1 1

Chapter 3

The z-Transform

2 2

Consider a sequence x[n] = u[n]. Its Fourier transform

does not converge.

Consider that, instead of ej, we use rej in the transform, where r ≥ 0 is a real number. Then we have

This transform will converge to , provided r>1.

0

][)(n

nj

n

njj eenxeX

z-Transform (1)

0

)]([)(n

njn

n

njj errenxreX

1)(1

1 jre

3 3

In general, we have

Let z=rej. Then, the z-transform of a sequence x[n] is defined as

with z being a complex variable.

z-Transform (2)

n

nznxzX ][)(

)(][ zXnxz

n

njn

n

njj ernxrenxreX ][)]([)(

4 4

Because z is a complex number, we often use the z-plane.

When |z|=1, that is, z takes value from the unit circle, the

z-transform reduces to the Fourier transform.

z-Transform (3)

Unit circle

Im

Re

1

z-planejez

5 5

Properties of the system can be easily studied and characterized in the z-domain (stability, causality, …).

The solution process reduces to a simple algebraic procedure.

In the temporal domain, the output sequence is y[n]=x[n]*h[n] (convolution), whereas in the z-domain it becomes Y(z)=X(z)H(z) (multiplication).

Why z-Transform

6 6

Absolute summability of z-transform

It is possible for the z-transform to converge even if the Fourier transform does not.

Convergence depends only on |z|. The region of convergence (ROC) consists of all values of z such that the last inequality holds.

If the ROC includes the unit circle, the Fourier transform of the sequence converges.

Region of Convergence

n

n

n

n znxznx ][][

7 7

Consider x[n]=anu[n]. Because it is nonzero only for n ≥ 0, this is an example of a right-sided sequence.

For convergence of X(z), we require

Thus, the ROC is the range of values of z for which |az-1|<1, or equivalently, |z|>|a|. Inside the ROC, the infinite series converges to

Example – Right-Sided Exponential Sequence (1)

0

1)(][)(n

n

n

nn azznuazX

n

naz 1

az

z

azzX

11

1)( |||| az

8 8

The infinite sum becomes a simple rational function of z inside the ROC.

Such a z-transform is determined to within a constant multiplier by its zeros and its poles.

For this example, one zero: z=0 (plotted as o);one pole: z=a (plotted as x).

When |a|<1, the ROC includes the unit circle.

Unit circle

Im

Re1x

z-planeROC

a

Example – Right-Sided Exponential Sequence (2)

9 9

Consider x[n]= –anu[–n –1]. Because it is nonzero only for n ≤ –1, this is an example of a left-sided sequence.

ROC

and

Example – Left-Sided Exponential Sequence

0

1

1

1

)(1

]1[)(

n

n

n

nn

n

nn

n

nn

zaza

zaznuazX

11 za

az

z

azzazX

11 1

1

1

11)(

az

Unit circle

Im

Re1x

z-plane

ROC

a

az

10 10

As can be seen from the two examples, the algebraic expression or pole-zero pattern does not completely specify the z-transform of a sequence; i.e., the ROC must also be specified.

Notes on ROC

az

z

azzX

11

1)( |||| az

az

z

azzX

11

1)( |||| az

][][ nuanx n

]1[][ nuanx n

11 11

Consider the sequence

ROC

and

Example – Two-Sided Exponential Sequence

))((

)(21

1

1

1)(

2

1

3

112

1

12

113

1

zz

zzzz

zX

2

1

3

1 z

]1[)(][)(][2

1

3

1 nununx nn

13

11

1 z

3

1z

12

11

1 z

2

1zIm

Re

1/2x

z-plane

ROC

x–1/3 1/12

2

1

3

1 z

12 12

Consider the sequence

Then

The ROC is determined by

which requires

Example – Finite-Length Sequence

az

az

zaz

az

azzazX

NN

N

N

N

n

nN

n

nn

11

1

1

0

11

0

1

1

)(1

)()(

otherwise,0

10,][

Nnanx

n

1

0

1 ||N

n

naz

0 and || za

Unit circle

Im

Re1

x

z-plane11th order pole

a

ROC

In this example N = 12 and 0< a <1.

x

13 13

Some Common z-Transform Pairs (1)

|a||z|az

aznuna

|a||z|az

aznuna

|a| |z|az

nua

|a| |z|az

nua

m

mzzmn

zz

nu

zz

nu

zn

z

n

n

n

n

m

1

]1[

1

][

1

1]1[

1

1][

0) (if or

0) (if 0except all ][

1|| 1

1]1[

1|| 1

1][

all 1][

ROC Transform Sequence

21

1

21

1

1

1

1

1

14 14

Some Common z-Transform Pairs (2)

0 1

1

otherwise,0

10,

|| ]cos2[1

]sin[][]sin[

|| ]cos2[1

]cos[1][]cos[

1|| ]cos2[1

][sin][][sin

1|| ]cos2[1

][cos1][][cos

ROC Transform Sequence

1

2210

10

0

2210

10

0

210

10

0

210

10

0

|z|az

zaNna

rzzrzr

zrnunr

rzzrzr

zrnunr

zzz

znun

zzz

znun

z

NNn

n

n

15 15

Property 1: The ROC is a ring or disk in the z-plane centered at the origin; i.e., 0 ≤rR<|z|<rL ≤ ∞.

Property 2: The Fourier transform of x[n] converges absolutely if and only if the ROC of the z-transform of x[n] includes the unit circle.

Property 3: The ROC cannot contain any pole.

Property 4: If x[n] is a finite-duration sequence, i.e., a sequence that is zero except in a finite interval –∞ < N1 ≤ n

≤ N2 < ∞, then the ROC is the entire z-plane, except

possibly z = 0 or z = ∞.

Properties of the ROC (1)

16 16

Property 5: If x[n] is a right-sided sequence, i.e., a sequence that is zero for n < N1 < ∞, the ROC extends

outward from the outmost (i.e., largest magnitude) finite pole in X(z) to (and possibly include) z = ∞.

Property 6: If x[n] is a left-sided sequence, i.e., a sequence that is zero for n > N2 > –∞, the ROC extends

inward from the innermost (smallest magnitude) nonzero pole in X(z) to (and possibly include) z = 0.

Properties of the ROC (2)

17 17

Property 7: A two-sided sequence is an infinite-duration sequence that is neither right sided nor left sided. If x[n] is a two-sided sequence, the ROC will consist of a ring in the z-plane, bounded on the interior and exterior by a pole and, consistent with property 3, not containing any poles.

Property 8: The ROC must be a connected region.

Properties of the ROC (3)

18 18

z-Transform with Different ROC (1)

Im

Rex

z-plane

Unit circlexx

For a system whose poles are shown in the figure, consider the stability and causality of the system.

19 19

z-Transform with Different ROC (2)

Im

Rex

z-planeROC

xx

Right-sided sequence

Im

Rex

z-plane

ROCxx

Left-sided sequence

20 20

z-Transform with Different ROC (3)

Im

Rex

z-plane

ROC

xx

Two-sided sequence

Im

Rex

z-plane

ROCxx

Two-sided sequence

21 21

Stability, Causality, and the ROC (1)

Im

Rex

z-planeROC

xx

Unit circle

Im

Rex

z-plane

xx

If the system is stable

h[n] is absolutely summable, and therefore has Fourier transform, the ROC must include the unit circle.

h[n] is two-sided, and therefore, the system is not causal.

22 22

Stability, Causality, and the ROC (2)

h[n] is right-sided.

h[n] is not stable.

Im

Rex

z-planeROC

xx

Unit circle

Im

Rex

z-plane

xx

If the system is causal

23 23

Im

Rex

z-plane

Unit circle

xx

Stability, Causality, and the ROC (3)

From the above discussion…

There is no ROC that would imply that the system is both stable and causal.

24 24

25 25

Proof of the Initial-Value Theorem

Since x[n]=0 for n<0, we have

]0[][lim]0[

][lim)(lim

1

0

xznxx

znxzX

n

n

z

n

n

zz

If x[n]=0 for n>0, then we have

]0[][lim]0[

][lim)(lim

10

0

00

xznxx

znxzX

n

n

z

n

n

zz

26 26

Inspection Method – use your familiar transform pairs

The Inverse z-Transform - Inspection Method

We know that

then we would recognize

||||,1

1][

1az

aznuan

][)(][2

1 nunx n

2

1

12

1 ||,1

1)(

z

zzX

Example: If we need to find the inverse z-transform of

27 27

When X(z) is expressed as as a ratio of polynomials in z-1; i.e.,

Such z-transforms arise frequently in the study of LTI systems.

Partial Fraction Expansion – obtain an alternative expression for X(z) as a sum of simple terms

The Inverse z-Transform -Partial Fraction Expansion (1)

N

k

kk

M

k

kk

za

zbzX

0

0)(

28 28

Zeros and poles for

There are M zeros and N poles at nonzero positions. In addition, if M > N, there are M – N poles at z = 0, or if N > M, there are N – M zeros at z = 0.

For the above expression, the z-transforms always have the same number of zeros and poles in the finite z-plane, and there are no poles or zeros at z = ∞.

N

k

kNk

M

M

k

kMk

N

N

k

kk

M

k

kk

zaz

zbz

za

zbzX

0

0

0

0)(

The Inverse z-Transform -Partial Fraction Expansion (2)

29 29

Note that X(z) can be written in the form

where ck’s are nonzero zeros of X(z), and dk’s are nonzero poles of X(z). (case 1) If M < N and the poles are all first order, we have

Multiplying both sides by (1 – dkz –1) and evaluating at z = dk,

N

kk

M

kk

zd

zc

a

bzX

1

1

1

1

0

0

)1(

)1()(

kdzkk zXzdA )()1( 1

N

k k

k

zd

AzX

111

)(

The Inverse z-Transform -Partial Fraction Expansion (3)

30 30

Example:

Rewrite X(z) as

where

2

1||,

11

1)(

12

114

1

zzz

zX

1)()1(4

11

4

11

zzXzA

12

12

14

11

11)(

z

A

z

AzX

2)()1(2

11

2

12

zzXzA

The Inverse z-Transform -Partial Fraction Expansion (4)

31 31

Therefore

and

Zeros and poles:

Two zeros at z = 0, and first order poles at z = 1/4 and 1/2.

12

114

1 1

2

1

1)(

zz

zX

][)(][)(2][4

1

2

1 nununx nn

2

1

4

1

2)(

z

z

z

zzX

The Inverse z-Transform -Partial Fraction Expansion (5)

32 32

(case 2) If M ≥ N and the poles are all first order, we have

The Br’s can be obtained by long division of the numerator by the denominator, with the division processing terminating when the remainder is of lower degree than the denominator.

N

kk

M

kk

zd

zc

a

bzX

1

1

1

1

0

0

)1(

)1()(

N

k k

kNM

r

rr zd

AzBzX

11

0 1)(

The Inverse z-Transform -Partial Fraction Expansion (6)

33 33

Example:

Since M=N=2, X(z) can be represented as

The constant B0 can be found by long division

1||,

11

1

1

21)(

112

1

21

22

112

3

21

zzz

z

zz

zzzX

12

12

11

0 11)(

z

A

z

ABzX

2

15

23

121

1

12

1212

322

1

z

zz

zzzz

The Inverse z-Transform -Partial Fraction Expansion (7)

34 34

(case 3) If X(z) has multiple-order poles. If X(z) has a pole of order s at z=di and all other poles are first-order, then (there are no Br terms if M < N)

with

N

kk

M

kk

zd

zc

a

bzX

1

1

1

1

0

0

)1(

)1()(

s

mm

i

mN

ikk k

kNM

r

rr zd

C

zd

AzBzX

11

,11

0 )1(1)(

1

)]()1[()()!(

1 1

idw

sims

ms

msi

m wXwddw

d

dmsC

The Inverse z-Transform -Partial Fraction Expansion (8)

35 35

The defining expression for the z-transform is a Laurent series where the sequence values x[n] are the coefficients of z-n. Thus, if the z-transform is given as a power series in the form

we can determine any particular value of the sequence by finding the coefficient of the appropriate power of z-1.

- We have already used this approach in finding the inverse transform of the polynomial part of the partial fraction expansion when M ≥ N.

- This method is also very useful for finite-length sequences where X(z) may have no simpler form than a polynomial in z-1.

212 ]2[]1[]0[]1[]2[

][)(

zxzxxzxzx

znxzXn

n

The Inverse z-Transform -Power Series Expansion (1)

36 36

Example 1 (Finite-length sequence)

Express X(z) as

we have

)1)(1)(1()( 1112

12 zzzzzX

The Inverse z-Transform -Power Series Expansion (2)

12

1

2

12 1)( zzzzX

]1[][]1[]2[][2

1

2

1 nnnnnx

37 37

Example 2 (Inverse transform by power series expansion)

Using the power series expansion for log(1+x), with |x|<1, we obtain

Therefore

||||),1log()( 1 azazzX

The Inverse z-Transform -Power Series Expansion (3)

1

1)1()(

n

nnn

n

zazX

0,0

1,)1(][1

n

nn

anx

nn

38 38

Obtain the inverse z-transform of

by a different method.

1

2

1

)(

az

az

dz

zdX

||||),1log()( 1 azazzX

||||,1

)(][

1

1

azaz

az

dz

zdXznnx

]1[)(][ 1 nuaannx n

11

1][)(

az

nua n

]1[)1(]1[)(][ 1111 nuanuaanx nn

nnn

Since

then

We know

therefore

39 39

Example 3 (Power series expansion by long division)

Carrying out long division

Therefore, x[n]= anu[n].

||||,1

1)(

1az

azzX

The Inverse z-Transform -Power Series Expansion (4)

221

22

221

1

1

11

1

11zaaz

za

zaaz

az

az

az

40 40

Example 3 (Power series expansion for a left-sided sequence)

Because X(z) at z=0 is infinite, the sequence is zero for n>0. Thus, we divide, so as to obtain a series in powers of z as follows

Therefore, x[n]= –anu[–n–1].

||||,1

1)(

1az

azzX

The Inverse z-Transform -Power Series Expansion (5)

221

2

2

zaza

az

azz

zza

41 41

Homework (3)

3.73.83.173.27