	0.7071	97.11
zero, o	-0.0875 - j0.7017	0.7071	-97.11
pole, x	0.3	0.3	0
pole, x	0	0	0

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 0.3, so for a causal system the ROC is $| z | > 0.3$ , which contains the unit circle, so the system is stable.

\begin{aligned} N (e^{j ω}) & = 4 e^j 2 ω + 0.7 e^ jw + 2 \\ = 4 cos (2 ω) + j 4 sin (2 ω) + 0.7 cos (ω) + j 0.7 sin (ω) + 2 \\ D (e^{j ω}) & = e^{j 2 ω} - 0.3 e^ jw \\ = \cos (2 ω) + jsin (2 ω) - 0.3 cos (ω) - j 0.3 sin (ω) \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	6.7	0.7	9.5714	9.5714
$\frac{π}{4}$	2.495 + j4.495	-0.2121 + j0.7879	4.5246 - j4.385	6.3008
$\frac{π}{2}$	-2 + j0.7	-1 - j0.3	1.6422 - j1.1927	2.0296
$\frac{3 π}{4}$	1.505 - j3.505	0.2121 - j1.2121	3.0165 + j0.7137	3.0998
$π$	5.3	1.3	4.0769	4.0769
$\frac{5 π}{4}$	1.505 + j3.505	0.2121 + j1.2121	3.0165 - j0.7137	3.0998
$\frac{3 π}{2}$	-2 - j0.7	-1 + j0.3	1.6422 + j1.1927	2.0296
$\frac{7 π}{4}$	2.495 - j4.495	-0.2121 - j0.7879	4.5246 + j4.385	6.3008
$2 π$	6.7	0.7	9.5714	9.5714

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	9.5714
$ω = \frac{π}{2}$ , the point z = j	1.5708	2.0296
$ω = π$ , the point $z = - 1$	3.1416	4.0769

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response falls monotonically, with $| H |$ = 9.5714 at $ω = 0$ and 4.0769 at $ω = π$ . That makes it a low pass shape.

What is different in this one

A pole lies on or outside the unit circle, at magnitude 1.1516, so this system is unstable. Say so: it is a mark, and it also tells you the magnitude response will be dominated by that pole.
The poles are a conjugate pair at an angle of 1.169 rad, so the response has a peak near that frequency and the sketch is symmetric about $ω = 0$ .

\begin{aligned} H (z) & = \frac{1 - 1.16 z^{- 1} + 4.58 z^{- 2}}{1 - 0.9 z^{- 1} + 1.3261 z^{- 2}} & from the difference equation \\ = \frac{z^{2} - 1.16 z + 4.58}{z^{2} - 0.9 z + 1.3261} & \times z^{2} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	0.58 + j2.06	2.1401	74.28
zero, o	0.58 - j2.06	2.1401	-74.28
pole, x	0.45 + j1.06	1.1516	67
pole, x	0.45 - j1.06	1.1516	-67

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 1.1516, so for a causal system the ROC is $| z | > 1.1516$ , which does not contain the unit circle, so the system is unstable.

\begin{aligned} N (e^{j ω}) & = e^{j 2 ω} - 1.16 e^ jw + 4.58 \\ = \cos (2 ω) + jsin (2 ω) - 1.16 cos (ω) - j 1.16 sin (ω) + 4.58 \\ D (e^{j ω}) & = e^{j 2 ω} - 0.9 e^ jw + 1.3261 \\ = \cos (2 ω) + jsin (2 ω) - 0.9 cos (ω) - j 0.9 sin (ω) + 1.3261 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	4.42	1.4261	3.0994	3.0994
$\frac{π}{4}$	3.7598 + j0.1798	0.6897 + j0.3636	4.3732 - j2.0449	4.8277
$\frac{π}{2}$	3.58 - j1.16	0.3261 - j0.9	2.4133 + j3.1033	3.9313
$\frac{3 π}{4}$	5.4002 - j1.8202	1.9625 - j1.6364	2.0794 + j0.8063	2.2302
$π$	6.74	3.2261	2.0892	2.0892
$\frac{5 π}{4}$	5.4002 + j1.8202	1.9625 + j1.6364	2.0794 - j0.8063	2.2302
$\frac{3 π}{2}$	3.58 + j1.16	0.3261 + j0.9	2.4133 - j3.1033	3.9313
$\frac{7 π}{4}$	3.7598 - j0.1798	0.6897 - j0.3636	4.3732 + j2.0449	4.8277
$2 π$	4.42	1.4261	3.0994	3.0994

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	3.0994
$ω = 1.169$ , the angle of the poles	1.1693	11.4875
$ω = \frac{π}{2}$ , the point z = j	1.5708	3.9313
$ω = π$ , the point $z = - 1$	3.1416	2.0892

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response rises to a peak of 11.4954 near $ω = 1.165$ , with $| H |$ = 3.0994 at $ω = 0$ and 2.0892 at $ω = π$ . That makes it a low pass shape.

What is different in this one

A pole lies on or outside the unit circle, at magnitude 2.0224, so this system is unstable. Say so: it is a mark, and it also tells you the magnitude response will be dominated by that pole.
There is a zero at the origin, which appears only after multiplying by $z^{4}$ . It changes the phase and not the magnitude, since its distance to every point of the unit circle is 1, and forgetting it is the usual lost mark.
The poles are a conjugate pair at an angle of 1.042 rad, so the response has a peak near that frequency and the sketch is symmetric about $ω = 0$ .

\begin{aligned} H (z) & = \frac{1}{1 + 3.1 z^{- 1} + 1.2854 z^{- 2} - 0.4994 z^{- 3} + 3.2532 z^{- 4}} & from the difference equation \\ = \frac{z^{4}}{z^{4} + 3.1 z^{3} + 1.2854 z^{2} - 0.4994 z + 3.2532} & \times z^{4} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	0	0	0
zero, o	0	0	0
zero, o	0	0	0
zero, o	0	0	0
pole, x	-2 + j0.3	2.0224	171.47
pole, x	-2 - j0.3	2.0224	-171.47
pole, x	0.45 + j0.77	0.8919	59.7
pole, x	0.45 - j0.77	0.8919	-59.7

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 2.0224, so for a causal system the ROC is $| z | > 2.0224$ , which does not contain the unit circle, so the system is unstable.

\begin{aligned} N (e^{j ω}) & = e^{j 4 ω} \\ = \cos (4 ω) + jsin (4 ω) \\ D (e^{j ω}) & = e^{j 4 ω} + 3.1 e^j 3 ω + 1.2854 e^j 2 ω - 0.4994 e^ jw + 3.2532 \\ = \cos (4 ω) + jsin (4 ω) + 3.1 cos (3 ω) + j 3.1 sin (3 ω) + 1.2854 cos (2 ω) + j 1.2854 sin (2 ω) - 0.4994 cos (ω) - j 0.4994 sin (ω) + 3.2532 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	1	8.1392	0.1229	0.1229
$\frac{π}{4}$	-1	-0.292 + j3.1243	0.0297 + j0.3173	0.3187
$\frac{π}{2}$	1	2.9678 - j3.5994	0.1364 + j0.1654	0.2144
$\frac{3 π}{4}$	-1	4.7983 + j0.5535	-0.2057 + j0.0237	0.207
$π$	1	2.938	0.3404	0.3404
$\frac{5 π}{4}$	-1	4.7983 - j0.5535	-0.2057 - j0.0237	0.207
$\frac{3 π}{2}$	1	2.9678 + j3.5994	0.1364 - j0.1654	0.2144
$\frac{7 π}{4}$	-1	-0.292 - j3.1243	0.0297 - j0.3173	0.3187
$2 π$	1	8.1392	0.1229	0.1229

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	0.1229
$ω = 1.042$ , the angle of the poles	1.0419	0.7981
$ω = \frac{π}{2}$ , the point z = j	1.5708	0.2144
$ω = π$ , the point $z = - 1$	3.1416	0.3404

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response rises to a peak of 0.7983 near $ω = 1.045$ , with $| H |$ = 0.1229 at $ω = 0$ and 0.3404 at $ω = π$ . That makes it a high pass shape.

Same as 2081 Bhadra Q42074 Ashwin · Q43+7 marksPlot the pole-zero in z-plane and Draw Magnitude Response (not to the scale) of the system described by difference equation

y [n]

- 0.

4 y [n - 1]

+ 0.

2 y [n - 2] = x [n]

+ 0.

5 x [n - 1]

+ 0.

6 x [n - 2]

+ 0.

8 x [n - 3]

What is different in this one

The poles are a conjugate pair at an angle of 1.107 rad, so the response has a peak near that frequency and the sketch is symmetric about $ω = 0$ .

\begin{aligned} H (z) & = \frac{1 + 0.5 z^{- 1} + 0.6 z^{- 2} + 0.8 z^{- 3}}{1 - 0.4 z^{- 1} + 0.2 z^{- 2}} & from the difference equation \\ = \frac{z^{3} + 0.5 z^{2} + 0.6 z + 0.8}{z^{3} - 0.4 z^{2} + 0.2 z} & \times z^{3} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	0.1845 + j0.9416	0.9595	78.91
zero, o	0.1845 - j0.9416	0.9595	-78.91
zero, o	-0.869	0.869	180
pole, x	0.2 + j0.4	0.4472	63.43
pole, x	0.2 - j0.4	0.4472	-63.43
pole, x	0	0	0

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 0.4472, so for a causal system the ROC is $| z | > 0.4472$ , which contains the unit circle, so the system is stable.

\begin{aligned} N (e^{j ω}) & = e^{j 3 ω} + 0.5 e^j 2 ω + 0.6 e^ jw + 0.8 \\ = \cos (3 ω) + jsin (3 ω) + 0.5 cos (2 ω) + j 0.5 sin (2 ω) + 0.6 cos (ω) + j 0.6 sin (ω) + 0.8 \\ D (e^{j ω}) & = e^{j 3 ω} - 0.4 e^j 2 ω + 0.2 e^ jw \\ = \cos (3 ω) + jsin (3 ω) - 0.4 cos (2 ω) - j 0.4 sin (2 ω) + 0.2 cos (ω) + j 0.2 sin (ω) \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	2.9	0.8	3.625	3.625
$\frac{π}{4}$	0.5172 + j1.6314	-0.5657 + j0.4485	0.8426 - j2.2158	2.3706
$\frac{π}{2}$	0.3 - j0.4	0.4 - j0.8	0.55 + j0.1	0.559
$\frac{3 π}{4}$	1.0828 + j0.6314	0.5657 + j1.2485	0.7456 - j0.5295	0.9145
$π$	-0.3	-1.6	0.1875	0.1875
$\frac{5 π}{4}$	1.0828 - j0.6314	0.5657 - j1.2485	0.7456 + j0.5295	0.9145
$\frac{3 π}{2}$	0.3 + j0.4	0.4 + j0.8	0.55 - j0.1	0.559
$\frac{7 π}{4}$	0.5172 - j1.6314	-0.5657 - j0.4485	0.8426 + j2.2158	2.3706
$2 π$	2.9	0.8	3.625	3.625

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	3.625
$ω = 1.107$ , the angle of the poles	1.1071	1.0815
$ω = \frac{π}{2}$ , the point z = j	1.5708	0.559
$ω = π$ , the point $z = - 1$	3.1416	0.1875

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response falls monotonically, with $| H |$ = 3.625 at $ω = 0$ and 0.1875 at $ω = π$ . That makes it a low pass shape.

Same as 2081 Bhadra Q42072 Chaitra · Q52+4 marksPlot the pole-zero in z-plane and Draw Magnitude Response (not to the scale) of the system described by difference equation.

y [n]

- 0.

4 y [n - 1]

+ 0.

1 y [n - 2] = x [n]

+ 0.

6 x [n - 1]

What is different in this one

There is a zero at the origin, which appears only after multiplying by $z^{2}$ . It changes the phase and not the magnitude, since its distance to every point of the unit circle is 1, and forgetting it is the usual lost mark.
The poles are a conjugate pair at an angle of 0.886 rad, so the response has a peak near that frequency and the sketch is symmetric about $ω = 0$ .

\begin{aligned} H (z) & = \frac{1 + 0.6 z^{- 1}}{1 - 0.4 z^{- 1} + 0.1 z^{- 2}} & from the difference equation \\ = \frac{z^{2} + 0.6 z}{z^{2} - 0.4 z + 0.1} & \times z^{2} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	-0.6	0.6	180
zero, o	0	0	0
pole, x	0.2 + j0.2449	0.3162	50.77
pole, x	0.2 - j0.2449	0.3162	-50.77

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 0.3162, so for a causal system the ROC is $| z | > 0.3162$ , which contains the unit circle, so the system is stable.

\begin{aligned} N (e^{j ω}) & = e^{j 2 ω} + 0.6 e^ jw \\ = \cos (2 ω) + jsin (2 ω) + 0.6 cos (ω) + j 0.6 sin (ω) \\ D (e^{j ω}) & = e^{j 2 ω} - 0.4 e^ jw + 0.1 \\ = \cos (2 ω) + jsin (2 ω) - 0.4 cos (ω) - j 0.4 sin (ω) + 0.1 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	1.6	0.7	2.2857	2.2857
$\frac{π}{4}$	0.4243 + j1.4243	-0.1828 + j0.7172	1.7231 - j1.0309	2.008
$\frac{π}{2}$	-1 + j0.6	-0.9 - j0.4	0.6804 - j0.9691	1.1841
$\frac{3 π}{4}$	-0.4243 - j0.5757	0.3828 - j1.2828	0.3215 - j0.4267	0.5342
$π$	0.4	1.5	0.2667	0.2667
$\frac{5 π}{4}$	-0.4243 + j0.5757	0.3828 + j1.2828	0.3215 + j0.4267	0.5342
$\frac{3 π}{2}$	-1 - j0.6	-0.9 + j0.4	0.6804 + j0.9691	1.1841
$\frac{7 π}{4}$	0.4243 - j1.4243	-0.1828 - j0.7172	1.7231 + j1.0309	2.008
$2 π$	1.6	0.7	2.2857	2.2857

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	2.2857
$ω = 0.886$ , the angle of the poles	0.8861	1.9223
$ω = \frac{π}{2}$ , the point z = j	1.5708	1.1841
$ω = π$ , the point $z = - 1$	3.1416	0.2667

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response falls monotonically, with $| H |$ = 2.2857 at $ω = 0$ and 0.2667 at $ω = π$ . That makes it a low pass shape.

Same as 2081 Bhadra Q42072 Kartik · Q56 marksPlot Magnitude Response (not to the scale) of the system described by difference equation.

y [n]

- 0.

5 y [n - 1]

+ 0.

3 y [n - 2] = x [n]

+ 0.

7 x [n - 1]

What is different in this one

There is a zero at the origin, which appears only after multiplying by $z^{2}$ . It changes the phase and not the magnitude, since its distance to every point of the unit circle is 1, and forgetting it is the usual lost mark.
The poles are a conjugate pair at an angle of 1.097 rad, so the response has a peak near that frequency and the sketch is symmetric about $ω = 0$ .

\begin{aligned} H (z) & = \frac{1 + 0.7 z^{- 1}}{1 - 0.5 z^{- 1} + 0.3 z^{- 2}} & from the difference equation \\ = \frac{z^{2} + 0.7 z}{z^{2} - 0.5 z + 0.3} & \times z^{2} above and below \end{aligned}

	Location	Magnitude	Angle, degrees
zero, o	-0.7	0.7	180
zero, o	0	0	0
pole, x	0.25 + j0.4873	0.5477	62.84
pole, x	0.25 - j0.4873	0.5477	-62.84

Poles marked x, zeros marked o, and the unit circle dashed.

The largest pole magnitude is 0.5477, so for a causal system the ROC is $| z | > 0.5477$ , which contains the unit circle, so the system is stable.

\begin{aligned} N (e^{j ω}) & = e^{j 2 ω} + 0.7 e^ jw \\ = \cos (2 ω) + jsin (2 ω) + 0.7 cos (ω) + j 0.7 sin (ω) \\ D (e^{j ω}) & = e^{j 2 ω} - 0.5 e^ jw + 0.3 \\ = \cos (2 ω) + jsin (2 ω) - 0.5 cos (ω) - j 0.5 sin (ω) + 0.3 \end{aligned}

$ω$	$N (e^{j ω})$	$D (e^{j ω})$	$H (e^{j ω})$	$\| H (e^{j ω}) \|$
0	1.7	0.8	2.125	2.125
$\frac{π}{4}$	0.495 + j1.495	-0.0536 + j0.6464	2.2338 - j0.9507	2.4277
$\frac{π}{2}$	-1 + j0.7	-0.7 - j0.5	0.473 - j1.3378	1.419
$\frac{3 π}{4}$	-0.495 - j0.505	0.6536 - j1.3536	0.1594 - j0.4426	0.4705
$π$	0.3	1.8	0.1667	0.1667
$\frac{5 π}{4}$	-0.495 + j0.505	0.6536 + j1.3536	0.1594 + j0.4426	0.4705
$\frac{3 π}{2}$	-1 - j0.7	-0.7 + j0.5	0.473 + j1.3378	1.419
$\frac{7 π}{4}$	0.495 - j1.495	-0.0536 - j0.6464	2.2338 + j0.9507	2.4277
$2 π$	1.7	0.8	2.125	2.125

Where	$ω$	$\| H (e^{j ω}) \|$
$ω = 0$ , the point $z = 1$	0	2.125
$ω = 1.097$ , the angle of the poles	1.0968	2.3168
$ω = \frac{π}{2}$ , the point z = j	1.5708	1.419
$ω = π$ , the point $z = - 1$	3.1416	0.1667

| H (e^{j ω}) |

, drawn from the computed response. The shape is what the question wants; the scale is a bonus.

The response rises to a peak of 2.4438 near $ω = 0.877$ , with $| H |$ = 2.125 at $ω = 0$ and 0.1667 at $ω = π$ . That makes it a low pass shape.

Circular convolution TOP 17/19

Ch 7 · DFT and FFT17 questions, from 17 of the 19 sittings3 to practise

The procedure, the same for every question in this topic

y [n] = x_{1} (N) x_{2} = \sum_{m = 0}^{N - 1} x_{1} [m] x_{2} [(n - m) mod N]

N is the DFT length. For unequal lengths N is the larger, the shorter sequence is zero padded up to it, and the answer has N samples.

Method	How it goes
Concentric circles	$x_{1}$ on the inner circle anticlockwise, $x_{2}$ on the outer clockwise. Multiply aligned pairs and add for $y [0]$ ; rotate the outer circle one step anticlockwise for $y [1]$ ; repeat N times.
Matrix	A circulant matrix of $x_{2}$ , each column the previous one rotated down, times the column vector $x_{1}$ .

"Find $x_{3} [n]$ if X3(k) = X1(k) X2(k)" IS this question. By the convolution property a product of DFTs is a circular convolution, so no DFT need ever be computed. Eleven of the sittings word it that way.
The modulo is the whole difference from linear convolution: what runs off the end wraps round to the front.

The checks that cost nothing

The sums multiply, exactly as in linear convolution.
N samples out, no more and no fewer.
Compare against the linear answer when N is smaller than $N_{1} + N_{2} - 1$ : the difference is precisely the wrapped tail, and seeing it is worth a line.

Practise this +8 like it2082 Bhadra · Q107 marksCompute the circular convolution of the following sequences:

h (n) = {1, 2, 1, - 1, 1}

and

x (n) = {1, 2, 3, 1}

What is different hereone sequence is shorter than N, so it is zero padded first

What is different in this one

The two sequences are different lengths, 5 and 4, so the shorter is zero padded to $N = 5$ before anything else happens. Getting N wrong makes every sample wrong.
$N = 5$ is smaller than $N_{1} + N_{2} - 1 = 8$ , so this circular answer is not the linear convolution: the last 3 samples have wrapped round and added to the front. That wrap is the whole point of the question.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=1	$x_{1}$ [1]=2	$x_{1}$ [2]=1	$x_{1}$ [3]=-1	$x_{1}$ [4]=1	sum
$y [0]$	1	0	1	3	2	1
$y [1]$	2	1	0	1	3	6
$y [2]$	3	2	1	0	1	9
$y [3]$	1	3	2	1	0	8
$y [4]$	0	1	3	2	1	4

y [0]

= 1(1) + 1(1) + -1(3) + 1(2) = 1

y [1]

= 1(2) + 2(1) + -1(1) + 1(3) = 6

y [2]

= 1(3) + 2(2) + 1(1) + 1(1) = 9

y [3]

= 1(1) + 2(3) + 1(2) + -1(1) = 8

y [4]

= 2(1) + 1(3) + -1(2) + 1(1) = 4

$y [n] = {1, 6, 9, 8, 4}$ . The check holds: 4 × 7 = 28, the sum of y.

Same as 2082 Bhadra Q102082 Baishakh · Q106 marksFind the circular convolution of the sequences

x_{1} [n] = {1, 1, - 1, - 1, 2}

and

x_{2} [n] = {1, - 1, - 2}

What is different in this one

The two sequences are different lengths, 5 and 3, so the shorter is zero padded to $N = 5$ before anything else happens. Getting N wrong makes every sample wrong.
$N = 5$ is smaller than $N_{1} + N_{2} - 1 = 7$ , so this circular answer is not the linear convolution: the last 2 samples have wrapped round and added to the front. That wrap is the whole point of the question.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=1	$x_{1}$ [1]=1	$x_{1}$ [2]=-1	$x_{1}$ [3]=-1	$x_{1}$ [4]=2	sum
$y [0]$	1	0	0	-2	-1	1
$y [1]$	-1	1	0	0	-2	-4
$y [2]$	-2	-1	1	0	0	-4
$y [3]$	0	-2	-1	1	0	-2
$y [4]$	0	0	-2	-1	1	5

y [0]

= 1(1) + -1(-2) + 2(-1) = 1

y [1]

= 1(-1) + 1(1) + 2(-2) = -4

y [2]

= 1(-2) + 1(-1) + -1(1) = -4

y [3]

= 1(-2) + -1(-1) + -1(1) = -2

y [4]

= -1(-2) + -1(-1) + 2(1) = 5

$y [n] = {1, - 4, - 4, - 2, 5}$ . The check holds: 2 × -2 = -4, the sum of y.

Same as 2082 Bhadra Q102081 Bhadra · Q106 marksIf X1(k) and X2(k) are the 5-point DFT of

x_{1}

(n) =

3^{n}

for

0 \leq n \leq 3

and

x_{2}

(n) =

2^{n}

for

0 \leq n \leq 4

. Find

x_{3}

(n) if X3(k) = X1(k)X2(k)

What is different in this one

The two sequences are different lengths, 4 and 5, so the shorter is zero padded to $N = 5$ before anything else happens. Getting N wrong makes every sample wrong.
$N = 5$ is smaller than $N_{1} + N_{2} - 1 = 8$ , so this circular answer is not the linear convolution: the last 3 samples have wrapped round and added to the front. That wrap is the whole point of the question.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=1	$x_{1}$ [1]=3	$x_{1}$ [2]=9	$x_{1}$ [3]=27	$x_{1}$ [4]=0	sum
$y [0]$	1	16	8	4	2	229
$y [1]$	2	1	16	8	4	365
$y [2]$	4	2	1	16	8	451
$y [3]$	8	4	2	1	16	65
$y [4]$	16	8	4	2	1	130

y [0]

= 1(1) + 3(16) + 9(8) + 27(4) = 229

y [1]

= 1(2) + 3(1) + 9(16) + 27(8) = 365

y [2]

= 1(4) + 3(2) + 9(1) + 27(16) = 451

y [3]

= 1(8) + 3(4) + 9(2) + 27(1) = 65

y [4]

= 1(16) + 3(8) + 9(4) + 27(2) = 130

$y [n] = {229, 365, 451, 65, 130}$ . The check holds: 40 × 31 = 1240, the sum of y.

Practise this +6 like it2081 Baishakh · Q106 marksCompute circular convolution of the following two sequences using DFT.

x (n) = {1, 2, 4, 5}

and

h (n) = {2, 1, 6, 8}

What is different in this one

$N = 4$ is smaller than $N_{1} + N_{2} - 1 = 7$ , so this circular answer is not the linear convolution: the last 3 samples have wrapped round and added to the front. That wrap is the whole point of the question.

The DFT length is $N = 4$ . The circular convolution is $y [n] = \sum_{m} x_{1} [m] x_{2}$ [(n - m) mod N], and the wrap is what makes it circular rather than linear.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=1	$x_{1}$ [1]=2	$x_{1}$ [2]=4	$x_{1}$ [3]=5	sum
$y [0]$	2	8	6	1	47
$y [1]$	1	2	8	6	67
$y [2]$	6	1	2	8	56
$y [3]$	8	6	1	2	34

y [0]

= 1(2) + 2(8) + 4(6) + 5(1) = 47

y [1]

= 1(1) + 2(2) + 4(8) + 5(6) = 67

y [2]

= 1(6) + 2(1) + 4(2) + 5(8) = 56

y [3]

= 1(8) + 2(6) + 4(1) + 5(2) = 34

$y [n] = {47, 67, 56, 34}$ . The check holds: 12 × 17 = 204, the sum of y.

Same as 2081 Baishakh Q102080 Bhadra · Q106 marksObtain the circular convolution of the following sequences:

x_{1}

(n) =

{1, 2, 3, 1}

and

x_{2}

(n) =

{4, 3, 2, 2}

What is different in this one

$N = 4$ is smaller than $N_{1} + N_{2} - 1 = 7$ , so this circular answer is not the linear convolution: the last 3 samples have wrapped round and added to the front. That wrap is the whole point of the question.

The DFT length is $N = 4$ . The circular convolution is $y [n] = \sum_{m} x_{1} [m] x_{2}$ [(n - m) mod N], and the wrap is what makes it circular rather than linear.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=1	$x_{1}$ [1]=2	$x_{1}$ [2]=3	$x_{1}$ [3]=1	sum
$y [0]$	4	2	2	3	17
$y [1]$	3	4	2	2	19
$y [2]$	2	3	4	2	22
$y [3]$	2	2	3	4	19

y [0]

= 1(4) + 2(2) + 3(2) + 1(3) = 17

y [1]

= 1(3) + 2(4) + 3(2) + 1(2) = 19

y [2]

= 1(2) + 2(3) + 3(4) + 1(2) = 22

y [3]

= 1(2) + 2(2) + 3(3) + 1(4) = 19

$y [n] = {17, 19, 22, 19}$ . The check holds: 7 × 11 = 77, the sum of y.

Same as 2081 Baishakh Q102080 Baishakh · Q122+5 marksWrite the complexity of DFT and FFT? Obtain the circular convolution of the following sequences:

X_{1} [n] = {1, 2, 3, 1}

and

X_{2} [n] = {4, 3, 2, 2}

What is different in this one

$N = 4$ is smaller than $N_{1} + N_{2} - 1 = 7$ , so this circular answer is not the linear convolution: the last 3 samples have wrapped round and added to the front. That wrap is the whole point of the question.

The DFT length is $N = 4$ . The circular convolution is $y [n] = \sum_{m} x_{1} [m] x_{2}$ [(n - m) mod N], and the wrap is what makes it circular rather than linear.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=1	$x_{1}$ [1]=2	$x_{1}$ [2]=3	$x_{1}$ [3]=1	sum
$y [0]$	4	2	2	3	17
$y [1]$	3	4	2	2	19
$y [2]$	2	3	4	2	22
$y [3]$	2	2	3	4	19

y [0]

= 1(4) + 2(2) + 3(2) + 1(3) = 17

y [1]

= 1(3) + 2(4) + 3(2) + 1(2) = 19

y [2]

= 1(2) + 2(3) + 3(4) + 1(2) = 22

y [3]

= 1(2) + 2(2) + 3(3) + 1(4) = 19

$y [n] = {17, 19, 22, 19}$ . The check holds: 7 × 11 = 77, the sum of y.

Same as 2082 Bhadra Q102079 Bhadra · Q117 marksIf X1(k) and X2(k) are DFT of sequence

x_{1} [n] = {1, 0, 0, 1}

and

x_{2} [n] = {2, 0, 2}

respectively then find the sequence

x_{3} [n]

; if DFT of

x_{3} [n]

is given by X3(k) = X1(k) X2(k).

What is different in this one

The two sequences are different lengths, 4 and 3, so the shorter is zero padded to $N = 4$ before anything else happens. Getting N wrong makes every sample wrong.
$N = 4$ is smaller than $N_{1} + N_{2} - 1 = 6$ , so this circular answer is not the linear convolution: the last 2 samples have wrapped round and added to the front. That wrap is the whole point of the question.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=1	$x_{1}$ [1]=0	$x_{1}$ [2]=0	$x_{1}$ [3]=1	sum
$y [0]$	2	0	2	0	2
$y [1]$	0	2	0	2	2
$y [2]$	2	0	2	0	2
$y [3]$	0	2	0	2	2

y [0]

= 1(2) = 2

y [1]

= 1(2) = 2

y [2]

= 1(2) = 2

y [3]

= 1(2) = 2

$y [n] = {2, 2, 2, 2}$ . The check holds: 2 × 4 = 8, the sum of y.

Same as 2082 Bhadra Q102079 Baishakh · Q117 marksFind the circular convolution of the sequences

x_{1} [n] = {1, - 1, - 2, 3, - 1}

and

x_{2} [n] = {1, 2, 3}

What is different in this one

The two sequences are different lengths, 5 and 3, so the shorter is zero padded to $N = 5$ before anything else happens. Getting N wrong makes every sample wrong.
$N = 5$ is smaller than $N_{1} + N_{2} - 1 = 7$ , so this circular answer is not the linear convolution: the last 2 samples have wrapped round and added to the front. That wrap is the whole point of the question.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=1	$x_{1}$ [1]=-1	$x_{1}$ [2]=-2	$x_{1}$ [3]=3	$x_{1}$ [4]=-1	sum
$y [0]$	1	0	0	3	2	8
$y [1]$	2	1	0	0	3	-2
$y [2]$	3	2	1	0	0	-1
$y [3]$	0	3	2	1	0	-4
$y [4]$	0	0	3	2	1	-1

y [0]

= 1(1) + 3(3) + -1(2) = 8

y [1]

= 1(2) + -1(1) + -1(3) = -2

y [2]

= 1(3) + -1(2) + -2(1) = -1

y [3]

= -1(3) + -2(2) + 3(1) = -4

y [4]

= -2(3) + 3(2) + -1(1) = -1

$y [n] = {8, - 2, - 1, - 4, - 1}$ . The check holds: 0 × 6 = 0, the sum of y.

Same as 2081 Baishakh Q102078 Bhadra · Q102+6 marksDifferentiate between DFT and DTFT. Find the circular convolution of

x_{1} [n] = {2, 1, 2, 1}

and

x_{2} [n] = {1, 2, 3, 4}

What is different in this one

$N = 4$ is smaller than $N_{1} + N_{2} - 1 = 7$ , so this circular answer is not the linear convolution: the last 3 samples have wrapped round and added to the front. That wrap is the whole point of the question.

The DFT length is $N = 4$ . The circular convolution is $y [n] = \sum_{m} x_{1} [m] x_{2}$ [(n - m) mod N], and the wrap is what makes it circular rather than linear.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=2	$x_{1}$ [1]=1	$x_{1}$ [2]=2	$x_{1}$ [3]=1	sum
$y [0]$	1	4	3	2	14
$y [1]$	2	1	4	3	16
$y [2]$	3	2	1	4	14
$y [3]$	4	3	2	1	16

y [0]

= 2(1) + 1(4) + 2(3) + 1(2) = 14

y [1]

= 2(2) + 1(1) + 2(4) + 1(3) = 16

y [2]

= 2(3) + 1(2) + 2(1) + 1(4) = 14

y [3]

= 2(4) + 1(3) + 2(2) + 1(1) = 16

$y [n] = {14, 16, 14, 16}$ . The check holds: 6 × 10 = 60, the sum of y.

Same as 2082 Bhadra Q102076 Chaitra · Q125 marksFind

x_{3} [n]

if DFT of

x_{3} [n]

is given by X3(k) = X1(k) * X2(k) where X1(k) and X2(k) are 4-point DFT of

x_{1} [n] = {1, 2, - 2}

and

x_{2} [n] = {1, 2, 3, - 1}

respectively.

What is different in this one

The two sequences are different lengths, 3 and 4, so the shorter is zero padded to $N = 4$ before anything else happens. Getting N wrong makes every sample wrong.
$N = 4$ is smaller than $N_{1} + N_{2} - 1 = 6$ , so this circular answer is not the linear convolution: the last 2 samples have wrapped round and added to the front. That wrap is the whole point of the question.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=1	$x_{1}$ [1]=2	$x_{1}$ [2]=-2	$x_{1}$ [3]=0	sum
$y [0]$	1	-1	3	2	-7
$y [1]$	2	1	-1	3	6
$y [2]$	3	2	1	-1	5
$y [3]$	-1	3	2	1	1

y [0]

= 1(1) + 2(-1) + -2(3) = -7

y [1]

= 1(2) + 2(1) + -2(-1) = 6

y [2]

= 1(3) + 2(2) + -2(1) = 5

y [3]

= 1(-1) + 2(3) + -2(2) = 1

$y [n] = {- 7, 6, 5, 1}$ . The check holds: 1 × 5 = 5, the sum of y.

Practise this2075 Chaitra · Q93+7 marksDifferentiate between linear convolution and circular convolution compute circular convolution of signals

X_{1} [n] = {0, 0, 1, 1}

and

X_{2} [n] = {1, 1, 1, 1}

What is different hereOne sequence is a constant, so every output sample is the same number: the sum of the other sequence. A five second answer once seen.

What is different in this one

$N = 4$ is smaller than $N_{1} + N_{2} - 1 = 7$ , so this circular answer is not the linear convolution: the last 3 samples have wrapped round and added to the front. That wrap is the whole point of the question.

The DFT length is $N = 4$ . The circular convolution is $y [n] = \sum_{m} x_{1} [m] x_{2}$ [(n - m) mod N], and the wrap is what makes it circular rather than linear.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=0	$x_{1}$ [1]=0	$x_{1}$ [2]=1	$x_{1}$ [3]=1	sum
$y [0]$	1	1	1	1	2
$y [1]$	1	1	1	1	2
$y [2]$	1	1	1	1	2
$y [3]$	1	1	1	1	2

y [0]

= 1(1) + 1(1) = 2

y [1]

= 1(1) + 1(1) = 2

y [2]

= 1(1) + 1(1) = 2

y [3]

= 1(1) + 1(1) = 2

$y [n] = {2, 2, 2, 2}$ . The check holds: 2 × 4 = 8, the sum of y.

Same as 2081 Baishakh Q102075 Ashwin · Q91+5 marksState the circular convolution property of DFT. Find the circular convolution of: (a)

{1, 2, - 1, 1}

and

x_{2}

(n) =

{1, 3, 5, 7}

What is different in this one

$N = 4$ is smaller than $N_{1} + N_{2} - 1 = 7$ , so this circular answer is not the linear convolution: the last 3 samples have wrapped round and added to the front. That wrap is the whole point of the question.

The DFT length is $N = 4$ . The circular convolution is $y [n] = \sum_{m} x_{1} [m] x_{2}$ [(n - m) mod N], and the wrap is what makes it circular rather than linear.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=1	$x_{1}$ [1]=2	$x_{1}$ [2]=-1	$x_{1}$ [3]=1	sum
$y [0]$	1	7	5	3	13
$y [1]$	3	1	7	5	3
$y [2]$	5	3	1	7	17
$y [3]$	7	5	3	1	15

y [0]

= 1(1) + 2(7) + -1(5) + 1(3) = 13

y [1]

= 1(3) + 2(1) + -1(7) + 1(5) = 3

y [2]

= 1(5) + 2(3) + -1(1) + 1(7) = 17

y [3]

= 1(7) + 2(5) + -1(3) + 1(1) = 15

$y [n] = {13, 3, 17, 15}$ . The check holds: 3 × 16 = 48, the sum of y.

Same as 2081 Baishakh Q102074 Ashwin · Q101+7 marksWhat is a zero padding? If X1(k) and X2(k) are DFT of sequence

x_{1} [n] = {1, 2, 0, 1, - 2}

and

x_{2} [n] = {1, 0, 1, 1, 2}

respectively then find the sequence

x_{3} [n]

; If DFT of

x_{3} [n]

is given by X3(k) = X1(k) X2(k).

What is different in this one

$N = 5$ is smaller than $N_{1} + N_{2} - 1 = 9$ , so this circular answer is not the linear convolution: the last 4 samples have wrapped round and added to the front. That wrap is the whole point of the question.

The DFT length is $N = 5$ . The circular convolution is $y [n] = \sum_{m} x_{1} [m] x_{2}$ [(n - m) mod N], and the wrap is what makes it circular rather than linear.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=1	$x_{1}$ [1]=2	$x_{1}$ [2]=0	$x_{1}$ [3]=1	$x_{1}$ [4]=-2	sum
$y [0]$	1	2	1	1	0	6
$y [1]$	0	1	2	1	1	1
$y [2]$	1	0	1	2	1	1
$y [3]$	1	1	0	1	2	0
$y [4]$	2	1	1	0	1	2

y [0]

= 1(1) + 2(2) + 1(1) = 6

y [1]

= 2(1) + 1(1) + -2(1) = 1

y [2]

= 1(1) + 1(2) + -2(1) = 1

y [3]

= 1(1) + 2(1) + 1(1) + -2(2) = 0

y [4]

= 1(2) + 2(1) + -2(1) = 2

$y [n] = {6, 1, 1, 0, 2}$ . The check holds: 2 × 5 = 10, the sum of y.

Same as 2082 Bhadra Q102073 Shrawan · Q117 marksFind

x_{3} [n]

if DFT of

x_{3} [n]

is given by X3(k) = X1(k) X2(k) where X1(k) and X2(k) are 5-point DFT of

x_{1} [n] = {1, - 2, 2, 1, 4}

and

x_{2} [n] = {2, 1, - 3, - 1}

respectively.

What is different in this one

The two sequences are different lengths, 5 and 4, so the shorter is zero padded to $N = 5$ before anything else happens. Getting N wrong makes every sample wrong.
$N = 5$ is smaller than $N_{1} + N_{2} - 1 = 8$ , so this circular answer is not the linear convolution: the last 3 samples have wrapped round and added to the front. That wrap is the whole point of the question.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=1	$x_{1}$ [1]=-2	$x_{1}$ [2]=2	$x_{1}$ [3]=1	$x_{1}$ [4]=4	sum
$y [0]$	2	0	-1	-3	1	1
$y [1]$	1	2	0	-1	-3	-16
$y [2]$	-3	1	2	0	-1	-5
$y [3]$	-1	-3	1	2	0	9
$y [4]$	0	-1	-3	1	2	5

y [0]

= 1(2) + 2(-1) + 1(-3) + 4(1) = 1

y [1]

= 1(1) + -2(2) + 1(-1) + 4(-3) = -16

y [2]

= 1(-3) + -2(1) + 2(2) + 4(-1) = -5

y [3]

= 1(-1) + -2(-3) + 2(1) + 1(2) = 9

y [4]

= -2(-1) + 2(-3) + 1(1) + 4(2) = 5

$y [n] = {1, - 16, - 5, 9, 5}$ . The check holds: 6 × -1 = -6, the sum of y.

Same as 2082 Bhadra Q102072 Chaitra · Q127 marksCompute Circular Convolution of

h (n) = {1, 2, 1, - 1, 1}

and

x [n] = {1, 2, 3, 1}

What is different in this one

The two sequences are different lengths, 5 and 4, so the shorter is zero padded to $N = 5$ before anything else happens. Getting N wrong makes every sample wrong.
$N = 5$ is smaller than $N_{1} + N_{2} - 1 = 8$ , so this circular answer is not the linear convolution: the last 3 samples have wrapped round and added to the front. That wrap is the whole point of the question.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=1	$x_{1}$ [1]=2	$x_{1}$ [2]=1	$x_{1}$ [3]=-1	$x_{1}$ [4]=1	sum
$y [0]$	1	0	1	3	2	1
$y [1]$	2	1	0	1	3	6
$y [2]$	3	2	1	0	1	9
$y [3]$	1	3	2	1	0	8
$y [4]$	0	1	3	2	1	4

y [0]

= 1(1) + 1(1) + -1(3) + 1(2) = 1

y [1]

= 1(2) + 2(1) + -1(1) + 1(3) = 6

y [2]

= 1(3) + 2(2) + 1(1) + 1(1) = 9

y [3]

= 1(1) + 2(3) + 1(2) + -1(1) = 8

y [4]

= 2(1) + 1(3) + -1(2) + 1(1) = 4

$y [n] = {1, 6, 9, 8, 4}$ . The check holds: 4 × 7 = 28, the sum of y.

Same as 2082 Bhadra Q102072 Kartik · Q112+6 marksWhy we need DFT? If X1(k) and X2(k) are DFT of sequence

x_{1} [n] = {1, 2, 4}

and

x_{2} [n] = {- 1, 2, 3, 1}

respectively, then find the sequence

x_{3} [n]

, if DFT of

x_{3} [n]

is given by X3(k) = X1(k) X2(k).

What is different in this one

The two sequences are different lengths, 3 and 4, so the shorter is zero padded to $N = 4$ before anything else happens. Getting N wrong makes every sample wrong.
$N = 4$ is smaller than $N_{1} + N_{2} - 1 = 6$ , so this circular answer is not the linear convolution: the last 2 samples have wrapped round and added to the front. That wrap is the whole point of the question.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=1	$x_{1}$ [1]=2	$x_{1}$ [2]=4	$x_{1}$ [3]=0	sum
$y [0]$	-1	1	3	2	13
$y [1]$	2	-1	1	3	4
$y [2]$	3	2	-1	1	3
$y [3]$	1	3	2	-1	15

y [0]

= 1(-1) + 2(1) + 4(3) = 13

y [1]

= 1(2) + 2(-1) + 4(1) = 4

y [2]

= 1(3) + 2(2) + 4(-1) = 3

y [3]

= 1(1) + 2(3) + 4(2) = 15

$y [n] = {13, 4, 3, 15}$ . The check holds: 7 × 5 = 35, the sum of y.

Same as 2081 Baishakh Q102071 Shrawan · Q97 marksA system has input signal

x [n] = {1, 2, 3, 4}

and impulse response

h [n] = {1, 3, 5, 7}

and the DFT of

x [n]

X [k]

and the DFT of

h [n]

H [k]

. Find the output of the system

y [n]

G [k] = X [k]

H [k]

What is different in this one

$N = 4$ is smaller than $N_{1} + N_{2} - 1 = 7$ , so this circular answer is not the linear convolution: the last 3 samples have wrapped round and added to the front. That wrap is the whole point of the question.

The DFT length is $N = 4$ . The circular convolution is $y [n] = \sum_{m} x_{1} [m] x_{2}$ [(n - m) mod N], and the wrap is what makes it circular rather than linear.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=1	$x_{1}$ [1]=2	$x_{1}$ [2]=3	$x_{1}$ [3]=4	sum
$y [0]$	1	7	5	3	42
$y [1]$	3	1	7	5	46
$y [2]$	5	3	1	7	42
$y [3]$	7	5	3	1	30

y [0]

= 1(1) + 2(7) + 3(5) + 4(3) = 42

y [1]

= 1(3) + 2(1) + 3(7) + 4(5) = 46

y [2]

= 1(5) + 2(3) + 3(1) + 4(7) = 42

y [3]

= 1(7) + 2(5) + 3(3) + 4(1) = 30

$y [n] = {42, 46, 42, 30}$ . The check holds: 10 × 16 = 160, the sum of y.

The inverse z-transform TOP 15/19

Ch 2 · The z-transform15 questions, from 15 of the 19 sittings11 to practise

The procedure, the same for every question in this topic

Improper? If the degree of the numerator reaches the degree of the denominator, long divide first. The quotient inverts to impulses and the remainder carries on through the steps below.
Expand $X (z)$ /z, not $X (z)$ . Dividing by z first is what makes every term come back as $z / (z - p)$ , which is the standard pair; expanding $X (z)$ itself gives $1 / (z - p)$ , which is that pair delayed by one and a lost mark.
The residue at each pole.
$A_{k} = {[(z - p_{k}) \frac{X (z)}{z}]}_{z = p_{k}}$
Invert term by term, against the ROC. The algebra is the same either way; the ROC alone decides which of these two a term becomes.
$\begin{aligned} ROC OUTSIDE this pole & ⟹ A_{k} (p_{k})^{n} u [n] & right sided \\ ROC INSIDE this pole & ⟹ - A_{k} (p_{k})^{n} u [- n - 1] & left sided \end{aligned}$
A repeated pole of order m needs a term for every power up to $(z - p)^{- m}$ , and the squared one inverts like this.
$\frac{z}{(z - p)^{2}} ⟶ n p^{n - 1} u [n]$

The ROC, not the algebra, is the question. The same $X (z)$ with three different ROCs gives three different sequences, and only the one whose ROC contains the unit circle is stable.
"Causal" replaces a stated ROC and means outside every pole.
A pole and a zero can cancel, dropping the order. Look for a common factor before grinding.

The checks that cost nothing

Initial value: for a causal $x [n]$ , $x [0]$ is the limit of $X (z)$ as z goes to infinity.
Run the difference equation the transform encodes and compare the first few samples against your closed form.
Wrong on sight: a right sided answer containing $u [- n - 1]$ , or a causal answer with samples at negative n.

Practise this +1 like it2082 Bhadra · Q31+5 marksDefine z-transform for a discrete time signal. Find the inverse z-transform for

H (z) = z / (3 z^{2} - 4 z + 1)

using partial fraction method for

1 / 3 < | z | < 1

What is different in this one

The numerator is of lower degree than the denominator, so it is already proper: no division, straight to F(z) = $X (z)$ /z.
The ROC is a ring, so the answer is two sided: the poles inside the ring give right sided terms and the ones outside it give left sided terms. This is the case that separates a real answer from a memorised one.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{z}{3 z^{2} - 4 z + 1}

2. Nothing to divide. Degree 1 over degree 2, so the fraction is already low enough on top.

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = \frac{1}{3 z^{2} - 4 z + 1}

The numerator carried a z of its own, and it cancels the z we just divided by, so there is no term at the origin at all.

3 z^{2} - 4 z + 1 = 3 (z - 0.3333) (z - 1) = 0

So the poles are $0.3333$ and $1$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{1}{3 (z - 0.3333) (z - 1)} \\ = \frac{A}{(z - 0.3333)} + \frac{B}{(z - 1)} \end{aligned}

Put them back over the common denominator $3 (z - 0.3333) (z - 1)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z - 1) + B (z - 0.3333) \\ A (z - 1) & = A z - A \\ B (z - 0.3333) & = B z - 0.3333 B \end{aligned}

Now collect the powers of z and set each one equal to the same power in $0.3333$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + B & = 0 & from z \\ - A - 0.3333 B & = 0.3333 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = - 0.5 \\ B & = 0.5 \end{aligned}

Putting those back and adding the terms returns $1$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = \frac{- 0.5 z}{(z - 0.3333)} + \frac{0.5 z}{(z - 1)}

6. Invert term by term. The ROC $0.3333 < | z | < 1$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
0.3333	0.3333	-0.5	the ROC lies outside it	-0.5 ${(0.3333)}^{n} u [n]$
1	1	0.5	the ROC lies inside it	-0.5 ${(1)}^{n} u [- n - 1]$

With the ROC $0.3333 < | z | < 1$ , $x [n] = - 0.5 {(0.3333)}^{n} u [n] - 0.5 {(1)}^{n} u [- n - 1]$ .

Practise this2082 Baishakh · Q31+5 marksDefine ROC of a Z-transform. Determine

x [n]

for

X (z) = 1 / (1 - 1.5 z^{- 1} + 0.5 z^{- 2})

if: a)

| z | > 1

| z | < 0.5

0.5 < | z | < 1

using partial fraction method.What is different hereOne

X (z)

, three answers. This is the question the whole chapter exists for, and the partial fractions are done ONCE: only the side each pole falls on changes.

What is different in this one

The same $X (z)$ is given with 3 different regions, so it stands for 3 different signals. The partial fractions are worked once: the poles and the constants do not know about the ROC, and only the very last step, which side each pole falls on, changes between the parts.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{z^{2}}{z^{2} - 1.5 z + 0.5}

2. Nothing to divide. Degree 2 over degree 2, so the fraction is already low enough on top.

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = \frac{z}{z^{2} - 1.5 z + 0.5}

The numerator carried a z of its own, and it cancels the z we just divided by, so there is no term at the origin at all.

z^{2} - 1.5 z + 0.5 = (z - 0.5) (z - 1) = 0

So the poles are $0.5$ and $1$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{z}{(z - 0.5) (z - 1)} \\ = \frac{A}{(z - 0.5)} + \frac{B}{(z - 1)} \end{aligned}

Put them back over the common denominator $(z - 0.5) (z - 1)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z - 1) + B (z - 0.5) \\ A (z - 1) & = A z - A \\ B (z - 0.5) & = B z - 0.5 B \end{aligned}

Now collect the powers of z and set each one equal to the same power in $z$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + B & = 1 & from z \\ - A - 0.5 B & = 0 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = - 1 \\ B & = 2 \end{aligned}

Putting those back and adding the terms returns $z$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = \frac{- 1 z}{(z - 0.5)} + \frac{2 z}{(z - 1)}

$| z | > 1$

Pole	$\| p \|$	Constant	Which side	Inverts to
0.5	0.5	-1	the ROC lies outside it	- ${(0.5)}^{n} u [n]$
1	1	2	the ROC lies outside it	2 ${(1)}^{n} u [n]$

With the ROC $| z | > 1$ , $x [n] = - {(0.5)}^{n} u [n] + 2 {(1)}^{n} u [n]$ .

$| z | < 0.5$

Pole	$\| p \|$	Constant	Which side	Inverts to
0.5	0.5	-1	the ROC lies inside it	${(0.5)}^{n} u [- n - 1]$
1	1	2	the ROC lies inside it	-2 ${(1)}^{n} u [- n - 1]$

With the ROC $| z | < 0.5$ , $x [n] = {(0.5)}^{n} u [- n - 1] - 2 {(1)}^{n} u [- n - 1]$ .

$0.5 < | z | < 1$

Pole	$\| p \|$	Constant	Which side	Inverts to
0.5	0.5	-1	the ROC lies outside it	- ${(0.5)}^{n} u [n]$
1	1	2	the ROC lies inside it	-2 ${(1)}^{n} u [- n - 1]$

With the ROC $0.5 < | z | < 1$ , $x [n] = - {(0.5)}^{n} u [n] - 2 {(1)}^{n} u [- n - 1]$ .

Side by side, the same transform read 3 ways:

ROC	Which side each pole falls	$x [n]$	Stable?
$\| z \| > 1$	0.5 is right sided, 1 is right sided	- ${(0.5)}^{n} u [n] + 2 {(1)}^{n} u [n]$	no
$\| z \| < 0.5$	0.5 is left sided, 1 is left sided	${(0.5)}^{n} u [- n - 1] - 2 {(1)}^{n} u [- n - 1]$	no
$0.5 < \| z \| < 1$	0.5 is right sided, 1 is left sided	- ${(0.5)}^{n} u [n] - 2 {(1)}^{n} u [- n - 1]$	no

Practise this +1 like it2081 Bhadra · Q31+5 marksDefine ROC of z-transform. Find the inverse z-transform using partial fraction expansion of

X (z) = (2 z^{3} - 5 z^{2} + z + 3) / ((z - 1) (z - 2))

for ROC

| z | < 1

What is different in this one

The numerator is of higher degree than the denominator (3 against 2), so it has to be long divided first. The quotient is a polynomial in z, which inverts to impulses at negative n, and only the remainder goes through partial fractions. Divide until the remainder reaches the degree of the divisor and stop there, not further.
The ROC is the inside of the innermost pole, so every term is left sided and the whole answer carries $u [- n - 1]$ with a sign change.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{2 z^{3} - 5 z^{2} + z + 3}{z^{2} - 3 z + 2}

			$2 z$
$z^{2} - 3 z + 2$ )	$2 z^{3}$	$- 5 z^{2}$	$+ z$	$+ 3$
	$2 z^{3}$	$- 6 z^{2}$	$+ 4 z$
		$z^{2}$	$- 3 z$	$+ 3$

X (z) = 2 z + \frac{z^{2} - 3 z + 3}{z^{2} - 3 z + 2}

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = 2 + \frac{z^{2} - 3 z + 3}{z^{3} - 3 z^{2} + 2 z}

The poles are the roots of that denominator. Note the z we just divided by is one of them, sitting at the origin.

z^{3} - 3 z^{2} + 2 z = z (z - 1) (z - 2) = 0

So the poles are $0$ , $1$ and $2$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{z^{2} - 3 z + 3}{z (z - 1) (z - 2)} \\ = \frac{A}{z} + \frac{B}{(z - 1)} + \frac{C}{(z - 2)} \end{aligned}

Put them back over the common denominator $z (z - 1) (z - 2)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z - 1) (z - 2) + B z (z - 2) + C z (z - 1) \\ A (z - 1) (z - 2) & = A z^{2} - 3 A z + 2 A \\ B z (z - 2) & = B z^{2} - 2 B z \\ C z (z - 1) & = C z^{2} - C z \end{aligned}

Now collect the powers of z and set each one equal to the same power in $z^{2} - 3 z + 3$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + B + C & = 1 & from z^{2} \\ - 3 A - 2 B - C & = - 3 & from z \\ 2 A & = 3 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = 1.5 \\ B & = - 1 \\ C & = 0.5 \end{aligned}

Putting those back and adding the terms returns $z^{2} - 3 z + 3$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = 2 z + 1.5 + \frac{- 1 z}{(z - 1)} + \frac{0.5 z}{(z - 2)}

6. Invert term by term. The ROC $| z | < 1$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
none, from the division		2	a polynomial term	2 $δ [n + 1]$
0, the origin	0	1.5	a polynomial term	1.5 $δ [n]$
1	1	-1	the ROC lies inside it	${(1)}^{n} u [- n - 1]$
2	2	0.5	the ROC lies inside it	-0.5 ${(2)}^{n} u [- n - 1]$

With the ROC $| z | < 1$ , $x [n] = 2 δ [n + 1] + 1.5 δ [n] + {(1)}^{n} u [- n - 1] - 0.5 {(2)}^{n} u [- n - 1]$ .

Practise this2081 Baishakh · Q32+2+3 marksDetermine the inverse z-transform of

X (z) = 1 / (1 - 0.8 z^{- 1} + 0.12 z^{- 2}) . (i)

if ROC is

| z | > 0.6

(ii) if ROC is

| z | < 0.2

(iii) if ROC is

0.2 < | z | < 0.6

What is different in this one

The same $X (z)$ is given with 3 different regions, so it stands for 3 different signals. The partial fractions are worked once: the poles and the constants do not know about the ROC, and only the very last step, which side each pole falls on, changes between the parts.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{z^{2}}{z^{2} - 0.8 z + 0.12}

2. Nothing to divide. Degree 2 over degree 2, so the fraction is already low enough on top.

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = \frac{z}{z^{2} - 0.8 z + 0.12}

The numerator carried a z of its own, and it cancels the z we just divided by, so there is no term at the origin at all.

z^{2} - 0.8 z + 0.12 = (z - 0.2) (z - 0.6) = 0

So the poles are $0.2$ and $0.6$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{z}{(z - 0.2) (z - 0.6)} \\ = \frac{A}{(z - 0.2)} + \frac{B}{(z - 0.6)} \end{aligned}

Put them back over the common denominator $(z - 0.2) (z - 0.6)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z - 0.6) + B (z - 0.2) \\ A (z - 0.6) & = A z - 0.6 A \\ B (z - 0.2) & = B z - 0.2 B \end{aligned}

Now collect the powers of z and set each one equal to the same power in $z$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + B & = 1 & from z \\ - 0.6 A - 0.2 B & = 0 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = - 0.5 \\ B & = 1.5 \end{aligned}

Putting those back and adding the terms returns $z$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = \frac{- 0.5 z}{(z - 0.2)} + \frac{1.5 z}{(z - 0.6)}

$| z | > 0.6$

Pole	$\| p \|$	Constant	Which side	Inverts to
0.2	0.2	-0.5	the ROC lies outside it	-0.5 ${(0.2)}^{n} u [n]$
0.6	0.6	1.5	the ROC lies outside it	1.5 ${(0.6)}^{n} u [n]$

With the ROC $| z | > 0.6$ , $x [n] = - 0.5 {(0.2)}^{n} u [n] + 1.5 {(0.6)}^{n} u [n]$ .

$| z | < 0.2$

Pole	$\| p \|$	Constant	Which side	Inverts to
0.2	0.2	-0.5	the ROC lies inside it	0.5 ${(0.2)}^{n} u [- n - 1]$
0.6	0.6	1.5	the ROC lies inside it	-1.5 ${(0.6)}^{n} u [- n - 1]$

With the ROC $| z | < 0.2$ , $x [n] = 0.5 {(0.2)}^{n} u [- n - 1] - 1.5 {(0.6)}^{n} u [- n - 1]$ .

$0.2 < | z | < 0.6$

Pole	$\| p \|$	Constant	Which side	Inverts to
0.2	0.2	-0.5	the ROC lies outside it	-0.5 ${(0.2)}^{n} u [n]$
0.6	0.6	1.5	the ROC lies inside it	-1.5 ${(0.6)}^{n} u [- n - 1]$

With the ROC $0.2 < | z | < 0.6$ , $x [n] = - 0.5 {(0.2)}^{n} u [n] - 1.5 {(0.6)}^{n} u [- n - 1]$ .

Side by side, the same transform read 3 ways:

ROC	Which side each pole falls	$x [n]$	Stable?
$\| z \| > 0.6$	0.2 is right sided, 0.6 is right sided	-0.5 ${(0.2)}^{n} u [n] + 1.5 {(0.6)}^{n} u [n]$	yes
$\| z \| < 0.2$	0.2 is left sided, 0.6 is left sided	0.5 ${(0.2)}^{n} u [- n - 1] - 1.5 {(0.6)}^{n} u [- n - 1]$	no
$0.2 < \| z \| < 0.6$	0.2 is right sided, 0.6 is left sided	-0.5 ${(0.2)}^{n} u [n] - 1.5 {(0.6)}^{n} u [- n - 1]$	no

$| z | > 0.6$ , so start the divisor at its highest term.

	$1$	$+ 0.8 z^{- 1}$	$+ 0.52 z^{- 2}$	$+ 0.32 z^{- 3}$	$+ 0.1936 z^{- 4}$	$+ 0.1165 z^{- 5}$	$+ 0.07 z^{- 6}$	$+ 0.042 z^{- 7}$
$1 - 0.8 z^{- 1} + 0.12 z^{- 2}$ )	$1$
	$1$	$- 0.8 z^{- 1}$	$+ 0.12 z^{- 2}$
		$0.8 z^{- 1}$	$- 0.12 z^{- 2}$
		$0.8 z^{- 1}$	$- 0.64 z^{- 2}$	$+ 0.096 z^{- 3}$
			$0.52 z^{- 2}$	$- 0.096 z^{- 3}$
			$0.52 z^{- 2}$	$- 0.416 z^{- 3}$	$+ 0.0624 z^{- 4}$
				$0.32 z^{- 3}$	$- 0.0624 z^{- 4}$
				$0.32 z^{- 3}$	$- 0.256 z^{- 4}$	$+ 0.0384 z^{- 5}$
					$0.1936 z^{- 4}$	$- 0.0384 z^{- 5}$
					$0.1936 z^{- 4}$	$- 0.1549 z^{- 5}$	$+ 0.0232 z^{- 6}$
						$0.1165 z^{- 5}$	$- 0.0232 z^{- 6}$
						$0.1165 z^{- 5}$	$- 0.0932 z^{- 6}$	$+ 0.014 z^{- 7}$
							$0.07 z^{- 6}$	$- 0.014 z^{- 7}$
							$0.07 z^{- 6}$	$- 0.056 z^{- 7}$	$+ 0.0084 z^{- 8}$
								$0.042 z^{- 7}$	$- 0.0084 z^{- 8}$
								$0.042 z^{- 7}$	$- 0.0336 z^{- 8}$	$+ 0.005 z^{- 9}$
									$0.0252 z^{- 8}$	$- 0.005 z^{- 9}$

$| z | < 0.2$ , so start the divisor at its lowest term.

	$8.3333 z^{2}$	$+ 55.5556 z^{3}$	$+ 300.9259 z^{4}$	$+ 1543.2099 z^{5}$	$+ 7780.3498 z^{6}$	$+ 39008.9163 z^{7}$
$0.12 - 0.8 z + z^{2}$ )	$z^{2}$
	$z^{2}$	$- 6.6667 z^{3}$	$+ 8.3333 z^{4}$
		$6.6667 z^{3}$	$- 8.3333 z^{4}$
		$6.6667 z^{3}$	$- 44.4444 z^{4}$	$+ 55.5556 z^{5}$
			$36.1111 z^{4}$	$- 55.5556 z^{5}$
			$36.1111 z^{4}$	$- 240.7407 z^{5}$	$+ 300.9259 z^{6}$
				$185.1852 z^{5}$	$- 300.9259 z^{6}$
				$185.1852 z^{5}$	$- 1234.5679 z^{6}$	$+ 1543.2099 z^{7}$
					$933.642 z^{6}$	$- 1543.2099 z^{7}$
					$933.642 z^{6}$	$- 6224.2798 z^{7}$	$+ 7780.3498 z^{8}$
						$4681.07 z^{7}$	$- 7780.3498 z^{8}$
						$4681.07 z^{7}$	$- 31207.1331 z^{8}$	$+ 39008.9163 z^{9}$
							$23426.7833 z^{8}$	$- 39008.9163 z^{9}$

Practise this2080 Bhadra · Q31+6 marksDefine the Region of Convergence (ROC). Find the inverse of

H (z) = (1 + 2 z^{- 1} + z^{- 2}) / (1 - 0.75 z^{- 1} + 0.125 z^{- 2})

; ROC

0.25 < | z | < 0.5

What is different herethe numerator reaches the degree of the denominator, so dividing by z leaves a pole AT THE ORIGIN, and that term inverts to a plain impulse on top of the exponentials

What is different in this one

The numerator reaches the degree of the denominator. Dividing by z is enough to make it proper, so no long division is needed, but the expansion will carry a term at the origin that inverts to an impulse.
The ROC is a ring, so the answer is two sided: the poles inside the ring give right sided terms and the ones outside it give left sided terms. This is the case that separates a real answer from a memorised one.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{z^{2} + 2 z + 1}{z^{2} - 0.75 z + 0.125}

2. Nothing to divide. Degree 2 over degree 2, so the fraction is already low enough on top.

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = \frac{z^{2} + 2 z + 1}{z^{3} - 0.75 z^{2} + 0.125 z}

The poles are the roots of that denominator. Note the z we just divided by is one of them, sitting at the origin.

z^{3} - 0.75 z^{2} + 0.125 z = z (z - 0.25) (z - 0.5) = 0

So the poles are $0$ , $0.25$ and $0.5$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{z^{2} + 2 z + 1}{z (z - 0.25) (z - 0.5)} \\ = \frac{A}{z} + \frac{B}{(z - 0.25)} + \frac{C}{(z - 0.5)} \end{aligned}

Put them back over the common denominator $z (z - 0.25) (z - 0.5)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z - 0.25) (z - 0.5) + B z (z - 0.5) + C z (z - 0.25) \\ A (z - 0.25) (z - 0.5) & = A z^{2} - 0.75 A z + 0.125 A \\ B z (z - 0.5) & = B z^{2} - 0.5 B z \\ C z (z - 0.25) & = C z^{2} - 0.25 C z \end{aligned}

Now collect the powers of z and set each one equal to the same power in $z^{2} + 2 z + 1$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + B + C & = 1 & from z^{2} \\ - 0.75 A - 0.5 B - 0.25 C & = 2 & from z \\ 0.125 A & = 1 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = 8 \\ B & = - 25 \\ C & = 18 \end{aligned}

Putting those back and adding the terms returns $z^{2} + 2 z + 1$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = 8 + \frac{- 25 z}{(z - 0.25)} + \frac{18 z}{(z - 0.5)}

6. Invert term by term. The ROC $0.25 < | z | < 0.5$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
0, the origin	0	8	a polynomial term	8 $δ [n]$
0.25	0.25	-25	the ROC lies outside it	-25 ${(0.25)}^{n} u [n]$
0.5	0.5	18	the ROC lies inside it	-18 ${(0.5)}^{n} u [- n - 1]$

With the ROC $0.25 < | z | < 0.5$ , $x [n] = 8 δ [n] - 25 {(0.25)}^{n} u [n] - 18 {(0.5)}^{n} u [- n - 1]$ .

Practise this +1 like it2080 Baishakh · Q31+5 marksDefine ROC. Find inverse z-transform of

X (z) = (1 + 2 z^{- 1} + z^{- 2}) / (1 - 1.5 z^{- 1} + 0.5 z^{- 2})

, ROC:

| z | > 1

What is different in this one

The numerator reaches the degree of the denominator. Dividing by z is enough to make it proper, so no long division is needed, but the expansion will carry a term at the origin that inverts to an impulse.
The ROC is the outside of the outermost pole, so every term is right sided and the answer is causal.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{z^{2} + 2 z + 1}{z^{2} - 1.5 z + 0.5}

2. Nothing to divide. Degree 2 over degree 2, so the fraction is already low enough on top.

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = \frac{z^{2} + 2 z + 1}{z^{3} - 1.5 z^{2} + 0.5 z}

The poles are the roots of that denominator. Note the z we just divided by is one of them, sitting at the origin.

z^{3} - 1.5 z^{2} + 0.5 z = z (z - 0.5) (z - 1) = 0

So the poles are $0$ , $0.5$ and $1$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{z^{2} + 2 z + 1}{z (z - 0.5) (z - 1)} \\ = \frac{A}{z} + \frac{B}{(z - 0.5)} + \frac{C}{(z - 1)} \end{aligned}

Put them back over the common denominator $z (z - 0.5) (z - 1)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z - 0.5) (z - 1) + B z (z - 1) + C z (z - 0.5) \\ A (z - 0.5) (z - 1) & = A z^{2} - 1.5 A z + 0.5 A \\ B z (z - 1) & = B z^{2} - B z \\ C z (z - 0.5) & = C z^{2} - 0.5 C z \end{aligned}

Now collect the powers of z and set each one equal to the same power in $z^{2} + 2 z + 1$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + B + C & = 1 & from z^{2} \\ - 1.5 A - B - 0.5 C & = 2 & from z \\ 0.5 A & = 1 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = 2 \\ B & = - 9 \\ C & = 8 \end{aligned}

Putting those back and adding the terms returns $z^{2} + 2 z + 1$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = 2 + \frac{- 9 z}{(z - 0.5)} + \frac{8 z}{(z - 1)}

6. Invert term by term. The ROC $| z | > 1$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
0, the origin	0	2	a polynomial term	2 $δ [n]$
0.5	0.5	-9	the ROC lies outside it	-9 ${(0.5)}^{n} u [n]$
1	1	8	the ROC lies outside it	8 ${(1)}^{n} u [n]$

With the ROC $| z | > 1$ , $x [n] = 2 δ [n] - 9 {(0.5)}^{n} u [n] + 8 {(1)}^{n} u [n]$ .

Check it without trusting any of the algebra: run the difference equation the transform encodes and compare the first samples. Both routes give ${1, 3.5, 5.75, 6.875, 7.4375, 7.71875}$ .

Practise this +1 like it2079 Bhadra · Q36 marksFind inverses Z-transform of

X (z) = (2 z^{4} + 2 z^{3} - 3 z + 2) / (z^{2} - 1.5 z - 1)

, ROC:

| z | < 0.5

What is different in this one

The numerator is of higher degree than the denominator (4 against 2), so it has to be long divided first. The quotient is a polynomial in z, which inverts to impulses at negative n, and only the remainder goes through partial fractions. Divide until the remainder reaches the degree of the divisor and stop there, not further.
The ROC is the inside of the innermost pole, so every term is left sided and the whole answer carries $u [- n - 1]$ with a sign change.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{2 z^{4} + 2 z^{3} - 3 z + 2}{z^{2} - 1.5 z - 1}

			$2 z^{2}$	$+ 5 z$
$z^{2} - 1.5 z - 1$ )	$2 z^{4}$	$+ 2 z^{3}$	$+ 0 z^{2}$	$- 3 z$	$+ 2$
	$2 z^{4}$	$- 3 z^{3}$	$- 2 z^{2}$
		$5 z^{3}$	$+ 2 z^{2}$	$- 3 z$	$+ 2$
		$5 z^{3}$	$- 7.5 z^{2}$	$- 5 z$
			$9.5 z^{2}$	$+ 2 z$	$+ 2$

X (z) = 2 z^{2} + 5 z + \frac{9.5 z^{2} + 2 z + 2}{z^{2} - 1.5 z - 1}

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = 2 z + 5 + \frac{9.5 z^{2} + 2 z + 2}{z^{3} - 1.5 z^{2} - z}

The poles are the roots of that denominator. Note the z we just divided by is one of them, sitting at the origin.

z^{3} - 1.5 z^{2} - z = z (z + 0.5) (z - 2) = 0

So the poles are $0$ , $- 0.5$ and $2$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{9.5 z^{2} + 2 z + 2}{z (z + 0.5) (z - 2)} \\ = \frac{A}{z} + \frac{B}{(z + 0.5)} + \frac{C}{(z - 2)} \end{aligned}

Put them back over the common denominator $z (z + 0.5) (z - 2)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z + 0.5) (z - 2) + B z (z - 2) + C z (z + 0.5) \\ A (z + 0.5) (z - 2) & = A z^{2} - 1.5 A z - A \\ B z (z - 2) & = B z^{2} - 2 B z \\ C z (z + 0.5) & = C z^{2} + 0.5 C z \end{aligned}

Now collect the powers of z and set each one equal to the same power in $9.5 z^{2} + 2 z + 2$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + B + C & = 9.5 & from z^{2} \\ - 1.5 A - 2 B + 0.5 C & = 2 & from z \\ - A & = 2 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = - 2 \\ B & = 2.7 \\ C & = 8.8 \end{aligned}

Putting those back and adding the terms returns $9.5 z^{2} + 2 z + 2$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = 2 z^{2} + 5 z - 2 + \frac{2.7 z}{(z + 0.5)} + \frac{8.8 z}{(z - 2)}

6. Invert term by term. The ROC $| z | < 0.5$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
none, from the division		2	a polynomial term	2 $δ [n + 2]$
none, from the division		5	a polynomial term	5 $δ [n + 1]$
0, the origin	0	-2	a polynomial term	-2 $δ [n]$
-0.5	0.5	2.7	the ROC lies inside it	-2.7 ${(- 0.5)}^{n} u [- n - 1]$
2	2	8.8	the ROC lies inside it	-8.8 ${(2)}^{n} u [- n - 1]$

With the ROC $| z | < 0.5$ , $x [n] = 2 δ [n + 2] + 5 δ [n + 1] - 2 δ [n] - 2.7 {(- 0.5)}^{n} u [- n - 1] - 8.8 {(2)}^{n} u [- n - 1]$ .

Same as 2082 Bhadra Q32079 Baishakh · Q31+5 marksDefine z-transform for a discrete time signal. Find the inverse z-transform for

H (z) = z / (3 z^{2} - 4 z + 1)

using partial fraction method for

1 / 3 < | z | < 1

What is different in this one

The numerator is of lower degree than the denominator, so it is already proper: no division, straight to F(z) = $X (z)$ /z.
The ROC is a ring, so the answer is two sided: the poles inside the ring give right sided terms and the ones outside it give left sided terms. This is the case that separates a real answer from a memorised one.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{z}{3 z^{2} - 4 z + 1}

2. Nothing to divide. Degree 1 over degree 2, so the fraction is already low enough on top.

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = \frac{1}{3 z^{2} - 4 z + 1}

The numerator carried a z of its own, and it cancels the z we just divided by, so there is no term at the origin at all.

3 z^{2} - 4 z + 1 = 3 (z - 0.3333) (z - 1) = 0

So the poles are $0.3333$ and $1$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{1}{3 (z - 0.3333) (z - 1)} \\ = \frac{A}{(z - 0.3333)} + \frac{B}{(z - 1)} \end{aligned}

Put them back over the common denominator $3 (z - 0.3333) (z - 1)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z - 1) + B (z - 0.3333) \\ A (z - 1) & = A z - A \\ B (z - 0.3333) & = B z - 0.3333 B \end{aligned}

Now collect the powers of z and set each one equal to the same power in $0.3333$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + B & = 0 & from z \\ - A - 0.3333 B & = 0.3333 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = - 0.5 \\ B & = 0.5 \end{aligned}

Putting those back and adding the terms returns $1$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = \frac{- 0.5 z}{(z - 0.3333)} + \frac{0.5 z}{(z - 1)}

6. Invert term by term. The ROC $0.3333 < | z | < 1$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
0.3333	0.3333	-0.5	the ROC lies outside it	-0.5 ${(0.3333)}^{n} u [n]$
1	1	0.5	the ROC lies inside it	-0.5 ${(1)}^{n} u [- n - 1]$

With the ROC $0.3333 < | z | < 1$ , $x [n] = - 0.5 {(0.3333)}^{n} u [n] - 0.5 {(1)}^{n} u [- n - 1]$ .

Practise this2078 Bhadra · Q31+5 marksDefine ROC. Find inverse z-transform of

X (z) = (z^{3} + z^{2} + 1.5 z + 0.5) / (z^{3} + 1.5 z^{2} + 0.5 z)

, ROC:

| z | < 1 / 2

What is different in this one

The numerator reaches the degree of the denominator. Dividing by z is enough to make it proper, so no long division is needed, but the expansion will carry a term at the origin that inverts to an impulse.
There is a repeated pole at 0. It needs a term for every power up to its order, and the extra one inverts with an n in front: $A z / (z - p)^{2}$ is (A/p) n $p^{n} u [n]$ , not A $p^{n} u [n]$ . Cover up cannot reach the lower constant at all, which is why the constants are found by comparing coefficients here.
The ROC is the inside of the innermost pole, so every term is left sided and the whole answer carries $u [- n - 1]$ with a sign change.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{z^{3} + z^{2} + 1.5 z + 0.5}{z^{3} + 1.5 z^{2} + 0.5 z}

2. Nothing to divide. Degree 3 over degree 3, so the fraction is already low enough on top.

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = \frac{z^{3} + z^{2} + 1.5 z + 0.5}{z^{4} + 1.5 z^{3} + 0.5 z^{2}}

The poles are the roots of that denominator. Note the z we just divided by is one of them, sitting at the origin.

z^{4} + 1.5 z^{3} + 0.5 z^{2} = z^{2} (z + 0.5) (z + 1) = 0

So the poles are $0$ , repeated 2 times, $- 0.5$ and $- 1$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{z^{3} + z^{2} + 1.5 z + 0.5}{z^{2} (z + 0.5) (z + 1)} \\ = \frac{A}{z} + \frac{B}{z^{2}} + \frac{C}{(z + 0.5)} + \frac{D}{(z + 1)} \end{aligned}

Put them back over the common denominator $z^{2} (z + 0.5) (z + 1)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A z (z + 0.5) (z + 1) + B (z + 0.5) (z + 1) + C z^{2} (z + 1) + D z^{2} (z + 0.5) \\ A z (z + 0.5) (z + 1) & = A z^{3} + 1.5 A z^{2} + 0.5 A z \\ B (z + 0.5) (z + 1) & = B z^{2} + 1.5 B z + 0.5 B \\ C z^{2} (z + 1) & = C z^{3} + C z^{2} \\ D z^{2} (z + 0.5) & = D z^{3} + 0.5 D z^{2} \end{aligned}

Now collect the powers of z and set each one equal to the same power in $z^{3} + z^{2} + 1.5 z + 0.5$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + C + D & = 1 & from z^{3} \\ 1.5 A + B + C + 0.5 D & = 1 & from z^{2} \\ 0.5 A + 1.5 B & = 1.5 & from z \\ 0.5 B & = 0.5 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = 0 \\ B & = 1 \\ C & = - 1 \\ D & = 2 \end{aligned}

Putting those back and adding the terms returns $z^{3} + z^{2} + 1.5 z + 0.5$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = \frac{1}{z} + \frac{- 1 z}{(z + 0.5)} + \frac{2 z}{(z + 1)}

6. Invert term by term. The ROC $| z | < 0.5$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
0, the origin	0	1	a polynomial term	$δ [n - 1]$
-0.5	0.5	-1	the ROC lies inside it	${(- 0.5)}^{n} u [- n - 1]$
-1	1	2	the ROC lies inside it	-2 ${(- 1)}^{n} u [- n - 1]$

The constant at 0 came out zero: a zero of the numerator sits on that pole and cancels it, so it contributes nothing.

With the ROC $| z | < 0.5$ , $x [n] = δ [n - 1] + {(- 0.5)}^{n} u [- n - 1] - 2 {(- 1)}^{n} u [- n - 1]$ .

Same as 2079 Bhadra Q32076 Chaitra · Q42+6 marksDefine ROC of z-transform. Find inverse z-transform using partial fraction expansion of

X (z) = (z^{4} + 5 z^{3} - 3 z + 4) / (z^{2} - 1.5 z - 1)

, ROC:

| z | < 0.5

How this is read $X (z) = (z^{4} + 5 z^{3} - 3 z + 4) / (z^{2} - 1.5 z - 1)$ . The numerator is two degrees higher, so the long division runs for two terms before the partial fractions start.

What is different in this one

The numerator is of higher degree than the denominator (4 against 2), so it has to be long divided first. The quotient is a polynomial in z, which inverts to impulses at negative n, and only the remainder goes through partial fractions. Divide until the remainder reaches the degree of the divisor and stop there, not further.
The ROC is the inside of the innermost pole, so every term is left sided and the whole answer carries $u [- n - 1]$ with a sign change.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{z^{4} + 5 z^{3} - 3 z + 4}{z^{2} - 1.5 z - 1}

			$z^{2}$	$+ 6.5 z$
$z^{2} - 1.5 z - 1$ )	$z^{4}$	$+ 5 z^{3}$	$+ 0 z^{2}$	$- 3 z$	$+ 4$
	$z^{4}$	$- 1.5 z^{3}$	$- z^{2}$
		$6.5 z^{3}$	$+ z^{2}$	$- 3 z$	$+ 4$
		$6.5 z^{3}$	$- 9.75 z^{2}$	$- 6.5 z$
			$10.75 z^{2}$	$+ 3.5 z$	$+ 4$

X (z) = z^{2} + 6.5 z + \frac{10.75 z^{2} + 3.5 z + 4}{z^{2} - 1.5 z - 1}

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = z + 6.5 + \frac{10.75 z^{2} + 3.5 z + 4}{z^{3} - 1.5 z^{2} - z}

The poles are the roots of that denominator. Note the z we just divided by is one of them, sitting at the origin.

z^{3} - 1.5 z^{2} - z = z (z + 0.5) (z - 2) = 0

So the poles are $0$ , $- 0.5$ and $2$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{10.75 z^{2} + 3.5 z + 4}{z (z + 0.5) (z - 2)} \\ = \frac{A}{z} + \frac{B}{(z + 0.5)} + \frac{C}{(z - 2)} \end{aligned}

Put them back over the common denominator $z (z + 0.5) (z - 2)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z + 0.5) (z - 2) + B z (z - 2) + C z (z + 0.5) \\ A (z + 0.5) (z - 2) & = A z^{2} - 1.5 A z - A \\ B z (z - 2) & = B z^{2} - 2 B z \\ C z (z + 0.5) & = C z^{2} + 0.5 C z \end{aligned}

Now collect the powers of z and set each one equal to the same power in $10.75 z^{2} + 3.5 z + 4$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + B + C & = 10.75 & from z^{2} \\ - 1.5 A - 2 B + 0.5 C & = 3.5 & from z \\ - A & = 4 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = - 4 \\ B & = 3.95 \\ C & = 10.8 \end{aligned}

Putting those back and adding the terms returns $10.75 z^{2} + 3.5 z + 4$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = z^{2} + 6.5 z - 4 + \frac{3.95 z}{(z + 0.5)} + \frac{10.8 z}{(z - 2)}

6. Invert term by term. The ROC $| z | < 0.5$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
none, from the division		1	a polynomial term	$δ [n + 2]$
none, from the division		6.5	a polynomial term	6.5 $δ [n + 1]$
0, the origin	0	-4	a polynomial term	-4 $δ [n]$
-0.5	0.5	3.95	the ROC lies inside it	-3.95 ${(- 0.5)}^{n} u [- n - 1]$
2	2	10.8	the ROC lies inside it	-10.8 ${(2)}^{n} u [- n - 1]$

With the ROC $| z | < 0.5$ , $x [n] = δ [n + 2] + 6.5 δ [n + 1] - 4 δ [n] - 3.95 {(- 0.5)}^{n} u [- n - 1] - 10.8 {(2)}^{n} u [- n - 1]$ .

Practise this2076 Ashwin · Q33+6 marksState Convolution property of Z-transform. Find inverse Z-transform of

X (z) = z

(z - 0.6) (z + 0.5)^{2}

}, ROC:

| z | > 0.6

What is different in this one

The numerator is of lower degree than the denominator, so it is already proper: no division, straight to F(z) = $X (z)$ /z.
There is a repeated pole at -0.5. It needs a term for every power up to its order, and the extra one inverts with an n in front: $A z / (z - p)^{2}$ is (A/p) n $p^{n} u [n]$ , not A $p^{n} u [n]$ . Cover up cannot reach the lower constant at all, which is why the constants are found by comparing coefficients here.
The ROC is the outside of the outermost pole, so every term is right sided and the answer is causal.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{z}{z^{3} + 0.4 z^{2} - 0.35 z - 0.15}

2. Nothing to divide. Degree 1 over degree 3, so the fraction is already low enough on top.

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = \frac{1}{z^{3} + 0.4 z^{2} - 0.35 z - 0.15}

The numerator carried a z of its own, and it cancels the z we just divided by, so there is no term at the origin at all.

z^{3} + 0.4 z^{2} - 0.35 z - 0.15 = {(z + 0.5)}^{2} (z - 0.6) = 0

So the poles are $- 0.5$ , repeated 2 times and $0.6$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{1}{{(z + 0.5)}^{2} (z - 0.6)} \\ = \frac{A}{(z + 0.5)} + \frac{B}{{(z + 0.5)}^{2}} + \frac{C}{(z - 0.6)} \end{aligned}

Put them back over the common denominator ${(z + 0.5)}^{2} (z - 0.6)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z + 0.5) (z - 0.6) + B (z - 0.6) + C {(z + 0.5)}^{2} \\ A (z + 0.5) (z - 0.6) & = A z^{2} - 0.1 A z - 0.3 A \\ B (z - 0.6) & = B z - 0.6 B \\ C {(z + 0.5)}^{2} & = C z^{2} + C z + 0.25 C \end{aligned}

Now collect the powers of z and set each one equal to the same power in $1$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + C & = 0 & from z^{2} \\ - 0.1 A + B + C & = 0 & from z \\ - 0.3 A - 0.6 B + 0.25 C & = 1 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = - 0.8264 \\ B & = - 0.9091 \\ C & = 0.8264 \end{aligned}

Putting those back and adding the terms returns $1$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = \frac{- 0.8264 z}{(z + 0.5)} + \frac{- 0.9091 z}{{(z + 0.5)}^{2}} + \frac{0.8264 z}{(z - 0.6)}

6. Invert term by term. The ROC $| z | > 0.6$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
-0.5	0.5	-0.8264	the ROC lies outside it	-0.8264 ${(- 0.5)}^{n} u [n]$
-0.5, order 2	0.5	-0.9091	the ROC lies outside it	1.8182 n ${(- 0.5)}^{n} u [n]$
0.6	0.6	0.8264	the ROC lies outside it	0.8264 ${(0.6)}^{n} u [n]$

With the ROC $| z | > 0.6$ , $x [n] = - 0.8264 {(- 0.5)}^{n} u [n]$ + 1.8182 n ${(- 0.5)}^{n} u [n] + 0.8264 {(0.6)}^{n} u [n]$ .

Check it without trusting any of the algebra: run the difference equation the transform encodes and compare the first samples. Both routes give ${0, 0, 1, - 0.4, 0.51, - 0.194}$ .

Same as 2081 Bhadra Q32074 Chaitra · Q31+3+5 marksDefine a ROC. What are the properties of ROC of z-transform? Find the inverse Z-transform of

X (z) = (2 z^{2} + 2 z^{2} + 3 z + 5) / (z^{2} - 0.1 z - 0.2)

, ROC:

| z | < 0.4

What is different in this one

The numerator is of higher degree than the denominator (3 against 2), so it has to be long divided first. The quotient is a polynomial in z, which inverts to impulses at negative n, and only the remainder goes through partial fractions. Divide until the remainder reaches the degree of the divisor and stop there, not further.
The ROC is the inside of the innermost pole, so every term is left sided and the whole answer carries $u [- n - 1]$ with a sign change.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{2 z^{3} + 2 z^{2} + 3 z + 5}{z^{2} - 0.1 z - 0.2}

			$2 z$
$z^{2} - 0.1 z - 0.2$ )	$2 z^{3}$	$+ 2 z^{2}$	$+ 3 z$	$+ 5$
	$2 z^{3}$	$- 0.2 z^{2}$	$- 0.4 z$
		$2.2 z^{2}$	$+ 3.4 z$	$+ 5$

X (z) = 2 z + \frac{2.2 z^{2} + 3.4 z + 5}{z^{2} - 0.1 z - 0.2}

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = 2 + \frac{2.2 z^{2} + 3.4 z + 5}{z^{3} - 0.1 z^{2} - 0.2 z}

The poles are the roots of that denominator. Note the z we just divided by is one of them, sitting at the origin.

z^{3} - 0.1 z^{2} - 0.2 z = z (z + 0.4) (z - 0.5) = 0

So the poles are $0$ , $- 0.4$ and $0.5$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{2.2 z^{2} + 3.4 z + 5}{z (z + 0.4) (z - 0.5)} \\ = \frac{A}{z} + \frac{B}{(z + 0.4)} + \frac{C}{(z - 0.5)} \end{aligned}

Put them back over the common denominator $z (z + 0.4) (z - 0.5)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z + 0.4) (z - 0.5) + B z (z - 0.5) + C z (z + 0.4) \\ A (z + 0.4) (z - 0.5) & = A z^{2} - 0.1 A z - 0.2 A \\ B z (z - 0.5) & = B z^{2} - 0.5 B z \\ C z (z + 0.4) & = C z^{2} + 0.4 C z \end{aligned}

Now collect the powers of z and set each one equal to the same power in $2.2 z^{2} + 3.4 z + 5$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + B + C & = 2.2 & from z^{2} \\ - 0.1 A - 0.5 B + 0.4 C & = 3.4 & from z \\ - 0.2 A & = 5 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = - 25 \\ B & = 11.0889 \\ C & = 16.1111 \end{aligned}

Putting those back and adding the terms returns $2.2 z^{2} + 3.4 z + 5$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = 2 z - 25 + \frac{11.0889 z}{(z + 0.4)} + \frac{16.1111 z}{(z - 0.5)}

6. Invert term by term. The ROC $| z | < 0.4$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
none, from the division		2	a polynomial term	2 $δ [n + 1]$
0, the origin	0	-25	a polynomial term	-25 $δ [n]$
-0.4	0.4	11.0889	the ROC lies inside it	-11.0889 ${(- 0.4)}^{n} u [- n - 1]$
0.5	0.5	16.1111	the ROC lies inside it	-16.1111 ${(0.5)}^{n} u [- n - 1]$

With the ROC $| z | < 0.4$ , $x [n] = 2 δ [n + 1] - 25 δ [n] - 11.0889 {(- 0.4)}^{n} u [- n - 1] - 16.1111 {(0.5)}^{n} u [- n - 1]$ .

Practise this2074 Ashwin · Q33+3 marksState and explain the properties of a Region of Convergence (ROC). Find the inverse z-transform of

X (z) = z^{2} [1 - (3 / 2) z^{- 1}] (1 + z^{- 1}) (1 - z^{- 1})

What is different hereNo ROC is given because none is needed: a finite length sequence converges everywhere except possibly at

z = 0

and z = infinity.

What is different in this one

There is nothing to invert: no partial fractions, no ROC decision. $X (z)$ is already a finite polynomial, so its coefficients are the sequence. Thirty seconds if you spot it.
A finite length sequence converges everywhere except possibly at $z = 0$ and z = infinity, which is why the question does not give you an ROC.
The factor $z^{2}$ is an advance, so the sequence starts at $n = - 2$ , before the origin.

Multiply the factors out, then read each coefficient as an impulse: a power of $z^{- k}$ is a delay of k and a power of $z^{k}$ an advance.

\begin{aligned} X (z) & = z^{2} (1 - 1.5 z^{- 1}) (1 + z^{- 1}) (1 - z^{- 1}) \\ = z^{2} (1 - 1.5 z^{- 1}) \\ = z^{2} (1 - 0.5 z^{- 1} - 1.5 z^{- 2}) \\ = z^{2} (1 - 1.5 z^{- 1} - z^{- 2} + 1.5 z^{- 3}) \end{aligned}

$x [n] = 1 δ [n + 2] - 1.5 δ [n + 1] - 1 δ [n] + 1.5 δ [n - 1]$ , that is {1, -1.5, -1, 1.5} for $n = - 2$ to 1.

Practise this2073 Shrawan · Q31+5 marksDefine a ROC. Find inverse Z-transform of

X (z) = z

(z - 0.4) (z + 1.5)^{2}

}, ROC:

| z | < 0.4

How this is read $X (z) = z / ((z - 0.4) (z + 1.5)^{2})$ , whose denominator multiplies out to $z^{3} + 2.6 z^{2} + 1.05 z - 0.9$ .

What is different in this one

The numerator is of lower degree than the denominator, so it is already proper: no division, straight to F(z) = $X (z)$ /z.
There is a repeated pole at -1.5. It needs a term for every power up to its order, and the extra one inverts with an n in front: $A z / (z - p)^{2}$ is (A/p) n $p^{n} u [n]$ , not A $p^{n} u [n]$ . Cover up cannot reach the lower constant at all, which is why the constants are found by comparing coefficients here.
The ROC is the inside of the innermost pole, so every term is left sided and the whole answer carries $u [- n - 1]$ with a sign change.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{z}{z^{3} + 2.6 z^{2} + 1.05 z - 0.9}

2. Nothing to divide. Degree 1 over degree 3, so the fraction is already low enough on top.

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = \frac{1}{z^{3} + 2.6 z^{2} + 1.05 z - 0.9}

The numerator carried a z of its own, and it cancels the z we just divided by, so there is no term at the origin at all.

z^{3} + 2.6 z^{2} + 1.05 z - 0.9 = (z - 0.4) {(z + 1.5)}^{2} = 0

So the poles are $0.4$ and $- 1.5$ , repeated 2 times.

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{1}{(z - 0.4) {(z + 1.5)}^{2}} \\ = \frac{A}{(z - 0.4)} + \frac{B}{(z + 1.5)} + \frac{C}{{(z + 1.5)}^{2}} \end{aligned}

Put them back over the common denominator $(z - 0.4) {(z + 1.5)}^{2}$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A {(z + 1.5)}^{2} + B (z - 0.4) (z + 1.5) + C (z - 0.4) \\ A {(z + 1.5)}^{2} & = A z^{2} + 3 A z + 2.25 A \\ B (z - 0.4) (z + 1.5) & = B z^{2} + 1.1 B z - 0.6 B \\ C (z - 0.4) & = C z - 0.4 C \end{aligned}

Now collect the powers of z and set each one equal to the same power in $1$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + B & = 0 & from z^{2} \\ 3 A + 1.1 B + C & = 0 & from z \\ 2.25 A - 0.6 B - 0.4 C & = 1 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = 0.277 \\ B & = - 0.277 \\ C & = - 0.5263 \end{aligned}

Putting those back and adding the terms returns $1$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = \frac{0.277 z}{(z - 0.4)} + \frac{- 0.277 z}{(z + 1.5)} + \frac{- 0.5263 z}{{(z + 1.5)}^{2}}

6. Invert term by term. The ROC $| z | < 0.4$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
0.4	0.4	0.277	the ROC lies inside it	-0.277 ${(0.4)}^{n} u [- n - 1]$
-1.5	1.5	-0.277	the ROC lies inside it	0.277 ${(- 1.5)}^{n} u [- n - 1]$
-1.5, order 2	1.5	-0.5263	the ROC lies inside it	-0.3509 n ${(- 1.5)}^{n} u [- n - 1]$

With the ROC $| z | < 0.4$ , $x [n] = - 0.277 {(0.4)}^{n} u [- n - 1] + 0.277 {(- 1.5)}^{n} u [- n - 1]$ - 0.3509 n ${(- 1.5)}^{n} u [- n - 1]$ .

Same as 2080 Baishakh Q32071 Shrawan · Q26 marksFind the inverse z transform of:

X (Z) = (1 + 2 z^{- 1} + z^{- 2}) / (1 + 1.5 z^{- 1} + 0.5 z^{- 2})

| z | > 1

using partial fraction method.

What is different in this one

The numerator reaches the degree of the denominator. Dividing by z is enough to make it proper, so no long division is needed, but the expansion will carry a term at the origin that inverts to an impulse.
The ROC is the outside of the outermost pole, so every term is right sided and the answer is causal.

1. Write $X (z)$ in positive powers of z. Multiply the top and the bottom by the highest power of z that appears, and the transform becomes a ratio of ordinary polynomials.

X (z) = \frac{z^{2} + 2 z + 1}{z^{2} + 1.5 z + 0.5}

2. Nothing to divide. Degree 2 over degree 2, so the fraction is already low enough on top.

3. Take F(z) = $X (z)$ /z. Every standard pair in the table has a z on top, so carrying that z aside now is what lets the answer be read straight off the table at the end.

F (z) = \frac{X (z)}{z} = \frac{z^{2} + 2 z + 1}{z^{3} + 1.5 z^{2} + 0.5 z}

The poles are the roots of that denominator. Note the z we just divided by is one of them, sitting at the origin.

z^{3} + 1.5 z^{2} + 0.5 z = z (z + 0.5) (z + 1) = 0

So the poles are $0$ , $- 0.5$ and $- 1$ .

4. Split F(z) and find the constants.

Split it into one term per pole. A pole of order two needs both powers, not just the square: leave the lower one out and the algebra cannot be satisfied.

\begin{aligned} F (z) & = \frac{z^{2} + 2 z + 1}{z (z + 0.5) (z + 1)} \\ = \frac{A}{z} + \frac{B}{(z + 0.5)} + \frac{C}{(z + 1)} \end{aligned}

Put them back over the common denominator $z (z + 0.5) (z + 1)$ . The numerator has to come back to what it was, so multiply each constant by the factors its own term is missing, and expand.

\begin{aligned} numerator & = A (z + 0.5) (z + 1) + B z (z + 1) + C z (z + 0.5) \\ A (z + 0.5) (z + 1) & = A z^{2} + 1.5 A z + 0.5 A \\ B z (z + 1) & = B z^{2} + B z \\ C z (z + 0.5) & = C z^{2} + 0.5 C z \end{aligned}

Now collect the powers of z and set each one equal to the same power in $z^{2} + 2 z + 1$ . That is one equation per power, and as many equations as there are constants.

\begin{aligned} A + B + C & = 1 & from z^{2} \\ 1.5 A + B + 0.5 C & = 2 & from z \\ 0.5 A & = 1 & from t h e c o n s t a n t \end{aligned}

Solving those together:

\begin{aligned} A & = 2 \\ B & = - 1 \\ C & = 0 \end{aligned}

Putting those back and adding the terms returns $z^{2} + 2 z + 1$ , the numerator we started with, so the constants are right.

5. Multiply the z back in. $X (z) = z$ F(z), which puts a z on top of every term and turns each one into a standard pair.

X (z) = 2 + \frac{- 1 z}{(z + 0.5)}

6. Invert term by term. The ROC $| z | > 1$ , not the algebra, decides each pole: the ROC lying outside a pole makes it causal, lying inside makes it anticausal.

Pole	$\| p \|$	Constant	Which side	Inverts to
0, the origin	0	2	a polynomial term	2 $δ [n]$
-0.5	0.5	-1	the ROC lies outside it	- ${(- 0.5)}^{n} u [n]$

The constant at -1 came out zero: a zero of the numerator sits on that pole and cancels it, so it contributes nothing.

With the ROC $| z | > 1$ , $x [n] = 2 δ [n] - {(- 0.5)}^{n} u [n]$ .

Check it without trusting any of the algebra: run the difference equation the transform encodes and compare the first samples. Both routes give ${1, 0.5, - 0.25, 0.125, - 0.0625, 0.03125}$ .

The FFT, decimation in time TOP 11/19

Ch 7 · DFT and FFT11 questions, from 11 of the 19 sittings3 to practise

The procedure, the same for every question in this topic

\begin{aligned} X [k] & = G [k] + W_{N}^{k} H [k] & G from the EVEN samples \\ X [k + \frac{N}{2}] & = G [k] - W_{N}^{k} H [k] & H from the ODD samples \end{aligned}

BUTTERFLY a --------o------> a + WN^r b X the twiddle multiplies the b --[W]---o------> a - WN^r b SECOND input, BEFORE the add and subtract

Order: the input is bit reversed and the output comes out natural. For $N = 8$ that is $x [0], x [4], x [2], x [6], x [1], x [5], x [3], x [7]$ .

Stages: $\log_{2} N$ of them, each of $N / 2$ butterflies. The first stage twiddles are all 1, so it is pure add and subtract.

W_{8}^{0} = 1, W_{8}^{1} = 0.707 - j 0.707, W_{8}^{2} = - j, W_{8}^{3} = - 0.707 - j 0.707

Pad to a power of two first. Six or seven given samples means an 8 point transform with zeros appended, and saying so is part of the answer.

The checks that cost nothing

$X [0]$ is the sum of all the samples. One addition, and it catches a stage error instantly.
Conjugate symmetry for a real input: $X [N - k] = X^{*} [k]$ , so $X [5] = X^{*} [3]$ and $X [6] = X^{*} [2]$ . Half the transform checks the other half.
Parseval: sum $| x [n] |^{2} = \frac{1}{N}$ sum | $X [k]$ |^2, if you have time.

Practise this +8 like it2081 Bhadra · Q92+7 marksHow does FFT reduce computational complexity compared to direct calculation of DFT? Compute 8-point DFT of

x [n] = {1, - 2, - 4, 3, 0, 1}

using DIT-FFT algorithm.

What is different in this one

Decimation in time: 8 points, 3 stages of 4 butterflies, with the input in bit reversed order.
The sequence has 6 samples, not a power of two, so it is zero padded to $N = 8$ before the algorithm starts. Say so in the answer: a radix 2 FFT cannot run on 6 points.

The sequence has 6 samples, and a radix 2 FFT needs a power of two, so zero pad to $N = 8$ and say so: $x [n] = {1, - 2, - 4, 3, 0, 1, 0, 0}$ .

Stage 1, 4 butterflies with twiddles $W_{2}^{0}$ :

\begin{aligned} 1 \\ 1 \\ - 4 \\ - 4 \\ - 1 \\ - 3 \\ 3 \\ 3 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ :

\begin{aligned} - 3 \\ 1 + j 4 \\ 5 \\ 1 - j 4 \\ 2 \\ - 3 - j 3 \\ - 4 \\ - 3 + j 3 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ :

\begin{aligned} - 1 \\ - 3.2426 + j 4 \\ 5 + j 4 \\ 5.2426 - j 4 \\ - 5 \\ 5.2426 + j 4 \\ 5 - j 4 \\ - 3.2426 - j 4 \end{aligned}

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	-1	1	180
$X [1]$	-3.2426 + j4	5.1492	129.03
$X [2]$	5 + j4	6.4031	38.66
$X [3]$	5.2426 - j4	6.5943	-37.34
$X [4]$	-5	5	180
$X [5]$	5.2426 + j4	6.5943	37.34
$X [6]$	5 - j4	6.4031	-38.66
$X [7]$	-3.2426 - j4	5.1492	-129.03

Practise this2081 Baishakh · Q92+6 marksWhy Decimation in Time Fast Fourier Transform (DITFFT) Algorithm is better than direct computation of DFT? Find 4 point DFT of the sequence

x (n) = {2, 2, 4}

using DITFFT algorithm.What is different hereonly four points, so two stages, and the only twiddles are 1 and -j

What is different in this one

Decimation in time: 4 points, 2 stages of 2 butterflies, with the input in bit reversed order.
The sequence has 3 samples, not a power of two, so it is zero padded to $N = 4$ before the algorithm starts. Say so in the answer: a radix 2 FFT cannot run on 3 points.
This is a 4 point transform, so there are only 2 stages of 2 butterflies and the only twiddles are $W 4^{0} = 1$ and $W 4^{1} = - j$ . Much shorter than the usual 8 point question.

The sequence has 3 samples, and a radix 2 FFT needs a power of two, so zero pad to $N = 4$ and say so: $x [n] = {2, 2, 4, 0}$ .

Decimation in time: the input goes in bit reversed order and the output comes out in natural order. For $N = 4$ that input order is $x [0]$ , $x [2]$ , $x [1]$ , $x [3]$ .

Stage 1, 2 butterflies with twiddles $W_{2}^{0}$ :

\begin{aligned} 6 \\ - 2 \\ 2 \\ 2 \end{aligned}

Stage 2, 2 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ :

\begin{aligned} 8 \\ - 2 - j 2 \\ 4 \\ - 2 + j 2 \end{aligned}

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	8	8	0
$X [1]$	-2 - j2	2.8284	-135
$X [2]$	4	4	0
$X [3]$	-2 + j2	2.8284	135

$X [k]$ = {8, -2 - j2, 4, -2 + j2}. Two free checks: $X [0] = 8$ , which is the sum of the samples; $X [3] = X^{*} [1]$ and $X [2] = X^{*} [2]$ , the conjugate symmetry a real input forces.

Same as 2081 Bhadra Q92080 Baishakh · Q112+6 marksHow fast is FFT? Find 8-point DFT of sequence

x [n] = {1, 1, 0, 0, 1, 1, 2}

using Decimation in Time Fast Fourier Transform (DITFFT) algorithm.

What is different in this one

Decimation in time: 8 points, 3 stages of 4 butterflies, with the input in bit reversed order.
The sequence has 7 samples, not a power of two, so it is zero padded to $N = 8$ before the algorithm starts. Say so in the answer: a radix 2 FFT cannot run on 7 points.

The sequence has 7 samples, and a radix 2 FFT needs a power of two, so zero pad to $N = 8$ and say so: $x [n] = {1, 1, 0, 0, 1, 1, 2, 0}$ .

Stage 1, 4 butterflies with twiddles $W_{2}^{0}$ :

\begin{aligned} 2 \\ 0 \\ 2 \\ - 2 \\ 2 \\ 0 \\ 0 \\ 0 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ :

\begin{aligned} 4 \\ j2 \\ 0 \\ - j 2 \\ 2 \\ 0 \\ 2 \\ 0 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ :

\begin{aligned} 6 \\ j2 \\ - j 2 \\ - j 2 \\ 2 \\ j2 \\ j2 \\ - j 2 \end{aligned}

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	6	6	0
$X [1]$	j2	2	90
$X [2]$	-j2	2	-90
$X [3]$	-j2	2	-90
$X [4]$	2	2	0
$X [5]$	j2	2	90
$X [6]$	j2	2	90
$X [7]$	-j2	2	-90

Same as 2081 Bhadra Q92079 Bhadra · Q102+6 marksWhy we need FFT? Find the 8-point DFT of the following sequence using radix-2 DITFFT algorithm.

What is different in this one

Decimation in time: 8 points, 3 stages of 4 butterflies, with the input in bit reversed order.

$N = 8$ , which is $2^{3}$ , so the transform takes 3 stages of 4 butterflies.

Stage 1, 4 butterflies with twiddles $W_{2}^{0}$ :

\begin{aligned} 1 \\ 1 \\ 3 - j 1 \\ 3 + j 1 \\ j2 \\ j2 \\ j4 \\ j4 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ :

\begin{aligned} 4 - j 1 \\ 2 - j 3 \\ - 2 + j 1 \\ j3 \\ j6 \\ 4 + j 2 \\ - j 2 \\ - 4 + j 2 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ :

\begin{aligned} 4 + j 5 \\ 6.2426 - j 4.4142 \\ - 4 + j 1 \\ 4.2426 + j 4.4142 \\ 4 - j 7 \\ - 2.2426 - j 1.5858 \\ j1 \\ - 4.2426 + j 1.5858 \end{aligned}

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	4 + j5	6.4031	51.34
$X [1]$	6.2426 - j4.4142	7.6456	-35.26
$X [2]$	-4 + j1	4.1231	165.96
$X [3]$	4.2426 + j4.4142	6.1225	46.14
$X [4]$	4 - j7	8.0623	-60.26
$X [5]$	-2.2426 - j1.5858	2.7467	-144.74
$X [6]$	j1	1	90
$X [7]$	-4.2426 + j1.5858	4.5293	159.51

$X [k]$ = {4 + j5, 6.2426 - j4.4142, -4 + j1, 4.2426 + j4.4142, 4 - j7, -2.2426 - j1.5858, j1, -4.2426 + j1.5858}. Two free checks: $X [0] = 4$ + j5, which is the sum of the samples.

Same as 2081 Bhadra Q92078 Bhadra · Q117 marksFind the 8-point DFT of

x [n] = u [n] - u [n - 4]

using FFT DIT algorithm.

How this is read $x [n] = u [n] - u [n - 4]$ is ${1, 1, 1, 1}$ for $n = 0$ to 3, and zero to n = 7.

What is different in this one

Decimation in time: 8 points, 3 stages of 4 butterflies, with the input in bit reversed order.

$N = 8$ , which is $2^{3}$ , so the transform takes 3 stages of 4 butterflies.

Stage 1, 4 butterflies with twiddles $W_{2}^{0}$ :

\begin{aligned} 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ :

\begin{aligned} 2 \\ 1 - j 1 \\ 0 \\ 1 + j 1 \\ 2 \\ 1 - j 1 \\ 0 \\ 1 + j 1 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ :

\begin{aligned} 4 \\ 1 - j 2.4142 \\ 0 \\ 1 - j 0.4142 \\ 0 \\ 1 + j 0.4142 \\ 0 \\ 1 + j 2.4142 \end{aligned}

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	4	4	0
$X [1]$	1 - j2.4142	2.6131	-67.5
$X [2]$	0	0	0
$X [3]$	1 - j0.4142	1.0824	-22.5
$X [4]$	0	0	0
$X [5]$	1 + j0.4142	1.0824	22.5
$X [6]$	0	0	0
$X [7]$	1 + j2.4142	2.6131	67.5

Practise this2075 Chaitra · Q82+8 marksHow fast is FFT? Find X(3) and X(5) for given sequence

x [n] = {1, - 2, 3, 2}

k = 0

to 3. X(5) is read by periodicity:

X [k + N] = X [k]

, so X(5) = X(1). Saying that is the question.

What is different in this one

Decimation in time: 4 points, 2 stages of 2 butterflies, with the input in bit reversed order.
This is a 4 point transform, so there are only 2 stages of 2 butterflies and the only twiddles are $W 4^{0} = 1$ and $W 4^{1} = - j$ . Much shorter than the usual 8 point question.
The question asks for X(3) and X(5) of a FOUR point sequence, and a 4 point DFT only has $k = 0$ to 3. X(5) is read by periodicity: $X [k + N] = X [k]$ , so X(5) = X(1). Saying that is the question.

$N = 4$ , which is $2^{2}$ , so the transform takes 2 stages of 2 butterflies.

Decimation in time: the input goes in bit reversed order and the output comes out in natural order. For $N = 4$ that input order is $x [0]$ , $x [2]$ , $x [1]$ , $x [3]$ .

Stage 1, 2 butterflies with twiddles $W_{2}^{0}$ :

\begin{aligned} 4 \\ - 2 \\ 0 \\ - 4 \end{aligned}

Stage 2, 2 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ :

\begin{aligned} 4 \\ - 2 + j 4 \\ 4 \\ - 2 - j 4 \end{aligned}

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	4	4	0
$X [1]$	-2 + j4	4.4721	116.57
$X [2]$	4	4	0
$X [3]$	-2 - j4	4.4721	-116.57

$X [k]$ = {4, -2 + j4, 4, -2 - j4}. Two free checks: $X [0] = 4$ , which is the sum of the samples; $X [3] = X^{*} [1]$ and $X [2] = X^{*} [2]$ , the conjugate symmetry a real input forces.

Same as 2081 Bhadra Q92075 Ashwin · Q83+7 marksHow fast is FFT compare to DFT? Draw the butterfly diagram of 8-point DFT of a sequence as

x [n] = n

+ 1 using Decimation in Time FFT algorithm.

How this is read $x [n] = n$ + 1 over $n = 0$ to 7 is ${1, 2, 3, 4, 5, 6, 7, 8}$ .

What is different in this one

Decimation in time: 8 points, 3 stages of 4 butterflies, with the input in bit reversed order.

$N = 8$ , which is $2^{3}$ , so the transform takes 3 stages of 4 butterflies.

Stage 1, 4 butterflies with twiddles $W_{2}^{0}$ :

\begin{aligned} 6 \\ - 4 \\ 10 \\ - 4 \\ 8 \\ - 4 \\ 12 \\ - 4 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ :

\begin{aligned} 16 \\ - 4 + j 4 \\ - 4 \\ - 4 - j 4 \\ 20 \\ - 4 + j 4 \\ - 4 \\ - 4 - j 4 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ :

\begin{aligned} 36 \\ - 4 + j 9.6569 \\ - 4 + j 4 \\ - 4 + j 1.6569 \\ - 4 \\ - 4 - j 1.6569 \\ - 4 - j 4 \\ - 4 - j 9.6569 \end{aligned}

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	36	36	0
$X [1]$	-4 + j9.6569	10.4525	112.5
$X [2]$	-4 + j4	5.6569	135
$X [3]$	-4 + j1.6569	4.3296	157.5
$X [4]$	-4	4	180
$X [5]$	-4 - j1.6569	4.3296	-157.5
$X [6]$	-4 - j4	5.6569	-135
$X [7]$	-4 - j9.6569	10.4525	-112.5

Same as 2081 Bhadra Q92073 Shrawan · Q102+6 marksWhy do we need DFT? Find 8-point DFT of sequence

x [n] = {1, - 1, 2, 2, 1, 1, 2, 2}

using Fast Fourier Transform algorithm.

What is different in this one

Decimation in time: 8 points, 3 stages of 4 butterflies, with the input in bit reversed order.

$N = 8$ , which is $2^{3}$ , so the transform takes 3 stages of 4 butterflies.

Stage 1, 4 butterflies with twiddles $W_{2}^{0}$ :

\begin{aligned} 2 \\ 0 \\ 4 \\ 0 \\ 0 \\ - 2 \\ 4 \\ 0 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ :

\begin{aligned} 6 \\ 0 \\ - 2 \\ 0 \\ 4 \\ - 2 \\ - 4 \\ - 2 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ :

\begin{aligned} 10 \\ - 1.4142 + j 1.4142 \\ - 2 + j 4 \\ 1.4142 + j 1.4142 \\ 2 \\ 1.4142 - j 1.4142 \\ - 2 - j 4 \\ - 1.4142 - j 1.4142 \end{aligned}

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	10	10	0
$X [1]$	-1.4142 + j1.4142	2	135
$X [2]$	-2 + j4	4.4721	116.57
$X [3]$	1.4142 + j1.4142	2	45
$X [4]$	2	2	0
$X [5]$	1.4142 - j1.4142	2	-45
$X [6]$	-2 - j4	4.4721	-116.57
$X [7]$	-1.4142 - j1.4142	2	-135

Same as 2081 Bhadra Q92072 Chaitra · Q118 marksFind the FFT of the signal

x [n] {1, 1, 2, 4, 3, 1, 2, 1}

using DIT-FFT algorithm.

What is different in this one

Decimation in time: 8 points, 3 stages of 4 butterflies, with the input in bit reversed order.

$N = 8$ , which is $2^{3}$ , so the transform takes 3 stages of 4 butterflies.

Stage 1, 4 butterflies with twiddles $W_{2}^{0}$ :

\begin{aligned} 4 \\ - 2 \\ 4 \\ 0 \\ 2 \\ 0 \\ 5 \\ 3 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ :

\begin{aligned} 8 \\ - 2 \\ 0 \\ - 2 \\ 7 \\ - j 3 \\ - 3 \\ j3 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ :

\begin{aligned} 15 \\ - 4.1213 - j 2.1213 \\ j3 \\ 0.1213 - j 2.1213 \\ 1 \\ 0.1213 + j 2.1213 \\ - j 3 \\ - 4.1213 + j 2.1213 \end{aligned}

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	15	15	0
$X [1]$	-4.1213 - j2.1213	4.6352	-152.76
$X [2]$	j3	3	90
$X [3]$	0.1213 - j2.1213	2.1248	-86.73
$X [4]$	1	1	0
$X [5]$	0.1213 + j2.1213	2.1248	86.73
$X [6]$	-j3	3	-90
$X [7]$	-4.1213 + j2.1213	4.6352	152.76

Same as 2081 Bhadra Q92072 Kartik · Q107 marksFind 8-point DFT of sequence

x [n] = {1, 1, 0, 1, 0, 1, 2}

using Decimation in Time Fast Fourier Transform (DITFFT) algorithm.

What is different in this one

Decimation in time: 8 points, 3 stages of 4 butterflies, with the input in bit reversed order.
The sequence has 7 samples, not a power of two, so it is zero padded to $N = 8$ before the algorithm starts. Say so in the answer: a radix 2 FFT cannot run on 7 points.

The sequence has 7 samples, and a radix 2 FFT needs a power of two, so zero pad to $N = 8$ and say so: $x [n] = {1, 1, 0, 1, 0, 1, 2, 0}$ .

Stage 1, 4 butterflies with twiddles $W_{2}^{0}$ :

\begin{aligned} 1 \\ 1 \\ 2 \\ - 2 \\ 2 \\ 0 \\ 1 \\ 1 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ :

\begin{aligned} 3 \\ 1 + j 2 \\ - 1 \\ 1 - j 2 \\ 3 \\ - j 1 \\ 1 \\ j1 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ :

\begin{aligned} 6 \\ 0.2929 + j 1.2929 \\ - 1 - j 1 \\ 1.7071 - j 2.7071 \\ 0 \\ 1.7071 + j 2.7071 \\ - 1 + j 1 \\ 0.2929 - j 1.2929 \end{aligned}

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	6	6	0
$X [1]$	0.2929 + j1.2929	1.3257	77.24
$X [2]$	-1 - j1	1.4142	-135
$X [3]$	1.7071 - j2.7071	3.2004	-57.76
$X [4]$	0	0	0
$X [5]$	1.7071 + j2.7071	3.2004	57.76
$X [6]$	-1 + j1	1.4142	135
$X [7]$	0.2929 - j1.2929	1.3257	-77.24

Same as 2081 Bhadra Q92071 Shrawan · Q106+2 marksFind DFT for

{1, 1, 2, 0, 1, 2, 0, 1}

using FFT DIT butterfly algorithm and plot the spectrum.

What is different in this one

Decimation in time: 8 points, 3 stages of 4 butterflies, with the input in bit reversed order.

$N = 8$ , which is $2^{3}$ , so the transform takes 3 stages of 4 butterflies.

Stage 1, 4 butterflies with twiddles $W_{2}^{0}$ :

\begin{aligned} 2 \\ 0 \\ 2 \\ 2 \\ 3 \\ - 1 \\ 1 \\ - 1 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ :

\begin{aligned} 4 \\ - j 2 \\ 0 \\ j2 \\ 4 \\ - 1 + j 1 \\ 2 \\ - 1 - j 1 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ :

\begin{aligned} 8 \\ - j 0.5858 \\ - j 2 \\ j3.4142 \\ 0 \\ - j 3.4142 \\ j2 \\ j0.5858 \end{aligned}

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	8	8	0
$X [1]$	-j0.5858	0.5858	-90
$X [2]$	-j2	2	-90
$X [3]$	j3.4142	3.4142	90
$X [4]$	0	0	0
$X [5]$	-j3.4142	3.4142	-90
$X [6]$	j2	2	90
$X [7]$	j0.5858	0.5858	90

The spectrum. The question asks for it plotted, so draw the magnitude against k. It is symmetric about the middle for a real input, which is the check.

X [k]

|, drawn from the computed samples.

The IIR lattice ladder TOP 10/19

Ch 4 · Filter structures10 questions, from 10 of the 19 sittings6 to practise

The procedure, the same for every question in this topic

From the denominator $A (z)$ , step the order down. The last coefficient of the polynomial is the reflection coefficient of that order.
$K_{m} = α_{m} (m)$

$α_{m - 1} (k) = \frac{α_{m} (k) - K_{m} α_{m} (m - k)}{1 - K_{m}^{2}}, k = 1 \dots m - 1$
Keep every intermediate polynomial: the ladder needs them.
The ladder taps come from the numerator, top down.
$\begin{aligned} C_{M} & = b_{M} \\ C_{m} & = b_{m} - \sum_{i = m + 1}^{M} C_{i} α_{i} (i - m) \end{aligned}$
Draw the all pole lattice, take a tap from every backward node $g_{m}$ , multiply by $C_{m}$ , and sum them into $y [n]$ .
$\begin{aligned} f_{m - 1} [n] & = f_{m} [n] - K_{m} g_{m - 1} [n - 1] & note the minus \\ g_{m} [n] & = K_{m} f_{m - 1} [n] + g_{m - 1} [n - 1] \end{aligned}$

$a_{0}$ must be 1. If the denominator starts with anything else, divide the whole fraction through by it before starting.
An all pole filter has no ladder beyond $C_{0} = 1$ , so the answer is the bare lattice.
The minus sign is the difference between the all pole lattice and the all zero one. Getting it backwards is the standard lost mark.

The checks that cost nothing

$| K m | < 1$ for every m if and only if the filter is stable. That single test replaces finding the poles, and it is worth stating whether the question asks or not.
$1 - K m^{2}$ must not be zero. If it is, a pole sits on the unit circle and no lattice exists: report the breakdown, do not invent a number.
Run the recursion forwards from the K values and you must get $A (z)$ back.

Practise this +1 like it2082 Bhadra · Q66 marksCompute Lattice ladder coefficients and draw lattice structure for the given system

H (z) = (2 - 0.7 z^{- 1} + 0.5 z^{- 2}) / (1 - 0.3 z^{- 1} + 0.25 z^{- 2})

What is different herea second order denominator, the usual case

What is different in this one

A denominator of order 2, so 2 reflection coefficients and 2 lattice stages.
A numerator of order 2 as well, so the ladder taps $C_{0}$ to $C_{2}$ have to be computed too, from the intermediate polynomials the recursion throws off.

\begin{aligned} A (z) & = 1 - 0.3 z^{- 1} + 0.25 z^{- 2} \\ B (z) & = 2 - 0.7 z^{- 1} + 0.5 z^{- 2} \end{aligned}

Denominator at this order	Reflection coefficient
$A 2 (z) = 1 - 0.3 z^{- 1} + 0.25 z^{- 2}$	K2 = 0.25
$A 1 (z) = 1 - 0.24 z^{- 1}$	K1 = -0.24

\begin{aligned} M \\ CM & = bM  Cm = bm - \sum Ci α_{i} (i - m) m = M - 1 \dots 0 \\ i & = m + 1 \end{aligned}

Ladder tap	b value	Value
$C_{2}$	0.5	0.5
$C_{1}$	-0.7	-0.55
$C_{0}$	2	1.743

K1 = -0.24, K2 = 0.25, and $C_{0} = 1.743$ , $C_{1} = - 0.55$ , $C_{2} = 0.5$ . Every $| K |$ is below 1, so the filter is stable, and that test replaces finding the poles.

Practise this +3 like it2082 Baishakh · Q510 marksA third order low pass filter has a system function

H (z) = (0.2759 + 0.5121 z^{- 1} + 0.5121 z^{- 2} + 0.2759 z^{- 3}) / (1 - 0.0010 z^{- 1} + 0.6546 z^{- 2} - 0.0775 z^{- 3})

. Implement the filter using lattice-ladder structure.

What is different in this one

A denominator of order 3, so 3 reflection coefficients and 3 lattice stages.
A numerator of order 3 as well, so the ladder taps $C_{0}$ to $C_{3}$ have to be computed too, from the intermediate polynomials the recursion throws off.

\begin{aligned} A (z) & = 1 - 0.001 z^{- 1} + 0.6546 z^{- 2} - 0.0775 z^{- 3} \\ B (z) & = 0.2759 + 0.5121 z^{- 1} + 0.5121 z^{- 2} + 0.2759 z^{- 3} \end{aligned}

Denominator at this order	Reflection coefficient
$A 3 (z) = 1 - 0.001 z^{- 1} + 0.6546 z^{- 2} - 0.0775 z^{- 3}$	K3 = -0.0775
$A 2 (z) = 1 + 0.05 z^{- 1} + 0.6585 z^{- 2}$	K2 = 0.6585
$A 1 (z) = 1 + 0.0302 z^{- 1}$	K1 = 0.0302

\begin{aligned} M \\ CM & = bM  Cm = bm - \sum Ci α_{i} (i - m) m = M - 1 \dots 0 \\ i & = m + 1 \end{aligned}

Ladder tap	b value	Value
$C_{3}$	0.2759	0.2759
$C_{2}$	0.5121	0.5124
$C_{1}$	0.5121	0.3059
$C_{0}$	0.2759	-0.0493

Same as 2082 Baishakh Q52081 Bhadra · Q56+4 marksCompute the lattice and ladder coefficients and draw lattice-ladder structure for the given IIR system

H (z) = (1 + z^{- 1} + z^{- 2}) / ((1 + 0.5 z^{- 1}) (1 + 0.3 z^{- 1}) (1 + 0.4 z^{- 1}))

How this is read $(1 + 0.5 z^{- 1}) (1 + 0.3 z^{- 1}) (1 + 0.4 z^{- 1})$ multiplies out to $1 + 1.2 z^{- 1} + 0.47 z^{- 2} + 0.06 z^{- 3}$ .

What is different in this one

A denominator of order 3, so 3 reflection coefficients and 3 lattice stages.
A numerator of order 2 as well, so the ladder taps $C_{0}$ to $C_{2}$ have to be computed too, from the intermediate polynomials the recursion throws off.

\begin{aligned} A (z) & = 1 + 1.2 z^{- 1} + 0.47 z^{- 2} + 0.06 z^{- 3} \\ B (z) & = 1 + z^{- 1} + z^{- 2} \end{aligned}

Denominator at this order	Reflection coefficient
$A 3 (z) = 1 + 1.2 z^{- 1} + 0.47 z^{- 2} + 0.06 z^{- 3}$	K3 = 0.06
$A 2 (z) = 1 + 1.176 z^{- 1} + 0.3994 z^{- 2}$	K2 = 0.3994
$A 1 (z) = 1 + 0.8404 z^{- 1}$	K1 = 0.8404

\begin{aligned} M \\ CM & = bM  Cm = bm - \sum Ci α_{i} (i - m) m = M - 1 \dots 0 \\ i & = m + 1 \end{aligned}

Ladder tap	b value	Value
$C_{2}$	1	1
$C_{1}$	1	-0.176
$C_{0}$	1	0.7485

K1 = 0.8404, K2 = 0.3994, K3 = 0.06, and $C_{0} = 0.7485$ , $C_{1} = - 0.176$ , $C_{2} = 1$ . Every $| K |$ is below 1, so the filter is stable, and that test replaces finding the poles.

Practise this2080 Bhadra · Q510 marksDraw the Lattice structure from the following system function

. H (z) = 1 / (1 - 0.2 z^{- 1} + 0.4 z^{- 2} + 0.6 z^{- 3})

What is different herean all pole system, so the lattice has no ladder taps

What is different in this one

A denominator of order 3, so 3 reflection coefficients and 3 lattice stages.
The numerator is a constant, so this is an all pole filter: the only ladder tap is $C_{0}$ and the answer is the bare lattice.

\begin{aligned} A (z) & = 1 - 0.2 z^{- 1} + 0.4 z^{- 2} + 0.6 z^{- 3} \\ B (z) & = 1 \end{aligned}

Denominator at this order	Reflection coefficient
$A 3 (z) = 1 - 0.2 z^{- 1} + 0.4 z^{- 2} + 0.6 z^{- 3}$	K3 = 0.6
$A 2 (z) = 1 - 0.6875 z^{- 1} + 0.8125 z^{- 2}$	K2 = 0.8125
$A 1 (z) = 1 - 0.3793 z^{- 1}$	K1 = -0.3793

\begin{aligned} M \\ CM & = bM  Cm = bm - \sum Ci α_{i} (i - m) m = M - 1 \dots 0 \\ i & = m + 1 \end{aligned}

Ladder tap	b value	Value
$C_{0}$	1	1

K1 = -0.3793, K2 = 0.8125, K3 = 0.6, and $C_{0} = 1$ . Every $| K |$ is below 1, so the filter is stable, and that test replaces finding the poles.

Same as 2082 Bhadra Q62080 Baishakh · Q56 marksCompute Lattice-ladder coefficients and draw lattice structure for given system

H (z) = (2 - 0.7 z^{- 1} + 0.5 z^{- 2}) / (1 - 0.3 z^{- 1} + 0.25 z^{- 2})

What is different in this one

A denominator of order 2, so 2 reflection coefficients and 2 lattice stages.
A numerator of order 2 as well, so the ladder taps $C_{0}$ to $C_{2}$ have to be computed too, from the intermediate polynomials the recursion throws off.

\begin{aligned} A (z) & = 1 - 0.3 z^{- 1} + 0.25 z^{- 2} \\ B (z) & = 2 - 0.7 z^{- 1} + 0.5 z^{- 2} \end{aligned}

Denominator at this order	Reflection coefficient
$A 2 (z) = 1 - 0.3 z^{- 1} + 0.25 z^{- 2}$	K2 = 0.25
$A 1 (z) = 1 - 0.24 z^{- 1}$	K1 = -0.24

\begin{aligned} M \\ CM & = bM  Cm = bm - \sum Ci α_{i} (i - m) m = M - 1 \dots 0 \\ i & = m + 1 \end{aligned}

Ladder tap	b value	Value
$C_{2}$	0.5	0.5
$C_{1}$	-0.7	-0.55
$C_{0}$	2	1.743

K1 = -0.24, K2 = 0.25, and $C_{0} = 1.743$ , $C_{1} = - 0.55$ , $C_{2} = 0.5$ . Every $| K |$ is below 1, so the filter is stable, and that test replaces finding the poles.

Practise this2079 Baishakh · Q56+1 marksCompute Lattice-ladder coefficients and draw lattice structure for given system

H (z) = (1 - 0.4 z^{- 1} + 0.25 z^{- 2}) / (1 - 0.3 z^{- 1} + 0.5 z^{- 2})

. Also check the stability of given system.

The lattice ladder

What is different in this one

A denominator of order 2, so 2 reflection coefficients and 2 lattice stages.
A numerator of order 2 as well, so the ladder taps $C_{0}$ to $C_{2}$ have to be computed too, from the intermediate polynomials the recursion throws off.

\begin{aligned} A (z) & = 1 - 0.3 z^{- 1} + 0.5 z^{- 2} \\ B (z) & = 1 - 0.4 z^{- 1} + 0.25 z^{- 2} \end{aligned}

Denominator at this order	Reflection coefficient
$A 2 (z) = 1 - 0.3 z^{- 1} + 0.5 z^{- 2}$	K2 = 0.5
$A 1 (z) = 1 - 0.2 z^{- 1}$	K1 = -0.2

\begin{aligned} M \\ CM & = bM  Cm = bm - \sum Ci α_{i} (i - m) m = M - 1 \dots 0 \\ i & = m + 1 \end{aligned}

Ladder tap	b value	Value
$C_{2}$	0.25	0.25
$C_{1}$	-0.4	-0.325
$C_{0}$	1	0.81

K1 = -0.2, K2 = 0.5, and $C_{0} = 0.81$ , $C_{1} = - 0.325$ , $C_{2} = 0.25$ . Every $| K |$ is below 1, so the filter is stable, and that test replaces finding the poles.

The rest of this questionThis topic works the part that belongs to it. The other part is worked under And the stability check, the last mark, where it belongs.

Same as 2082 Baishakh Q52076 Chaitra · Q66+4 marksCompute Lattice and Ladder coefficients and Draw lattice-ladder structure for given IIR system

H (z) = (0.5 - 2 z^{- 1} + 3 z^{- 2}) / (1 - 0.5 z^{- 1} - 0.7 z^{- 2} + 0.3 z^{- 3})

What is different in this one

A denominator of order 3, so 3 reflection coefficients and 3 lattice stages.
A numerator of order 2 as well, so the ladder taps $C_{0}$ to $C_{2}$ have to be computed too, from the intermediate polynomials the recursion throws off.

\begin{aligned} A (z) & = 1 - 0.5 z^{- 1} - 0.7 z^{- 2} + 0.3 z^{- 3} \\ B (z) & = 0.5 - 2 z^{- 1} + 3 z^{- 2} \end{aligned}

Denominator at this order	Reflection coefficient
$A 3 (z) = 1 - 0.5 z^{- 1} - 0.7 z^{- 2} + 0.3 z^{- 3}$	K3 = 0.3
$A 2 (z) = 1 - 0.3187 z^{- 1} - 0.6044 z^{- 2}$	K2 = -0.6044
$A 1 (z) = 1 - 0.8056 z^{- 1}$	K1 = -0.8056

\begin{aligned} M \\ CM & = bM  Cm = bm - \sum Ci α_{i} (i - m) m = M - 1 \dots 0 \\ i & = m + 1 \end{aligned}

Ladder tap	b value	Value
$C_{2}$	3	3
$C_{1}$	-2	-1.044
$C_{0}$	0.5	1.4722

K1 = -0.8056, K2 = -0.6044, K3 = 0.3, and $C_{0} = 1.4722$ , $C_{1} = - 1.044$ , $C_{2} = 3$ . Every $| K |$ is below 1, so the filter is stable, and that test replaces finding the poles.

Same as 2082 Baishakh Q52076 Ashwin · Q56+3 marksCompute Lattice and Ladder coefficients and Draw lattice-ladder structure for given IIR system

H (z) = (0.7 - 1.5 z^{- 1} + 0.5 z^{- 2}) / (1 - 0.5 z^{- 1} - 0.7 z^{- 2} + 0.3 z^{- 3})

What is different in this one

A denominator of order 3, so 3 reflection coefficients and 3 lattice stages.
A numerator of order 2 as well, so the ladder taps $C_{0}$ to $C_{2}$ have to be computed too, from the intermediate polynomials the recursion throws off.

\begin{aligned} A (z) & = 1 - 0.5 z^{- 1} - 0.7 z^{- 2} + 0.3 z^{- 3} \\ B (z) & = 0.7 - 1.5 z^{- 1} + 0.5 z^{- 2} \end{aligned}

Denominator at this order	Reflection coefficient
$A 3 (z) = 1 - 0.5 z^{- 1} - 0.7 z^{- 2} + 0.3 z^{- 3}$	K3 = 0.3
$A 2 (z) = 1 - 0.3187 z^{- 1} - 0.6044 z^{- 2}$	K2 = -0.6044
$A 1 (z) = 1 - 0.8056 z^{- 1}$	K1 = -0.8056

\begin{aligned} M \\ CM & = bM  Cm = bm - \sum Ci α_{i} (i - m) m = M - 1 \dots 0 \\ i & = m + 1 \end{aligned}

Ladder tap	b value	Value
$C_{2}$	0.5	0.5
$C_{1}$	-1.5	-1.3407
$C_{0}$	0.7	-0.0778

Practise this2075 Chaitra · Q49 marksDraw the lattice structure from the following system function

H (t) = 1 / (1 + (2 / 3) z^{- 1} + (5 / 8) z^{- 2} + (2 / 3) z^{- 3} + z^{- 4})

What is different herefourth order, so the recursion runs twice more; an all pole system, so the lattice has no ladder taps

What is different in this one

This one has no lattice at all. A reflection coefficient comes out with magnitude exactly 1, so the step down divides by zero, and the reason is a pole sitting on the unit circle.

Step the order down on the denominator as usual.

A4(z) = 1 +

0.6667 z^{- 1} + 0.625 z^{- 2} + 0.6667 z^{- 3} + z^{- 4}

K4 = 1, so 1

- K^{2}

= 0

\begin{aligned} pole                       magnitude \\ 0.4429 + j 0.89661 \\ 0.4429 - j 0.89661 \\ - 0.7762 + j 0.63051 \\ - 0.7762 - j 0.63051 \end{aligned}

Practise this2075 Ashwin · Q57+3 marksDraw the Lattice structure from the following system function:

1 / (3 + (39 / 24) Z^{- 1} + (15 / 8) Z^{- 2} + (3 / 9) Z^{- 3})

. And represent 5/8 and -5/8 in sign magnitude, 1's complement and 2's complement format.

The lattice structure

The denominator does not start at 1, so divide numerator and denominator through by 3 first. Every recursion in this chapter assumes $a_{0} = 1$ .

What is different in this one

A denominator of order 3, so 3 reflection coefficients and 3 lattice stages.
The numerator is a constant, so this is an all pole filter: the only ladder tap is $C_{0}$ and the answer is the bare lattice.

\begin{aligned} A (z) & = 1 + 0.5417 z^{- 1} + 0.625 z^{- 2} + 0.1111 z^{- 3} \\ B (z) & = 0.3333 \end{aligned}

Denominator at this order	Reflection coefficient
$A 3 (z) = 1 + 0.5417 z^{- 1} + 0.625 z^{- 2} + 0.1111 z^{- 3}$	K3 = 0.1111
$A 2 (z) = 1 + 0.4781 z^{- 1} + 0.5719 z^{- 2}$	K2 = 0.5719
$A 1 (z) = 1 + 0.3042 z^{- 1}$	K1 = 0.3042

\begin{aligned} M \\ CM & = bM  Cm = bm - \sum Ci α_{i} (i - m) m = M - 1 \dots 0 \\ i & = m + 1 \end{aligned}

Ladder tap	b value	Value
$C_{0}$	0.3333	0.3333

K1 = 0.3042, K2 = 0.5719, K3 = 0.1111, and $C_{0} = 0.3333$ . Every $| K |$ is below 1, so the filter is stable, and that test replaces finding the poles.

The rest of this questionThis topic works the part that belongs to it. The other part is worked under Fixed point number formats (5/8 and -5/8 in the three formats), where it belongs.

Direct form realization TOP 9/19

Ch 4 · Filter structures9 questions, from 9 of the 19 sittings2 to practise

The procedure, the same for every question in this topic

y [n] = \sum_{k = 0}^{M} b_{k} x [n - k] - \sum_{k = 1}^{N} a_{k} y [n - k], a_{0} = 1

Form	How it is drawn	Delays
Direct form I	The all zero section first, then the all pole section. Two separate delay lines.	$M + N$
Direct form II	The all pole section first, then the all zero, sharing one delay line through the node $w [n]$ .	$max (M, N)$ , canonic

The shared node is what makes the second form canonic, and it is worth writing both of its equations out:

\begin{aligned} w [n] & = x [n] - a_{1} w [n - 1] - a_{2} w [n - 2] - \dots \\ y [n] & = b_{0} w [n] + b_{1} w [n - 1] + b_{2} w [n - 2] + \dots \end{aligned}

Rearrange first, so that $y [n]$ stands alone. The feedback coefficients change sign as they cross the equals sign; that is where the marks go.
$a_{0}$ must be 1. $3 y [n]$ + ... means dividing the whole equation by 3 first.
Why form II is allowed: $H (z) = B (z)$ . 1/ $A (z)$ and an LTI cascade commutes, so 1/ $A (z)$ may go first; both halves then see the same $w [n]$ , so one delay chain serves both. Write that sentence beside the drawing.

The checks that cost nothing

Count the delays and say the number: M + N against max(M, N). It is a mark.
Label every multiplier with its value and every box with $z^{- 1}$ . An unlabelled drawing scores nothing.
Read the graph back. Walk the drawing and re-derive the difference equation; if it does not come back, a sign is wrong.

Practise this +5 like it2082 Bhadra · Q52+2 marksObtain the direct form I and direct form II realization of the following system:

y [n]

- 0.

75 y [n - 1]

- 0.

25 y [n - 2] = x [n]

+ 0.

5 x [n - 1]

What is different in this one

A feed forward part of order 1 and a feedback part of order 2, so direct form I needs 3 delays and direct form II needs 2.
The two halves are of different order, 1 and 2, so direct form I needs 3 delays and direct form II needs only 2. Saying both numbers is a mark.

Rearrange so that $y [n]$ stands alone. The feedback coefficients change sign as they cross the equals sign, and that is where the marks are lost.

H (z) = \frac{1 + 0.5 z^{- 1}}{1 - 0.75 z^{- 1} - 0.25 z^{- 2}}

\begin{aligned} y [n] & = x [n] + 0.5 x [n - 1] + 0.75 y [n - 1] + 0.25 y [n - 2] \\ feed  forward b_{0} & = 1, b_{1} = 0.5 \\ feedback a_{1} & = - 0.75, so  the  multiplier  is 0.75 \\ a_{2} & = - 0.25, so  the  multiplier  is 0.25 \end{aligned}

Direct form I: the all zero half first, then the all pole half, 1 + 2 = 3 delays.

Direct form II: one delay chain shared by both halves, 2 delays, which is the fewest possible.

Direct form I needs 1 + 2 = 3 delays; direct form II needs max(1, 2) = 2, and is therefore the canonic form. Both realize the same $H (z)$ .

Same as 2082 Bhadra Q52079 Bhadra · Q52+2 marksDraw direct form I and Direct form II realization of the following system.

y [n]

- 0.

25 y [n - 2] + x [n]

+ 0.

4 x [n - 1]

+ 0.

5 x [n - 2]

How this is readThe printed equation has a plus where the equals sign belongs. It reads $y [n]$ - 0. $25 y [n - 2] = x [n]$ + 0. $4 x [n - 1]$ + 0. $5 x [n - 2]$ .

What is different in this one

A feed forward part of order 2 and a feedback part of order 2, so direct form I needs 4 delays and direct form II needs 2.
One of the coefficients is zero. That delay still exists in the chain, it simply has no multiplier hanging off it: leaving it out shortens the structure and changes the filter.

Rearrange so that $y [n]$ stands alone. The feedback coefficients change sign as they cross the equals sign, and that is where the marks are lost.

H (z) = \frac{1 + 0.4 z^{- 1} + 0.5 z^{- 2}}{1 - 0.25 z^{- 2}}

\begin{aligned} y [n] & = x [n] + 0.4 x [n - 1] + 0.5 x [n - 2] + 0.25 y [n - 2] \\ feed  forward b_{0} & = 1, b_{1} = 0.4, b_{2} = 0.5 \\ feedback a_{1} & = 0, so  the  multiplier  is 0 \\ a_{2} & = - 0.25, so  the  multiplier  is 0.25 \end{aligned}

Direct form I: the all zero half first, then the all pole half, 2 + 2 = 4 delays.

Direct form II: one delay chain shared by both halves, 2 delays, which is the fewest possible.

Direct form I needs 2 + 2 = 4 delays; direct form II needs max(2, 2) = 2, and is therefore the canonic form. Both realize the same $H (z)$ .

Same as 2082 Bhadra Q52079 Baishakh · Q64 marksObtain the Direct Form I and Direct Form II realization of the following system:

y [n]

- 0.

75 y [n - 1]

- 0.

25 y [n - 2] = x [n]

+ 0.

5 x [n - 1]

What is different in this one

A feed forward part of order 1 and a feedback part of order 2, so direct form I needs 3 delays and direct form II needs 2.
The two halves are of different order, 1 and 2, so direct form I needs 3 delays and direct form II needs only 2. Saying both numbers is a mark.

Rearrange so that $y [n]$ stands alone. The feedback coefficients change sign as they cross the equals sign, and that is where the marks are lost.

H (z) = \frac{1 + 0.5 z^{- 1}}{1 - 0.75 z^{- 1} - 0.25 z^{- 2}}

\begin{aligned} y [n] & = x [n] + 0.5 x [n - 1] + 0.75 y [n - 1] + 0.25 y [n - 2] \\ feed  forward b_{0} & = 1, b_{1} = 0.5 \\ feedback a_{1} & = - 0.75, so  the  multiplier  is 0.75 \\ a_{2} & = - 0.25, so  the  multiplier  is 0.25 \end{aligned}

Direct form I: the all zero half first, then the all pole half, 1 + 2 = 3 delays.

Direct form II: one delay chain shared by both halves, 2 delays, which is the fewest possible.

Direct form I needs 1 + 2 = 3 delays; direct form II needs max(1, 2) = 2, and is therefore the canonic form. Both realize the same $H (z)$ .

Practise this +2 like it2078 Bhadra · Q63+7 marksThe system function of a filter is

H (z) = 1 + (13 / 24) z^{- 1} + (5 / 8) z^{- 2} + (1 / 3) z^{- 3}

. Draw the Direct Form and Lattice Structure implementation of the above filter.

The direct form

What is different in this one

A feed forward part of order 3 and a feedback part of order 0, so direct form I needs 3 delays and direct form II needs 3.
There is no feedback: this is an FIR filter, so direct form I and direct form II are the same drawing and there is no all pole section.

Rearrange so that $y [n]$ stands alone. The feedback coefficients change sign as they cross the equals sign, and that is where the marks are lost.

H (z) = \frac{1 + 0.5417 z^{- 1} + 0.625 z^{- 2} + 0.3333 z^{- 3}}{1}

\begin{aligned} y [n] & = x [n] + 0.5417 x [n - 1] + 0.625 x [n - 2] + 0.3333 x [n - 3] \\ feed  forward b_{0} & = 1, b_{1} = 0.5417, b_{2} = 0.625, b_{3} = 0.3333 \\ feedback \end{aligned}

Direct form II: one delay chain shared by both halves, 3 delays, which is the fewest possible.

Direct form I needs 3 + 0 = 3 delays; direct form II needs max(3, 0) = 3, and is therefore the canonic form. Both realize the same $H (z)$ .

The rest of this questionThis topic works the part that belongs to it. The other part is worked under The FIR lattice (The lattice structure), where it belongs.

Same as 2082 Bhadra Q52074 Chaitra · Q55 marksObtain the Direct Form I and Direct Form II realization of the following system.

3 y [n] + y [n - 1] + 2 y [n - 4] = 2 x [n] + x [n - 3]

What is different in this one

A feed forward part of order 3 and a feedback part of order 4, so direct form I needs 7 delays and direct form II needs 4.
The leading coefficient is 3, not 1, so the whole equation is divided through by it before anything is drawn. Every structure in this chapter assumes $a_{0} = 1$ .
The two halves are of different order, 3 and 4, so direct form I needs 7 delays and direct form II needs only 4. Saying both numbers is a mark.
One of the coefficients is zero. That delay still exists in the chain, it simply has no multiplier hanging off it: leaving it out shortens the structure and changes the filter.

The leading coefficient is 3, so divide the whole equation through by it first: $a_{0}$ must be 1 before any structure is drawn.

Rearrange so that $y [n]$ stands alone. The feedback coefficients change sign as they cross the equals sign, and that is where the marks are lost.

H (z) = \frac{0.6667 + 0.3333 z^{- 3}}{1 + 0.3333 z^{- 1} + 0.6667 z^{- 4}}

\begin{aligned} y [n] & = 0.6667 x [n] + 0.3333 x [n - 3] - 0.3333 y [n - 1] - 0.6667 y [n - 4] \\ feed  forward b_{0} & = 0.6667, b_{1} = 0, b_{2} = 0, b_{3} = 0.3333 \\ feedback a_{1} & = 0.3333, so  the  multiplier  is - 0.3333 \\ a_{2} & = 0, so  the  multiplier  is 0 \\ a_{3} & = 0, so  the  multiplier  is 0 \\ a_{4} & = 0.6667, so  the  multiplier  is - 0.6667 \end{aligned}

Direct form I: the all zero half first, then the all pole half, 3 + 4 = 7 delays.

Direct form II: one delay chain shared by both halves, 4 delays, which is the fewest possible.

Direct form I needs 3 + 4 = 7 delays; direct form II needs max(3, 4) = 4, and is therefore the canonic form. Both realize the same $H (z)$ .

Same as 2078 Bhadra Q62074 Ashwin · Q53+7 marksDraw the direct form and Lattice structure of a filter with system function

H (z) = 1 + 0.7 z^{- 1} + 1.2 z^{- 2} - z^{- 3}

The direct form

What is different in this one

A feed forward part of order 3 and a feedback part of order 0, so direct form I needs 3 delays and direct form II needs 3.
There is no feedback: this is an FIR filter, so direct form I and direct form II are the same drawing and there is no all pole section.

Rearrange so that $y [n]$ stands alone. The feedback coefficients change sign as they cross the equals sign, and that is where the marks are lost.

H (z) = \frac{1 + 0.7 z^{- 1} + 1.2 z^{- 2} - z^{- 3}}{1}

\begin{aligned} y [n] & = x [n] + 0.7 x [n - 1] + 1.2 x [n - 2] - x [n - 3] \\ feed  forward b_{0} & = 1, b_{1} = 0.7, b_{2} = 1.2, b_{3} = - 1 \\ feedback \end{aligned}

Direct form II: one delay chain shared by both halves, 3 delays, which is the fewest possible.

Direct form I needs 3 + 0 = 3 delays; direct form II needs max(3, 0) = 3, and is therefore the canonic form. Both realize the same $H (z)$ .

The rest of this questionThis topic works the part that belongs to it. The other part is worked under The FIR lattice (The lattice structure), where it belongs.

Same as 2078 Bhadra Q62073 Shrawan · Q53+7 marksThe system function of a filter is

H (z) = 2 + 1.8 z^{- 1} - 1.6 z^{- 2} + z^{- 3}

. Draw the Direct Form and Lattice Structure implementation of the above filter.

The direct form

What is different in this one

A feed forward part of order 3 and a feedback part of order 0, so direct form I needs 3 delays and direct form II needs 3.
There is no feedback: this is an FIR filter, so direct form I and direct form II are the same drawing and there is no all pole section.

Rearrange so that $y [n]$ stands alone. The feedback coefficients change sign as they cross the equals sign, and that is where the marks are lost.

H (z) = \frac{2 + 1.8 z^{- 1} - 1.6 z^{- 2} + z^{- 3}}{1}

\begin{aligned} y [n] & = 2 x [n] + 1.8 x [n - 1] - 1.6 x [n - 2] + x [n - 3] \\ feed  forward b_{0} & = 2, b_{1} = 1.8, b_{2} = - 1.6, b_{3} = 1 \\ feedback \end{aligned}

Direct form II: one delay chain shared by both halves, 3 delays, which is the fewest possible.

Direct form I needs 3 + 0 = 3 delays; direct form II needs max(3, 0) = 3, and is therefore the canonic form. Both realize the same $H (z)$ .

The rest of this questionThis topic works the part that belongs to it. The other part is worked under The FIR lattice (The lattice structure), where it belongs.

Same as 2082 Bhadra Q52072 Chaitra · Q65 marksDetermine the Direct Form I and Direct Form II realization of the following system.

y (n) = - 0.1 y (n - 1) + 0.2 y (n - 2) + 3 x (n) + 3.6 x (n - 2) + 0.6 x (n - 2)

How this is readThe paper prints $x (n - 2)$ twice. The second is $x (n - 1)$ , giving $y (n) = - 0.1 y (n - 1) + 0.2 y (n - 2) + 3 x (n) + 3.6 x (n - 1) + 0.6 x (n - 2)$ .

What is different in this one

A feed forward part of order 2 and a feedback part of order 2, so direct form I needs 4 delays and direct form II needs 2.

Rearrange so that $y [n]$ stands alone. The feedback coefficients change sign as they cross the equals sign, and that is where the marks are lost.

H (z) = \frac{3 + 3.6 z^{- 1} + 0.6 z^{- 2}}{1 + 0.1 z^{- 1} - 0.2 z^{- 2}}

\begin{aligned} y [n] & = 3 x [n] + 3.6 x [n - 1] + 0.6 x [n - 2] - 0.1 y [n - 1] + 0.2 y [n - 2] \\ feed  forward b_{0} & = 3, b_{1} = 3.6, b_{2} = 0.6 \\ feedback a_{1} & = 0.1, so  the  multiplier  is - 0.1 \\ a_{2} & = - 0.2, so  the  multiplier  is 0.2 \end{aligned}

Direct form I: the all zero half first, then the all pole half, 2 + 2 = 4 delays.

Direct form II: one delay chain shared by both halves, 2 delays, which is the fewest possible.

Direct form I needs 2 + 2 = 4 delays; direct form II needs max(2, 2) = 2, and is therefore the canonic form. Both realize the same $H (z)$ .

Same as 2082 Bhadra Q52072 Kartik · Q64 marksDetermine the Direct Form II realization of the following system

y (n) = - 0.1 y (n - 1) + 0.72 y (n - 2) + 0.7 x (n) - 0.252 x (n - 2)

What is different in this one

A feed forward part of order 2 and a feedback part of order 2, so direct form I needs 4 delays and direct form II needs 2.
One of the coefficients is zero. That delay still exists in the chain, it simply has no multiplier hanging off it: leaving it out shortens the structure and changes the filter.

Rearrange so that $y [n]$ stands alone. The feedback coefficients change sign as they cross the equals sign, and that is where the marks are lost.

H (z) = \frac{0.7 - 0.252 z^{- 2}}{1 + 0.1 z^{- 1} - 0.72 z^{- 2}}

\begin{aligned} y [n] & = 0.7 x [n] - 0.252 x [n - 2] - 0.1 y [n - 1] + 0.72 y [n - 2] \\ feed  forward b_{0} & = 0.7, b_{1} = 0, b_{2} = - 0.252 \\ feedback a_{1} & = 0.1, so  the  multiplier  is - 0.1 \\ a_{2} & = - 0.72, so  the  multiplier  is 0.72 \end{aligned}

Direct form I: the all zero half first, then the all pole half, 2 + 2 = 4 delays.

Direct form II: one delay chain shared by both halves, 2 delays, which is the fewest possible.

Direct form I needs 2 + 2 = 4 delays; direct form II needs max(2, 2) = 2, and is therefore the canonic form. Both realize the same $H (z)$ .

FIR design by windowing TOP 8/19

Ch 5 · FIR design8 questions, from 8 of the 19 sittings5 to practise

The procedure, the same for every question in this topic

The ideal response, which is what the window is going to truncate.
$h_{d} [n] = \frac{\sin (ω_{c} (n - τ))}{π (n - τ)}, h_{d} [τ] = \frac{ω_{c}}{π}, τ = \frac{N - 1}{2}$
Read the specification. The transition width $Δ ω = ω_{s} - ω_{p}$ , and the attenuation $A_{s}$ in decibels.
Choose the window from the ATTENUATION, never from the width. The width is then fixed for you, and it is N that absorbs it.
Find N from that window's transition rule, round up, and take the next odd value for a Type I linear phase filter.
The cut off sits in the middle of the transition.
$ω_{c} = \frac{ω_{p} + ω_{s}}{2}$
Multiply, sample by sample.
$h [n] = h_{d} [n] w [n], n = 0 \dots N - 1$

Window	Stopband attenuation	Transition width
Rectangular	21 dB	$1.8 π / N$
Bartlett	25 dB	$6.1 π / N$
Hanning	44 dB	$6.2 π / N$
Hamming	53 dB	$6.6 π / N$
Blackman	74 dB	$11 π / N$

Ripple into decibels first: $| H | \leq 0.01$ is As = 40 dB, so Hanning at 44 dB is the smallest window that does it. Saying why that window is worth a mark on its own.
Take the smallest window that reaches the attenuation. Choosing Blackman for a 40 dB filter is not wrong, it is wasteful, and the order roughly doubles.

The checks that cost nothing

$h [n]$ must be symmetric about $τ$ . It is automatic, and checking it catches an arithmetic slip immediately.
h[ $τ$ ] = $ω_{c} / π$ , the largest coefficient, and it is the 0/0 limit of the formula.
The window is zero at both ends for Hanning and Blackman, so $h [0] = h [N - 1] = 0$ . That is expected, not an error.

How this is read $ω = 2 π$ f / fs turns the edges into $2 π$ (2000)/20000 = $0.2 π$ and $2 π$ (5000)/20000 = $0.5 π$ radians per sample.

What is different in this one

The stopband asks for 42 dB, and Hanning is the smallest window that reaches it, at 44 dB. Choosing a bigger one is not wrong, it is wasteful: the order roughly doubles.

Convert the edges to radians per sample with $ω = 2 π$ f / fs, with fs = 20000 Hz.

The window is chosen by the stopband attenuation, never by the transition width: take the smallest window that reaches 42 dB.

Window	Stopband attenuation	Transition width
Rectangular	21 dB	$1.8 π$ / N	too weak
Bartlett	25 dB	$6.1 π$ / N	too weak
Hanning	44 dB	$6.2 π$ / N	chosen
Hamming	53 dB	$6.6 π$ / N
Blackman	74 dB	$11 π$ / N

Hanning gives 44 dB, which meets the 42 dB asked for, and is the smallest that does.

\begin{aligned} transition  width  dw & = ω_{s} - ω_{p} = 1.5708 - 0.6283 = 0.9425 rad \\ N \geq 6.2 π / dw & = \frac{6.2 π}{0.9425} = 20.67 \\ round  up, and  take  the  next  ODD  value : N & = 21 \end{aligned}

\begin{aligned} cut  off ω_{c} & = (ω_{p} + ω_{s}) / 2 = (0.6283 + 1.5708) / 2 = 1.0996 rad \\ delay τ & = (N - 1) / 2 = 10 samples \\ ideal  response h_{d} [n] & = \sin (ω_{c} (n - t a u)) / (π (n - τ)) \\ h_{d} [τ] & = ω_{c} / π = 0.35 \\ window ω [n] & = 0.5 - 0.5 \cos (2 π n / (N - 1)) \end{aligned}

n	$h_{d} [n]$	$w [n]$	$h [n] = h_{d} [n] w [n]$
0	-0.03183	0	0
1	-0.01606	0.02447	-0.00039
2	0.02339	0.09549	0.00223
3	0.04491	0.20611	0.00926
4	0.01639	0.34549	0.00566
5	-0.04502	0.5	-0.02251
6	-0.07568	0.65451	-0.04953
7	-0.0166	0.79389	-0.01318
8	0.12876	0.90451	0.11646
9	0.28362	0.97553	0.27668
10	0.35	1	0.35
11	0.28362	0.97553	0.27668
12	0.12876	0.90451	0.11646

The remaining coefficients follow from the symmetry $h [n] = h [20 - n]$ , which every linear phase filter has.

Practise this +1 like it2082 Baishakh · Q62+6 marksDifferentiate between analog and digital filter. Design an FIR lowpass digital filter that will have a -3dB cut-off at

30 π

rad/s and an attenuation of 50 dB at

45 π

rad/s. The filter is required to have a linear phase and the system uses a sampling rate of 100 Hz.

How this is readA -3 dB cut off at $30 π$ rad/s with a 100 Hz sampling rate is $ω = \frac{30 π}{100} = 0.3 π$ , and $45 π$ rad/s is $0.45 π$ .

What is different in this one

The stopband asks for 50 dB, and Hamming is the smallest window that reaches it, at 53 dB. Choosing a bigger one is not wrong, it is wasteful: the order roughly doubles.

The window is chosen by the stopband attenuation, never by the transition width: take the smallest window that reaches 50 dB.

Window	Stopband attenuation	Transition width
Rectangular	21 dB	$1.8 π$ / N	too weak
Bartlett	25 dB	$6.1 π$ / N	too weak
Hanning	44 dB	$6.2 π$ / N	too weak
Hamming	53 dB	$6.6 π$ / N	chosen
Blackman	74 dB	$11 π$ / N

Hamming gives 53 dB, which meets the 50 dB asked for, and is the smallest that does.

\begin{aligned} transition  width  dw & = ω_{s} - ω_{p} = 1.4137 - 0.9425 = 0.4712 rad \\ N \geq 6.6 π / dw & = \frac{6.6 π}{0.4712} = 44 \\ round  up, and  take  the  next  ODD  value : N & = 45 \end{aligned}

\begin{aligned} cut  off ω_{c} & = (ω_{p} + ω_{s}) / 2 = (0.9425 + 1.4137) / 2 = 1.1781 rad \\ delay τ & = (N - 1) / 2 = 22 samples \\ ideal  response h_{d} [n] & = \sin (ω_{c} (n - t a u)) / (π (n - τ)) \\ h_{d} [τ] & = ω_{c} / π = 0.375 \\ window ω [n] & = 0.54 - 0.46 \cos (2 π n / (N - 1)) \end{aligned}

n	$h_{d} [n]$	$w [n]$	$h [n] = h_{d} [n] w [n]$
0	0.01023	0.08	0.00082
1	-0.0058	0.08468	-0.00049
2	-0.01592	0.09863	-0.00157
3	-0.00641	0.12157	-0.00078
4	0.0125	0.15302	0.00191
5	0.0173	0.19236	0.00333
6	0	0.23876	0
7	-0.01961	0.29131	-0.00571
8	-0.01608	0.34891	-0.00561
9	0.00937	0.4104	0.00385
10	0.02653	0.47454	0.01259
11	0.01107	0.54	0.00598
12	-0.02251	0.60546	-0.01363

The remaining coefficients follow from the symmetry $h [n] = h [44 - n]$ , which every linear phase filter has.

Practise this +2 like it2081 Bhadra · Q68 marksDesign a low pass FIR filter using suitable window to meet following specifications:

0.99 \leq | H (e^{j ω}) | \leq 1.01

for

0 \leq | w | \leq 0.3 π

;

| H (e^{j ω}) | \leq 0.01

for

0.35 π

≤ |

ω

| ≤

π

What is different herethe ripple is given as a number rather than in dB, so it is converted first

What is different in this one

The stopband asks for 40 dB, and Hanning is the smallest window that reaches it, at 44 dB. Choosing a bigger one is not wrong, it is wasteful: the order roughly doubles.

Turn the ripple into decibels first: passband ripple 0.01 means Ap = -20 $\log_{10} (1 - 0.01) = 0.09$ dB; stopband ripple 0.01 means As = -20 $\log_{10} (0.01)$ = 40 dB.

The window is chosen by the stopband attenuation, never by the transition width: take the smallest window that reaches 40 dB.

Window	Stopband attenuation	Transition width
Rectangular	21 dB	$1.8 π$ / N	too weak
Bartlett	25 dB	$6.1 π$ / N	too weak
Hanning	44 dB	$6.2 π$ / N	chosen
Hamming	53 dB	$6.6 π$ / N
Blackman	74 dB	$11 π$ / N

Hanning gives 44 dB, which meets the 40 dB asked for, and is the smallest that does.

\begin{aligned} transition  width  dw & = ω_{s} - ω_{p} = 1.0996 - 0.9425 = 0.1571 rad \\ N \geq 6.2 π / dw & = \frac{6.2 π}{0.1571} = 124 \\ round  up, and  take  the  next  ODD  value : N & = 125 \end{aligned}

\begin{aligned} cut  off ω_{c} & = (ω_{p} + ω_{s}) / 2 = (0.9425 + 1.0996) / 2 = 1.021 rad \\ delay τ & = (N - 1) / 2 = 62 samples \\ ideal  response h_{d} [n] & = \sin (ω_{c} (n - t a u)) / (π (n - τ)) \\ h_{d} [τ] & = ω_{c} / π = 0.325 \\ window ω [n] & = 0.5 - 0.5 \cos (2 π n / (N - 1)) \end{aligned}

n	$h_{d} [n]$	$w [n]$	$h [n] = h_{d} [n] w [n]$
0	0.00233	0	0
1	-0.00273	0.00064	0
2	-0.00531	0.00257	-0.00001
3	-0.00282	0.00577	-0.00002
4	0.00249	0.01024	0.00003
5	0.00557	0.01596	0.00009
6	0.00334	0.02293	0.00008
7	-0.00221	0.03112	-0.00007
8	-0.00582	0.04052	-0.00024
9	-0.0039	0.0511	-0.0002
10	0.00189	0.06283	0.00012
11	0.00607	0.07568	0.00046
12	0.0045	0.08962	0.0004

The remaining coefficients follow from the symmetry $h [n] = h [124 - n]$ , which every linear phase filter has.

Same as 2081 Bhadra Q62081 Baishakh · Q610 marksDesign a linear phase FIR filter using suitable window to meet following specifications:

0.99 \leq | H (e^{j ω}) | \leq 1.01

for

0 \leq | w | \leq 0.3 π

;

| H (e^{j ω}) | \leq 0.01

for

0.35 π

≤ |

ω

| ≤

π

What is different in this one

The stopband asks for 40 dB, and Hanning is the smallest window that reaches it, at 44 dB. Choosing a bigger one is not wrong, it is wasteful: the order roughly doubles.

Turn the ripple into decibels first: passband ripple 0.01 means Ap = -20 $\log_{10} (1 - 0.01) = 0.09$ dB; stopband ripple 0.01 means As = -20 $\log_{10} (0.01)$ = 40 dB.

The window is chosen by the stopband attenuation, never by the transition width: take the smallest window that reaches 40 dB.

Window	Stopband attenuation	Transition width
Rectangular	21 dB	$1.8 π$ / N	too weak
Bartlett	25 dB	$6.1 π$ / N	too weak
Hanning	44 dB	$6.2 π$ / N	chosen
Hamming	53 dB	$6.6 π$ / N
Blackman	74 dB	$11 π$ / N

Hanning gives 44 dB, which meets the 40 dB asked for, and is the smallest that does.

\begin{aligned} transition  width  dw & = ω_{s} - ω_{p} = 1.0996 - 0.9425 = 0.1571 rad \\ N \geq 6.2 π / dw & = \frac{6.2 π}{0.1571} = 124 \\ round  up, and  take  the  next  ODD  value : N & = 125 \end{aligned}

\begin{aligned} cut  off ω_{c} & = (ω_{p} + ω_{s}) / 2 = (0.9425 + 1.0996) / 2 = 1.021 rad \\ delay τ & = (N - 1) / 2 = 62 samples \\ ideal  response h_{d} [n] & = \sin (ω_{c} (n - t a u)) / (π (n - τ)) \\ h_{d} [τ] & = ω_{c} / π = 0.325 \\ window ω [n] & = 0.5 - 0.5 \cos (2 π n / (N - 1)) \end{aligned}

n	$h_{d} [n]$	$w [n]$	$h [n] = h_{d} [n] w [n]$
0	0.00233	0	0
1	-0.00273	0.00064	0
2	-0.00531	0.00257	-0.00001
3	-0.00282	0.00577	-0.00002
4	0.00249	0.01024	0.00003
5	0.00557	0.01596	0.00009
6	0.00334	0.02293	0.00008
7	-0.00221	0.03112	-0.00007
8	-0.00582	0.04052	-0.00024
9	-0.0039	0.0511	-0.0002
10	0.00189	0.06283	0.00012
11	0.00607	0.07568	0.00046
12	0.0045	0.08962	0.0004

The remaining coefficients follow from the symmetry $h [n] = h [124 - n]$ , which every linear phase filter has.

Practise this2080 Bhadra · Q610 marksDesign the symmetric FIR Low Pass Filter (LPF) for which the desired frequency response is expressed as Hd(W) =

e^{- j W t a u}

for

| W | \leq W c

and 0 elsewhere. The length of the filter should be 7 and

Ω_{c} = 1

rad/sample. Make use of the Hanning window.What is different herethe window and the length are given, so the choosing step is skipped

What is different in this one

The stopband asks for 44 dB, and Hanning is the smallest window that reaches it, at 44 dB. Choosing a bigger one is not wrong, it is wasteful: the order roughly doubles.
The question fixes the length at $N = 7$ , so there is no order calculation: go straight to the cut off and the coefficients.
The question names the window, so the choosing step is skipped. Most of these do not, and picking the smallest window that reaches the attenuation is the first mark.
The cut off is given as 1 rad per sample rather than derived from the band edges.

The question names the Hanning window, so there is no choosing to do.

The question fixes the length at $N = 7$ .

\begin{aligned} cut  off ω_{c} & = 1 rad, given  by  the  question \\ delay τ & = (N - 1) / 2 = 3 samples \\ ideal  response h_{d} [n] & = \sin (ω_{c} (n - t a u)) / (π (n - τ)) \\ h_{d} [τ] & = ω_{c} / π = 0.3183 \\ window ω [n] & = 0.5 - 0.5 \cos (2 π n / (N - 1)) \end{aligned}

n	$h_{d} [n]$	$w [n]$	$h [n] = h_{d} [n] w [n]$
0	0.01497	0	0
1	0.14472	0.25	0.03618
2	0.26785	0.75	0.20089
3	0.31831	1	0.31831
4	0.26785	0.75	0.20089
5	0.14472	0.25	0.03618
6	0.01497	0	0

Same as 2082 Baishakh Q62079 Baishakh · Q75+3 marksDesign a low pass digital FIR filter having Pass band edge frequency w_p =

0.2 π

, Stop band edge frequency w_s =

0.45 π

and Stop band attenuation a_s = 51 dB using any appropriate window function.

What is different in this one

The stopband asks for 51 dB, and Hamming is the smallest window that reaches it, at 53 dB. Choosing a bigger one is not wrong, it is wasteful: the order roughly doubles.

The window is chosen by the stopband attenuation, never by the transition width: take the smallest window that reaches 51 dB.

Window	Stopband attenuation	Transition width
Rectangular	21 dB	$1.8 π$ / N	too weak
Bartlett	25 dB	$6.1 π$ / N	too weak
Hanning	44 dB	$6.2 π$ / N	too weak
Hamming	53 dB	$6.6 π$ / N	chosen
Blackman	74 dB	$11 π$ / N

Hamming gives 53 dB, which meets the 51 dB asked for, and is the smallest that does.

\begin{aligned} transition  width  dw & = ω_{s} - ω_{p} = 1.4137 - 0.6283 = 0.7854 rad \\ N \geq 6.6 π / dw & = \frac{6.6 π}{0.7854} = 26.4 \\ round  up, and  take  the  next  ODD  value : N & = 27 \end{aligned}

\begin{aligned} cut  off ω_{c} & = (ω_{p} + ω_{s}) / 2 = (0.6283 + 1.4137) / 2 = 1.021 rad \\ delay τ & = (N - 1) / 2 = 13 samples \\ ideal  response h_{d} [n] & = \sin (ω_{c} (n - t a u)) / (π (n - τ)) \\ h_{d} [τ] & = ω_{c} / π = 0.325 \\ window ω [n] & = 0.54 - 0.46 \cos (2 π n / (N - 1)) \end{aligned}

n	$h_{d} [n]$	$w [n]$	$h [n] = h_{d} [n] w [n]$
0	0.0159	0.08	0.00127
1	-0.0082	0.09337	-0.00077
2	-0.02814	0.13269	-0.00373
3	-0.02251	0.19569	-0.0044
4	0.00826	0.27869	0.0023
5	0.03784	0.37688	0.01426
6	0.03458	0.48455	0.01675
7	-0.0083	0.59545	-0.00494
8	-0.05882	0.70312	-0.04135
9	-0.06438	0.80131	-0.05159
10	0.00832	0.88431	0.00736
11	0.14181	0.94731	0.13434
12	0.2714	0.98663	0.26778

The remaining coefficients follow from the symmetry $h [n] = h [26 - n]$ , which every linear phase filter has.

Same as 2081 Bhadra Q62076 Chaitra · Q86 marksDesign the FIR filter using suitable window for the specifications:

0.899 \leq | H (e^{j ω}) | \leq 1

for

| w | \leq 0.2 π

;

| H (e^{j ω}) | \leq 0.01

for

0.4 π \leq ω \leq π

How this is read $0.899 \leq | H | \leq 1$ gives a passband ripple of 0.101 and $| H | \leq 0.01$ gives As = 40 dB, which is what chooses the window.

What is different in this one

The stopband asks for 40 dB, and Hanning is the smallest window that reaches it, at 44 dB. Choosing a bigger one is not wrong, it is wasteful: the order roughly doubles.
The two ripples differ, 0.101 in the passband and 0.01 in the stopband, so it is the stopband figure that chooses the window.

Turn the ripple into decibels first: passband ripple 0.101 means Ap = -20 $\log_{10} (1 - 0.101) = 0.92$ dB; stopband ripple 0.01 means As = -20 $\log_{10} (0.01)$ = 40 dB.

The window is chosen by the stopband attenuation, never by the transition width: take the smallest window that reaches 40 dB.

Window	Stopband attenuation	Transition width
Rectangular	21 dB	$1.8 π$ / N	too weak
Bartlett	25 dB	$6.1 π$ / N	too weak
Hanning	44 dB	$6.2 π$ / N	chosen
Hamming	53 dB	$6.6 π$ / N
Blackman	74 dB	$11 π$ / N

Hanning gives 44 dB, which meets the 40 dB asked for, and is the smallest that does.

\begin{aligned} transition  width  dw & = ω_{s} - ω_{p} = 1.2566 - 0.6283 = 0.6283 rad \\ N \geq 6.2 π / dw & = \frac{6.2 π}{0.6283} = 31 \\ round  up, and  take  the  next  ODD  value : N & = 31 \end{aligned}

\begin{aligned} cut  off ω_{c} & = (ω_{p} + ω_{s}) / 2 = (0.6283 + 1.2566) / 2 = 0.9425 rad \\ delay τ & = (N - 1) / 2 = 15 samples \\ ideal  response h_{d} [n] & = \sin (ω_{c} (n - t a u)) / (π (n - τ)) \\ h_{d} [τ] & = ω_{c} / π = 0.3 \\ window ω [n] & = 0.5 - 0.5 \cos (2 π n / (N - 1)) \end{aligned}

n	$h_{d} [n]$	$w [n]$	$h [n] = h_{d} [n] w [n]$
0	0.02122	0	0
1	0.01336	0.01093	0.00015
2	-0.00757	0.04323	-0.00033
3	-0.02523	0.09549	-0.00241
4	-0.02341	0.16543	-0.00387
5	0	0.25	0
6	0.02861	0.34549	0.00989
7	0.03784	0.44774	0.01694
8	0.01405	0.55226	0.00776
9	-0.03118	0.65451	-0.02041
10	-0.06366	0.75	-0.04775
11	-0.04677	0.83457	-0.03904
12	0.03279	0.90451	0.02966

The remaining coefficients follow from the symmetry $h [n] = h [30 - n]$ , which every linear phase filter has.

0.2 π

, Stop band edge frequency w_s =

0.5 π

What is different in this one

The stopband asks for 41 dB, and Hanning is the smallest window that reaches it, at 44 dB. Choosing a bigger one is not wrong, it is wasteful: the order roughly doubles.

The window is chosen by the stopband attenuation, never by the transition width: take the smallest window that reaches 41 dB.

Window	Stopband attenuation	Transition width
Rectangular	21 dB	$1.8 π$ / N	too weak
Bartlett	25 dB	$6.1 π$ / N	too weak
Hanning	44 dB	$6.2 π$ / N	chosen
Hamming	53 dB	$6.6 π$ / N
Blackman	74 dB	$11 π$ / N

Hanning gives 44 dB, which meets the 41 dB asked for, and is the smallest that does.

\begin{aligned} transition  width  dw & = ω_{s} - ω_{p} = 1.5708 - 0.6283 = 0.9425 rad \\ N \geq 6.2 π / dw & = \frac{6.2 π}{0.9425} = 20.67 \\ round  up, and  take  the  next  ODD  value : N & = 21 \end{aligned}

\begin{aligned} cut  off ω_{c} & = (ω_{p} + ω_{s}) / 2 = (0.6283 + 1.5708) / 2 = 1.0996 rad \\ delay τ & = (N - 1) / 2 = 10 samples \\ ideal  response h_{d} [n] & = \sin (ω_{c} (n - t a u)) / (π (n - τ)) \\ h_{d} [τ] & = ω_{c} / π = 0.35 \\ window ω [n] & = 0.5 - 0.5 \cos (2 π n / (N - 1)) \end{aligned}

n	$h_{d} [n]$	$w [n]$	$h [n] = h_{d} [n] w [n]$
0	-0.03183	0	0
1	-0.01606	0.02447	-0.00039
2	0.02339	0.09549	0.00223
3	0.04491	0.20611	0.00926
4	0.01639	0.34549	0.00566
5	-0.04502	0.5	-0.02251
6	-0.07568	0.65451	-0.04953
7	-0.0166	0.79389	-0.01318
8	0.12876	0.90451	0.11646
9	0.28362	0.97553	0.27668
10	0.35	1	0.35
11	0.28362	0.97553	0.27668
12	0.12876	0.90451	0.11646

The remaining coefficients follow from the symmetry $h [n] = h [20 - n]$ , which every linear phase filter has.

FIR design with the Kaiser window TOP 8/19

Ch 5 · FIR design8 questions, from 8 of the 19 sittings4 to practise

The procedure, the same for every question in this topic

\begin{aligned} δ & = min (δ_{p}, δ_{s}) & take the SMALLER ripple \\ A & = - 20 \log_{10} δ & in decibels \\ Δ ω & = ω_{s} - ω_{p} & in radians \end{aligned}

The shape parameter, in three cases by the attenuation:

\begin{aligned} A > 50 & β = 0.1102 (A - 8.7) \\ 21 \leq A \leq 50 & β = 0.5842 (A - 21)^{0.4} + 0.07886 (A - 21) \\ A < 21 & β = 0 \end{aligned}

N \geq \frac{A - 8}{2.285 Δ ω} + 1 round up, next ODD

w [n] = \frac{I_{0} (β \sqrt{1 - {[\frac{2 n - (N - 1)}{N - 1}]}^{2}})}{I_{0} (β)}

Why Kaiser rather than a fixed window: $β$ tunes the attenuation to exactly what is asked, so N comes out smaller than for any fixed window meeting the same specification.
A band pass specification has two transition bands, and the narrower one sets N, because one filter has to meet both.

The checks that cost nothing

$β$ lands between 0 and about 9 for every specification these papers set. A wild $β$ means A was computed from the wrong ripple.
The smaller ripple wins. Using $δ_{p}$ when $δ_{s}$ is smaller gives too short a filter, and the stopband then fails.
A very narrow transition gives a very long filter. $N = 225$ for a $0.02 π$ transition is correct, not a mistake: state it and move on.

Practise this +3 like it2080 Bhadra · Q72+2+2 marksKaiser window is to be used to design a linear phase FIR filter that meets following specification:

| H (e^{j ω}) | \leq 0.01

for

0.21 π

≤ |

ω

| ≤

π

;

0.95 \leq | H (e^{j ω}) | \leq 1.05

for

0 \leq | w | \leq 0.19 π

. Calculate the optimum value of ripple, attenuation and window length.

What is different in this one

The ripples are different, 0.05 and 0.01, and the design takes the smaller one. Taking the wrong one gives too short a filter and the stopband then fails.
The transition band is only $0.02 π$ wide, which is very narrow. A filter length in the hundreds is the right answer here, not a mistake.

Take the smaller of the two ripples, delta = 0.01, and turn it into decibels: $A = - 20 \log_{10} (0.01)$ = 40 dB.

\begin{aligned} 21 \leq A \leq 50, so β & = 0.5842 {(A - 21)}^{0.4} + 0.07886 (A - 21) \\ = 0.5842 {(19)}^{0.4} + 0.07886 (19) = 3.3953 \\ transition  dw & = ω_{s} - ω_{p} = 0.6597 - 0.5969 = 0.0628 rad \\ order M \geq (A - 8) / (2.285 dw) \\ = (40 - 8) / (2.285 x 0.0628) = 222.89 \\ round  up : M & = 223, length N = M + 1 = 224 \\ α & = τ = M / 2 = 111.5, ω_{c} = (ω_{p} + ω_{s}) / 2 = 0.6283 rad \end{aligned}

w [n] = \frac{I_{0} (β \sqrt{1 - {[\frac{2 n - (N - 1)}{N - 1}]}^{2}})}{I_{0} (β)}

n	$h_{d} [n]$	$w [n]$	$h [n]$
0	0.00231	0.14797	0.00034
1	0.00089	0.15568	0.00014
2	-0.0009	0.16352	-0.00015
3	-0.00237	0.17149	-0.00041
4	-0.00296	0.17957	-0.00053
5	-0.00242	0.18778	-0.00045
6	-0.00093	0.19611	-0.00018
7	0.00094	0.20455	0.00019
8	0.00249	0.21311	0.00053
9	0.00311	0.22178	0.00069
10	0.00254	0.23056	0.00058
11	0.00098	0.23944	0.00023
12	-0.00099	0.24842	-0.00025

The rest follow from $h [n] = h [223 - n]$ .

Same as 2080 Bhadra Q72080 Baishakh · Q72+8 marksIn which case do we choose FIR filter and IIR filter? Design a linear phase FIR filter using Kaiser Window to meet the following specifications:

0.99 \leq | H (e^{j ω}) | \leq 1.01

for

0 \leq w \geq 0.016 π

;

| H (e^{j ω}) | \leq 0.01

for

0.08 π \leq ω \leq 2 π

What is different in this one

Both ripples are 0.01, so there is no choosing to do: delta is that number.

Take the smaller of the two ripples, delta = 0.01, and turn it into decibels: $A = - 20 \log_{10} (0.01)$ = 40 dB.

\begin{aligned} 21 \leq A \leq 50, so β & = 0.5842 {(A - 21)}^{0.4} + 0.07886 (A - 21) \\ = 0.5842 {(19)}^{0.4} + 0.07886 (19) = 3.3953 \\ transition  dw & = ω_{s} - ω_{p} = 0.2513 - 0.0503 = 0.2011 rad \\ order M \geq (A - 8) / (2.285 dw) \\ = (40 - 8) / (2.285 x 0.2011) = 69.65 \\ round  up : M & = 70, length N = M + 1 = 71 \\ α & = τ = M / 2 = 35, ω_{c} = (ω_{p} + ω_{s}) / 2 = 0.1508 rad \end{aligned}

w [n] = \frac{I_{0} (β \sqrt{1 - {[\frac{2 n - (N - 1)}{N - 1}]}^{2}})}{I_{0} (β)}

n	$h_{d} [n]$	$w [n]$	$h [n]$
0	-0.00768	0.14797	-0.00114
1	-0.00857	0.17298	-0.00148
2	-0.00931	0.19923	-0.00186
3	-0.00988	0.22666	-0.00224
4	-0.01026	0.25516	-0.00262
5	-0.01042	0.28466	-0.00297
6	-0.01035	0.31504	-0.00326
7	-0.01003	0.34621	-0.00347
8	-0.00945	0.37803	-0.00357
9	-0.0086	0.4104	-0.00353
10	-0.00748	0.44319	-0.00332
11	-0.0061	0.47625	-0.0029
12	-0.00444	0.50946	-0.00226

The rest follow from $h [n] = h [70 - n]$ .

Practise this2079 Bhadra · Q72+8 marksDefine Gibb's phenomenon. Design the FIR filter using Kaiser window technique for the specifications:

0.899 \leq | H (e^{j ω}) | \leq 1

for

| w | \leq 0.2 π

;

| H (e^{j ω}) | \leq 0.01

for

0.4 π \leq ω \leq π

What is different in this one

The ripples are different, 0.101 and 0.01, and the design takes the smaller one. Taking the wrong one gives too short a filter and the stopband then fails.

Take the smaller of the two ripples, delta = 0.01, and turn it into decibels: $A = - 20 \log_{10} (0.01)$ = 40 dB.

\begin{aligned} 21 \leq A \leq 50, so β & = 0.5842 {(A - 21)}^{0.4} + 0.07886 (A - 21) \\ = 0.5842 {(19)}^{0.4} + 0.07886 (19) = 3.3953 \\ transition  dw & = ω_{s} - ω_{p} = 1.2566 - 0.6283 = 0.6283 rad \\ order M \geq (A - 8) / (2.285 dw) \\ = (40 - 8) / (2.285 x 0.6283) = 22.29 \\ round  up : M & = 23, length N = M + 1 = 24 \\ α & = τ = M / 2 = 11.5, ω_{c} = (ω_{p} + ω_{s}) / 2 = 0.9425 rad \end{aligned}

w [n] = \frac{I_{0} (β \sqrt{1 - {[\frac{2 n - (N - 1)}{N - 1}]}^{2}})}{I_{0} (β)}

n	$h_{d} [n]$	$w [n]$	$h [n]$
0	-0.02734	0.14797	-0.00405
1	-0.01376	0.22787	-0.00314
2	0.01521	0.31772	0.00483
3	0.03699	0.41466	0.01534
4	0.03001	0.51524	0.01546
5	-0.00766	0.6156	-0.00472
6	-0.05157	0.71165	-0.0367
7	-0.06303	0.79931	-0.05038
8	-0.01423	0.87475	-0.01245
9	0.09003	0.93459	0.08414
10	0.20959	0.97608	0.20458
11	0.28902	0.99732	0.28825
12	0.28902	0.99732	0.28825

The rest follow from $h [n] = h [23 - n]$ .

Practise this +1 like it2078 Bhadra · Q78 marksDesign a linear phase FIR filter using KAISER window to meet the following specifications:

| H (e^{j ω}) | \leq 0.01

for

0 \leq | w | \leq 0.25 π

;

0.95 \leq | H (e^{j ω}) | \leq 1.05

for

0.35 π \leq | w | \leq 0.6 π

;

| H (e^{j ω}) | \leq 0.01

for

0.65 π

≤ |

ω

| ≤

π

What is different in this one

This is a band pass specification, not low pass: a stopband, a passband, then a second stopband. The ideal response is therefore the difference of two low pass responses.
There are two transition bands, $0.1 π$ and $0.05 π$ wide, and the narrower one sets the length, because one filter has to meet both.

The narrower transition is what the order formula is fed.

\begin{aligned} lower  transition 1.0996 - 0.7854 & = 0.3142 rad \\ upper  transition 2.042 - 1.885 & = 0.1571 rad \\ take  the  narrower  dw & = 0.1571 rad \\ delta & = min (0.05, 0.01) = 0.01 \\ A & = - 20 \log_{10} (0.01) = 40 dB \\ β & = 3.3953 \\ N \geq (A - 8) / (2.285 dw) + 1 & = 90.15, so N = 91, τ = 45 \\ cut  offsω_{c} 1 & = (0.7854 + 1.0996) / 2 = 0.9425 rad \\ ω_{c} 2 & = (1.885 + 2.042) / 2 = 1.9635 rad \end{aligned}

\begin{aligned} ideal  band  pass h_{d} [n] & = [\sin (ω c 2 (n - t a u)) - \sin (ω c 1 (n - t a u))] / (π (n - τ)) \\ h_{d} [τ] & = (ω_{c} 2 - ω_{c} 1) / π = 0.325 \end{aligned}

n	$h_{d} [n]$	$w [n]$	$h [n]$
0	0.00978	0.14797	0.00145
1	-0.00298	0.16731	-0.0005
2	0.00055	0.18742	0.0001
3	-0.00185	0.20825	-0.00039
4	-0.01345	0.22977	-0.00309
5	0	0.25194	0
6	0.01414	0.27472	0.00389
7	0.00204	0.29806	0.00061
8	-0.00063	0.3219	-0.0002
9	0.00364	0.34621	0.00126
10	-0.01257	0.37091	-0.00466
11	-0.01212	0.39596	-0.0048
12	0.01189	0.42129	0.00501

The rest follow from $h [n] = h [90 - n]$ .

delta = 0.01, $A = 40$ dB, $β = 3.3953$ , $N = 91$ , with cut offs wc1 = 0.9425 and wc2 = 1.9635 rad and a group delay of 45 samples.

Practise this2076 Ashwin · Q92+4+4 marksIn which case do we choose FIR filter and IIR filter? Design a Kaiser Window to meet the following specifications:

0.99 \leq | H (e^{j ω}) | \leq 1.01

for

0 \leq w \leq 0.16 π

;

| H (e^{j ω}) | \leq 0.01

for

0.18 π \leq ω \leq 2 π

. Draw the flow chart for Remez-Exchange algorithmWhat is different hereThe transition band is only

0.02 π

wide, which is what makes the filter long. A number in the hundreds is the right answer here, not a mistake.

How this is readThe stopband is printed as $0.18 π$ to $2 π$ , which runs past $π$ . A digital spectrum stops at $π$ , so the stopband is $0.18 π$ to $π$ and it is worked that way.

What is different in this one

Both ripples are 0.01, so there is no choosing to do: delta is that number.
The transition band is only $0.02 π$ wide, which is very narrow. A filter length in the hundreds is the right answer here, not a mistake.

Take the smaller of the two ripples, delta = 0.01, and turn it into decibels: $A = - 20 \log_{10} (0.01)$ = 40 dB.

\begin{aligned} 21 \leq A \leq 50, so β & = 0.5842 {(A - 21)}^{0.4} + 0.07886 (A - 21) \\ = 0.5842 {(19)}^{0.4} + 0.07886 (19) = 3.3953 \\ transition  dw & = ω_{s} - ω_{p} = 0.5655 - 0.5027 = 0.0628 rad \\ order M \geq (A - 8) / (2.285 dw) \\ = (40 - 8) / (2.285 x 0.0628) = 222.89 \\ round  up : M & = 223, length N = M + 1 = 224 \\ α & = τ = M / 2 = 111.5, ω_{c} = (ω_{p} + ω_{s}) / 2 = 0.5341 rad \end{aligned}

w [n] = \frac{I_{0} (β \sqrt{1 - {[\frac{2 n - (N - 1)}{N - 1}]}^{2}})}{I_{0} (β)}

n	$h_{d} [n]$	$w [n]$	$h [n]$
0	0.0004	0.14797	0.00006
1	0.0018	0.15568	0.00028
2	0.00272	0.16352	0.00044
3	0.00289	0.17149	0.0005
4	0.00225	0.17957	0.0004
5	0.00097	0.18778	0.00018
6	-0.00061	0.19611	-0.00012
7	-0.00205	0.20455	-0.00042
8	-0.00294	0.21311	-0.00063
9	-0.00302	0.22178	-0.00067
10	-0.00225	0.23056	-0.00052
11	-0.00084	0.23944	-0.0002
12	0.00084	0.24842	0.00021

The rest follow from $h [n] = h [223 - n]$ .

$A = 40$ dB, $β = 3.3953$ , $N = 224$ , cut off $ω_{c} = 0.5341$ rad, group delay 111.5 samples. The first coefficients are ${0.00006, 0.00028, 0.00044, 0.0005, 0.0004, 0.00018}$ .

Same as 2080 Bhadra Q72075 Ashwin · Q78+4 marksDesign a linear phase FIR filter using Kaiser Window to meet the following specifications:

0.99 \leq | H (e^{j ω}) | \leq 1.01

for

0 \leq w \leq 0.19 π

;

| H (e^{j ω}) | \leq 0.01

for

0.21 π \leq ω \leq π

. Draw the flow chart for Optimum filter design.

What is different in this one

Both ripples are 0.01, so there is no choosing to do: delta is that number.
The transition band is only $0.02 π$ wide, which is very narrow. A filter length in the hundreds is the right answer here, not a mistake.

Take the smaller of the two ripples, delta = 0.01, and turn it into decibels: $A = - 20 \log_{10} (0.01)$ = 40 dB.

\begin{aligned} 21 \leq A \leq 50, so β & = 0.5842 {(A - 21)}^{0.4} + 0.07886 (A - 21) \\ = 0.5842 {(19)}^{0.4} + 0.07886 (19) = 3.3953 \\ transition  dw & = ω_{s} - ω_{p} = 0.6597 - 0.5969 = 0.0628 rad \\ order M \geq (A - 8) / (2.285 dw) \\ = (40 - 8) / (2.285 x 0.0628) = 222.89 \\ round  up : M & = 223, length N = M + 1 = 224 \\ α & = τ = M / 2 = 111.5, ω_{c} = (ω_{p} + ω_{s}) / 2 = 0.6283 rad \end{aligned}

w [n] = \frac{I_{0} (β \sqrt{1 - {[\frac{2 n - (N - 1)}{N - 1}]}^{2}})}{I_{0} (β)}

n	$h_{d} [n]$	$w [n]$	$h [n]$
0	0.00231	0.14797	0.00034
1	0.00089	0.15568	0.00014
2	-0.0009	0.16352	-0.00015
3	-0.00237	0.17149	-0.00041
4	-0.00296	0.17957	-0.00053
5	-0.00242	0.18778	-0.00045
6	-0.00093	0.19611	-0.00018
7	0.00094	0.20455	0.00019
8	0.00249	0.21311	0.00053
9	0.00311	0.22178	0.00069
10	0.00254	0.23056	0.00058
11	0.00098	0.23944	0.00023
12	-0.00099	0.24842	-0.00025

The rest follow from $h [n] = h [223 - n]$ .

Same as 2078 Bhadra Q72074 Chaitra · Q89+3 marksDesign a FIR linear phase filter using Kaiser window that meets the following specifications:

| H (e^{j ω}) | \leq 0.01

for

0 \leq | w | \leq 0.25 π

;

0.95 \leq | H (e^{j ω}) | \leq 1.05

for

0.35 π \leq | w | \leq 0.6 π

;

| H (e^{j ω}) | \leq 0.01

for

0.65 π

≤ |

ω

| ≤

π

. Also determine the minimum length (M+1) of the impulse response and Kaiser window parameter

β

What is different in this one

This is a band pass specification, not low pass: a stopband, a passband, then a second stopband. The ideal response is therefore the difference of two low pass responses.
There are two transition bands, $0.1 π$ and $0.05 π$ wide, and the narrower one sets the length, because one filter has to meet both.

The narrower transition is what the order formula is fed.

\begin{aligned} lower  transition 1.0996 - 0.7854 & = 0.3142 rad \\ upper  transition 2.042 - 1.885 & = 0.1571 rad \\ take  the  narrower  dw & = 0.1571 rad \\ delta & = min (0.05, 0.01) = 0.01 \\ A & = - 20 \log_{10} (0.01) = 40 dB \\ β & = 3.3953 \\ N \geq (A - 8) / (2.285 dw) + 1 & = 90.15, so N = 91, τ = 45 \\ cut  offsω_{c} 1 & = (0.7854 + 1.0996) / 2 = 0.9425 rad \\ ω_{c} 2 & = (1.885 + 2.042) / 2 = 1.9635 rad \end{aligned}

\begin{aligned} ideal  band  pass h_{d} [n] & = [\sin (ω c 2 (n - t a u)) - \sin (ω c 1 (n - t a u))] / (π (n - τ)) \\ h_{d} [τ] & = (ω_{c} 2 - ω_{c} 1) / π = 0.325 \end{aligned}

n	$h_{d} [n]$	$w [n]$	$h [n]$
0	0.00978	0.14797	0.00145
1	-0.00298	0.16731	-0.0005
2	0.00055	0.18742	0.0001
3	-0.00185	0.20825	-0.00039
4	-0.01345	0.22977	-0.00309
5	0	0.25194	0
6	0.01414	0.27472	0.00389
7	0.00204	0.29806	0.00061
8	-0.00063	0.3219	-0.0002
9	0.00364	0.34621	0.00126
10	-0.01257	0.37091	-0.00466
11	-0.01212	0.39596	-0.0048
12	0.01189	0.42129	0.00501

The rest follow from $h [n] = h [90 - n]$ .

delta = 0.01, $A = 40$ dB, $β = 3.3953$ , $N = 91$ , with cut offs wc1 = 0.9425 and wc2 = 1.9635 rad and a group delay of 45 samples.

0.99 \leq | H (e^{j ω}) | \leq 1.01

for

0 \geq w \geq 0.19 π

;

| H (e^{j ω}) | \leq 0.01

for

0.21 π \leq ω \leq π

How this is readThe paper prints the passband as $0 \geq w \geq 0.19 π$ , its inequalities reversed by the scan. It means 0 to $0.19 π$ , and it is worked that way.

What is different in this one

Both ripples are 0.01, so there is no choosing to do: delta is that number.
The transition band is only $0.02 π$ wide, which is very narrow. A filter length in the hundreds is the right answer here, not a mistake.

Take the smaller of the two ripples, delta = 0.01, and turn it into decibels: $A = - 20 \log_{10} (0.01)$ = 40 dB.

\begin{aligned} 21 \leq A \leq 50, so β & = 0.5842 {(A - 21)}^{0.4} + 0.07886 (A - 21) \\ = 0.5842 {(19)}^{0.4} + 0.07886 (19) = 3.3953 \\ transition  dw & = ω_{s} - ω_{p} = 0.6597 - 0.5969 = 0.0628 rad \\ order M \geq (A - 8) / (2.285 dw) \\ = (40 - 8) / (2.285 x 0.0628) = 222.89 \\ round  up : M & = 223, length N = M + 1 = 224 \\ α & = τ = M / 2 = 111.5, ω_{c} = (ω_{p} + ω_{s}) / 2 = 0.6283 rad \end{aligned}

w [n] = \frac{I_{0} (β \sqrt{1 - {[\frac{2 n - (N - 1)}{N - 1}]}^{2}})}{I_{0} (β)}

n	$h_{d} [n]$	$w [n]$	$h [n]$
0	0.00231	0.14797	0.00034
1	0.00089	0.15568	0.00014
2	-0.0009	0.16352	-0.00015
3	-0.00237	0.17149	-0.00041
4	-0.00296	0.17957	-0.00053
5	-0.00242	0.18778	-0.00045
6	-0.00093	0.19611	-0.00018
7	0.00094	0.20455	0.00019
8	0.00249	0.21311	0.00053
9	0.00311	0.22178	0.00069
10	0.00254	0.23056	0.00058
11	0.00098	0.23944	0.00023
12	-0.00099	0.24842	-0.00025

The rest follow from $h [n] = h [223 - n]$ .

The FFT, decimation in frequency TOP 8/19

Ch 7 · DFT and FFT8 questions, from 8 of the 19 sittings3 to practise

The procedure, the same for every question in this topic

\begin{aligned} a [n] & = x [n] + x [n + \frac{N}{2}] & feeds the EVEN outputs \\ b [n] & = (x [n] - x [n + \frac{N}{2}]) W_{N}^{n} & feeds the ODD outputs \end{aligned}

BUTTERFLY a --------o------> a + b X the twiddle multiplies the b --------o--[W]-> (a - b) WN^r DIFFERENCE, AFTER the add and subtract

Order: the input is natural and the output comes out bit reversed, which is the mirror of decimation in time. For $N = 8$ the outputs land as $X [0], X [4], X [2], X [6], X [1], X [5], X [3], X [7]$ .

Stages: $\log_{2} N$ of them, each of $N / 2$ butterflies. Here it is the LAST stage whose twiddles are all 1.

DIF is DIT with the twiddle moved. Same cost, same butterflies, the multiply after the subtraction instead of before, and the bit reversal at the other end.
Label the outputs. Writing them down the page in natural order is the classic lost mark here.

The checks that cost nothing

$X [0]$ is the sum of the samples, and it comes out of the top wire.
Conjugate symmetry for a real input, as in DIT.
Compare against the DIT answer if the same sequence appears in another paper: the two must agree exactly, because they compute the same DFT.

Practise this +5 like it2082 Bhadra · Q118 marksFind 8-point DFT using DIF-FFT of the sequence

x [n]

= {1,

\frac{1}{2}

, -1,

- \frac{1}{2}

, 2, -3/2}.What is different heredecimation in frequency, where the input stays in order and the output comes out bit reversed

What is different in this one

Decimation in frequency: 8 points, 3 stages of 4 butterflies, with the output in bit reversed order.
The sequence has 6 samples, not a power of two, so it is zero padded to $N = 8$ before the algorithm starts. Say so in the answer: a radix 2 FFT cannot run on 6 points.

The sequence has 6 samples, and a radix 2 FFT needs a power of two, so zero pad to $N = 8$ and say so: $x [n] = {1, 0.5, - 1, - 0.5, 2, - 1.5, 0, 0}$ .

Decimation in frequency: the input stays in natural order and the output comes out bit reversed, so the last step is to relabel it.

Stage 1, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ applied to the differences:

\begin{aligned} 3 \\ - 1 \\ - 1 \\ - 0.5 \\ - 1 \\ 1.4142 - j 1.4142 \\ j1 \\ 0.3536 + j 0.3536 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ applied to the differences:

\begin{aligned} 2 \\ - 1.5 \\ 4 \\ j0.5 \\ - 1 + j 1 \\ 1.7678 - j 1.0607 \\ - 1 - j 1 \\ - 1.7678 - j 1.0607 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{2}^{0}$ applied to the differences:

\begin{aligned} 0.5 \\ 3.5 \\ 4 + j 0.5 \\ 4 - j 0.5 \\ 0.7678 - j 0.0607 \\ - 2.7678 + j 2.0607 \\ - 2.7678 - j 2.0607 \\ 0.7678 + j 0.0607 \end{aligned}

Read the output in bit reversed order: $X [0]$ , $X [4]$ , $X [2]$ , $X [6]$ , $X [1]$ , $X [5]$ , $X [3]$ , $X [7]$ .

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	0.5	0.5	0
$X [1]$	0.7678 - j0.0607	0.7702	-4.52
$X [2]$	4 + j0.5	4.0311	7.13
$X [3]$	-2.7678 - j2.0607	3.4506	-143.33
$X [4]$	3.5	3.5	0
$X [5]$	-2.7678 + j2.0607	3.4506	143.33
$X [6]$	4 - j0.5	4.0311	-7.13
$X [7]$	0.7678 + j0.0607	0.7702	4.52

x [n] = {1, 4, - 1}

W 4^{0} = 1

and

W 4^{1} = - j

. Most sittings set the 8 point version.

What is different in this one

Decimation in frequency: 4 points, 2 stages of 2 butterflies, with the output in bit reversed order.
The sequence has 3 samples, not a power of two, so it is zero padded to $N = 4$ before the algorithm starts. Say so in the answer: a radix 2 FFT cannot run on 3 points.
This is a 4 point transform, so there are only 2 stages of 2 butterflies and the only twiddles are $W 4^{0} = 1$ and $W 4^{1} = - j$ . Much shorter than the usual 8 point question.
Only 4 points, so there are two stages of two butterflies and the only twiddles are $W 4^{0} = 1$ and $W 4^{1} = - j$ . Most sittings set the 8 point version.

The sequence has 3 samples, and a radix 2 FFT needs a power of two, so zero pad to $N = 4$ and say so: $x [n] = {1, 4, - 1, 0}$ .

Decimation in frequency: the input stays in natural order and the output comes out bit reversed, so the last step is to relabel it.

Stage 1, 2 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ applied to the differences:

\begin{aligned} 0 \\ 4 \\ 2 \\ - j 4 \end{aligned}

Stage 2, 2 butterflies with twiddles $W_{2}^{0}$ applied to the differences:

\begin{aligned} 4 \\ - 4 \\ 2 - j 4 \\ 2 + j 4 \end{aligned}

Read the output in bit reversed order: $X [0]$ , $X [2]$ , $X [1]$ , $X [3]$ .

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	4	4	0
$X [1]$	2 - j4	4.4721	-63.43
$X [2]$	-4	4	180
$X [3]$	2 + j4	4.4721	63.43

$X [k]$ = {4, 2 - j4, -4, 2 + j4}. Two free checks: $X [0] = 4$ , which is the sum of the samples; $X [3] = X^{*} [1]$ and $X [2] = X^{*} [2]$ , the conjugate symmetry a real input forces.

Same as 2082 Bhadra Q112080 Bhadra · Q98 marksCompute 8-point DIF-FFT of sequence

x (n) = {2, 1, 2, 1, 1, 2, 1, 2}

What is different in this one

Decimation in frequency: 8 points, 3 stages of 4 butterflies, with the output in bit reversed order.

$N = 8$ , which is $2^{3}$ , so the transform takes 3 stages of 4 butterflies.

Decimation in frequency: the input stays in natural order and the output comes out bit reversed, so the last step is to relabel it.

Stage 1, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ applied to the differences:

\begin{aligned} 3 \\ 3 \\ 3 \\ 3 \\ 1 \\ - 0.7071 + j 0.7071 \\ - j 1 \\ 0.7071 + j 0.7071 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ applied to the differences:

\begin{aligned} 6 \\ 6 \\ 0 \\ 0 \\ 1 - j 1 \\ j1.4142 \\ 1 + j 1 \\ j1.4142 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{2}^{0}$ applied to the differences:

\begin{aligned} 12 \\ 0 \\ 0 \\ 0 \\ 1 + j 0.4142 \\ 1 - j 2.4142 \\ 1 + j 2.4142 \\ 1 - j 0.4142 \end{aligned}

Read the output in bit reversed order: $X [0]$ , $X [4]$ , $X [2]$ , $X [6]$ , $X [1]$ , $X [5]$ , $X [3]$ , $X [7]$ .

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	12	12	0
$X [1]$	1 + j0.4142	1.0824	22.5
$X [2]$	0	0	0
$X [3]$	1 + j2.4142	2.6131	67.5
$X [4]$	0	0	0
$X [5]$	1 - j2.4142	2.6131	-67.5
$X [6]$	0	0	0
$X [7]$	1 - j0.4142	1.0824	-22.5

Same as 2082 Bhadra Q112079 Baishakh · Q102+6 marksWhy we need DFT? Find 8-point DFT of sequence

x [n] = {1, 2, 4, 3, 5, - 1, 3}

using Decimation in Frequency Fast Fourier Transform (DIFFFT) algorithm.

What is different in this one

Decimation in frequency: 8 points, 3 stages of 4 butterflies, with the output in bit reversed order.
The sequence has 7 samples, not a power of two, so it is zero padded to $N = 8$ before the algorithm starts. Say so in the answer: a radix 2 FFT cannot run on 7 points.

The sequence has 7 samples, and a radix 2 FFT needs a power of two, so zero pad to $N = 8$ and say so: $x [n] = {1, 2, 4, 3, 5, - 1, 3, 0}$ .

Decimation in frequency: the input stays in natural order and the output comes out bit reversed, so the last step is to relabel it.

Stage 1, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ applied to the differences:

\begin{aligned} 6 \\ 1 \\ 7 \\ 3 \\ - 4 \\ 2.1213 - j 2.1213 \\ - j 1 \\ - 2.1213 - j 2.1213 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ applied to the differences:

\begin{aligned} 13 \\ 4 \\ - 1 \\ j2 \\ - 4 - j 1 \\ - j 4.2426 \\ - 4 + j 1 \\ - j 4.2426 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{2}^{0}$ applied to the differences:

\begin{aligned} 17 \\ 9 \\ - 1 + j 2 \\ - 1 - j 2 \\ - 4 - j 5.2426 \\ - 4 + j 3.2426 \\ - 4 - j 3.2426 \\ - 4 + j 5.2426 \end{aligned}

Read the output in bit reversed order: $X [0]$ , $X [4]$ , $X [2]$ , $X [6]$ , $X [1]$ , $X [5]$ , $X [3]$ , $X [7]$ .

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	17	17	0
$X [1]$	-4 - j5.2426	6.5943	-127.34
$X [2]$	-1 + j2	2.2361	116.57
$X [3]$	-4 - j3.2426	5.1492	-140.97
$X [4]$	9	9	0
$X [5]$	-4 + j3.2426	5.1492	140.97
$X [6]$	-1 - j2	2.2361	-116.57
$X [7]$	-4 + j5.2426	6.5943	127.34

Same as 2082 Bhadra Q112076 Chaitra · Q117 marksFind 8-point DFT of sequence

x [n] = {1, 2, 3, 3, 5, 0, 4, 6}

using Decimation in frequency Fast Fourier Transform (DIFFFT) algorithm.

What is different in this one

Decimation in frequency: 8 points, 3 stages of 4 butterflies, with the output in bit reversed order.

$N = 8$ , which is $2^{3}$ , so the transform takes 3 stages of 4 butterflies.

Decimation in frequency: the input stays in natural order and the output comes out bit reversed, so the last step is to relabel it.

Stage 1, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ applied to the differences:

\begin{aligned} 6 \\ 2 \\ 7 \\ 9 \\ - 4 \\ 1.4142 - j 1.4142 \\ j1 \\ 2.1213 + j 2.1213 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ applied to the differences:

\begin{aligned} 13 \\ 11 \\ - 1 \\ j7 \\ - 4 + j 1 \\ 3.5355 + j 0.7071 \\ - 4 - j 1 \\ - 3.5355 + j 0.7071 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{2}^{0}$ applied to the differences:

\begin{aligned} 24 \\ 2 \\ - 1 + j 7 \\ - 1 - j 7 \\ - 0.4645 + j 1.7071 \\ - 7.5355 + j 0.2929 \\ - 7.5355 - j 0.2929 \\ - 0.4645 - j 1.7071 \end{aligned}

Read the output in bit reversed order: $X [0]$ , $X [4]$ , $X [2]$ , $X [6]$ , $X [1]$ , $X [5]$ , $X [3]$ , $X [7]$ .

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	24	24	0
$X [1]$	-0.4645 + j1.7071	1.7692	105.22
$X [2]$	-1 + j7	7.0711	98.13
$X [3]$	-7.5355 - j0.2929	7.5412	-177.77
$X [4]$	2	2	0
$X [5]$	-7.5355 + j0.2929	7.5412	177.77
$X [6]$	-1 - j7	7.0711	-98.13
$X [7]$	-0.4645 - j1.7071	1.7692	-105.22

Same as 2082 Bhadra Q112076 Ashwin · Q82+8 marksWhy we need DFT? Find 8-point DFT of sequence

x [n] = {1, 2, 3, 3, 5, 1, 4, 2}

using Decimation in frequency Fast Fourier Transform (DIFFFT) algorithm.

What is different in this one

Decimation in frequency: 8 points, 3 stages of 4 butterflies, with the output in bit reversed order.

$N = 8$ , which is $2^{3}$ , so the transform takes 3 stages of 4 butterflies.

Decimation in frequency: the input stays in natural order and the output comes out bit reversed, so the last step is to relabel it.

Stage 1, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ applied to the differences:

\begin{aligned} 6 \\ 3 \\ 7 \\ 5 \\ - 4 \\ 0.7071 - j 0.7071 \\ j1 \\ - 0.7071 - j 0.7071 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ applied to the differences:

\begin{aligned} 13 \\ 8 \\ - 1 \\ j2 \\ - 4 + j 1 \\ - j 1.4142 \\ - 4 - j 1 \\ - j 1.4142 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{2}^{0}$ applied to the differences:

\begin{aligned} 21 \\ 5 \\ - 1 + j 2 \\ - 1 - j 2 \\ - 4 - j 0.4142 \\ - 4 + j 2.4142 \\ - 4 - j 2.4142 \\ - 4 + j 0.4142 \end{aligned}

Read the output in bit reversed order: $X [0]$ , $X [4]$ , $X [2]$ , $X [6]$ , $X [1]$ , $X [5]$ , $X [3]$ , $X [7]$ .

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	21	21	0
$X [1]$	-4 - j0.4142	4.0214	-174.09
$X [2]$	-1 + j2	2.2361	116.57
$X [3]$	-4 - j2.4142	4.6721	-148.89
$X [4]$	5	5	0
$X [5]$	-4 + j2.4142	4.6721	148.89
$X [6]$	-1 - j2	2.2361	-116.57
$X [7]$	-4 + j0.4142	4.0214	174.09

Practise this2074 Chaitra · Q93+7 marksWhy do we need DFT? Draw the butterfly structure to compute the DFT of the following signal using Radix-2 DIFFFT algorithm, and compute X(2) and X(1) only

x [n] = {1.5, - 1, 1.8, 0.6, 3, 1.7}

What is different in this one

Decimation in frequency: 8 points, 3 stages of 4 butterflies, with the output in bit reversed order.
The sequence has 6 samples, not a power of two, so it is zero padded to $N = 8$ before the algorithm starts. Say so in the answer: a radix 2 FFT cannot run on 6 points.
The question asks only for X(2) and X(1), but the butterfly diagram has to be drawn in full to reach them, so the whole transform is here.

The sequence has 6 samples, and a radix 2 FFT needs a power of two, so zero pad to $N = 8$ and say so: $x [n] = {1.5, - 1, 1.8, 0.6, 3, 1.7, 0, 0}$ .

Decimation in frequency: the input stays in natural order and the output comes out bit reversed, so the last step is to relabel it.

Stage 1, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ applied to the differences:

\begin{aligned} 4.5 \\ 0.7 \\ 1.8 \\ 0.6 \\ - 1.5 \\ - 1.9092 + j 1.9092 \\ - j 1.8 \\ - 0.4243 - j 0.4243 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ applied to the differences:

\begin{aligned} 6.3 \\ 1.3 \\ 2.7 \\ - j 0.1 \\ - 1.5 - j 1.8 \\ - 2.3335 + j 1.4849 \\ - 1.5 + j 1.8 \\ 2.3335 + j 1.4849 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{2}^{0}$ applied to the differences:

\begin{aligned} 7.6 \\ 5 \\ 2.7 - j 0.1 \\ 2.7 + j 0.1 \\ - 3.8335 - j 0.3151 \\ 0.8335 - j 3.2849 \\ 0.8335 + j 3.2849 \\ - 3.8335 + j 0.3151 \end{aligned}

Read the output in bit reversed order: $X [0]$ , $X [4]$ , $X [2]$ , $X [6]$ , $X [1]$ , $X [5]$ , $X [3]$ , $X [7]$ .

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	7.6	7.6	0
$X [1]$	-3.8335 - j0.3151	3.8464	-175.3
$X [2]$	2.7 - j0.1	2.7019	-2.12
$X [3]$	0.8335 + j3.2849	3.389	75.76
$X [4]$	5	5	0
$X [5]$	0.8335 - j3.2849	3.389	-75.76
$X [6]$	2.7 + j0.1	2.7019	2.12
$X [7]$	-3.8335 + j0.3151	3.8464	175.3

Same as 2082 Bhadra Q112074 Ashwin · Q97 marksCompute the 8-point DFT of the sequence

x [n]

= {

\frac{1}{2}

\frac{1}{2}

\frac{1}{2}

\frac{1}{2}

, 0, 0, 0, 0} using Decimation in Frequency Fast Fourier Transform (DIF-FFT) algorithm.

What is different in this one

Decimation in frequency: 8 points, 3 stages of 4 butterflies, with the output in bit reversed order.

$N = 8$ , which is $2^{3}$ , so the transform takes 3 stages of 4 butterflies.

Decimation in frequency: the input stays in natural order and the output comes out bit reversed, so the last step is to relabel it.

Stage 1, 4 butterflies with twiddles $W_{8}^{0}$ , $W_{8}^{1}$ , $W_{8}^{2}$ , $W_{8}^{3}$ applied to the differences:

\begin{aligned} 0.5 \\ 0.5 \\ 0.5 \\ 0.5 \\ 0.5 \\ 0.3536 - j 0.3536 \\ - j 0.5 \\ - 0.3536 - j 0.3536 \end{aligned}

Stage 2, 4 butterflies with twiddles $W_{4}^{0}$ , $W_{4}^{1}$ applied to the differences:

\begin{aligned} 1 \\ 1 \\ 0 \\ 0 \\ 0.5 - j 0.5 \\ - j 0.7071 \\ 0.5 + j 0.5 \\ - j 0.7071 \end{aligned}

Stage 3, 4 butterflies with twiddles $W_{2}^{0}$ applied to the differences:

\begin{aligned} 2 \\ 0 \\ 0 \\ 0 \\ 0.5 - j 1.2071 \\ 0.5 + j 0.2071 \\ 0.5 - j 0.2071 \\ 0.5 + j 1.2071 \end{aligned}

Read the output in bit reversed order: $X [0]$ , $X [4]$ , $X [2]$ , $X [6]$ , $X [1]$ , $X [5]$ , $X [3]$ , $X [7]$ .

k	$X [k]$	Magnitude	Phase, degrees
$X [0]$	2	2	0
$X [1]$	0.5 - j1.2071	1.3066	-67.5
$X [2]$	0	0	0
$X [3]$	0.5 - j0.2071	0.5412	-22.5
$X [4]$	0	0	0
$X [5]$	0.5 + j0.2071	0.5412	22.5
$X [6]$	0	0	0
$X [7]$	0.5 + j1.2071	1.3066	67.5

The FIR lattice HOT 7/19

Ch 4 · Filter structures7 questions, from 7 of the 19 sittings2 to practise

The procedure, the same for every question in this topic

H (z) = A_{m} (z) = 1 + α_{m} (1) z^{- 1} + \dots + α_{m} (m) z^{- m}

The last coefficient is the reflection coefficient, always.
$K_{m} = α_{m} (m)$
Step the order down.
$α_{m - 1} (k) = \frac{α_{m} (k) - K_{m} α_{m} (m - k)}{1 - K_{m}^{2}}, k = 1 \dots m - 1$
Repeat until $K_{1}$ , then draw the structure.
$\begin{aligned} f_{m} [n] & = f_{m - 1} [n] + K_{m} g_{m - 1} [n - 1] \\ g_{m} [n] & = K_{m} f_{m - 1} [n] + g_{m - 1} [n - 1] \\ y [n] & = f_{M} [n] \end{aligned}$

The leading coefficient must be 1. $H (z) = 2 + 1.8 z^{- 1} . .$ . means dividing by 2 first and putting the gain 2 at the output.
An FIR filter is always stable, whatever the K values, because its impulse response is finite. Say that before anything else when stability is asked.

The checks that cost nothing

$| K m | < 1$ for every m if and only if the filter is minimum phase, that is, every zero inside the unit circle. A $| K |$ above 1 is a legitimate answer, not an error.
$1 - K m^{2} = 0$ stops the recursion. $| K | = 1$ means a zero exactly on the unit circle and no lattice exists. Report it.
Run it forwards from the K values: $A (z)$ must come back.

Practise this +3 like it2081 Baishakh · Q5b5 marksDraw the lattice structure for the given FIR filter and also check whether the system is stable

. H (z) = 1 + (13 / 24) z^{- 1} + (5 / 8) z^{- 2} + (1 / 3) z^{- 3}

What is different in this one

An order 3 filter, so there are 3 reflection coefficients and 3 lattice stages to draw.
Every $| K |$ comes out below 1, so the filter is minimum phase as well as stable.

The last coefficient of the polynomial is the reflection coefficient of that order. Step the order down with the Levinson Durbin recursion and read the next one off.

\begin{aligned} K_{m} & = α_{m} (m) \end{aligned}

α_{m - 1} (k) = \frac{α_{m} (k) - K_{m} α_{m} (m - k)}{1 - K_{m}^{2}}, k = 1 \dots m - 1

Polynomial at this order	Reflection coefficient	$1 - K^{2}$
$A 3 (z) = 1 + 0.5417 z^{- 1} + 0.625 z^{- 2} + 0.3333 z^{- 3}$	K3 = 0.3333	0.8889
$A 2 (z) = 1 + 0.375 z^{- 1} + 0.5 z^{- 2}$	K2 = 0.5	0.75
$A 1 (z) = 1 + 0.25 z^{- 1}$	K1 = 0.25	0.9375

Also worked under Direct form realizationPractise this +2 like it2078 Bhadra · Q63+7 marksThe system function of a filter is

H (z) = 1 + (13 / 24) z^{- 1} + (5 / 8) z^{- 2} + (1 / 3) z^{- 3}

. Draw the Direct Form and Lattice Structure implementation of the above filter.

The lattice structure

What is different in this one

An order 3 filter, so there are 3 reflection coefficients and 3 lattice stages to draw.
Every $| K |$ comes out below 1, so the filter is minimum phase as well as stable.

The last coefficient of the polynomial is the reflection coefficient of that order. Step the order down with the Levinson Durbin recursion and read the next one off.

\begin{aligned} K_{m} & = α_{m} (m) \end{aligned}

α_{m - 1} (k) = \frac{α_{m} (k) - K_{m} α_{m} (m - k)}{1 - K_{m}^{2}}, k = 1 \dots m - 1

Polynomial at this order	Reflection coefficient	$1 - K^{2}$
$A 3 (z) = 1 + 0.5417 z^{- 1} + 0.625 z^{- 2} + 0.3333 z^{- 3}$	K3 = 0.3333	0.8889
$A 2 (z) = 1 + 0.375 z^{- 1} + 0.5 z^{- 2}$	K2 = 0.5	0.75
$A 1 (z) = 1 + 0.25 z^{- 1}$	K1 = 0.25	0.9375

The rest of this questionThis topic works the part that belongs to it. The other part is worked under Direct form realization (The direct form), where it belongs.

Same as 2081 Baishakh Q5b2074 Chaitra · Q65 marksDetermine the lattice coefficients coefficients corresponding to the FIR filter with the system function:

H (z) = A 3 (z) = 1 + (52 / 96) z^{- 1} + (25 / 40) z^{- 2} + (1 / 3) z^{- 3}

How this is read52/96 is 13/24 and 25/40 is 5/8, so this is the same filter the other sittings set in lowest terms.

What is different in this one

An order 3 filter, so there are 3 reflection coefficients and 3 lattice stages to draw.
Every $| K |$ comes out below 1, so the filter is minimum phase as well as stable.

The last coefficient of the polynomial is the reflection coefficient of that order. Step the order down with the Levinson Durbin recursion and read the next one off.

\begin{aligned} K_{m} & = α_{m} (m) \end{aligned}

α_{m - 1} (k) = \frac{α_{m} (k) - K_{m} α_{m} (m - k)}{1 - K_{m}^{2}}, k = 1 \dots m - 1

Polynomial at this order	Reflection coefficient	$1 - K^{2}$
$A 3 (z) = 1 + 0.5417 z^{- 1} + 0.625 z^{- 2} + 0.3333 z^{- 3}$	K3 = 0.3333	0.8889
$A 2 (z) = 1 + 0.375 z^{- 1} + 0.5 z^{- 2}$	K2 = 0.5	0.75
$A 1 (z) = 1 + 0.25 z^{- 1}$	K1 = 0.25	0.9375

Also worked under Direct form realizationSame as 2078 Bhadra Q62074 Ashwin · Q53+7 marksDraw the direct form and Lattice structure of a filter with system function

H (z) = 1 + 0.7 z^{- 1} + 1.2 z^{- 2} - z^{- 3}

The lattice structure

What is different in this one

This one has no lattice at all. The step down divides by $1 - K^{2}$ , and here a reflection coefficient comes out with magnitude exactly 1, so that divisor is zero. Reporting the breakdown and saying why IS the answer.

Step the order down as usual and read off the reflection coefficients.

A3(z) = 1 +

0.7 z^{- 1} + 1.2 z^{- 2} - z^{- 3}

K3 =

α_{3} (3)

= -1 1

- K^{2} = 1 - {(- 1)}^{2}

= 0

\begin{aligned} zeros  of H (z) : - 0.6183 + j 1.2171, - 0.6183 - j 1.2171, 0.5366 \\ magnitudes:      1.3651,   1.3651,   0.5366 \end{aligned}

The rest of this questionThis topic works the part that belongs to it. The other part is worked under Direct form realization (The direct form), where it belongs.

Also worked under Direct form realizationSame as 2078 Bhadra Q62073 Shrawan · Q53+7 marksThe system function of a filter is

H (z) = 2 + 1.8 z^{- 1} - 1.6 z^{- 2} + z^{- 3}

. Draw the Direct Form and Lattice Structure implementation of the above filter.

The lattice structure

The leading coefficient is 2, and the recursion needs it to be 1. Divide through by it, find the lattice of $1 + 0.9 z^{- 1} - 0.8 z^{- 2} + 0.5 z^{- 3}$ , and put the gain 2 at the output.

What is different in this one

An order 3 filter, so there are 3 reflection coefficients and 3 lattice stages to draw.
One of the reflection coefficients has magnitude 2.6, above 1, so this filter is not minimum phase: at least one zero lies outside the unit circle. It is still perfectly stable, being FIR.

The last coefficient of the polynomial is the reflection coefficient of that order. Step the order down with the Levinson Durbin recursion and read the next one off.

\begin{aligned} K_{m} & = α_{m} (m) \end{aligned}

α_{m - 1} (k) = \frac{α_{m} (k) - K_{m} α_{m} (m - k)}{1 - K_{m}^{2}}, k = 1 \dots m - 1

Polynomial at this order	Reflection coefficient	$1 - K^{2}$
$A 3 (z) = 1 + 0.9 z^{- 1} - 0.8 z^{- 2} + 0.5 z^{- 3}$	K3 = 0.5	0.75
$A 2 (z) = 1 + 1.7333 z^{- 1} - 1.6667 z^{- 2}$	K2 = -1.6667	-1.7778
$A 1 (z) = 1 - 2.6 z^{- 1}$	K1 = -2.6	-5.76

The rest of this questionThis topic works the part that belongs to it. The other part is worked under Direct form realization (The direct form), where it belongs.

Same as 2081 Baishakh Q5b2072 Chaitra · Q75 marksCompute the lattice coefficients and draw the lattice structure of following FIR system

. H (z) = 1 + 2 z^{- 1} + z^{- 2}

What is different in this one

This one has no lattice at all. The step down divides by $1 - K^{2}$ , and here a reflection coefficient comes out with magnitude exactly 1, so that divisor is zero. Reporting the breakdown and saying why IS the answer.

Step the order down as usual and read off the reflection coefficients.

A2(z) = 1 +

2 z^{- 1} + z^{- 2}

K2 =

α_{2} (2)

= 1 1

- K^{2} = 1 - {(1)}^{2}

= 0

\begin{aligned} zeros  of H (z) : - 1, - 1 \\ magnitudes:      1,   1 \end{aligned}

Same as 2081 Baishakh Q5b2072 Kartik · Q76 marksCompute the lattice coefficients and draw the lattice structure of following FIR system

H (z) = 1 + 2 z^{- 1} - 3 z^{- 2} + 4 z^{- 3}

What is different in this one

An order 3 filter, so there are 3 reflection coefficients and 3 lattice stages to draw.
One of the reflection coefficients has magnitude 4, above 1, so this filter is not minimum phase: at least one zero lies outside the unit circle. It is still perfectly stable, being FIR.

The last coefficient of the polynomial is the reflection coefficient of that order. Step the order down with the Levinson Durbin recursion and read the next one off.

\begin{aligned} K_{m} & = α_{m} (m) \end{aligned}

α_{m - 1} (k) = \frac{α_{m} (k) - K_{m} α_{m} (m - k)}{1 - K_{m}^{2}}, k = 1 \dots m - 1

Polynomial at this order	Reflection coefficient	$1 - K^{2}$
$A 3 (z) = 1 + 2 z^{- 1} - 3 z^{- 2} + 4 z^{- 3}$	K3 = 4	-15
$A 2 (z) = 1 - 0.9333 z^{- 1} + 0.7333 z^{- 2}$	K2 = 0.7333	0.4622
$A 1 (z) = 1 - 0.5385 z^{- 1}$	K1 = -0.5385	0.7101

Periodicity and the fundamental period HOT 6/19

Ch 1 · Signals and systems6 questions, from 6 of the 19 sittings3 to practise

The procedure, the same for every question in this topic

periodic ⟺ \frac{ω_{0}}{2 π} is RATIONAL

\begin{aligned} \frac{ω_{0}}{2 π} & = \frac{k}{N} in LOWEST terms ⟹ fundamental period N \\ N & = LCM (N_{1}, N_{2}, \dots) & for several components \end{aligned}

One aperiodic component makes the whole signal aperiodic, however well behaved the others are.

Only the coefficient of n matters. The phase never affects periodicity.
No $π$ in $ω_{0}$ means aperiodic: $\sin (0.8 n)$ has $ω_{0} / 2 π = 0.4 / π$ , irrational.
A product is handled like a sum: find each component's period and take the LCM.

The checks that cost nothing

N must be a positive integer. That is the entire difference from continuous time.
Lowest terms. Failing to reduce k/N gives a period that is a multiple of the fundamental one, which is periodic but not the answer.
Substitute back: $x [n + N]$ must equal $x [n]$ . One sample is enough to catch an error.

Practise this +3 like it2081 Bhadra · Q14 marksDetermine if the signal

x [n] = \cos (2 π n / 5) + \sin (π n / 3)

is periodic or not. If the signal is periodic, find its fundamental period.

What is different in this one

2 components, so each period is found separately and the answer is their LCM.
The phase plays no part: only the coefficient of n decides periodicity.

A discrete time sinusoid is periodic only if $ω_{0} / (2 π)$ is rational. Reduce it to k/N in lowest terms and N is the period of that component; for several components take the LCM.

Component	$ω_{0}$	$ω_{0} / 2 π$ , lowest terms	Period
$\cos (2 π n / 5)$	$\frac{2 π}{5}$	1 / 5	5
$\sin (π n / 3)$	$\frac{π}{3}$	1 / 6	6

N = LCM(5, 6) = 30.

$x [n] = \cos (2 π n / 5) + \sin (π n / 3)$ is periodic with fundamental period $N = 30$ samples.

Same as 2081 Bhadra Q12080 Bhadra · Q25 marksDetermine whether the given signal is periodic or not. If the signal is periodic, determine the fundamental period

x [n] = e^{j π n / 16} \cos (n π / 17)

What is different in this one

2 components, so each period is found separately and the answer is their LCM.
The phase plays no part: only the coefficient of n decides periodicity.

A discrete time sinusoid is periodic only if $ω_{0} / (2 π)$ is rational. Reduce it to k/N in lowest terms and N is the period of that component; for several components take the LCM.

Component	$ω_{0}$	$ω_{0} / 2 π$ , lowest terms	Period
$e^{j π n / 16}$	$\frac{π}{16}$	1 / 32	32
$\cos (n π / 17)$	$\frac{π}{17}$	1 / 34	34

N = LCM(32, 34) = 544.

$x [n] = e^{j π n / 16} \cos (n π / 17)$ is periodic with fundamental period $N = 544$ samples.

Practise this2080 Baishakh · Q12+2 marksCheck whether following signals are periodic or not. If yes, state their periodic time. a)

x [n]

= Sin(n

π

) + Cos(n

π

) b)

x [n]

= Sin(3n

\frac{π}{5}

) + Cos(4n

\frac{π}{7}

)

a) $\sin (n π) + \cos (n π)$

What is different in this one

2 components, so each period is found separately and the answer is their LCM.
The phase plays no part: only the coefficient of n decides periodicity.

A discrete time sinusoid is periodic only if $ω_{0} / (2 π)$ is rational. Reduce it to k/N in lowest terms and N is the period of that component; for several components take the LCM.

Component	$ω_{0}$	$ω_{0} / 2 π$ , lowest terms	Period
$\sin (n π)$	$π$	1 / 2	2
$\cos (n π)$	$π$	1 / 2	2

N = LCM(2, 2) = 2.

$x [n] = \sin (n π) + \cos (n π)$ is periodic with fundamental period $N = 2$ samples.

b) $\sin (3 n π / 5) + \cos (4 n π / 7)$

What is different in this one

2 components, so each period is found separately and the answer is their LCM.
The phase plays no part: only the coefficient of n decides periodicity.

A discrete time sinusoid is periodic only if $ω_{0} / (2 π)$ is rational. Reduce it to k/N in lowest terms and N is the period of that component; for several components take the LCM.

Component	$ω_{0}$	$ω_{0} / 2 π$ , lowest terms	Period
$\sin (3 π n / 5)$	$\frac{3 π}{5}$	3 / 10	10
$\cos (4 π n / 7)$	$\frac{4 π}{7}$	2 / 7	7

N = LCM(10, 7) = 70.

$x [n] = \sin (3 n π / 5) + \cos (4 n π / 7)$ is periodic with fundamental period $N = 70$ samples.

Same as 2081 Bhadra Q12079 Bhadra · Q12+2 marksDefine energy and power signal. Determine whether the signal

x [n] = \cos (2 π n / 5) + \sin (π n / 3)

is periodic or non-periodic and if it is periodic, find its fundamental period.

What is different in this one

2 components, so each period is found separately and the answer is their LCM.
The phase plays no part: only the coefficient of n decides periodicity.

A discrete time sinusoid is periodic only if $ω_{0} / (2 π)$ is rational. Reduce it to k/N in lowest terms and N is the period of that component; for several components take the LCM.

Component	$ω_{0}$	$ω_{0} / 2 π$ , lowest terms	Period
$\cos (2 π n / 5)$	$\frac{2 π}{5}$	1 / 5	5
$\sin (π n / 3)$	$\frac{π}{3}$	1 / 6	6

N = LCM(5, 6) = 30.

$x [n] = \cos (2 π n / 5) + \sin (π n / 3)$ is periodic with fundamental period $N = 30$ samples.

Same as 2081 Bhadra Q12078 Bhadra · Q14 marksDetermine whether the signal

x [n] = \cos (π n / 2) \cos (π n / 4)

is periodic or non periodic and if it is periodic, find its fundamental period.

What is different in this one

2 components, so each period is found separately and the answer is their LCM.
The phase plays no part: only the coefficient of n decides periodicity.

A discrete time sinusoid is periodic only if $ω_{0} / (2 π)$ is rational. Reduce it to k/N in lowest terms and N is the period of that component; for several components take the LCM.

Component	$ω_{0}$	$ω_{0} / 2 π$ , lowest terms	Period
$\cos (π n / 2)$	$\frac{π}{2}$	1 / 4	4
$\cos (π n / 4)$	$\frac{π}{4}$	1 / 8	8

N = LCM(4, 8) = 8.

$x [n] = \cos (π n / 2) \cos (π n / 4)$ is periodic with fundamental period $N = 8$ samples.

Practise this2074 Ashwin · Q12+2 marksDefine Energy and Power type discrete time signal. Check whether signal

x [n] = e^{j (π n / 3 + π / 4)}

is periodic or not. If it is periodic, state its periodic time.What is different hereThe phase

\frac{π}{4}

plays no part: only the coefficient of n decides periodicity.

What is different in this one

1 component, so the period is read straight off.
The phase plays no part: only the coefficient of n decides periodicity.

A discrete time sinusoid is periodic only if $ω_{0} / (2 π)$ is rational. Reduce it to k/N in lowest terms and N is the period of that component; for several components take the LCM.

Component	$ω_{0}$	$ω_{0} / 2 π$ , lowest terms	Period
$e^{j (π n / 3 + π / 4)}$	$\frac{π}{3}$	1 / 6	6

$x [n] = e^{j (π n / 3 + π / 4)}$ is periodic with fundamental period $N = 6$ samples.

Testing a system for its properties HOT 5/19

Ch 1 · Signals and systems5 questions, from 5 of the 19 sittings5 to practise

The procedure, the same for every question in this topic

Take them one at a time, and show the test for each. The marks are for the test, not for the verdict.

Property	The test
Static	Does $y [n]$ use x at any other index, or a past output? If not, it is static.
Linear	Compute $T {a_{1} x_{1} + a_{2} x_{2}}$ and $a_{1} T {x_{1}} + a_{2} T {x_{2}}$ , and compare them.
Time invariant	Compute $y [n, k] = T {x [n - k]}$ and $y [n - k]$ , and compare. Replace every n in $y [n]$ , including the explicit ones.
Causal	Does $y [n]$ ever need x at a future index?
Stable	Bound $\| y [n] \|$ given $\| x [n] \| \leq M_{x}$ , or find one bounded input that blows the output up.

An explicit n multiplying the input is the classic time varying flag, and so is a folded index: delaying the input does not delay the n or the fold with it.
To disprove stability, one input is enough. A constant or $u [n]$ almost always works, and it is a shorter answer than any bound.
Marks are awarded per property, so answer every part asked even when one is obvious. A bare yes or no with no working scores nothing.

The checks that cost nothing

Squaring, exponentiating or taking a logarithm of the input is non linear, but still time invariant and still memoryless. Getting one of those right does not settle the others.
No $x [n]$ in the expression at all means a zero input gives a non zero output, so linearity fails at the first line.
Any FIR system is stable, whatever else it is.

Practise this2082 Baishakh · Q12+2 marksA discrete time system has input

x [n]

and output

y [n]

and its input-output relation is

y [n] = x [n] + x [- n]

. Is the system causal, time invariant or not?What is different herea different system:

y [n] = x [n] + x [- n]

. The fold is what breaks both time invariance and causality: at

n = - 2

the output needs

x [2]

, which has not arrived, and delaying the input does not delay the folded half with it.

What is different in this one

4 properties asked here: linear, time invariant, causal, stable. Marks are given per property, so answer every one even when it is obvious.
The linear, time invariant, stable verdicts below were checked by RUNNING the system, not asserted: on a weighted sum of two random inputs, on a shifted input, and on a bounded step.
The fold is what breaks both time invariance and causality: at $n = - 2$ the output needs $x [2]$ , which has not arrived, and delaying the input does not delay the folded half with it.

Take the properties one at a time and show the test for each. A bare yes or no with no working scores nothing.

\begin{aligned} y [n] & = x [n] + x [- n] \end{aligned}

Property	The test	Verdict
Linear	is T{a $x_{1} + b x_{2}$ } the same as a T{ $x_{1}$ } + b T{ $x_{2}$ }?	yes [tested]
Time invariant	is T{ $x [n - k]$ } the same as $y [n - k]$ , the explicit n included?	no [tested]
Causal	does $y [n]$ ever need an input from the future?	no
Stable	does every bounded input give a bounded output?	yes [tested]

Linear?

\begin{aligned} T (a x_{1} [n] + b x_{2} [n]) \\ = (a x_{1} [n] + b x_{2} [n]) + (a x_{1} [- n] + b x_{2} [- n]) \\ = a (x_{1} [n] + x_{1} [- n]) + b (x_{2} [n] + x_{2} [- n]) \\ = a y_{1} [n] + b y_{2} [n] \end{aligned}

The two sides are identical for every a and b, so it is LINEAR.

Time invariant?

\begin{aligned} delay the input \\ T (x [n - k]) [n] & = x [n - k] + x [- n - k] \\ delay the output \\ y [n - k] & = x [n - k] + x [- (n - k)] = x [n - k] + x [- n + k] \\ compare the second terms \\ x [- n - k] \\ x [- n + k] \end{aligned}

They agree only when $k = 0$ , so it is NOT time invariant. The fold reverses the shift instead of carrying it.

Causal?

\begin{aligned} y [- 2] & = x [- 2] + x [- (- 2)] = x [- 2] + x [2] \end{aligned}

The output at $n = - 2$ needs $x [2]$ , two samples into the future, so it is NOT causal.

Stable?

\begin{aligned} bound the input \\ | x [n] | \leq B & for every n \\ then \\ | y [n] | & = | x [n] + x [- n] | \leq | x [n] | + | x [- n] | \leq 2 B \end{aligned}

A bound in gives a bound out, so it is STABLE.

The fold is what breaks both time invariance and causality: at $n = - 2$ the output needs $x [2]$ , which has not arrived, and delaying the input does not delay the folded half with it.

Practise this2080 Bhadra · Q12+2+2 marksA discrete time system has input

x (n)

and output

y (n)

. The input output relation of the system is given by

y (n)

= sum from k=0 to n of

x (k)

u [n]

gives

y [n] = n

+ 1, which is the disproof to write down.

What is different in this one

5 properties asked here: static, linear, time invariant, causal, stable. Marks are given per property, so answer every one even when it is obvious.
The linear, time invariant, stable verdicts below were checked by RUNNING the system, not asserted: on a weighted sum of two random inputs, on a shifted input, and on a bounded step.

Take the properties one at a time and show the test for each. A bare yes or no with no working scores nothing.

\begin{aligned} y [n] & = sum  from k = 0 to n of x [k] & the accumulator \end{aligned}

Property	The test	Verdict
Static	does $y [n]$ use x at any other index, or a past output?	no
Linear	is T{a $x_{1} + b x_{2}$ } the same as a T{ $x_{1}$ } + b T{ $x_{2}$ }?	yes [tested]
Time invariant	is T{ $x [n - k]$ } the same as $y [n - k]$ , the explicit n included?	no [tested]
Causal	does $y [n]$ ever need an input from the future?	yes
Stable	does every bounded input give a bounded output?	no [tested]

Static?

\begin{aligned} y [3] & = x [0] + x [1] + x [2] + x [3] \end{aligned}

It needs every sample since 0, not $x [3]$ alone, so it has MEMORY and is not static.

Linear?

\begin{aligned} T (a x_{1} + b x_{2}) [n] \\ = \sum over k = 0 to n of (a x_{1} [k] + b x_{2} [k]) \\ = a \sumx_{1} [k] + b \sumx_{2} [k] \\ = a y_{1} [n] + b y_{2} [n] \end{aligned}

A sum splits over addition, so it is LINEAR.

Time invariant?

\begin{aligned} delay the input \\ T (x [n - d]) [n] & = \sum over k = 0 to n of x [k - d] \\ = \sum over m = - d to n - d of x [m] \\ delay the output \\ y [n - d] & = \sum over k = 0 to n - d of x [k] \\ compare the lower limits \\ - d \\ 0 \end{aligned}

The delayed input drags d extra samples into the sum, so it is NOT time invariant. The lower limit pinned at 0 is what breaks it.

Causal?

\begin{aligned} y [n] & = x [0] + x [1] + \dots + x [n] \\ the largest index used is n \end{aligned}

Nothing beyond n is ever read, so it is CAUSAL.

Stable?

\begin{aligned} take  the  bounded  input x [n] & = u [n], so | x [n] | \leq 1 \\ y [0] & = 1, y [1] = 2, y [2] = 3, \dots, y [n] = n + 1 \end{aligned}

A bounded input gives an output that grows without bound, so it is NOT stable.

Practise this2076 Chaitra · Q25 marksDetermine whether the following system are: a)

y [n] = x [- n]

is time-invariant or not. b)

y [n]

= x[

n^{2}

] is linear or not.

a) is $y [n] = x [- n]$ time invariant?

What is different in this one

4 properties asked here: linear, time invariant, causal, stable. Marks are given per property, so answer every one even when it is obvious.
The linear, time invariant, stable verdicts below were checked by RUNNING the system, not asserted: on a weighted sum of two random inputs, on a shifted input, and on a bounded step.

Take the properties one at a time and show the test for each. A bare yes or no with no working scores nothing.

\begin{aligned} y [n] & = x [- n] \end{aligned}

Property	The test	Verdict
Linear	is T{a $x_{1} + b x_{2}$ } the same as a T{ $x_{1}$ } + b T{ $x_{2}$ }?	yes [tested]
Time invariant	is T{ $x [n - k]$ } the same as $y [n - k]$ , the explicit n included?	no [tested]
Causal	does $y [n]$ ever need an input from the future?	no
Stable	does every bounded input give a bounded output?	yes [tested]

Linear?

\begin{aligned} T (a x_{1} [n] + b x_{2} [n]) & = a x_{1} [- n] + b x_{2} [- n] \\ = a y_{1} [n] + b y_{2} [n] \end{aligned}

Identical for every a and b, so it is LINEAR.

Time invariant?

\begin{aligned} delay the input \\ T (x [n - k]) [n] & = x [- n - k] \\ delay the output \\ y [n - k] & = x [- (n - k)] = x [- n + k] \end{aligned}

$X [- n - k]$ against $x [- n + k]$ : equal only when $k = 0$ , so it is NOT time invariant.

Causal?

\begin{aligned} y [- 1] & = x [- (- 1)] = x [1] \end{aligned}

The output at $n = - 1$ needs the input one sample ahead, so it is NOT causal.

Stable?

\begin{aligned} bound the input \\ | x [n] | \leq B & for every n \\ then \\ | y [n] | & = | x [- n] | \leq B \end{aligned}

The fold moves samples about and never makes one larger, so it is STABLE.

b) is $y [n] = x [n^{2}]$ linear?

What is different in this one

2 properties asked here: linear, time invariant. Marks are given per property, so answer every one even when it is obvious.
The linear, time invariant verdicts below were checked by RUNNING the system, not asserted: on a weighted sum of two random inputs, on a shifted input, and on a bounded step.

Take the properties one at a time and show the test for each. A bare yes or no with no working scores nothing.

\begin{aligned} y [n] & = x [n^{2}] \end{aligned}

Property	The test	Verdict
Linear	is T{a $x_{1} + b x_{2}$ } the same as a T{ $x_{1}$ } + b T{ $x_{2}$ }?	yes [tested]
Time invariant	is T{ $x [n - k]$ } the same as $y [n - k]$ , the explicit n included?	no [tested]

Linear?

\begin{aligned} T (a x_{1} [n] + b x_{2} [n]) & = a x_{1} [n^{2}] + b x_{2} [n^{2}] \\ = a y_{1} [n] + b y_{2} [n] \end{aligned}

Reading the input at a different index is still linear in the input, so it is LINEAR.

Time invariant?

\begin{aligned} delay the input \\ T (x [n - k]) [n] & = x [n^{2} - k] \\ delay the output \\ y [n - k] & = x [{(n - k)}^{2}] = x [n^{2} - 2 kn + k^{2}] \end{aligned}

The two indices differ by -2kn + $k^{2}$ , which depends on n, so it is NOT time invariant.

Practise this2075 Ashwin · Q13+3 marksDetermine whether the following sequences are linear or not: a)

y [n] = x^{2} [n]

y [n] = \cos (5 π n / 8 + π / 4)

What is different herea different system:

y [n] = x^{2} [n]

; a different system:

y [n] = \cos (5 π n / 8 + π / 4)

How this is readThe second has no $x [n]$ in it at all, so a zero input gives a non zero output and superposition fails at the first line. That is the whole answer.

a) is $y [n] = x^{2} [n]$ linear?

What is different in this one

4 properties asked here: linear, time invariant, causal, stable. Marks are given per property, so answer every one even when it is obvious.
The linear, time invariant, stable verdicts below were checked by RUNNING the system, not asserted: on a weighted sum of two random inputs, on a shifted input, and on a bounded step.

Take the properties one at a time and show the test for each. A bare yes or no with no working scores nothing.

\begin{aligned} y [n] & = x^{2} [n] \end{aligned}

Property	The test	Verdict
Linear	is T{a $x_{1} + b x_{2}$ } the same as a T{ $x_{1}$ } + b T{ $x_{2}$ }?	no [tested]
Time invariant	is T{ $x [n - k]$ } the same as $y [n - k]$ , the explicit n included?	yes [tested]
Causal	does $y [n]$ ever need an input from the future?	yes
Stable	does every bounded input give a bounded output?	yes [tested]

Linear?

\begin{aligned} T (a x_{1} [n] + b x_{2} [n]) & = {(a x 1 [n] + b x 2 [n])}^{2} \\ = a^{2} x_{1}^{2} + 2 ab x_{1} x_{2} + b^{2} x_{2}^{2} \\ a y_{1} [n] + b y_{2} [n] & = a x_{1}^{2} + b x_{2}^{2} \\ one  counterexample  settles  it, with x_{1} & = x_{2} = 1 and a = b = 1 \\ T (1 + 1) & = 2^{2} = 4 T (1) + T (1) = 1 + 1 = 2 \end{aligned}

4 is not 2, so it is NOT linear. Squaring is the classic nonlinearity.

Time invariant?

\begin{aligned} delay the input \\ T (x [n - k]) [n] & = x^{2} [n - k] \\ delay the output \\ y [n - k] & = x^{2} [n - k] \end{aligned}

The same expression on both sides, so it is TIME INVARIANT. There is no explicit n anywhere in it.

Causal?

\begin{aligned} y [n] & = x^{2} [n] \\ the only index used is n \end{aligned}

It uses the present input alone, so it is CAUSAL, and static as well.

Stable?

\begin{aligned} bound the input \\ | x [n] | \leq B & for every n \\ then \\ | y [n] | & = | x [n] |^{2} \leq B^{2} \end{aligned}

A finite bound squared is still finite, so it is STABLE.

b) is $y [n] = \cos (5 π n / 8 + π / 4)$ linear?

What is different in this one

2 properties asked here: linear, time invariant. Marks are given per property, so answer every one even when it is obvious.
The linear, time invariant verdicts below were checked by RUNNING the system, not asserted: on a weighted sum of two random inputs, on a shifted input, and on a bounded step.

Take the properties one at a time and show the test for each. A bare yes or no with no working scores nothing.

\begin{aligned} y [n] & = \cos (5 π n / 8 + π / 4) \end{aligned}

Property	The test	Verdict
Linear	is T{a $x_{1} + b x_{2}$ } the same as a T{ $x_{1}$ } + b T{ $x_{2}$ }?	no [tested]
Time invariant	is T{ $x [n - k]$ } the same as $y [n - k]$ , the explicit n included?	no [tested]

Linear?

\begin{aligned} the input never appears on the right hand side at all \\ put x [n] & = 0 \\ y [n] & = \cos (5 π n / 8 + π / 4), which  is  not 0 \end{aligned}

A linear system must give zero out for zero in, and this one does not, so it is NOT linear.

Time invariant?

\begin{aligned} delay the input, and nothing happens, since x is ignored \\ T (x [n - k]) [n] & = \cos (5 π n / 8 + π / 4) \\ delay the output, and the cosine moves \\ y [n - k] & = \cos (5 π (n - k) / 8 + π / 4) \end{aligned}

Delaying the input does nothing while delaying the output shifts the cosine, so it is NOT time invariant. The explicit n is what breaks it.

Practise this2074 Chaitra · Q23+3 marksDetermine whether the following system are: a)

y [n] = y [n - 4] + x [n - 4]

is Time-invariant or not b)

y [n] = x^{2} [n]

is Linear or Non-linearWhat is different herea different system:

y [n] = y [n - 4] + x [n - 4]

; a different system:

y [n] = x^{2} [n]

How this is readThe recursion $y [n] = y [n - 4] + x [n - 4]$ with zero initial conditions accumulates every fourth past input, which is how it is written for the test.

a) is $y [n] = y [n - 4] + x [n - 4]$ time invariant?

What is different in this one

4 properties asked here: linear, time invariant, causal, stable. Marks are given per property, so answer every one even when it is obvious.
The linear, time invariant, stable verdicts below were checked by RUNNING the system, not asserted: on a weighted sum of two random inputs, on a shifted input, and on a bounded step.

Take the properties one at a time and show the test for each. A bare yes or no with no working scores nothing.

\begin{aligned} y [n] & = y [n - 4] + x [n - 4] \end{aligned}

Property	The test	Verdict
Linear	is T{a $x_{1} + b x_{2}$ } the same as a T{ $x_{1}$ } + b T{ $x_{2}$ }?	yes [tested]
Time invariant	is T{ $x [n - k]$ } the same as $y [n - k]$ , the explicit n included?	yes [tested]
Causal	does $y [n]$ ever need an input from the future?	yes
Stable	does every bounded input give a bounded output?	no [tested]

Linear?

\begin{aligned} every term is a constant times one sample of x or of y \\ nothing is squared, multiplied together, or passed \\ through a function, and the system starts at rest \end{aligned}

A constant coefficient difference equation at rest is LINEAR.

Time invariant?

\begin{aligned} the coefficients are constants and both delays are \\ fixed at 4, and there is no explicit n anywhere \end{aligned}

Delaying the input delays the whole recursion with it, so it is TIME INVARIANT.

Causal?

\begin{aligned} y [n] & = y [n - 4] + x [n - 4] \\ the  largest  index  used  is n - 4 \end{aligned}

It reads only the past, so it is CAUSAL.

Stable?

\begin{aligned} take  the  bounded  input x [n] & = δ [n] \\ y [4] & = 1, y [8] = 1, y [12] = 1, \dots & for ever \\ \sum | h [n] | & = 1 + 1 + 1 + \dots = infinity \end{aligned}

The impulse response never decays, so it is NOT stable. The pole sits ON the unit circle.

b) is $y [n] = x^{2} [n]$ linear?

What is different in this one

2 properties asked here: linear, time invariant. Marks are given per property, so answer every one even when it is obvious.
The linear, time invariant verdicts below were checked by RUNNING the system, not asserted: on a weighted sum of two random inputs, on a shifted input, and on a bounded step.

Take the properties one at a time and show the test for each. A bare yes or no with no working scores nothing.

\begin{aligned} y [n] & = x^{2} [n] \end{aligned}

Property	The test	Verdict
Linear	is T{a $x_{1} + b x_{2}$ } the same as a T{ $x_{1}$ } + b T{ $x_{2}$ }?	no [tested]
Time invariant	is T{ $x [n - k]$ } the same as $y [n - k]$ , the explicit n included?	yes [tested]

Linear?

\begin{aligned} T (a x_{1} [n] + b x_{2} [n]) & = {(a x 1 [n] + b x 2 [n])}^{2} \\ = a^{2} x_{1}^{2} + 2 ab x_{1} x_{2} + b^{2} x_{2}^{2} \\ a y_{1} [n] + b y_{2} [n] & = a x_{1}^{2} + b x_{2}^{2} \\ one  counterexample  settles  it, with x_{1} & = x_{2} = 1 and a = b = 1 \\ T (1 + 1) & = 2^{2} = 4 T (1) + T (1) = 1 + 1 = 2 \end{aligned}

4 is not 2, so it is NOT linear. Squaring is the classic nonlinearity.

Time invariant?

\begin{aligned} delay the input \\ T (x [n - k]) [n] & = x^{2} [n - k] \\ delay the output \\ y [n - k] & = x^{2} [n - k] \end{aligned}

The same expression on both sides, so it is TIME INVARIANT. There is no explicit n anywhere in it.

The output for an exponential input HOT 5/19

Ch 3 · Frequency domain5 questions, from 5 of the 19 sittings3 to practise

The procedure, the same for every question in this topic

A complex exponential goes through an LTI system unchanged except for a complex gain, and that gain is the frequency response at its own frequency.

x [n] = A e^{j ω_{0} n} ⟶ y [n] = A H (e^{j ω_{0}}) e^{j ω_{0} n}

A real cosine is the same statement in two halves: the magnitude scales it and the angle shifts it.

\begin{aligned} x [n] & = A \cos (ω_{0} n + ϕ) \\ ⟶ y [n] & = A | H (e^{j ω_{0}}) | \cos (ω_{0} n + ϕ + ∠ H (e^{j ω_{0}})) \end{aligned}

For the impulse response the paper keeps setting:

h [n] = a^{n} u [n] ⟹ H (e^{j ω}) = \frac{1}{1 - a e^{- j ω}}

Do not convolve. The input runs over all n, so it is not a finite sequence and graphical convolution is the wrong tool. This is the cheapest question on the paper once that is seen.
A sum of components is done one at a time and added: superposition. A constant is the $ω = 0$ case.

The checks that cost nothing

The frequency never changes. If your answer has a frequency the input did not, something is wrong.
$| H |$ at $ω = 0$ is the sum of $h [n]$ , which is a one line check on the algebra.
Give the answer in both forms, magnitude and angle, and as a single exponential. It costs a line and it is what the marks are for.

Practise this +2 like it2082 Baishakh · Q25 marksFind the output of an LTI system having impulse response

h [n] = {(1 / 2)}^{n} u [n]

excited by an input

x [n] = 5 e^{j π n / 3}

What is different in this one

1 component at one frequency, so H is evaluated once and the results added: superposition.
Do not convolve. The input runs over all n, so it is not a finite sequence and the graphical method is the wrong tool.

x [n] = e^{j ω_{0} n} ⟶ y [n] = H (e^{j ω_{0}}) e^{j ω_{0} n}, A \cos (ω_{0} n) ⟶ A | H | \cos (ω_{0} n + ∠ H)

\begin{aligned} H (e^{j ω}) & = 1 / (1 - 0.5 e^{- j ω}), the  transform  of h [n] = {(1 / 2)}^{n} u [n] \end{aligned}

Component	$ω_{0}$	$H (e^{j ω_{0}})$	in polar form	Output amplitude
$x [n] = 5 e^{j π n / 3}$	$0.3333 π$	1 - j0.5774	1.1547 angle -30 deg	5.7735

Where	By convolution	By the formula
$n = 0$	5	5
$n = 1$	5	5
$n = 2$	0	0
$n = 5$	0	0

$y [n] = 5.7735 e^{j (0.3333 π n - 30 d e g)}$ .

Same as 2082 Baishakh Q22080 Baishakh · Q25 marksFind the output of LTI system having impulse response

h [n] = {(1 / 2)}^{n} u [n]

and input

x [n] = 5 e^{j π n / 3}

for -infinity < n < infinity.

What is different in this one

1 component at one frequency, so H is evaluated once and the results added: superposition.
Do not convolve. The input runs over all n, so it is not a finite sequence and the graphical method is the wrong tool.

x [n] = e^{j ω_{0} n} ⟶ y [n] = H (e^{j ω_{0}}) e^{j ω_{0} n}, A \cos (ω_{0} n) ⟶ A | H | \cos (ω_{0} n + ∠ H)

\begin{aligned} H (e^{j ω}) & = 1 / (1 - 0.5 e^{- j ω}), the  transform  of h [n] = {(1 / 2)}^{n} u [n] \end{aligned}

Component	$ω_{0}$	$H (e^{j ω_{0}})$	in polar form	Output amplitude
$x [n] = 5 e^{j π n / 3}$	$0.3333 π$	1 - j0.5774	1.1547 angle -30 deg	5.7735

Where	By convolution	By the formula
$n = 0$	5	5
$n = 1$	5	5
$n = 2$	0	0
$n = 5$	0	0

$y [n] = 5.7735 e^{j (0.3333 π n - 30 d e g)}$ .

Practise this2073 Shrawan · Q43+7 marksState linear constant coefficient difference equation and corresponding system function. Determine the output sequence of the system with impulse response

h [n] = {(1 / 2)}^{n} u [n]

when the input signal is

x [n] = 10

- 5sin(

π

n/2) + 20cos(

π

What is different in this one

3 components at 3 different frequencies, so H is evaluated 3 times and the results added: superposition.
One of them is a constant, which is simply the $ω = 0$ case: the input times the DC gain.
The input is real, so the answer must be real. A sine goes in and a sine of the same frequency comes out, scaled by $| H |$ and shifted by the angle of H. Leaving the answer as complex exponentials is not finishing the question.
Do not convolve. The input runs over all n, so it is not a finite sequence and the graphical method is the wrong tool.

x [n] = e^{j ω_{0} n} ⟶ y [n] = H (e^{j ω_{0}}) e^{j ω_{0} n}, A \cos (ω_{0} n) ⟶ A | H | \cos (ω_{0} n + ∠ H)

\begin{aligned} H (e^{j ω}) & = 1 / (1 - 0.5 e^{- j ω}), the  transform  of h [n] = {(1 / 2)}^{n} u [n] \end{aligned}

Component	$ω_{0}$	$H (e^{j ω_{0}})$	in polar form	Output amplitude
the constant 10, which is the $ω = 0$ case	$0 π$	2	2 angle 0 deg	20
-5 $\sin (π n / 2)$ , taken at $ω = \frac{π}{2}$	$0.5 π$	0.8 - j0.4	0.8944 angle -26.57 deg	4.4721
20 $\cos (π n)$ , taken at $ω = π$	$1 π$	0.6667	0.6667 angle 0 deg	13.3333

Where	By convolution	By the formula
$n = 0$	35.3333	35.3333
$n = 1$	2.6667	2.6667
$n = 2$	31.3333	31.3333
$n = 5$	2.6667	2.6667

Same as 2082 Baishakh Q22072 Kartik · Q44 marksFind the output of LTI System having impulse response

h [n] = {(1 / 2)}^{n} u [n]

and input signal

x [n] = 5 e^{j π n / 3}

for -infinity < n < infinity.

What is different in this one

1 component at one frequency, so H is evaluated once and the results added: superposition.
Do not convolve. The input runs over all n, so it is not a finite sequence and the graphical method is the wrong tool.

x [n] = e^{j ω_{0} n} ⟶ y [n] = H (e^{j ω_{0}}) e^{j ω_{0} n}, A \cos (ω_{0} n) ⟶ A | H | \cos (ω_{0} n + ∠ H)

\begin{aligned} H (e^{j ω}) & = 1 / (1 - 0.5 e^{- j ω}), the  transform  of h [n] = {(1 / 2)}^{n} u [n] \end{aligned}

Component	$ω_{0}$	$H (e^{j ω_{0}})$	in polar form	Output amplitude
$x [n] = 5 e^{j π n / 3}$	$0.3333 π$	1 - j0.5774	1.1547 angle -30 deg	5.7735

Where	By convolution	By the formula
$n = 0$	5	5
$n = 1$	5	5
$n = 2$	0	0
$n = 5$	0	0

$y [n] = 5.7735 e^{j (0.3333 π n - 30 d e g)}$ .

h [n] = {(1 / 2)}^{n} u [n]

. Determine

y [n]

, if the input is

x [n] = A e^{j π n}

.What is different hereAt

ω = π

the exponential

e^{- j ω}

is exactly -1, so H is

1 / (1 + 0.5) = 2 / 3

with no complex arithmetic at all. The answer is

y [n]

= (2A/3)

e^{j π n}

What is different in this one

1 component at one frequency, so H is evaluated once and the results added: superposition.
Do not convolve. The input runs over all n, so it is not a finite sequence and the graphical method is the wrong tool.

x [n] = e^{j ω_{0} n} ⟶ y [n] = H (e^{j ω_{0}}) e^{j ω_{0} n}, A \cos (ω_{0} n) ⟶ A | H | \cos (ω_{0} n + ∠ H)

\begin{aligned} H (e^{j ω}) & = 1 / (1 - 0.5 e^{- j ω}), the  transform  of h [n] = {(1 / 2)}^{n} u [n] \end{aligned}

Component	$ω_{0}$	$H (e^{j ω_{0}})$	in polar form	Output amplitude
$x [n] = A e^{j π n}$ , with A factored out	$1 π$	0.6667	0.6667 angle 0 deg	0.6667

Where	By convolution	By the formula
$n = 0$	0.6667	0.6667
$n = 1$	-0.6667	-0.6667
$n = 2$	0.6667	0.6667
$n = 5$	-0.6667	-0.6667

$y [n] = 0.6667 e^{j (π n)}$ .

Energy signal or power signal PIN 3/19

Ch 1 · Signals and systems3 questions, from 3 of the 19 sittings3 to practise

The procedure, the same for every question in this topic

\begin{aligned} E & = \sum_{n = - \infty}^{\infty} | x [n] |^{2} & the energy \\ P & = lim_{N \to \infty} \frac{1}{2 N + 1} \sum_{n = - N}^{N} | x [n] |^{2} & the average power \end{aligned}

\begin{aligned} energy signal & : 0 < E < \infty, and then P = 0 \\ power signal & : 0 < P < \infty, and then E = \infty \end{aligned}

Try the energy first. If the sum converges you are done and the power is zero. If it diverges, form the power.
A periodic signal is always a power signal, and the limit collapses to the average over one period.
A complex exponential needs no calculation: | $e^{j θ}$ | = 1 at every n, so E is infinite and P = 1.

The checks that cost nothing

A signal is never both, and it can be neither: n $u [n]$ has infinite energy and infinite power.
P for $u [n]$ is $\frac{1}{2}$ , not 1. The window holds $2 N + 1$ samples of which only $N + 1$ are non zero.
Quote the units: a voltage signal gives energy in $V^{2}$ s and power in $V^{2}$ .

Practise this2082 Bhadra · Q12+2 marksCompare between energy signal and power signal. Determine whether the signal

x [n] = e^{j (π / 2 n + 4 π / 7)}

What is different in this one

Try the energy first. If the sum settles the signal is an energy signal and its power is zero; if it grows without bound, form the power instead. Every question in this topic is that one decision.

Form the energy first. If it settles on a finite value the signal is an energy signal and its power is zero; if it grows without bound, form the power instead.

\begin{aligned} inf                                 1        N \\ E & = \sum | x [n] |^{2} P = lim - - - - - - \sum | x [n] |^{2} \\ n & = - inf N \to inf 2 N + 1 n = - N \end{aligned}

Carried out over windows of growing width, with $x [n] = e^{j (π n / 2 + 4 π / 7)}$ :

N	E over -N to N	$P = E / (2 N + 1)$
50	101	1
200	401	1
1000	2001	1
5000	10001	1

$x [n] = e^{j (π n / 2 + 4 π / 7)}$ is a power signal: the energy grows without bound while the power settles at 1.

Practise this2079 Baishakh · Q12+2 marksCompare between energy signal and power signal. Determine whether the signal

x [n] = e^{j (π / 2 n + 4 π / 7)}

is energy signal or power signal.

What is different in this one

Try the energy first. If the sum settles the signal is an energy signal and its power is zero; if it grows without bound, form the power instead. Every question in this topic is that one decision.

Form the energy first. If it settles on a finite value the signal is an energy signal and its power is zero; if it grows without bound, form the power instead.

\begin{aligned} inf                                 1        N \\ E & = \sum | x [n] |^{2} P = lim - - - - - - \sum | x [n] |^{2} \\ n & = - inf N \to inf 2 N + 1 n = - N \end{aligned}

Carried out over windows of growing width, with $x [n] = e^{j (π n / 2 + 4 π / 7)}$ :

N	E over -N to N	$P = E / (2 N + 1)$
50	101	1
200	401	1
1000	2001	1
5000	10001	1

$x [n] = e^{j (π n / 2 + 4 π / 7)}$ is a power signal: the energy grows without bound while the power settles at 1.

Practise this2072 Kartik · Q12+3 marksDefine energy and power signal. Check the signal

x [n] = u [n]

and

x [n] = δ [n]

is Energy or Power type.

a) $x [n] = u [n]$

What is different in this one

Try the energy first. If the sum settles the signal is an energy signal and its power is zero; if it grows without bound, form the power instead. Every question in this topic is that one decision.

Form the energy first. If it settles on a finite value the signal is an energy signal and its power is zero; if it grows without bound, form the power instead.

\begin{aligned} inf                                 1        N \\ E & = \sum | x [n] |^{2} P = lim - - - - - - \sum | x [n] |^{2} \\ n & = - inf N \to inf 2 N + 1 n = - N \end{aligned}

Carried out over windows of growing width, with $x [n] = u [n]$ :

N	E over -N to N	$P = E / (2 N + 1)$
50	51	0.505
200	201	0.5012
1000	1001	0.5002
5000	5001	0.5

$x [n] = u [n]$ is a power signal: the energy grows without bound while the power settles at 0.5.

b) $x [n] = δ [n]$

What is different in this one

Try the energy first. If the sum settles the signal is an energy signal and its power is zero; if it grows without bound, form the power instead. Every question in this topic is that one decision.

Form the energy first. If it settles on a finite value the signal is an energy signal and its power is zero; if it grows without bound, form the power instead.

\begin{aligned} inf                                 1        N \\ E & = \sum | x [n] |^{2} P = lim - - - - - - \sum | x [n] |^{2} \\ n & = - inf N \to inf 2 N + 1 n = - N \end{aligned}

Carried out over windows of growing width, with $x [n] = δ [n]$ :

N	E over -N to N	$P = E / (2 N + 1)$
50	1	0.0099
200	1	0.0025
1000	1	0.0005
5000	1	0.0001

$x [n] = δ [n]$ is a energy signal: E settles at 1 and the power falls to zero.

Cascade form PIN 3/19

Ch 4 · Filter structures3 questions, from 3 of the 19 sittings1 to practise

The procedure, the same for every question in this topic

\begin{aligned} cascade & H (z) = b_{0} \prod_{k} H_{k} (z) & each a real biquad \\ parallel & H (z) = C + \sum_{k} H_{k} (z) & from a partial fraction expansion \end{aligned}

H_{k} (z) = \frac{1 + b_{1 k} z^{- 1} + b_{2 k} z^{- 2}}{1 + a_{1 k} z^{- 1} + a_{2 k} z^{- 2}}

A conjugate pair multiplies out to a real quadratic, which is the whole reason the sections can be built at all:

(1 - r e^{j θ} z^{- 1}) (1 - r e^{- j θ} z^{- 1}) = 1 - 2 r \cos θ z^{- 1} + r^{2} z^{- 2}

The one hard rule: a complex root must sit in the same section as its conjugate, or the section has complex coefficients and cannot be built.
Why not one long direct form: quantizing the coefficients of a high order polynomial moves its roots a long way, and a pole can leave the unit circle. Sections do not suffer that.
In practice pair each pole with its nearest zero, and put the poles closest to the unit circle last.

The checks that cost nothing

Every section must have real coefficients. A complex number left in a section means a conjugate was separated from its partner.
Multiply the sections back out and the original $H (z)$ must return.
Count the sections: half the order, rounded up.

Practise this +2 like it2081 Baishakh · Q5a5 marksDraw the cascaded form structure of

H (z)

= 10(1

- 0.25 z^{- 1}

)(1

- 0.667 z^{- 1}

)(1 +

2 z^{- 1}

)/((1

- 0.75 z^{- 1}

)(1

- 0.125 z^{- 1}

){

1 - (0.5 + j 0.5) z^{- 1}

}{

1 - (0.5 - j 0.5) z^{- 1}

})

What is different in this one

3 zeros and 4 poles, so 2 second order sections in the chain.
2 of the roots are complex, so each must stay with its conjugate: that pairing is the one hard rule, and a section with a complex coefficient left in it cannot be built.
There is an overall gain of 10 to place, and it is usually split across the sections rather than all put at the front, so that no section overflows.

\begin{aligned} (1 - r e^{j θ} z^{- 1}) (1 - r e^{- j θ} z^{- 1}) & = 1 - 2 r \cos (θ) z^{- 1} + r^{2} z^{- 2} \end{aligned}

Section	Numerator	Denominator
H1(z)	$1 - 0.917 z^{- 1} + 0.1668 z^{- 2}$	$1 - 0.875 z^{- 1} + 0.0938 z^{- 2}$
H2(z)	$1 + 2 z^{- 1}$	$1 - z^{- 1} + 0.5 z^{- 2}$

Section 1 of 2, in direct form II

Direct form II: one delay chain shared by both halves, 2 delays, which is the fewest possible.

Section 2 of 2, in direct form II

Direct form II: one delay chain shared by both halves, 2 delays, which is the fewest possible.

The chain. The input enters the gain, the gain feeds section 1, and each section feeds the next. Nothing is summed between them: that is the parallel form, not this one.

\begin{aligned} x [n] - \to [10] - \to [section 1] - \to [section 2] - \to y [n] \end{aligned}

Same as 2081 Baishakh Q5a2080 Baishakh · Q64 marksRealize the given system in Cascade Form of 2nd order section flow graph representation.

H (z) = \frac{(1 - 0.4 z^{- 1}) (1 + 0.2 z^{- 1}) (1 - 0.3 e^{j π / 6} z^{- 1}) (1 - 0.3 e^{- j π / 6} z^{- 1})}{(1 - 0.5 e^{j π / 3} z^{- 1}) (1 - 0.5 e^{- j π / 3} z^{- 1}) (1 + 0.7 e^{j π / 4} z^{- 1}) (1 + 0.7 e^{- j π / 4} z^{- 1})}

What is different in this one

4 zeros and 4 poles, so 2 second order sections in the chain.
6 of the roots are complex, so each must stay with its conjugate: that pairing is the one hard rule, and a section with a complex coefficient left in it cannot be built.

\begin{aligned} (1 - r e^{j θ} z^{- 1}) (1 - r e^{- j θ} z^{- 1}) & = 1 - 2 r \cos (θ) z^{- 1} + r^{2} z^{- 2} \end{aligned}

Section	Numerator	Denominator
H1(z)	$1 - 0.2 z^{- 1} - 0.08 z^{- 2}$	$1 - 0.5 z^{- 1} + 0.25 z^{- 2}$
H2(z)	$1 - 0.5196 z^{- 1} + 0.09 z^{- 2}$	$1 + 0.9899 z^{- 1} + 0.49 z^{- 2}$

Section 1 of 2, in direct form II

Direct form II: one delay chain shared by both halves, 2 delays, which is the fewest possible.

Section 2 of 2, in direct form II

Direct form II: one delay chain shared by both halves, 2 delays, which is the fewest possible.

The chain. The input enters the gain, the gain feeds section 1, and each section feeds the next. Nothing is summed between them: that is the parallel form, not this one.

\begin{aligned} x [n] - \to [1] - \to [section 1] - \to [section 2] - \to y [n] \end{aligned}

Same as 2081 Baishakh Q5a2076 Chaitra · Q74 marksRealize the given system in Cascade form of 2nd order section in signal flow graph representation.

H (z) = \frac{(1 - 0.5 z^{- 1}) (1 + 0.35 z^{- 1}) (1 - 0.3 e^{j 2 n π / 5} z^{- 1}) (1 - 0.3 e^{- j 2 n π / 5} z^{- 1})}{(1 - 0.6 e^{j n π / 3} z^{- 1}) (1 - 0.6 e^{- j n π / 3} z^{- 1}) (1 + 0.5 e^{j 2 n π / 7} z^{- 1}) (1 + 0.5 e^{- j 2 n π / 7} z^{- 1})}

How this is readThe n in the printed exponents $e^{j 2 n π / 5}$ is a scanning artifact: these are fixed angles, $\frac{2 π}{5}$ , $\frac{π}{3}$ and $\frac{2 π}{7}$ .

What is different in this one

4 zeros and 4 poles, so 2 second order sections in the chain.
6 of the roots are complex, so each must stay with its conjugate: that pairing is the one hard rule, and a section with a complex coefficient left in it cannot be built.

\begin{aligned} (1 - r e^{j θ} z^{- 1}) (1 - r e^{- j θ} z^{- 1}) & = 1 - 2 r \cos (θ) z^{- 1} + r^{2} z^{- 2} \end{aligned}

Section	Numerator	Denominator
H1(z)	$1 - 0.15 z^{- 1} - 0.175 z^{- 2}$	$1 - 0.6 z^{- 1} + 0.36 z^{- 2}$
H2(z)	$1 - 0.1854 z^{- 1} + 0.09 z^{- 2}$	$1 + 0.6235 z^{- 1} + 0.25 z^{- 2}$

Section 1 of 2, in direct form II

Direct form II: one delay chain shared by both halves, 2 delays, which is the fewest possible.

Section 2 of 2, in direct form II

Direct form II: one delay chain shared by both halves, 2 delays, which is the fewest possible.

The chain. The input enters the gain, the gain feeds section 1, and each section feeds the next. Nothing is summed between them: that is the parallel form, not this one.

\begin{aligned} x [n] - \to [1] - \to [section 1] - \to [section 2] - \to y [n] \end{aligned}

Plotting a shifted, folded or scaled sequence PIN 2/19

Ch 1 · Signals and systems2 questions, from 2 of the 19 sittings1 to practise

The procedure, the same for every question in this topic

Tabulate the original. Write n against $x [n]$ for every n where $x [n]$ is non zero, and mark where $n = 0$ is. Everything that follows is index arithmetic on that table, and the table is what stops you guessing.
Find which n survive. The new sequence is $y [n] = x [a n + b]$ . It is non zero only while an + b lands inside the range where x is non zero, so solve that at both ends.
$n_{low} \leq \frac{k_{low} - b}{a}, n_{high} \geq \frac{k_{high} - b}{a} (the inequalities swap when a < 0)$
Evaluate sample by sample. For each surviving n compute an + b and read x at that index. Do not try to shift and fold the picture in your head: the order of the operations matters and the arithmetic does not care about the order.
Mark the new origin and plot. The answer is a sequence AND its position, so an unmarked list of numbers is only half an answer.

The checks that cost nothing

Count the samples. Folding and shifting keep the count; decimation by a throws away roughly a - 1 of every a, and interpolation multiplies it. If your count is wrong the index arithmetic is wrong.
Check one sample by hand against the original table, preferably the one at the new $n = 0$ . It catches a sign error in b immediately.
A fold reverses the order. If the answer reads in the same direction as the original and there was a minus in front of n, you have folded nothing.

Practise this +1 like it2076 Chaitra · Q12+3 marksDefine even and odd type discrete time signals with suitable example. Plot the signal

x [- 2 n + 3]

where

x [n] = {1, 2, 0, - 1, - 3, - 4}

What is different in this one

The answer runs from $n = - 1$ to 1, 3 samples, and it does not start at the origin, so the $n = 0$ sample must be marked.
Work the index arithmetic, never the picture: solve for the n that put the argument inside the range where the original is non zero, then evaluate sample by sample.

Work the index arithmetic, never the picture. x is non zero for $0 \leq - 2 n + 3 \leq 5$ , that is $- 1 \leq n \leq 1.5$ , so only $n = - 1$ , 0, 1 survive.

\begin{aligned} n & = - 1 \to x [- 2 (- 1) + 3] = x [5] = - 4 \\ n & = 0 \to x [3] = - 1 \\ n & = 1 \to x [1] = 2 \\ n & = 2 \to x [- 1], outside  the  sequence, so  nothing \end{aligned}

y [n] = x [- 2 n + 3]

, drawn from the computed samples.

$y [n] = x [- 2 n + 3]$ = {-4, -1, 2} for $n = - 1$ to 1. Folding and decimating by 2 keeps only the odd indexed samples of the original, which is why three of the six survive.

Same as 2076 Chaitra Q12074 Chaitra · Q13+2 marksPlot the sequence

x [n] = u [n] - u [n - 3] + 5 δ [n - 4] = n u [n - 6]

. List out the properties of LTI system.

What is different in this one

The answer runs from $n = 0$ to 8, 9 samples, and it starts at the origin.
Work the index arithmetic, never the picture: solve for the n that put the argument inside the range where the original is non zero, then evaluate sample by sample.

Draw each piece on the same axis and add. Note the printed equation uses an equals sign where the last plus belongs.

\begin{aligned} u [n] - u [n - 3] a rectangle  of  height 1 over n & = 0, 1, 2 \\ 5 δ [n - 4] a single  sample  of  height 5 at n & = 4 \\ n u [n - 6] a ramp  starting  at n & = 6, so 6, 7, 8 \dots \end{aligned}

x [n]

, drawn from the computed samples.

$x [n]$ = {1, 1, 1, 0, 5, 0, 6, 7, 8} for $n = 0$ to 8. The three pieces never overlap, so no sample is the sum of two of them.

Locating the region of convergence PIN 2/19

Ch 2 · The z-transform2 questions, from 2 of the 19 sittings2 to practise

The procedure, the same for every question in this topic

Transform each term with its own ROC. A term does not have one ROC, it has the ROC that makes its own series converge.
$\begin{aligned} a^{n} u [n] & ⟶ \frac{z}{z - a}, | z | > | a | & right sided \\ - b^{n} u [- n - 1] & ⟶ \frac{z}{z - b}, | z | < | b | & left sided \end{aligned}$
Mind the sign. The standard pair carries a minus on the left sided term, so a plus in the question picks up a minus in the transform.
Intersect the regions. The ROC of the whole transform is where every term converges at once, and an empty intersection means there is no z-transform at all.

The intersection can be empty, and when it is, the correct answer is that the z-transform does not exist. That is a complete answer, not a failure.
A ring exists only if the right sided pole is the inner one.

The checks that cost nothing

Does it contain the unit circle? If yes the signal is absolutely summable and has a DTFT; if no it does not. One line, and it is usually a mark.
An ROC never contains a pole, so if your region does, you have made an error or the question has a misprint.

Practise this2075 Chaitra · Q64+6 marksList out the properties of Region of convergence and locate the ROC of the following signal

x [n] = {(0.1)}^{n} u [n] + {(0.3)}^{n} u [- n - 1]

What is different in this one

2 terms, 1 right sided and 1 left sided, so the ROC will be a ring if it exists at all.
Watch the sign on the left sided term: the standard pair is $- b^{n} u [- n - 1]$ , so a plus in the question picks up a minus in the transform.

Transform each term with its own ROC, then intersect them.

Term	Transform	Its own ROC
${(0.1)}^{n} u [n]$	$z / (z - 0.1)$	$\| z \| > 0.1$
${(0.3)}^{n} u [- n - 1]$	$- z / (z - 0.3)$	$\| z \| < 0.3$

Practise this2075 Ashwin · Q33+6 marksList out the properties of Region of convergence and locate the ROC of the following signal.

x [n] = {(0.6)}^{n} u [n] + {(0.25)}^{n} u [n]

What is different in this one

2 terms, 2 right sided and 0 left sided, so the ROC is simply the outside of the outermost pole.
Both terms are right sided, so no sign trap here.

Transform each term with its own ROC, then intersect them.

Term	Transform	Its own ROC
${(0.6)}^{n} u [n]$	$z / (z - 0.6)$	$\| z \| > 0.6$
${(0.25)}^{n} u [n]$	$z / (z - 0.25)$	$\| z \| > 0.25$

From lattice coefficients back to $H (z)$ PIN 2/19

Ch 4 · Filter structures2 questions, from 2 of the 19 sittings1 to practise

The procedure, the same for every question in this topic

A_{m} (z) = A_{m - 1} (z) + K_{m} z^{- m} A_{m - 1} (z^{- 1})

where $A_{m - 1} (z^{- 1})$ is the same coefficient list reversed. Start from $A_{0} (z) = 1$ and apply it once per K, in the order $K_{1}, K_{2}, K_{3}$ .

What the question asks	What $H (z)$ is
All zero	$H (z) = A_{M} (z)$ , and the FIR coefficients are its own.
All pole	$H (z) = 1 / A_{M} (z)$ , and there are no FIR coefficients at all.

Read which of the two the question wants. The arithmetic up to AM(z) is identical, and the two wordings alternate between sittings.

The checks that cost nothing

Step back down from the answer and the original K values must return.
The first coefficient stays 1 at every order.
The last coefficient of Am is $K_{m}$ , by construction: check it at each step.

Practise this +1 like it2079 Bhadra · Q66 marksGiven a 3-stage lattice filter for all zero polynomial with coefficients K1 = 1/4, K2 =

\frac{1}{2}

and K3 = 1/3. Obtain the system function and FIR filter coefficients of this filter.

What is different in this one

3 reflection coefficients, so the recursion runs 3 times and the answer is a polynomial of order 3.
Read the question for which it wants: all zero means $H (z) = A (z)$ and the FIR coefficients are its own; all pole means $H (z) = 1 / A (z)$ and there are no FIR coefficients to give. The arithmetic is identical either way.

Run the order update once per reflection coefficient, starting from A0(z) = 1.

\begin{aligned} Am (z) & = A (m - 1) (z) + K_{m} z^{- m} A (m - 1) (z^{- 1}) \\ where A (m - 1) (z^{- 1}) is  the  same  coefficient  list  REVERSED . \end{aligned}

Order	Polynomial
A0(z)	1
A1(z)	$1 + 0.25 z^{- 1}$
A2(z)	$1 + 0.375 z^{- 1} + 0.5 z^{- 2}$
A3(z)	$1 + 0.5417 z^{- 1} + 0.625 z^{- 2} + 0.3333 z^{- 3}$

Same as 2079 Bhadra Q62071 Shrawan · Q45 marksIf a 3 stage lattice filter for all pole polynomial has coefficients K1 = 1/4, K2 =

\frac{1}{2}

and K3 = 1/3. Obtain the system function of this filter.

What is different in this one

3 reflection coefficients, so the recursion runs 3 times and the answer is a polynomial of order 3.
Read the question for which it wants: all zero means $H (z) = A (z)$ and the FIR coefficients are its own; all pole means $H (z) = 1 / A (z)$ and there are no FIR coefficients to give. The arithmetic is identical either way.

Run the order update once per reflection coefficient, starting from A0(z) = 1.

\begin{aligned} Am (z) & = A (m - 1) (z) + K_{m} z^{- m} A (m - 1) (z^{- 1}) \\ where A (m - 1) (z^{- 1}) is  the  same  coefficient  list  REVERSED . \end{aligned}

Order	Polynomial
A0(z)	1
A1(z)	$1 + 0.25 z^{- 1}$
A2(z)	$1 + 0.375 z^{- 1} + 0.5 z^{- 2}$
A3(z)	$1 + 0.5417 z^{- 1} + 0.625 z^{- 2} + 0.3333 z^{- 3}$

Even and odd parts PIN 1/19

Ch 1 · Signals and systems1 questions, from 1 of the 19 sittings1 to practise

The procedure, the same for every question in this topic

\begin{aligned} x_{e} [n] & = \frac{x [n] + x [- n]}{2} & the even part, symmetric about n = 0 \\ x_{o} [n] & = \frac{x [n] - x [- n]}{2} & the odd part, antisymmetric \\ x [n] & = x_{e} [n] + x_{o} [n] & and they add back, which is the check \end{aligned}

Tabulate. Write n across, $x [n]$ under it, then $x [- n]$ by reversing about $n = 0$ , not about the first entry. That reversal is the whole question.

The checks that cost nothing

$x_{o} [0] = 0$ , always, since an odd signal must vanish at the origin.
xe is symmetric about $n = 0$ .
They add back to $x [n]$ at every n.

Practise this2071 Shrawan · Q14+5 marksFind the odd and even part of the following signal: [a stem plot of

x [n]

with values 1 at

n = - 4

, -3, -2, -1 and 0, and 2 at

n = 1

, 2, 3, 4]. A discrete time LTI system has input signal and impulse response as,

x [n] = 1

for

- 1 \leq n \leq 1

and 0 elsewhere, and

h [n] = 1

for

- 1 \leq n \leq 1

and 0 elsewhere. Find the output of the system using graphical method.

How this is readThe stem plot is $x [n] = 1$ for $n = - 4$ to 0 and 2 for $n = 1$ to 4. The second half convolves two identical three sample rectangles, each from $n = - 1$ to 1.

The even and odd parts

Tabulate $x [n]$ against n, write $x [- n]$ under it by reversing about $n = 0$ , then halve the sum and the difference.

n	-4	-3	-2	-1	0	1	2	3	4
$x [n]$	1	1	1	1	1	2	2	2	2
$x [- n]$	2	2	2	2	1	1	1	1	1
$x_{e} [n]$	1.5	1.5	1.5	1.5	1	1.5	1.5	1.5	1.5
$x_{o} [n]$	-0.5	-0.5	-0.5	-0.5	0	0.5	0.5	0.5	0.5

The rest of this questionThis topic works the part that belongs to it. The other part is worked under Convolution (The output, by the graphical method), where it belongs.

Checking a system for BIBO stability PIN 1/19

Ch 3 · Frequency domain1 questions, from 1 of the 19 sittings1 to practise

The procedure, the same for every question in this topic

Get $h [n]$ from the difference equation. Put $x [n] = δ [n]$ and run the recursion forwards until the pattern is clear. For a first order system it is a geometric sequence after the first sample.
Write the absolute sum. Stability is one condition and this is it.
$S = \sum_{n = - \infty}^{\infty} | h [n] | < \infty ⟺ BIBO stable$
Recognise the series. Almost every one of these is geometric, and a geometric series converges when, and only when, its ratio is below one in magnitude.
$\sum_{n = 0}^{\infty} r^{n} = \frac{1}{1 - r} for | r | < 1, divergent otherwise$
Say the condition, not just yes or no. When the coefficient is a symbol the answer IS the condition on it, and the marks are for stating it and for saying what happens at the boundary.
Then say it in the other two languages, which costs one line and shows you know they are the same statement: the pole lies inside the unit circle, and the region of convergence contains the unit circle.

The checks that cost nothing

The boundary case is not stable. At $| r | = 1$ the terms never shrink, the sum grows without bound, and the system is an accumulator. Examiners set this deliberately.
A bounded input is not enough on its own. BIBO asks for EVERY bounded input to give a bounded output, so one example proves nothing; the absolute sum proves it for all of them at once.
Check the pole against the circle as a second opinion. If the two answers disagree you have made an arithmetic slip, not a discovery.

Practise this2081 Baishakh · Q15 marksA discrete-time LTI system is given by difference equation

y (n) = x (n) + e^{a} y (n - 1)

. Check this system for BIBO stability.

What is different in this one

The coefficient is a symbol, not a number, so the answer is a condition on it rather than a yes or a no.
Every other stability question on the paper hands you numbers. The test is identical; only the last line changes.

Find $h [n]$ first. Put $x [n] = δ [n]$ and run the recursion: $h [0] = 1$ , and each later sample is $e^{a}$ times the one before it.

\begin{aligned} h [0] & = 1 \\ h [1] & = e^{a} \\ h [2] & = {(e^{a})}^{2} \\ h [n] & = {(e^{a})}^{n} u [n] \end{aligned}

Now the BIBO test, which is the same one every time: the system is stable when the impulse response is absolutely summable.

\begin{aligned} S & = \sum | h [n] | for n = 0 to  infinity \\ = \sum | e^{a} |^n \\ a geometric  series  with  ratio | e^{a} | \end{aligned}

A geometric series converges when, and only when, its ratio is below one in magnitude, so the whole question is that one comparison.

a	\| $e^{a}$ \|	the sum
$a = - 1$	0.3679	stable, $S = 1.582$
$a = - 0.1$	0.9048	stable, $S = 10.5083$
$a = 0$	1	not stable, the terms never shrink
$a = 1$	2.7183	not stable, the terms grow

The zero input response PIN 1/19

Ch 3 · Frequency domain1 questions, from 1 of the 19 sittings1 to practise

The procedure, the same for every question in this topic

Set the input to zero. $x [n] = 0$ , which leaves the homogeneous equation: the system running on its stored energy alone.
Assume a geometric solution and divide through by the lowest power.
$y [n] = λ^{n} ⟹ λ^{N} + a_{1} λ^{N - 1} + \dots + a_{N} = 0$
That is the characteristic equation, and its roots are the poles of $H (z)$ : a root outside the unit circle is an unstable system.
Solve it, then write the matching form. Which of the three you get is decided by the roots, and using the wrong one is the usual lost mark.
$\begin{aligned} distinct real & y [n] = C_{1} λ_{1}^{n} + C_{2} λ_{2}^{n} \\ repeated & y [n] = (C_{1} + C_{2} n) λ^{n} \\ complex pair & y [n] = r^{n} (C_{1} \cos θ n + C_{2} \sin θ n) \end{aligned}$
Fit the constants to the initial conditions the question gives, usually $y [- 1]$ and $y [- 2]$ .

The checks that cost nothing

Read the stability off the roots. A root outside the unit circle means the natural response grows without bound. One sentence, one mark.
Substitute a root back into the characteristic equation.
The total response is this plus the zero state response; say which one the question asked for.

Practise this2078 Bhadra · Q44 marksDetermine the zero-input response for a second order system given by:

y [n] - 3 y [n - 1] - 4 y [n - 2] = x [n]

What is different in this one

An order 2 system, so the characteristic equation is a quadratic and there are 2 constants to fit to the initial conditions.
The roots are real and distinct, so the response is a plain sum of two exponentials.

1. The characteristic equation. Set the input to zero and assume $y [n] = λ^{n}$ , which is the one shape that survives a constant coefficient recursion. Divide through by $λ^{n - N}$ .

\begin{aligned} with x [n] & = 0 : 1 - 3 λ^- 1 - 4 λ^- 2 = 0 \\ multiply  through  by λ^{2} : \\ λ^{2} - 3 λ - 4 & = 0 \end{aligned}

2. Solve it. Factorise, or use the quadratic formula.

λ

= ( 3 +/-

\sqrt{< m > {(- 3)}^{2}}

- 4(-4) } ) / 2 = ( 3 +/-

\sqrt{25}

) / 2 = ( 3 +/- 5 ) / 2

λ = 4

, -1

3. The form of the response. One term per root, each with its own constant.

\begin{aligned} y [n] & = C_{1} {(4)}^{n} + C_{2} {(- 1)}^{n} \end{aligned}

\begin{aligned} y [- 2] & = {(4)}^{- 2} C_{1} + {(- 1)}^{- 2} C_{2} \\ = 0.0625 C_{1} + 1 C_{2} \\ y [- 1] & = {(4)}^{- 1} C_{1} + {(- 1)}^{- 1} C_{2} \\ = 0.25 C_{1} - 1 C_{2} \\ with y [- 2] & = 0, y [- 1] = 1 \end{aligned}

\begin{aligned} solving : C_{1} & = 3.2, C_{2} = - 0.2 \end{aligned}

Fixed point number formats PIN 1/19

Ch 4 · Filter structures1 questions, from 1 of the 19 sittings1 to practise

The procedure, the same for every question in this topic

Format	How a negative number is written
Sign magnitude	A sign bit, and the magnitude unchanged.
One's complement	Complement every bit of the positive form.
Two's complement	Complement every bit, then add 1 in the last place.

With the binary point straight after the sign bit the weights run $0.5, 0.25, 0.125 \dots$ , so

\frac{5}{8} = 0.625 = {0.101}_{2}

The checks that cost nothing

Read the two's complement back with the sign bit weighted -1: it must give the negative number you started from.
All three agree on the positive value and differ only in the negative one.

Also worked under The IIR lattice ladderPractise this2075 Ashwin · Q57+3 marksDraw the Lattice structure from the following system function:

1 / (3 + (39 / 24) Z^{- 1} + (15 / 8) Z^{- 2} + (3 / 9) Z^{- 3})

. And represent 5/8 and -5/8 in sign magnitude, 1's complement and 2's complement format.

5/8 and -5/8 in the three formats

With the binary point just after the sign bit the weights are 0.5, 0.25, 0.125 and so on, so 0.625 = 0.101 in binary.

Format	+0.625	-0.625
Sign magnitude	0.101	1.101, only the sign bit changes
One's complement	0.101	1.010, every bit flipped
Two's complement	0.101	1.011, flipped then plus one in the last place

Check the two's complement by reading the sign bit with weight -1: 1.011 = -1 + 0.375 = -0.625, as required.

The rest of this questionThis topic works the part that belongs to it. The other part is worked under The IIR lattice ladder (The lattice structure), where it belongs.

Linear convolution through the DFT PIN 1/19

Ch 7 · DFT and FFT1 questions, from 1 of the 19 sittings1 to practise

The procedure, the same for every question in this topic

x has $N_{1}$ samples and h has $N_{2}$ . The linear convolution of the two is longer than either, and that length is the whole point.

\begin{aligned} linear convolution has & N_{1} + N_{2} - 1 samples \\ pad BOTH to N \geq & N_{1} + N_{2} - 1 & and the circular one EQUALS it \end{aligned}

Without the padding the tail wraps round and adds to the front, which is time domain aliasing, and the first few samples come out wrong.

The checks that cost nothing

Show the unpadded answer too if there is time: the difference is exactly the wrapped tail, and seeing it is the point of the question.
The sums still multiply, padded or not.

Practise this2074 Chaitra · Q101+5 marksDefine zero padding. Find the linear convolution through circular convolution with padding of zeros for the following sequences:

x [n] = {1, 1, 1, 1}

and

h [n] {2, 3}

What is different in this one

x has 4 samples and h has 2, so the linear convolution has 5 and the DFT length must be at least that. Pad both to N = 5.
At the unpadded length of 4 the answer would be wrong: the last 1 samples wrap round and add to the front. Showing that is the point of the question.

x has 4 samples and h has 2, so the linear convolution has 4 + 2 - 1 = 5 samples and the DFT length must be at least that. Pad both sequences to N = 5.

Now do the circular convolution at that padded length. This is where the numbers come from, and at $N = 5$ it is the linear answer exactly.

The DFT length is $N = 5$ . The circular convolution is $y [n] = \sum_{m} x_{1} [m] x_{2}$ [(n - m) mod N], and the wrap is what makes it circular rather than linear.

Write it as a circulant matrix: each row is the previous row of $x_{2}$ rotated by one, and each row is multiplied by the $x_{1}$ values in the heading.

	$x_{1}$ [0]=1	$x_{1}$ [1]=1	$x_{1}$ [2]=1	$x_{1}$ [3]=1	$x_{1}$ [4]=0	sum
$y [0]$	2	0	0	0	3	2
$y [1]$	3	2	0	0	0	5
$y [2]$	0	3	2	0	0	5
$y [3]$	0	0	3	2	0	5
$y [4]$	0	0	0	3	2	3

y [0]

= 1(2) = 2

y [1]

= 1(3) + 1(2) = 5

y [2]

= 1(3) + 1(2) = 5

y [3]

= 1(3) + 1(2) = 5

y [4]

= 1(3) = 3

$y [n] = {2, 5, 5, 5, 3}$ . The check holds: 4 × 5 = 20, the sum of y.

What the padding bought. Do the same convolution at the unpadded length of 4 and compare, sample by sample: the tail has nowhere to go and lands on the front.

Sample	Linear answer	At $N = 4$	What happened
$n = 0$	2	5	wrapped: $y [0] + y [4] = 2$ + 3
$n = 1$	5	5	unchanged
$n = 2$	5	5	unchanged
$n = 3$	5	5	unchanged
$n = 4$	3	not there	lost off the end

$y [n] = {2, 5, 5, 5, 3}$ , which is the linear convolution exactly.

Every question from 19 papers · 2071 Shrawan to 2082 Bhadra

The complete question bank

All 200 questions set on this subject, reproduced verbatim from the scanned papers. Read them by paper, newest first, or by chapter, where repeats are merged and counted. Every question is tagged with its chapter and linked to the card that answers it, and the 171 that carry a computation are marked, because every one of them is worked in full in the numerical solutions.

How to use the bank

By paper: work backwards from the newest. This paper repeats its own blueprint, so the last three sittings are the best guide to the next one.
By chapter: revise a chapter, then answer its questions. A question set in several sittings appears once, with how many times it was set and when, and every other wording listed under it.
Every link goes to the answer. A question with two links needs both cards for a full answer.
Numericals are tagged. The theory half of a mixed question is answered in the solutions section; the computation is not.

Regular2082 Bhadra

2082 Bhadra · Regular · BEI, BCT · 11 questions

Ch 1Numerical2+2Energy signals and power signals

Q1. Compare between energy signal and power signal. Determine whether the signal $x [n] = e^{j (π / 2 n + 4 π / 7)}$ is energy signal or power signal.

Ch 1Numerical5The convolution sum, and four ways to do it

Q2. Find the output of an LTI system whose impulse response is given by ${0.5}^{n} {u [n] - u [n - 3]}$ and input is given by ${2, 1, 0.5, - 1}$ .

Ch 2Numerical1+5The definition Inverting

X (z)

by long division

Q3. Define z-transform for a discrete time signal. Find the inverse z-transform for $H (z) = z / (3 z^{2} - 4 z + 1)$ using partial fraction method for $1 / 3 < | z | < 1$ .

Ch 3Numerical2+8Plotting poles and zeros and sketching the magnitud...

Q4. Draw the poles and zeros in the z-plane for the system with poles at 0.45 +/- j1.6 and zeros at 0.58 +/- j2.06. Also plot the magnitude response (not in scale) of the system.

Ch 4Numerical2+2FIR structures

Q5. Obtain the direct form I and direct form II realization of the following system: $y [n]$ - 0. $75 y [n - 1]$ - 0. $25 y [n - 2] = x [n]$ + 0. $5 x [n - 1]$

Ch 4Numerical6The IIR lattice ladder

Q6. Compute Lattice ladder coefficients and draw lattice structure for the given system $H (z) = (2 - 0.7 z^{- 1} + 0.5 z^{- 2}) / (1 - 0.3 z^{- 1} + 0.25 z^{- 2})$

Ch 5Numerical10The window method, step by step

Q7. Design a linear phase FIR system to meet the following specifications: Passband edge = 2 kHz, Stopband edge = 5 kHz, Stopband attenuation = 42 dB, Sampling frequency = 20 kHz

Ch 51+4The optimum filter and the Remez exchange algorithm

Q8. What is an optimal filter? Explain Remez exchange algorithm for FIR filter design with the help of a flow chart.

Ch 6Numerical3+12The design route

Q9. Explain frequency warping effect in detail. The passband and stopband frequencies are 350Hz and 1,000Hz respectively. The attenuation at passband and stopband are -3dB and -10dB respectively. The sampling frequency is 5,000Hz. Design a digital low-pass Butterworth filter using bilinear transformation technique.

Ch 7Numerical7Circular convolution, and how it differs from linear

Q10. Compute the circular convolution of the following sequences: $h (n) = {1, 2, 1, - 1, 1}$ and $x (n) = {1, 2, 3, 1}$

Ch 7Numerical8Decimation in frequency

Q11. Find 8-point DFT using DIF-FFT of the sequence $x [n]$ = {1, $\frac{1}{2}$ , -1, $- \frac{1}{2}$ , 2, -3/2}.

Back2082 Baishakh

2082 Baishakh · Back · BCT, BEI · 10 questions

Ch 1Numerical2+2The five system properties, and how to test each one

Q1. A discrete time system has input $x [n]$ and output $y [n]$ and its input-output relation is $y [n] = x [n] + x [- n]$ . Is the system causal, time invariant or not?

Ch 3Numerical5The frequency response, and why exponentials pass s...The convolution sum, and four ways to do it

Q2. Find the output of an LTI system having impulse response $h [n] = {(1 / 2)}^{n} u [n]$ excited by an input $x [n] = 5 e^{j π n / 3}$ .

Ch 2Numerical1+5The region of convergence, and its properties Inverting

X (z)

by long division

Q3. Define ROC of a Z-transform. Determine $x [n]$ for $X (z) = 1 / (1 - 1.5 z^{- 1} + 0.5 z^{- 2})$ if: a) $| z | > 1$ b) $| z | < 0.5$ c) $0.5 < | z | < 1$ using partial fraction method.

Ch 3Numerical3+7Plotting poles and zeros and sketching the magnitud...

Q4. The poles of a system are located at 0.45 +/- j1.6 and the zeros at 0.58 +/- j2.06. Map the poles and zeros in the z-plane and plot the magnitude and phase response (not to scale) of the system.

Ch 4Numerical10The IIR lattice ladder

Q5. A third order low pass filter has a system function $H (z) = (0.2759 + 0.5121 z^{- 1} + 0.5121 z^{- 2} + 0.2759 z^{- 3}) / (1 - 0.0010 z^{- 1} + 0.6546 z^{- 2} - 0.0775 z^{- 3})$ . Implement the filter using lattice-ladder structure.

Ch 5Numerical2+6Why FIR, and when to choose it The window method, step by step

Q6. Differentiate between analog and digital filter. Design an FIR lowpass digital filter that will have a -3dB cut-off at $30 π$ rad/s and an attenuation of 50 dB at $45 π$ rad/s. The filter is required to have a linear phase and the system uses a sampling rate of 100 Hz.

Ch 52+5Gibbs' phenomenon The optimum filter and the Remez exchange algorithm

Q7. Explain Gibbs phenomenon in FIR filter design. Describe Remez exchange algorithm for FIR filter design along with flowchart.

Ch 6Numerical12+3The design route Bilinear transformation against impulse invariance

Q8. Using bilinear transformation technique, design a Butterworth filter which satisfies the following conditions: $0.8 \leq | H (e^{j ω}) | \leq 1$ for $0 \leq w \leq 0.2 π$ ; $| H (e^{j ω}) | \leq 0.2$ for $0.6 π \leq ω \leq π$ . Consider sampling frequency 1 Hz. Compare Impulse Invariance Method and Bilinear Transformation technique for analog filter to digital filter conversion.

Ch 7Numerical1+2+6Why the FFT is fast Decimation in frequency

Q9. What is Fast Fourier Transform algorithm? How Decimation in Frequency Fast Fourier Transform algorithm reduces the calculation complexity of DFT? Find 4 point DFT of the sequence $x [n] = {1, 4, - 1}$ using Decimation in Frequency Fast Fourier Transform algorithm.

Ch 7Numerical6Circular convolution, and how it differs from linear

Q10. Find the circular convolution of the sequences $x_{1} [n] = {1, 1, - 1, - 1, 2}$ and $x_{2} [n] = {1, - 1, - 2}$ .

Regular2081 Bhadra

2081 Bhadra · Regular · BCT, BEI · 10 questions

Ch 1Numerical4Testing a signal for periodicity

Q1. Determine if the signal $x [n] = \cos (2 π n / 5) + \sin (π n / 3)$ is periodic or not. If the signal is periodic, find its fundamental period.

Ch 1Numerical5The convolution sum, and four ways to do it

Q2. Find the output of an LTI System with impulse response $h [n] = 2 δ [n + 1] + 2 δ [n - 1]$ when an input of $x [n] = δ [n] + 2 δ [n - 1] - δ [n - 3]$ is applied to it.

Ch 2Numerical1+5The region of convergence, and its properties Inverting

X (z)

by long division

Q3. Define ROC of z-transform. Find the inverse z-transform using partial fraction expansion of $X (z) = (2 z^{3} - 5 z^{2} + z + 3) / ((z - 1) (z - 2))$ for ROC $| z | < 1$

Ch 3Numerical3+7Plotting poles and zeros and sketching the magnitud...

Q4. Plot the pole zero in z plane and draw magnitude response (not to the scale) of the system described by difference equation $y [n] - 0.4 y [n - 1] + 0.25 y [n - 2] = x [n] - 0.4 x [n - 1]$

Ch 4Numerical6+4The IIR lattice ladder

Q5. Compute the lattice and ladder coefficients and draw lattice-ladder structure for the given IIR system $H (z) = (1 + z^{- 1} + z^{- 2}) / ((1 + 0.5 z^{- 1}) (1 + 0.3 z^{- 1}) (1 + 0.4 z^{- 1}))$

Ch 5Numerical8The window method, step by step

Q6. Design a low pass FIR filter using suitable window to meet following specifications: $0.99 \leq | H (e^{j ω}) | \leq 1.01$ for $0 \leq | w | \leq 0.3 π$ ; $| H (e^{j ω}) | \leq 0.01$ for $0.35 π$ ≤ | $ω$ | ≤ $π$

Ch 51+6The optimum filter and the Remez exchange algorithm

Q7. What do you mean by optimum filter? Describe the Remez exchange algorithm for FIR filter design along with the flowchart.

Ch 6Numerical12+3The design route Bilinear transformation against impulse invariance

Q8. Design a digital low pass IIR filter using Bilinear transformation to meet following specifications: $0.9 \leq | H (e^{j ω}) | \leq 1$ for $0 \leq | w | \leq \frac{π}{2}$ ; $| H (e^{j ω}) | \leq 0.2$ for $3 π / 4 \leq | w | \leq π$ . Use Butterworth approximation for design with sampling frequency of 1Hz. Compare Impulse Invariance Method and Bilinear Transformation method.

Ch 7Numerical2+7Why the FFT is fast Decimation in time

Q9. How does FFT reduce computational complexity compared to direct calculation of DFT? Compute 8-point DFT of $x [n] = {1, - 2, - 4, 3, 0, 1}$ using DIT-FFT algorithm.

Ch 7Numerical6Circular convolution, and how it differs from linear

Q10. If X1(k) and X2(k) are the 5-point DFT of $x_{1}$ (n) = $3^{n}$ for $0 \leq n \leq 3$ and $x_{2}$ (n) = $2^{n}$ for $0 \leq n \leq 4$ . Find $x_{3}$ (n) if X3(k) = X1(k)X2(k)

Back2081 Baishakh

2081 Baishakh · Back · BEI, BCT · 11 questions

Ch 35The BIBO stability test

Q1. A discrete-time LTI system is given by difference equation $y (n) = x (n) + e^{a} y (n - 1)$ . Check this system for BIBO stability.

Ch 1Numerical6The convolution sum, and four ways to do it

Q2. Find the output $y (n)$ of LTI system having impulse response $h (n) = {(1 / 3)}^{n} u (n - 3)$ and input $x (n) = {(1 / 6)}^{n - 6} u (n)$ .

Ch 2Numerical2+2+3The region of convergence, and its properties Inverting

X (z)

by long division

Q3. Determine the inverse z-transform of $X (z) = 1 / (1 - 0.8 z^{- 1} + 0.12 z^{- 2}) . (i)$ if ROC is $| z | > 0.6$ (ii) if ROC is $| z | < 0.2$ (iii) if ROC is $0.2 < | z | < 0.6$

Ch 3Numerical3+7Plotting poles and zeros and sketching the magnitud...

Q4. Plot the pole-zero in z plane and draw the magnitude response (not to the scale) of the system described by difference equation $y (n) = 0.67 x (n) - 0.3 x (n - 1) + 2.75 y (n - 1)$

Ch 4Numerical5Cascade and parallel forms

Q5a. Draw the cascaded form structure of $H (z)$ = 10(1 $- 0.25 z^{- 1}$ )(1 $- 0.667 z^{- 1}$ )(1 + $2 z^{- 1}$ )/((1 $- 0.75 z^{- 1}$ )(1 $- 0.125 z^{- 1}$ ){ $1 - (0.5 + j 0.5) z^{- 1}$ }{ $1 - (0.5 - j 0.5) z^{- 1}$ })

Ch 4Numerical5The FIR lattice

Q5b. Draw the lattice structure for the given FIR filter and also check whether the system is stable $. H (z) = 1 + (13 / 24) z^{- 1} + (5 / 8) z^{- 2} + (1 / 3) z^{- 3}$

Ch 5Numerical10The window method, step by step

Q6. Design a linear phase FIR filter using suitable window to meet following specifications: $0.99 \leq | H (e^{j ω}) | \leq 1.01$ for $0 \leq | w | \leq 0.3 π$ ; $| H (e^{j ω}) | \leq 0.01$ for $0.35 π$ ≤ | $ω$ | ≤ $π$

Ch 53+3Gibbs' phenomenon The Kaiser window

Q7. What is Gibbs phenomenon and how can it be minimized? Why Kaiser window is better than other fixed windows in FIR filter design?

Ch 6Numerical2+10Bilinear transformation against impulse invariance The design route

Q8. Differentiate between bilinear transformation and impulse invariance. Design a Butterworth digital IIR lowpass filter using bilinear transformation by taking $T = 0.1$ second, to satisfy the following specifications: $0.6 \leq | H (e^{j ω}) | \leq 1$ for $0 \leq w \leq 0.35 π$ ; $| H (e^{j ω}) | \leq 0.1$ for $0.7 π \leq ω \leq π$

Ch 7Numerical2+6Why the FFT is fast Decimation in time

Q9. Why Decimation in Time Fast Fourier Transform (DITFFT) Algorithm is better than direct computation of DFT? Find 4 point DFT of the sequence $x (n) = {2, 2, 4}$ using DITFFT algorithm.

Ch 7Numerical6Circular convolution, and how it differs from linear

Q10. Compute circular convolution of the following two sequences using DFT. $x (n) = {1, 2, 4, 5}$ and $h (n) = {2, 1, 6, 8}$

Regular2080 Bhadra

2080 Bhadra · Regular · BEI, BCT · 10 questions

Ch 1Numerical2+2+2The five system properties, and how to test each one

Q1. A discrete time system has input $x (n)$ and output $y (n)$ . The input output relation of the system is given by $y (n)$ = sum from k=0 to n of $x (k)$ . Check whether the system is memory less, time invariant and stable or not?

Ch 1Numerical5Testing a signal for periodicity

Q2. Determine whether the given signal is periodic or not. If the signal is periodic, determine the fundamental period $x [n] = e^{j π n / 16} \cos (n π / 17)$ .

Ch 2Numerical1+6The region of convergence, and its properties Inverting

X (z)

by long division

Q3. Define the Region of Convergence (ROC). Find the inverse of $H (z) = (1 + 2 z^{- 1} + z^{- 2}) / (1 - 0.75 z^{- 1} + 0.125 z^{- 2})$ ; ROC $0.25 < | z | < 0.5$

Ch 3Numerical3+7Plotting poles and zeros and sketching the magnitud...

Q4. Plot the pole-zero on the z-plane and draw Magnitude response (not to the scale) of an LTI system described by the equation, $y (n) = x (n) + 0.8 x (n - 1) + 0.8 x (n - 2) + 0.49 y (n - 2)$ .

Ch 4Numerical10The IIR lattice ladder

Q5. Draw the Lattice structure from the following system function $. H (z) = 1 / (1 - 0.2 z^{- 1} + 0.4 z^{- 2} + 0.6 z^{- 3})$

Ch 5Numerical10The linear phase condition The window method, step by step

Q6. Design the symmetric FIR Low Pass Filter (LPF) for which the desired frequency response is expressed as Hd(W) = $e^{- j W t a u}$ for $| W | \leq W c$ and 0 elsewhere. The length of the filter should be 7 and $Ω_{c} = 1$ rad/sample. Make use of the Hanning window.

Ch 5Numerical2+2+2The Kaiser window

Q7. Kaiser window is to be used to design a linear phase FIR filter that meets following specification: $| H (e^{j ω}) | \leq 0.01$ for $0.21 π$ ≤ | $ω$ | ≤ $π$ ; $0.95 \leq | H (e^{j ω}) | \leq 1.05$ for $0 \leq | w | \leq 0.19 π$ . Calculate the optimum value of ripple, attenuation and window length.

Ch 6Numerical12The design route

Q8. Using Bilinear transformation, design a Butterworth low pass filter which satisfies following conditions: $0.9 \leq | H (e^{j ω}) | \leq 1$ for $0 \leq ω \leq \frac{π}{2}$ ; $| H (e^{j ω}) | \leq 0.2$ for $\frac{3 π}{4} \leq ω \leq π$ . Consider sampling frequency of 1 Hz.

Ch 7Numerical8Decimation in frequency

Q9. Compute 8-point DIF-FFT of sequence $x (n) = {2, 1, 2, 1, 1, 2, 1, 2}$ .

Ch 7Numerical6Circular convolution, and how it differs from linear

Q10. Obtain the circular convolution of the following sequences: $x_{1}$ (n) = ${1, 2, 3, 1}$ and $x_{2}$ (n) = ${4, 3, 2, 2}$

Back2080 Baishakh

2080 Baishakh · Back · BEI, BCT · 12 questions

Ch 1Numerical2+2Testing a signal for periodicity

Q1. Check whether following signals are periodic or not. If yes, state their periodic time. a) $x [n]$ = Sin(n $π$ ) + Cos(n $π$ ) b) $x [n]$ = Sin(3n $\frac{π}{5}$ ) + Cos(4n $\frac{π}{7}$ )

Ch 3Numerical5The frequency response, and why exponentials pass s...The convolution sum, and four ways to do it

Q2. Find the output of LTI system having impulse response $h [n] = {(1 / 2)}^{n} u [n]$ and input $x [n] = 5 e^{j π n / 3}$ for -infinity < n < infinity.

Ch 2Numerical1+5The region of convergence, and its properties Inverting

X (z)

by long division

Q3. Define ROC. Find inverse z-transform of $X (z) = (1 + 2 z^{- 1} + z^{- 2}) / (1 - 1.5 z^{- 1} + 0.5 z^{- 2})$ , ROC: $| z | > 1$ .

Ch 3Numerical4+6FIR against IIR Plotting poles and zeros and sketching the magnitud...

Q4. Differentiate between FIR system and IIR System. The poles of a system are located at 0.45 +/- j1.6 and zeros at 0.58 +/- j2.06. Map the poles and zeros in the z-plane and plot the magnitude response (not in scale) of the system.

Ch 4Numerical6The IIR lattice ladder

Q5. Compute Lattice-ladder coefficients and draw lattice structure for given system $H (z) = (2 - 0.7 z^{- 1} + 0.5 z^{- 2}) / (1 - 0.3 z^{- 1} + 0.25 z^{- 2})$ .

Ch 4Numerical4Cascade and parallel forms

Q6. Realize the given system in Cascade Form of 2nd order section flow graph representation. $H (z) = \frac{(1 - 0.4 z^{- 1}) (1 + 0.2 z^{- 1}) (1 - 0.3 e^{j π / 6} z^{- 1}) (1 - 0.3 e^{- j π / 6} z^{- 1})}{(1 - 0.5 e^{j π / 3} z^{- 1}) (1 - 0.5 e^{- j π / 3} z^{- 1}) (1 + 0.7 e^{j π / 4} z^{- 1}) (1 + 0.7 e^{- j π / 4} z^{- 1})}$

Ch 5Numerical2+8Why FIR, and when to choose it The Kaiser window

Q7. In which case do we choose FIR filter and IIR filter? Design a linear phase FIR filter using Kaiser Window to meet the following specifications: $0.99 \leq | H (e^{j ω}) | \leq 1.01$ for $0 \leq w \geq 0.016 π$ ; $| H (e^{j ω}) | \leq 0.01$ for $0.08 π \leq ω \leq 2 π$

Ch 55Gibbs' phenomenon

Q8. Explain in detail about how Gibb's oscillation arise while using the rectangular window in FIR filter design.

Ch 6Numerical11The design route

Q9. Design a low pass digital IIR filter by Bilinear Transformation method to an approximate Butterworth low pass filter, if passband edge frequency is $0.26 π$ radians and maximum deviation of 0.99 dB below 0 dB gain in the passband. The maximum gain of -14.99 dB and frequency is $0.58 π$ radians in stopband, Consider sampling frequency 0.5 Hz.

Ch 64Digital domain spectral transformation

Q10. Describe digital domain Spectral Transformation features and parameters for low pass to high pass filter in IIR Filter design.

Ch 7Numerical2+6Why the FFT is fast Decimation in time

Q11. How fast is FFT? Find 8-point DFT of sequence $x [n] = {1, 1, 0, 0, 1, 1, 2}$ using Decimation in Time Fast Fourier Transform (DITFFT) algorithm.

Ch 7Numerical2+5Why the FFT is fast Circular convolution, and how it differs from linear

Q12. Write the complexity of DFT and FFT? Obtain the circular convolution of the following sequences: $X_{1} [n] = {1, 2, 3, 1}$ and $X_{2} [n] = {4, 3, 2, 2}$

Regular2079 Bhadra

2079 Bhadra · Regular · BEI, BCT · 11 questions

Ch 1Numerical2+2Energy signals and power signals Testing a signal for periodicity

Q1. Define energy and power signal. Determine whether the signal $x [n] = \cos (2 π n / 5) + \sin (π n / 3)$ is periodic or non-periodic and if it is periodic, find its fundamental period.

Ch 1Numerical5The convolution sum, and four ways to do it

Q2. Find the output of LTI system having input signal $x [n] = δ [n] + 2 δ [n - 1] - δ [n - 3]$ and $h [n] = 2 δ [n + 1] + 2 δ [n - 1]$ .

Ch 2Numerical6Inverting

X (z)

by long division

Q3. Find inverses Z-transform of $X (z) = (2 z^{4} + 2 z^{3} - 3 z + 2) / (z^{2} - 1.5 z - 1)$ , ROC: $| z | < 0.5$ , using partial fraction expansion method.

Ch 3Numerical3+7Plotting poles and zeros and sketching the magnitud...

Q4. Plot the pole-zero in z-plane and draw the magnitude response (not to the scale) of the equation of the system describe by difference equation: $y [n]$ - 0. $35 y [n - 1]$ + 0. $25 y [n - 2] = x [n]$ - 0. $75 x [n - 1]$ .

Ch 4Numerical2+2FIR structures

Q5. Draw direct form I and Direct form II realization of the following system. $y [n]$ - 0. $25 y [n - 2] + x [n]$ + 0. $4 x [n - 1]$ + 0. $5 x [n - 2]$

Ch 4Numerical6From lattice coefficients back to the system function

Q6. Given a 3-stage lattice filter for all zero polynomial with coefficients K1 = 1/4, K2 = $\frac{1}{2}$ and K3 = 1/3. Obtain the system function and FIR filter coefficients of this filter.

Ch 5Numerical2+8Gibbs' phenomenon The Kaiser window

Q7. Define Gibb's phenomenon. Design the FIR filter using Kaiser window technique for the specifications: $0.899 \leq | H (e^{j ω}) | \leq 1$ for $| w | \leq 0.2 π$ ; $| H (e^{j ω}) | \leq 0.01$ for $0.4 π \leq ω \leq π$

Ch 55The optimum filter and the Remez exchange algorithm

Q8. Discuss the Remez exchange algorithm for FIR filter design.

Ch 6Numerical11+4The design route Digital domain spectral transformation

Q9. Design a low pass discrete time Butterworth filter using bilinear transformation having following specifications: Passband frequency ( $Ω_{p}$ ) = $0.25 π$ radians, Stopband frequency ( $Ω_{s}$ ) = $0.55 π$ radians, Passband ripple (deltaP) = 0.11, Stopband ripple (deltaS) = 0.21. Consider sampling frequency of 0.5 Hz. Also, convert the obtained digital low pass filter to high pass filter with new pass band frequency, W'p = $0.45 π$ using digital domain transformation.

Ch 7Numerical2+6Why the FFT is fast Decimation in time

Q10. Why we need FFT? Find the 8-point DFT of the following sequence using radix-2 DITFFT algorithm.

Ch 7Numerical7Circular convolution, and how it differs from linear

Q11. If X1(k) and X2(k) are DFT of sequence $x_{1} [n] = {1, 0, 0, 1}$ and $x_{2} [n] = {2, 0, 2}$ respectively then find the sequence $x_{3} [n]$ ; if DFT of $x_{3} [n]$ is given by X3(k) = X1(k) X2(k).

Back2079 Baishakh

2079 Baishakh · Back · BCT · 11 questions

Ch 1Numerical2+2Energy signals and power signals

Q1. Compare between energy signal and power signal. Determine whether the signal $x [n] = e^{j (π / 2 n + 4 π / 7)}$ is energy signal or power signal.

Ch 1Numerical5The convolution sum, and four ways to do it

Q2. Find the output of LTI system having impulse response $h [n] = {(1 / 2)}^{n} {u [n + 2] - u [n - 2]}$ to the input $x [n] = {2, 1, 0, - 1, 4}$ .

Ch 2Numerical1+5The definition Inverting

X (z)

by long division

Q3. Define z-transform for a discrete time signal. Find the inverse z-transform for $H (z) = z / (3 z^{2} - 4 z + 1)$ using partial fraction method for $1 / 3 < | z | < 1$ .

Ch 3Numerical3+7Plotting poles and zeros and sketching the magnitud...

Q4. Plot pole-zero in z-plane and draw magnitude response (not to the scale) of the system described by difference equation $y [n] - 0.3 y [n - 1]$ + 0. $2 y [n - 2] = x [n]$ - 0. $5 x [n - 1]$ .

Ch 4Numerical6+1The IIR lattice ladder

Q5. Compute Lattice-ladder coefficients and draw lattice structure for given system $H (z) = (1 - 0.4 z^{- 1} + 0.25 z^{- 2}) / (1 - 0.3 z^{- 1} + 0.5 z^{- 2})$ . Also check the stability of given system.

Ch 4Numerical4FIR structures

Q6. Obtain the Direct Form I and Direct Form II realization of the following system: $y [n]$ - 0. $75 y [n - 1]$ - 0. $25 y [n - 2] = x [n]$ + 0. $5 x [n - 1]$

Ch 5Numerical5+3The window method, step by step

Q7. Design a low pass digital FIR filter having Pass band edge frequency w_p = $0.2 π$ , Stop band edge frequency w_s = $0.45 π$ and Stop band attenuation a_s = 51 dB using any appropriate window function.

Ch 51+6The optimum filter and the Remez exchange algorithm

Q8. What do you understand by optimum filter? Describe Remez exchange algorithm for FIR filter design along with the flowchart.

Ch 6Numerical11+3The design route Bilinear transformation against impulse invariance

Q9. Design a low pass digital IIR filter by Bilinear Transformation method to an approximate Butterworth low pass filter, if passband edge frequency is $0.24 π$ radians and maximum deviation of 0.98 dB below 0 dB gain in the passband. The maximum gain of -14.95 dB and frequency is $0.57 π$ radians in stopband, consider sampling frequency 0.5 Hz. Compare impulse invariance method with bilinear transformation method.

Ch 7Numerical2+6The DFT and the IDFT Decimation in frequency

Q10. Why we need DFT? Find 8-point DFT of sequence $x [n] = {1, 2, 4, 3, 5, - 1, 3}$ using Decimation in Frequency Fast Fourier Transform (DIFFFT) algorithm.

Ch 7Numerical7Circular convolution, and how it differs from linear

Q11. Find the circular convolution of the sequences $x_{1} [n] = {1, - 1, - 2, 3, - 1}$ and $x_{2} [n] = {1, 2, 3}$ .

Regular2078 Bhadra

2078 Bhadra · Regular · BCT · 11 questions

Ch 1Numerical4Testing a signal for periodicity

Q1. Determine whether the signal $x [n] = \cos (π n / 2) \cos (π n / 4)$ is periodic or non periodic and if it is periodic, find its fundamental period.

Ch 1Numerical5The convolution sum, and four ways to do it

Q2. Find the output of LTI system having impulse response $h [n] = u [n] - u [n - 4]$ and input signal $x [n] = {(1 / 2)}^{n} u [n]$ .

Ch 2Numerical1+5The region of convergence, and its properties Inverting

X (z)

by long division

Q3. Define ROC. Find inverse z-transform of $X (z) = (z^{3} + z^{2} + 1.5 z + 0.5) / (z^{3} + 1.5 z^{2} + 0.5 z)$ , ROC: $| z | < 1 / 2$ .

Ch 3Numerical4The zero input response

Q4. Determine the zero-input response for a second order system given by: $y [n] - 3 y [n - 1] - 4 y [n - 2] = x [n]$

Ch 3Numerical2+4Plotting poles and zeros and sketching the magnitud...

Q5. Plot the pole-zero in z-plane and draw magnitude response (not to the scale) of the system described by difference equation. $y [n] - 0.4 y [n - 1] + 0.25 y [n - 2] = x [n]$ - 0. $4 x [n - 1]$

Ch 4Numerical3+7FIR structures The FIR lattice

Q6. The system function of a filter is $H (z) = 1 + (13 / 24) z^{- 1} + (5 / 8) z^{- 2} + (1 / 3) z^{- 3}$ . Draw the Direct Form and Lattice Structure implementation of the above filter.

Ch 5Numerical8The Kaiser window

Q7. Design a linear phase FIR filter using KAISER window to meet the following specifications: $| H (e^{j ω}) | \leq 0.01$ for $0 \leq | w | \leq 0.25 π$ ; $0.95 \leq | H (e^{j ω}) | \leq 1.05$ for $0.35 π \leq | w | \leq 0.6 π$ ; $| H (e^{j ω}) | \leq 0.01$ for $0.65 π$ ≤ | $ω$ | ≤ $π$

Ch 51+6The optimum filter and the Remez exchange algorithm

Q8. What is optimum filter? Show mathematical expression of Remez exchange algorithm for FIR filter design.

Ch 6Numerical12+3Impulse invariance The design route

Q9. Design a LPF Butterworth filter using Impulse Invariance Method (IIM) method with passband and stopband frequencies 200Hz and 500Hz respectively. The passband and stopband attenuations are 5dB and 12dB respectively. The sampling frequency is 5000Hz. What is pre-warping and why it is necessary? Explain.

Ch 7Numerical2+6DFT against DTFT Circular convolution, and how it differs from linear

Q10. Differentiate between DFT and DTFT. Find the circular convolution of $x_{1} [n] = {2, 1, 2, 1}$ and $x_{2} [n] = {1, 2, 3, 4}$

Ch 7Numerical7Decimation in time

Q11. Find the 8-point DFT of $x [n] = u [n] - u [n - 4]$ using FFT DIT algorithm.

Regular2076 Chaitra

2076 Chaitra · Regular · BCT · 12 questions

Ch 1Numerical2+3Even and odd parts of a signal Discrete time signals and the elementary sequences

Q1. Define even and odd type discrete time signals with suitable example. Plot the signal $x [- 2 n + 3]$ where $x [n] = {1, 2, 0, - 1, - 3, - 4}$ .

Ch 1Numerical5The five system properties, and how to test each one

Q2. Determine whether the following system are: a) $y [n] = x [- n]$ is time-invariant or not. b) $y [n]$ = x[ $n^{2}$ ] is linear or not.

Ch 1Numerical6The convolution sum, and four ways to do it

Q3. Find the output of LTI system having input signal $x [n] = u [n + 1] - u [n - 4]$ and impulse response $h [n] = {(1 / 2)}^{n} u [n - 1]$ .

Ch 2Numerical2+6The region of convergence, and its properties Inverting

X (z)

by long division

Q4. Define ROC of z-transform. Find inverse z-transform using partial fraction expansion of $X (z) = (z^{4} + 5 z^{3} - 3 z + 4) / (z^{2} - 1.5 z - 1)$ , ROC: $| z | < 0.5$ .

Ch 3Numerical2+6Plotting poles and zeros and sketching the magnitud...

Q5. Draw the pole-zero in the z-plane for a system with poles at 0.45 +/- j1.06 and zeroes at 0.58 +/- j2.06. Also plot the magnitude response (not to the scale) of the system.

Ch 4Numerical6+4The IIR lattice ladder

Q6. Compute Lattice and Ladder coefficients and Draw lattice-ladder structure for given IIR system $H (z) = (0.5 - 2 z^{- 1} + 3 z^{- 2}) / (1 - 0.5 z^{- 1} - 0.7 z^{- 2} + 0.3 z^{- 3})$ .

Ch 4Numerical4Cascade and parallel forms

Q7. Realize the given system in Cascade form of 2nd order section in signal flow graph representation. $H (z) = \frac{(1 - 0.5 z^{- 1}) (1 + 0.35 z^{- 1}) (1 - 0.3 e^{j 2 n π / 5} z^{- 1}) (1 - 0.3 e^{- j 2 n π / 5} z^{- 1})}{(1 - 0.6 e^{j n π / 3} z^{- 1}) (1 - 0.6 e^{- j n π / 3} z^{- 1}) (1 + 0.5 e^{j 2 n π / 7} z^{- 1}) (1 + 0.5 e^{- j 2 n π / 7} z^{- 1})}$

Ch 5Numerical6The window method, step by step

Q8. Design the FIR filter using suitable window for the specifications: $0.899 \leq | H (e^{j ω}) | \leq 1$ for $| w | \leq 0.2 π$ ; $| H (e^{j ω}) | \leq 0.01$ for $0.4 π \leq ω \leq π$

Ch 51+5The optimum filter and the Remez exchange algorithm

Q9. What is optimum filter? Show mathematical expression of Remez exchange algorithm for FIR filter design.

Ch 6Numerical10The design route

Q10. Design a digital low pass Butterworth filter by applying bilinear transformation techniques for the given specifications: Passband peak to peak ripple ≤ 1dB, Passband edge frequency = 1.2KHz, Stopband Attenuation ≥ 40dB, Stopband edge frequency = 2.5 KHz, Sample rate = 8KHz

Ch 7Numerical7Decimation in frequency

Q11. Find 8-point DFT of sequence $x [n] = {1, 2, 3, 3, 5, 0, 4, 6}$ using Decimation in frequency Fast Fourier Transform (DIFFFT) algorithm.

Ch 7Numerical5Circular convolution, and how it differs from linear

Q12. Find $x_{3} [n]$ if DFT of $x_{3} [n]$ is given by X3(k) = X1(k) * X2(k) where X1(k) and X2(k) are 4-point DFT of $x_{1} [n] = {1, 2, - 2}$ and $x_{2} [n] = {1, 2, 3, - 1}$ respectively.

Back2076 Ashwin

2076 Ashwin · Back · BCT · 9 questions

Ch 14+3The discrete time Fourier transform, and its proper...The discrete time Fourier series, and its properties

Q1. Explain Fourier transform multiplication property for two sequences. Write Drichlet's conditions for Fourier series.

Ch 1Numerical6The convolution sum, and four ways to do it

Q2. Find convolution between two signals $x [n] = 2^{n}$ 4[-n], $0 < a < 1$ and $h [n]$ = 4[n]

Ch 2Numerical3+6The convolution property, derived Inverting

X (z)

by long division

Q3. State Convolution property of Z-transform. Find inverse Z-transform of $X (z) = z$ /{ $(z - 0.6) (z + 0.5)^{2}$ }, ROC: $| z | > 0.6$

Ch 34+3Stability and causality, three ways of saying the s...

Q4. Describe stability and causality characteristics of LTI system in terms of Impulse Response and ROC of its transfer function with suitable examples.

Ch 4Numerical6+3The IIR lattice ladder

Q5. Compute Lattice and Ladder coefficients and Draw lattice-ladder structure for given IIR system $H (z) = (0.7 - 1.5 z^{- 1} + 0.5 z^{- 2}) / (1 - 0.5 z^{- 1} - 0.7 z^{- 2} + 0.3 z^{- 3})$

Ch 3Numerical2+8Plotting poles and zeros and sketching the magnitud...

Q6. For the system described by the following difference equation: $y [n] = 0$ . $67 x [n]$ - 0. $3 x [n - 1]$ + 2. $75 y [n - 1]$ . Map the poles and zero in the z-plane and plot the phase response of the system.

Ch 6Numerical12The design route

Q7. Design a low pass discrete IIR filter by Bilinear Transformation method to an approximate Butterworth filter having specifications as below: Pass bandedge frequency ( $ω_{p}$ ) = $0.22 π$ radians, Stop bandedge frequency ( $ω_{s}$ ) = $0.54 π$ radians, Passband ripple (deltap) = 0.11, Stopband ripple (deltas) = 0.22, Consider sampling frequency 0.5 Hz.

Ch 7Numerical2+8The DFT and the IDFT Decimation in frequency

Q8. Why we need DFT? Find 8-point DFT of sequence $x [n] = {1, 2, 3, 3, 5, 1, 4, 2}$ using Decimation in frequency Fast Fourier Transform (DIFFFT) algorithm.

Ch 5Numerical2+4+4Why FIR, and when to choose it The Kaiser window The optimum filter and the Remez exchange algorithm

Q9. In which case do we choose FIR filter and IIR filter? Design a Kaiser Window to meet the following specifications: $0.99 \leq | H (e^{j ω}) | \leq 1.01$ for $0 \leq w \leq 0.16 π$ ; $| H (e^{j ω}) | \leq 0.01$ for $0.18 π \leq ω \leq 2 π$ . Draw the flow chart for Remez-Exchange algorithm

Regular and Back2075 Chaitra

2075 Chaitra · Regular and Back · BCT · 9 questions

Ch 13+4Energy signals and power signals The discrete time Fourier transform, and its proper...

Q1. Define Power and Energy type discrete time signal with suitable example. Differentiate between Fourier Series and Fourier Transform.

Ch 1Numerical5+2The convolution sum, and four ways to do it

Q2. Find the output of LTI system having impulse response $h [n]$ with $h [- 2] = 3$ , $h [0] = 2$ , $h [1] = 1$ and input signal $x [n] = {(2)}^{n}$ , for $- 1 \leq n \leq 3$ . Also check the answer.

Ch 3Numerical2+7Plotting poles and zeros and sketching the magnitud...

Q3. Plot the pole-zero in z-plane and draw magnitude response (not to scale) of the system described by differential equation $y (n) - 0.3 y (n - 1) = 2 x (n - 2) + 0.7 x (n - 1) + 4 x (n)$

Ch 4Numerical9The IIR lattice ladder

Q4. Draw the lattice structure from the following system function $H (t) = 1 / (1 + (2 / 3) z^{- 1} + (5 / 8) z^{- 2} + (2 / 3) z^{- 3} + z^{- 4})$

Ch 52+6The optimum filter and the Remez exchange algorithm

Q5. What is optimum filter? Show mathematical expression of Remez exchange algorithm for FIR filter design.

Ch 2Numerical4+6The region of convergence, and its properties

Q6. List out the properties of Region of convergence and locate the ROC of the following signal $x [n] = {(0.1)}^{n} u [n] + {(0.3)}^{n} u [- n - 1]$

Ch 6Numerical10The design route

Q7. Using bilinear transformation, design a digital filter using Butterworth approximation which satisfies the following conditions: $0.8 \leq | H (e^{j ω}) | \leq 1$ for $0 \leq W \leq 0.2 π$ ; $| H (e^{j ω}) | \leq 0.2$ for $0.6 π \leq W \leq π$

Ch 7Numerical2+8Why the FFT is fast Decimation in time

Q8. How fast is FFT? Find X(3) and X(5) for given sequence $x [n] = {1, - 2, 3, 2}$ using DITFFT algorithm.

Ch 7Numerical3+7Circular convolution, and how it differs from linear

Q9. Differentiate between linear convolution and circular convolution compute circular convolution of signals $X_{1} [n] = {0, 0, 1, 1}$ and $X_{2} [n] = {1, 1, 1, 1}$

Back2075 Ashwin

2075 Ashwin · Back · BCT · 9 questions

Ch 1Numerical3+3The five system properties, and how to test each one

Q1. Determine whether the following sequences are linear or not: a) $y [n] = x^{2} [n]$ b) $y [n] = \cos (5 π n / 8 + π / 4)$

Ch 1Numerical5The convolution sum, and four ways to do it

Q2. Find the output of LTI system having impulse response $h [n] = 2^{n} {u [n] - u [n - 3]}$ and input signal $x [n] = δ [n] + δ [n - 1] + δ [n - 2]$ .

Ch 2Numerical3+6The region of convergence, and its properties

Q3. List out the properties of Region of convergence and locate the ROC of the following signal. $x [n] = {(0.6)}^{n} u [n] + {(0.25)}^{n} u [n]$

Ch 3Numerical2+8Plotting poles and zeros and sketching the magnitud...

Q4. Draw the poles and zeros in the z-plane for a system with poles at 0.45 +/- j1.06 and zeros at 0.58 +/- j2.06. Also plot the magnitude response of the system.

Ch 4Numerical7+3The IIR lattice ladder Fixed point number formats

Q5. Draw the Lattice structure from the following system function: $1 / (3 + (39 / 24) Z^{- 1} + (15 / 8) Z^{- 2} + (3 / 9) Z^{- 3})$ . And represent 5/8 and -5/8 in sign magnitude, 1's complement and 2's complement format.

Ch 6Numerical12The design route

Q6. Design a digital low-pass filter with the following specification: i) Pass-band magnitude constant to 0.7 dB below the frequency of $0.15 π$ ii) Stop-band attenuation at least 14 dB for the frequencies between $0.6 π$ to $π$ . Use Butterworth approximation as a prototype and use bilinear transformation method to obtain the digital filter.

Ch 5Numerical8+4The Kaiser window The optimum filter and the Remez exchange algorithm

Q7. Design a linear phase FIR filter using Kaiser Window to meet the following specifications: $0.99 \leq | H (e^{j ω}) | \leq 1.01$ for $0 \leq w \leq 0.19 π$ ; $| H (e^{j ω}) | \leq 0.01$ for $0.21 π \leq ω \leq π$ . Draw the flow chart for Optimum filter design.

Ch 7Numerical3+7Why the FFT is fast Decimation in time

Q8. How fast is FFT compare to DFT? Draw the butterfly diagram of 8-point DFT of a sequence as $x [n] = n$ + 1 using Decimation in Time FFT algorithm.

Ch 7Numerical1+5The properties Circular convolution, and how it differs from linear

Q9. State the circular convolution property of DFT. Find the circular convolution of: (a) ${1, 2, - 1, 1}$ and $x_{2}$ (n) = ${1, 3, 5, 7}$

Regular2074 Chaitra

2074 Chaitra · Regular · BCT · 10 questions

Ch 1Numerical3+2Discrete time signals and the elementary sequences The convolution sum, and four ways to do it

Q1. Plot the sequence $x [n] = u [n] - u [n - 3] + 5 δ [n - 4] = n u [n - 6]$ . List out the properties of LTI system.

Ch 1Numerical3+3The five system properties, and how to test each one

Q2. Determine whether the following system are: a) $y [n] = y [n - 4] + x [n - 4]$ is Time-invariant or not b) $y [n] = x^{2} [n]$ is Linear or Non-linear

Ch 2Numerical1+3+5The region of convergence, and its properties Inverting

X (z)

by long division

Q3. Define a ROC. What are the properties of ROC of z-transform? Find the inverse Z-transform of $X (z) = (2 z^{2} + 2 z^{2} + 3 z + 5) / (z^{2} - 0.1 z - 0.2)$ , ROC: $| z | < 0.4$ .

Ch 3Numerical2+8Plotting poles and zeros and sketching the magnitud...

Q4. The poles of a system are located at: 0.45 - 0.77i and -2 +/- 0.3i. Map the poles and zero in the z-plane and plot the magnitude response of the system.

Ch 4Numerical5FIR structures

Q5. Obtain the Direct Form I and Direct Form II realization of the following system. $3 y [n] + y [n - 1] + 2 y [n - 4] = 2 x [n] + x [n - 3]$

Ch 4Numerical5The FIR lattice

Q6. Determine the lattice coefficients coefficients corresponding to the FIR filter with the system function: $H (z) = A 3 (z) = 1 + (52 / 96) z^{- 1} + (25 / 40) z^{- 2} + (1 / 3) z^{- 3}$

Ch 6Numerical12Impulse invariance The design route

Q7. Design a digital low-pass filter with the following specification: i) Pass-band magnitude constant to 0.7 dB below the frequency of $0.15 π$ ii) Stop-band attenuation at least 14 dB for the frequencies between $0.6 π$ to $π$ . Use Butterworth approximation as a prototype and use impulse invariance method to obtain the digital filter.

Ch 5Numerical9+3The Kaiser window

Q8. Design a FIR linear phase filter using Kaiser window that meets the following specifications: $| H (e^{j ω}) | \leq 0.01$ for $0 \leq | w | \leq 0.25 π$ ; $0.95 \leq | H (e^{j ω}) | \leq 1.05$ for $0.35 π \leq | w | \leq 0.6 π$ ; $| H (e^{j ω}) | \leq 0.01$ for $0.65 π$ ≤ | $ω$ | ≤ $π$ . Also determine the minimum length (M+1) of the impulse response and Kaiser window parameter $β$ .

Ch 7Numerical3+7The DFT and the IDFT Decimation in frequency

Q9. Why do we need DFT? Draw the butterfly structure to compute the DFT of the following signal using Radix-2 DIFFFT algorithm, and compute X(2) and X(1) only $x [n] = {1.5, - 1, 1.8, 0.6, 3, 1.7}$ .

Ch 7Numerical1+5Zero padding

Q10. Define zero padding. Find the linear convolution through circular convolution with padding of zeros for the following sequences: $x [n] = {1, 1, 1, 1}$ and $h [n] {2, 3}$ .

Back2074 Ashwin

2074 Ashwin · Back · BCT · 10 questions

Ch 1Numerical2+2Energy signals and power signals Testing a signal for periodicity

Q1. Define Energy and Power type discrete time signal. Check whether signal $x [n] = e^{j (π n / 3 + π / 4)}$ is periodic or not. If it is periodic, state its periodic time.

Ch 1Numerical3+2The convolution sum, and four ways to do it

Q2. Find the output of LTI system having impulse response $h [n] = {(1 / 2)}^{n} {u [n + 2] - u [n - 2]}$ and input signal $x [n] = {2, 1, 0.5, - 1}$ . Also check the answer.

Ch 2Numerical3+3The region of convergence, and its properties Inverting

X (z)

by long division

Q3. State and explain the properties of a Region of Convergence (ROC). Find the inverse z-transform of $X (z) = z^{2} [1 - (3 / 2) z^{- 1}] (1 + z^{- 1}) (1 - z^{- 1})$

Ch 3Numerical3+7Plotting poles and zeros and sketching the magnitud...

Q4. Plot the pole-zero in z-plane and Draw Magnitude Response (not to the scale) of the system described by difference equation $y [n]$ - 0. $4 y [n - 1]$ + 0. $2 y [n - 2] = x [n]$ + 0. $5 x [n - 1]$ + 0. $6 x [n - 2]$ + 0. $8 x [n - 3]$

Ch 4Numerical3+7FIR structures The FIR lattice

Q5. Draw the direct form and Lattice structure of a filter with system function $H (z) = 1 + 0.7 z^{- 1} + 1.2 z^{- 2} - z^{- 3}$ .

Ch 5Numerical2+6The Kaiser window The window method, step by step

Q6. Why Kaiser window is better than other fixed windows in FIR filter design? Find out first six coefficients of impulse response of a low pass FIR filter having Pass band edge frequency w_p = $0.2 π$ , Stop band edge frequency w_s = $0.5 π$ and Stop band attenuation a_s = 41dB using any appropriate window function.

Ch 51+6The optimum filter and the Remez exchange algorithm

Q7. What is an optimum filter? Show mathematical expression of the Remez exchange algorithm for FIR filter design with flow chart.

Ch 6Numerical15The design route

Q8. Design a low pass discrete IIR filter by Bilinear Transformation method to an approximate Butterworth filter having specifications as below: Pass bandedge frequency ( $ω_{p}$ ) = $0.27 π$ radians, Stop bandedge frequency ( $ω_{s}$ ) = $0.58 π$ radians, Passband ripple (deltap) = 0.11, Stopband ripple (deltas) = 0.21, Consider sampling frequency 0.5 Hz.

Ch 7Numerical7Decimation in frequency

Q9. Compute the 8-point DFT of the sequence $x [n]$ = { $\frac{1}{2}$ , $\frac{1}{2}$ , $\frac{1}{2}$ , $\frac{1}{2}$ , 0, 0, 0, 0} using Decimation in Frequency Fast Fourier Transform (DIF-FFT) algorithm.

Ch 7Numerical1+7Zero padding Circular convolution, and how it differs from linear

Q10. What is a zero padding? If X1(k) and X2(k) are DFT of sequence $x_{1} [n] = {1, 2, 0, 1, - 2}$ and $x_{2} [n] = {1, 0, 1, 1, 2}$ respectively then find the sequence $x_{3} [n]$ ; If DFT of $x_{3} [n]$ is given by X3(k) = X1(k) X2(k).

New Back2073 Shrawan

2073 Shrawan · New Back · BCT · 11 questions

Ch 13The discrete time Fourier series, and its properties

Q1. Explain the process of calculating fourier series coefficients.

Ch 1Numerical6The convolution sum, and four ways to do it

Q2. Determine the system output $y (n)$ of the following signals: $h (n) = {1, 1, 1}$ and $x (n) = {1, 1, 1, 1}$

Ch 2Numerical1+5The region of convergence, and its properties Inverting

X (z)

by long division

Q3. Define a ROC. Find inverse Z-transform of $X (z) = z$ /{ $(z - 0.4) (z + 1.5)^{2}$ }, ROC: $| z | < 0.4$

Ch 3Numerical3+7Difference equations, and the system function The frequency response, and why exponentials pass s...The convolution sum, and four ways to do it

Q4. State linear constant coefficient difference equation and corresponding system function. Determine the output sequence of the system with impulse response $h [n] = {(1 / 2)}^{n} u [n]$ when the input signal is $x [n] = 10$ - 5sin( $π$ n/2) + 20cos( $π$ n), -infinity < n < infinity.

Ch 4Numerical3+7FIR structures The FIR lattice

Q5. The system function of a filter is $H (z) = 2 + 1.8 z^{- 1} - 1.6 z^{- 2} + z^{- 3}$ . Draw the Direct Form and Lattice Structure implementation of the above filter.

Ch 56The window method, step by step Gibbs' phenomenon

Q6. Explain in detail about how rectangular window is used in FIR filter design. How Gibb's oscillations arise in this process.

Ch 59The optimum filter and the Remez exchange algorithm

Q7. Explain about Remaz exchange algorithm with suitable derivation and flow chart.

Ch 6Numerical12The design route

Q8. Using bilinear transformation, design a butterworth low pass filter which satisfies the following Magnitude Response: $0.89125 \leq | H (e^{j ω}) | \leq 1$ for $0 \leq w \leq 0.2 π$ ; $| H (e^{j ω}) | \leq 0.17783$ for $0.3 π \leq ω \leq π$

Ch 63The design route

Q9. Explain briefly about bilinear transformation method of IIR filter design.

Ch 7Numerical2+6The DFT and the IDFT Decimation in time

Q10. Why do we need DFT? Find 8-point DFT of sequence $x [n] = {1, - 1, 2, 2, 1, 1, 2, 2}$ using Fast Fourier Transform algorithm.

Ch 7Numerical7Circular convolution, and how it differs from linear

Q11. Find $x_{3} [n]$ if DFT of $x_{3} [n]$ is given by X3(k) = X1(k) X2(k) where X1(k) and X2(k) are 5-point DFT of $x_{1} [n] = {1, - 2, 2, 1, 4}$ and $x_{2} [n] = {2, 1, - 3, - 1}$ respectively.

Regular2072 Chaitra

2072 Chaitra · Regular · BCT · 12 questions

Ch 14The discrete time Fourier series, and its properties

Q1. How fourier series coefficients are calculated? Explain.

Ch 1Numerical3+2The convolution sum, and four ways to do it

Q2. Find the output of LTI system having impulse response $h [n]$ with $h [- 2] = 1$ , $h [0] = 2$ , $h [1] = 3$ and input signal $x [n]$ with $x [0] = \frac{1}{2}$ , $x [2] = 2$ , $x [3] = 3$ . Also check the answer.

Ch 26The region of convergence, and its properties

Q3. Explain the properties of Region of Convergence with examples.

Ch 34Stability and causality, three ways of saying the s...

Q4. Describe stability and causality characteristics of LTI system in terms of Impulse Response and ROC of its transfer function with suitable examples.

Ch 3Numerical2+4Plotting poles and zeros and sketching the magnitud...

Q5. Plot the pole-zero in z-plane and Draw Magnitude Response (not to the scale) of the system described by difference equation. $y [n]$ - 0. $4 y [n - 1]$ + 0. $1 y [n - 2] = x [n]$ + 0. $6 x [n - 1]$

Ch 4Numerical5FIR structures

Q6. Determine the Direct Form I and Direct Form II realization of the following system. $y (n) = - 0.1 y (n - 1) + 0.2 y (n - 2) + 3 x (n) + 3.6 x (n - 2) + 0.6 x (n - 2)$

Ch 4Numerical5The FIR lattice

Q7. Compute the lattice coefficients and draw the lattice structure of following FIR system $. H (z) = 1 + 2 z^{- 1} + z^{- 2}$

Ch 55+3The window method, step by step The Kaiser window

Q8. Describe how digital FIR filter can be design by window method. Why Kaiser window is better than other fixed windows in FIR filter design?

Ch 51+6The optimum filter and the Remez exchange algorithm

Q9. What is an optimum filter? Show mathematical expression of Remez exchange algorithm for FIR filter design.

Ch 6Numerical3+12Bilinear transformation against impulse invariance Impulse invariance

Q10. Explain about the advantages of selecting bilinear transformation method over impulse invariance method (I I M). Design a digital low pass Butterworth filter using impulse invariant transformation with pass band and stop band frequencies 200Hz and 500Hz respectively. The pass band and stop band attenuation are -5dB and -12dB respectively. The sampling frequency is 5kHz. Use IIM method.

Ch 7Numerical8Decimation in time

Q11. Find the FFT of the signal $x [n] {1, 1, 2, 4, 3, 1, 2, 1}$ using DIT-FFT algorithm.

Ch 7Numerical7Circular convolution, and how it differs from linear

Q12. Compute Circular Convolution of $h (n) = {1, 2, 1, - 1, 1}$ and $x [n] = {1, 2, 3, 1}$ .

New Back2072 Kartik

2072 Kartik · New Back · BCT · 11 questions

Ch 1Numerical2+3Energy signals and power signals

Q1. Define energy and power signal. Check the signal $x [n] = u [n]$ and $x [n] = δ [n]$ is Energy or Power type.

Ch 1Numerical5The convolution sum, and four ways to do it

Q2. Find the output of LTI system having impulse response $h [n] = {(1 / 3)}^{n} {u [n + 1] - u [n - 2]}$ and input signal $x [n] = {2, 1, 0.5, 3}$ .

Ch 23+3The region of convergence, and its properties The convolution property, derived

Q3. State the properties of region of convergence (ROC). Drive the convolution property of Z-transform.

Ch 3Numerical4The frequency response, and why exponentials pass s...The convolution sum, and four ways to do it

Q4. Find the output of LTI System having impulse response $h [n] = {(1 / 2)}^{n} u [n]$ and input signal $x [n] = 5 e^{j π n / 3}$ for -infinity < n < infinity.

Ch 3Numerical6Plotting poles and zeros and sketching the magnitud...

Q5. Plot Magnitude Response (not to the scale) of the system described by difference equation. $y [n]$ - 0. $5 y [n - 1]$ + 0. $3 y [n - 2] = x [n]$ + 0. $7 x [n - 1]$

Ch 4Numerical4FIR structures

Q6. Determine the Direct Form II realization of the following system $y (n) = - 0.1 y (n - 1) + 0.72 y (n - 2) + 0.7 x (n) - 0.252 x (n - 2)$

Ch 4Numerical6The FIR lattice

Q7. Compute the lattice coefficients and draw the lattice structure of following FIR system $H (z) = 1 + 2 z^{- 1} - 3 z^{- 2} + 4 z^{- 3}$

Ch 5Numerical6+8The optimum filter and the Remez exchange algorithm The Kaiser window

Q8. Draw the flowchart of Remez-Exchange theorem and explain it. Design an FIR linear phase filter using Kaiser window to meet the following specifications: $0.99 \leq | H (e^{j ω}) | \leq 1.01$ for $0 \geq w \geq 0.19 π$ ; $| H (e^{j ω}) | \leq 0.01$ for $0.21 π \leq ω \leq π$

Ch 6Numerical15The design route

Q9. Design a low pass digital filter by Bilinear Transformation method to an approximate Butterworth filter, if passband edge frequency is $0.25 π$ radians and maximum deviation of 1 dB below 0 dB gain in the passband. The maximum gain of -15 dB and frequency is $0.45 π$ radians in stopband, Consider sampling frequency 1Hz.

Ch 7Numerical7Decimation in time

Q10. Find 8-point DFT of sequence $x [n] = {1, 1, 0, 1, 0, 1, 2}$ using Decimation in Time Fast Fourier Transform (DITFFT) algorithm.

Ch 7Numerical2+6The DFT and the IDFT Circular convolution, and how it differs from linear

Q11. Why we need DFT? If X1(k) and X2(k) are DFT of sequence $x_{1} [n] = {1, 2, 4}$ and $x_{2} [n] = {- 1, 2, 3, 1}$ respectively, then find the sequence $x_{3} [n]$ , if DFT of $x_{3} [n]$ is given by X3(k) = X1(k) X2(k).

New Back2071 Shrawan

2071 Shrawan · New Back · BCT · 10 questions

Ch 1Numerical4+5Even and odd parts of a signal The convolution sum, and four ways to do it

Q1. Find the odd and even part of the following signal: [a stem plot of $x [n]$ with values 1 at $n = - 4$ , -3, -2, -1 and 0, and 2 at $n = 1$ , 2, 3, 4]. A discrete time LTI system has input signal and impulse response as, $x [n] = 1$ for $- 1 \leq n \leq 1$ and 0 elsewhere, and $h [n] = 1$ for $- 1 \leq n \leq 1$ and 0 elsewhere. Find the output of the system using graphical method.

Ch 2Numerical6Inverting

X (z)

by long division

Q2. Find the inverse z transform of: $X (Z) = (1 + 2 z^{- 1} + z^{- 2}) / (1 + 1.5 z^{- 1} + 0.5 z^{- 2})$ , $| z | > 1$ using partial fraction method.

Ch 3Numerical2+3+5Difference equations, and the system function The frequency response, and why exponentials pass s...The convolution sum, and four ways to do it

Q3. Why do we need difference equation? State linear constant coefficient difference equation and corresponding system function. Consider an LTI system with impulse response $h [n] = {(1 / 2)}^{n} u [n]$ . Determine $y [n]$ , if the input is $x [n] = A e^{j π n}$ .

Ch 4Numerical5From lattice coefficients back to the system function

Q4. If a 3 stage lattice filter for all pole polynomial has coefficients K1 = 1/4, K2 = $\frac{1}{2}$ and K3 = 1/3. Obtain the system function of this filter.

Ch 41+1+2+1Quantisation, limit cycles and the dead band

Q5. What is the importance of quantization in Digital Signal Processing? Which one is better rounding or truncation? Explain about limit cycles in recursive system? Define dead band.

Ch 56The window method, step by step Gibbs' phenomenon

Q6. Explain in detail about how rectangular window is used in FIR filter design. How Gibb's oscillations arise in this process.

Ch 59The optimum filter and the Remez exchange algorithm

Q7. What is a Remez exchange algorithm? Derive its equation and draw its flow chart.

Ch 6Numerical15The design route

Q8. Design a low pass digital filter by Bilinear Transformation method to an approximate Butterworth filter it passband frequency is $0.2 π$ radians and maximum deviation of 1 db below 0 dB gain in the pass band. The maximum gain of -15 db and frequency is $0.4 π$ radians in stop band, consider sampling frequency 1 Hz.

Ch 7Numerical7Circular convolution, and how it differs from linear

Q9. A system has input signal $x [n] = {1, 2, 3, 4}$ and impulse response $h [n] = {1, 3, 5, 7}$ and the DFT of $x [n]$ is $X [k]$ and the DFT of $h [n]$ is $H [k]$ . Find the output of the system $y [n]$ if $G [k] = X [k]$ . $H [k]$

Ch 7Numerical6+2Decimation in time

Q10. Find DFT for ${1, 1, 2, 0, 1, 2, 0, 1}$ using FFT DIT butterfly algorithm and plot the spectrum.

Chapter 1 · Discrete-time signals and systems 11 distinct questions, from 19 of the 19 sittings

TOP 14/19Numerical2082 Bhadra Q2, 2081 Bhadra Q2, 2081 Baishakh Q2, 2079 Bhadra Q2, 2079 Baishakh Q2, 2078 Bhadra Q2, 2076 Chaitra Q3, 2076 Ashwin Q2, 2075 Chaitra Q2, 2075 Ashwin Q2, 2074 Ashwin Q2, 2073 Shrawan Q2, 2072 Chaitra Q2, 2072 Kartik Q2

Find the output of an LTI system whose impulse response is given by ${0.5}^{n} {u [n] - u [n - 3]}$ and input is given by ${2, 1, 0.5, - 1}$ .

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Find the output of an LTI System with impulse response $h [n] = 2 δ [n + 1] + 2 δ [n - 1]$ when an input of $x [n] = δ [n] + 2 δ [n - 1] - δ [n - 3]$ is applied to it. (2081 Bhadra Q2)
Find the output $y (n)$ of LTI system having impulse response $h (n) = {(1 / 3)}^{n} u (n - 3)$ and input $x (n) = {(1 / 6)}^{n - 6} u (n)$ . (2081 Baishakh Q2)
Find the output of LTI system having input signal $x [n] = δ [n] + 2 δ [n - 1] - δ [n - 3]$ and $h [n] = 2 δ [n + 1] + 2 δ [n - 1]$ . (2079 Bhadra Q2)
Find the output of LTI system having impulse response $h [n] = {(1 / 2)}^{n} {u [n + 2] - u [n - 2]}$ to the input $x [n] = {2, 1, 0, - 1, 4}$ . (2079 Baishakh Q2)
Find the output of LTI system having impulse response $h [n] = u [n] - u [n - 4]$ and input signal $x [n] = {(1 / 2)}^{n} u [n]$ . (2078 Bhadra Q2)
Find the output of LTI system having input signal $x [n] = u [n + 1] - u [n - 4]$ and impulse response $h [n] = {(1 / 2)}^{n} u [n - 1]$ . (2076 Chaitra Q3)
Find convolution between two signals $x [n] = 2^{n}$ 4[-n], $0 < a < 1$ and $h [n]$ = 4[n] (2076 Ashwin Q2)
Find the output of LTI system having impulse response $h [n]$ with $h [- 2] = 3$ , $h [0] = 2$ , $h [1] = 1$ and input signal $x [n] = {(2)}^{n}$ , for $- 1 \leq n \leq 3$ . Also check the answer. (2075 Chaitra Q2)
Find the output of LTI system having impulse response $h [n] = 2^{n} {u [n] - u [n - 3]}$ and input signal $x [n] = δ [n] + δ [n - 1] + δ [n - 2]$ . (2075 Ashwin Q2)
Find the output of LTI system having impulse response $h [n] = {(1 / 2)}^{n} {u [n + 2] - u [n - 2]}$ and input signal $x [n] = {2, 1, 0.5, - 1}$ . Also check the answer. (2074 Ashwin Q2)
Determine the system output $y (n)$ of the following signals: $h (n) = {1, 1, 1}$ and $x (n) = {1, 1, 1, 1}$ (2073 Shrawan Q2)
Find the output of LTI system having impulse response $h [n]$ with $h [- 2] = 1$ , $h [0] = 2$ , $h [1] = 3$ and input signal $x [n]$ with $x [0] = \frac{1}{2}$ , $x [2] = 2$ , $x [3] = 3$ . Also check the answer. (2072 Chaitra Q2)
Find the output of LTI system having impulse response $h [n] = {(1 / 3)}^{n} {u [n + 1] - u [n - 2]}$ and input signal $x [n] = {2, 1, 0.5, 3}$ . (2072 Kartik Q2)

The convolution sum, and four ways to do it

HOT 5/19Numerical2082 Baishakh Q1, 2080 Bhadra Q1, 2076 Chaitra Q2, 2075 Ashwin Q1, 2074 Chaitra Q2

A discrete time system has input $x [n]$ and output $y [n]$ and its input-output relation is $y [n] = x [n] + x [- n]$ . Is the system causal, time invariant or not?

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A discrete time system has input $x (n)$ and output $y (n)$ . The input output relation of the system is given by $y (n)$ = sum from k=0 to n of $x (k)$ . Check whether the system is memory less, time invariant and stable or not? (2080 Bhadra Q1)
Determine whether the following system are: a) $y [n] = x [- n]$ is time-invariant or not. b) $y [n]$ = x[ $n^{2}$ ] is linear or not. (2076 Chaitra Q2)
Determine whether the following sequences are linear or not: a) $y [n] = x^{2} [n]$ b) $y [n] = \cos (5 π n / 8 + π / 4)$ (2075 Ashwin Q1)
Determine whether the following system are: a) $y [n] = y [n - 4] + x [n - 4]$ is Time-invariant or not b) $y [n] = x^{2} [n]$ is Linear or Non-linear (2074 Chaitra Q2)

The five system properties, and how to test each one

HOT 4/19Numerical2081 Bhadra Q1, 2080 Bhadra Q2, 2080 Baishakh Q1, 2078 Bhadra Q1

Determine if the signal $x [n] = \cos (2 π n / 5) + \sin (π n / 3)$ is periodic or not. If the signal is periodic, find its fundamental period.

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Determine whether the given signal is periodic or not. If the signal is periodic, determine the fundamental period $x [n] = e^{j π n / 16} \cos (n π / 17)$ . (2080 Bhadra Q2)
Check whether following signals are periodic or not. If yes, state their periodic time. a) $x [n]$ = Sin(n $π$ ) + Cos(n $π$ ) b) $x [n]$ = Sin(3n $\frac{π}{5}$ ) + Cos(4n $\frac{π}{7}$ ) (2080 Baishakh Q1)
Determine whether the signal $x [n] = \cos (π n / 2) \cos (π n / 4)$ is periodic or non periodic and if it is periodic, find its fundamental period. (2078 Bhadra Q1)

Testing a signal for periodicity

PIN 3/19Numerical2082 Bhadra Q1, 2079 Baishakh Q1, 2072 Kartik Q1

Compare between energy signal and power signal. Determine whether the signal $x [n] = e^{j (π / 2 n + 4 π / 7)}$ is energy signal or power signal.

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Define energy and power signal. Check the signal $x [n] = u [n]$ and $x [n] = δ [n]$ is Energy or Power type. (2072 Kartik Q1)

Energy signals and power signals

PIN 2/19Numerical2079 Bhadra Q1, 2074 Ashwin Q1

Define energy and power signal. Determine whether the signal $x [n] = \cos (2 π n / 5) + \sin (π n / 3)$ is periodic or non-periodic and if it is periodic, find its fundamental period.

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Define Energy and Power type discrete time signal. Check whether signal $x [n] = e^{j (π n / 3 + π / 4)}$ is periodic or not. If it is periodic, state its periodic time. (2074 Ashwin Q1)

Energy signals and power signals Testing a signal for periodicity

PIN 2/192073 Shrawan Q1, 2072 Chaitra Q1

Explain the process of calculating fourier series coefficients.

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How fourier series coefficients are calculated? Explain. (2072 Chaitra Q1)

The discrete time Fourier series, and its properties

PIN 1/19Numerical2076 Chaitra Q1

Define even and odd type discrete time signals with suitable example. Plot the signal $x [- 2 n + 3]$ where $x [n] = {1, 2, 0, - 1, - 3, - 4}$ .

Even and odd parts of a signal Discrete time signals and the elementary sequences

PIN 1/192076 Ashwin Q1

Explain Fourier transform multiplication property for two sequences. Write Drichlet's conditions for Fourier series.

The discrete time Fourier series, and its properties The discrete time Fourier transform, and its proper...

PIN 1/192075 Chaitra Q1

Define Power and Energy type discrete time signal with suitable example. Differentiate between Fourier Series and Fourier Transform.

Energy signals and power signals The discrete time Fourier transform, and its proper...

PIN 1/19Numerical2074 Chaitra Q1

Plot the sequence $x [n] = u [n] - u [n - 3] + 5 δ [n - 4] = n u [n - 6]$ . List out the properties of LTI system.

The convolution sum, and four ways to do it Discrete time signals and the elementary sequences

PIN 1/19Numerical2071 Shrawan Q1

Find the odd and even part of the following signal: [a stem plot of $x [n]$ with values 1 at $n = - 4$ , -3, -2, -1 and 0, and 2 at $n = 1$ , 2, 3, 4]. A discrete time LTI system has input signal and impulse response as, $x [n] = 1$ for $- 1 \leq n \leq 1$ and 0 elsewhere, and $h [n] = 1$ for $- 1 \leq n \leq 1$ and 0 elsewhere. Find the output of the system using graphical method.

Even and odd parts of a signal The convolution sum, and four ways to do it

Chapter 2 · The z-transform 8 distinct questions, from 19 of the 19 sittings

HOT 7/19Numerical2082 Baishakh Q3, 2081 Bhadra Q3, 2080 Bhadra Q3, 2080 Baishakh Q3, 2078 Bhadra Q3, 2076 Chaitra Q4, 2073 Shrawan Q3

Define ROC of a Z-transform. Determine $x [n]$ for $X (z) = 1 / (1 - 1.5 z^{- 1} + 0.5 z^{- 2})$ if: a) $| z | > 1$ b) $| z | < 0.5$ c) $0.5 < | z | < 1$ using partial fraction method.

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Define ROC of z-transform. Find the inverse z-transform using partial fraction expansion of $X (z) = (2 z^{3} - 5 z^{2} + z + 3) / ((z - 1) (z - 2))$ for ROC $| z | < 1$ (2081 Bhadra Q3)
Define the Region of Convergence (ROC). Find the inverse of $H (z) = (1 + 2 z^{- 1} + z^{- 2}) / (1 - 0.75 z^{- 1} + 0.125 z^{- 2})$ ; ROC $0.25 < | z | < 0.5$ (2080 Bhadra Q3)
Define ROC. Find inverse z-transform of $X (z) = (1 + 2 z^{- 1} + z^{- 2}) / (1 - 1.5 z^{- 1} + 0.5 z^{- 2})$ , ROC: $| z | > 1$ . (2080 Baishakh Q3)
Define ROC. Find inverse z-transform of $X (z) = (z^{3} + z^{2} + 1.5 z + 0.5) / (z^{3} + 1.5 z^{2} + 0.5 z)$ , ROC: $| z | < 1 / 2$ . (2078 Bhadra Q3)
Define ROC of z-transform. Find inverse z-transform using partial fraction expansion of $X (z) = (z^{4} + 5 z^{3} - 3 z + 4) / (z^{2} - 1.5 z - 1)$ , ROC: $| z | < 0.5$ . (2076 Chaitra Q4)
Define a ROC. Find inverse Z-transform of $X (z) = z$ /{ $(z - 0.4) (z + 1.5)^{2}$ }, ROC: $| z | < 0.4$ (2073 Shrawan Q3)

Inverting

X (z)

by long division The region of convergence, and its properties

PIN 3/19Numerical2075 Chaitra Q6, 2075 Ashwin Q3, 2072 Chaitra Q3

List out the properties of Region of convergence and locate the ROC of the following signal $x [n] = {(0.1)}^{n} u [n] + {(0.3)}^{n} u [- n - 1]$

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List out the properties of Region of convergence and locate the ROC of the following signal. $x [n] = {(0.6)}^{n} u [n] + {(0.25)}^{n} u [n]$ (2075 Ashwin Q3)
Explain the properties of Region of Convergence with examples. (2072 Chaitra Q3)

The region of convergence, and its properties

PIN 2/19Numerical2082 Bhadra Q3, 2079 Baishakh Q3

Define z-transform for a discrete time signal. Find the inverse z-transform for $H (z) = z / (3 z^{2} - 4 z + 1)$ using partial fraction method for $1 / 3 < | z | < 1$ .

Inverting

X (z)

by long division The definition

PIN 2/19Numerical2081 Baishakh Q3, 2074 Ashwin Q3

Determine the inverse z-transform of $X (z) = 1 / (1 - 0.8 z^{- 1} + 0.12 z^{- 2}) . (i)$ if ROC is $| z | > 0.6$ (ii) if ROC is $| z | < 0.2$ (iii) if ROC is $0.2 < | z | < 0.6$

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State and explain the properties of a Region of Convergence (ROC). Find the inverse z-transform of $X (z) = z^{2} [1 - (3 / 2) z^{- 1}] (1 + z^{- 1}) (1 - z^{- 1})$ (2074 Ashwin Q3)

Inverting

X (z)

by long division The region of convergence, and its properties

PIN 2/19Numerical2079 Bhadra Q3, 2071 Shrawan Q2

Find inverses Z-transform of $X (z) = (2 z^{4} + 2 z^{3} - 3 z + 2) / (z^{2} - 1.5 z - 1)$ , ROC: $| z | < 0.5$ , using partial fraction expansion method.

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Find the inverse z transform of: $X (Z) = (1 + 2 z^{- 1} + z^{- 2}) / (1 + 1.5 z^{- 1} + 0.5 z^{- 2})$ , $| z | > 1$ using partial fraction method. (2071 Shrawan Q2)

Inverting

X (z)

by long division

PIN 1/19Numerical2076 Ashwin Q3

State Convolution property of Z-transform. Find inverse Z-transform of $X (z) = z$ /{ $(z - 0.6) (z + 0.5)^{2}$ }, ROC: $| z | > 0.6$

Inverting

X (z)

by long division The convolution property, derived

PIN 1/19Numerical2074 Chaitra Q3

Define a ROC. What are the properties of ROC of z-transform? Find the inverse Z-transform of $X (z) = (2 z^{2} + 2 z^{2} + 3 z + 5) / (z^{2} - 0.1 z - 0.2)$ , ROC: $| z | < 0.4$ .

Inverting

X (z)

by long division The region of convergence, and its properties

PIN 1/192072 Kartik Q3

State the properties of region of convergence (ROC). Drive the convolution property of Z-transform.

The region of convergence, and its properties The convolution property, derived

Chapter 3 · LTI systems in the frequency domain 7 distinct questions, from 19 of the 19 sittings

TOP 16/19Numerical2082 Bhadra Q4, 2082 Baishakh Q4, 2081 Bhadra Q4, 2081 Baishakh Q4, 2080 Bhadra Q4, 2079 Bhadra Q4, 2079 Baishakh Q4, 2078 Bhadra Q5, 2076 Chaitra Q5, 2076 Ashwin Q6, 2075 Chaitra Q3, 2075 Ashwin Q4, 2074 Chaitra Q4, 2074 Ashwin Q4, 2072 Chaitra Q5, 2072 Kartik Q5

Draw the poles and zeros in the z-plane for the system with poles at 0.45 +/- j1.6 and zeros at 0.58 +/- j2.06. Also plot the magnitude response (not in scale) of the system.

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The poles of a system are located at 0.45 +/- j1.6 and the zeros at 0.58 +/- j2.06. Map the poles and zeros in the z-plane and plot the magnitude and phase response (not to scale) of the system. (2082 Baishakh Q4)
Plot the pole zero in z plane and draw magnitude response (not to the scale) of the system described by difference equation $y [n] - 0.4 y [n - 1] + 0.25 y [n - 2] = x [n] - 0.4 x [n - 1]$ (2081 Bhadra Q4)
Plot the pole-zero in z plane and draw the magnitude response (not to the scale) of the system described by difference equation $y (n) = 0.67 x (n) - 0.3 x (n - 1) + 2.75 y (n - 1)$ (2081 Baishakh Q4)
Plot the pole-zero on the z-plane and draw Magnitude response (not to the scale) of an LTI system described by the equation, $y (n) = x (n) + 0.8 x (n - 1) + 0.8 x (n - 2) + 0.49 y (n - 2)$ . (2080 Bhadra Q4)
Plot the pole-zero in z-plane and draw the magnitude response (not to the scale) of the equation of the system describe by difference equation: $y [n]$ - 0. $35 y [n - 1]$ + 0. $25 y [n - 2] = x [n]$ - 0. $75 x [n - 1]$ . (2079 Bhadra Q4)
Plot pole-zero in z-plane and draw magnitude response (not to the scale) of the system described by difference equation $y [n] - 0.3 y [n - 1]$ + 0. $2 y [n - 2] = x [n]$ - 0. $5 x [n - 1]$ . (2079 Baishakh Q4)
Draw the pole-zero in the z-plane for a system with poles at 0.45 +/- j1.06 and zeroes at 0.58 +/- j2.06. Also plot the magnitude response (not to the scale) of the system. (2076 Chaitra Q5)
For the system described by the following difference equation: $y [n] = 0$ . $67 x [n]$ - 0. $3 x [n - 1]$ + 2. $75 y [n - 1]$ . Map the poles and zero in the z-plane and plot the phase response of the system. (2076 Ashwin Q6)
Plot the pole-zero in z-plane and draw magnitude response (not to scale) of the system described by differential equation $y (n) - 0.3 y (n - 1) = 2 x (n - 2) + 0.7 x (n - 1) + 4 x (n)$ (2075 Chaitra Q3)
Draw the poles and zeros in the z-plane for a system with poles at 0.45 +/- j1.06 and zeros at 0.58 +/- j2.06. Also plot the magnitude response of the system. (2075 Ashwin Q4)
The poles of a system are located at: 0.45 - 0.77i and -2 +/- 0.3i. Map the poles and zero in the z-plane and plot the magnitude response of the system. (2074 Chaitra Q4)
Plot the pole-zero in z-plane and Draw Magnitude Response (not to the scale) of the system described by difference equation $y [n]$ - 0. $4 y [n - 1]$ + 0. $2 y [n - 2] = x [n]$ + 0. $5 x [n - 1]$ + 0. $6 x [n - 2]$ + 0. $8 x [n - 3]$ (2074 Ashwin Q4)
Plot the pole-zero in z-plane and Draw Magnitude Response (not to the scale) of the system described by difference equation. $y [n]$ - 0. $4 y [n - 1]$ + 0. $1 y [n - 2] = x [n]$ + 0. $6 x [n - 1]$ (2072 Chaitra Q5)
Plot Magnitude Response (not to the scale) of the system described by difference equation. $y [n]$ - 0. $5 y [n - 1]$ + 0. $3 y [n - 2] = x [n]$ + 0. $7 x [n - 1]$ (2072 Kartik Q5)

Plotting poles and zeros and sketching the magnitud...

PIN 3/19Numerical2082 Baishakh Q2, 2080 Baishakh Q2, 2072 Kartik Q4

Find the output of an LTI system having impulse response $h [n] = {(1 / 2)}^{n} u [n]$ excited by an input $x [n] = 5 e^{j π n / 3}$ .

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Find the output of LTI system having impulse response $h [n] = {(1 / 2)}^{n} u [n]$ and input $x [n] = 5 e^{j π n / 3}$ for -infinity < n < infinity. (2080 Baishakh Q2)
Find the output of LTI System having impulse response $h [n] = {(1 / 2)}^{n} u [n]$ and input signal $x [n] = 5 e^{j π n / 3}$ for -infinity < n < infinity. (2072 Kartik Q4)

The frequency response, and why exponentials pass s...

PIN 2/192076 Ashwin Q4, 2072 Chaitra Q4

Describe stability and causality characteristics of LTI system in terms of Impulse Response and ROC of its transfer function with suitable examples.

Stability and causality, three ways of saying the s...

PIN 2/19Numerical2073 Shrawan Q4, 2071 Shrawan Q3

State linear constant coefficient difference equation and corresponding system function. Determine the output sequence of the system with impulse response $h [n] = {(1 / 2)}^{n} u [n]$ when the input signal is $x [n] = 10$ - 5sin( $π$ n/2) + 20cos( $π$ n), -infinity < n < infinity.

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Why do we need difference equation? State linear constant coefficient difference equation and corresponding system function. Consider an LTI system with impulse response $h [n] = {(1 / 2)}^{n} u [n]$ . Determine $y [n]$ , if the input is $x [n] = A e^{j π n}$ . (2071 Shrawan Q3)

Difference equations, and the system function The frequency response, and why exponentials pass s...

PIN 1/192081 Baishakh Q1

A discrete-time LTI system is given by difference equation $y (n) = x (n) + e^{a} y (n - 1)$ . Check this system for BIBO stability.

The BIBO stability test

PIN 1/19Numerical2080 Baishakh Q4

Differentiate between FIR system and IIR System. The poles of a system are located at 0.45 +/- j1.6 and zeros at 0.58 +/- j2.06. Map the poles and zeros in the z-plane and plot the magnitude response (not in scale) of the system.

FIR against IIR Plotting poles and zeros and sketching the magnitud...

PIN 1/19Numerical2078 Bhadra Q4

Determine the zero-input response for a second order system given by: $y [n] - 3 y [n - 1] - 4 y [n - 2] = x [n]$

The zero input response

Chapter 4 · Discrete filter structures 8 distinct questions, from 19 of the 19 sittings

TOP 9/19Numerical2082 Bhadra Q6, 2082 Baishakh Q5, 2081 Bhadra Q5, 2080 Bhadra Q5, 2080 Baishakh Q5, 2079 Baishakh Q5, 2076 Chaitra Q6, 2076 Ashwin Q5, 2075 Chaitra Q4

Compute Lattice ladder coefficients and draw lattice structure for the given system $H (z) = (2 - 0.7 z^{- 1} + 0.5 z^{- 2}) / (1 - 0.3 z^{- 1} + 0.25 z^{- 2})$

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A third order low pass filter has a system function $H (z) = (0.2759 + 0.5121 z^{- 1} + 0.5121 z^{- 2} + 0.2759 z^{- 3}) / (1 - 0.0010 z^{- 1} + 0.6546 z^{- 2} - 0.0775 z^{- 3})$ . Implement the filter using lattice-ladder structure. (2082 Baishakh Q5)
Compute the lattice and ladder coefficients and draw lattice-ladder structure for the given IIR system $H (z) = (1 + z^{- 1} + z^{- 2}) / ((1 + 0.5 z^{- 1}) (1 + 0.3 z^{- 1}) (1 + 0.4 z^{- 1}))$ (2081 Bhadra Q5)
Draw the Lattice structure from the following system function $. H (z) = 1 / (1 - 0.2 z^{- 1} + 0.4 z^{- 2} + 0.6 z^{- 3})$ (2080 Bhadra Q5)
Compute Lattice-ladder coefficients and draw lattice structure for given system $H (z) = (2 - 0.7 z^{- 1} + 0.5 z^{- 2}) / (1 - 0.3 z^{- 1} + 0.25 z^{- 2})$ . (2080 Baishakh Q5)
Compute Lattice-ladder coefficients and draw lattice structure for given system $H (z) = (1 - 0.4 z^{- 1} + 0.25 z^{- 2}) / (1 - 0.3 z^{- 1} + 0.5 z^{- 2})$ . Also check the stability of given system. (2079 Baishakh Q5)
Compute Lattice and Ladder coefficients and Draw lattice-ladder structure for given IIR system $H (z) = (0.5 - 2 z^{- 1} + 3 z^{- 2}) / (1 - 0.5 z^{- 1} - 0.7 z^{- 2} + 0.3 z^{- 3})$ . (2076 Chaitra Q6)
Compute Lattice and Ladder coefficients and Draw lattice-ladder structure for given IIR system $H (z) = (0.7 - 1.5 z^{- 1} + 0.5 z^{- 2}) / (1 - 0.5 z^{- 1} - 0.7 z^{- 2} + 0.3 z^{- 3})$ (2076 Ashwin Q5)
Draw the lattice structure from the following system function $H (t) = 1 / (1 + (2 / 3) z^{- 1} + (5 / 8) z^{- 2} + (2 / 3) z^{- 3} + z^{- 4})$ (2075 Chaitra Q4)

The IIR lattice ladder

HOT 6/19Numerical2082 Bhadra Q5, 2079 Bhadra Q5, 2079 Baishakh Q6, 2074 Chaitra Q5, 2072 Chaitra Q6, 2072 Kartik Q6

Obtain the direct form I and direct form II realization of the following system: $y [n]$ - 0. $75 y [n - 1]$ - 0. $25 y [n - 2] = x [n]$ + 0. $5 x [n - 1]$

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Draw direct form I and Direct form II realization of the following system. $y [n]$ - 0. $25 y [n - 2] + x [n]$ + 0. $4 x [n - 1]$ + 0. $5 x [n - 2]$ (2079 Bhadra Q5)
Obtain the Direct Form I and Direct Form II realization of the following system. $3 y [n] + y [n - 1] + 2 y [n - 4] = 2 x [n] + x [n - 3]$ (2074 Chaitra Q5)
Determine the Direct Form I and Direct Form II realization of the following system. $y (n) = - 0.1 y (n - 1) + 0.2 y (n - 2) + 3 x (n) + 3.6 x (n - 2) + 0.6 x (n - 2)$ (2072 Chaitra Q6)
Determine the Direct Form II realization of the following system $y (n) = - 0.1 y (n - 1) + 0.72 y (n - 2) + 0.7 x (n) - 0.252 x (n - 2)$ (2072 Kartik Q6)

FIR structures

HOT 4/19Numerical2081 Baishakh Q5b, 2074 Chaitra Q6, 2072 Chaitra Q7, 2072 Kartik Q7

Draw the lattice structure for the given FIR filter and also check whether the system is stable $. H (z) = 1 + (13 / 24) z^{- 1} + (5 / 8) z^{- 2} + (1 / 3) z^{- 3}$

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Determine the lattice coefficients coefficients corresponding to the FIR filter with the system function: $H (z) = A 3 (z) = 1 + (52 / 96) z^{- 1} + (25 / 40) z^{- 2} + (1 / 3) z^{- 3}$ (2074 Chaitra Q6)
Compute the lattice coefficients and draw the lattice structure of following FIR system $. H (z) = 1 + 2 z^{- 1} + z^{- 2}$ (2072 Chaitra Q7)
Compute the lattice coefficients and draw the lattice structure of following FIR system $H (z) = 1 + 2 z^{- 1} - 3 z^{- 2} + 4 z^{- 3}$ (2072 Kartik Q7)

The FIR lattice

PIN 3/19Numerical2081 Baishakh Q5a, 2080 Baishakh Q6, 2076 Chaitra Q7

Draw the cascaded form structure of $H (z)$ = 10(1 $- 0.25 z^{- 1}$ )(1 $- 0.667 z^{- 1}$ )(1 + $2 z^{- 1}$ )/((1 $- 0.75 z^{- 1}$ )(1 $- 0.125 z^{- 1}$ ){ $1 - (0.5 + j 0.5) z^{- 1}$ }{ $1 - (0.5 - j 0.5) z^{- 1}$ })

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Realize the given system in Cascade Form of 2nd order section flow graph representation. $H (z) = \frac{(1 - 0.4 z^{- 1}) (1 + 0.2 z^{- 1}) (1 - 0.3 e^{j π / 6} z^{- 1}) (1 - 0.3 e^{- j π / 6} z^{- 1})}{(1 - 0.5 e^{j π / 3} z^{- 1}) (1 - 0.5 e^{- j π / 3} z^{- 1}) (1 + 0.7 e^{j π / 4} z^{- 1}) (1 + 0.7 e^{- j π / 4} z^{- 1})}$ (2080 Baishakh Q6)
Realize the given system in Cascade form of 2nd order section in signal flow graph representation. $H (z) = \frac{(1 - 0.5 z^{- 1}) (1 + 0.35 z^{- 1}) (1 - 0.3 e^{j 2 n π / 5} z^{- 1}) (1 - 0.3 e^{- j 2 n π / 5} z^{- 1})}{(1 - 0.6 e^{j n π / 3} z^{- 1}) (1 - 0.6 e^{- j n π / 3} z^{- 1}) (1 + 0.5 e^{j 2 n π / 7} z^{- 1}) (1 + 0.5 e^{- j 2 n π / 7} z^{- 1})}$ (2076 Chaitra Q7)

Cascade and parallel forms

PIN 3/19Numerical2078 Bhadra Q6, 2074 Ashwin Q5, 2073 Shrawan Q5

The system function of a filter is $H (z) = 1 + (13 / 24) z^{- 1} + (5 / 8) z^{- 2} + (1 / 3) z^{- 3}$ . Draw the Direct Form and Lattice Structure implementation of the above filter.

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Draw the direct form and Lattice structure of a filter with system function $H (z) = 1 + 0.7 z^{- 1} + 1.2 z^{- 2} - z^{- 3}$ . (2074 Ashwin Q5)
The system function of a filter is $H (z) = 2 + 1.8 z^{- 1} - 1.6 z^{- 2} + z^{- 3}$ . Draw the Direct Form and Lattice Structure implementation of the above filter. (2073 Shrawan Q5)

FIR structures The FIR lattice

PIN 2/19Numerical2079 Bhadra Q6, 2071 Shrawan Q4

Given a 3-stage lattice filter for all zero polynomial with coefficients K1 = 1/4, K2 = $\frac{1}{2}$ and K3 = 1/3. Obtain the system function and FIR filter coefficients of this filter.

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If a 3 stage lattice filter for all pole polynomial has coefficients K1 = 1/4, K2 = $\frac{1}{2}$ and K3 = 1/3. Obtain the system function of this filter. (2071 Shrawan Q4)

From lattice coefficients back to the system function

PIN 1/19Numerical2075 Ashwin Q5

Draw the Lattice structure from the following system function: $1 / (3 + (39 / 24) Z^{- 1} + (15 / 8) Z^{- 2} + (3 / 9) Z^{- 3})$ . And represent 5/8 and -5/8 in sign magnitude, 1's complement and 2's complement format.

The IIR lattice ladder Fixed point number formats

PIN 1/192071 Shrawan Q5

What is the importance of quantization in Digital Signal Processing? Which one is better rounding or truncation? Explain about limit cycles in recursive system? Define dead band.

Quantisation, limit cycles and the dead band

Chapter 5 · FIR filter design 13 distinct questions, from 19 of the 19 sittings

TOP 11/192082 Bhadra Q8, 2081 Bhadra Q7, 2079 Bhadra Q8, 2079 Baishakh Q8, 2078 Bhadra Q8, 2076 Chaitra Q9, 2075 Chaitra Q5, 2074 Ashwin Q7, 2073 Shrawan Q7, 2072 Chaitra Q9, 2071 Shrawan Q7

What is an optimal filter? Explain Remez exchange algorithm for FIR filter design with the help of a flow chart.

Also set as:

What do you mean by optimum filter? Describe the Remez exchange algorithm for FIR filter design along with the flowchart. (2081 Bhadra Q7)
Discuss the Remez exchange algorithm for FIR filter design. (2079 Bhadra Q8)
What do you understand by optimum filter? Describe Remez exchange algorithm for FIR filter design along with the flowchart. (2079 Baishakh Q8)
What is optimum filter? Show mathematical expression of Remez exchange algorithm for FIR filter design. (2078 Bhadra Q8)
What is an optimum filter? Show mathematical expression of the Remez exchange algorithm for FIR filter design with flow chart. (2074 Ashwin Q7)
Explain about Remaz exchange algorithm with suitable derivation and flow chart. (2073 Shrawan Q7)
What is an optimum filter? Show mathematical expression of Remez exchange algorithm for FIR filter design. (2072 Chaitra Q9)
What is a Remez exchange algorithm? Derive its equation and draw its flow chart. (2071 Shrawan Q7)

The optimum filter and the Remez exchange algorithm

HOT 5/19Numerical2082 Bhadra Q7, 2081 Bhadra Q6, 2081 Baishakh Q6, 2079 Baishakh Q7, 2076 Chaitra Q8

Design a linear phase FIR system to meet the following specifications: Passband edge = 2 kHz, Stopband edge = 5 kHz, Stopband attenuation = 42 dB, Sampling frequency = 20 kHz

Also set as:

Design a low pass FIR filter using suitable window to meet following specifications: $0.99 \leq | H (e^{j ω}) | \leq 1.01$ for $0 \leq | w | \leq 0.3 π$ ; $| H (e^{j ω}) | \leq 0.01$ for $0.35 π$ ≤ | $ω$ | ≤ $π$ (2081 Bhadra Q6)
Design a linear phase FIR filter using suitable window to meet following specifications: $0.99 \leq | H (e^{j ω}) | \leq 1.01$ for $0 \leq | w | \leq 0.3 π$ ; $| H (e^{j ω}) | \leq 0.01$ for $0.35 π$ ≤ | $ω$ | ≤ $π$ (2081 Baishakh Q6)
Design a low pass digital FIR filter having Pass band edge frequency w_p = $0.2 π$ , Stop band edge frequency w_s = $0.45 π$ and Stop band attenuation a_s = 51 dB using any appropriate window function. (2079 Baishakh Q7)
Design the FIR filter using suitable window for the specifications: $0.899 \leq | H (e^{j ω}) | \leq 1$ for $| w | \leq 0.2 π$ ; $| H (e^{j ω}) | \leq 0.01$ for $0.4 π \leq ω \leq π$ (2076 Chaitra Q8)

The window method, step by step

PIN 3/19Numerical2080 Bhadra Q7, 2078 Bhadra Q7, 2074 Chaitra Q8

Kaiser window is to be used to design a linear phase FIR filter that meets following specification: $| H (e^{j ω}) | \leq 0.01$ for $0.21 π$ ≤ | $ω$ | ≤ $π$ ; $0.95 \leq | H (e^{j ω}) | \leq 1.05$ for $0 \leq | w | \leq 0.19 π$ . Calculate the optimum value of ripple, attenuation and window length.

Also set as:

Design a linear phase FIR filter using KAISER window to meet the following specifications: $| H (e^{j ω}) | \leq 0.01$ for $0 \leq | w | \leq 0.25 π$ ; $0.95 \leq | H (e^{j ω}) | \leq 1.05$ for $0.35 π \leq | w | \leq 0.6 π$ ; $| H (e^{j ω}) | \leq 0.01$ for $0.65 π$ ≤ | $ω$ | ≤ $π$ (2078 Bhadra Q7)
Design a FIR linear phase filter using Kaiser window that meets the following specifications: $| H (e^{j ω}) | \leq 0.01$ for $0 \leq | w | \leq 0.25 π$ ; $0.95 \leq | H (e^{j ω}) | \leq 1.05$ for $0.35 π \leq | w | \leq 0.6 π$ ; $| H (e^{j ω}) | \leq 0.01$ for $0.65 π$ ≤ | $ω$ | ≤ $π$ . Also determine the minimum length (M+1) of the impulse response and Kaiser window parameter $β$ . (2074 Chaitra Q8)

The Kaiser window

PIN 2/192081 Baishakh Q7, 2079 Bhadra Q7

What is Gibbs phenomenon and how can it be minimized? Why Kaiser window is better than other fixed windows in FIR filter design?

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Define Gibb's phenomenon. Design the FIR filter using Kaiser window technique for the specifications: $0.899 \leq | H (e^{j ω}) | \leq 1$ for $| w | \leq 0.2 π$ ; $| H (e^{j ω}) | \leq 0.01$ for $0.4 π \leq ω \leq π$ (2079 Bhadra Q7)

Gibbs' phenomenon The Kaiser window

PIN 2/19Numerical2075 Ashwin Q7, 2072 Kartik Q8

Design a linear phase FIR filter using Kaiser Window to meet the following specifications: $0.99 \leq | H (e^{j ω}) | \leq 1.01$ for $0 \leq w \leq 0.19 π$ ; $| H (e^{j ω}) | \leq 0.01$ for $0.21 π \leq ω \leq π$ . Draw the flow chart for Optimum filter design.

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Draw the flowchart of Remez-Exchange theorem and explain it. Design an FIR linear phase filter using Kaiser window to meet the following specifications: $0.99 \leq | H (e^{j ω}) | \leq 1.01$ for $0 \geq w \geq 0.19 π$ ; $| H (e^{j ω}) | \leq 0.01$ for $0.21 π \leq ω \leq π$ (2072 Kartik Q8)

The Kaiser window The optimum filter and the Remez exchange algorithm

PIN 2/19Numerical2074 Ashwin Q6, 2072 Chaitra Q8

Why Kaiser window is better than other fixed windows in FIR filter design? Find out first six coefficients of impulse response of a low pass FIR filter having Pass band edge frequency w_p = $0.2 π$ , Stop band edge frequency w_s = $0.5 π$ and Stop band attenuation a_s = 41dB using any appropriate window function.

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Describe how digital FIR filter can be design by window method. Why Kaiser window is better than other fixed windows in FIR filter design? (2072 Chaitra Q8)

The Kaiser window The window method, step by step

PIN 2/192073 Shrawan Q6, 2071 Shrawan Q6

Explain in detail about how rectangular window is used in FIR filter design. How Gibb's oscillations arise in this process.

Gibbs' phenomenon The window method, step by step

PIN 1/19Numerical2082 Baishakh Q6

Differentiate between analog and digital filter. Design an FIR lowpass digital filter that will have a -3dB cut-off at $30 π$ rad/s and an attenuation of 50 dB at $45 π$ rad/s. The filter is required to have a linear phase and the system uses a sampling rate of 100 Hz.

Why FIR, and when to choose it The window method, step by step

PIN 1/192082 Baishakh Q7

Explain Gibbs phenomenon in FIR filter design. Describe Remez exchange algorithm for FIR filter design along with flowchart.

Gibbs' phenomenon The optimum filter and the Remez exchange algorithm

PIN 1/19Numerical2080 Bhadra Q6

Design the symmetric FIR Low Pass Filter (LPF) for which the desired frequency response is expressed as Hd(W) = $e^{- j W t a u}$ for $| W | \leq W c$ and 0 elsewhere. The length of the filter should be 7 and $Ω_{c} = 1$ rad/sample. Make use of the Hanning window.

The linear phase condition The window method, step by step

PIN 1/19Numerical2080 Baishakh Q7

In which case do we choose FIR filter and IIR filter? Design a linear phase FIR filter using Kaiser Window to meet the following specifications: $0.99 \leq | H (e^{j ω}) | \leq 1.01$ for $0 \leq w \geq 0.016 π$ ; $| H (e^{j ω}) | \leq 0.01$ for $0.08 π \leq ω \leq 2 π$

Why FIR, and when to choose it The Kaiser window

PIN 1/192080 Baishakh Q8

Explain in detail about how Gibb's oscillation arise while using the rectangular window in FIR filter design.

Gibbs' phenomenon

PIN 1/19Numerical2076 Ashwin Q9

In which case do we choose FIR filter and IIR filter? Design a Kaiser Window to meet the following specifications: $0.99 \leq | H (e^{j ω}) | \leq 1.01$ for $0 \leq w \leq 0.16 π$ ; $| H (e^{j ω}) | \leq 0.01$ for $0.18 π \leq ω \leq 2 π$ . Draw the flow chart for Remez-Exchange algorithm

Why FIR, and when to choose it The Kaiser window The optimum filter and the Remez exchange algorithm

Chapter 6 · IIR filter design 8 distinct questions, from 19 of the 19 sittings

TOP 11/19Numerical2082 Bhadra Q9, 2080 Bhadra Q8, 2080 Baishakh Q9, 2076 Chaitra Q10, 2076 Ashwin Q7, 2075 Chaitra Q7, 2075 Ashwin Q6, 2074 Ashwin Q8, 2073 Shrawan Q8, 2072 Kartik Q9, 2071 Shrawan Q8

Explain frequency warping effect in detail. The passband and stopband frequencies are 350Hz and 1,000Hz respectively. The attenuation at passband and stopband are -3dB and -10dB respectively. The sampling frequency is 5,000Hz. Design a digital low-pass Butterworth filter using bilinear transformation technique.

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Using Bilinear transformation, design a Butterworth low pass filter which satisfies following conditions: $0.9 \leq | H (e^{j ω}) | \leq 1$ for $0 \leq ω \leq \frac{π}{2}$ ; $| H (e^{j ω}) | \leq 0.2$ for $\frac{3 π}{4} \leq ω \leq π$ . Consider sampling frequency of 1 Hz. (2080 Bhadra Q8)
Design a low pass digital IIR filter by Bilinear Transformation method to an approximate Butterworth low pass filter, if passband edge frequency is $0.26 π$ radians and maximum deviation of 0.99 dB below 0 dB gain in the passband. The maximum gain of -14.99 dB and frequency is $0.58 π$ radians in stopband, Consider sampling frequency 0.5 Hz. (2080 Baishakh Q9)
Design a digital low pass Butterworth filter by applying bilinear transformation techniques for the given specifications: Passband peak to peak ripple ≤ 1dB, Passband edge frequency = 1.2KHz, Stopband Attenuation ≥ 40dB, Stopband edge frequency = 2.5 KHz, Sample rate = 8KHz (2076 Chaitra Q10)
Design a low pass discrete IIR filter by Bilinear Transformation method to an approximate Butterworth filter having specifications as below: Pass bandedge frequency ( $ω_{p}$ ) = $0.22 π$ radians, Stop bandedge frequency ( $ω_{s}$ ) = $0.54 π$ radians, Passband ripple (deltap) = 0.11, Stopband ripple (deltas) = 0.22, Consider sampling frequency 0.5 Hz. (2076 Ashwin Q7)
Using bilinear transformation, design a digital filter using Butterworth approximation which satisfies the following conditions: $0.8 \leq | H (e^{j ω}) | \leq 1$ for $0 \leq W \leq 0.2 π$ ; $| H (e^{j ω}) | \leq 0.2$ for $0.6 π \leq W \leq π$ (2075 Chaitra Q7)
Design a digital low-pass filter with the following specification: i) Pass-band magnitude constant to 0.7 dB below the frequency of $0.15 π$ ii) Stop-band attenuation at least 14 dB for the frequencies between $0.6 π$ to $π$ . Use Butterworth approximation as a prototype and use bilinear transformation method to obtain the digital filter. (2075 Ashwin Q6)
Design a low pass discrete IIR filter by Bilinear Transformation method to an approximate Butterworth filter having specifications as below: Pass bandedge frequency ( $ω_{p}$ ) = $0.27 π$ radians, Stop bandedge frequency ( $ω_{s}$ ) = $0.58 π$ radians, Passband ripple (deltap) = 0.11, Stopband ripple (deltas) = 0.21, Consider sampling frequency 0.5 Hz. (2074 Ashwin Q8)
Using bilinear transformation, design a butterworth low pass filter which satisfies the following Magnitude Response: $0.89125 \leq | H (e^{j ω}) | \leq 1$ for $0 \leq w \leq 0.2 π$ ; $| H (e^{j ω}) | \leq 0.17783$ for $0.3 π \leq ω \leq π$ (2073 Shrawan Q8)
Design a low pass digital filter by Bilinear Transformation method to an approximate Butterworth filter, if passband edge frequency is $0.25 π$ radians and maximum deviation of 1 dB below 0 dB gain in the passband. The maximum gain of -15 dB and frequency is $0.45 π$ radians in stopband, Consider sampling frequency 1Hz. (2072 Kartik Q9)
Design a low pass digital filter by Bilinear Transformation method to an approximate Butterworth filter it passband frequency is $0.2 π$ radians and maximum deviation of 1 db below 0 dB gain in the pass band. The maximum gain of -15 db and frequency is $0.4 π$ radians in stop band, consider sampling frequency 1 Hz. (2071 Shrawan Q8)

The design route

HOT 4/19Numerical2082 Baishakh Q8, 2081 Bhadra Q8, 2081 Baishakh Q8, 2079 Baishakh Q9

Using bilinear transformation technique, design a Butterworth filter which satisfies the following conditions: $0.8 \leq | H (e^{j ω}) | \leq 1$ for $0 \leq w \leq 0.2 π$ ; $| H (e^{j ω}) | \leq 0.2$ for $0.6 π \leq ω \leq π$ . Consider sampling frequency 1 Hz. Compare Impulse Invariance Method and Bilinear Transformation technique for analog filter to digital filter conversion.

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Design a digital low pass IIR filter using Bilinear transformation to meet following specifications: $0.9 \leq | H (e^{j ω}) | \leq 1$ for $0 \leq | w | \leq \frac{π}{2}$ ; $| H (e^{j ω}) | \leq 0.2$ for $3 π / 4 \leq | w | \leq π$ . Use Butterworth approximation for design with sampling frequency of 1Hz. Compare Impulse Invariance Method and Bilinear Transformation method. (2081 Bhadra Q8)
Differentiate between bilinear transformation and impulse invariance. Design a Butterworth digital IIR lowpass filter using bilinear transformation by taking $T = 0.1$ second, to satisfy the following specifications: $0.6 \leq | H (e^{j ω}) | \leq 1$ for $0 \leq w \leq 0.35 π$ ; $| H (e^{j ω}) | \leq 0.1$ for $0.7 π \leq ω \leq π$ (2081 Baishakh Q8)
Design a low pass digital IIR filter by Bilinear Transformation method to an approximate Butterworth low pass filter, if passband edge frequency is $0.24 π$ radians and maximum deviation of 0.98 dB below 0 dB gain in the passband. The maximum gain of -14.95 dB and frequency is $0.57 π$ radians in stopband, consider sampling frequency 0.5 Hz. Compare impulse invariance method with bilinear transformation method. (2079 Baishakh Q9)

Bilinear transformation against impulse invariance The design route

PIN 1/192080 Baishakh Q10

Describe digital domain Spectral Transformation features and parameters for low pass to high pass filter in IIR Filter design.

Digital domain spectral transformation

PIN 1/19Numerical2079 Bhadra Q9

Design a low pass discrete time Butterworth filter using bilinear transformation having following specifications: Passband frequency ( $Ω_{p}$ ) = $0.25 π$ radians, Stopband frequency ( $Ω_{s}$ ) = $0.55 π$ radians, Passband ripple (deltaP) = 0.11, Stopband ripple (deltaS) = 0.21. Consider sampling frequency of 0.5 Hz. Also, convert the obtained digital low pass filter to high pass filter with new pass band frequency, W'p = $0.45 π$ using digital domain transformation.

The design route Digital domain spectral transformation

PIN 1/19Numerical2078 Bhadra Q9

Design a LPF Butterworth filter using Impulse Invariance Method (IIM) method with passband and stopband frequencies 200Hz and 500Hz respectively. The passband and stopband attenuations are 5dB and 12dB respectively. The sampling frequency is 5000Hz. What is pre-warping and why it is necessary? Explain.

The design route Impulse invariance

PIN 1/19Numerical2074 Chaitra Q7

Design a digital low-pass filter with the following specification: i) Pass-band magnitude constant to 0.7 dB below the frequency of $0.15 π$ ii) Stop-band attenuation at least 14 dB for the frequencies between $0.6 π$ to $π$ . Use Butterworth approximation as a prototype and use impulse invariance method to obtain the digital filter.

The design route Impulse invariance

PIN 1/192073 Shrawan Q9

Explain briefly about bilinear transformation method of IIR filter design.

The design route

PIN 1/19Numerical2072 Chaitra Q10

Explain about the advantages of selecting bilinear transformation method over impulse invariance method (I I M). Design a digital low pass Butterworth filter using impulse invariant transformation with pass band and stop band frequencies 200Hz and 500Hz respectively. The pass band and stop band attenuation are -5dB and -12dB respectively. The sampling frequency is 5kHz. Use IIM method.

Bilinear transformation against impulse invariance Impulse invariance

Chapter 7 · The discrete Fourier transform 13 distinct questions, from 19 of the 19 sittings

TOP 12/19Numerical2082 Bhadra Q10, 2082 Baishakh Q10, 2081 Bhadra Q10, 2081 Baishakh Q10, 2080 Bhadra Q10, 2079 Bhadra Q11, 2079 Baishakh Q11, 2076 Chaitra Q12, 2075 Chaitra Q9, 2073 Shrawan Q11, 2072 Chaitra Q12, 2071 Shrawan Q9

Compute the circular convolution of the following sequences: $h (n) = {1, 2, 1, - 1, 1}$ and $x (n) = {1, 2, 3, 1}$

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Find the circular convolution of the sequences $x_{1} [n] = {1, 1, - 1, - 1, 2}$ and $x_{2} [n] = {1, - 1, - 2}$ . (2082 Baishakh Q10)
If X1(k) and X2(k) are the 5-point DFT of $x_{1}$ (n) = $3^{n}$ for $0 \leq n \leq 3$ and $x_{2}$ (n) = $2^{n}$ for $0 \leq n \leq 4$ . Find $x_{3}$ (n) if X3(k) = X1(k)X2(k) (2081 Bhadra Q10)
Compute circular convolution of the following two sequences using DFT. $x (n) = {1, 2, 4, 5}$ and $h (n) = {2, 1, 6, 8}$ (2081 Baishakh Q10)
Obtain the circular convolution of the following sequences: $x_{1}$ (n) = ${1, 2, 3, 1}$ and $x_{2}$ (n) = ${4, 3, 2, 2}$ (2080 Bhadra Q10)
If X1(k) and X2(k) are DFT of sequence $x_{1} [n] = {1, 0, 0, 1}$ and $x_{2} [n] = {2, 0, 2}$ respectively then find the sequence $x_{3} [n]$ ; if DFT of $x_{3} [n]$ is given by X3(k) = X1(k) X2(k). (2079 Bhadra Q11)
Find the circular convolution of the sequences $x_{1} [n] = {1, - 1, - 2, 3, - 1}$ and $x_{2} [n] = {1, 2, 3}$ . (2079 Baishakh Q11)
Find $x_{3} [n]$ if DFT of $x_{3} [n]$ is given by X3(k) = X1(k) * X2(k) where X1(k) and X2(k) are 4-point DFT of $x_{1} [n] = {1, 2, - 2}$ and $x_{2} [n] = {1, 2, 3, - 1}$ respectively. (2076 Chaitra Q12)
Differentiate between linear convolution and circular convolution compute circular convolution of signals $X_{1} [n] = {0, 0, 1, 1}$ and $X_{2} [n] = {1, 1, 1, 1}$ (2075 Chaitra Q9)
Find $x_{3} [n]$ if DFT of $x_{3} [n]$ is given by X3(k) = X1(k) X2(k) where X1(k) and X2(k) are 5-point DFT of $x_{1} [n] = {1, - 2, 2, 1, 4}$ and $x_{2} [n] = {2, 1, - 3, - 1}$ respectively. (2073 Shrawan Q11)
Compute Circular Convolution of $h (n) = {1, 2, 1, - 1, 1}$ and $x [n] = {1, 2, 3, 1}$ . (2072 Chaitra Q12)
A system has input signal $x [n] = {1, 2, 3, 4}$ and impulse response $h [n] = {1, 3, 5, 7}$ and the DFT of $x [n]$ is $X [k]$ and the DFT of $h [n]$ is $H [k]$ . Find the output of the system $y [n]$ if $G [k] = X [k]$ . $H [k]$ (2071 Shrawan Q9)

Circular convolution, and how it differs from linear

HOT 6/19Numerical2081 Bhadra Q9, 2081 Baishakh Q9, 2080 Baishakh Q11, 2079 Bhadra Q10, 2075 Chaitra Q8, 2075 Ashwin Q8

How does FFT reduce computational complexity compared to direct calculation of DFT? Compute 8-point DFT of $x [n] = {1, - 2, - 4, 3, 0, 1}$ using DIT-FFT algorithm.

Also set as:

Why Decimation in Time Fast Fourier Transform (DITFFT) Algorithm is better than direct computation of DFT? Find 4 point DFT of the sequence $x (n) = {2, 2, 4}$ using DITFFT algorithm. (2081 Baishakh Q9)
How fast is FFT? Find 8-point DFT of sequence $x [n] = {1, 1, 0, 0, 1, 1, 2}$ using Decimation in Time Fast Fourier Transform (DITFFT) algorithm. (2080 Baishakh Q11)
Why we need FFT? Find the 8-point DFT of the following sequence using radix-2 DITFFT algorithm. (2079 Bhadra Q10)
How fast is FFT? Find X(3) and X(5) for given sequence $x [n] = {1, - 2, 3, 2}$ using DITFFT algorithm. (2075 Chaitra Q8)
How fast is FFT compare to DFT? Draw the butterfly diagram of 8-point DFT of a sequence as $x [n] = n$ + 1 using Decimation in Time FFT algorithm. (2075 Ashwin Q8)

Decimation in time Why the FFT is fast

HOT 4/19Numerical2082 Bhadra Q11, 2080 Bhadra Q9, 2076 Chaitra Q11, 2074 Ashwin Q9

Find 8-point DFT using DIF-FFT of the sequence $x [n]$ = {1, $\frac{1}{2}$ , -1, $- \frac{1}{2}$ , 2, -3/2}.

Also set as:

Compute 8-point DIF-FFT of sequence $x (n) = {2, 1, 2, 1, 1, 2, 1, 2}$ . (2080 Bhadra Q9)
Find 8-point DFT of sequence $x [n] = {1, 2, 3, 3, 5, 0, 4, 6}$ using Decimation in frequency Fast Fourier Transform (DIFFFT) algorithm. (2076 Chaitra Q11)
Compute the 8-point DFT of the sequence $x [n]$ = { $\frac{1}{2}$ , $\frac{1}{2}$ , $\frac{1}{2}$ , $\frac{1}{2}$ , 0, 0, 0, 0} using Decimation in Frequency Fast Fourier Transform (DIF-FFT) algorithm. (2074 Ashwin Q9)

Decimation in frequency

HOT 4/19Numerical2078 Bhadra Q11, 2072 Chaitra Q11, 2072 Kartik Q10, 2071 Shrawan Q10

Find the 8-point DFT of $x [n] = u [n] - u [n - 4]$ using FFT DIT algorithm.

Also set as:

Find the FFT of the signal $x [n] {1, 1, 2, 4, 3, 1, 2, 1}$ using DIT-FFT algorithm. (2072 Chaitra Q11)
Find 8-point DFT of sequence $x [n] = {1, 1, 0, 1, 0, 1, 2}$ using Decimation in Time Fast Fourier Transform (DITFFT) algorithm. (2072 Kartik Q10)
Find DFT for ${1, 1, 2, 0, 1, 2, 0, 1}$ using FFT DIT butterfly algorithm and plot the spectrum. (2071 Shrawan Q10)

Decimation in time

PIN 3/19Numerical2079 Baishakh Q10, 2076 Ashwin Q8, 2074 Chaitra Q9

Why we need DFT? Find 8-point DFT of sequence $x [n] = {1, 2, 4, 3, 5, - 1, 3}$ using Decimation in Frequency Fast Fourier Transform (DIFFFT) algorithm.

Also set as:

Why we need DFT? Find 8-point DFT of sequence $x [n] = {1, 2, 3, 3, 5, 1, 4, 2}$ using Decimation in frequency Fast Fourier Transform (DIFFFT) algorithm. (2076 Ashwin Q8)
Why do we need DFT? Draw the butterfly structure to compute the DFT of the following signal using Radix-2 DIFFFT algorithm, and compute X(2) and X(1) only $x [n] = {1.5, - 1, 1.8, 0.6, 3, 1.7}$ . (2074 Chaitra Q9)

The DFT and the IDFT Decimation in frequency

PIN 1/19Numerical2082 Baishakh Q9

What is Fast Fourier Transform algorithm? How Decimation in Frequency Fast Fourier Transform algorithm reduces the calculation complexity of DFT? Find 4 point DFT of the sequence $x [n] = {1, 4, - 1}$ using Decimation in Frequency Fast Fourier Transform algorithm.

Decimation in frequency Why the FFT is fast

PIN 1/19Numerical2080 Baishakh Q12

Write the complexity of DFT and FFT? Obtain the circular convolution of the following sequences: $X_{1} [n] = {1, 2, 3, 1}$ and $X_{2} [n] = {4, 3, 2, 2}$

Circular convolution, and how it differs from linear Why the FFT is fast

PIN 1/19Numerical2078 Bhadra Q10

Differentiate between DFT and DTFT. Find the circular convolution of $x_{1} [n] = {2, 1, 2, 1}$ and $x_{2} [n] = {1, 2, 3, 4}$

Circular convolution, and how it differs from linear DFT against DTFT

PIN 1/19Numerical2075 Ashwin Q9

State the circular convolution property of DFT. Find the circular convolution of: (a) ${1, 2, - 1, 1}$ and $x_{2}$ (n) = ${1, 3, 5, 7}$

Circular convolution, and how it differs from linear The properties

PIN 1/19Numerical2074 Chaitra Q10

Define zero padding. Find the linear convolution through circular convolution with padding of zeros for the following sequences: $x [n] = {1, 1, 1, 1}$ and $h [n] {2, 3}$ .

Zero padding

PIN 1/19Numerical2074 Ashwin Q10

What is a zero padding? If X1(k) and X2(k) are DFT of sequence $x_{1} [n] = {1, 2, 0, 1, - 2}$ and $x_{2} [n] = {1, 0, 1, 1, 2}$ respectively then find the sequence $x_{3} [n]$ ; If DFT of $x_{3} [n]$ is given by X3(k) = X1(k) X2(k).

Circular convolution, and how it differs from linear Zero padding

PIN 1/19Numerical2073 Shrawan Q10

Why do we need DFT? Find 8-point DFT of sequence $x [n] = {1, - 1, 2, 2, 1, 1, 2, 2}$ using Fast Fourier Transform algorithm.

The DFT and the IDFT Decimation in time

PIN 1/19Numerical2072 Kartik Q11

Why we need DFT? If X1(k) and X2(k) are DFT of sequence $x_{1} [n] = {1, 2, 4}$ and $x_{2} [n] = {- 1, 2, 3, 1}$ respectively, then find the sequence $x_{3} [n]$ , if DFT of $x_{3} [n]$ is given by X3(k) = X1(k) X2(k).

Circular convolution, and how it differs from linear The DFT and the IDFT

39 formulas · one scroll through all seven chapters · click any formula for its derivation

The formula sheet

Every formula this subject asks you to carry, in syllabus order, as one continuous scroll. The bar below narrows it to a single chapter when you are revising one. Click a formula to see where it comes from: a formula you cannot derive is a formula you will misremember under pressure, and several of these questions award the derivation on its own.

How to use this page

Read it as one scroll the week before, and as one chapter the night before.
Open the derivation once per formula, then close it and see whether you can rebuild it. That is the difference between a formula you have read and one you own.
Each formula says what it earns, so none of them is abstract: the line underneath names the question it is for and how often that question has been set.
Test yourself on these in the flashcards section, which asks these same formulas and comes back more often to the ones you miss.

1Discrete time signals and systems

The energy of a discrete time signal

E = \sum_{n = - \infty}^{\infty} | x [n] |^{2}

Deciding whether a signal is an energy signal, which every sitting pairs with the periodicity test.

Where it comes from

The instantaneous power of a sample is $| x [n] |^{2}$ , by analogy with $v^{2}$ /R in a one ohm resistor. Energy is power accumulated over time, and in discrete time accumulation is a sum, so the energy is the sum of $| x [n] |^{2}$ over every n at which the signal exists.

The modulus matters: for a complex signal, $x [n]^{2}$ is not the power and can even be negative. Writing $| x [n] |^{2} = x [n] x^{*} [n]$ is what keeps it real and positive.

The average power of a discrete time signal

\begin{aligned} P & = lim_{N \to \infty} \frac{1}{2 N + 1} \sum_{n = - N}^{N} | x [n] |^{2} \\ periodic: P & = \frac{1}{N} \sum_{n = 0}^{N - 1} | x [n] |^{2} \end{aligned}

The other half of the energy or power question, and the reason a complex exponential is a power signal with

P = 1

in one line.

Where it comes from

Average power is energy per sample. Over the window -N to N there are $2 N + 1$ samples, so the average is the energy in that window divided by $2 N + 1$ , and the limit widens the window to the whole signal.

The $2 N + 1$ , rather than 2N, is where the answer P = $\frac{1}{2}$ for the unit step comes from: the window holds $2 N + 1$ samples of which only $N + 1$ are non zero, and $(N + 1) / (2 N + 1)$ tends to $\frac{1}{2}$ .

For a periodic signal every period contributes the same amount, so the limit collapses to the average over one period and no limit needs taking.

The condition for a discrete time sinusoid to be periodic

\begin{aligned} periodic & ⟺ \frac{ω_{0}}{2 π} is rational \\ \frac{ω_{0}}{2 π} & = \frac{k}{N} in lowest terms ⟹ period N \\ several: N & = LCM (N_{1}, N_{2}, \dots) \end{aligned}

Question 1 in six of the nineteen sittings, on its own or beside the energy test.

Where it comes from

Periodicity means $x [n + N] = x [n]$ for an integer N. For $x [n] = e^{j ω_{0} n}$ that demands $e^{j ω_{0} (n + N)} = e^{j ω_{0} n}$ , so $e^{j ω_{0} N} = 1$ , so $ω_{0}$ N = $2 π$ k for some integer k.

Rearranged, $ω_{0} / (2 π) = k$ /N: a ratio of integers. If $ω_{0} / (2 π)$ is irrational no integer N can satisfy it and the signal is aperiodic, however smooth it looks.

This is the one real difference from continuous time, where any period is allowed and every sinusoid is periodic.

The even and odd parts of a signal

x_{e} [n] = \frac{x [n] + x [- n]}{2}, x_{o} [n] = \frac{x [n] - x [- n]}{2}, x [n] = x_{e} [n] + x_{o} [n]

A two or three mark opener, and the reason every real spectrum is conjugate symmetric.

Where it comes from

Assume the split exists: $x [n] = x_{e} [n] + x_{o} [n]$ . Replace n by -n and use the definitions $x_{e} [- n] = x_{e} [n]$ and $x_{o} [- n] = - x_{o} [n]$ , which gives $x [- n] = x_{e} [n] - x_{o} [n]$ .

Add the two equations and the odd parts cancel, leaving 2 $x_{e} [n]$ . Subtract them and the even parts cancel, leaving 2 $x_{o} [n]$ . Halving each gives the formulas, and the derivation also proves the split is unique.

The convolution sum

\begin{aligned} y [n] & = \sum_{k = - \infty}^{\infty} x [k] h [n - k] = x [n] * h [n] \\ starts at n_{1} + m_{1}, length N_{x} + N_{h} - 1 \end{aligned}

The most set question in the subject: sixteen of nineteen sittings ask for a

y [n]

Where it comes from

Three steps, and each uses one property:

1. Sifting. Any sequence is a sum of scaled, shifted impulses: $x [n] = \sum_{k} x [k] δ [n - k]$ .

2. Linearity. The response to that sum is the sum of the responses: $y [n] = \sum_{k} x [k]$ T{ $δ [n - k]$ }.

3. Time invariance. T{ $δ [n - k]$ } = $h [n - k]$ , because the system does not care when the impulse arrives.

Putting them together gives the sum, and it also proves the claim that makes the subject work: $h [n]$ alone describes the system completely.

The discrete time Fourier series pair

\begin{aligned} x [n] & = \sum_{k = 0}^{N - 1} c_{k} e^{j 2 π k n / N} & synthesis \\ c_{k} & = \frac{1}{N} \sum_{n = 0}^{N - 1} x [n] e^{- j 2 π k n / N} & analysis \end{aligned}

Explaining the process of finding the coefficients, which two sittings ask outright.

Where it comes from

Multiply the synthesis equation by $e^{- j 2 π m n / N}$ and sum over one period in n, then interchange the two sums.

The inner sum is $\frac{1}{N}$ times the sum over one period of $e^{j 2 π (k - m) n / N}$ . It is a geometric series with ratio $e^{j 2 π (k - m) / N}$ , whose numerator 1 $- e^{j 2 π (k - m)}$ is zero unless k = m, where every term is 1 and the sum is N. That is orthogonality.

Every term therefore dies except k = m, leaving N c[m], and dividing by N gives the analysis equation.

The discrete time Fourier transform pair

\begin{aligned} X (e^{j ω}) & = \sum_{n = - \infty}^{\infty} x [n] e^{- j ω n} & analysis \\ x [n] & = \frac{1}{2 π} \int_{- π}^{π} X (e^{j ω}) e^{j ω n} d ω & synthesis \\ exists when \sum | x [n] | < \infty \end{aligned}

Assumed knowledge in every frequency domain question, and the object every filter design shapes.

Where it comes from

Take the Fourier series of a periodic signal and let the period N grow without bound. The spacing between the harmonics, $2 π$ /N, shrinks to zero, the discrete lines N $c_{k}$ merge into a continuous function of $ω$ , and the sum in the synthesis equation becomes an integral.

Periodic in $2 π$ , always, because $e^{- j (ω + 2 π) n} = e^{- j ω n}$ for integer n. All the information is in one period, which is why every spectrum in this subject is drawn over $- π$ to $π$ .

The multiplication property of the DTFT

x_{1} [n] x_{2} [n] ⟷ \frac{1}{2 π} \int_{- π}^{π} X_{1} (e^{j θ}) X_{2} (e^{j (ω - θ)}) d θ

The whole explanation of the window method and of Gibbs oscillation, and one sitting asks for the property itself.

Where it comes from

Write the transform of the product: $X (e^{j ω}) = \sum_{n} x_{1} [n] x_{2} [n] e^{- j ω n}$ .

Replace $x_{1} [n]$ by its inverse transform, (1/ $2 π$ ) times the integral of X1( $e^{j}$ th) $e^{j θ n}$ d th, and interchange the sum and the integral.

What is left inside is $\sum_{n} x_{2} [n] e^{- j (ω - θ) n}$ , which is X2( $e^{j}$ ( $ω$ - th)).

The convolution is periodic, over one span of $2 π$ , because both spectra are: an ordinary infinite convolution of two periodic functions would diverge.

The spectrum of a sampled signal

\begin{aligned} X (e^{j ω}) & = \frac{1}{T} \sum_{k = - \infty}^{\infty} X_{a} (j \frac{ω - 2 π k}{T}) \\ ω & = Ω T = \frac{2 π F}{F_{s}}, F_{s} > 2 F_{max} \end{aligned}

The theory behind every sampling question, and the reason an anti alias filter exists.

Where it comes from

Sampling is multiplication by an impulse train of period T. The transform of an impulse train is another impulse train, spaced $2 π$ /T apart.

By the multiplication property, multiplying in time convolves in frequency, and convolving anything with an impulse train makes copies of it at the impulse positions. Hence the sum of shifted copies, scaled by $\frac{1}{T}$ .

If the copies are spaced closer than the signal is wide they overlap, high frequencies fold down and are indistinguishable from low ones: that is aliasing, and the condition that prevents it is the sampling theorem.

2The z-transform

The z-transform

X (z) = \sum_{n = - \infty}^{\infty} x [n] z^{- n} (quote the ROC with it, always)

Two sittings ask for the definition, and every question in chapters 2 to 6 uses it.

Where it comes from

Write z = r $e^{j ω}$ and the sum becomes sum of $x [n] r^{- n} e^{- j ω n}$ : the DTFT of the sequence $x [n] r^{- n}$ . The extra factor $r^{- n}$ is a convergence aid, and it lets the transform exist for sequences whose DTFT does not, such as the unit step or a growing exponential.

On the unit circle, $r = 1$ , the z-transform is the DTFT. That is why the unit circle appears in every stability statement.

The region of convergence, and what it tells you

\begin{aligned} finite length & : the whole plane, bar z = 0 or \infty \\ right sided & : | z | > r_{max} \\ left sided & : | z | < r_{min} \\ two sided & : r_{1} < | z | < r_{2} \\ stable & ⟺ the ROC contains | z | = 1 \\ causal & ⟺ the ROC is outside the outermost pole \end{aligned}

Two marks in eight sittings for the definition, seven more for the properties, and the deciding step of every inverse transform.

Where it comes from

Convergence of sum | $x [n] z^{- n}$ | depends only on $| z |$ , so the ROC can only be a set of radii: a ring centred on the origin. It contains no pole, since $X (z)$ is infinite there.

A right sided sequence has only negative powers of z, which shrink as $| z |$ grows, so it converges outward. A left sided one has only positive powers and converges inward. A two sided one needs both at once, which is a ring.

The ROC containing $| z | = 1$ is precisely the statement that sum $| x [n] |$ converges, which is absolute summability, which is BIBO stability.

Inverting

X (z)

by partial fractions

\begin{aligned} 1. divide first if \deg (num) \geq \deg (den) \\ 2. expand \frac{X (z)}{z}, not X (z) \\ A_{k} = {[(z - p_{k}) \frac{X (z)}{z}]}_{z = p_{k}} \\ 3. ROC outside p_{k} : A_{k} p_{k}^{n} u [n] \\ ROC inside p_{k} : - A_{k} p_{k}^{n} u [- n - 1] \end{aligned}

Question 3 in fifteen of the nineteen sittings.

Where it comes from

The standard pair is $z / (z - a)$ , not $1 / (z - a)$ , so every term must arrive with a z on top. Expanding $X (z)$ /z and multiplying back by z at the end is what guarantees that; expanding $X (z)$ directly leaves terms of the form $A / (z - a)$ , whose inverse is a shifted version and a common source of lost marks.

The two possible inverses come from the geometric series read in opposite directions: $z / (z - a) = 1 / (1 - a z^{- 1})$ expands in powers of $a z^{- 1}$ when $| z | > | a |$ , which is a causal sequence, and in powers of z/a when $| z | < | a |$ , which is anticausal and carries the minus sign.

The convolution property of the z-transform

x_{1} [n] * x_{2} [n] ⟷ X_{1} (z) X_{2} (z) ⟹ Y (z) = H (z) X (z)

It defines

H (z)

, which is the object every remaining chapter manipulates.

Where it comes from

$Y (z) = \sum_{n}$ [ $\sum_{k} x_{1} [k] x_{2}$ [n-k] ] $z^{- n}$ . Interchange the sums, valid inside the common ROC, and substitute m = n - k so that $z^{- n} = z^{- m} z^{- k}$ .

The two sums then separate completely: $\sum_{k} x_{1} [k] z^{- k}$ times $\sum_{m} x_{2} [m] z^{- m}$ , which is X1(z) X2(z).

The ROC is at least the intersection of the two, and can be larger if a pole of one cancels a zero of the other.

The z-transform pairs worth memorising

\begin{aligned} δ [n] & ⟷ 1 & all z \\ u [n] & ⟷ \frac{z}{z - 1} & | z | > 1 \\ a^{n} u [n] & ⟷ \frac{z}{z - a} & | z | > | a | \\ - a^{n} u [- n - 1] & ⟷ \frac{z}{z - a} & | z | < | a | \\ n a^{n - 1} u [n] & ⟷ \frac{z}{(z - a)^{2}} & | z | > | a | \\ δ [n - k] & ⟷ z^{- k} \end{aligned}

Every inverse transform question, and the check that a partial fraction answer is the right shape.

Where it comes from

All of them come from one geometric series. For $a^{n} u [n]$ the transform is sum from $n = 0$ of $(a z^{- 1})^{n}$ , which converges to $1 / (1 - a z^{- 1}) = z / (z - a)$ when $| a z^{- 1} | < 1$ , that is $| z | > | a |$ .

For the left sided version, sum from $n = - 1$ down of $- a^{n} z^{- n}$ = -sum from $m = 1$ of $(z / a)^{m} = z / (z - a)$ when $| z | < | a |$ : the same algebra, a different ROC, which is the point of the whole chapter.

The repeated pole pair follows by differentiating the first with respect to a, and the cosine pair by writing the cosine as two exponentials.

3LTI systems in the frequency domain

The frequency response, and the response to an exponential

\begin{aligned} H (e^{j ω}) & = \sum_{n = - \infty}^{\infty} h [n] e^{- j ω n} \\ x [n] = A e^{j ω_{0} n} & ⟹ y [n] = A H (e^{j ω_{0}}) e^{j ω_{0} n} \\ x [n] = A \cos (ω_{0} n + ϕ) & ⟹ y [n] = A | H (e^{j ω_{0}}) | \cos (ω_{0} n + ϕ + ∠ H (e^{j ω_{0}})) \end{aligned}

Five sittings ask for a

y [n]

this way, and it is the cheapest five marks on the paper once you stop trying to convolve.

Where it comes from

Put A $e^{j ω_{0} (n - k)}$ into the convolution sum. The factor A $e^{j ω_{0} n}$ does not involve k, so it comes outside, leaving $\sum_{k} h [k] e^{- j ω_{0} k}$ , which is $H (e^{j ω_{0}})$ .

So the exponential comes out unchanged in shape, multiplied by one complex number: it is an eigenfunction and $H (e^{j ω_{0}})$ is the eigenvalue.

The real sinusoid follows by writing it as two exponentials at plus and minus $ω_{0}$ and using the conjugate symmetry $H (e^{- j ω}) = H^{*} (e^{j ω})$ that a real $h [n]$ forces.

Reading the magnitude response off the pole zero map

| H (e^{j ω}) | = | K | \frac{\prod_{k} | e^{j ω} - z_{k} |}{\prod_{k} | e^{j ω} - p_{k} |}

The most set question in the subject: seventeen sittings want this sketch.

Where it comes from

Factor $H (z) = K$ times the product of (z - zk) over the product of (z - pk). Evaluate on the unit circle, z = $e^{j ω}$ , and take the modulus: the modulus of a product is the product of the moduli, and | $e^{j ω}$ - zk| is precisely the distance from the point $e^{j ω}$ to the zero zk.

Sliding the point round the circle therefore moves it near each pole and each zero in turn. A pole close to the circle makes its distance small, and a small number on the bottom makes $| H |$ large: a peak. A zero close to the circle makes a small number on top: a dip.

Something at the origin sits at distance 1 from every point of the unit circle, so it contributes a constant factor of 1 and affects only the phase.

The difference equation and its system function

\begin{aligned} \sum_{k = 0}^{N} a_{k} y [n - k] & = \sum_{k = 0}^{M} b_{k} x [n - k], a_{0} = 1 \\ H (z) & = \frac{b_{0} + b_{1} z^{- 1} + \dots + b_{M} z^{- M}}{1 + a_{1} z^{- 1} + \dots + a_{N} z^{- N}} \end{aligned}

The bridge between chapters 2 and 4, and the first step of every realization question.

Where it comes from

Take the z-transform of both sides, term by term, using the shift property $x [n - k]$ maps to $z^{- k} X (z)$ . Every delay becomes a power of $z^{- 1}$ , and the equation becomes $Y (z)$ sum $a_{k} z^{- k} = X (z)$ sum $b_{k} z^{- k}$ .

Dividing gives $H (z) = Y (z) / X (z)$ , a ratio of polynomials in $z^{- 1}$ whose coefficients are exactly the numbers in the difference equation. That correspondence is what lets a structure be drawn straight from the equation.

BIBO stability

BIBO ⟺ \sum_{n = - \infty}^{\infty} | h [n] | < \infty ⟺ the ROC contains | z | = 1 ⟺ | p_{k} | < 1 \forall k

Asked outright once, and needed as a one line justification in most filter questions.

Where it comes from

Sufficiency. If $| x [n] |$ ≤ Mx then | $y [n]$ | = |sum $h [k] x [n - k]$ | ≤ sum $| h [k] | | x [n - k] |$ ≤ Mx sum $| h [k] |$ , by the triangle inequality. A finite sum bounds the output.

Necessity. Choose the bounded input $x [n]$ = sign( $h [- n]$ ). Then $y [0] = \sum_{k} h [k]$ sign( $h [k]$ ) = sum $| h [k] |$ , so if that sum diverges a bounded input has produced an unbounded output.

The ROC statement is the same condition written for z-transforms, since $| z | = 1$ makes | $x [n] z^{- n}$ | = $| x [n] |$ .

Linear phase, and the condition on

h [n]

\begin{aligned} h [n] & = \pm h [N - 1 - n] \\ H (e^{j ω}) & = e^{- j ω (N - 1) / 2} H_{r} (ω), H_{r} real \\ τ_{g} & = \frac{N - 1}{2} samples, at every ω \end{aligned}

The property every FIR design in chapter 5 is built to have.

Where it comes from

Pair the terms of the transform sum that the symmetry relates, n and N-1-n. Factor out $e^{- j ω (N - 1) / 2}$ from each pair and what is left inside is e^(jw(n $- (N - 1) / 2$ )) + e^(-jw(n $- (N - 1) / 2$ )), which is 2 cos of a real quantity.

The sum of those cosines is real, so the whole transform is a real function multiplied by $e^{- j ω (N - 1) / 2}$ : the phase is exactly $- ω$ (N-1)/2, a straight line, and the group delay, minus its derivative, is the constant $(N - 1) / 2$ .

An antisymmetric $h [n]$ gives sines instead, which adds a constant $\frac{π}{2}$ to the phase and leaves it linear still.

4Discrete filter structures

Direct form II, and why it is allowed

\begin{aligned} w [n] & = x [n] - \sum_{k = 1}^{N} a_{k} w [n - k] \\ y [n] & = \sum_{k = 0}^{M} b_{k} w [n - k] \\ delays max (M, N) : canonic \end{aligned}

Nine sittings ask for direct form I and II, and it is a drawing, not a calculation.

Where it comes from

$H (z) = B (z)$ times 1/ $A (z)$ , and an LTI cascade is commutative, so the all pole section 1/ $A (z)$ may be placed first without changing the overall response.

Once it is first, both sections are driven by the same internal signal $w [n]$ , so the two delay chains hold identical values at every instant and one chain can serve both. That is the entire argument, and it is worth writing beside the drawing.

The step down recursion: coefficients to lattice

\begin{aligned} K_{m} & = α_{m} (m) \\ α_{m - 1} (k) & = \frac{α_{m} (k) - K_{m} α_{m} (m - k)}{1 - K_{m}^{2}}, k = 1 \dots m - 1 \end{aligned}

Seven sittings for the FIR lattice and ten more for the lattice ladder.

Where it comes from

Start from the order update Am $(z) = A (m - 1) (z) + K m z^{-} m A (m - 1) (z^{- 1})$ . Reading the coefficient of $z^{- m}$ on both sides gives $α_{m} (m) = K_{m}$ directly, since A(m-1) has no $z^{- m}$ term and the reversed polynomial contributes its leading 1.

Reading the coefficient of $z^{- k}$ gives $α_{m} (k) = α_{m - 1} (k) + K_{m} α_{m - 1} (m - k)$ . Writing the same relation with k replaced by m-k gives a second equation in the same two unknowns; solving the pair eliminates $α_{m - 1} (m - k)$ and leaves the recursion, with the $1 - K m^{2}$ appearing as the determinant of that two by two system.

The determinant is why | $K_{m}$ | = 1 stops the recursion: the two equations become dependent, and no lattice exists.

The order update: lattice back to coefficients

A_{m} (z) = A_{m - 1} (z) + K_{m} z^{- m} A_{m - 1} (z^{- 1})

Two sittings ask for

H (z)

from three given K values, and it is the check on every lattice answer.

Where it comes from

The lattice stage computes fm = $f (m - 1) + K m z^{- 1} g (m - 1)$ and gm = $K m f (m - 1) + z^{- 1} g (m - 1)$ . In transform terms, with Am(z) the transfer function from the input to fm and Bm(z) that to gm, this reads Am = $A (m - 1) + K m z^{- 1} B (m - 1)$ and Bm = $K m A (m - 1) + z^{- 1} B (m - 1)$ .

Induction shows Bm(z) = $z^{- m}$ Am $(z^{- 1})$ , the reversed polynomial, at every order: it is true at $m = 0$ and the pair of relations preserves it. Substituting that into the first relation gives the update.

The ladder coefficients

\begin{aligned} C_{M} & = b_{M} \\ C_{m} & = b_{m} - \sum_{i = m + 1}^{M} C_{i} α_{i} (i - m), m = M - 1 \dots 0 \\ y [n] & = \sum_{m = 0}^{M} C_{m} g_{m} [n] \end{aligned}

The second half of every lattice ladder question.

Where it comes from

The backward signals g0 ... gM of the lattice are the outputs of the reversed polynomials Bm(z) = $z^{- m}$ Am $(z^{- 1})$ , which form a basis: any polynomial of degree M can be written as a weighted sum of them.

Writing $B (z) = \sum_{m}$ Cm Bm(z) and matching the coefficient of $z^{- M}$ gives CM = bM immediately, since only BM reaches that power. Matching the next power down gives C(M-1) with one correction term, and so on, which is the recursion.

Quantization noise and the dead band

\begin{aligned} q & = 2^{- B}, σ_{e}^{2} = \frac{q^{2}}{12}, SNR = 6.02 B + 1.76 dB \\ dead band: | y [n] | & \leq \frac{q}{2 (1 - | a |)} \end{aligned}

The one quantization question, and the reason a sharp IIR filter suffers most.

Where it comes from

Model the rounding error as uniformly distributed over -q/2 to +q/2. The variance of a uniform distribution of width q is $q^{2} / 12$ , which is the noise power.

For a full scale sinusoid the signal power is $A^{2}$ /2 with A = $2^{B - 1}$ q, and forming 10 log10 of the ratio gives 6.02B + 1.76 dB: about 6 dB per bit.

The dead band follows from asking when rounding cannot reduce the output further: the filter multiplies by a and the quantizer rounds the result back up whenever $| a y | \geq | y | - q / 2$ , which rearranges to the bound shown.

5FIR filter design

The ideal low pass impulse response

h_{d} [n] = \frac{\sin (ω_{c} (n - τ))}{π (n - τ)}, h_{d} [τ] = \frac{ω_{c}}{π}, τ = \frac{N - 1}{2}

The first line of every window design, eleven sittings of them.

Where it comes from

Take the inverse DTFT of the ideal response, which is $e^{- j ω t a u}$ inside $| w | \leq w c$ and zero outside. The integral runs only over the passband, so

$h_{d} [n]$ = (1/ $2 π$ ) times the integral from $- ω_{c}$ to $ω_{c}$ of $e^{j ω (n - t a u)}$ dw, which evaluates to [ $e^{j ω_{c} (n - t a u)} - e^{- j ω_{c} (n - t a u)}$ ] / ( $2 π$ j (n $- τ$ )).

The bracket is 2j $\sin (ω_{c} (n - t a u))$ , and the j cancels, leaving the sinc. At n = $τ$ the expression is 0/0; the limit of $\sin (x)$ /x is 1, so $h_{d} [τ] = ω_{c} / π$ .

The factor $e^{- j ω t a u}$ in the desired response is the delay that makes the filter causal, and it is exactly what makes the result symmetric about $τ$ .

The windows: attenuation and transition width

\begin{aligned} Rectangular & : 21 dB, & Δ ω & = 1.8 π / N \\ Bartlett & : 25 dB, & Δ ω & = 6.1 π / N \\ Hanning & : 44 dB, & Δ ω & = 6.2 π / N \\ Hamming & : 53 dB, & Δ ω & = 6.6 π / N \\ Blackman & : 74 dB, & Δ ω & = 11 π / N \end{aligned}

Choosing the window is the first mark of every window design question.

Where it comes from

By the multiplication property, the designed response is the ideal brick wall convolved with the window's own spectrum.

The main lobe of that spectrum smears the band edge, so its width becomes the transition band: a longer window has a narrower main lobe, which is why the width goes as 1/N.

The side lobes ride over the discontinuity and become the passband and stopband ripple, so their height sets the attenuation. Side lobe height depends on the window's shape, not its length, which is why the attenuation column has no N in it.

The whole table is that one trade: a smoother taper buys lower side lobes and pays with a wider main lobe.

The Kaiser window design equations

\begin{aligned} δ & = min (δ_{p}, δ_{s}), A = - 20 \log_{10} δ \\ β & = 0.1102 (A - 8.7) & if A > 50 \\ β & = 0.5842 (A - 21)^{0.4} + 0.07886 (A - 21) & if 21 \leq A \leq 50 \\ β & = 0 & if A < 21 \\ N & \geq \frac{A - 8}{2.285 Δ ω} + 1 \\ w [n] & = \frac{I_{0} (β \sqrt{1 - {(\frac{2 n - (N - 1)}{N - 1})}^{2}})}{I_{0} (β)} \end{aligned}

Eleven sittings, usually for the full fifteen marks.

Where it comes from

These are empirical fits, not derivations, and saying so is honest: Kaiser found them by computing the achievable attenuation for a range of $β$ and fitting curves.

What can be derived is the shape. The Kaiser window approximates the prolate spheroidal sequence, the function with the greatest possible fraction of its energy in the main lobe for a given side lobe level, which is why it is close to the best a window can do.

The reason it beats a fixed window in practice is simpler: a fixed window overshoots the required attenuation, and every decibel of overshoot is paid for in filter length.

The Remez exchange algorithm and the alternation theorem

\begin{aligned} E (ω) & = W (ω) [H_{d} (ω) - H_{r} (ω)], min max_{ω} | E (ω) | \\ alternation: E (ω_{i}) = - E (ω_{i + 1}) = \pm δ at L + 2 points \\ H_{r} (ω_{i}) & + (- 1)^{i} \frac{δ}{W (ω_{i})} = H_{d} (ω_{i}), i = 1 \dots L + 2 \end{aligned}

Fifteen of the nineteen sittings, and it is almost entirely bookwork.

Where it comes from

The theorem is what turns "minimise the worst error" into a finite set of equations. If there were fewer than L + 2 alternations, a polynomial of the available degree could be added to reduce the peak error, so the approximation would not be optimal; conversely, with L + 2 alternations no such polynomial exists, because it would have to change sign L + 1 times and therefore have L + 1 roots, which exceeds its degree.

The algorithm is then: guess L + 2 extremal frequencies, solve the equations for delta in closed form, interpolate Hr onto a dense grid, find the new extrema, exchange, and repeat. delta increases at every exchange and is bounded above by the true minimax error, so it converges.

Gibbs' phenomenon

overshoot \approx 9 % of the jump, for every N

Six sittings, usually as the theory half of a window design question.

Where it comes from

Truncating the ideal response is multiplying by a rectangular window, whose spectrum is the Dirichlet kernel $\sin (w N / 2) / \sin (ω / 2)$ , with a first side lobe only 13 dB below the main lobe.

As the convolution sweeps that kernel past the discontinuity, the side lobes ride over the edge and produce an overshoot. Lengthening the window compresses the kernel horizontally but leaves the ratio of side lobe to main lobe unchanged, so the overshoot keeps its size.

The cure is therefore not a longer window but a smoother one: a taper that falls to zero at both ends has much lower side lobes and much less overshoot.

6IIR filter design

The Butterworth magnitude and its order

\begin{aligned} | H_{a} (j Ω) |^{2} & = \frac{1}{1 + {(Ω / Ω_{c})}^{2 N}} \\ N & \geq \frac{\log_{10} [\frac{10^{0.1 A_{s}} - 1}{10^{0.1 A_{p}} - 1}]}{2 \log_{10} (Ω_{s} / Ω_{p})} \\ Ω_{c} & = \frac{Ω_{p}}{{(10^{0.1 A_{p}} - 1)}^{1 / 2 N}} \end{aligned}

Eighteen of the nineteen sittings set this design.

Where it comes from

Put the two specification points into the magnitude squared. At W = $Ω_{p}$ it must be at least $10^{- A p / 10}$ , which rearranges to $(W p / W c)^{2 N} \leq 10^{0.1 A p} - 1$ . At W = $Ω_{s}$ it must be at most $10^{- A s / 10}$ , which gives $(W s / W c)^{2 N} \geq 10^{0.1 A s} - 1$ .

Divide the second by the first and $Ω_{c}$ cancels, leaving $(W s / W p)^{2 N} \geq [10^{0.1 A s} - 1] / [10^{0.1 A p} - 1]$ . Taking logarithms gives the order formula, and it must be rounded up because N is an integer and rounding down fails the specification.

Putting N back into either inequality, as an equality, gives the two expressions for $Ω_{c}$ : one meets the passband exactly, the other the stopband.

The Butterworth pole positions

\begin{aligned} s_{k} & = Ω_{c} \exp (j π \frac{2 k + N + 1}{2 N}), k = 0 \dots N - 1 \\ N = 1 : H_{a} (s) & = \frac{Ω_{c}}{s + Ω_{c}} \\ N = 2 : H_{a} (s) & = \frac{Ω_{c}^{2}}{s^{2} + 1.4142 Ω_{c} s + Ω_{c}^{2}} \\ N = 3 : H_{a} (s) & = \frac{Ω_{c}^{3}}{(s + Ω_{c}) (s^{2} + Ω_{c} s + Ω_{c}^{2})} \end{aligned}

The step between the order and the transfer function, every time.

Where it comes from

Continue $| H_{a} (j Ω) |$ ^2 into the s plane by putting W = s/j, giving $H a (s) H a (- s) = 1 / [1 + (s / (j W c))^{2 N}]$ . Its poles satisfy $(s / (j W c))^{2 N} = - 1$ , so $s^{2 N} = - {(j Ω_{c})}^{2 N}$ , whose 2N roots are evenly spaced on a circle of radius $Ω_{c}$ .

The roots come in pairs symmetric about the origin, one in each half plane. Assigning the left half plane ones to $H_{a} (s)$ gives a stable, causal filter; the right half plane ones belong to Ha(-s) and are discarded. The angle formula picks exactly the left half set.

No pole ever lands on the jW axis, since the spacing is $π$ /N and the offset places them between the axis directions.

The bilinear transformation and frequency warping

\begin{aligned} s & = \frac{2}{T} \frac{1 - z^{- 1}}{1 + z^{- 1}} \\ Ω & = \frac{2}{T} \tan (\frac{ω}{2}) (prewarp every critical ω) \end{aligned}

Thirteen sittings, and "explain frequency warping" is a question in its own right.

Where it comes from

Put z = $e^{j ω}$ and s = jW into the substitution. Factor $e^{j ω / 2}$ from the top and bottom of ( $e^{j ω} - 1) / (e^{j ω}$ + 1): what remains is ( $e^{j ω / 2} - e^{- j ω / 2}$ )/( $e^{j ω / 2} + e^{- j ω / 2}$ ), which is 2j $\sin (ω / 2)$ over 2 $\cos (ω / 2)$ , that is j $\tan (ω / 2)$ .

So jW = (2/T) j $\tan (ω / 2)$ , giving the warping relation. It is monotonic and one to one, so the entire infinite jW axis maps onto the unit circle exactly once and nothing aliases; but it is non linear, which compresses the high frequencies.

Prewarping undoes that at the band edges: design the analog filter at the warped frequencies and the transformation pulls them back to exactly the digital edges asked for.

Impulse invariance

\begin{aligned} h [n] & = T h_{a} (n T) \\ \frac{A_{k}}{s - p_{k}} & ⟶ \frac{T A_{k}}{1 - e^{p_{k} T} z^{- 1}} \\ H (e^{j ω}) & = \frac{1}{T} \sum_{k = - \infty}^{\infty} H_{a} (j \frac{ω - 2 π k}{T}) aliasing \end{aligned}

Three sittings set the design, and five more ask how it compares with the bilinear route.

Where it comes from

The analog impulse response of a single pole term Ak/(s - pk) is Ak $e^{p k t} u (t)$ . Sampling it gives Ak $e^{p k n T} u [n]$ = Ak ( $e^{p k T}$ )^n $u [n]$ , whose z-transform is the standard pair Ak/(1 $- e^{p k T} z^{- 1}$ ), and the factor T comes from the $h [n] = T$ ha(nT) convention.

The frequency relation is the sampling result of chapter 1 applied to ha(t): sampling makes the spectrum repeat every $2 π$ /T and the copies add. If $H_{a} (j Ω)$ has not decayed by W = $π$ /T the copies overlap, which is why this method is useless for a high pass filter.

Only the poles map. The zeros of $H (z)$ fall wherever the sum of the mapped terms puts them, which is the other difference from the bilinear transformation.

Digital spectral transformation, low pass to high pass

\begin{aligned} z^{- 1} & ⟶ - \frac{z^{- 1} + a}{1 + a z^{- 1}} \\ a & = - \frac{\cos ((ω_{c} + ω_{c}^{'}) / 2)}{\cos ((ω_{c} - ω_{c}^{'}) / 2)} \end{aligned}

Two sittings ask for the low pass to high pass case by name.

Where it comes from

The substitution must be all pass, $| G (e^{j ω}) | = 1$ , so that the magnitude values are only relabelled in frequency and never changed in size: the passband stays a passband and the ripple is preserved.

The first order all pass function has one parameter. Demanding that the prototype cut off $ω_{c}$ be mapped to the wanted cut off $ω_{c}$ ' fixes it, and solving that one condition gives the expression for a.

Because an all pass function maps the unit disc to itself, poles inside the circle stay inside: stability survives the transformation.

7The discrete Fourier transform

The DFT and the IDFT

\begin{aligned} X [k] & = \sum_{n = 0}^{N - 1} x [n] W_{N}^{k n}, W_{N} = e^{- j 2 π / N} \\ x [n] & = \frac{1}{N} \sum_{k = 0}^{N - 1} X [k] W_{N}^{- k n} \\ W_{N}^{N} & = 1, W_{N}^{k + N / 2} = - W_{N}^{k}, W_{N}^{2} = W_{N / 2} \end{aligned}

Five sittings ask why we need it, and every FFT question is this transform computed quickly.

Where it comes from

The DFT is the DTFT sampled at N equally spaced points, $ω = 2 π$ k/N, around the unit circle. That is the whole definition: a finite list in, a finite list out, which is the only kind of spectrum a machine can hold.

The inverse follows from orthogonality, exactly as for the Fourier series: substitute the analysis equation into the synthesis equation, interchange the sums, and the inner sum is N when the indices match and zero otherwise.

Sampling in frequency makes the time sequence periodic with period N, and that single consequence is why DFT convolution comes out circular.

Circular convolution

\begin{aligned} x_{1} [n] ⊛_{N} x_{2} [n] & = \sum_{m = 0}^{N - 1} x_{1} [m] x_{2} [(n - m) mod N] \\ X_{1} [k] X_{2} [k] & ⟷ circular, not linear \\ equal to linear when N \geq N_{1} + N_{2} - 1 \end{aligned}

Seventeen sittings, often disguised as "find

x_{3}

if X3 = X1 X2".

Where it comes from

Take the inverse DFT of the product $X_{1} [k] X_{2} [k]$ . Write $X_{1} [k]$ as its own sum over m, interchange the two sums, and the inner sum becomes the inverse DFT of X2 evaluated at n - m. Because $W_{N}^{- k .}$ is periodic with period N, that index is taken modulo N.

The modulo is the entire difference from linear convolution: the shifted sequence wraps around instead of running off the end, because the DFT sees both sequences as periodic.

Zero padding both to $N \geq N 1 + N 2 - 1$ leaves the wrapped part nowhere to land, so the two convolutions agree exactly.

The cost of the FFT

\begin{aligned} DFT & : N^{2} complex multiplications \\ radix-2 FFT & : \frac{N}{2} \log_{2} N \\ speed up & = \frac{2 N}{\log_{2} N} \end{aligned}

Six sittings ask how fast, and it is the opening two marks of most FFT questions.

Where it comes from

Split the N point DFT into two N/2 point DFTs. Each costs $(N / 2)^{2}$ , so the pair costs $N^{2}$ /2, half the original, plus N/2 twiddle multiplications to combine them.

Recurse. Each split halves the work again, and the sequence can be halved log2 N times before reaching transforms of length 1, which need no arithmetic. That leaves log2 N stages of N/2 butterflies, one complex multiplication each.

The saving comes entirely from the two properties of $W_{N}$ : the symmetry $W_{N}^{k + N / 2} = - W_{N}^{k}$ , which lets one product serve two outputs, and $W N^{2} = W (N / 2)$ , which makes a long transform's twiddles into a short transform's twiddles.

Decimation in time

\begin{aligned} X [k] & = G [k] + W_{N}^{k} H [k] \\ X [k + \frac{N}{2}] & = G [k] - W_{N}^{k} H [k], k = 0 \dots \frac{N}{2} - 1 \end{aligned}

Eleven sittings want an eight point butterfly diagram.

Where it comes from

Split the defining sum by the parity of n, putting n = 2r in one half and n = 2r+1 in the other. The even half gives sum of $x [2 r] W_{N}^{2 r k}$ , and since $W N^{2} = W (N / 2)$ that is an N/2 point DFT, $G [k]$ . The odd half gives $W_{N}^{k}$ times another N/2 point DFT, $H [k]$ .

For the second half of the outputs, replace k by k + N/2. G and H are periodic with period N/2 so they are unchanged, while $W_{N}^{k + N / 2}$ = - $W_{N}^{k}$ . The two lines therefore share one multiplication, which is the butterfly.

Recursing the split reorders the input by the reversed bits of its index, which is where the bit reversal comes from: it is bookkeeping, not arithmetic.

Decimation in frequency

\begin{aligned} X [2 r] & = \sum_{n = 0}^{N / 2 - 1} (x [n] + x [n + \frac{N}{2}]) W_{N / 2}^{r n} \\ X [2 r + 1] & = \sum_{n = 0}^{N / 2 - 1} ((x [n] - x [n + \frac{N}{2}]) W_{N}^{n}) W_{N / 2}^{r n} \end{aligned}

Eight sittings, and the papers name which of the two they want.

Where it comes from

Split the defining sum in half by position instead of parity: the first N/2 samples and the second N/2. In the second sum put n = m + N/2, which brings out a factor $W_{N}^{k N / 2} = {(- 1)}^{k}$ .

For even k that factor is +1, so the two halves add; for odd k it is -1, so they subtract, and the leftover $W_{N}^{n}$ rides on the difference. Each resulting sum is an N/2 point DFT.

The even outputs and the odd outputs are computed separately, which is what leaves the output in bit reversed order. Reverse every arrow of a decimation in time flow graph and this one appears: they are transposes.

88 cards · 39 formulas and 49 theory answers · what you miss comes back sooner

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Digital Signal Analysis and Processing

Where the marks actually areThe same paper, every year

The blueprint of the paper

What repeats, in order

How this reader is put together

Five things that quietly cost marks

Notation used here

The whole subject on one page

How to read this reader

Discrete-time signals and systems

1.1Discrete time signal, basic signal types

Discrete time signals and the elementary sequences PIN 2/19

Even and odd parts of a signal PIN 2/19

The even and odd parts

1.2Energy signal, power signal

Energy signals and power signals HOT 6/19

a) x[n]=u[n]

b) x[n]=δ[n]

1.3Periodicity of a discrete time signal

Testing a signal for periodicity HOT 6/19

a) sin(nπ)+cos(nπ)

b) sin(3nπ/5)+cos(4nπ/7)

1.4Transformation of the independent variable

Shifting, folding and scaling a sequence PIN 2/19

1.5Discrete time Fourier series and properties

The discrete time Fourier series, and its properties PIN 3/19

1.6Discrete time Fourier transform and properties

The discrete time Fourier transform, and its properties PIN 2/19

1.7Discrete time system properties

The five system properties, and how to test each one HOT 5/19

Linear?

Time invariant?

Causal?

Stable?

Static?

Linear?

Time invariant?

Causal?

Stable?

a) is y[n]=x[−n] time invariant?

Linear?

Time invariant?

Causal?

Stable?

b) is y[n]=x[n2] linear?

Linear?

Time invariant?

a) is y[n]=x2[n] linear?

Linear?

Time invariant?

Causal?

Stable?

b) is y[n]=cos(5πn/8+π/4) linear?

Linear?

Time invariant?

a) is y[n]=y[n−4]+x[n−4] time invariant?

Linear?

Time invariant?

Causal?

Stable?

b) is y[n]=x2[n] linear?

Linear?

Time invariant?

1.8LTI systems: the convolution sum and its properties

The convolution sum, and four ways to do it TOP 18/19

The output, by the graphical method

1.9Frequency response of an LTI system

The frequency response of an LTI system

1.10Sampling a continuous time signal, spectral properties

Sampling a continuous time signal, and the spectrum of the samples

1.11Last minute recall

Last minute recall, chapter 1

The z-transform

2.1Definition, convergence and the region of convergence

The definition PIN 2/19

The region of convergence, and its properties TOP 14/19

The standard pairs PIN 2/19

2.2Properties of the z-transform

The properties HOT 7/19

The convolution property, derived PIN 2/19

a) $x [n] = u [n]$

b) $x [n] = δ [n]$

a) $\sin (n π) + \cos (n π)$

b) $\sin (3 n π / 5) + \cos (4 n π / 7)$

a) is $y [n] = x [- n]$ time invariant?

b) is $y [n] = x [n^{2}]$ linear?

a) is $y [n] = x^{2} [n]$ linear?

b) is $y [n] = \cos (5 π n / 8 + π / 4)$ linear?

a) is $y [n] = y [n - 4] + x [n - 4]$ time invariant?

b) is $y [n] = x^{2} [n]$ linear?

Inverting $X (z)$ by long division TOP 15/19

$| z | > 1$

$| z | < 0.5$

$0.5 < | z | < 1$

$| z | > 0.6$

$| z | < 0.2$

$0.2 < | z | < 0.6$

Inverting $X (z)$ by partial fractions TOP 15/19

$| z | > 1$

$| z | < 0.5$

$0.5 < | z | < 1$

$| z | > 0.6$

$| z | < 0.2$

$0.2 < | z | < 0.6$