Lecture Notes on Signals and Systems II ECE 320

University of Cyprus

Department of Electrical and Computer Engineering

Lecture Notes on Signals and Systems IIECE 3201

Charalambos D. Charalambous

1Revised: 4, September 2017

ii

Chapter 1

INTRODUCTION

The main objective of this chapter is to give some examples of systems that are analyzed in

this course. These are building block for more compex systems.

1.1 Signals and Systems

Signal. A signal is physical quantity that can be measured or cannot be measured, and is

represented in mathematical modeling, statistical modeling, engineering, and experimental

sciences, by y = f(x), where x is called the independent variable chosen from some set, and

y is called the dependent variable taking values in some set, according to the specific rule

f(·) that relates x to y. The dependent variable y represents the output or outcome whose

variation is being studied by varying the independent variable x.

Signals are often categorized into

• Analog signals- both x and y take continuous values, say real number,

• Discrete-Time signals- x takes discrete values from a countable set and y takes

continuous values,

• Digital Signals- x can take continuous or discrete values while y takes only quantized

values that can be represented in binary arithmetic using finite number of bits,

• ...

System. A system is a machine that is represented by a transformation, say, F (·), that

operates on one signal to generate another signal. The machine or transformation acts on an

input signal, that can be an analog signal x(t) , a discrete-time sigal, x(n), etc., to generate

1

2 CHAPTER 1. INTRODUCTION

via the map F (·) an output signal that can be analog y(t), a discrete-time signal y(n), etc.

The input signal is the cause and the output signal is the effect, due to the machine acting

on the input signal.

F(.)y(t)x(t)

x(n) y(n)

Figure 1.1.1: General representation of a system.

Systems are catergorized according to their properties, such as,

• Linear or nonlinear,

• Time-varying or time-invariant,

• Static or dynamic,

• causal or noncausal,

• stable or unstable,

• ...

1.2 Preface

This course is concerned with discrete-time input signals and systems operating on them to

produce discrete-time output signals. These signas are described by functions with domain

that is specified by a a discrete-time set, with emphasis on countable sets represented by the

set of integers, and range which is the set of real numbers or complex numbers.

Discrete-time signals emerge in at least two ways:

• When the physical quantity called a source generates information at discrete-time

intervals, such as, the value of the stocks at the end of the day, the temperature of a

room that is measures every one hour, etc.

1.3. EXAMPLES 3

• When the physical quantity, the source generates information continuously in time,

but the information signal is sampled by applying Nyquist theorem, at a sampling rate

of T seconds. That is, the analog signal x(t), t ∈ (−∞,∞) is replaced by another

discrete-time signal x(nT ), n = ...− 2,−1, 0, 1, 2, . . .. The main reason for sampling is

to prepare the signal x(t) to be processed by a digital computer.

As time progresses more and more digital processing is required or processing of discrete-

time signals, due to the rapid growth of digital processors that are cheap to build and very

reliable. Examples of such processors are the following:

• Applications of digital filtering techniques in association with Analog-to-Digital (AD)

and Digital-to-Analog (D/A) conversions in communication systems,

• Sample data systems (1940 - 1950) to integrate systems described by laws of physics

and digital computers, such as, automatic pilot control, control of aircrafts, etc.,

Mathematical models based on nature’s laws of Newtonian physics are integrated with

digital computers, because computations are more efficient using the binary representations

of signals.

1.3 Examples

(a) Sampled Data Control Systems

Digital

ComputerController

Continuous

plantTs

y(t)x(t) ek(t)

Ts

Discrete Continuous

Figure 1.3.2: Block diagram of a continuous plant or system that is controlled using a digitalprocessor or computer

The various blocks and elements are the following.


1. Continuous plant: may be the dynamical model of an aircraft, car, etc,

2. Controller: the system that forces the dynamical plant to follow a desired trajectory,

3. Digital Computer: the processor or computer that gives commands to the controller,

4. Ts : the sampling period of signals,

5. e(t) = y(t)− x(t): the error signal between the output signal y(t) and the desired one

x(t). For example, if y(t) is the actual speed of a car and x(t) is the desired speed of

a car, then the controller is the actual auto-pilot control system.

(b) Digital Signal Processing SystemsSmoothes

out

Presampling

analog filterA/D

Digital

FilterD/A

Analog

Filter

)(~ tx

Ensures

Nyquist reconstruction

Samples the analog signal

Converts to binary number

Converts to

analog signal

0 1 2 3

)(ˆ ty)(tx

k

Figure 1.3.3: A/D conversion-processing-D/A conversion

For example, x(t) may be a speech signal which is corrupted or distorted by noise and a

computed is used to remove the noise to recover a clear speech signal.

(c) Macroeconometric Model

The following example illustrates how one can develop a discrete-time microeconomic

model.

Let

Y [n]4= Income in year n,

I[n]4= Investment in year n,

C[n]4= Consumption in year n.

1.3. EXAMPLES 5

Then from above the above definitions it follows that

Y [n] = I[n] + C[n]

We make the following assumption for simplicity:

1. Consumption in year n, C[n], is proportional to the income in year n− 1, i.e.,

C[n] = αY [n− 1], α > 0

2. Investment in year n, I[n], is proportional to the increase in income in the previous

year over the year before

I[n] = β(Y [n− 1]− Y [n− 2]), β > 0

Then the income in year n is given by the discrete-time equation

Y [n] = (α + β)Y [n− 1]− βY [n− 2], n ≥ 2

It is assumed that the initial conditions are specified. i.e.,

Y (−1), Y (−2) are known

This course will develop several techniques to analyze with (a)-(b), and much more

complex systems.

The material presented in this course are analogous to the material presented in Signals

and Systems I.


Chapter 2

DISCRETE-TIME SIGNALS &SYSTEMS

The main objective of this chapter is to

1. establish the mathematical notation,

2. introduce the mathematical definition of discrete-time signals,

3. give the basic properties of discrete-time signals,

4. introduce basic linear spaces or vector spaces of signals, and the notion of distance and

metric on elements of these spaces,

5. introduce the mathematical definition of discrete-time systems,

6. give the basic properties of discrete-time systems

7. analyze linear discrete-time systems.

2.1 Definitions

The following notation is used.

R4= (−∞,∞): the set of real numbers

I4= {...,−2,−1, 0, 1, 2, . . .} : the set of integer numbers

Q4= the set of rational numbers, that is, q ∈ Q is represented by q = m

n,m ∈ I, n ∈ I

7

8 CHAPTER 2. DISCRETE-TIME SIGNALS & SYSTEMS

C : the set of complex numbers, that is any c ∈ C is represented by

c = α + jβ, α ∈ R, β ∈ R, j2 = −1

2.2 Discrete-Time Signals

A signal is a function f(·) or map, defined by

f : T −→ R

where

T is called the Domain of f , and

R is called the Range of f .

For example,

R ⊂ R which means f is a real signal,

R ⊂ C which means f is a complex signal,

T ⊂ R which means the independent variable varies over the set of real numbers,

T = {t1, t2, . . .} ⊂ R which means the independent variable varies over the discrete-time

set {t1, t2, . . .}, that is, f is a discrete-time signal. One may replace {t1, t2, . . .} by tn = nδ,

n ∈ I, where δ is the sampling interval.

Discrete-Time Signal. A discrete time signal is defined by

f : I −→ R, with R which is either R or C

We usually write f [n], n ∈ I.

Examples.

• Brightness of points in an image.

• Price of stocks taken at arbitrary discrete time intervals

2.2. DISCRETE-TIME SIGNALS 9

f[n]

nt2t1 t3 t4

Figure 2.2.1: Example of a discrete-time signal.

2.2.1 Transformations of Time Variables

Next, we present several transformations on discrete-time signals and explain their meaning.

• Reflection

f [n]

k n

f [-n]

-k n

This may represent a recorded signal f [n] that is played backward in time.

• Change of Time Scale

Correction: α < 1 should be replaced by 0 < α < 1.

This may represent a recorded signal f [n] that is played in fast forward mode (α > 1) or it

is played in slow mode (0 < α < 1)


f [ n] , < 1f [ n], > 1 f [n]

n

e.g =2

nn

• Time Shift

f [ n ] f [ n-k ]f [ n+k ], k>0

n1+k nn2+knn n1n2n1-kn2-k

2.2.2 Even and Odd Signals

Even Signal.

f is even ⇔ (if and only if) f [−n] = f [n], ∀ n ∈ I


f [n]

f [-n1] f [n1]

n

Figure 2.2.2: The graph of an even signal

Odd Signal.

f is odd ⇔ f [−n] = −f [n], ∀ n ∈ I

Note the following identities:

even{f [n]} =1

2

{f [n] + f [−n]

}

odd{f [n]} =1

2

{f [n]− f [−n]

}

2.2.3 Periodic Signals

A signal f is called periodic⇔{

there exists N ∈ I such thatf [n] = f [n+N ], ∀ n ∈ I

Note that 2N, 3N, . . . are also periods

If N is a period then nN, n ∈ I are also periods

Fundamental Period.

N0 is called the fundamental period of a period signal⇔ N0 is the smallest period for which

f [n] = f [n+N ],∀n ∈ I

ωo4= 2π

Nois called the fundamental frequency in radians/second.


f [n1]

-f [n1]

Figure 2.2.3: Graph of an odd signal

. . .

f [n]

. . .

n

2N N

Figure 2.2.4: Example of a periodic signal.


2.2.4 Harmonic Signals/Complex Exponentials

A harmonic signal is defined by

f [n] = c an, n ∈ I, c, a ∈ C (general case)

= c ezn, a = ez, z ∈ C

• Real Exponential Signals are specified by a = ez, c, a, z ∈ R

cezn,z>0

| |>1

cezn,z<0

| |<0

f[n] = c ezn

nn

Figure 2.2.5: Example of a real exponential signal

Correction: |α| < 0 should be replaced by |α| < 1.

• Imaginary Exponential Signals are specified by c∈ R, c>0, z=jωo, ωo ∈ R

f [n] = c ejωon.

Next, we derive Euler’s Formula/Identity:

z ∈ C⇒ ez4= 1 + z +

z2

2!+z3

3!+ · · ·

⇒ ejωon = 1 + jωon−ωo

2n2

2!− j ωo

3n3

3!+ωo

4n4

4!+ · · ·

= (1− ωo2n2

2!+ωo

4n4

4!+ · · ·) +

+ j(ωon−ωo

3n3

3!+ωo

5n5

5!+ · · ·)

= cos(ωon) + jsin(ωon)


Periodic with N = 2πω0m,m ∈ I, provided 2π

ω0is a rational number

N04= Fundamental Period = 2π

ω0corresponds to the smallest m ∈ I such that N = 2π

ω0m is a

rational number

We show the above statements below.

ejω0(n+N) = ejω0n

⇔ ejω0N = 1

⇔ ω0N = 2πm,m ∈ I

⇔ ω0

2π=m

N

• Fundamental Differences between f(t) = ejωot, f [n] = ejωon

A. Continuous-Time Signals. f(t) = ejωot, ωo ∈ R, t ∈ R

1. The larger the magnitude of ωo, the higher is the rate of oscillation in the signal

2. ejωot is periodic for any ωo.

3. f(t) = ejωot are all distinct for distinct values of ωo.

B. Discrete-Time Signals. f [n] = ej(ωo+2π)n = ej2πnejωon = ejωon, ωo ∈ R, n ∈ I

1. f [n] = ejωon at ωo → ωo + 2π is the same as that at ωo.

2. f [n] = ejωon = ej(ωo±kπ)n, k = 2, 4, 6, · · · ⇒{consider only ωo in any interval of length

2π, e.g. 0 ≤ ωo ≤ 2π, −π ≤ ωo ≤ π}

3. ejωon periodic

⇔ ejωo(n+N)=ejωon,N ∈ I,∀n ∈ I

⇔ ejωoN = 1

⇔ ωoN = 2πm,m ∈ I

⇔ N = 2πmωo

(integer)

⇔ ωo2π

= mN

⇔ periodic iff ωo2π

is a rational number

⇔ Fundamental frequency = 2πN

= ωom


f [n]=cos(0•n)=1 f [n]=cos( n /8)=1

n

. . .

n

. . .. . .

f [n]=cos( n /4)=1 f [n]=cos( n /2)=1

n

. . .

n

. . .. . . . . .

Figure 2.2.6: Sketch of a periodic signal

Example 2.2.1 Several examples of periodic and nonperiodic signals are given below.

• x[n] = cos(8π n31

) ⇒ ωo = 8π31

⇒8π31

2π= 4

31

⇒ periodic

• x[n] = cos( n31

) ⇒ ωo = 131

⇒131

2πnot rational

⇒ not periodic

• x[n] = ej(2π3)n + ej(

3π4)n

ej(2π3)n ⇒ No

1 = 3

[No1 = m2π

ωo= m 2π

2π3

= m · 3, m→ 1]

ej(3π4)n ⇒ No

2 = 8

[No2 = m2π

ωo= m 2π

3π4

= m · 83, m→ 3]


[or ej3π4(n+N) = ej

3π4n · ej 3π4 N ⇒ No

2 = 8]

Fundamental Period of x[n] is No = 24

General Rule. If pNo1 = qNo

2 = N, p, q ∈ I

⇒ x[n+N ] = x1[n+ pNo1] + x2[n+ qNo

2]

= x1[n] + x2[n]

• Harmonically Related Signals

fk[n] = ejk(2πNo

)n, ωo4= 2π

No, k = 0,±1,±2, . . .

fundamental periods:

Nko = 2π

kωo=

⇒ k ·Nok are also periods

⇒ No = 2πωo

is a common period for all fk

fo[n] = 1 = fNo [n]

f1[n] = fNo+1[n] = ej2πNo

n

f2[n] = fNo+2[n] = e2j2πNo

n

...

f−1[n] = fNo−1[n] = ej2πNo

(No−1)n = e−j2πNo

n

⇒ fk[n] =∑N−1i=0 aifi[n] =

∑N−1i=0 aie

ji 2πNo

n

General Rule. fk[n] = fk+rNo [n] for any r = . . . , 2,−1, 0, 1, 2, . . .


2.2.5 The Unit step and Sample (Impulse)

Impulse Function.

δ[n]4=

{0, n 6= 01, n = 0

[n]

1

n

Figure 2.2.7: Graph of Impulse Function

Unit Step Function.

u[n]4=

{1, n ≥ 00, n < 0

u[n]

n

Figure 2.2.8: Graph of Unit Step Functoon

• First Difference of Unit Step

δ[n] = u[n]− u[n− 1]


• Running Sum of Unit Sample

u[n] =n∑

m=−∞δ[m], ∀m ∈ I

=0∑

k=∞δ[n− k],∀k ∈ I [k = n−m]

=∞∑k=0

δ[n− k],∀k ∈ I [k = n−m]

Interval of

summation

mn 0

[m]

n < 0

mn

n > 0

[m]

0

Interval of

summation

Figure 2.2.9: Graph of Running Sum of Unit Sample

• Sampling Property of Unit Sample

1. f [n]δ[n] = f [0]δ[n]

2. f [n]δ[n− k] = f [k]δ[n− k]

2.3. LINEAR SPACES OF SIGNALS AND NORMS 19

x (t)

tx

xp [n]

Ts 2Ts. . .

xp [n]

0

)(k

skTn

Figure 2.2.10: Sampling using train of impluse functions

2.3 Linear Spaces of Signals and Norms

• Definition of a Linear Space (General)

A linear space is defined by (F ,X ,+, •), where

1. F - Field of numbers (R or C)

2. X - Some set of elements x ∈ X (called vectors)

3. “•” - Multiplication of vectors by scalar

α · x, α ∈ F , x ∈ X

4. “+” - Addition of vectors in X

x+ y, x, y ∈ X


Examples

(a) F = R, X = R2

x1

x2

R2

x1+x2 , x1x2 = x1T x2

Figure 2.3.11: Vector in a two dimensional space

(b) F = {0, 1}, X = {0, 1}-Binary operations

+ 0 10 0 11 1 0

,

• 0 10 0 01 0 1

Conditions under which (X , •,+) is a Linear Space over the field F are the fol-

lowing:

1. (x+ y) + z = x+ (y + z), ∀x, y, z ∈ X [Associative]

2. x+ y = y + x, ∀x, y ∈ X [Commutative]

3. There exists ∅ ∈ X such that ∅+ x = x,∀x ∈ X [Additive identity]

4. ∀x ∈ X there exists −x ∈ X , such that x+ (−x) = ∅, ∀x ∈ X [Additive Inverse]

5. α(β · x) = (α · β)x, ∀x ∈ X , α, β ∈ F [Ass. Scalar]


6. 1 · x = x, 0 · x = ∅, ∀x ∈ X [Mult. Ident.]

7. α(x+ y) = αx+ βy, ∀x, y ∈ X [Scalar Mult. Distributive w.r.t.vector Addition]

8. (α + β)x = αx+ βy , ∀α, β ∈ F [Scalar Mult. is Distr. w.r.t scalar addition]

Example 2.3.1 Linear Space L = ({0, 1}, •,+),F = {0, 1}

Check of 3.+ 0 10 0 11 1 0

⇒ ∅+ x = x,∀x ∈ X , ∅ ∈ X⇒ ∅ = 0

Check of 4.+ 0 10 0 11 1 0

∃ − x ∈ X 3 x+ (−x) = ∅,∀x ∈ X1 + (−x) = 0, if − x = 1

0 + (−x) = 0, if − x = 0

Example 2.3.2 For the Linear Space L of continuous time signals on R (F = R), defined

by

X 4={

f : (−∞,∞)→ R|f is continuous signal}

with operations

“•” Defined pointwise by(αf)(t) = α · f(t), ∀t, ∀f ∈ L, ∀α ∈ R

“+” Defined pointwise by

(f1 + f2)(t) = f1(t) + f2(t), ∀f1, f2 ∈ L, ∀t

“∅” is ∅(t) = 0, ∀t Additive identity.

“−x” is (−f)(t) = −f(t), ∀t Additive Inverse


Example 2.3.3 The Linear Space ` of Discrete-Time Signals on R (F = R) defined by

X 4={f : I → R

}with operations

“•” Defined by(αf)[n] = αf [n], ∀n ∈ I, f ∈ `, α ∈ R

“+” Defined by

(f1 + f2)[n] = f1[n] + f2[n], ∀n ∈ I, f1, f2 ∈ `

“∅” is ∅[n] = 0, ∀n ∈ I

“−x” is (−f)[n] = −f [n], ∀n ∈ I, f ∈ `

• “Distance” between elements of L

d(x, y)4=‖ x− y ‖, x, y ∈ X

x-y = length

xy

e.g.

Figure 2.3.12: Distance in two-dimensional space

“‖ x ‖” is called the norm of x ∈ X

⇒ ‖ x ‖= distance of x ∈ X from ∅ ∈ X

• Given a Linear Space (F ,X , •,+) then ‖ · ‖ is a “Norm” iff the following conditions

hold.

1. ‖ x ‖≥ 0, ∀x ∈ X


2. ‖ x ‖= 0⇔ x = ∅

3. ‖ αx ‖=| α | ‖ x ‖, ∀x ∈ X , α ∈ F

4. ‖ x+ y ‖≤‖ x ‖ + ‖ y ‖, ∀x, y ∈ X [Triangle in.]

IR2

2 2

1 2x x xx=(x1,x2)

x2e.g.

x1

Figure 2.3.13: Norm in two-dimensional space

y

x

yx

Figure 2.3.14: Triangle inequality

• The norm ‖ · ‖ induces an inner product on elements of Linear Space denoted by

< x, y >, ∀x, y ∈ X , so we can multiply vectors.

e.g. x ∈ R2, y ∈ R2 : < x, y >= xTy


• Given a Linear Space then < ·, · > is an inner product on this space if and only if the

following conditions hold.

1. < x, y >= (< y, x >)∗, ∀x, y ∈ X

2. < x, x >≥ 0;< x, x >= 0⇔ x = ∅,∀x ∈ X

3. < αx+ βy, z >= α∗ < x, z > +β∗ < y, z >, ∀x, y, z ∈ X , α, β ∈ Fwhere “∗” means complex conjugate transpose.

• Normed Linear Space

A normed linear space consists of

a Linear Space with ‖ · ‖,, i.e.,(F ,X , ·,+, ‖ · ‖), where X issuch that x ∈ X ⇔‖ x ‖<∞

Below, we give some of the important examples.

1. Norms

(α) ‖ x ‖∞= max(|x1|, |x2|)

IR2

x

x

Figure 2.3.15: Norm on two-dimensional space.

(β) L and ` can be normed linear spaces by introducing different norms ‖ · ‖ on them

e.g. X = L : ‖ x ‖p4= (

∫∞−∞ |x(τ)|p)

1p , p = 1, 2, 3, . . .


(γ) ‖ x ‖∞4= supt∈R | x(t)|. This is called the supremum norm, i.e., the smallest upper

bound of |x(t)|, t ∈ R

Checking that ‖ · ‖∞ is a norm:

1.− 3. Obvious

4. ‖ x+ y ‖∞= supt∈R |x(t) + y(t)|≤ supt∈R(|x(t)|+ |y(t)|)≤ supt∈R |x(t)|+ supt∈R |y(t)|= ‖ x ‖∞ + ‖ y ‖∞

Checking that ‖ · ‖1 is a norm:

1. Obvious

2.∫∞−∞ |f(τ)|dτ = 0 ⇔ f(τ) = 0, ∀τ

3.∫∞−∞ |αf(τ)|d(τ) = |α|

∫∞−∞ |f(τ)|d(τ)

4.∫∞−∞ |x(τ) + y(τ)|dτ ≤

∫∞−∞(|x(τ)|+ |y(τ)|)dτ

=∫∞−∞ |x(τ)|dτ +

∫∞−∞ |y(τ)|dτ

• ‖ · ‖p and ‖ · ‖∞ generate different spaces :

Lp4= {f ∈ L | ‖ f ‖p<∞}

L∞4= {f ∈ L | ‖ f ‖∞<∞}


e.g. L = `

‖ x ‖p= (∑∞n=−∞ |x[n]|p)

1p , p = 1, 2, . . .

‖ x ‖∞= supn∈I |x[n]|

⇒ `p4= {f ∈ ` | ‖ f ‖p<∞}

`∞4= {f ∈ ` | ‖ f ‖∞<∞}

Signals are often classified into the following categories:

1. ‖ f ‖22 is called the energy of the signal f

2. ‖ f ‖∞ - is called the amplitude of the signal f

Note that {f is uniformly bounded} ⇔‖ f ‖∞<∞

f [n]

Figure 2.3.16: A uniformly bounded signal.

2.4. DISCRETE-TIME SYSTEMS 27

2.4 Discrete-Time Systems

A System is a processing device, machine or transformation that transforms input signals to

output signals.

A discrete-time system is defined by the input-output map as follows:

SD︸︷︷︸system

: x[n]︸︷︷︸input signal

−→ y[n]︸︷︷︸output signal

DSx[n] ][][ nxSny D=

Figure 2.4.17: Block diagram of a discrete-time system that maps signals into signals

2.4.1 Block Diagram of Systems

Types of Interconnection of systems

• Serial or Cascade

• Paralel

• Feedback

• Combinations


1Ds2Ds

xSSxSSySy DDDDD 12122)(

1x [n] y1

Radio Receiver Amplifier

Figure 2.4.18: Block diagram of a serial connection of systems

1Ds

2Ds

+

+xSxSy DD

21

Microphone feedback

into amplifier

Figure 2.4.19: Block diagram of a paralel connection of systems


Controller System

1Ds2Ds+

-

x e y

][][

][][][

12neSSny

nynxne

DD

( )

1Ds

2Ds

+

-

x ye

][][

][][][

1

2

neSny

nySnxne

D

D(b)

Figure 2.4.20: Block diagram of a feedback connection

1Ds

2D

s

3D

s

4D

s

++

+

6D

s

5D

s-

yx

Figure 2.4.21: Block diagram of a combination


2.4.2 Basic Properties of Systems

• Memoryless Systems (Instantaneous)

y[n] (value of y at instant n) depends only on x[n] (value of x at instant n)

y[n] = F (x[n]), x[n] ∈ <n, y[n] ∈ <n, F : <n → <n

Ix[n] y[n]=x[n]

e.g.

Figure 2.4.22: A memoryless system

y[n] = (x[n] + x5[n])2√

y[n] = x[n− 1] ×

• Systems with Memory

y[n] depends on some x[k], k 6= n

e.g. y[n] = x[n− 1], y[n] =n∑

k=−∞x[k]

= y[n− 1] + x[n]


• Invertible Systems

x yis invertible{ DS

{

⇔ { } { }

==

≠⇔≠

22

11

2121

,where

yy

whene.g.,;x[.]

determinesuniquely

functionaasy[.]

xSy

xSy

xx

D

D

outputDinstict

input toDistinct

map,1-1aisDS

≡

x xyDS 1−

DS⇔invertibleisDS

I

1such

exists there,

−DS

Figure 2.4.23: Definition and block diagram

Examples:

1. y[n] = 5x[n]

2. y[n] =∑nk=−∞ u[k] = y[n− 1] + u[n]

⇒ S−1D is z[n] = y[n]− y[n− 1]


5x[n] x[n]y[n]

DS 1−

DS

⇒

5

1

Figure 2.4.24: The block diagram of inverse of 1.

1u[n] y [n]-y [n-1]=u[n]y [n ]

DS

1−

DS

∑−∞=

n

k

.

Delay

one unit

-

+

y[n-1]

Figure 2.4.25: The block diagram of inverse of 2.


• Causal Systems (Nonanticipative)

y[n] (value of y at instant n) depends only on past andcurrent values of x[k], k ≤ n (but not the future values)

The mathematical definition is the following:{if y1 = SDx1, y2 = SDx2, and

x1[k] = x2[k], ∀ k ∈ (−∞, n]}

⇒ {y1[k] = y2[k], ∀ k ∈ (−∞, n]}

Examples.

1. y[n] = x[n]− x[n + 1] : is not Causal

2. y[n] = x[−n] : is not Causal

3. y[n] = x[n + 1] : is not Causal

Check: y[n] = x[n + 1]

-1 0 k

]1[][1

+= kkx δ

-1 0 k

]1[][2

+= kukx

x1[k] = x2[k], ∀ k ∈ (−∞,−1]

y1[k] = δ[k + 2]y2[k] = u[k + 2]

⇒ y1[k] 6= y2[k], ∀ k ∈ (−∞,−1]


Applications: Suppose we are interested in defining aslowly varying trend in data that also contain high-frequencyfluctuations about the trend (e.g. stock market)

y[n] =1

2M + 1

M∑k=−M

x[n− k]︸︷︷︸

: is not Causal

average over an intervalM to smooth out

fluctuations and keeponly the trend


• Bounded-Input-Bounded-Output (BIBO) Stable Sys-tems

DSx y

is B.I.B.O. stable )⇔

)

{If ||x||∞ ≤ k <∞} ⇒ {exists a k2 <∞ s.t.|y||∞ ≤ k2}

Examples.

1. SD1 : y[n] = ex[n] : |x[n]| ≤ k1, ∀n ∈ I,⇒ e−k1 ≤ |y[n]| ≤ ek1

⇒ SD1 is B.I.B.O. stable

2. SD2 : y[n] = ∑nk=−∞ u[k]

⇒ y[0] = 1, y[1] = 2, y[2] = 3, ...⇒ y[n] = (n + 1)u[n]

Counter example: y[n] = (n + 1)u[n] = (n + 1).Is |y[n]| ≤ k2 for some finite k2, ∀n.NO!. ∃ some n which exceeds any constant k2 < ∞ weimpose on |y[n]| ≤ k2


• Time-Invariant Systems

Systems such that

DSx[n] y[n]

then DS

][ 0nnx − 0

],[

0

0

≠

−

n

nny )if

)

Figure 2.4.26: Graphical representation of time-invariant system

This means, time-shift in input causes same time-shift inoutput

Example 2.4.1

1. SD : y[n] = nx[n]

DS][ 0nnxn −][ 0nnx −

}is nottime-invariant

][)(][ 000 nnxnnnny −−=−

2. SD : y[n] = x[2n]

DS]2[ 0nnx −][ 0nnx −

}is time-varying

)](2[][ oo nnxnny −=−


• Linear Systems (Superposition holds)

Systems satisfying

DS1y

2y

1x

2x

DS{ DS(i)

{

{21 xx +

21 yy +⇒

{

additive

≡

y1 = SDx1

y2 = SDx2

⇒ {y1 + y2 = SD(x1 + x2)}

DS(ii) {xα yα

⇒

{scaling

≡

DS{ x y

{∈α

y = SDxα ∈ C

⇒ {αy = S(αx)}


Note: (i) and (ii) are equivalent to the single conditioncalled Superposition

DS1y

2y

1x

2x

DS{ DS

{

{ 21 bxx +α 21 byy +α⇒

{

≡

y1 = SDx1, α ∈ Cy2 = SDx2, b ∈ C

⇒

y︷︸︸︷αy1 + by2 = αSDx1 + bSDx2

= S(αx1 + bx2︸︷︷︸x

)

The principle of Superposition States that theresponse of a linear combination of inputs is the linearcombination of corresponding outputs

Important Consequences:

“For all Linear Systems”: Response to an input zero forall times give output zero, that is, SD(0)=0

Follows from (ii) : take α=0 and any

DSx y

then SD(αx) = S(0) = αy = 0y = 0


e.g. y[n] = 2x[n] + 3

if x = 0 ⇒ y[n] = 3 ⇒ Not a linear system

Linear

System

x[n]

3)(][0 =ny

+

output of a

linear system

y[n] (= 2x[n]+3)

[

[

[zero-input resp .]

Example 2.4.2

1. y[n] = (x[n])2

DS( )2

11 ][][ nxny =

( )

( )

( ) ( )

][][2][][

][][

][][

][][

2121

2

2

2

1

2

21

2

33

nxnxnyny

nxnx

nxnx

nxny

++≠

+=

+=

=

][1 nx

][2 nx

DS

DS

Let

][][][ 213 nxnxnx +=

⇒

( )2

22 ][][ nxny =

Additivity

fails

2. y[n] + ny[n− 1] = x[n]


DS

][1 ny

][3 ny

][1 nx

][2 nx

DS

DS

Let

][][][ 22113 nxnxnx αα +=

⇒

][2 ny

)]([]1[][: 111 ∗=−+ nxnnyny

)]([]1[][: 222 ∗∗=−+ nxnnyny

α1 × (∗) + α2 × (∗) ⇒α1y1[n] + α1ny1[n− 1] + α2y2[n] + α2ny2[n− 1]

= α1x1[n] + α2x2[n]

⇒ (α1y1[n] + α2y2[n]) + n(α1y1[n− 1] + α2y2[n− 1])

= α1x1[n] + α2x2[n]

⇒ y[n] + ny[n− 1] = x[n] it is a Linear System

Note : y[n]+ny[n-1]+3=x[n] it is not a Linear System

2.5. LINEAR TIME INVARIANT SYSTEMS (LTI SYS.)41

2.5 Linear Time Invariant Systems (LTI Sys.)

DS

][1 ny][1 nx

][ 01 nnx −

DS

if

][ 01 nny −

then

2.5.1 Representations of signals in terms of Impulses

0 n

][nδ

-2 -1 0 1 2 3 4 n

][nx

1 x[2]

......

Figure 2.5.27: Sifting property of impulse function

δ[n] can be used to construct any x[n]:

x[n] = · · · x[−2]δ[n + 2] + x[−1]δ[n + 1] + x[0]δ[n]

+ x[1]δ[n− 1] + x[2]δ[n− 2] + · · ·=

∞∑k=−∞

x[k]δ[n− k]


[Linear combination of shifted unit impulses δ[n-k]]

x[k]δ[n− k] =

x[k], n = k

0, n 6= k

-1 0 1 2 3 4

2kknkx =− ],[][ δ

x[2]

......

n

e.g.

x[n] = u[n]⇒ x[n] =∞∑

k=−∞u[k]δ[n− k]

=∞∑k=0

δ[n− k]

=−∞∑m=n

δ[m] ,m = n− k

Shifting property of δ[n]:

x[n] =∞∑

k=−∞x[k]δ[n− k] (2.5.1)


2.5.2 General Representations of a LTI System

DS

where

In],[][][ ∑∞

−∞=

∈∀−=m

kxknhny

[n]δ

⇒DS{

][nx ][ny

{

k]-[nδ k]-h[n

h[n]

k]-[n x[n]a

toSofresponsek]-h[. D

δ=

DS

Proof. Consider the response of a linear system (pos-sibly time-varying) to an arbitrary x[n]

DS

x[n]

][ny

Linear

∑∞

−∞=

−k

knkx ][][ δ

== · · · x[−1]δ[n + 1] + x[0]δ[n] + x[1]δ[n− 1] · · ·

Let h[n, k] denote the value of the output of SD at timen, as a response to δ[.− k]


DS

Ik

],[h

∈∀

kn][ kn −δ

Figure 2.5.28: Impulse response of linear time-varying system

The principle of superposition states:

x[n] = ∑∞m=−∞ x[k]δ[.− k]

is a linear combinationof δ[.− k] inputs

ANDh[., k] is a response to

δ[., k] for any k

⇒

y[.] the response to x[.]is a linear combinationof outputs h[., k] :

y[.] = ∑∞k=−∞ x[k]h[., k]

(2.5.2)

6

under the assumption that Σ(.) is finite

Since SD is time-invariant

DS

][h n][nδD

S

][h kn −][ kn −δ

Figure 2.5.29: Impulse response of linear time-invariant system


“if h[.] is the response

to δ[.]′′

⇒

“Then h[.− k] isthe response to δ[.− k]′′

But h[.-k] is the response to δ[.− k]

Hence h[n,k]=h[n-k], ∀ n,k ∈ I and the proof follows

Important Conclusions:

• Linear TI systems are fully determined by their im-pulse response h[.] to δ[.]. If we know h[.] we cancalculate the response to any x[.].

NOT TRUE FOR NONLINEAR SYSTEMS

• Linear time-varying systems are fully determined bythe response h[.,k] to δ[.− k]

y[n] =∞∑

k=−∞x[k]h[n, k] (2.5.3)


2.5.3 Convolution Sum (only for LTI systems)

(x ∗ y)[n]4=

∞∑k=−∞

x[k]y[n− k] (2.5.4)

• Sufficient conditions for the existence of x ∗ y(any of the following)

(i) exists no such that y = 0 outside [−no, no] or elsex = 0 outside [−no, no]

y[k] x[k]

OR

0n− k k0n− 0n0n

(ii) exist no such that x = 0 and y = 0 outside (−∞, no]

y[k] x[k]

k k0n0n

... ... ... ...

(iii) exist no such that x = 0 and y = 0 outside [no,∞)


Example 2.5.1 (calculation of x ∗ y)

1. x[n] = δ[n], y[n] = u[n]

⇒ (x ∗ y)[n] =∞∑

k=−∞δ[k]u[n− k]

=∞∑

k=−∞u[k]δ[n− k]

δ[n− k] =

0, k 6= n1, k = n

⇒ (x ∗ y)[n] = 0 + 0 + · · · + u[n]δ[n− k] + 0

+ 0 + 0 + · · ·= u[n], ∀ n ∈ I

2. (δ ∗ δ)[n] = δ[n]


• Properties of Convolution

(a) (x ∗ y)[n] = (y ∗ x)[n], ∀ x, y [Commutative]

(x ∗ y)[n] =∞∑

k=−∞x[k]y[n− k]

=−∞∑k=∞

x[n− k]y[k], k4= n− k

=∞∑

k=−∞y[k]x[n− k] = (y ∗ x)[n]

(b) (x ∗ y) ∗ z = x ∗ (y ∗ z), ∀ x, y, z [Associativity]

[(x ∗ y) ∗ z] =∞∑

k=−∞

∞∑k=−∞

x[k]y[k − k]z[n− k]

=∞∑

k=−∞x[k]

∞∑k=−∞

y[k − k]z[n− k]

n4= k − k

=∞∑

k=−∞x[k]

∞∑n=−∞

y[n]z[n− n− k]

=∞∑

k=−∞x[k](y ∗ z)[n− k] = [x ∗ (y ∗ z)][n]

(c) x ∗ (y + z) = x ∗ y + x ∗ z, ∀ x, y, z [Distributive]

[x ∗ (y + z)][n] =∞∑

k=−∞x[k][y[n− k] + z[n− k]]

=∞∑

k=−∞x[k]y[n− k] +

∞∑k=−∞

x[k]z[n− k]

= (x ∗ y)[n] + (x ∗ z)[n]

(d) α(x ∗ y) = (αx) ∗ y = x ∗ (αy), ∀ x, y, ∀ α ∈ C


[Commutative of scalar multiplication and convolution]

α(x ∗ y)[n] =∞∑

k=−∞αx[k]y[n− k]

=∞∑

k=∞x[k](αy[n− k]) = [x ∗ (αy)][n]

(e) (x∗y)(.−η) = x(.−η)∗y = x∗y(.−η),∀x, y,∀η ∈ <

(x ∗ y)[n− η] =∞∑

k=−∞x[k]y[n− η − k]

=∞∑

ξ=−∞x[ξ − η]y[n− ξ] ξ

4= k + η

= [x(.− η) ∗ y][n] = x(.− η) ∗ y

(f) 1. Associativity

If h1 = impulse resp.of sys. SD1

h2 = impulse resp.of sys. SD2

⇒

then h1 ∗ h2 =impulse resp. ofa cascade SD2SD1

1DSx

2DSy

21 hh ∗y

1h 2h 12 DD SS

≡x

Figure 2.5.30: Cascade connection of two systems


2. Distributivity

If h1 = impulse resp.of sys. SD1

h2 = impulse resp.of sys. SD2

⇒

h1 + h2 = impulseresp. ofSD1 + SD2

1DSx

2DS

y

21 hh +

y

1h

2h

21 DD SS +≡+

x

Figure 2.5.31: Paralel connection of two systems

Example 2.5.2 Suppose the impulse response of anLTI system is h[n] = u[n] and the input is x[n] =αnu[n], 0 < α < 1. Calculate y(n).

][nhx[n] ∑

∞

−∞=

−=k

knhnxny ][][][

Case (i) : n < 0 ⇒ y[n] = 0


... ...

x[k]Solution:

k0 1 2

... ...

h[k]

k0 1 2

... ...

h[-k]

k -2 -1 0

... ...

h[n-k]

kn

, n<0

0

Figure 2.5.32: Graphical approach


Case (ii) : n ≥ 0 ⇒ y[n] =n∑k=0

αk1 =1− αn+1

1− α

⇒ y[n] =1− αn+1

1− αu[n]

......

n

α−1

1

][1

1][

1

nunyn

α

α

−

−=

+

1

Figure 2.5.33: Graph of the solution

*


2.5.4 The Step Response of Convolution

LTIh[.][.]δ

Step Response

S[n] =∞∑

k=−∞h[n− k]u[k]

=∞∑k=0

h[n− k], ∀ n ∈ I

=n∑

k=−∞h[k], k = n− k

Impulse Response h[n]=S[n]-S[n-1], ∀ n ∈ I

2.5.5 Invertibility of LTI Systems

LTIh[.][.]δ

The system withimpulse response

hI [n] is theinverse of the

system with responseh[n]

if (h[n] ∗ hI [n] = δ[n])

if h[n] = u[n]

h[n] ∗ hI [n] = u[n] ∗ (δ[n]− δ[n− 1])

= u[n]− u[n− 1] = δ[n]


][nhx[n] x[n]

][nhI

x[n]][nδ

x[n]if

][][ nunh =x[n] x[n]

]1[

][][

−

−=

n

nnhI

δ

δ∑

−∞=

=n

k

kxny ][][

e.g.

2.5.6 Causality of Convolution Systems

Proposition 2.5.3 Any causal convolution system hasthe representation

y[n] =n∑

k=−∞h[n− k]x[k]

=∞∑k=0

h[k]x[n− k], ∀ n ∈ I (2.5.5)

y[.][.]h

x[.]

e.g. , h[n] = 0, ∀ n < 0


ProofBy def. ofcausality

⇒y[n] does not dependon u[k] for k > n

From y[n] =∞∑

k=−∞h[n− k]x[k] =

∞∑k=−∞

x[n− k]h[k]

⇒ h[n− k] = 0, ∀ n− k < 0

⇒ h[n] = 0, ∀ n < 0

Example 2.5.4

1. Accumulator h[n] = u[n], hI [n] = δ[n]− δ[n− 1]

are causal

2. h[n] = u[n + 1] is not causal

2.5.7 B.I.B.O. Stability of Conv. Systems

h[.][.]δ

DS

LTI

Proposition 2.5.5 The convolution system SD is BIBOstable if and only if

∞∑k=−∞

|h[k]| <∞ (2.5.6)

impulse response is absolutely summable

(||h||1 <∞) (2.5.7)


Proof. (⇐) Sufficiency: Suppose (2.5.6) holds. We shallshow that any Bounded Input results in a Bounded Out-put

(e.g. if supn∈I|x[n]| ≤ k1, ∃ k2 3 sup

n∈I|y[k]| ≤ k2)

y[n] =∞∑

k=−∞h[k]x[n− k]

Consider x[.] such that ||x||∞4= sup

n∈I|x[n]| ≤ k1

|y[n]| = |∞∑

k=−∞h[k]x[n− k]|

≤∞∑

k=−∞|h[k]||x[n− k]|

≤ k1

∞∑k=−∞

|h[k]|

∀ n, supk∈I|x[n− k]| = sup

k

|x[k]| ≤ k1

⇒ ||y||∞ = supn∈I|y[n]| ≤ k1

∞∑k=−∞

|h[k]|︸︷︷︸

k2

<∞

⇒ BIBO stability

(⇒) Necessity: We shall show that B.I.B.O. implies(2.5.6) ≡ equivalent to showing that if (2.5.6) is falsethen the system is not B.I.B.O. (e.g. there are boundedinputs which give unbounded outputs)

Suppose∞∑

k=−∞|h[k]| =∞


Pick an input x[n] =

0, if h[−n] = 0

h[−n]|h[−n]|, if h[−n] 6= 0

Then |x[n]| ≤ 1, ∀ n ∈ I and so it is bounded. Consider

y[0] =∞∑

k=−∞x[−k]h[k]

=∞∑

k=−∞

h2[k]

|h[k]|=

∞∑k=−∞

|h[k]| → ∞

Hence, the output is not bounded. Thus the system is notstable and therefore absolute summability is necessary.


2.6 General Difference Equations

Input-Output Relation:

aNy[n−N ] + aN−1y[n−N + 1] + . . . + a0y[n] =

bMx[n−M ] + bN−1x[n−M + 1] + . . . + b0x[n],

n ∈ I (2.6.8)

with auxiliary conditions given at no ∈ I

y[n0 −N ] = 0y[n0 −N + 1] = 0

...y[n0 − 1] = 0

(2.6.9)

or

N∑k=0

aky[n− k] =M∑k=0

bkx[n− k] (2.6.10)

Note that the coefficients are constants.Note: Difference equations with non-zero auxiliary con-ditions are not linear systems, e.g.,(2.6.9) with

y[n0 −N ]y[n0 −N + 1]

...y[n0 − 1]

= y0 6= 0 in place of(2.6.9)(2.6.11)

2.6. GENERAL DIFFERENCE EQUATIONS 59

is not linear because if y1[.] and y2[.] satisfy (2.6.8), (2.6.9)then y1[.] + y2[.] satisfiesy1[n0 −N ] + y2[n0 −N ]

...y1[n0 − 1] + y2[n0 − 1]

= 2y0 6= y0 unless y0 = 0

However if y[.] solves (2.6.10) with non-zero auxiliary con-ditions (2.6.11) ⇒

y[.] = yH [.] + yU [.], y0 6= 0 (2.6.12)

where

yH [.] ≡ solution to the homogeneous difference equation

N∑k=0

akyH [n− k] = 0, y0 6= 0 (2.6.13)

With non-zero auxiliary conditions (2.6.11)

yU [.] ≡ solution of the Linear system (2.6.10),(2.6.9)

(2.6.14)


homogeneous

part

non

homogeneous

part

+

HyUy

x

xy

Figure 2.6.34: Representation of solution

Equivalently y[.] can be found as

y[.] = yG[.] + yp[.] (2.6.15)

yG[.] is the general solution of the homogeneous equation(2.6.13) in terms of an unknown auxiliary condition C.

yP [.] is any particular solution to the controlled equation(2.6.10).


C is calculated after a particular solution is found so thaty[n0 −N ]

...y[n0 − 1]

= C +

yP [n0 −N ]

...yP [n0 − 1]

= y0 (2.6.16)

y0 is the original auxiliary condition

• General Solution of Homogeneous equation of (2.6.10)[Free or Natural Response]

The homogeneous Equation corresponding to (2.6.10) is

N∑k=0

akyh[n− k] = 0 (2.6.17)

We assume solutions of the form

yG[n] = Arn, A ∈ C (2.6.18)

where r = em is a constant to be determined

Substituting (2.6.18) into (2.6.17)

N∑k=0

akArn−k = 0 = Arn

N∑k=0

akr−k

⇒ a0 + a1r−1 + a2r

−2 + . . .+ aN−1r−N+1 + aNr

−N = 0

Characteristic Polynomial: [ Roots are the Natural Modesof the System]


a0rN + a1r

N−1 + . . . + aN−1r + aN = 0 (2.6.19)

Each real root rk of multiplicity mk generates ”basis so-lutions”

yi,kh,R = nirnk , i = 0, 1, ...,mk − 1

Each pair of complex conjugate roots rk, r∗k, rk = σk+jωk

of multiplicity mk generates ”basis solutions”

yi,kh,C = nirnk , yi,kh,C∗ = nir∗k

n, i = 0, 1, ...,mk − 1

The general homogeneous difference equation, yG[.], is alinear combination of all basis solutions

yG[n] =K1∑k=1

mk−1∑i=0

ci,kyi,kh,R[n] +

K2∑k=1

mk−1∑i=0

(cci,kyi,kh,c[n] + cc

∗i,ky

i,kh,c∗[n])

K1:number of real roots of (2.6.19)K2:number of complex roots of (2.6.19)

c = [ci,k, cci,k, c

c∗i,k], i = 0, ...,mk − 1, k = 1, ..., N


Note: The basis are linearly independent and span RN

⇒ for any n0 and y0= [y1,0, ..., yN,0] a value of C canalways be found such that

yG[n0 −N ]...

yG[n0 − 1]

= y0

Example 2.6.1

1. Consider

2y[n] + 5y[n− 1] + 4.5y[n− 2] + 1.25y[n− 3] = 51n,

n ≥ 0

y[0] = 0, y[1] = 1, y[2] = 1 (2.6.20)

Characteristic Pol.: 2r3 + 5r2 + 4.5r + 1.25 = 0

Roots of Char. Pol.: r1 = −0.5, r2 = −1 + j0.5,r∗2 = −1− j0.5

Homogeneous Solution

yG[n] = C0,1(−0.5)n +

+Cc0.1(−1 + j0.5)n + Cc∗

0.1(−1− j0.5)n, n ≥ 0

(2.6.21)

2. Consider:

9y[n] + 3y[n− 1]− 5y[n− 2] + y[n− 3] =1

3, n ≥ 0

y[0] =1

9, y[1] = 0, y[2] =

2

27(2.6.22)


Characteristic Pol.: 9r3 + 3r2 − 5r + 1 = 0

Roots of Char. Pol.: r1 = −1, r2 =1

3of multiplicity 2

Homogeneous solution:

yG[n] = C0,1(−1)n + C0,2(1

3)n + C1,2(

1

3)n.n, n ≥ 0

• Particular Solution of (2.6.10)-Method ofUndetermined Coefficients.

When the input does not contain terms which also appear

in the homogeneous solution, the particular solution is

determined by the forcing function, and it is called the

forced response of the system

The method of undetermined coefficients can be used to

obtain the particular solution corresponding to the spe-

cific input function.


Table to determine the form of particular so-lution

x[n] yP[n]

1. nk A1nk + A2n

k−1 + ... + Ak+1

2. an Aan

3. sin bn or cos bn A1 sin bn + A2 cos bn4. nkan an(A1n

k + A2nk−1 + ... + Akk + Ak+1

5. an sin bn or an cos bn an(A1 sin bn + A2 cos bn)

For example, if the input signal has a term of the formnk, then the corresponding Particular Solution is of theform:

yP [n] = A1nk + A2n

k−1 + ... + Ak+1

where A’s are the constant to be determined from substi-

tuting into the difference equation.

Similarly, for the other choices 2.− 5.

Remark 2.6.2 If the input function has terms which

also appear in the homogeneous solution the particu-

lar solution can be affected by the characteristic roots.

In such a case the natural modes are said to be ex-

cited by the input. The corresponding terms in the

complete solution cannot be classified as belonging to

the naturale response or the forced response.


Example 2.6.3 Consider:

2y[n] + 5y[n− 1] + 4.5y[n− 2] + 1.25y[n− 3] = 51n,

n ≥ 0 (2.6.23)

y[0] = 0, y[1] = 1, y[2] = 1

Particular Solution: yP [n] = A1n + A2, n ≥ 0

Substituting yP [n] into (2.6.23)

2(A1n + A2) + 5(A1[n− 1] + A2) +

+4.5(A1[n− 2] + A2) + 1.25(A1[n− 3] + A2) = 51n

⇒ 12.75A1n− 17.75A1 + 12.75A2 = 51n

Equating Coefficients of Powers of n

n0 : A2 =284

51n1 : A1 = 4

Complete Solution (Using (2.6.21))

y[n] = C0,1(−0.5)n + Cc0,1(−1 + j0.5)n +

+Cc∗0,1(−1− j0.5)n + 4n +

284

51, n ≥ 0


Determination of constants:

y[0] = 0⇒ C0,1 + Cc0,1 + Cc∗

0,1 + 28451 = 0

y[1] = 1⇒ −0.5C0,1 + (−1 + j0.5)Cc0,1+

+(−1− j0.5)Cc∗0,1 + 4 + 284

51 = 0y[2] = 1⇒ 0.25C0,1 + (0.75− j1)Cc

0,1+

+(0.75 + j1)Cc∗0,1 + 8 + 284

51 = 0

⇒

⇒

C0,1 = −73.333Cc

0,1 = 33.882− j22.529 = 40.688 6 − 0.5868Cc,∗

0,1 = 33.882 + j22.529 = 40.688 6 0.5868

Complete Solution:

y[n] = −73.333(−0, 5)n +

+81.376(1.118)n. cos(2.6779n− 0.5868)

+4n + 5.5686, n ≥ 0

Example 2.6.4 Consider:

9y[n] + 3y[n− 1]− 5y[n− 2] + y[n− 3] = (1

3)n

y[0] =1

9, y[1] = 0, y[2] =

2

27, n ≥ 0

Particular Solution:

yP [n] = An2(1

3)n, n ≥ 0

Complete Solution: Follow the above procedure.


Remark: Not all difference systems are causal. If we

make the following additional assumption about the sys-

tem:

{Whenever x[k] = 0, ∀k ≤ n0} ⇒{y[k] = 0, ∀k ≤ n0} ← so that y[n0] = 0

then the system will be causal. This property referred to

as the system being initially at REST.

e.g. The System

2y[n] + 5y[n− 1] + 4.5y[n− 2] + 1.25y[n− 3] = x[k],

k ≥ 0

with

x[k] = 51k, y[0] 6= 0, y[1], y[2] arbitrary

is not causal.

2.7. RECURSIVE SOLUTIONS OF DIFF. EQUATIONS69

2.7 Recursive Solutions of Diff. Equations

General Case - Constant Coefficients

Recall (2.6.10)N∑k=0


bkx[n− k]

⇒ y[n]︸︷︷︸ =1

a0{

M∑k=0

bkx[n− k]−N∑k=0

aky[n− k]}

↑Solution at time n depends on past values of input and

output

If x[n] is known ∀n and y[−N ], y[−N+1], . . . , y[−1] are

known, then

y[0] =1

a0{

M∑k=0

bkx[−k]−N∑k=0

aky[−k]}...

y[2] = Recursively...

y[n] = Recursively

Example 2.7.1

y[n]− 1

2y[n− 1] = δ[n], ∀n ∈ I

y[−1] = a [Initial Condition]


Solution:

n ≥ 0:

y[n] =1

2y[n− 1] + δ[n]

y[0] =1

2y[−1] + 1 =

1

2a + 1

y[1] =1

2y[0] + δ[1] = (

1

2)2a +

1

2...

y[n] = (1

2)n+1a + (

1

2)n n ≥ 0 (2.7.24)

n < 0:

y[n− 1] = 2y[n]− 2δ[n], (δ[n] = 0,∀n < 0)

y[−2] = 2y[−1] = 2a

y[−3] = 2y[−2] = 22a...

y[−n] = 2n−1a = (1

2)−n+1a, (2.7.25)

Combining (2.7.24) and (2.7.25)

y[n] = (1

2)n+1a + (

1

2)nu[n]. ∀n ∈ I

2.8. FINITE IMPULSE RESPONSE AND INFINITE IMPULSE RESPONSE OF LTI DEFERENCE EQUATIONS71

2.8 Finite Impulse Response and Infinite Impulse Re-

sponse of LTI Deference Equations

Recall (2.6.10)N∑k=0


bkx[n− k]

• Finite Impulse Response (FIR): Set N = 0 in (2.6.10)

y[n] =M∑k=0

(bkao

)x[n− k]

(no initial conditions are required)

⇒ h[n] =

bna0, 0 ≤ n ≤M

0, otherwise

• Infinite Impulse Response (IIR)

{Consider (2.6.10), with N ≥ 1, x[n] = δ[n] and initial

Rest }⇒ {In general, the Impulse Response contains infinite

terms)}e.g.

y[n]− 1

2y[n− 1] = x[n]

⇒ y[n] = x[n] +1

2y[n− 1]

h[n] = δ[n] +1

2h[n− 1]


Assume initial Rest

⇒ Since x[n] = δ[n] = 0, ∀n ≤ −1 then y[n] =

0, ∀n ≤ −1

⇒ y[−1] = 0

h[0] = δ[0] +1

2h[−1] = 1, h[−1] = 0

h[1] = δ[1] +1

2h[0] =

1

2

h[2] = δ[2] +1

2h[1] = (

1

2)2

...

h[n] = δ[n] +1

2h[n− 1] = (

1

2)n, n ≥ 0

⇒ h[n] = (1

2)nu[n]← Impulse Response of infinite

duration

2.9. BLOCKDIAGRAMREPRESENTATIONS OF LTI SYSTEMS (SIMULATION DIAGRAMS)73

2.9 Block Diagram Representations of LTI Systems

(Simulation Diagrams)

• Multiplier

x[n] x[n]

Figure 2.9.35: Block diagram of multiplier

• Adder

x_1[n]

x_2[n]

x_3[n]

+

-

x_1[n]+x_2[n]-x_3[n]

Figure 2.9.36: Block diagram of adder


• Unit Delay

x[n]D

x[n-1]

Figure 2.9.37: Block diagram of unit delay

2.9. BLOCKDIAGRAMREPRESENTATIONS OF LTI SYSTEMS (SIMULATION DIAGRAMS)75

• IIR

N

K

K

M

K

K KnyaKnxbny00

][][][

D D D D Dx[n]

D

b_0 b_1 b_2

x[n-1] x[n-2] x[n-N]

b_M+

+++ - -

-

y[n]

y[n-1]

y[n-2] y[n-N]

_1 _2 _N

Figure 2.9.38: Realization of solution y[n]


Example. Consider the equation

y[n] = bx[n]− αy[n− 1], y[−1] = 0.

The realization of y[n] is

+

D

x[n]b

–

y[n]=bx[n]- y[n-1]

Chapter 3

Fourier Representation ofDiscrete-Time Periodic and AperiodicSignals

The main objective of this chapter is to

1. introduce the Fourier series representation of periodic

signals,

2. introduce the Fourier transform of periodic and ape-

riodic signals,

3. give the basic properties of Fourier series representa-

tion and Fourier transform,

4. analyze linear time-invariant discrete-time systems.

3.1 Response of a Convolution System to Harmonic

Signal

• Response to a Harmonic signal

77

78CHAPTER 3. FOURIER REPRESENTATION OF DISCRETE-TIME PERIODIC ANDAPERIODIC SIGNALS

h y[n] = h[n]*ej nx[n] = ej n

Impulse Resp.

Figure 3.1.1: Response of LTI system to a harmonic signal.

By convolution we have

y[n] =∞∑

k=−∞h[k]ejω[n−k] (3.1.1)

= ejωn∞∑

k=−∞h[k]e−jωk

︸︷︷︸4=H(ejω)

= H(ejω)ejωn (3.1.2)

Hence, y[·] is again harmonic with “amplitude”H(ejω).

Any signal x for which the output y isy = c · x (c ∈ C or c ∈ R)

↑constant times the input

thenx is called an ”eigenfunction” of the system, andc is called an eigenvalue

H(·) ≡ is called the frequency response of a system

3.1. RESPONSE OF A CONVOLUTION SYSTEM TO HARMONIC SIGNAL 79

• If the systemis BIBO stable

then

H(ejω) isfinite ∀ω ∈ R

To see this notice that|H(ejω)| = |

∞∑k=−∞

h[k] e−jωk|

≤∞∑

k=−∞|h[k]| |e−jωk| , |e−jωk| = 1

=∞∑

k=−∞|h[k]| = ‖ h ‖1<∞

if and only if the system is B.I.B.O stable

• Response of a LTI System to a Linear Combination of Har-monic Signals

Consider the signal:

x[n] =N∑j=1

αj ejωjn , αi ∈ C, ωi ∈ R

Then y[n] = (h ∗ x)[n] =N∑i=1αi (h ∗ ejωin)

⇒ y[n] =N∑j=1

αj H(ejωj) ejωj n (3.1.3)


Hence, the output y has the same harmonics as the inputx with different amplitides.

• Frequency Response Difference Equations with Con-stant Coefficients

Consider the LTI system:

N∑k=0

αk y[n− k] =M∑k=0

bk y[n− k] (3.1.4)

Then by the convolution property of LTI systems wehave:

njenx ][ njj

eeHny )(][

SD

Figure 3.1.2: Response of LTI system to harmonic signals

By the time-invariant property of SD we have:

N∑k=0

αkH(ejω)ejω(n−k) =M∑k=0

bkejω(n−k)

⇒ H(ejω)N∑k=0

αk e−jωk =

M∑k=0

bk e−jωk

Hence, the frequency response is given by


H(ejω) =

M∑k=0

bke−jωk

N∑k=0

αke−jωk

↑ Exists if the systemis BIBO stable

Example 3.1.1(a) Consider the first-order recursive filter

SD : y[n]− αy[n− 1] = x[n], |α| < 1 (3.1.5)

and assume SD is at initial rest.

SD

nje njj

eeH )(

Figure 3.1.3: Response of LTI system to harmonic signals

Then

H(ejω)ejωn − αH(ejω)ejω(n−1) = ejωn

⇒ H(ejω) =1

1− αe−jω⇒ h[n] = αnu[n] [unstable if |α| ≥ 1]


x [n] y [n]h= n u [n]

SD

Figure 3.1.4: Input-output representation of SD

The frequency response of SD is given by

H(ejω) =∞∑

k=−∞h[k] e−jωk

=∞∑

k=−∞αku[k] e−jωk =

∞∑k=0

(αe−jω)k

⇒ H(ejω) =1

1− αe−jωThe step response of SD is given by

s[n] = u[n] ∗ h[n]

=∞∑

k=−∞u[k]h[n− k] =

∞∑k=−∞

u[k]αn−ku[n− k]

=∞∑k=0

αn−ku[n− k]

= αnn∑k=0

(α−1)k = αn1− (α−1)n+1

1− α−1, n ≥ 0


⇒ s[n] =1− αn+1

1− αu[n]

Note: |α| controls speed of long-term value of h[n], s[n]

The magintude and phase of the frequency responseH(ejω) are shown below.

-

2 Increase

atten.

cos21

1)(

2

jeH

0 - 0

= 0.6

Low-pass filter

(Minim.Atten.

at low freq.)

)( jeH

)( jeH

Max

atten.

2

-

Min

atten.

- 00

)( jeH

a = -0.6

High-pass filter

Figure 3.1.5: magintude and phase of the frequency response H(ejω): Low pass and highpass filters.

|α| controls size of filter passband as |α| ↑

Positive α < 1 corresponds to Low-Pass FilterNegative α > −1 corresponds to High-Pass Filter


(b) Non-recursive Filters. Consider the LTI system

y[n] =M∑

k=−Nbkx[n− k]

↑ weighted average of (N + M + 1) values of x[n].(Good for frequency selective filtering)

e.g. Moving-average filter: y[n] for any n, say no,is an average of values of x[n], near n0.

⇒ Rapid high-freq. components of the input are aver-aged out and lower freq.variations are retained, corre-sponding to smoothing or lowpass filtering the originalsequence.

3-point moving Average Filter:

y[n] = 13(x[n− 1] + x[n] + x[n + 1])

⇒ h[n] = 13(δ[n− 1] + δ[n] + δ[n + 1])

⇒ H(ejω) = 13(ejω + 1 + e−jω) = 1

3(1 + 2cosω)

N + m + 1 Moving Average Filter:


)( jeH

-2 - 2

3

2

3

1

Figure 3.1.6: Graph of magnitude of frequency response of Low-pass filter.

y[n] = 1N+M+1

M∑k=−N

x[n− k]

. . .. . .

2 2

( )j

H e

2

if N>0 non-cansal


High-Pass Filter:


y[n] = x[n]−x[n−1]2 [caunsal]

⇒ h[n] = 12[δ[n]− δ[n− 1]]

⇒ H(ejω) = 12(1− e−jω) = je

jω2 sin(ω2 )

1

-

Figure 3.1.8: Graph of magnitude of frequency response of High-pass filter.


Low-Pass Filter:

y[n] = 12(x[n] + x[n− 1]) [causal]

⇒ h[n] = 12(δ[n] + δ[n− 1])

⇒ H(ejω) = 12(1 + e−jω) = e

−jω2 cos(ω2 )

- 0

( )j

H e


Recall: Low frequency occurs near ω = 0,±2π,±4π, . . .High frequency occurs near ω = ±π,±3π, . . .

• Linear Combination of Harmonically Related Com-plex Exponentials

Periodic Signal x[n] : x[n] = x[n + N ] (3.1.6)

Fundamental Period: Smallest N ∈ I such that


(3.1.6) holdsFundamental Freq: ωo = 2π

N

Discrete-Time Complex Exponential Signals:

ϕk[n] = ejkωon (3.1.7)

= ejk(2πN )n, k = 0,±1,±2, . . . (3.1.8)

Fact: ϕk[n] = ϕk+rN [n], r ∈ I

Any periodic signal is a linear combination of complexexp:

x[n] =∑kαkϕk[n] =

∑kαke

jk2πN n

=∑

k=<N>αkϕk[n],

Note: There are N distinct signals ϕk[n] in one periodN .

3.2. FOURIER REPRESENTATION OF DISCRETE-TIME PERIODIC SIGNALS 89

3.2 Fourier Representation of Discrete-Time Periodic

Signals

Let x[·] be a periodic DT signal with fundamental periodN ∈ I (freq. ωo = 2π

N )

Then x[·] has the following spectral representation:

x[n] =∑

k=<N>αke

jkωon =∑

k=<N>αke

jk2πN n, n ∈ I(3.2.9)

↑ denotes summation as kvaries of a range of Nsuccesive integers beginningfrom any integer k = ko ∈ I

• (3.2.9) is finite sum, hence there is no problem withconvergence unlike the continuous-time case

Justification:

x[n] =∞∑

k=−∞cke

jkωon, ∀n ∈ I (3.2.10)

But ∀r ∈ I and ∀n ∈ I we have

ej(k+rN)ωon = cos(k2π

Nn + r2πn) + jsin(k

2π

Nn + r2πn)


= cos(k2π

Nn) + jsin(

2π

Nn) = ejkωon

There are only N different functions in the set:

ϕk[n] = ejkωon, k ∈ I

Hence, x[·] is a finite sum as in (3.2.10)

Calculation of Spectral Coefficients αk, k ∈ I :

x[n] · e−jr2πN n =

∑k=<N>

αk · ejk2πN n · e−jr

2πN n

=∑

k=<N>αk · ej(k−r)

2πN n (3.2.11)

⇒ ∑n=<N>

x[n] · e−j2πN n =

∑n=<N>

∑k=<N>

αkej(k−r)2π

N n

[sum (∗) over one period]

=∑

k=<N>αk

∑n=<N>

ej(k−r)2πN n

︸︷︷︸(1)

(3.2.12)

Recall

N−1∑n=0

αn =

1−αN1−α , if α 6= 1N , if α = 1


⇒ (1) =

∑n=<N>

ej(k−r)2πN n =

1−ej(k−r)2πN N

1−ej(k−r)2πN

= 0, if k − r 6= mN,

m = 0,±1,±2, . . .N , if k − r = mN,

m = 0,±1,±2, . . .

Substituting (1) to (3.2.12) :

∑n=<N>

x[n] · e−jr2πN n =

∑k=<N>

αk∑

n=<N>ej(k−r)

2πN n

=∑

k=<N>αkδ[k − r −mN ]N

⇒ ∑n=<N>

x[n]e−jr2πN n =

∑k=<N>

Nαkδ[k − r −mN ]

=∑

k=<N>Nαk δ[k − r]︸︷︷︸

=

1,if k=r0,otherwise

= Nαr

Hence, the specral coefficients are given by


αk =1

N

∑n=<N>

x[n]e−jk2πN n

• Fourier Series Representation Pair of Periodic Signalsx[n]

x[n] =∑

k=<N>αke

jk2πN n, ωo =

2π

Nαk = 1

N

∑n=<N>

x[n]e−jk2πN n

Note the following:

• The DTFS (of periodic signals) is finite, hence it al-ways converges

• αk+N = αk, because xk+N [n] = xk[n]


Example 3.2.1 (Examples of FS representation)

1. x[n] = ejk2πN n, 2π

N = ωo, k ∈ I

⇒ x[n] periodic with fundamental period N

⇒ x[n] =∑

k=<N>αke

jk2πN n

with αk = 1, αj = 0, for all other j ∈ [0, N − 1]

spectrum(line spectrum)

i

(N+k)0 N

. . .. . .

- -(N-k)

Figure 3.2.10: Line spectrum of FS coefficients.

↑ αk+Nr , r ∈ I

2. x[n] = sinωon. Hence, it is periodic if 2πωo

= N isan integer or ratio of integers

(a) Let 2πωo

= N ⇒ x[n] periodic


with period N, ωo = 2πN

⇒ x[n] = 12je

j 2πN n − 1

2je−j 2π

N n

⇒ α 1 = 12j , α−1 = − 1

2j , αk = 0,∀ k 6= 1,−1over an interval

k , =5 [ k+N= k]

-7 -6 -1 92-5 -4 -3 -2 40 1 3 5 6 7 8 10

j2

1

j2

1j2

1

j2

1

j2

1

j2

1

j2

1



3. Periodic Square Wave x[n] = 1 for −N1 ≤ n ≤ N1

-N -N1

N10

here N1=2

N-2 N

Figure 3.2.12: Graph of periodic square wave.

The Fourier coefficients are computed as follows.

αk =1

N

∑n=<N>

x[k]ejk2πN n

=1

N

n=N1∑n=−N1

1 · ejk2πN n

=1

N

2N1∑m=0

e−jk2πN (m−N1) [m = n + N1]

=1

Nejk

2πN N1

2N1∑m=0

e−jk2πNm

=1

Nejk

2πN N1

(1− e−jk2πN (2N1+1)

1− e−jk2πN

)

=1

Ne−jk( 2π

2N ) [ejk2π

N (N1+12) − e−jk2π

N (N1+12)]

e−jk2πN

12 [ejk

2πN

12 − e−jk2π

N12 ]


αk = 1N

sin[2πkN (N1+1

2)]

sin[(2πN )k2 ]

, k 6= 0,±N,±2N, . . .

k = 1, 2, . . . , N − 1

⇒ αk = 2N1+1N , k = 0,±N,±2N, . . .

Next, we determine how to plot the Fourier coeffi-cients using the definition of the sinc function.

Let

f (ω) =sin[ω2 (2N1 + 1)]

sin(ω2 )(3.2.13)

Then f (ω) is like the sinc function sinxx , but it is pe-

riodic with period 2π

Then we can express αk as follows:

αk =1

Nf (

2πk

N) =

1

Nf (ωk)|ωk=2πk

N(3.2.14)

That is, αk are computed from the sample values off (ω) at ωk = 2πk

N

The zero crossing are found by

ω

2(2N1 + 1) = kπ, k = ±1,±2, . . . (3.2.15)

The samples are ωk = k2πN


N1=3, =10

02

7

4

7

8

7

10

7

12

7

14

7

0

2

10

1

2

3 4

5

6 7

8

9

2

6

7


The general formulae of FS representation is

x[n] =

(N−1)2∑

k=−(N−1)2

αkejk(2π

N )n, if N is odd

N/2∑k=−N2 +1

αkejk2π

N n, if N is even


Table 3.1 Properties of Continuous-Time Fourier SeriesProperty Section Periodic Signal Fourier Ser. Coeff.

- -x(t)y(t)

}Periodic with per. T andfund. frequency ωo = 2π

T

akbk

Linearity 3.5.1 Ax(t) +By(t) Aak +Bbk

Time Shifting 3.5.2 x(t− to) ake−jkωoto = ake

−jk( 2πT

)to

Frequency Shifting ejMωot = ejM( 2πT

)tx(t) ak−MConjugation 3.5.6 x∗(t) a∗−kTime Reversal 3.5.3 x(−t) a−kTime Scaling 3.5.4 x(at), a > 0(periodic with T

a ) akPeriodic Convolution

∫T x(τ)y(t− τ)dτ Takbk

Multiplication 3.5.5 x(t)y(t)+∞∑`=−∞

a`bk−`

Differentiation dx(t)dt jkωoak = jk 2π

T ak

Integration∫ t−∞ x(t)dt

finite valued andperiodiconly ifao = 0

( 1jkωo

)ak = ( 1jk( 2π

T))ak

Conjugate Symmetry forReal signals

3.5.6 x(t)real

ak = a∗−kReak = Rea−kImak = −Ima−k|ak| = |a−k|6 ak = −6 a−k

Real and Even Signals 3.5.6 x(t) real and even ak real and even

Real and Odd Signals 3.5.6 x(t) real and oddakpurely imaginaryand odd

Even-Odd Decompositionof real Signals

{xe(t) = Ev{x(t)} [x(t) real]xo(t) = Od{x(t)} [x(t) real]

Re{ak}jIm{ak}

Parseval’s Relation for Periodic Signals

1T

∫T |x(t)|2dt =

+∞∑k=−∞

|ak|2


Table 3.1 Properties of Discrete-Time Fourier SeriesProperty Periodic Signal Fourier Ser. Coeff.

-x[n]y[n]

}Periodic with period N andfundamental frequency ωo = 2π

N

akbk

}Periodic withperiod N

Linearity Ax[n] +By[n] Aak +Bbk

Time Shifting x[n− no] ake−jk( 2π

N)no

Frequency Shifting ejM( 2πN

)nx[n] ak−MConjugation x∗[n] a∗−kTime Reversal x[−n] a−k

Time Scaling x(m)[n] =

{x[n/m], if n is a multiple of m0, if n is not a multiple of m

(periodic with period mN) 1mak

viewed as periodicwith period mN

Periodic Convolution∑

r=<N>

x[r]y[n− r] Nakbk

Multiplication x[n]y[n]∑

`=<N>

a`bk−`

First Difference x[n]− x[n− 1] (1− ejk(2πN

))ak

Running Sumn∑

k=−∞x[k]

finite valued and periodiconly if ao = 0

( 1

(1−ejk(2πN

)))ak

Conjugate Symmetry forReal signals

x[n] real

ak = a∗−kReak = Rea−kImak = −Ima−k|ak| = |a−k|6 ak = −6 a−k

Real and Even Signals x[n] real and even ak real and even

Real and Odd Signals x[n] real and oddakpurely imaginaryand odd

Even-Odd Decompositionof real Signals

{xe(t) = Ev{x(t)} [x[n] real]xo(t) = Od{x(t)} [x[n] real]

Re{ak}jIm{ak}

Parseval’s Relation for Periodic Signals1N

∑n=<N>

|x[n]|2dt =∑

k=<N>

|ak|2


• Properties of Discrete-Time FS Representation

Consider the periodic signals

x[n] = x[n + N ], y[n] = y[n + N ]

x[n] F.S←→ αk, y[n] F.S←→ βk

1. Multiplication: x[n]y[n] F.S←→ ck =∑

`=<N>α`βk−`

[Both periodic with same period N ]

Proof.

x[n]y[n] =N−1∑k=0

N−1∑`=0

αkβ`ej 2πN (k+`)n

=N−1∑k=0

k+N−1∑`′=k

αkβ`′−kej 2πN `′n [`

′= k + `]

=N−1∑k=0

N−1∑`′=0

αkβ`′−kej 2πN `′n [β`′−k and e

j 2πN `′nperiodic,N ]

=N−1∑`′=0

[N−1∑k=0

αkβ`′−k]︸︷︷︸c`′

ej2πN `′n

2. Time Shifting: x[n− no] F.S←→ αke−jk2π

N no

βk =1

N

N−1∑k=0

x[k − no]e−jk2πN n


=N−no−1∑k′=−no

x[k′]e−jk

′ 2πN n · e−j

2πN kno [k

′= k − no]

= e−j2πN kno

N−no−1∑k′=−no

x[k′]e−jk

′ 2πN n

⇒ βk = αke−j 2π

N kno

3. Time Reversal: x[−n] F.S←→ α−k

Proof.

x[−n] =∑

k=<N>αke

jk2πN (−n)

=k′=−N+1∑k′=0

α−k′ejk′ 2πN n k

′= −k

=N−1∑k=0

α−kejk2π

N n

4. Conjugation: x∗[n] F.S←→ α∗−k

Proof.

(x[n])∗ = (∑

k=<N>αke

jk2πN n)∗


⇒ x∗[n] =∑

k=<N>α∗ke

−jk2πN N [Let k

′= −k]

⇒ x∗[n] =∑

k′=<N>

α∗−k′ejk2π

N n

5. Parseval’s Identity: 1N

∑n=<N>

|x[n]|2 =∑

k=<N>|αk|2

Proof.

ck =∑

`=<N>α`βk−` =

1

N

∑n=<N>

x[n]y[n]e−j2πN kn [by. 1]

Let k=0:∑

`=<N>α`β−` =

1

N

∑n=<N>

x[n]y[n]

Let y[n] = x∗[n]⇒ β` = α∗−` :∑

`=<N>α`α

∗` =

1

N

∑n=<N>

x[n]x∗[n] [by 4.]

⇒ ∑`=<N>

|α`|2 =1

N

∑n=<N>

|x[n]|∗


3.2.1 Applications of FS Representations to LTISystems

Consider the LTI system with impulse response h[n]: .

y[n]∑

>=<

=Nk

nN

jk

kenx

π

α

2

][

][nh

L.T.I. , B.I.B.O. stable

Figure 3.2.14: Output of a LTI system when the input is a perioic signal x[n]

Since the systemis LTI we apply convolution as follows:

y[n] =∞∑

`=−∞h[`]x[n− `]

=∞∑

`=−∞h[`]

∑k=<N>

αk ejk2π

N (n−`)

=∑

k=<N>αk e

jk2πN n

∞∑`=−∞

h[`] e−jk2πN `

We define the frequency response by

H(ejω)4=

∞∑`=−∞

h[`] e−jω`

Then we have

y[n] =∑

k=<N>αkH(ejω)|ω=2π

N kejk

2πN n (3.2.16)

Hence, y[n] is periodic with same period as x[n]


y[n]

= n

Nnx

π2cos][

][][ nunh nα=

11 <<− α

Figure 3.2.15: LTI system of Example 3.2.2.

Example 3.2.2 Consider Figure 3.2.15. Then

x[n] =1

2ej

2πN n +

1

2e−j

2πN n

H(ejω) =∞∑

k=∞h[k]e−jωk

=∞∑

k=−∞αku[k]e−jωk

⇒ H(ejω) =1

1− α e−jω

By (3.2.16):

y[n] =∑

k=<N>αkH(ejω)|ω=2π

N kejk

2πN n

⇒ y[n] =1

2H(ejω)|ω=2π

Nej

2πN n +

1

2H(ejω)|ω=−2π

Nej−2πN n

=1

2(

1

1− α e−j2πN

) ej2πN n +

1

2(

1

1− α ej2πN

) e−j2πN n

Write:

1

1− α e−j2πN

= rejθ N = 4 ⇒ r = 1

1+α2

θ = −tan−1(α)

3.3. FOURIER REPRESENTATION OF APERIODIC SIGNALS: THE FOURIER TRANSFORM105

⇒ y[n] = r cos(2π

N+ θ)

3.3 Fourier Representation of Aperiodic Signals: The

Fourier Transform

Suppose x[n] is aperiodicUnder certain conditions x[n] can be represented by aFourier integral as follows:

x[n] =1

2π

∫<2π> e

jωn ∞∑k=−∞

x[k] e−jωk

︸︷︷︸X(ejω)

dω (3.3.17)

So that

x[n] =1

2π

∫<2π>X(ejω) ejωndω (3.3.18)

is the Inverse Discrete Fourier Transform (IDFT) ofX(ejω),wherethe Fourier Tranform (FT) of x[n] is

X(ejω) =∞∑

n=−∞x[n]e−jωn (3.3.19)

Discrete Fourier Transform (DFT) of x[n]

We use the following notation:

x[n] = F−1{X(ejω)}, X(ejω) = F{x[n]}


Justification:Passage from a spectral Representation to the FourierIntegral.

x[n]

n

aperiodic

1N− 2N

N

x[n] :

Figure 3.3.16: Graph of aperiodic signal x[n].

n

periodic

1N− 2N

][~

nx

:][~

nx

... ...

Figure 3.3.17: Construction of graph of periodic signal x[n] from x[n].

x[.] has fundamental period N (ωo = 2πN )

x[n] = x[n], ∀ n ∈ [−N1, N2]


Spectral Representation of x[.]:

x[n] =∑

k=<N>αk e

jk2πN n (3.3.20)

αk =1

N

∑n=<N>

x[n] e−jk2πN n︸︷︷︸

periodic with period N

Since x[n] = x[n] over one period that include [−N1, N2]

⇒ αk =1

N

N2∑n=−N1

x[n]e−jk2πN n

=1

N

∞∑n=−∞

x[n]e−j2πN nk

[x[n] = outside −N1 ≤ n ≤ N2]Define:

X(ejω)4=

∞∑n=−∞

x[k] ejωn

αk =1

NX(ejkωo), ωo =

2π

N(3.3.21)

Note that αk are proportional to the samples of X(ejω,ω ≡ spacing of the samples in frequency domain.Then we have

x[n] =1

2π

∑k=<N>

X(ejkωo)ejkωonωo, ωo =2π

NDefine

ωk4= kωo, ∀ k ∈ I

ωo = ωk+1 − ωk4= ∆ωk

⇒ N →∞⇔ ∆ωk → 0


Note that 2πN = ωo = ∆ωk. Then we have

x[n] =1

2π

∑k=<N>

X(ejωk)ejωkn︸︷︷︸periodic: 2π period

∆ωk

Notice that as N →∞, then x[n] = x[n], and we have:

⇒ x[n] = limN→∞

x[n] = lim∆ωk→0

x[n] =1

2π

∫2πX(ejω)ejωndω

2

2period:inperiodicnjjk eeX

00 njkjkeeX

0k0)1(k

0

Figure 3.3.18: Grapg of convergence of sum to an integral.

x[n] =1

2π

∑∆ωk=<2π>

X(ejωk)ejωkn∆ωk


x[n] = limN→∞

x[n]︸︷︷︸∑k=<N>

X(ejkωo)∆ωk︸︷︷︸represents area of rectangles

= lim∆ωk→0

x[n]

=1

2π

∫<2π>X(ejω)ejωndω

• Derivation of Discrete-Time Fourier Transform FromContinuous-Time Fourier Transform.

Let x(t) be aperiodic. Then the Fourier Transform is

X(ω) = F{x(t)} =∫ ∞−∞ x(t)e−jωtdt

XSampling:x(t) ∑

∞

−∞=

−==n

s nTttxnxtx )()(][)( δ

∑∞

−∞=

−n

nTt )(δ

Figure 3.3.19: Sampling of a continuous-time signal.

(a) Fourier Transform of Discrete-Time Signals

⇒ Xs(ω) = F{xs(t)}=

∫ ∞−∞ xs(t)e

jωtdt

=∫ ∞−∞

∞∑n=−∞

x(t)δ(t− nT )e−jωtdt

⇒ Xs(ω) =∞∑

n=−∞x(nT )e−jωTn


Let ω4= ωT

⇒ X(ejω) = F{x[n]} =∞∑

n=−∞x[n]ejωn (3.3.22)

ejωn is periodicwith period 2πin the variable ω

⇒

X(ejω) is also periodicwith period 2π

⇒ X(ej(ω+2π)) =∞∑

n=−∞x[n]e−j(ω+2π)n

=∞∑

n=−∞x[n]e−jωn = X(ejω)

(b) Inverse Fourier Transform of Discrete-Time Signals

From(2.17) : X(ejω) =∞∑

p=−∞x[p]e−jωp

⇒∫<0,2π>X(ejω)ejωdω =

∫<2π>

∞∑p=−∞

x[p]e−jωpejωndω

⇒∫<2π>X(ejω)ejωdω =

∫<2π>

∞∑p=−∞

x[p]e−jω(n−p)dω

Recall :∫ 2π0 ejω(n−p)dω =

2π , if n = p

0 , n 6= p

⇒∫<2π>X(ejω)ejωdω =

∞∑p=−∞

∫<2π> x[p]e−jω(n−p)dω

= x[n]2π

⇒ x[n] =1

2π

∫<2π>X(ejω)ejωn


3.3.1 Convergence of the Discrete-Time FourierTransform

Question: When does x[n] has a Fourier Integral Rep-resentation?

Conditions:

(i)∞∑

n=−∞|x[n]| <∞ (x[n] is absoluteley summable)

e.g. ||x||1 <∞Or

(ii)∞∑

n=−∞|x[n]|2 <∞ (x[n] is finite energy)

e.g. ||x||2 <∞

Examples:(a) Consider the signal

x[n] = δ[n]; then X(ejω) = 1

Also

x[n] =1

2π

∫ ω−ω e

jωndω

= δ(n) if ω = π ⇒ No Gibb’s phenomenon

Hence,

x[n] =ω

π

sin ωn

ωn=ω

π

sin(ωn)

ωn, ωn = ωn


Note: Let x[n] be aperiodic

Approximation : x[n] =1

2π

∫ ω−ωX(ejω)ejωnδω, |ω| ≤ ω

Then x[n] = x[n], for ω = π

⇒ No Gibbs Phenomenon


x

xsin1

2 23

0 ,...2,1, kkx

0n

n

][^

nx

)(nn

4

321 ,2

3,

2,

4

4nn

n

n=1

n=0

n=2

n=3

n2

n

1/4

Figure 3.3.20: Graph of sinc function and x[n].


0

][^

nx

)(nn

nnn

2,

2

n2

n

1/2

n=2 n=4

3n

n=0

n=1

1

^

][][

when

nxnx

nnn

n

][^

nx

2n

)(nn

n=0

Figure 3.3.21: Graph of x[n].


(b) Consider the signal

x[n] = αnu[n], |α| < 1

Then

X(ejω) =∞∑

n=−∞αnu[n]e−jωn

=∞∑n=0

(αe−jω)n =1

1− αe−jω


2 2

1

1

1

1

passLow

periodic

0)( jeX

22

1

1

1

1

passHigh

0

)( jeX

Figure 3.3.22: Graph of frequency response


3.3.2 The Fourier Transform of a Periodic Sig-nal

Now, we determine the FT of periodic signals.Suppose x[n] is periodic, with fundamental frequencyωo = 2π

N and FS representation

x[n] =∑

k=<N>αke

jk2πN n

Then the Fourier transform is

F{x[n]} =∞∑

k=−∞2παkδ(ω − kωo), 0 ≤ ω ≤ 2π

Justification:

Consider : x[n] = ejωon

Claim : X(ejω) =∞∑

k=−∞2πδ(ω − ωo − 2πk)

...0ω− 0ω πω 20 +πω 20 −

...ω

)(~

ωjeX

π2

π20

Verification:1

2π

∫2π X(ejω)ejωndω

=1

2π

∫<2π>

∞∑`=−∞

2πδ(ω − ωo − 2π`)ejωndω


Hence,

x[n] = ejωon

Consider

x[n] =∑

k=<N>αke

jk(2πN )n

F{x[n]} =∑

k=<N>αkF{ejk(2π

N )n}

In the range [0, 2π] : F{ejk(2πN )n} = 2πδ(ω − k2π

N)

⇒ X(ejω) =∑

k=<N>αk2πδ(ω − k

2π

N), 0 ≤ ω ≤ 2π

Since X(ejω) is periodic with period 2π, it follows thatX(ejω) consists of a set of N impulses of strength 2παk,k=0,1,2,...,N-1, repeated at intervals N ωo︸︷︷︸

2πN

= 2π. Thus

X(ejω) =N−1∑k=0

2παkδ(ω − kωo), ∀ ω

OR

X(ejω) =∞∑

k=−∞2παkδ(ω −

2π

Nk), ∀ ω

Alternatively Derivation:

x[n] =∑

k=<N>αke

jk(2πN )n

= α0 + α1ej(2πN )n + α2e

j2(2πN )n +

+ · · · + αN−1ej(N−1)(2π

N )n


...

0ω−

3

2

0

π

ω

N

point2π

...

ω

02πα

3=N

002ω−

3

22

2 0

π

ω

3

23

3 0

π

ω

12πα 22πα 02πα 12πα 22πα22πα

12πα

1−N

Figure 3.3.23: Graph of Fourier Transform when N = 3.

F{α0} = F{ejω0n}∣∣∣∣ω0=0

=∞∑

`=−∞2πα0δ(ω − 2π`)

...

NK+= αα

k

...

ωπ2− 0 π2

N−= παπα 22

0 02πα

Nπαπα 22

0=

F{α1ej(2πN )n} =

∞∑`=−∞

2πα1δ(ω − 2π

N− 2π`)

...

NK+= αα

k

...

ω( )

NN

π21+−

1122

+−=

Nπαπα 1

2πα11

22+

=N

παπα

( )N

Nπ2

1+N

π2


F{α2ej2(2π

N )n} =∞∑

`=−∞2πα2δ(ω − 2

2π

N− 2π`)

... ...

ω( )

NN

π22+−

2222 +−=

Nπαπα 2

2πα22

22 +=N

παπα

( )N

Nπ2

2+

N

π22

F{αN−1ej(N−1)(2π

N )n} =∞∑

`=−∞2παN−1δ(ω − (N − 1)

2π

N− 2π`)

... ...

ω

( )N

Nπ2

1−−

1122

−−−=

NNπαπα

1122

−−= παπα

N 12

−Nπα

( )N

Nπ2

1−N

π2

... ...

02πα

0π2−

N

π2

N

π22

12πα

Nπα2

12 +Nπα1

2 −παN−πα2

... ... ...

π2− ( )

+

N

π21N

ω


Examples:

1. x[n] = cosω0n =1

2ejω0n +

1

2e−jω0n, ω0 =

2π

5

⇒ X(ejω) =∞∑

`=−∞πδ(ω − ω0 − 2π`)

+∞∑

`=−∞πδ(ω + ω0 − 2π`)

⇒ X(ejω) = πδ(ω − ω0) + πδ(ω + ω0), − π ≤ ω < π

and X(ejω) repeats periodically with period 2π

0ω ω0 πω 20 +−

π2− π2 02 ωπ +πω 20 −−02 ωπ +− 0ω−

π

)(ωj

eX

Figure 3.3.24: Graph of FT of x[n] = cosω0n.

2. A Periodic Pulse Train is x[n] = ∑∞k=−∞ δ[n− kN ]

The FS coefficients are found as follows:

αk =1

N

∑n=<N>

x[n]e−jk2πN n

=1

N

∑n=<N>

∞∑`=−∞

δ[n− `N ]e−jk2πN n

⇒ αk =1

N, [Interval of summation 0 ≤ n ≤ N − 1]


n0 N2N− N

x[n]

Figure 3.3.25: Graph of periodic pulse train.

⇒ x[n] =∑

k=<N>

1

Nejk

2πN n

Hence, the FT of the pulse train is

X(ejω) =∞∑

k=−∞

2π

Nδ(ω − 2π

Nk)

...

N

π2

...

ω002ω

N

π2

)( ωjeX

Figure 3.3.26: Graph of FT of periodic pulse train.


3.3.3 Properties of DT Fourier Transform

Notation:

F{x[n]} = X(ejω)

x[n] = F−1{X(ejω)}

x[n]FT←→ X(ejω)

• Periodicity: X(ej(ω+2π)) = X(ejω)

Proof. X(ej(ω+2π)) =∞∑

n−∞x[n]e−jωn

︸︷︷︸X(ejω)

e−j2πn

• Linearity:

F{α1x1[n] + α2x2[n]} =

α1X1(ejω) + α2X2(ejω), ∀ α1, α2 ∈ C

Proof.

F{α1x1[n] + α2x2[n]} =∞∑

n=−∞(α1x1[n] + α2x2[n])

= α1

∞∑n=−∞

x1[n]e−jωn + α2

∞∑n=−∞

x2[n]e−jωn

• Time Shifting and Frequency Shifting

(a) F{x[n− n0]} = e−jωn0X(ejω), ∀ n0 ∈ I

(b) F{ejω0nx[n]} = X(ej(ω−ω0)), ∀ ω0 ∈ R

Proof.

(a) F{x[n− n0]} =∞∑

n=−∞x[n− n0]e−jωn


=n=∞∑n=−∞

x[n]e−jωn

︸︷︷︸X(ejω)

e−jωn0, n = n− n0

(b) F{ejω0nx[n]} =∞∑

n=−∞ejω0nx[n]ejωn

=∞∑

n=−∞x[n]e−j(ω−ω0)n = X(ejω)

∣∣∣∣ω→ω−ω0

e.g. Let F{hL[n]} = HL(ejω)

π2−cω

π2ππ− cω−cωπ +− 2cωπ −− 2 cωπ −2

cωπ +2

)(ωj

L eH

Figure 3.3.27: Fraph of ideal Low-Pass filter.

Construction of a High-Pass filter:

π− πcωπ +−cωπ −− cωπ −

cωπ +

)(ωj

H eH

ω

Figure 3.3.28: Fraph of ideal High-Pass filter.


Since HH(ejω) = HL(ej(ω−π))

e.g. HH(ejπ) = HL(ej(π−π)) = HL(ej0)

⇒ hH [n] = ejπnhL = (−1)nhL[n]

• Conjugation and Conjugate Symmetry:

(a) x∗[n]FT←→X∗(e−jω)

(b)

If x[n] is real,e.g. x[n] = x∗[n]

FT←→

X(ejω) = X∗(e−jω)

(c) Real{X(ejω)} is even function of ω

Im{X(ejω)} is odd function of ωif x(n) is real

Proof.

(a) F{x∗[n]} =∞∑

k=−∞x∗[n]e−jωn

=( ∞∑k=−∞

x[n]ejωn)∗

=( ∞∑k=−∞

x[n]e−j(−ω)n)∗

= X∗(e−jωn)∣∣∣∣ω→−ω

(b) F{x[n]} =∞∑

k=−∞x[n]e−jωn

=( ∞∑k=−∞

x[n]ejωn)∗


= X∗(e−jωn)

(c) X(e−jωn) =∞∑

k=−∞x[n]e−jωn

=∞∑

k=−∞x[n] cos(ωn)− j

∞∑k=−∞

x[n] sin(ωn)

Re{X(e−jωn)} =∞∑

k=−∞x[n] cos(ωn)⇒ even function of ω

Im{X(e−jωn)} =∞∑

k=−∞x[n] sin(ωn)⇒ odd function of ω

• Differencing and Accumulation

(a) Differencing

x[n]− x[n− 1] F.T.←→ (1− e−jω)X(ejω)

(b) Accumulation

n∑m=−∞

x[m] F.T.←→ 1

(1− e−jω)X(ejω) +

+ πX(ejω)∞∑

k=−∞δ(ω − 2πk)

Proof. (a)

F{x[n]} − F{x[n− 1]} = X(ejω)− e−jωX(ejω)


e.g.

x[n] = u[n] F.T.←→ 1

(1− e−jω).1 + π

∞∑k=−∞

δ(ω − 2πk)

Proof.

δ[n] F.T.←→ G(ejω) = 1

u[n] =n∑

m=−∞δ[m]

⇒ F{u[n]} = F{n∑

m=−∞δ[m]}

=1

(1− e−jω).1 + π

∞∑k=−∞

δ(ω − 2πk)

• Time Reversal:

x[−n] FT←→ X(e−jω)

Proof:

F{x[−n]} =∞∑

n=−∞x[−n]e−jωn


=∞∑

m=−∞x[m]ejωm

=∞∑

m=−∞x[m]e−j(−ω)m

︸︷︷︸X(ejω)|ω→−ω

• Time Expansion:

Define

x(k)[n] =

x[nk ] ,if n is a multiple of k (n=rk)

0 ,if n is not a multiple of k

⇒ x(k)[n] slowed-down version of x[n]

and x(k)[n]←→ X(ejkω)

X(k)(ejω) =

∞∑n=−∞

x(k)[n]e−jωn [n = rk]

=∞∑

r=−∞x(k)[rk]︸︷︷︸

x[r]

e−jωrk

F{x(k)[n]} =∞∑

r=−∞x[r]e−j(ωk)r = X(ejkω)


-3 -2 -1 0 1 2 3 n

X[n]

0 1 2 3 4 5 6 n

… …

X_3[n]

Figure 3.3.29: Graph of time expansion k = 3.


Example 3.3.1 Consider the signal

x[n] =

1 if n ∈ {−2,−1, 0, 1, 2}0 if n /∈ {−2,−1, 0, 1, 2} (3.3.23)

x[n]

n-2 -1 0 1 2

. . . . . .

1

Figure 3.3.30: Graph on (3.3.23).

-2 - 0 2

X(e^j )

Figure 3.3.31: Graph on FT of (3.3.23).


Example 3.3.2 Consider x(2)[n] of (3.3.23).

X_(2)[n]

1

n-4 -3 - 2 -1 0 1 2 3 4

Figure 3.3.32: Graph of x(2)[n].

- - /2 0 /2

X_(2)(e^j )= (e^j2 )

Figure 3.3.33: Graph on FT of x(2)[n] .


Example 3.3.3 Consider x(3)[n] of (3.3.23).

X_(3)[n]

n-3 -2 -1 0 1 2 3 4 5 6

Figure 3.3.34: Graph of x(3)[n].

-2 /3 - /3 0 /3 2 /3

X_(3)(e^j )= (e^j3 )

Figure 3.3.35: Graph on FT of x(2)[n] .


• Differentiation in Frequency:

nx[n] F.T.←→ jd

dωX(ejω)

Proof:

X(ejω) =∞∑

n=−∞x[n]e−jωn

d

dωX(ejω) =

∞∑n=−∞

x[n](−jn)e−jωn

= −j∞∑

n=−∞nx[n]e−jωn

• Parseval’s Relation:

∞∑n=−∞

|x[n]|2 =1

2π

∫<2π> |X(ejω)|2dω

Equivalently

‖x‖22 =

1

2π‖X(ejω)‖2

2,[0,2π]

Proof.

∞∑n=−∞

|x[n]|2 =∞∑

n=−∞x[n]x∗[n]

=∞∑

n=−∞x[n](F−1{X(ejω)})∗


=∞∑

n=−∞x[n](

1

2π

∫<2π>X(ejω)ejωndω︸︷︷︸

x[n]

)∗

=1

2π

∞∑n=−∞

x[n]∫<2π>X

∗(ejω)e−jωndω

=1

2π

∫<2π>X

∗(ejω)∞∑


︸︷︷︸X(ejω)

dω

=1

2π

∫<2π> |X(ejω)|2dω

For example, we can compute the energy either in

time domain or frequency domain using the FT pair:

x[n] = cosω0n =1

2ejω0n +

1

2e−jω0n ←→

←→ πδ(ω − ω0) + πδ(ω + ω0),

−π ≤ ω ≤ π

• Convolution Property:

x[n] ∗ y[n]←→ X(ejω)Y (ejω)


Proof.

F(x[n] ∗ y[n]) =∞∑

n=−∞

∞∑k=−∞

x[k]y[n− k]e−jωn

=∞∑

k=−∞x[k]

∞∑n=−∞

y[n− k]e−jωn

︸︷︷︸F{y[n−k]}=Y (ejω)e−jωk

=∞∑

k=−∞x[k]e−jωk

︸︷︷︸X(ejω)

Y (ejω)


Example 3.3.4 Consider the LTI delay system

h[n] = δ[n− no], no ∈ I

LTIx[n] y[n]

h[n]= [n-n_0]

Figure 3.3.36: LTI system h[n] = δ[n− no].

Then the frequency response is

H(ejω) =∞∑

n=−∞δ[n− n0]e−jωn = e−jωn0

The FT of the output is

Y (ejω) = e−jωn0X(ejω)

The output is

y[n] = x[n− n0]


Example 3.3.5 Consider the Low-pass filter

-2 - - _0 0 _0 2

(e^j )

Figure 3.3.37: Graph of Low-pass filter.

Then the impulse response of the filter is given by

h[n] =1

2π

∫ π−πH(ejω)ejωndω

=1

2π

∫ ω0−ω0

ejωndω

=sinω0n

πn=ω0

π

sinω0n

ω0n

Figure 3.3.38: Graph of impulse response of Low-pass filter.


Example 3.3.6 Consider a filter and an input de-

scribed in frequency domain as follows.

X(ejω) =1

1− βe−jω,

H(ejω) =1

1− αe−jω, |α| < 1, |β| < 1

H(e^j )X(e^j ) Y(e^j )

Figure 3.3.39: System described in frequency domain.

Then the output y[n] is obtained by applying partial

fraction expansions as follows.

Y (ejω) =1

1− αe−jω1

1− βe−jω

=A

1− αe−jω+

B

1− βe−jω

where the coefficient are found from

A = Y (ejω)(1− αe−jω)|1/α=e−jω =α

α− β

B = Y (ejω)(1− βe−jω)|1/β=e−jω = − β

α− β


Hence, we have

Y (ejω) =α

α− β1

1− αe−jω− β

α− β1

1− βe−jω

F−1{Y (ejω)} =α

α− βF−1{ 1

1− αe−jω} −

− β

α− βF−1{ 1

1− βe−jω}

⇒ y[n] =α

α− βαnu[n]− β

α− ββnu[n]

• Modulation: Convolution in frequency domain:

If

y[n] = x1[n]x2[n]F{x1[n]} = X1(ejω)F{x2[n]} = X2(ejω)

then

F{y[n]} =1

2π

∫<2π>X1(ejθ)X2(ej(ω−θ))dθ

=1

2πX1(ejω) ∗X2(ejω)

Proof.

Y (ejω) =∞∑

n=−∞y[n]e−jωn

=∞∑

n=−∞x1[n]x2[n]e−jωn


=∞∑

n=−∞x2[n]{ 1

2π

∫<2π>X1(ejθ)ejθndθ}e−jωn

=1

2π

∫<2π>X1(ejθ)[

∞∑n=−∞

x2[n]e−j(ω−θ)n

︸︷︷︸X2(ej(ω−θ))

]dθ

=1

2π

∫<2π>X1(ejθ)X2(ej(ω−θ))dθ


x[n] = x1[n]x2[n]

where

x1[n] =sin(3πn

4 )

πn,

x2[n] =sin(πn2 )

πn

Then

X(ejω) =1

2π

∫<2π>=[−π,π] X1 (ejθ)X2(ej(ω−θ))dθ

(aperiodic convolution)

We perform the convolution as follows. Let

X1(ejω) =

X1(ejω), −π < ω ≤ π0, otherwise


Then we can write the convolution as follows.

X(ejω) =1

2π

∫ ∞−∞ X2(ejω)X1(ej(ω−θ))dθ

- - /2 /2

X_2(e^j )

Figure 3.3.40: Graph of H2(ejω).

- -3 /4 0 3 /4

_1(e^j )

Figure 3.3.41: Graph of H1(ejω).

Compute X(ejω),−π < ω ≤ π and extend by peri-

odicity.


3.3.4 Analysis of Convolution Systems via FourierTransforms

Recall the Fourier Transform of Impulse Response

H(ejω)4=

∞∑n=−∞

h[n]e−jωn

• Frequency Response of a LTI System is the Fourier

Transform of the System’s Impulse Response.

h[n]x[n] y[n]

Convolution System:

y[n] = (h ∗ x)[n]

⇒ F{y[n]} = F{(h ∗ x)[n]} = F{h[n]}F{x[n]}

⇒ Y (ejω) = H(ejω)X(ejω) (3.3.24)

H(ejω) is the frequency response of the system be-

cause for x = ejω0n we obtain

y[n] =∞∑

k=−∞h[k]ejω0(n−k) =

∞∑k=−∞

h[k]e−jω0kejω0n

= F{h[n]}ejω0n = H(ejω0)ejω0n


Conclusion: The input-output behavior of a convolution

system whose impulse response, h[n], has a Fourier trans-

form (e.g., ‖h‖1 < ∞ or ‖h‖2 < ∞) is fully determined

by its Frequency Response.

H(ejω) = F{h[n]}

Example 3.3.8 Consider the LIT system described

by difference equation

N∑k=0

αky[n− k] =M∑k=0

βkx[n− k]

Then

F{N∑k=0

αky[n− k]} = F{M∑k=0

βkx[n− k]}

⇒N∑k=0

αke−jkωY (ejω) =

M∑k=0

βke−jkωX(ejω)

(by linearity and Time Shifting)

Hence, the frequency response is

H(ejω)4=Y (ejω)

X(ejω)=

∑Mk=0 βke

−jkω

∑Nk=0 αke

−jkω


Example 3.3.9 Consider the impulse response

h[n] = αnu[n], |α| < 1

Then the frequency response is

H(ejω) =∞∑n=0

αne−jωn =1

1− αe−jω

|H(ejω)| =1√

1 + α2 − 2α cosω

6 H(ejω) = − tan−1 α sinω

1− α cosω

-2 - 2

1/(1+ )

1/(1- )

| (e^j )|

Figure 3.3.42: Graph of magnitude of frequency response.

-2 - 2

1/(1+ )

Phase of

(e^j )2

1

1

tan

a

a

2

1

1

tan

a

a

Figure 3.3.43: Graph of phase of frequency response.



x[n] = ejω0n, ω0 arbitrary [not necessarily periodic]

Claim:

X(ejω) =∞∑

m=−∞2πδ(ω − ω0 − 2πm)

[In [0, 2π], X(ejω) consists of a δ function of strength

= 2π occurring at ω = ω0]

⇒ X(ejω) is a periodic extension of its restriction to

[0, π]

Justification:

x[n] = F−1{X(ejω)}

=1

2π

∫ π−πX(ejω)ejωndω

=1

2π

∫ π−π

∞∑m=−∞

2πδ(ω − ω0 − 2πm)ejωndω

⇒ x[n] = ejωon


X(e^j )

_0-4 _0-2 0 _0 _0+2

2

Figure 3.3.44: Graph of frequency response.

Modification:

x[n] = ejkω0n, ω0 =2π

N[periodic]

⇒ X(ejω) =∞∑

m=−∞2πδ(ω − kω0 − 2πm)

⇒ X(ejω) =∞∑

m=−∞2πδ(ω − kω0 −Nω0m) [2π = Nω0]

X(e^j )

( -2 ) _0 ( - ) _0 _0 ( + ) _0 ( +2 ) _0

. . . . . .



Example 3.3.11 Consider the Low-pass filter

H(ejω) =

1, 0 ≤ |ω| ≤ ωc2, ωc < |ω| < π

-2 - _c _c - _c+2 2 _c+2

(e^j )

1

Figure 3.3.46: Graph of frequency response of filter.

Then h[n] =1

2π

∫ ωc−ωc e

jωndω

=1

2π

ejωn

jn|ωc−ωc =

1

2π

ejωcn − e−jωcn

jn

=sinωcn

πn

⇒ h[n] =ωcπ

sinωcn

ωcn


h[n]

_n(n)

_c= /4

_n= _cn

1

2

Figure 3.3.47: Graph of impulse response of filter.


Example 3.3.12 Consider the impulse response and

input signal given by

h[n] = δ[n]− sin(πn/8)πn

x[n] = cos(πn9 ) + sin(πn7 + 12)

h[n]x[n] y[n]

Figure 3.3.48: Block diagram of system.

We wish to compute y[n].

Solution: Each input signal is periodic

x1[n] = cos πn9 → periodic N1 = 18x2[n] = cos(πn7 + 1

2)→ periodic N2 = 14

⇒⇒ x[n]is periodic if x[n + N ] = x1[n + N ] + x2[n + N ]

⇒ If ∃p, q integers pN1 = qN2 = N , then x[n] is

periodic.

⇒ N = 126 [fundam. period]

⇒ ω0 =2π

126

x[n] =1

2[e

jπn9 + e

−jπn9 ] +

1

2j[ej

12+j πn7 + e−j

12−j

πn7 ]


Now, we determine αk = 0, 0 ≤ k ≤ 125 by using the

following.

αk = αrN+k, r, k ∈ I

7ω0 =π

9, 9ω0 =

π

7

α−7 = α119 = α7 =1

2

α−9 = α117 =1

2je−j

12 , α9 = α∗117

Hence, we have

x[n] =1

2ej7ω0n +

ej12

2jej9ω0n +

e−j12

2jej117ω0n +

1

2ej119ω0n

From

F{ejωon} =∞∑

k=−∞2πδ(ω − ω0 − 2πk)

X(ejω) = 2π{12δ(ω − 7ω0) +

ej12

2jδ(ω − 9ω0) +

− e−j12

2jδ(ω − 117ω0) +

1

2δ(ω − 119ω0)},

0 ≤ ω ≤ 2π

Now, we compute the FT of h[n]:

H(ejω) = 1− 1

8F{sin πn/8

πn}


= 1− 1

8F{sin πn/8

πn/8}

= 1− 1

8F{sin πn/8

πn/8} 4= 1−HN(ejω), −π ≤ |ω| ≤ π

where

HN(ejω) =

1, 0 ≤ |ω| ≤ π

80, π

8 ≤ |ω| < π

Hence,

H(ejω) =

1, π

8 ≤ |ω| <15π8

0, otherwise

Y (ejω) = H(ejω)X(ejω)

= 2π[ej

12

2jδ(ω − 9ω0)− e−j

12

2jδ(ω − 117ω0)]

⇒ y[n] =ej

12

2jej9ω0n − e−j

12

2jej117ω0n

⇒ y[n] = sin(πn

7+

1

2)


Example 3.3.13 Consider the second-order DT causal

difference equation

y[n]− 3

4y[n− 1] +

1

8y[n− 2] = 2x[n].

Then

H(ejω) =2

(1− 12e−jω)(1− 1

4e−jω)

By partial fraction expansion we have

H(ejω) =4

1− 12e−jω −

2

1− 14e−jω

Hence,

h[n] = 4(1

2)n

u[n]− 2(1

4)n

u[n]


• Connections

(a) Parallel connections in time and frequency domain

H_2

H_1

+Y=

(H_1+H_2)+

Xy[.]

+

+

h_2[.]

h_1[.]

x[n]


(b)Cascade connections in time and frequency do-

main.

||| |||

YXH_1.H_2

X Y H_2H_1h_2h_1

yx

xh_1*h_2

y


(c)Inverse system connections in time and frequency

domain

h_1 h_2

H_2Y

|||

H_1

x

X

x

21*hh

121

HH

|||


Example 3.3.14 Consider the block diagram

x(n)

w_3(n)

+

xx

H_lp(e^j )

H_lp(e^j )w_2(n)

w_4(n)

(-1)^n(-1)^n

y(n)

Figure 3.3.49: Block diagram of system.

Find the Frequency Response of the system.

- /4 \4

1

Figure 3.3.50: Graph of frequency response of filter HLP (ejω).

Solution:

w1(n) = (−1)nx(n) = ejπnx(n)←→

←→ W1(ejω) = X(ej(ω−π))

w2(n) = w1(n) ∗ hlp(n)←→


←→ W2(ejω) = W1(ejω)HLP (ejω)

= X(ej(ω−π))HLP (ejω)

w3(n) = (−1)nw2(n) = ejπnw2(n)←→

←→ W3(ejn) = W2(ej(ω−π)) = X(ej(ω−2π))HLP (ej(ω−π))

⇒ W3(ejω) = X(ejω)HLP (ej(ω−π))(by periodicity)

w4(n) = x(n) ∗ hlp(n)←→

←→ WH(ejω) = X(ejω)HLP (ejω)

⇒ y(n) = w3(n) + w4(n)←→

←→ Y (ejω) = W3(ejω) + W4(ejω)

⇒ Y (ejω) = X(ejω)[HLP (ej(ω−π)) + HLP (ejω)︸︷︷︸Frequency Response

]

- -3 /4 - /4 /4 3 /4 2

(e^j )



Chapter 4

THE Z-TRANSFORM

The Z-transform of DT signals is more general than the

Fourier transform, because the convergence of the sum

is defined with respect to a region of convergence in the

complex plane. It is the analog of the Laplace transform

of CT signals.

Consequently, the Z-transform of DT signals is appropri-

ate to analyze LTI systems that are not necessarily BIBO

stable, unlike the analysis of LTI systems using Fourier

transforms, which is limited tol BIBO stable systems.

159

160 CHAPTER 4. THE Z-TRANSFORM

4.1 Response of LTI Systems to Complex Exponentials

We consider the complex exponential

x[n] = czn, c, z ∈ C

= cesn, z4= es, s ∈ C

h[n]

LTI

x[n] y[n]

Figure 4.1.1: Block diagram of LTI system.

(a) Let x[n] = czn. Then by the convolution property of

LTI systems we have

y[n] =∞∑

k=−∞h[k]czn−k

=∞∑

k=−∞h[k]z−k

︸︷︷︸eigenvalue

czn︸︷︷︸eigenfunction

y[n] =∞∑

k=−∞h[k]z−k

︸︷︷︸H(z)

czn = H(z)czn

4.1. RESPONSE OF LTI SYSTEMS TO COMPLEX EXPONENTIALS 161

where H(z) is given by

H(z)4=

∞∑k=−∞

h[k]z−k (4.1.1)

and it is called

the Transfer Function (TF) of the LTI system, or

the Z-transform of h[·].(b) Let x[n] = ∑

k αkznk . Then

y[n] =∑kαkH(zk)z

nk

(c) Now, we show how to obtain the discrete-time Fourier

Transform (DTFT) from the Z-transform.

By definition, the DTFT corresponds to

z = ejω so that x[n] = cejωn which implies

H(z)∣∣∣∣z=ejω

=∞∑

k=−∞h[k]e−jωk

︸︷︷︸DTFT


(d) Now, we show that H(z) is also obtained from the

uniform sampling of a CT signal x(t), and then taking

the Laplace transform.

Xx(t)

p(t)

)(*

tx

Figure 4.1.2: Block diagram of sampling by periodic delat function.

x∗(t) =∞∑

k=−∞x(kTs)δ(t− kTs)

p(t) =∞∑

k=−∞δ(t− kTs)

Now, the Laplace transform of the sampled signal is by

x(t) p(t) )(* tx

sT sT0 0

...... ... ...

t t

Figure 4.1.3: Original CT signal, periodic delta function, and sampled signal.

definition:

X∗(s)4= L{x∗(t)} =

∫ ∞−∞ x

∗(t)e−stdt

4.2. TWO-SIDED Z-TRANSFORM 163

=∫ ∞−∞

∞∑k=−∞

x(kTs)δ(t− kTs)e−stdt

=∞∑

k=−∞

∫ ∞−∞ x(kTs)δ(t− kTs)e−stdt

Hence, we obtain

X∗(s) =∞∑

k=−∞x(kTs)e

−kTss

Now, we define z4= esTs. Then we have

X∗(s)∣∣∣∣z=esTs

=∞∑

k=−∞x(kTs)z

−k

X(z) = X∗(s)∣∣∣∣z=esTs

=∞∑

k=−∞x(kTs)z

−k

This shows the claim.

4.2 Two-Sided Z-Transform

• Definition of the Two-Sided Z-Transform:

Let x[n] be a DT signal. Then

Z{x[n]} 4= X(z)4=

∞∑n=−∞

x[n]z−n, for all z ∈ ROCx ⊂ C

is called the two-sided Z-Transorm, and

ROCx is called the Region of Convergence of the Two-

sided Z-Transform of x[n].


Note the following.if x[n] has a FTF{x[n]} = X(ejω)

then X(ejω) =∞∑


where z is complex variable that is expressed in polar

form z = rejω.

Hence,

X(z) = X(rejω) =∞∑

n=−∞x[n](rejω)−n

=∞∑

n=−∞(x[n]r−n)e−jωn

Therefore, the Z−transform is

X(z) = F{x[n]r−n (4.2.2)

Remark.

X(z) may exists while F{x[n]} may not.

e.g. x[n] = αnu[n] does not have F{x[n]} for |a| ≥ 1

X(z)∣∣∣∣z=ejω

= X(ejω) = F{x[n]} [e.g. r = 1]


Im

Re

z-plane

ω

ωjrez =

r

Im

Re

unit-circle

ω

ωjez =

1

Figure 4.2.4: Graph of complex plane.


Example 4.2.1 Now, we present a few examples.

(a)Unit Step: u[n] =

1, n ≥ 00, n < 0

U(z) =∞∑k=0

1.z−k

=∞∑k=0

(z−1)k = limN→∞

∞∑k=0

(z−1)k

= limN→∞

1− (z−1)N+1

1− z−1

The limit is finite if |z−1| < 1, which then implies

U(z) = limN→∞

1− (z−1)N+1

1− z−1, for |z−1| < 1

⇒ U(z) =1

1− z−1, for |z| > 1

The Region of Convergence of U(z) is

ROCu = {z ∈ C | |z| > 1}Im

Re

1

0

uRC

Figure 4.2.5: Graph of ROCu.


Note the following. Suppose z ∈ ROCu, i.e.,z = 1.

Then U(z) = ∑∞k=0(1)−n = ∞, hence z cannot be on

the circle that defines the boundary of ROCu.

(b) Replected Unit Step u[−n] =

1, n ≤ 00, n > 0

Then the Z−transform is

U(z) =∞∑

n=−∞u[−n]z−n =

0∑n=−∞

z−n

=∞∑n=0

zn

= limN→∞

1− zN+1

1− z, finite for |z| < 1

⇒ U(z) =1

1− z, for |z| < 1

⇒ ROCu = {z ∈ C | |z| < 1}

Im

Re1

uRC

Figure 4.2.6: Graph of ROCu.


(c) Right Sided Exponential Signal: x[n] = αnu[n],

α ∈ R. Then

X(z) =∞∑

n=−∞αnu[n]z−n =

∞∑n=0

αnz−n

=∞∑n=0

(αz−1)n

=1

1− αz−1, ROCx = {z ∈ C; |αz−1| < 1}

⇒ X(z) =z

z − α, ROCx = {z ∈ C; |z| > |α|}

Im

Re

xROC

α

Figure 4.2.7: Graph of ROCx.

Special case. If α = 1 then Z{u[n]} = zz−1, ROCu =

{z ∈ C; |z| > 1}, which means F{n[n]} does not con-

verge, because |z| = 1 is not part of ROCu


(d) Left Sided Exponential Signal: x[n] = −αnu[−n−1], α ∈ R. Then

X(z) =∞∑

n=−∞−αnu[−n− 1]z−n = −

−1∑n=−∞

αnz−n

= −∞∑n=1

α−nzn = 1−∞∑n=0

(α−1z)n

= 1− 1

1− α−1z, ROCx = {z ∈ C; |α−1z| < 1}

⇒ X(z) =z

z − α,ROCx = {z ∈ C, |z| < |α|}

Note that if, 0 < α < 1, then X(ejω) does not exist.

Nottice that (c),(d) have the same Z{·}, but they are

Im

Re0

xROC

α

Figure 4.2.8: Graph of ROCx.

distinguishable only from the Region of Convergence.


(e) Combination of (c) and (d) (Right and Left):

x[n] = αnu[n]− βnu[−n− 1]4= x1[n] + x2[n]

Then

Z{x[n]} =z

z − α+

z

z − β

ROCx1 = {z ∈ C; |z| > |α|} ROCx2 = {z ∈ C; |z| < |β|}

Im

Re

2x

ROC

α

Im

Re

2x

ROC

β

Figure 4.2.9: Graph of ROCx1 , ROCx2 .

We call α and β the the “poles” of Z{x[n]}.For Z{x[n]} to converge z must belong to both ROCx1

and ROCx2. Hence, we have

ROCx = ROCx1

⋂ROCx2 = {z ∈ C

∣∣∣∣ |z| < |β|, |z| > |α|}= {z ∈ C ; |α| < |z| < |β|}


• Rules for Determining the ROC of a Two-sided Z-

Transform

1. The ROC of X(z) consists of a ring in the z-plane

centered about the origin.

e.g. ROC consists of values of z = rejω for which

F{x[n]r−n} exists

⇒ROC consists of those z such that ∑∞n=−∞ |x[n]|r−n <

∞⇒ Convergence depends on r = |z|, not ω⇒ If z ∈ ROC then all values of z on same circle

∈ ROC⇒ ROC consists of concentric rings

2. The ROC does not contain any poles

e.g. X(z) is infinite at poles z = zp

3. If x[n] is of finite duration, then the ROC is the entire

z-plane except possibly z = 0 and/or z =∞e.g. X(z) = ∑N2

n=N1x[n]z−n

If z 6= 0,∞ then each term is finite, hence x(z) con-


verges

IfN1 < 0, N2 > 0, thenX(z) contains zpositive, znegative,

which implies if |z| → 0 then znegative → ∞, and if

|z| → ∞, then zpositive →∞. This implies the ROC

does not include z = 0 and z =∞.

If N1 ≥ 0 then z =∞ ∈ ROCIf N2 ≤ 0 then z = 0 ∈ ROC

Example 4.2.2

1.x[n] = δ[n]⇒ X(z) = 1, ROCx = {z ∈ C}

2.x[n] = δ[n− 1]⇒ X(z) = z−1, ROCx = {z ∈ C; z 6= 0},

3.x[n] = δ[n + 1]⇒ X(z) = z,ROCx = {z ∈ C; |z| 6=∞}

4. If x[n] is a right-sided seq., and if the circle |z| = r0 ∈ROC then all finite values of |z| > r0 ∈ ROCe.g.

x(z) =∞∑

n=N1

x[n]z−n

|z| = r0 ∈ ROC ⇒ x[n]r−n0 absolutely summable


|z| = r1 > r0 ⇒ x[n]r−n1 decays more rapidly than

x[n]r−n0 , for n > 0 ⇒ x[n]r−n1 absolutely summable

If N1 ≥ 0⇒ |z| → ∞ ∈ ROCIf N1 < 0⇒ |z| → ∞ not ∈ ROC

5. If x[n] is a left-sided sequence, and if the circle |z| =

r0 ∈ ROC then all values of z for which 0 < |z| <r0 ∈ ROCe.g.

X(z) =N2∑

n=−∞x[n]z−n , N2 ≥ 0 or N2 < 0

If N2 < 0⇒ z = 0 ∈ ROCIf N2 ≥ 0⇒ |z| → 0 not ∈ ROC

6. If x[n] is two-sided, and if the circle |z| = r0 ∈ ROCthen the ROC will consist of a ring in the z-plane that

includes the circle |z| = r0

e.g. x[n] = x1[n] + x2[n]

ROCx 3 ROCx1

⋂ROCx2


Im

Re

∞=zpossibly

1xROC

Im

Re

0zpossibly

2

=

xROC

Figure 4.2.10: Graph of ROC at extreme points.

Example 4.2.3 Now, we present several examples.

(a) Consider a finite duration signal

x[n] =

αn , 0 ≤ n ≤ N − 1, α > 00, otherwise

Then

X(z) =N−1∑n=0

αnz−n =N−1∑n=0

(αz−1)n

⇒ X(z) =1

zN−1

zN − αN

z − α,

ROCx = {z ∈ C|z 6= 0 and/or z 6=∞}

Next, we identify the zeros and poles.

zeros: zN = αN ⇒ zk = αej2πN k, k = 0, 1, · · · , N − 1

poles: z = 0 of order (N − 1) and z = α


⇒ z0 = α and z = α cancel

⇒ zk = αej2πN k, k = 1, 2, · · · , N − 1

⇒ X(z) = 1zN−1(z − z1)(z − z2) · · · (z − zN−1),

ROCx{z ∈ C|z 6= 0}

Im (z)

Re(z)

unit circle10 << α

++

cancel pole and zero(N-1)st order

Figure 4.2.11: Graph of poles and zeros.

(b) Sum of two right-sided exponentials. Consider

the signal

x[n] = 7(1

3)nu[n]− 6(

1

2)nu[n]

Then

X(z) =7

1− 13z−1− 6

1− 12z−1

=z(z − 3

2)

(z − 13)(z − 1

2)

ROC1 = {z ∈ C| |13z−1| < 1} ROC2 = {z ∈ C| |1

2z−1| < 1}


For convergence of X(z), then z must belong to

ROC1 and ROC2. This implies

ROCx = ROC1 ∩ROC2 = {z ∈ C| |z| > 12}

Im (z)

Re(z)++

xROC

31

21

23

Figure 4.2.12: Graph of region of coonvergence.


(c) Two-sided infinite sequence. Consider the sig-

nal x[n] = b|n|, b > 0

nbnx =][

......n

0<b<1

......n

b>1

Figure 4.2.13: Graph of two-sided infinite sequence

Then

x[n] = bnu[n] + b−nu[−n− 1]

Next, we compute the Z−transform of each signal.

x1[n]4= bnu[n]

Z←→

1

1− bz−1,

|z| > b (e.g. |bz−1| < 1))

x2[n]4= b−nu[−n− 1]

Z←→ − 1

1− b−1z−1,

|z| < 1

b(e.g |b−1z−1| > 1)

7. If the Z-transform X(z) is Rational, then its ROC is

bounded by poles or extends to infinity.


Im

Re

1x

ROC

b

Im

Re

2x

ROC

b1

+ +

unit circleunit circle

Im

Re

1x

ROC

b+

unit circle

Im

Re

2x

ROC

b1

+

Im

Re

xROC

b1

+

unit circle

+b

b>1

0<b<1

Figure 4.2.14: Graph different choices of ROC of two-sided infinite seq.


8. If the Z-transform is rational, and if x[n] is right-

sided, then the ROC is the region in the z-plane out-

side the outermost pole. Furthermore, if x[n] is causal(e.g.,

right-sided and x[n] = 0, ∀ n < 0), the the ROC

includes z =∞Im

Re

3z

++ +

2z

1z

∞=z

toextendscausal

Figure 4.2.15: Graph of ROC of causal signal.

9. If the Z-transform of x[n] is rational, and if x[n] is

left-sided, then the ROC is the region in the z-plane

inside the innermost non-zero pole and extends in-

wards, and possibly includes z = 0. If x[n] is anti-

causal (e.g.,left-sided and x[n] = 0, ∀ n ≥ 0), then

the ROC also includes z = 0

180 CHAPTER 4. THE Z-TRANSFORMIm

Re

3z

++ +

2z

1z

anticausalif

Figure 4.2.16: Graph of ROC of anticausal signal.

Example 4.2.4 Now, suppose the Z−transform is given

by

X(z) =1

(1− 13z−1)(1− 2z−1)

Then

X(z) =z2

(z − 13)(z − 2)

Im

Re2

+ +

31

circleunit

+

double

zero

Figure 4.2.17: Graph of poles and zeros.

Next, we give all possible Regions of Convergence.


(a) Suppose the signal is right-sided. Then

X(z) =z2

(z − 13)(z − 2)

↔ x[n]

Im

Re2

+ +

31

xROC

Figure 4.2.18: Graph of ROC.

(b) Suppose the signal is left-sided. Then

X(z) =z2

(z − 13)(z − 2)

↔ x[n]

Im

Re

2

+ +3

1

ROC



(c) Suppose the signal is a combination of left-sided

and right-sided signals. Then

X(z) =z2

(z − 13)(z − 2)

↔ x[n] = x1[n] + x2[n]

2

+

ROC

Im

Re+

31


4.3 Right Sided Z Transform

The right-sided Z−transform is useful for the analysis of

difference equations with non-zero initial conditions. We

introduce it below.

Let x[n] be a DT signal. Then the right-sidedZ−transform

is defined by

Z+{x[n]} =∞∑n=0

x[n]z−n (4.3.3)

4.3. RIGHT SIDED Z TRANSFORM 183

This is equivalent to

Z+{x[n]} = Z{x[n]u[n]} (4.3.4)

Example 4.3.1 Right-sided Z−transforms.

(a) Let x[n] = αn, α > 0. Then Z{x[n]} does not

exist. However,

Z+{x[n]} = Z{αnu[n]}

⇒ Z+{x[n]} =1

1− αz−1, |z| > |α|

(b) Let x[n] = αn+1u[n + 1].

Then the two-sided Z−transform is

Z{x[n]} =∞∑

n=−1αn+1z−n [n = n + 1]

= z∞∑n=0

(αz−1)n = z1

1− αz−1,

ROCx = {z ∈ C| |αz−1| < 1 and z 6=∞}

Hence,

X(z) =z

1− αz−1,

ROCx = {z ∈ C| |z| > |α|, z 6=∞} e.g., not causal signal


The right-sided Z−transform is

Z+{x[n]} =∞∑n=0

αn+1z−n = α∞∑n=0

(αz−1)n

=α

1− αz−1, |z| > |α|, e.g., causal signal

4.4. PROPERTIES OF THE Z-TRANSFORM 185

4.4 Properties of the Z-Transform

Suppose

Z{x1[n]} = X1(z), ROCx1

Z{x2[n]} = X2(z), ROCx2

andZ+{x1[n]} = X1+(z), ROC+

x1

Z+{x2[n]} = X2+(z), ROC+x2

exist for x1[.], x2[.]

• Linearity (Same for Z and Z+)

for α ∈ R, b ∈ R, then

Z{αx1[n] + βx2[n]} = αX1(z) + βX2(z)

with region of convergence for the combined signal

given by ROC ⊇ ROCx1 ∩ROCx2

meaning it contains at least the intersection of the

two, but it could be larger.

e.g.

x[n] = αnu[n] + αnu[n− 1]

⇒ X(z) =1

1− αz−1︸︷︷︸X1(z)

−∞∑n=1

αnz−n

︸︷︷︸∑∞n=0(αz−1)n−1

,


ROCx1 = {z ∈ C||z| > α}

=z

z − α− z

z − α+ 1,

ROCx2 = {z ∈ C||z| > α}

⇒ X(z) = 1,

ROCx = {z ∈ C}

ROCx1 ∩ROCx2 = {z ∈ C; |z| > α}

• Time Shifting

If x[n] Z←→ X(z), ROCx = R,

then

x[n− n0] Z←→ z−n0X(z), ROC = R

except for the possible addition or deletion of the ori-

gin or infinity

Proof.

Z{x[n− n0]} =∞∑

n=−∞x[n− n0]z−n

=∞∑

n′=−∞x[n′]z−n

′−n0 [n′ = n− n0]


= z−n0∞∑

n′=−∞x[n′]z−n

′

︸︷︷︸X(z)

,

ROC ⊇ ROCx ∩

ROC = {z ∈ C | possible except of z = 0 or z =∞}

If n0 > 0 then z = 0 is a pole which may cancel zeros

of X(z) at z = 0

⇒ z = 0 may be a pole of z−n0X(z) while it may not

be a pole of X(z).

⇒ ROC = ROCx deleting z = 0

If n0 < 0 then z = 0 is a zero which may cancel poles

of X(z) at z = 0

⇒ z = 0 may be a zero of z−n0X(z) while it may not

be a pole of X(z).

⇒ z =∞ is a pole of z−n0X(z)

⇒ ROC = ROCx deleting z =∞

• Convolution

{x1[n] Z←→ X1(z), ROCx1}and

{x2[n] Z←→ X2(z), ROCx2}


⇒x1[n] ∗ x2[n] Z←→ X1(z)X2(z),ROC ⊇ ROCx1 ∩ROCx2

For Z+, then

Z+{x1[n]u[n] ∗ x2[n]u[n]} = X1+(z)X2+(z),

ROC ⊇ ROC+x1∩ROC+

x2

Consider the example shown below Then

LTI

h[n]

x[n] y[n]=h[n]*x[n]

h[n]= [n]- [n-1]

h[n] Z←→ 1− z−1, ROCh = {z ∈ C|z 6= 0}

If x[n] Z←→ X(z), ROCx, then

y[n] Z←→ z − 1

zX(z), ROCy ⊇ ROCh ∩ROCx


This implies we may have a possible deletion from

ROCx at z = 0 and or addition at z = 1.

• Time Reversal (Same for Z , Z+)

x[n] Z←→ X(z),

ROCx = {z ∈ C;R1 < |z| < R2}

⇒

⇒x2[n]

4= x[−n] Z←→ X(z−1),

ROCx2{z ∈ C; 1R2< |z| < 1

R1}

ReRe

1R

2R

ImIm

1

1

R2

1

R

xROC

2x

ROC

Figure 4.4.21: Graph of change of ROC.

Proof.

Z{x[−n]} =∞∑

n=−∞x[−n]z−n =

∞∑n=−∞

x[n]zn

=∞∑

n=−∞x[n](z−1)−n = X(z−1),


ROCx2 = {z ∈ C|R1 < |1

z| < R2} =

= ROCx2 = {z ∈ C| 1

R2< |z| < 1

R1}

That is, if z0 ∈ ROCx, then 1z0∈ ROCx2

• Scaling in the Z-Domain (Same for Z , Z+) Suppose

x[n] Z←→ X(z),

ROCx = {z ∈ C|R1 < |z| < R2}

Then

x2[n]4= zn0x[n] Z←→ X(

z

z0),

ROCx2 = {z ∈ C|R1|z0| < |z| < R2|z0|}

If z0 = ejω0 then ejω0nx[n] Z←→ X(e−jω0z),

ROCx2 = ROCx

This means we have a rotation of poles and zeros

clockwise by ω0

e.g. a factor 1 − αz−1 becomes 1 − αejω0z−1, which

implies a pole at z = α is transformed to a pole at

z = αejω0


Re

Im

r

Re

Im

0

Rotation by _0

Figure 4.4.22: Graph of rotation.

Proof:

Z{z0nx[n]} =

∞∑n=−∞

z0nx[n]z−n

=∞∑

n=−∞x[n](

z

z0)n = X(

z

z0),

ROCx2 = {z ∈ C|R1 < |z

z0| < R2}

e.g. the Z−transform of αnx[n] is X(α−1z)

• Conjugation (Same for Z , Z+)

{x[n] Z←→ X(z), ROCx} ⇒

{x∗[n]ZX∗

(z∗), ROCx}


Proof:

Z{x∗[n]} =∞∑

n=−∞x∗[n]z−n

= (∞∑

n=−∞x[n](z∗)−n)∗

= X∗(z∗),ROCx∗ = {z ∈ C|z∗ ∈ ROCx = ROCx}

• Differentiation in the Z-Domain (Same for Z , Z+)

{x[n] Z←→ X(z), ROCx} ⇒

{nx[n] Z←→ −z ddzX(z),

ROCx with possible exception of z = 0}

Proof:

−z ddzX(z) = −z d

dz

∞∑n=−∞

x[n]z−n

= −z∞∑

n=−∞nx[n]z−n−1

=∞∑

n=−∞nx[n]z−n

︸︷︷︸Z{nx[n]}

e.g.

X(z) = log(1 + αz−1), |z| > |α|


⇒ nx[n] Z←→ −z ddzX(z) = +z

αz−2

1 + αz−1

=αz−1

1 + αz−1, |z| > |α|

• Time ExpansionDefine the time expanded signal

x(k)[n] =

x[n/k] of n is a multiple of k

0 if n is not a multiple of k

Then x[n] Z←→ X(z),

ROCx = {z ∈ C|R1 < |z| < R2}

⇒x(k)[n] Z←→ X(zk),

ROC = {z ∈ C|R1 < |zk| < R2}

Note that if X(z) has a pole at z = α then X(zk) hasa pole at z = α1/k.

• Time Shift: One-Sided: n0 > 0

(a) Z{x[n + n0]} = zn0X(z),

ROCx except z =∞ (shown before)

(b) Z+{x[n + n0]} = zn0{X+(z)−n0−1∑m=0

x[m]z−m},ROCx except z =∞


(c) Z+{x[n− n0]} = z−n0{X+(z) +−1∑

m=−n0

x[m]z−m},ROCx except z = 0

Proof:

(b) Z+{x[n + n0]} =∞∑n=0

x[n + n0]z−n

=∞∑

m=n0x[m]z−(m−n0), [m = n + n0]

= zn0∞∑

m=n0x[m]z−m

= zn0[∞∑m=0

x[m]z−m −n0−1∑m=0

x[m]z−m]

= zn0[X(z)−n0−1∑m=0

x[m]z−m]

(c) Z+{x[n− n0]} =

=∞∑n=0

x[n− n0]z−n

=∞∑

m=−n0

x[m]z−(m+n0), [m = n− n0]

= z−n0[∞∑m=0

x[m]z−m +−1∑

m=−n0

x[m]z−m]

Example 4.4.1

y[n]− 1

2y[n− 1] = δ[n], y[−1] = 3


⇒ Y+(z)− 1

2z−1[Y+(z) + y(−1)z] = 1

⇒ Y+(z) =5

2

z

z − 1/2, ROCY+ = {z ∈ C|z > 1

2}

⇒ Z−1{Y+(z)} =5

2(1

2)nu[n]

• The Initial Value Property: (Z+)

For a causal sequence, x[n], (e.g. x[n] = 0,∀n < 0) then

x[0] = limz→∞X(z)

Proof:

X(z) =∞∑n=0

x[n]z−n =

= x[0] ∗ 1 + x[1] ∗ z−1 + ... + x[m] ∗ z−m + ...

⇒ limz→∞X(z) = x[0]


x[n] = 7(1

3)nu[n]− 6(

1

2)nu[n]

Then

X(z) =z(z − 3/2)

(z − 1/3)(z − 1/2),

ROCx = {z ∈ C||z| > 1

2}


Note that x[0] = 1 and also limz→∞X(z) = 1 = x[0]

• Final Value Property: (only for Z+)

Let x[.] be a bounded seq. ∀n and

Z+{x[n]} = X+(z)

⇒{limn→∞ x[n] = limz→1(z − 1)X+(z)

}

Proof.

Z+{x[n + 1]− x[n]} = zX+(z)− zx(0)−X+(z)

OR∞∑n=0

[x[n + 1]− x[n]]z−n = (z − 1)X+(z)− zx(0)

⇒ (z − 1)X+(z) = zx(0) +

+∞∑k=0

[x[k + 1]− x[k]]z−k

⇒ limz→1

(z − 1)X+(z) = x(0) + (x[1]− x[0]) +

+... + (x[k + 1]− x[k]) + ...

= limk→∞

x[k]

Example 4.4.3 (Some applications)

(a) Consider the bounded signal.

x[n] =

0, n < 0αn, n ≥ 0

4.5. THE INVERSE Z-TRANSFORM 197

Then

X+(z) =z

z − α, |z| > α, |α| < 1

⇒ limn→∞x[n] = lim

z→1(z − 1)X+(z)

= limz→1

(z − 1)z

z − α= 0

this is consistent with limn→∞x[n]||α|<1 = 0

(b) Consider the unbounded signal.

x[n] = 2nu[n]←→ X+(z) =z

z − 2, |z| > 2

Then

limz→1

(z − 1)z

z − 2= 0 6=∞ (which the true limit)

4.5 The Inverse Z-Transform

The inverse Z-Transform is found from complex analysis

by performing contour integrations. However, we will not

use this method. Rather, we will apply partial fraction

expansions and properties of Z-Transforms.

X(z) |z=rejω = F{x[n]r−n}, for any r such that z ∈ ROC


⇒ x[n]r−n = F−1{X(rejω)} (Applying Inverse F.T.)

⇒ x[n] = rn1

2π

∫<2π>X(rejω)ejωndω

⇒ x[n] =1

2π

∫<2π>X(rejω)(rejω)ndω

Recover x[n] from Z-transform evaluated along the con-

tour z = rejω ∈ ROC)

n = fixed, ω ∈ [0, 2π]. Then

z = rejω

dz = jrejωdω = jzdω

⇒ dω =1

jz−1dz

z | over |z| = r

ω | over 2π interval in ω

⇒ x[n] =1

2πj

∮X(z)zn−1dz (z-plane integral)


Integration around a counterclockwise contour centered

at origin and radius r, such that X(z) converges. e.g.

|z| = r ∈ ROC

ROC

contour

Im

Re

⇒ Z−∞{X(z)} =1

2πj

∮X(z)zn−1dz

The integral requires the use of contour integration in

z−plane

Z−1+ {X+(z)} =

1

2πj

∮X+(z)zn−1dz

Now, we show how to calculate Z−1,Z−1+ by partial

fraction expansion and Transform Tables.


Example 4.5.1 (Inverse Z−transform by partial frac-

tions.)

(a) Consider

X(z) =1 + 0.5z−1 − 0.32z−2

(1− 0.5z−1)(1− 0.2z−1)2 , |z| > 0.5

By partial fraction expansions then

X(z) =A

(1− 0.5z−1)+

B + Cz−1

(1− 0.2z−1)2 (4.5.5)

Now, we determine A, as follows (see additional notes

on partial fractions).

A = X(z).(1− 0.5z−1)|z−1=2 = 2

Then from (4.5.5) we have

A(1− 0.2z−1)2

+ (B + Cz−1)(1− 0.5z−1)

= 1 + 0.5z−1 − 0.32z−2

Next, we determine B,C from the coefficients of pow-

ers of z.

Coefficient of z0: A + B = 1, hence B = 1 − A =

1− 2 = −1


Coefficient of z−1: A(−0.4) − 0.5B + C = 0.5, hence

C = 0.5 + 0.8− 0.5 = 0.8.

Finally,

X(z) =2

1− 0.5z−1+−1 + 0.8z−1

(1− 0.2z−1)2

=2

1− 0.5z−1− 1

1− 0.2z−1+

0.6z−1

(1− 0.2z−1)2

}5.0||/{ zZzROCx

1xROC

2xROC

3xROC

321 xxxROCROCROC

Correspond toRight-sided

xxxxx

0.2

0.20.2

0.50.5

2)2.0(

6.0

2.05.0

2)(

z

z

z

z

z

zzX

double


The inverse Z−transform gives

x[n] = 2(0.5)nu[n]− (0.2)nu[n] + 3n(0.2)nu[n]

(b) Consider

X(z) =1 + 0.5z−1 − 0.32z−2

(1− 0.5z−1)(1− 0.2z−1)2 , |z| < 0.2

By the partial Fraction of (a) then

xROC

1xROC

2xROC

3xROC

Corresponds toleft-sided

xx

2)2.0(

6.0

2.05.0

2)(

z

z

z

z

z

zzX

xx

0.2 0.5

L.S. L.S. L.S.


ROCx = {z ∈ C| |z| < 0.2} =

= ROCx1 ∩ROCx2 ∩ROCx3

⇒ x[n] = −2(0.5)nu[−n− 1] +

+(0.2)nu[−n− 1]− 3n(0.2)nu[−n− 1]


(c) Consider

X(z) =1 + 0.5z−1 − 0.32z−2

(1− 0.5z−1)(1− 0.2z−1)2 , 0.2 < |z| < 0.5

Then by partial fraction expansion we have

2)2.0(

6.0

2.05.0

2)(

z

z

z

z

z

zzX

1xROC

2xROC

3xROC

xROC

Combination of

left and right sided

signals

x0.2

x0.2xx

0.20.5

xx0.5

Right-Sided

Signal

Left-Sided

Signal

Right-Sided

Signal

ROCx = {z ∈ C| 0.2 < |z| < 0.5} =

= ROCx1 ∩ROCx2 ∩ROCx3

⇒ x[n] = −2(0.5)nu[−n− 1]− (0.2)nu[n] + 3n(0.2)nu[n]


(d) Consider

X+(z) =2z

z − 0.5− z

z − 0.2+

0.6z

(z − 0.2)2

This contains only Right-Sided Signals. Hence, ROCx+ =

{z ∈ C| |z| > 0.5}. The inverse is found as before.

• Inverse Z-Transform by long Division

Another method to find the inverse Z-transform is to

apply long division. We illustrate this below.

(a) Consider

X(z) =1 + 0.5z−1 − 0.32z−2

1− 0.9z−1 + 0.24z−2 − 0.2z−3, |z| > 0.5

=1 + 0.5z−1 − 0.32z−2

(1− 0.5z−1)(1− 0.2z−1)2

By long division we have

X(z) = 1 + 1.4z−1 + 0.7z−2 + 0.314z−3 + 0.1426z−4 + ...

⇒ x[0] = 1, x[1] = 1.4, x[2] = 0.7, x[3] = 0.314, ...

(agrees with example (a))


(b) Consider

X(z) =1

1− αz−1, |z| > |α|

⇒ 1

1− αz−1= 1 + αz−1 + α2z−2 + ...

(converges, since |αz−1| < 1)

⇒ x[0] = 1, x[1] = α, x[2] = α2 ⇒ x[n] = αnu[n]

4.6. ANALYSIS OF CONVOLUTION SYSTEMS USING Z-TRANSFORMS 207

4.6 Analysis of Convolution Systems Using Z-Transforms

Convolution System:

h[n]x[n]

LTI

])[*(][][][ nhxknhkxnyk

Figure 4.6.23: Convolution system.

Suppose Z{x[n]} exists in ROCx and Z{h[n]} in ROCh

⇒ Z{y[n]} = Z{x[n]}Z{h[n]}

Y (z)4= Z{y[n]} (exists in ROCy ⊇ ROCx ∩ROCh)

X(z)4= Z{x[n]}


The Transfer Function of the LTI System is

H(z)4= Z{h[n]} ⇒ Y (z) = X(z)H(z)

Let x[n] = zn then

y[n] =∞∑

k=−∞zn−kh[k] = zn

∞∑k=−∞

h[k]z−k

︸︷︷︸H(z)

⇒ y[n] = H(z)zn

Conclusion:

The input-output behavior of a convolution system whose

impulse response, h[n] has aZ-transformH(z) = Z{h[n]},is determined by its transfer function

4= H(z) and its Re-

gion of Convergence.

4.6. ANALYSIS OF CONVOLUTION SYSTEMS USING Z-TRANSFORMS 209

Example 4.6.1 Connections.

(a) Parallel Connection

h1[-]

h2[-]

H1,ROC1

H2,ROC2

+

-

y [n] X (z) Y (z)x [n]

x [n] y [n] X (z) Y (z)H1(z)+H2(z)

ROC ROC1 ROC2

h1[n]+h2[n]

Figure 4.6.24: Parallel connection.


(b) Cascade Connection

H1

ROC1

H2

ROC2

X (z) Y (z)h1 [n] h2 [n]

x [n] y [n]

H1(z)·H2(z)

ROC ROC1 ROC2

y [n](h1* h2) [n]

x [n] X (z) Y (z)

Figure 4.6.25: Cascade connection.

4.7 LTI Systems Characterized by Difference Equa-

tions

Now, we illustrate the different techniques that allow us to

calculate the impulse response, without assuming the LTI

system is BIBO stable, as done when we applied Fourier

Transfor methods.

4.7. LTI SYSTEMS CHARACTERIZED BY DIFFERENCE EQUATIONS 211

• Calculation of Impulse Response

N∑k=0

αky[n− k] =M∑k=0

βkx[n− k]

Assume zero Initial Conditions

(a) Convolution approach

x [n] = zn h [·] y [n] = H(z) zn

LTI

⇒N∑k=0

αkz−kY (z) =

M∑k=0

βkz−kX(z)

⇒ H(z) = Y (z)X(z) =

M∑k=0

βkz−k

N∑k=0

αkz−k

We need to specify the Region of Convergence using in-

formation about causallity and stability

(b) Z−transform approach

Z{N∑k=0

aky[n− k]} =Z{M∑k=0

bkx[n− k]}

⇒N∑k=0

akz−kY (z) =

M∑k=0

bkz−kX(z)


Since y[n] = x[n] ∗ h[n] then Y (z) = H(z)X(z)

⇒ H(z)4= Y (z)

X(z) =

M∑k=0

βkz−k

N∑k=0

αkz−k

Factorized Form:

H(z) =βMαN

M∏i=1

(z−1 − σ−1i )

N∏i=1

(z−1 − λ−1i )

=βMαN

M ′∏i=1

(z − σi)N ′∏i=1

(z − λi)

where

σi −→ zeros of system

λi −→ poles of system

Suppose H(z) exists in ROCh. Then

Y (z) exists in ROCy ⊇ ROCh ∩ROCx

The Impulse Response is h[n] =Z−1{H(z)


The Step Response is y[n] = s[n] =Z−1{H(z)U(z)},where U(z) =Z{u[n]}

• Causal LTI Systems with x[n] = 0,∀n < 0

x yh

LTI

Figure 4.7.26: Block diagram of LTI system.

If h[n] is Causal then h[n] = 0,∀n < 0. Hence, we have

H(z) = Z{h[n]} = Z{h[n]u[n]} = Z+{h[n]} = H+(z)

Since the input satisfies x[n] = 0,∀n < 0, then we have

X(z) =Z{x[n]} =Z{x[n]u[n]}=Z+{x[n]} = X+(z)

Then

y[n] = h[n] ∗ x[n] = h[n]u[n] ∗ x[n]u[n]


=∞∑

k=−∞h[k]u[k]x[n− k]u[n− k]

=∞∑k=0

h[k]x[n− k]u[n− k]

⇒ y[n] =n∑k=0

h[k]x[n− k]

⇒ y[n] = 0,∀n < 0

u[n− k] =

1, n > k0, otherwise

k > 0⇒ y[n] = 0,∀n < 0

⇒ Y (z) =Z{y[n]} =Z{y[n]u[n]} [y[n] = 0,∀n < 0]

=Z+{y[n]} = Y+(z)

⇒ Y (z) = Y+(z) =Z{h[n]u[n]∗x[n]u[n]} = H+(z)X+(z)

—bf Conclusion. If the LTI is Causal and x[n] = 0,∀n <0 then Z+{·} is sufficient for the analysis of the system.

If x(n) = 0, n < no, no < 0, then define x(n) = x(n−no) = 0, n < 0 and compute y(n).

Finally compute y(n) = y(n + no)


• Relationship Between Causality and Stability of Dif-

ference (LTI) Systems and the Region of Convergence

of H(z)

I. Suppose the LTI is Causal, e.g., h[n] = 0, ∀n < 0

⇒ H(z) is Right-sided

⇒ ROCh = {z ∈ C| |z| > maxi=1,...,N |pole zi|}

Im

Re0 x

x

x

|zi|

H(z) =∞∑n=0

h[n]z−n ⇒ ROC includes z =∞ [for causal

systems]

Suppose:N∑k=0


βkx[n− k]

⇒ H(z) = b0+b1z−1+...+bMz

−M

a0+a1z−1+...+aNz−N

⇒ ROCh is Right-sided and is to the right of the outmost


pole.

Conclusion.

(a) A discrete-time system is causal if and only if the ROC

of its system function is the exterior of a circle, including

infinity.

Justification:

(a) (⇒)Assume Causal. Then h[n] = 0,∀n < 0⇒ H(z)

is Right-sided. Since H(z) =∞∑n=0

h[n]z−n then ROC in-

cludes z =∞

(⇐) Assume ROC is the exterior of a circle. Then h[n] =

0, ∀n ≤ no, no < 0 e.g. Z{δ[n]}ROC={z∈C} = 1,Z{δ[n+

1]}ROC={z∈C|z 6=∞}

⇒ H(z) =∞∑

n=noh[n]z−n, no < 0 ↑

includes zpositivepower

If z = ∞ ∈ ROC then zpositivepower are excluded ⇒H(z) =

∞∑noh[n] for some no ≥ 0

⇒ ∑h[n]z−n = 0, ∀n < no

⇒ Causal

(b) A discrete-time LTI system with rational function


H(z) is causal if and only if:

(i) The ROC is the exterior of a circle outside the outer-

most pole

(ii) with H(z) expressed as a ratio of polynomials in z, the

order of the numerator cannot be greater than the order

of the denominator.

II. Suppose LTI is stable

⇒ ‖h‖1 <∞⇒ F{h[n]} exists

⇒ H(ejω) = F{h[n]} exists

⇒ROCh must include |z| = 1

z-plan

unit circle

included in ROC

Conclusion. An LTI system is B.I.B.O Stable if and

only if the region of its system function H(z) includes the

unit circle z=1.


⇐ Suppose z = 1 not ∈ ROC, then F{h(n)} =∞⇒ |H(ejω) = |

∞∑n=−∞

h(n)e−jωn| ≤∞∑

n=−∞|h(n)| · 1 =∞

e.g ‖h‖1 =∞ hence the system is not BIBO stable.

Main Conclusion.

A causal LTI system withrational H(z) is stable

if and only if

all poles of H(z) satisfy|zi| < 1, i = 1, . . . , N

Example 4.7.1 Now, we illustrate the concepts via

examples.

(a) H(z) = z(z−1)(z−2)(z−0.5)(z−0.8)

⇒ Not causal

[Numerator higher order than denominator]⇒ z =∞not ∈ ROC(b) H(z) = (0.5)nu[n]

⇒ H(z) = zz−0.5, ROCh = {z ∈ C| |z| > 0.5}


ROC - zero

x

Im

unit

circle

Rex

0.5

[Both Caunsal and stable]

(c) h[n] = (0.5)n+1u[n + 1]

⇒ h[n] = h[n]|n→n+1, h[n] = (0.5)nu[n]

But h[n] Z←→ zz−0.5, ROCh = {z ∈ C| |z| > 0.5}

⇒ h[n] Z←→ zH(z) = z zz−0.5, ROCh = {z ∈ C| |z| >

0.5, except z =∞}⇒ Not Causal


• Solving Difference Equations via Z+-Transforms

N∑k=0


bkx[n− k]

with initial conditions y[−1], y[−2], . . . , y[−N ].

We use

Z+{x(n− no)} = z−no{X+(z) +−1∑

m=−nox(m)z−m}

Then

Z+{N∑k=0

aky[n− k]} = Z+{M∑k=0

b kx[n− k]}

⇒ Y+(z)N∑k=0

akz−k +

N∑k=0

akz−k −1∑

m=−ky[m]z−m

= X+(z)M∑k=0

bkz−k +

M∑k=0

bkz−k −1∑

m=−kx[m]z−m

⇒ Y+(z)N∑k=0

akz−k =

X+(z)N∑k=0

bkz−k +

M∑k=0

bkz−k −1∑

m=−kx[m]z−m −

N∑k=0

akz−k −1∑

m=−ky[m]z−m

⇒ Y+(z) =

N∑k=0

bkz−k

N∑k=0

akz−k

︸︷︷︸H+(z)

X+(z)


+{M∑k=0

−1∑m=−k

bkz−kx[m]z−m −

N∑k=0

−1∑m=−k

akz−ky[m]z−m}

N∑k=0

akz−k

Y+(z) =

N∑k=0

bkz−kX+(z)

N∑k=0

akz−k

+

[M∑k=0

−1∑m=−k


N∑k=0

−1∑m=−k

akz−ky[m]z−m]

N∑k=0

akz−k

,

ROC+y ∩ROC+

x

Y+(z) is the superposition of zero-input and zero state

response, as follows.

N∑k=0

bkz−kX+(z)

N∑k=0

akz−k

←− Y+(z) zero-state Resp.

[M∑k=0

−1∑m=−k


N∑k=0

−1∑m=−k

akz−ky[m]z−m]

N∑k=0

akz−k

←−

zero-Input Resp.


Example 4.7.2 Consider

H+(z) =z + 0.5

(z + 0.2)(z + 0.4), |z| > | − 0.4|

X+(z) =z

z − 1, |z| > 1

Then

H+(z)X+(z) = Y+(z) = z(z+0.5)(z+0.2)(z+0.4)(z−1), |z| > 1

⇒ Y+(z) = −1.25zz+0.2 +

12.8zz+0.4 +

2528zz−1

(1) Natural Response: Z−1+ {·} of the terms which come

from the poles of TF

yh[n] = 12.8(−0.4)n − 1.25(−0.2)n, n ≥ 0

(2) Forced Response: Z−1+ {·} of the terms which come

from input forcing function

yp = 2528, n ≥ 0

(3) Complete Response:

y[n] = 12.8(−0.4)n − 1.25(−0.2)n + 25

28, n ≥ 0


Note also that from H+(z) we can determine the dif-

ference equation

y[n + 2] + 0.6y[n + 1] + 0.08y[n] = x[n + 1] + 0.5x[n]

Note also that the system is initially at Rest, e.g,

y[n] = 0,∀n < 0 before input is applied, because |z| =

∞ ∈ ROCh

When x[n] = u[n] ⇒ y[0] = 0, y[1] = 1 (letting n =

−2,−1)

which agrees with solution above.

Example 4.7.3 Consider example 4.7.2,with input

x(n) = (−0.4)n, n ≥ 0. Then

X+(z) = zz+0.4,ROCx = {z ∈ C| |z| > | − 0.4|}

⇒ Y+(z) = H+(z)X+(z) =z + 0.5

(z + 0.2)(z + 0.4)· z

z + 0.4

=7.5z

z + 0.2− 7.5z

z + 0.4− 0.5z

(z + 0.4)2︸︷︷︸Response due to x(n)

y(n) = (7.5)(−0.2)n − 7.5(−0.4)n︸︷︷︸Natural Response


+ 1.25n(−0.4)n︸︷︷︸Result of input excitingthe natural mode(-0.4)

, n ≥ 0

Note the following.

• The pole z = −0.4 of system T.F is excited by the

input function, resulting in a double pole which leads

to the term n(−0.4)n.

• Direct calculation using the difference equation yields

the same answer.

e.g.,

y(n+2)+0.6y(n+1)+0.08y(n) = x(n+1)+0.5x(n),

where x(n) = (−0.4)nu(n), and y(n) = 0, n < 0.

Example 4.7.4 Consider

y(n+ 2) + 0.6y(n+ 1) + 0.08y(n) = x(n+ 1) + 0.5x(n)

⇒ Y+(z)(z2 + 0.6z + 0.08)− y(0)z2 − y(1)z − 0.6y(0)z

= (z + 0.5)X+(z)− x(0)z


⇒ Y+(z) =z + 0.5

(z + 0.2)(z + 0.4)X+(z)

︸︷︷︸Represents the part ofthe output due to zero IC’s.All zero-state Response

+y(0)z2 + y(1)z + 0.6y(0)z − x(0)z

(z + 0.2)(z + 0.4)︸︷︷︸Represents the part ofoutput due to non-zero IC’s.All zero-input Response

Thus, Y+(z) consists of three components:

(a) The Natural Response under zero IC’s

(b) The Natural Response due to IC’s only

(c) The forced Response

(a),(c) zero-state response

Let y(0) = 0, y(1) = 1.5︸︷︷︸These do not correspondto initial Restassumption as before

, x(n) = u(n)


Then

Y+(z) = (z+0.5)z(z+0.2)(z+0.4)(z−1) + 1.5z−z

(z+0.2)(z+0.4)

Natural Responseunder zero IC’s

is 12.8(−0.4)n − 1.25(−0.2)n, n ≥ 0,

[example 4.7.2]

Forced Response is 2528, n ≥ 0 [example 4.7.2]

The Natural Response due to IC’s only is

1.5z − z(z + 0.2)(z + 0.4)

=2.5z

z + 0.2− 2.5z

z + 0.4

which implies 2.5(−0.2)n − 2.5(−0.4)n, n ≥ 0

Total Natural Response is

yh(n) = 12.8(−0.4)n − 1.25(−0.2)n

+ 2.5(−0.2)n − 2.5(−0.4)n, n ≥ 0

= (1.25)(−0.2)n − 157 (−0.4)n, n ≥ 0


Complete Response is y(n) = yh(n) + 2528, n ≥ 0

Note the following.

•y(0) = 0y(1) = 1.5

as expected

• Under zero IC’s, the natural response is y(1) = 1 as

in example 4.7.2.

• Hence, the contribution of the non-zero IC’s to the

natural response at n = 1 is 1.5−1 = 0.5. This agrees

with the value obtained from the Natural Response

due to IC’s only which is

2.5(−0.2)− 2.5(−0.4) = 0.5


• Numerical Solutions of Differential Equations

Let y(t) be continuous at t = kTs, k ∈ I

Let y(k)4= y(kTs)

y (t)

Ts(k-1) Tsk t

We approximate the differential equation by using Back-ward Difference. This is how differential equatiosn aresolved by computers.

(a)

d

dty(t)|t=kTs ≈

y(kTs)− y([k − 1]Ts)

Ts

=y(k)− y(k − 1)

Ts


(b)

d2

dt2y(t)|t=kTs =

d

dt[d

dty(t)]|t=kTs

≈ddty(t)|t=kTs − d

dty(t)|t=(k−1)Ts

Ts

=[y(k)− y(k − 1)]/Ts − [y(k − 1)− y(k − 2)]/Ts

Ts

=1

Ts2 [y(k)− 2y(k − 1) + y(k − 2)]

(c)

dn

dtny(t)|t=kTs ≈

1

Ts

n n∑i=0

(−1)i ni

y(k − i)

(d) ddty(t)|t=0 = dy(0)

dt = y(0)−y(−1)Ts

⇒ y(−1)− y(0) = −Tsdy(0)dt


Example 4.7.5

2

3 6

+

- -

++ +

--1F 10

=2.5V20=0Vx (t) y (t)1

8

F

The equation of the circuit is

d2

dt2y(t) + 3

d

dty(t) + 2y(t) = 2x(t), y(0) = 0,

d

dty(0) = 3

Using Backward Difference Approximation we have

( 1Ts

2 + 3Ts

+ 2)y(n)− ( 2Ts

2 + 3Ts

)y(n− 1)

+ 1Ts

2y(n− 2) = 2x(n), n ≥ 0

y(0) = 0, y(−1) = −3Ts

(∗)


t Exact Solution (Approx.)T=0.05 (Approx.)T=0.10.0 0 0 00.4 .7717 .7438 .71851.0 1.0972 1.0796 1.06251.5 1.1236 1.1168 1.10962.0 1.0987 1.0979 1.9065

Characteristic roots of (*):r1 = 1

1+Ts, r2 = 11 + 2Ts

Step Response: y(n) = c1rn1 + c2r

n2 + 1, n ≥ 0

c1 = 1, c2 = −2⇒ y(n) = ( 1

1+Ts)n − 2( 1

1+2Ts)n

+ 1, n ≥ 0

Exact step solution:ye(t) = e−t − 2e−2t + 1, t ≥ 0

⇒ ye(n) = e−nTs − 2e−2nTs + 1

= (1

eTs)n − 2(

1

e2Ts)n + 1

≈ (1

1 + Ts)n − 2(

1

1 + 2Ts)n + 1


Chapter 5

Frequency Analysis of Signals andSystems and Sampling Theorem

5.1 Magnitude and Phase Representations of DTFT

Recall that a complex function is expressed in terms of

magnitude and phase as follows.

X(ejω) = |X(ejω)|ej 6 X(ejω),

where 6 X(ejω) = tan−1 Im{X(ejω)}Re{X(ejω)} Provides information

about relative phases of complex expressions

|X(ejω)| Provides information about the relative mag-

nitude of complex expressions.

233

234CHAPTER 5. FREQUENCYANALYSIS OF SIGNALS AND SYSTEMS AND SAMPLING THEOREM

(a) Periodic:

x(n) =∑

k=<N>ake

jkωon, an =1

N

∑k=<N>

x(k)e−jkωon

1N

∑n=<N>

|x(n)|2 =∑

k=<N>|ak|2

|ak|2 = Average power in the kth harmonic content of

x(n)

(b) Aperiodic:∞∑

n=−∞|x(n)|2 =

1

2π

∫<0,2π> |X(ejω)|2︸︷︷︸

energy densityspectrum of x(n)

dω

|X(ejω)|2dω2π is the amount of energy of x(n) that lies in

the infinitesimal frequency band between [ω, ω + dω]

5.1. MAGNITUDE AND PHASE REPRESENTATIONS OF DTFT 235

Example 5.1.1 We discuss some examples.

(a) x(n)↔ |X(ejω)|ej 6 X(ejω)

x(−n)↔ |X(ejω)|e−j 6 X(ejω) if x(n) is real because

x(−n)↔ X(e−jω)

(c) Design an H(ejω) so that y(n) ≈ x(n) for low freq.

input signal x(t)

n (t) {high-freq. Noise n>> c}

+ )( jeH

x (t) y (t)

-3 C2--2

)( jeN

channel

3

⇒ Y (ejω) =

X(ejω),∀ω ∈ [−ωc, ωc]0,otherwise


5.2 Frequency Response of LTI Systems

x (n) y (n)=x(n)*h(n)h (n)

)( jeX

)( jeH

)()()( jjjeXeHeY

So, H(ejω) = |H(ejω)|ej 6 H(ejω)

Shapes the freq. characteristics of X(ejω)

⇒ |Y (ejω)| = |H(ejω)||X(ejω)| [Scale]

6 Y (ejω) = 6 H(ejω) + 6 X(ejω) [phase shift]

• Linear and Nonlinear phase

If 6 H(ejω) is linear in ω then phase distortion is linear

If 6 H(ejω) is nonlinear in ω then phase distortion is non-

5.2. FREQUENCY RESPONSE OF LTI SYSTEMS 237

linear

Assume Linear phase distortion of the form H(ejω) =

e−jωn

n

x (n)

0)()(njjj

eeXeY0)(

njjeeH

)( jeX

n00 n0

y (n)=x(n-n0)

Delay of n0

In 0, 0

Assume Nonlinear phase distortion of the form H(ejω) =

ejg(ω)n

)(j

eXnjg

e)(

njgjjeeXeY

)()()(

0 n

x (n)

0 n

y (n)


• Log-Magnitude Plot

log |Y (ejω)| = log |H(ejω)| + log |X(ejω)|

6 Y (ejω) = 6 H(ejω) + 6 X(ejω)

20 log10(·) 4= decibels(dB)

Thus, 0dB ⇒ (·) = 1

− 3dB = 20 log 1√2|H(ejω)|max

Here h(n) is Real so only 0 ≤ ω ≤ π is needed

|H(ejω)| = |H(e−jω)| and 6 H(e−jω) = −6 H(ejω)

5.2. FREQUENCY RESPONSE OF LTI SYSTEMS 239

-

)(log20 10

jeH

-

)( jeH


5.3 Ideal Frequency Selective Filters: Low Pass, High

Pass, Band Pass

Frequency Selective Filters are systems which select cer-

tain frequencies while they reject others.

• Low Pass

-2 - C 2-C0

)(j

eH

1

stopband Passband stopband

• High Pass

)(j

eH

--3

3

• Band Pass

)(j

eH

-2 2-

5.3. IDEAL FREQUENCY SELECTIVE FILTERS: LOWPASS, HIGH PASS, BAND PASS241

• FIR Filters (Non-Recursive Moving Average)

y(n) = 1M+N+1

M∑k=−N

bkx(n−k) [Freq.Selective Filter]

Y (ejω) = 1M+N+1

M∑k=−N

bke−jωkX(ejω)

The output at time n0, i.e., y(n0) is the average of val-

ues of x(n), near n0. Hence, high freq. components of

the input are average out and lower freq. variations are

retained, corresponding to smoothing or lowpass filtering

of the original sequence.

(a) M + N + 1 >> 1, bk = 1 [Low Pass Filter]

H(ejω) =Y (ejω)

X(ejω)=

1

M + N + 1

M∑k=−N

e−jωk

=1

M + N + 1

M∑k=−N

e−jωk

=1

M + N + 1

(1− (e−jω)N+M+1)

1− e−jω


⇒ H(ejω) = 1M+N+1e

jωN+M2

sin(ω(N+M+1)2 )

sin(ω2 )

- 2

…

)(j

eH

(b) M + N + 1 = 3, bk = 1

y(n) = 13{x(n− 1) + x(n) + x(n + 1)}

⇒ h(n) = 13δ(n− 1) + 1

3δ(n) + 13δ(n + 1)

⇒ H(ejω) = 13[ejω + 1 + e−jω] = 1

3(1 + 2 cosω)

(c) M + N + 1 = 2, bk = 1 [Low Pass Filter]

y(n) = 12(x(n) + x(n− 1))


-2 2- 0

)(j

eH

3

1

2

3

⇒ h(n) = 12δ(n) + 1

2δ(n− 1)

⇒ H(ejω) = 12[1 + e−jω] = e−

jω2 cos(ω2 )

)(j

eH

-2 - 20

……

-

)(j

eH

2

2

……

Low Freq.:ω = 0, 2π, . . .


High Freq.:ω = π, 3π, . . .

e.g

Suppose x(n) = kej0n ⇒ y(n) = H(ej0)kej0n = k =

x(n)

Suppose x(n) = kejπn ⇒ y(n) = H(ejπ)kejπn = 0

(d) M +N + 1 = 2,−bi = bk, i 6= k. [High Pass Filter]

y(n) = x(n)−x(n−1)2

⇒ h(n) = 12δ(n)− 1

2δ(n− 1)

⇒ H(ejω) = 12(1− e−jω) = e−j

ω2 +j π2 sin(ω2 )

-3 32--2 0

1

)( jeH


• IIR Filters

(a) Low-Pass

- -C C0

)(j

L eH

HL(ejω) =

1, |ω| ≤ ωc0, otherwise e.g. ωc < |ω| ≤ π

hL(n) = 12π

∫ π−π e

jωnH(ejω)dω =1

2π

∫ ωc−ωc e

jωndω

= 12πejωcn−e−jωcn

jn

⇒ hL(n) = sinωcnπn = ωc

πsinωcnωcn

ωn = kπ, k = ±1,±2, . . .

ωn = ωcn = kπ, k = ±1,±2, . . .

hL(n) is Non-Causal


h (n)

n (n)

…

C

4C

[ ]n C

C

n

22

4n C

n n

( )C

n

…

As ωc −→ π

n

hL(n)

C

C

C increases C

n

hAp (n)hL(n)

C

1

-

( )j

H e

- C C

C increases ( )j

H eC

( )j

H e

0-


(b) Delayed Version of Low Pass Impulse Resp.

[x(n− n0)↔ e−jωn0X(ejω)]

x (n) y (n)

( )j

L

x (n) y (n)

H e 0j ne

0( )Lh n n

Linear phase distortion

Almost Causal

(c) Delayed Version of Low Pass Freq.Resp. by “π”

ejω0nx(n)↔ X(ej(ω−π)) Gives the High-Pass Filter

HH(ejω) = HL(ejω)|ω→ω−π

- -( - C) ( - C)

( )j

HH e

0

[High-Pass]

hH(n) = F−1{HL(ej(ω−π))} = ejπnhL(n) = (−1)nhL(n)


(d) Combination of Low-Pass and High-Pass Filters

-

1

( )j

LHH e

0 C- C

x (n)

( )j

LH e

y (n)

( )j

LH e

x x

( )( ) ( ) ( )

j j j

LH L LH e H e H e

(e) Band Pass can be generated by

HB(ejω) = HL(ej(ω−ω0)), 0 < ω0 < π


(f) Step Response

hL(n)( ) ( )

n

L L

m

y n h n( )jU e

( )j

LH e( )

jY e u (n)

……

n

yL(n)

Undesired Properties:

1. Non-Causal and hence we need to approximateHL(ejω)

by a causal one.

2. Oscillations in the Step Response (ringing)

3. Order of Filter is very high


(g) Non-Ideal Frequency Selective Filters

( )j

H e

Passbandstopband

Passband Ripple

1- 1

1

1+ 1

-

2

0 sp

5.4. FREQUENCY RESPONSE OF FIRST AND SECOND-ORDER SYSTEMS 251

5.4 Frequency Response of First and Second-Order Sys-

tems

First-Order Discrete-Time Systems

( )j

H ex (n) y (n)

LTI, Cansal

• y(n)− ay(n− 1) = x(n) , |a| < 1

⇒ H(ejω) = 11−ae−jω ←→ h(n) = anu(n)

⇒ s(n) = h(n) ∗ u(n) = 1−an+1

1−a u(n)

⇒ H(ejω) = 1

[(1+a2)−2a cosω]12e−j6 a sinω

1−a cosω

Note:

H(ejω) = 11− a

ejω= ejω

ejω−a


⇒ y(n + 1)− ay(n) = x(n + 1), n ≥ 0

⇒ y(n + 1) = x(n + 1) + ay(n)

y(n) = − a1−aa

nu(n) + 11−au(n) = 1

1−a[1− an+1]u(n),

x(n) = u(n)

Initial Conditions: y(0) = 1,

yn(n) = canu(n)yp(n) = k ⇒ k − ak = 1

Step Resp.

⇒ k = 11−a ⇒ y(n) = canu(n) + 1

1−au(n)

y(0) = 1 = c + 11−a ⇒ c = 1− 1

1−a = 1−a−11−a = −a

1−a


Graph of h(n)

h (n) h (n)

…

17

8a

n

…

n

7

8a

Ringing-Oscillations

1

…

h (n) h (n)

n

3

4a1

3

4a1

…n

h (n) h (n)

1 1

…n

1

2a

…n

1

2a

h (n) h (n)1 1

…

n

1

4a

…

n

1

4a


Graph of s(n)

n

s (n)

1

4

3

s (n)

……

n

13

4

1

4a 1

4a

Ringing Oscillations

…

n

s (n)

1

2

1

2a

n

s (n)

1

1

2a

2

3…

n

s (n)

1

43

4a

…

n

s (n)

14

7

3

4a

s(n) = 1−an+1

1−a u(n) = 11−au(n) [as n→∞]

Note:

• |a| determines the rate at which the first-order system

responds to inputs.

• h(n), s(n) converge to their final value at the rate at


which |a|n converges to zero. Thus, impulse response

decays sharply and s(n) settles quickly for |a| small.

For |a| ≈ 1, the response is slow.

• Unlike continuous-time systems, first-order discrete-

time systems display oscillatory behaviour, for a < 0,

in which s(n) exhibits both overshoot of its final value

and ringing.

Graph of |H(ejω)|, 6 H(ejω), a > 0.[Low Pass Filter]

-

-2 20

4

8

16

20log ( )j

H e

7

8

3

4

1

4

Trade-off. a 1

Good Filtering lowpass

Slow response

--220

7

8

3

4

( )jH e

4

4


• System attenuates freq. at ω = ±π, 3π, . . .

•

@ω = 0 : |H(ejω)| = 11−a

@ω = π2 : |H(ejω)| = 1

1+a

1. for small a, |H(ejω)|max ≈ |H(ejω)|min

2. For a ≈ 1 |H(ejω)|max >> |H(ejω)|

Thus, for α ' 1, filtering and amplification is selective

over a narrow band of frequencies.

Graph of H(ejω), 6 H(ejω) α < 0 (High Pass Filter)

Same as that of Low Pass Filter, shifted by π freq.

units.


|)(|j

eH

- 0

4

3

8

7


• Second-Order Discrete-Time Systems

H(z)x(n) y(n)

LTI System, Causal

• H(z) = kz2+cz+dz2+az+b

⇒ Poles: zP1,2 =−α±

√a2 − 4b

2⇒ B.I.B.O.⇔ zP1,2 ∈ {z ∈ C| |z| < 1}

1

Re(z)

Im(z)


(a) Underdamped Case

(Poles are complex conjugate pairs e.g., b > a2/4)

B.I.B.O. ⇔ |zP1,P2| < 1⇒ b < 1

⇒ a2

4 < b < 1 (Underdamped and B.I.B.O. stable)

(b) Critically Damped Case

(Poles are identical e.g., b = a2/4)

zP1,P2 = −a2

B.I.B.O. ⇔ −2 < a < 2 )| − a

2| < 1⇒ |a| < 2)

⇒ a2

4 = b < 1 (Critically damped and B.I.B.O. stable)


(c) Overdamped Case

The two poles are distinct and real e.g., b < a2/4

B.I.B.O. and overdamped ⇔

⇔ b > a− 1, b > −a− 1, 2 > a > −2

-1

1

210-1

b=1

b=a-1b

underdamped

critically damped

overdamped

b=-a-1

a

4

2a

b

Region of Stability of 2nd Order Systems


• y(n)− 2r cos θy(n− 1) + r2y(n− 2) = x(n)

where

0 < r < 1, 0 ≤ θ ≤ π

⇒ H(ejω) =1

1− 2r cos θe−jω + r2e−j2ω

H(ejω) =1

(1− (rejθ)e−jω)(1− (re−jθ)e−jω)

For θ 6= 0, π :︸︷︷︸(underdamped)

H(ejω) =A

1− rejθe−jω+

B

1− re−jθe−jω

A =1

1− re−jθe−jω|e−jω= 1

rejθ=

ejθ

2j sin θ

B =1

1− re+jθe−jω|e−jω= 1

re−jθ=−e−jθ

2j sin θ

F−1{H(ejω)} = h(n) = [A(rejθn) + B(re−jθ)

n]u(n)

⇒ h(n) = rnsin[(n + 1)θ]

sin θu[n], θ 6= 0, π

For θ = 0, π :︸︷︷︸(critically damped)(a) θ = 0 :


H(ejω) =1

(1− re−jω)2

h(n) = (n + 1)rnu(n)

(b) θ = π :

H(ejω) =1

(1 + re−jω)2

h(n) = (n + 1)(−r)nu(n)

• The rate of decay of h(n) is controlled by r; r ≈ 1,

slow decay, r ≈ 0, fast decay.

n

h(n)

=0

r=1/4

h(n)

=0

r=3/4

n


• The frequency of oscillation is controlled by θ; θ ≈ 0,

no oscillations, θ ≈ π, rapid oscillations

h(n)h(n)

h(n) h(n)

n

nn

n

=

r=3/4

= \2

r=3/4

=

r=1/4

= \2

r=1/4

Step Response:

(a) θ 6= 0, π (Underdamped)

S(n) = {A[1− (rejθ)

n+1

1− rejθ] + B[

1− (re−jθ)n+1

1− re−jθ]}u(n)

(b) θ = 0 (Critically Damped)

S(n) = [1

(r − 1)2 −r

(r − 1)2rn +

r

r − 1(n + 1)rn]u(n)


(c) θ = π (Critically Damped)

S(n) = [1

(r + 1)2 +r

(r + 1)2(−r)n +

+r

r + 1(n + 1)(−r)n]u(n)

• For θ 6= 0, the impulse response has a damped oscil-

latory behavior, and the step response exhibits ring-

ing and overshooting. However, it reaches final value

faster.

• For θ = 0, no dumped oscillatory behavior, and the

step response does not exhibit ringing and overshoot.

It takes longer to reach final value.


Frequency Response:

[Low-Pass]

-2 0 2

r=1/2

r=3/4

=0:|)(|log20 j

eH

[Low-Pass]

-2 0 2

r=1/2

r=3/4

= /4: |)(|log20 jeH


[Band-Pass]

-2 0 2

r=1/2 r=3/4

= /2: )(log20 jeH

θ = 3π4 → similar to θ = −π

4 shifted by π (High Pass)

θ = π → similar to θ = 0 shifted by π (High Pass)

• y(n)− (d1 + d2)y(n− 1) + d1d2y(n− 2) = x(n)

where |d1| < 1, |d2| < 1

⇒ H(ejω) =1

(1− d1e−jω)(1− d2e−jω)(Overdamped)

⇒ F−1{H(ejω)} = (Adn1 + Bdn2)u(n)

where A =d1

d1 − d2, B =

d2

d2 − d1

⇒ h(n) = (A(d1)n + B(d2)n)u(n)


⇒ s(n) = [A(1− dn+1

1

1− d1) + B(

1− dn+12

1− d2)]u(n)

Facts:

)(j

eH )(1

jeH )(

2

jeH

where H1(ejω) = 11−d1e−jω

, H2(ejω) = 11−d2e−jω

• If d1 ≈ 0, d2 ≈ 0, the response is fast; if d1 ≈ 1,

d2 ≈ 1, the settling time is long

• If d1 and d2 are negative the response is oscillatory

• If d1 > 0, d2 > 0, impulse response and step response

settle without oscillations.


5.5 Sampling of Continuous-Time Signals

Signal processing is done using digital computers, hence

we need to convert continuous-time signals to discrete-

time and vise-versa.

C/D

Conversion

DiscreteTime

System

Conv. From continous to discrete and

from discrete to continous

D /C

Conversion

)(tyc)(txc

sT sT

)()( sc nxnxd

)()( sc nynyd

5.5. SAMPLING OF CONTINUOUS-TIME SIGNALS 269

We note that in general, a signal cannot be uniquely

reconstructed from its samples.

1T

2T 3

T4

T

)(1

tx

)(3

tx

)(txtrne

5.5.1 Impulse Train Sampling: C/D Conv.

X)(txc

Signal to be sampled

)()()( tPtxtxp

Sampled Signal

0 ,)()( s

n

s TnTttP


⇒ xP (t) = xc(t).p(t) =∞∑

n=−∞xc(t)δ(t− nTs)

=∞∑

n=−∞xc(nTs)δ(t− nTs) (5.5.1)

δ(t− nTs) = 0, for t 6= nTs

…

t

tt 0 Ts0

……

)(txc

)(txP

)(tP

0 Ts 2Ts

)0(c

x

)(sc

Tx

)2(sc

Tx


• Frequency Domain Representation and Reconstruc-

tion

Suppose F{x(t)} = X(jω) exists.

Then,

Xc(jω) = F{xc(t)} = F{xc(t).p(t)} =1

2πXc(jω) ∗ P (jω)

Since p(t) is periodic then we have the following Fourier

series.

p(t) =∞∑

k=−∞αke

jkωst (F.S. Repr.)

αn =1

Ts

∫ Ts2

−Ts2p(t)e−jnωst,∀n

=1

Ts

∫ Ts2

−Ts2δ(t)e−jnωst =

1

Ts,∀n

⇒ p(t) =1

Ts

∞∑k=−∞

ejkωst (5.5.2)

(5.5.3)

Also, F{ejkωst} = 2πδ(ω − kωs)

⇒ P (jω) =2π

Ts

∞∑n=−∞

δ(ω − nωs), ωs =2π

Ts(5.5.4)

⇒ XP (jω) =1

Ts

∞∑n=−∞

Xc(jω) ∗ δ(ω − nωs)


=1

Ts

∞∑n=−∞

Xc(j(ω − nωs)) ∗ δ(ω)

⇒ XP (jω) =1

Ts

∞∑n=−∞

Xc(j(ω − nωs)) (5.5.5)

h xx y h y

e.g.

L.T.I. L.T.I.

Suppose that xc(t) is bandlimited, as shown below.

)( of spectrum )( txjX c

00 0

cut-off frequency


From (5.5.5) we have the following.

(i) For ωs > 2ω0 then

)( jXP )( of plicasRe jX c

0-s00

sT

1

s- s-0 0-s 0s

......

(ii) For ωs < 2ω0 then

sT

1

)( jXP

“aliasing effect”

)( of plicasRelonger No jX c


So, xc(t) can be reproduced from xP (t) if ωs > 2ω0,

by using an Ideal Filter as follows.

H(j )tx

c

tP

txP

t ,txtxcr

s s0

RemovedRemovedT

H(j )Passes

Low-Pass ideal filter


The Sampling Theorem

Let xc(t) be a bandlimited signal with Xc(jω) =

0,∀|ω| < ω0, e.g.,)( jX c

0 00

Then xc(t) is uniquely determined by its samples

xc(nTs), n = 0,±1,±2, ... if the sampling frequency

is at least twice ω0, e.g.,

ωs =2π

Ts> 2ω0 (5.5.6)

Here ωs is called Nyquist sampling rate. The recon-

struction is performed by an ideal low pass filter, with

cut-off frequency, ωc which satisfies

ω0 < ωc < ωs − ω0 (5.5.7)


5.5.2 Zero-order hold Sampling (Sample-dataSystems)

Impulse train Sampling requires the generation of im-

pulses, which are idealizations of pulses with short

duration and very large amplitude, which are very

difficult to built.

An alternative device is the zero-order hold.

)(txc

t

0 ts

T

)(0

th

)(tP

)(0 tX

Zero-order Hold

...

sT0

sT2

Sample and Hold

till the next

sampling instant

x0(t) = xP (t) ∗ h0(t)

=∞∑

n=−∞xc(nTs)δ(t− nTs) ∗ h0(t)


=∞∑

n=−∞xc(nTs)δ(t) ∗ h0(t− nTs)(5.5.8)

⇒ x0(t) =∞∑

n=−∞x(nTs)h0(t− nTs) (5.5.9)

)(0

tx

)(txc

t0s

Ts

T2 ...

• Frequency Domain Representation (Reconstruction)

Suppose xc(t) is bandlimited, e.g., Xc(jω) = 0,

∀|ω| > ωc. From x0(t) we need to reconstruct xc(t),

e.g., we need to construct a filter Hr(jω) such that


0 sT

)(0 th

t

)(0 tx)(tx

)(tP

holdorder zero

)(thr

)(jHr

)()( txtr c

00with

passlowideal)(

sc

jH

)( jH

c c0Let H0(jω)4= F{h0(t)}

Then H0(jω) = e−jωTs2 [2sin(ωTs/2)

ω ]

↑represents the time-shift of a rectangular pulse cen-

tered at t = 0

⇒ H0(jω).Hr(jω) = H(jω)

⇒ Hr(jω) =H(jω)

H0(jω)=

ejωTs/2

[2sin(ωTs/2)ω ]

H(jω)


Graph of Hr(jω) for ωc = ωs2

|)(| jH r)( jH r

.1

.

.

2

s

2

s0

cc

sT

2

2

2

s

2

s


Documents

Lecture Notes on Signals and Systems II ECE 320