Chapter 5Chapter 5

ZZ Transform Transform

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Z transform– Representation, analysis, and design of discrete signal

– Similar to Laplace transform

– Conversion of digital signal into frequency domain

1. Introduction1. Introduction

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z transform– Two-sided z transform

– One-sided z transform• If n < 0, x(n) = 0

2. 2. zz transform transform

0

( ) ( ) n

n

X z x n z

( )

( )

n

n

n

X z x

x n z

(5-1)

(5-2)

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Example 5-1(1) Non-causal

(2) Non-causal

5 4 3 2( ) ( ) 2 2 2n

n

X z x n z z z z z z

Fig. 5-1.

2 1 2( ) ( ) 2 2 2n

n

X z x n z z z z z

Fig. 5-1.

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Example 5-1(3) Causal

(4)

• Geometric series with common ratio of

Fig. 5-1.

Fig. 5-1.

1 2 3 4 5( ) ( ) 2 2 2n

n

X z x n z z z z z z

1, 0( )

0, 0

nx n

n

1 2

0

( ) ( ) 1n n

n n

X z x n z z z z

1z

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• Convergence of series

• Region of convergence

1 2

1

( ) 1

1 (1 ) ( 1)

X z z z

z z z

Fig. 5-2.

Region of convergence

(5-3)

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• Z=2

• Z=1/2

2 3( ) 1 1 2 (1 2) (1 2)

2 (2 1) 2

X z

2 3( ) 1 1 0.5 (1 0.5) (1 0.5)

1 2 4 8

X z

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Example 5-2

– z transform

0 , 0( )

, 0, 0n

nx n

b n b

1

0

( ) ( )

1 ( ) ,

n n

n

n

n

X z b z

b zz b

z z b

Region of convergence

Fig. 5-3.

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Characteristic of z transform

(1) is a polynomial equation of z and determined from samples, (2) can be reconstructed by removing in

(3) is independent to sampling interval,

(4) z transform of delayed signal by samples is

• z transform of delayed signal

• Expression of difference equation

( )X z ( )x n

( )x n nz ( )X z

( )X z T

m ( )mz X z

( ) ( )

( ) ( )m

x n X z

x n m z X z

{ ( )} ( ) n

n

z x n m x n m z

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•

(5) Same to discrete Fourier transform by replacing to

( )( ) ( )

( )

( )

n k m

n k

m k

k

m

x n m z x k z

z x k z

z X z

n m k

z j Te

( ) ( )j TsX j X e (5-

4)

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Table of z transform

Table 5-1.

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Example 5-3

(1)

(2)

(3)

( ) ( ), ( ) 1x n n X z

( ) ( )x n u n

0

1 2

11

( )

1

1, 1

1

n

n

X z z

z z

zz

( ) , 0anTx n e n

0

1 2 2

11

( )

1

1, 1

1

anT n

n

aT aT

aTaT

X z e z

e z e z

e ze z

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Ideally sampled function,

– Laplace transform

3. Relation between Z transform 3. Relation between Z transform and Laplace transformand Laplace transform

( )sx t

0

( ) ( ) ( )sn

x t x t t nT

0

00

00

0

( ) ( )

( ) ( )

( ) ( )

( )

sts s

st

n

st

n

nTs

n

X s x t e dt

x t t nT e dt

x t e t nT dt

x n e

(5-5)

(5-6)

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– z transform

– Relation

( ) ( ) sTe zX z X s

( ) ( ) n

n

X z x n z

(5-7)

(5-8)

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Example 5-4 (1)

(2)

( ) ( )x t u t1

( )X ss

( ) ( )x n u n0

11

( )

1, 1

1

n

n

X z z

zz

( ) ( )atx t e u t 1( )X s

s a

( ) ( )anTx n e u n0

11

( )

1, 1

1

anT n

n

aTaT

X z e z

e ze z

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(3) ( ) (sin ) ( )x t t u t2 2

( )X ss

( ) (sin ) ( )x n nT u n0

0

1

2 1

( ) (sin )

2

sin

2 cos 1

n

n

j nT j nTn

n

X z nT z

e ez

j

z T

z z T

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– Periodicity

– s-plane and z-plane

( )ss j m TsTe e

Fig. 5-4.

(5-9)

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Corresponding points(1) Left side plane in s plane inside of unit circle in z plane

(2) Right side plane in s plane out of unit circle

(3) axis in s plane unit circle in z plane

(4) Increased frequency in s plane mapped on the unit circle in z plane

j

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– Corresponding points

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Definition of inverse z transform

– Power series of

4. Inverse Z transform4. Inverse Z transform

1( ) ( )x n z X z

0

1 2 3

( ) ( )

(0) (1) (2) (3)

n

n

X z x n z

x x z x z x z

( )X z

1 20 1 2

1 20 1 2

( )N

NM

M

b b z b z b zX z

a a z a z a z

(5-10)

(5-11)

(5-12)

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– Three methods to obtain inverse z transform• Power series expansion

• Partial fraction expansion

• Residue

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– Power series expansion• Long division

1 20 1 2

1 20 1 2

1 2 3

( )

(0) (1) (2) (3)

NN

MM

b b z b z b zX z

a a z a z a z

x x z x z x z

(5-13)

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Example 5-5– Inverse z transform using long division

1 2

1 2

1 2( )

1 2

z zX z

z z

1 2 3

1 2 1 2

1 2

1 2

1 2 3

2 3

2 3 4

3 4

1 3 2 41 2 1 2

1 2

3

3 3 6

2 6

2 2 4

4 4

z z zz z z z

z z

z z

z z z

z z

z z z

z z

1 2 3( ) 1 3 2 4X z z z z

(0) 1, (1) 3, (2) 2, (3) 4,x x x x

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– Partial fraction expansion

1 20 1 2

1 20 1 2

( )N

NM

M

b b z b z b zX z

a a z a z a z

1 20 1 1 1

1 2

1 20

1 2

01

0 11

( )1 1 1

1

M

M

M

M

Mk

k k

Mk

k k

C C CX z B

p z p z p z

C z C z C zB

z p z p z p

C zB

z p

CB

p z

where is poles of ,

is coefficients for partial fraction, and

kp ( )X z

kC

0 / .N NB b a

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• Partial fraction for N>M

10 1

( )1

N M Mkr

rr k k

CX z B z

p z

where is calculated using long division.rB

1

( )( )

(1 ) ( )

k

k

k kz p

k z p

X zC z p

z

p z X z

1

mi

i k

D

z p

1 ( )( )

( )!k

m im

i km iz p

d X zD z p

m i dz z

(5-17)

(5-18)

(5-19)

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Example 5-6– Inverse z transform

1

1 2( )

1 0.25 0.375

zX z

z z

2( )

0.25 0.375 ( 0.75)( 0.5)

z zX z

z z z z

1 2( )( 0.75)( 0.5) 0.75 0.5

C z C zzX z

z z z z

1 2( )

( 0.75)( 0.5) 0.75 0.5

C CX z z

z z z z z z

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10.75 0.75

0.75

( 0.75) ( ) ( 0.75)

( 0.75)( 0.5)

10.8

0.5

z z

z

z X z zC

z z z

z

20.5 0.5

( 0.5) ( ) ( 0.5)

( 0.75)( 0.5)

0.8z z

z X z zC

z z z

0.8 0.8( )

0.75 0.5

z zX z

z z

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• Inverse z transform using table 5-1

1 0.80.8(0.75)

0.75nz

zz

1 0.80.8( 0.5)

0.5nz

zz

( ) 0.8 (0.75) ( 0.5) , 0n nx n n

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Example 5-7– Inverse z transform

• poles

1 2

1 2

1 2( )

1 0.5

z zX z

z z

2

2

2

1 2

( ) 2 1( )

( ) 0.5

2 1

( )( )

N z z zX z

D z z z

z z

z p z p

1 2

1

1 (1 4 0.5)0.5 0.5

2p j

*2 1 0.5 0.5p p j

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• Partial fraction

1

0 11 2 11

2

( )( ) ( ) ( )

z p

B z pz p X z C z pC

z z z p

1

21 1

11 2

( ) ( ) ( )( 2 1)

( )( )

2 1.5

0.5 0.5

0.5 3.5

z p

z p X z z p z zC

z z z p z p

j

j

j

*2 1 0.5 3.5C C j

0 1 2

1 1

( ) B C CX z

z z z p z p

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• z transform

• Inverse z transform

1 2

1 1

( ) 2

( 0.5 3.5) ( 0.5 3.5)2

0.5 0.5 0.5 0.5

C z C zX z

z p z p

j z j z

z j z j

1 1 2

1 1

2 3.5355(0.7071) cos(45 98.13 )

7.071(0.7071) cos(45 98.13 )

n

n

C z C zz n

z p z p

n

1(2) 2 ( )z n

( ) 2 ( ) 7.071(0.7071) cos(45 98.13 ), 0nx n n n n

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Example 5-8– Inverse z transform

2

2( )

( 0.5)( 1)

zX z

z z

1 22

( )

0.5 ( 1) ( 1)

D DX z C

z z z z

22

2

0.5

( 0.5)0.5 / (0.5 1) 2

( 0.5)( 1)z

z zC

z z z

2 2 2

1 2

1 1

21 1

( 1) ( ) ( 1)

( 0.5)( 1)

0.52

0.5 ( 0.5)

z z

z z

d z X z d z zD

dz z dz z z z

d z z z

dz z z

2 2 2

2 2

1 1

( 1) ( ) ( 1)2

( 0.5)( 1)z z

z X z z zD

z z z z

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• Inverse z transform

2

2 2 2( )

0.5 ( 1) ( 1)

z z zX z

z z z

( ) 2(0.5) 2 2 2 ( 1) (0.5) , 0n nx n n n n

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– Residue• Cauchy’s theory using contour integral

• Calculation of residue

11( ) ( )

2n

Cx n z X z dz

j

where contour integral including all poles.C

where ,

m is order of poles.

1( ) ( )nF z z X z

11( ) ( )

2n

Cx n z X z dz

j

1

1

1Residue ( ), ( ) ( )

( 1)! k

mm

k km z p

dF z p z p F z

m dz

(5-20)

(5-21)

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• For single pole

Unit circle 1z

Fig. 5-5.

1

Residue ( ), ( ) ( )

( ) ( )k

k k

nk z p

F z p z p F z

z p z X z

(5-22)

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Example 5-9– Find discrete time signal

• If ,

• Inverse z transform

( )( 0.75)( 0.5)

zX z

z z

1

( )( 0.75)( 0.5)

( 0.75)( 0.5)

n

n

z zF z

z z

z

z z

( ) Residue ( ),0.75 Residue ( ), 0.5x n F z F z

1( ) ( )nF z z X z

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• Sum of residue

0.75

0.75

Residue ( ),0.75 ( 0.75) ( )

( 0.75)

( 0.75)( 0.5)

(0.75)

0.75 0.5

0.8(0.75)

z

n

z

n

n

F z z F z

z z

z z

0.5

0.5

Residue ( ), 0.5 ( 0.5) ( )

( 0.5)

( 0.75)( 0.5)

0.8( 0.5)

z

n

z

n

F z z F z

z z

z z

( ) 0.8 (0.75) ( 0.5)n nx n

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Example 5-10– Inverse z transform

where and as 1 0.5 0.5p j 1 0.5 0.5p j *2 1 .p p

2

2

2 1( )

0.5

z zX z

z z

2

1 2

2 1( )

( )( )

z zX z

z p z p

1 2 21

2 2

( 2 1) ( 2 1)( ) ( )

0.5 ( 0.5)

n nn z z z z z z

F z z X zz z z z z

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Fig. 5-5.

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• n=0,2

2

2 1( )

( 0.5)

z zF z

z z z

1 2(0) Residue ( ),0 Residue ( ), Residue ( ),x F z F z p F z p

0

2

2

0

Residue ( ),0 ( )

( 2 1)

( 0.5)

1/ 0.5 2

z

z

F z zF z

z z z

z z z

1

1

1 1

21

1 2

Residue ( ), ( ) ( )

( )( 2 1)

( )( )

0.5 3.5

z p

z p

F z p z p F z

z p z z

z z p z p

j

2Residue ( ), 0.5 3.5F z p j

1 2(0) Residue ( ),0 Residue ( ), Residue ( ),

2 0.5 3.5 0.5 3.5

1

x F z F z p F z p

j j

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• n>0

2

2

( 2 1)( )

( 0.5)

nz z zF z

z z z

1 2( ) Residue ( ), Residue ( ),x n F x p F z p

1

1

1 1

21

1 2

Residue ( ), ( ) ( )

( ) ( 2 1)

( )( )

z p

n

z p

F z p z p F z

z p z z z

z z p z p

45 98.131Residue ( ), (0.7071 ) (3.5355 )

3.5355(0.7071) cos(45 98.13 ) sin(45 98.13 )

j n j

n

F z p e e

n j n

2Residue ( ), 3.5355(0.7071) cos(45 98.13 ) sin(45 98.13 )nF z p n j n

1 2( ) Residue ( ), Residue ( ),

7.071(0.7071) cos(45 98.13 ), 0n

x n F z p F z p

n n

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Example 5-11

where F(z) has poles at z=0.5, and z=1.

2

2( )

( 0.5)( 1)

zX z

z z

( ) Residue ( ), kx n F z p

11

2( ) ( )

( 0.5)( 1)

nn z

F z z X zz z

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• Sum of residue

( ) 2 ( 1) (0.5)nx n n

2 1

2

1

1

2

1

2

( 1)Residue ( ),1

( 0.5)( 1)

( 0.5)( 1)

( 0.5)

0.5( 1) 1 / (0.5) 2( 1)

n

z

n n

z

d z zF z

dz z z

z n z z

z

n n

1 2( ) Residue ( ), Residue ( ),x n F z p F z p

1 1

2 2

0.5

2

( 0.5)Residue ( ),0.5)

( 0.5)( 1) ( 1)

0.5(0.5) / (0.5) 2(0.5)

n n

z

n n

z z zF z

z z z

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linearity

Convolution

Differentiation

5. Characteristic of z transform5. Characteristic of z transform

1 2 1 2( ) ( ) ( ) ( )ax n bx n aX z bX z

( ) ( ) ( )k

y n h k x n k

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

( ) ( )

( )( )

x n X z

dX znx n z

dz

(5-23)

(5-24)

(5-25)