j nne e

Chapter 10

The Z-TransformComplex Frequency Domain (Z-Domain) Analysis of LTI System

● Representation of Aperiodic Signals

● Response of LTI System to Aperiodic Signals

j ne §5

特例

推广§10

j nnr e

e

j zz e

r

nz

10.0 Introduction

[ ]y n[ ]x n L

( )jX e

F 1F

( ) ( )j jH e Y e

( )jH e

1

2

3

[ ] ( )jh n H eF

:Key [ ] ( )jx n X eF

2

1[ ] ( )

2

( ) [ ]

j j n

j j n

n

x n X e e d

X e x n e

﹡ Problems: [ ]n

x n

Frequency analysis√

?Frequency analysis

§5

Frequency Domain Analysis

[ ]n

x n

﹡ Condition:

j ne ﹡Cause: Basic signal:

﹡Measure: Basic signal:j nne e

n

nn

represent

√represent

n

( )j ne e ( )j nre nz

z z

Im

Re

②( ) [ ] n

nX z x n z

1[ ] [ ( )]x n Z X z11[ ] ( )

2n

z r

x n X z z dzj

①

10.1 The Z-Transform Pair

A. The Transform Pair [ ] ~ nx n z

[ ] ,n

nx n r

Under Condition we have

( ) [ ( )]X z Z x n

Z a br z r

r

j jz e e re

r e z

( )X z

z-plane

Integral line

j nnr e j nne e

jre

e

Z

ar br

反

正

B. Understanding of The Transform Pairs

11[ ] ( )

2n

z r

x n X z z dzj

2

1[ ] ( )

2j j nnx n X r e r e d

( )( )jj j X reX re e

﹡Inverse Transform

jz re jdz re j d z jd

Frequency

2

1[ ] ( )

2j j nnx n X r e r e d

1 1 2 2[ ] ( ) ( )j j n j j nn nx n X re r e X re r e …… …… ……

11 ( )( )

jj j X reX re e 2

2 ( )( )jj j X reX re e

1cosnr nRe

2cosnr nRe

2nrFrequencynr 1

n n

( ) [ ] n

nX z x n z

﹡The Transform

jz re

( ) [ ] ( )j j nn

nX re x n r e

[ ] n

nx n r

[ ] ~ j nnx n r e ﹡ Similarity :

[ ] ~ j nn nx n r e z ﹡ Similarity :

{ [ ] }nF x n r

Im

Rear

brROC

r

Integralline

[ ] n

nx n r

11[ ] ( )

2n

z r

x n X z z dzj

C. The Convergence Region of the Z-Transform

z

a br r r generally: ROC

点： jre j nnr e ：基本信号

2

1[ ] ( )

2j j nnx n X r e r e d

jz re ， dz z jd

jz e Let

z 1 if

ROC

11[ ] ( )

2

( ) [ ]

n

z r

n

n

x n X z z dzj

X z x n z

D. Relations Between Z-Transform and Discrete-Time Fourier Transform

2

1[ ] ( )

2

( ) [ ]

j j n

j j n

n

x n X e e d

X e x n e

Z-Transform on Unit Circle

1z

jdz j e d

＝ Discrete-Time Fourier Transformar

brROC

1r

Im

Rea

ROC

Unit Circle

( )Z zX z

z a

10.2 The Region of Convergence of The Z-Transform

10.2.1 The ROC.

<Examples 10.1>

[ ] [ ]nx n a u n

0 1a ( ) | jz eX z

Z

Condition=ROC

n

[ ]x n

z a

( )jX e

① if ,1a 1z Unit Circle ROC

Im

ReZ a

1a ( ) | jz eX z

② if ,1a

[ ] [ ] ( )n Zz a

zx n a u n X z

z a

( )jX e

一般：右边信号 收敛域向外Z

n

[ ]x n

1z Unit Circle ROC

UnitCircle

1a ( ) | jz eX z

< Examples 10.2> [ ] [ 1]nx n a u n

① if ,1a

Z

Im

Rean

[ ]x n

( )Z zX z

z a

z a

1z Unit Circle ROC

UnitCircle

( )jX e

( ) | jz eX z 1a

② if ,1a

[ ] [ 1] ( )n Zz a

zx n a u n X z

z a

( )jX e

Im

ReZ a

一般：左边信号 收敛域向内Z

n

[ ]x n

1z then the Unit Circle ROC

UnitCircle

<10.1+10.2>

2[ ]x n 1( )X nROC for

Left-sided Right-sided

z a z a1

1

1 az [ ]n Za u n

z

z a<10.1>

[ 1]nZ a u n

<10.2>

( )X Z 2[ ]x n1[ ]x n

1[ ]x nIntegral 2 ( )X nROC for

Im

Rea

Integral

[ ] n

nx n r

Im

Re

Im

Re

10.2.3. General Rule for ROC

, ( ) 'at poles X z doesn t exist

Im

Re

0can be can be

a br z r A. ROC :

B. Poles ROC

z

右边信号

双边信号

左边信号

?

0?

Im

Rear 0r br

0

[ ]x n is two sided

z r

C. ROC 0

a b

a b

r z r

r r r

: ROC

[ ]x n

n

双边信号

环形收敛域或无收敛域

Z

[ ]n

x n b

Im

Re

Unite Circle

b 1/b

<Example 10.7>

0 1b

nbnb

[ ]x n

n

Im

Re

Unite Circle

b1/b

n

[ ]x nnb nb

1b

双边信号 环形收敛域Z 双边信号 无收敛域Z

2

1

1

( )( )Z b z

b z b z b

1

b z b

2

1

( ) [ ]N

n

n NX z x n z

D. is finite duration

z ROC 0z ROC

Im

Re

nz

nz

nz

nz

[ ]x n

n

[ ]x n

n

[ ]x n

n

[ ]x n

Im

Re

Pole at Im

Re

Poles at

0

ROC0z z

“环形” “向内” “向外”

0

ROC: entire Z-plane, 0z

z

possibly except

<Example 10.6 >

,0 1, 0[ ]

0 ,

na n N ax n

else

2

: ( 0 ~ 1)j kNzero ae k N

0 ( 1):

N st orderpole

a

Im

Rea

ROC

0 1N

[ ]x n

n

z a

pole

zero

0

0

Zz z planeexcept z

1

1 N N

n

z a

z az

?

1r

Im

Re0r

E. 0( )r z ROC ( )z ROC?

0nr

1nr

0[ ] n

nx n r

0

[ ]x n is right sided

z r ROC

[ ]x n

n

右边信号 收敛域向外Z

1 0r r

1 0r r1[ ] n

nx n r

Im

Re0r1 0r r

1 0r r

F. 0(0 )z r ROC ( 0 )z ROC

?

1nr

0nr [ ]x n

n

0

[ ]x n is left sided

z r ROC

左边信号 收敛域向内Z

0[ ] n

nx n r

1[ ] n

nx n r

1r

Im

Re

Im

Re

G. Rational

Im

Re

﹡left-sided signal ﹡ Two-sided signal ﹡right-sided signal

( )X z ROC: Bounded by poles

0z ROC ?﹡ z ROC?﹡

10.4 Geometric Evaluation of The Fourier Transform From The Zero-Pole Plot

10.4.1 Geometric Evaluation of Z-Transform

A. The Method

( )( )

( )

B zX z

A z

1

1

0 0

| |( )

| |

( ) ( ) ( )

M

kkN

kk

M N

k kk k

z zX z K

z p

X z z z z p K

零点距离积

极点距离积零点相位和 极点相位和

zero

pole

1

1

( )

( )

M

kkN

kk

z zK

z p

B. Example

1

1( )

1X z

az

Im

Rea

Im

Rea

( )X z

0z

z a

for z a

Im

Re1/2

1

1( )

11

2

X zz

10.4.2 Geometric Evaluation of Fourier Transform

A. The Method

B. Example

( ) | ( )jj

z eX z X e

2 2

( )jX e

( )jX e

if Unit Circle ROC, as above, 1z Let

012

z

z

1

2for z

UnitCircle

jz e

10.5 Properties of Z-Transform

00[ ] ( )nZ

z Rx n n z X z

1 1

2 2

1

2

[ ] ( )

[ ] ( )

Zz R

Zz R

x n X z

x n X z

1 2 1 21 2,[ ] [ ] ( ) ( )Z

z R R R Rax n bx n aX z bX z

[ ] ( )Zz Rx n X z

（ 可能加入或去掉）0,z

10.5.1 Linearity k kk k

Za a

10.5.2 Time shifting shift 00

nZn z

11[ ] ( )

2n

z r

x n X z z dzj

[ ] 1Zz z planen

00

11[ ] ( )

2n n

z r

x n n X z z z dzj

<Example>

<Proof>

Z

00[ ] nZ

z z planen n z

Z

[ ] ( )Zz Rx n X z

0

( )z

Xz

0 0( / ) [ ] n n

nX z z x n z z

10.5.3 Scaling in the Z-Domain

( ) [ ] n

nX z x n z

Im

Re0 1z r 0 2z r

Scaling

0z R平移

Im

Re

R

1r 2r

Z

Z

0n Zz

0 1z 外扩

0 1z

or

内收

<Proof>

0 ''

0( )

[ ] Znz RR z R

z x n

[ ] ( )Zz Rx n X z

10.5.4 Time Reversal

'1( ' )

1[ ] ( )Z

z RR

R

x n Xz

( ) [ ] n

nX z x n z

1( ) [ ] n

nX x n z

z

Im

Re

1/R

21/ r 11/ r

Z

Z

Z 1( )z

<Proof>

Im

Re

R

1r 2r

[ ] n

nx n z

[ ] ( )Zz Rx n X z

10.5.5 Time Expansion

( )1/

'( ' )

[ ] ( )k

kk

Zz RR R

x n X z

Where

( )

[ / ][ ]k

x n kx n

， if n is a multiple of k

， else

（时域扩展）Im

Re

1/1

kr 1/2

kr

integer

Z

k=3

-4k -3k -2k -k 0 k 2k 3k 4k补零

( )[ ]kx n

n

[2]x

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

[ ]x n

n

[2]x

[ ] ( )Zz R

x n X z

10.5.6 Conjugation

* * *[ ] ( )Zz R

x n X z

For real signal : * *( ) ( )X z X z

Im

Re

Z

[ ] ( )Zz Rx n X z

10.5.7 The Convolution Property

1 1

2 2

1

2

[ ] ( )

[ ] ( )

Zz R

Zz R

x n X z

x n X z

1 2 1 2

1 2

''

[ ] [ ] ( ) ( )Zz RR R R

x n x n X z X z

10.5.8 Differentiation in Z- Domain

[ ] '( )Zz Rnx n zX z ( )

'( )dX z

X zdz

Z

<Example>

1 1

1 2[ ](1 )

n Zz a

n za u n

a az

1

1

1[ ]

1n Z

z aa u naz

1 2

2 1 3

( 1)[ ]

2 (1 )n Z

z an n z

a u na az

Differentiation'( )zX z a

Differentiation ， Linear'( )zX z a

Important : useful in Inverse Z-Transform

1[ 1]nZ

z a a u n

1[ 1]nZ

z ana u n

a

1

2

( 1)[ 1]

2nZ

z an n

a u na

<Example>

1( ) log(1 )X z az

1

1

( )

1

dX z azz

dz az

111

1 ( ) [ 1]1

nZz a

aza a u n

az

( )[ ] [ 1]

nax n u n

n

Linearity, Time-scaling

1

1

1( ) [ ]

1nZ

z a a u naz

1[ ] ?Z

z a x n

1[ ]Z

z a nx n

( ) [ 1]na u n

10.5.9 The Initial-value Theorem

10.5.10 Table 10.1 include all properties

10.6 Some Common Z-Transform Pairs

Table 10.2

For causal [ ]x n ( [ ] 0 0)x n for n ， we have

( 检验变换的正确性 )

[0] lim ( )z

x X z

，0

lim [ ] n

zn

x n z

1

[ ][0] lim

nzn

x nx

z

10.3 Inverse Z-Transform

1 ( )[ ]

2n

z r

X zx n z dz

j z

① Contour Integral

围线积分Im

Re

ROC

z rIntegral line: ② Partial-Fraction Expansion

部分分式展开

( )X z for any kind of

( )X z for rational

A. Partial-Fraction Expansion for Rational

1. Basic Z-Transform Pairs

( )X z

(10.5.8 example)

1

1

1 az

1

1 2(1 )

z

az

2

1 3(1 )

z

az

1[ ]n Z

z ana u n

a

1[ ]n Z

z aa u n

1

2

( 1)[ ]

2n Z

z an n

a u na

1[ 1]nZ

z a a u n

1[ 1]nZ

z ana u n

a

1

2

( 1)[ 1]

2nZ

z an n

a u na

12 11

1 1 2 1 111 (1 ) (1 ) 1

i i n

i i n

A z AA A

p z p z p z p z

…… ……

2. Idea

( )X z

z①

② Get by Formula in Appendix (Partial-Fraction Expansion)1 ~ nA A

( )X z③

1

1

z

④ [ ]x n

1Z

一阶极点 二阶极点 一阶极点( )

( )

D z

N z 2 11

21 ( )( )

i i n

i ni

A AA A

z p z p z pz p

…… ……

1Z ROC

1 1 2 2 1 1[ ] [ ] [ ] [ ]n ni i i iA x n A x n A x n A x n …… ……

0 1m

ma a z a z ……

2 10 1

mma z a z a z ……

0 1[ 1] [ 2] [ 1]ma n a n a n m ……

B. Examples2 3 8

( )( 2)( 2)( 3)

z zX z

z z z

①( )X z

z

②1 1 1

( )1 2 1 2 1 3

B C DX z A

z z z

1

1

z

1Z

2 / 3 3/ 20 9 / 4 26 /15

1[ ] ?Z x n

2 3 8

( 2)( 2)( 3)

z z

z z z z

2 2 3

A B C D

z z z z

Im

Re-3 2-2BCD

[ ] [ ] [ 2 ( 2) ] [ ] ( 3) [ 1]n n nx n A n B C u n D u n

[ ] [ ] [ 2 ( 2) ( 3) ] [ 1]n n nx n A n B C D u n

[ ] [ ] [ 2 ( 2) ( 3) ] [ ]n n nx n A n B C D u n

③

1Z

for ROC:

for ROC:

Im

Re-3 2-2

Im

Re-2 2-3

for ROC: 2z

2 3z

3 z

左 右

右

BCD

BCD

②1 1 1

( )1 2 1 2 1 3

B C DX z A

z z z

左

10.7 Analysis and Configuration of LTI systems using Z-Transform

10.7.1 System Function of LTI System :

A. Response of LTI System to nz

nz ( ) nH z zLTI

[ ]h n

( )H z

* ,where ( ) [ ] n

nH z h n z

System Function or Transfer Function

System Function

[ ] ( )Zh n H z

( ) ( )n L nH zz H z z

( )H z

B. Explanation of

nz

( )H z

对各衰减因子各频率的衰减复正弦信号的幅度调整和相位调整作用

（类似于 ）( )H s

r

je e

ror j nne e 函数集 的选择其中：

( ) ( )n L nH zz H z z

( )H z ( )jH e e

相频特性（给定 ）

2 2 2 2

( )jH e e ( )H z

幅频特性（给定 ）

j nne e ( )LTIH z [ ( )]( )

jj j n H e enH e e e e ( ) nH z z

<Example>

1

1( )

1H z

az

Im

Rea

Im

Rea

( )H z

Integral Line

jre j nnr e or

j nne e

0z

z a

for z a

r

C. The Method to Obtain

1. From

[ ] ( )Zh n H z ( ) [ ] n

nH z h n z

2. From the Linear-Coefficient Different Equation of LTI System

0 0[ ] [ ]k k

N M

k ka y n k b x n k

, Linearity, Time-Shifting

( )H z

[ ]h n

Z

:

[ ] [ ]Lx n y n

0 0( ) ( )

M

k k

Nk k

k ka z Y z b z X z

[ ] ( )Zx n X z

[ ] ( )kZx n k z X z

0

0

( ) N

Mk

kk

kk

k

b zH z

a z

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

Coefficient of right-side of Equ.

<Example>1 1

[ ] [ 1] [ ] [ 1]2 3

y n y n x n x n

1

1

11

3( )1

12

zH z

z

0 0( ) ( )

M

k k

Nk k

k ka z Y z b z X z

( )( )

( )

Y zH z

X z

Coefficient of left-side of Equ.

[ ]n k kz

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

10.7.2 System Performance vs.

A. Causality vs.

Causality ROC:

Causality

Rational ROC:

Im

Re

( )H z

( )H z

( )H z

2. including

Cross outermost pole

①

②

Im

Re

① ②

2. Including

1. exterior outside of a circle

1. exterior outside of a circle

B. Stability vs.

Stability 1z

[ ]n

h n

Fourier Transform 1z

( )H z

ROC

ROC( )jH e

jz e

Im

Re

1z Im

Re

1z Im

Re

1z

Stable Unstable Unstable

Im

Re1/2 2

Im

Re1/2 2

Im

Re21/2

1[ ] [( ) 2 ] [ 1]

2n nh n u n

1[ ] ( ) [ ] 2 [ 1]

2n nh n u n u n

1[ ] [( ) 2 ] [ ]

2n nh n u n

Unstable, noncausal Stable, noncausal

Unstable, causal

<Example>

11

1 1( )

1 1 212

H zzz

1z

1z 1z

C. Stable & Causal System ~

Rational

Causality

Stability

z ROCExterior to the circle Acrossing outer most pole

1z ROC

All poles lies inside unit circle 1z

Im

Re1

Unit circle

( )H z

( )H z

10.7.3 Z-Domain Analysis of LTI System

1. Idea

( ) ( )n nLTIH zz H z z

( )[ ] [ ]LTIH zx n y n

: Basic relation between input and output

: Relation between any input and output

1 ( )[ ]

2

1 ( )[ ] ( )

2

n

z r

n

X zx n z dz

j z

X zy n H z z dz

j z

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

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

① 信号分解 ②已知输入输

出

③ 响应合成

L L

( )Y z

2. Steps

LTI[ ]x n [ ]y n

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

[ ]h n

( )H zZ 1Z

① ③

②

[ ] ( )Zx n X zKey :

（类似于 域分析）s

① ③

(For zero-state response)

选择合成 的函数集

[ ( )]

2

1( )

2

jj j n Y renY r e r e dj

① :

3. Role of LTI System explained by Z-Domain Analysis

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

( ) ( ) ( )

X z H z Y z

X z H z Y z

jz r e

幅度调整

相位调整

[ ( )]

2

1[ ] ( )

2

jj j n X renx n X r e r e dj

[ ]y n

( )H z ( )jH r e

( )jH re

[ ]x nr

( )H z

j nnr e

调整幅度 调整相位L

② :

规定了每个函数集的幅相调整方法

4. Example

[ ] ( 3) [ ]nh n u n ， [ ] [ ]x n u n ：求 [ ] ?y n

<Solution>

LTI[ ]u n ?

11 z

1

1

1 3z

( 3) [ ]nu n

1

1

1 3z Z 1Z

① ③

②

( 3)z 1z 3z

3 1

[ ( 3) ] [ ]4 4

n u n

1 1(1 3 )(1 )z z

1Z 3z

1 1

(3 / 4) (1/ 4)

1 3 1z z

10.8 System Function Block Program of LTI System

10.9 The Unilateral Laplace Transform

10.9.1 Definition

0| |[ ] ( )uuZ

z rx n X z 0

1

| |

( ) [ ]

1[ ] ( )

2

nu

n

nu

Z r

X z x n z

x n X z z dzj

i.e.

( 0)n

[ ]x n

[ ], 0[ ]

0 ,u

x n nx n

else

uZ

单边化

0| |z rZ

, 0[ ]

?, 0

nx n

n

√

( )uX s ( )uX s

1

uZ

[ ] [ ]x n u n

单边化①

②

<10.32>

[ ] [ ]nx n a u n

[ ] [ ]ux n x nuZ

| | | |z aZ

1

1( )

1uX zaz

causal

For causal signal

Z

uZ ( )uX z

( )X z

[ ]ux n

[ ]x n

<10.33>

For non-causal signal

Z

uZ ( )uX z

( )X z

1[ ] [ 1]nx n a u n

[ ] [ ]nux n a a u n u

Z

单边化

| | | |z aZ

1( )

1u

aX z

az

non-causal

Z

[ ]x n

11

z

az

| | | |z a

1| | | |1

[ ]1

nz aZa u n

az

| | | |z aZ

1| | | | ( )1uz a

Z aX z

az

2

1

uZ

[ ]ux n

[ ]x n

0n

[ ]x n1 [ 1]na u n

n

n

[ ]ux n 1[ ]na a u n

①

②

1( )

1

zX z

az

Im

Re

Im

Re

10.9.2 Properties of Unilateral Z-Transform

Table 10.3 (Compared to Table 10.1) Difference

A. Roc:

( )uX z ( )urational X zB. Time Reversal:

Don’t exist

C. Convolution:

11

22

1

2|

| |

|

[ ] ( )

[ ] ( )

u

u

u

uz

z r

r

x n X z

x n X z

Z

Z

1 2[ ] [ ]if x n and x n are causal

1 21 2

1 2)| | max( ,[ ] [ ] ( ) ( )uu uz r rx n x n X z X zZ

D. Time Shifting:

0| |[ ] ( )uuz r

Zx n X z

0

1| |[ 1] ( ) [ 1]u

uz rZx n z X z x

0| |[ 1] ( ) [0]uuz r

Zx n zX z zx

10.9.3 Solving Difference Equation Using the Unilateral Z-Transform

<10.37> Causal LTI System

11( ) 3 ( ) 3 [ 1]

1u uY z z Y z yz

1 1 1

3( )

1 3 (1 3 )(1 )uY zz z z

uZ inputstate

Zero input response Zero state response

[ ] 3 [ 1] [ ]y n y n x n , 0

[ ]?, 0

nx n

n

[ 1]y , ,

Full Response

Causal

ROC

① If 8, 1

1 1

3 2( )

1 3 1uY zz z

[ ] [3 ( 3) 2] [ ]nuy n u n

[ ] 3 ( 3) 2ny n 0for n

1uZ

② If 0

1 1( )

(1 3 )(1 )uY zz z

3 1

[ ] [ ( 3) ] [ ]4 4

nuy n u n

0for n

1uZ

(zero-state)

3 1[ ] [ ( 3) ]

4 4ny n

Causal

ROC

Im

Re1-3

Im

Re1-3

(Full Response)

1 1

(3/ 4) (1/ 4)

1 3 1z z

* Alternative Way of Solving Zero-State Response: when [ ] [ ]x n u n

1

1( )

1 3H z

z

| | 3z

1( )

1X z

z

| | 1z

1 1( )

(1 3 )(1 )Y z

z z

| | 3z

3 1[ ] [ ( 3) ] [ ]

4 4ny n u n

1Z

Causal →ROC

实际未说明初始状态都是零状态

Im

Re-3 1

Im

Re1

Im

Re-3

[ ]x n for n