Chapter 4

97

Chapter 4

Chapter 4: Introduction to z-Transform ......................................... 98

4.1 z-Transform ...................................................................... 99

4.2 Properties of z-Transform................................................ 102

4.2.1 Linearity ................................................................... 102

4.2.2 Shifting property (Delay theorem) ............................. 102

4.2.3 Time reversal ............................................................ 103

4.2.4 Multiplication by exponential sequence ..................... 103

4.2.5 Differentiation in the z-domain .................................. 103

4.2.6 Discrete convolution.................................................. 103

4.2.7 Example of discrete convolution ................................ 108

4.3 Inverse z-Transform ......................................................... 111

4.3.1 Relationship between the z-transform and Laplace

transform. ..............................................................................112

4.4 Frequency Response Estimation .......................................113

4.5 Pole-Zero Description of Discrete-Time Systems ..............115

4.6 A Second Order Resonant System (Complex Poles) ......... 121

4.7 Summary ........................................................................ 124

Chapter 4: Problem Sheet 4

Chapter 4

98

Chapter 4: Introduction to z-

Transform

j

S-Plane

The primary role of the Laplace transform in engineering is

transient and stability analysis of causal LTI system described

by differential equations.

The primary roles of the z-transform are the study of system

characteristics and derivation of computational structures for

implementing discrete-time systems on computers. The

transform is also used solve difference equations.

Analogue Domain

(Continuous-time Domain)

x(t)

X(s)

Laplace

Transform

s=j

X()

-analogue

frequency

j

S-Plane Stable

Region

Discrete-time Domain

x[n]

X(z)

z-transform

z=ej

X(ej

)

-digital frequency

s

aa

f

fT 2

-

t = nT

Im(z) z-Plane

=

= - Re(z)

stable

region

|z|=1(unit

circle)

Chapter 4

99

4.1 z-Transform

The z-transform of a discrete-time signal

n

nznxzX )(

In causal systems, x[n] may be zero when n < 0

0

)(n

nznxzX one-sided transform

Clearly, the z-transform is a power series with an infinite

number of terms and so may not converge for all values of z.

The region where the z-transform converges is known as the

region of convergence (ROC) and in this region the values of

X(z) are finite.

The step sequence:

00

01

n

nnx

...11)( 21

0

zzzzXn

n

This is a geometric series with a common ratio of z-1

.

The series converges if |z-1

| < 1 or equivalently if |z| > 1

z = rej

, = digital frequency

where z is a complex variable

two-sided transform

Chapter 4

100

11

1...1)(

1

21

z

z

zzzzX

In this case, the z-transform is valid everywhere outside a circle

of unit radius whose centre is at the origin (see below)

|z| > 1 then X(z) converges and |z| < 1 then it diverges

Let z = 2 221

1...221)(

1

21

zX

Let z = 0.5

42111

211

21zX

So the Region Of Convergence (ROC) is seen to be bounded by

the circle |z| = 1, the radius of the pole of

1)(

z

zzX

Im(z)

region of

convergence Re(z)

|z|=1 is a circle

of unit radius

referred to as

the ‘unit circle’

|z|=1

Chapter 4

101

The delta sequence: [n]

1}{0

n

nznnZ

The geometric sequence: x[n] = an

1

1

1)(

0

z

afor

az

z

z

azazX

n

nn

or equivalently, |z| > |a|.

when a = 1, x[n] = 1 for n 0 ie. x[n] = u[n]

1)(

z

zzX ROC: |z|>1

The complex exponential sequence: x[n] = ejn

1cos2

)sincos(

1)(

)(

))((

)(

1

1

2

2

0

zz

jzz

zeez

ezz

ezez

ezz

ez

ez

ez

z

ez

z

z

ezeez

jj

j

jj

j

j

j

jjjn

nnjnj

Chapter 4

102

1cos2

sin

1cos2

)cos(}sin{cos

1cos2

sin

1cos2

)cos(}{

22

22

zz

zj

zz

zznjnZ

zz

zj

zz

zzeZ nj

Exploiting the linearity property

1cos2

sin}{sin

1cos2

)cos(}{cos

2

2

zz

znZ

zz

zznZ

4.2 Properties of z-Transform

4.2.1 Linearity

)()( zbYzaXnbynaxZ

4.2.2 Shifting property (Delay theorem)

)(zXzknx kZ

A very important property of the z-transform is the delay

theorem.

eg:

zXznxZ

zXznxZ

2

1

2

1

z-1

Unit delay

x[n]

X(z)

x[n-1]

z-1

X(z) T

x[n] x[n-1]

One sample delay

Chapter 4

103

4.2.3 Time reversal

11

zXor

zXnx

z

Example: Find the z-transform of x[n] = -u[n].

zz

zznu

z

zznu

1

1

1;

1 1

1

ROC: |z| < 1

4.2.4 Multiplication by exponential sequence

)( 1 azXnxa zn

As a special case if x[n] is multiplied by ejn

)( zeXz

nxe jnj

4.2.5 Differentiation in the z-domain

)(zXdz

dz

znnx

4.2.6 Discrete convolution

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

then Y(z) = X(z) . H(z)

Chapter 4

104

)()(

)()(

}*{

0 0

0

0

zHzX

zHzX

zmhzkx

zmhmuzkxku

zmhmukxku

knmLetzknhknukxku

zknhknukxku

knhknukxkuZnhnxZ

k m

mk

km

m

k

k

km

km

k

n

n

k

n

n

k

k

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

Y(z) = X(z) H(z)

h[n]

H(z)

x[n]

X(z)

y[n]

Y(z)

Chapter 4

105

Example: Concept of the transfer function

]2[]1[]2[]1[][][ 21210 nybnybnxanxanxany

Take the z-transform of both sides:

2

2

1

1

2

2

1

10

2

2

1

10

2

2

1

1

2

2

1

1

2

2

1

10

1

1

zbzb

zazaa

zX

zYzH

zazaazXzbzbzY

zzYbzzYbzzXazzXazXazY

Example:

Find the difference-equation of the following transfer function

23

25)(

2

zz

zzH

First rewrite H(z) as a ratio of polynomials in z-1

2121

21

21

2523

231

25)(

)(

)(

zzXzzXzzYzzYzY

zz

zzzH

zX

zY

Take inverse z-transform

22152213 nxnxnynyny

Chapter 4

106

Example:

If x[n] = u[n] – u[n-10], find X(z)?

1

109

0 1

1)1()(

z

zzzX

n

n

Example:

If y[n] = -nu[-n-1], find Y(z)?

k

k

k

k

n

n

n

nn

z

z

z

znuzY

0

1

1

1

1)(

22132215 nynynxnxny

Chapter 4

107

The sum converges provided 1

z ie. |z| < ||

||||

||||1

11)(

1

zz

z

zz

zY

Depict the ROC and pole and zero locations in the z-plane

-1

u[n+1]

n -1

u[-(n+1)]

n

Im(z)

Re(z)

Chapter 4

108

4.2.7 Example of discrete convolution

Compute the convolution y[n] of the digital signals given by

x1[n] = [1, -2, 1];

x2[n] = 1 for 0 n 5,

x2[n] = 0 elsewhere

y[n] = x1[n] * x2[n] Y(z) = X1(z) X2(z)

X1(z) = 1 –2z-1

+ z-2

X2(z) = 1 + z-1

+ z-1

+ z-3

+ z-4

+z-5

Y(z) = X1(z) X2(z)

= 1-z-1

– z-6

+ z-7

Inverse z-transform

y[n] = [1, -1, 0,0,0,0,-1,1]

Chapter 4

109

Example: Determine the system function H(z) of the system

shown below:

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

Y(z) = X(z) + az-1

Y(z)

11

1

)(

)()(

azzX

zYzH

+

T

x[n] y[n]

a ay[n-1]

+

z-1

X(z) Y(z)

a az-1

Y(z)

Chapter 4

110

Basic z-Transforms

Signal Transform ROC

[n] 1 all z

U[n] 11

1 z

|z| > 1

n u[n]

11

1 z

|z| > ||

nn u[n]

21

1

)1(

zα

zα

|z| > ||

cos(n) u[n] 21

1

cos21

cos1

zz

z

|z|>1

sin(n) u[n] 21

1

cos21

sin

zz

z

|z| > 1

rncos(n) u[n]

221

1

cos21

cos1

zrrz

rz

|z| > r

rnsin(n) u[n]

221

1

cos21

sin1

zrrz

rz

|z|>r

z-Transform Properties

Signal Transform

x[n] X(z)

ax[n] + by[n] aX(z) + bY(z)

x[n-k] z-k

X(z)

anx[n]

a

zX

x[-n]

zX

1

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

nx[n] )(zX

dz

dz

Chapter 4

111

Note: 1

)1(21

1

1...1

z

zzzzS

NN

N

If N 11

1

zS , |z

-1| < 1

4.3 Inverse z-Transform

Partial fraction method

The inverse z-transform allows us to recover the discrete-time

sequence.

x[n] = Z-1

[X(z)]

where X(z) is the z-transform of x[n].

Example: Find x[n] for the following:

5.0

5

4

75.0

5

4

5.0

5

4

75.0

5

4

5.075.0)5.0)(75.0(375.025.0

375.025.01)(

2

21

1

z

z

z

z

zzz

z

B

z

Az

zz

z

zz

z

zz

zzX

0,)5.0()75.0(5

4

5.0

1

5

4

75.0

1

5

4][

n

z

z

z

znx

nn

ZZ

Power series method

Residue method - evaluating the contour integral

Chapter 4

112

4.3.1 Relationship between the z-transform and Laplace

transform.

If we let z = esT

, then z = e( +j)T

(T is the sampling period)

z = e T ejT = e T ej -

Thus |z| = e T

and sf

fTz 2 ( = digital

frequency)

As varies from - to the s-plane is mapped to the z-plane as

shown in Figure 4.1.

The entire j axis in the s-plane is mapped onto the unit circle.

The left-hand s-plane is mapped to the inside of the unit circle

and the right-hand s-plane maps to the outside of the unit circle.

j

s-plane

2

z-plane

= 0

If =

2

sff

Unit circle

|z|=1

=

= -

(Half the

sampling

frequency)

Figure 4.1: Mapping of the s-plane to the z-plane.

Chapter 4

113

In terms of frequency response, the j axis is the most important

in the s-plane. In this case = 0 and the frequency points in the

s-plane are related to points on the z-plane unit circle by z =

eT

.ejT

= 1 ej

4.4 Frequency Response Estimation

There are many instances when it is necessary to evaluate the

frequency response of discrete-time systems. The frequency

response of a system can be readily obtained from its z-

transforms.

For example, if we set z = ej

, that is evaluate the z-transform

around the unit circle, we obtain the Fourier Transform of the

system.

Example:

11

1)(

azzH , 0 < a < 1, say a = 0.6

Find H(). {H()-the frequency response}

jez

jezzHH

|)()( , -

Fourier transform of the discrete-time system

jezzHH

|)(

Fourier transform of the discrete-time system.

Chapter 4

114

Example: Find H() if 11 zzH

cos22sincos1

1

22

H

eH j

-

sincos1

1

1

1|

jaaaezHH

jez j

22sincos1

1)(

aaH

2cos21

1

aa

= T; = 2 f/fs:; = f = fs/2

fs= sampling frequency

|H()|

a1

1

a1

1

-

-fs/2 fs/2 0 f (analogue frequency)

digital frequency

Chapter 4

115

4.5 Pole-Zero Description of Discrete-Time Systems

The zeros of a z-transform H(z) are the values of z for which

H(z)=0. The poles of a z-transform are the values of z for which

H(z)=∞ . If H(z) is a rational function , then

The complex quantities (or may be real) z1, z2, z3 …. are called

zeros of H(z) and the complex quantities (or may be real) p1, p2,

p3 … are called the poles of H(z). We thus see that H(z) is

completely determined , except for the constant a0, by the values

of poles and zeros.

The information contained in the z-transform can be

conveniently displayed as a poles-zero diagram (see figure

below)

In the diagram, ‘X’ marks the position of a pole and ‘O’ denotes

the position of a zero.

L

L

M

M

zbzb

zazaa

zX

zYzH

......1

...

)(

)()(

1

1

1

10

))......()((

)).......()((

21

210

L

M

pzpzpz

zzzzzza

0.75

|z|=1 0.5

-0.5

Im(z)

Re(z) -1

Chapter 4

116

The poles are located at z = 0.5 ± 0.5j and z = 0.75; a single zero

is at z = -1.

An important feature of the pole-zero diagram is the unit circle

|z|=1. The pole-zero diagram provides an insight into the

properties of a given discrete-time system.

From the locations of the poles and zeros we can infer the

frequency response of the system as well as its degree of

stability.

For a stable system, all the poles must lie inside the unit circle.

Zeros may lie inside, on, or outside the unit circle.

Example: Determine the transfer function H(z) of a discrete-

time system with the pole-zero diagram shown below:

21

2

5.01

)1(

)5.05.0()5.05.0(

)1()1()(

zz

zK

jzjz

jzjzKzH

0.5

|z| =1 0.5

-0.5

Im(z)

Re(z) -1

Chapter 4

117

Example: Determine the pole-zero plot: az

zzH

)(

Example: Determine the pole-zero plot:

1

21

1

)().........)((

)()(

MM

MM

z

Mzzzzz

azz

azzH

The zero z = a cancels the pole at z = a. Thus, H(z) has M-1

zeros and M-1 poles as shown in the diagram below for M = 8.

|z|=a

Im(z)

Re(z)

|z|=a

Im(z)

Re(z)

z= a

8 poles

Chapter 4

118

Consider a system, H(z) with two complex conjugate poles in

the z-plane :

jrep 1

jrep 2

)(01 zeroz

A typical transfer function might be:

jj

jjjj

rez

jr

rez

jrz

rez

B

rez

Az

rezrez

zzH

sin2

1

sin2

1

)(

11 1

1

1

1

)sin2(

1)(

zrezrerjzH

jj

Poles

|z|=1

Im(z)

Re(z)

r p1

p2

Chapter 4

119

njnjn

njnj

eerj

r

rererj

nh

sin2

sin2

1)(

This is the impulse response of the 2nd

order system with

complex poles.

We note that the impulse response will decay away to zero

provided r is less than one. [ r < 1]

Recall that r is also the distance from the origin in the z-plane to

the poles p1 or p2, so that system will be stable if the poles in the

z-plane lie inside the unit circle.

= frequency of

oscillation

nrnh n sin

sin

1)( 1

Chapter 4

120

Example:

θnrθ

nh n sinsin

1)( 1

Exponential decay

sinewave (r<1)

Stable system

rn

Poles inside

unit circle

-

Poles on the

unit circle

r = 1

-1

1

Poles outside

unit circle

-

Exponential increasing

(r>1)

rn

θnrθ

nh n sinsin

1)( 1

One real Pole inside

the unit circle rn, = 0

n

n

n

Chapter 4

121

Note:

A system that is both stable and causal must have all its

poles inside that unit circle within the z-plane.

We cannot have a pole outside the unit circle, since the

inverse transform of a pole located outside the circle will

contribute either a right sided increasing exponential term,

which is not stable, or a left-sided decaying exponential

term that is not causal.

4.6 A Second Order Resonant System (Complex

Poles)

21

2

2

2

2

1

11

1)(

bzbz

z

zbzbzH

(A)

z -1

z -1

+ y[n]

x[n]

-b1

-b2

0 r

002

001

sincos

sincos

0

0

jrrrep

jrrrep

j

j

All pole system has poles only (without counting the zeros at the origin)

Chapter 4

122

(B) cos2)(

)(

))(())(()(

20

2

2

22

2

2

21

2

21

12

2

00

00

rzrz

z

rzeerz

zzH

rezrez

z

pzpz

z

bzbz

zzH

jj

jj

Comparing (A) and (B), we obtain

We can derive H() and the magnitude from 2

2

1

11

1)(

zbzbzH

b1 b2 = r2

-0.94 0.5

-1.16 0.7

-1.34 0.9

-1.41 0.99

-20

-10

0

10

20

30

40

50

dB

01 cosr2b 2

2 rb 2

1

0b2

bCos

sf

f00

2

0 = resonant frequency

I II

III

IV

2

11

0

01

2

2

2cos

)cos(2

b

b

rb

rb

IV

I

40

4

dB

Chapter 4

123

Example: Sketch the magnitude response for the system

having the transfer function

)9.01()9.01(

1)(

1414

1

zeze

zzH

jj

The system has a zero at z = -1 & poles at 49.0

j

ez

Magnitude response will be zero at = and large at

0=± /4 because the poles are close to the unit circle.

Example: Sketch the approximate magnitude response from the

pole-zero map given below:

|z|= 1

+ 0.8

- 0.8 Re(z)

Im(z)

dB

- -/2 0 /2

-fs/2 fs/2

- -/4 /4

|H()|

Magnitude

Response

/4 0.9

=0

= -

=

Chapter 4

124

Example: Sketch an approximate magnitude response from the

pole-zero map given below:

4.7 Summary

At the end of this chapter, it is expected that you should know:

The properties of z-transforms and their application.

Discrete convolution in the time and z domains.

How to find the inverse z-transform, given a transfer

function.

The difference between a z-transform and a Laplace

transform and when to use each.

Estimation of frequency response from a transfer function

Hand-calculate magnitude and phase responses for simple

transfer functions and plot them.

The pole-zero description of a discrete time system.

Given a pole-zero diagram, transfer function or difference

equation, how to find any of the other representations and

discuss the system’s stability with reference to the pole-zero

diagram.

How to derive the resonance frequency equation for a

second-order resonance system (a complex pole pair).

- -/2 0 /2

|H()| in dB

+ 0.5

- 0.5 Re(z)

1

|z|=1

Im(z)