3 THE Z-TRANSFORM

3.0 INTRODUCTION We have seen that the Fourier transform plays a key role in

representing and analyzing discrete-time signals and

system .In this chapter ,we develop the z-transform

representation of a sequence and study how the properties of

a sequence are related to the propertied of its z-transform .

3.1 z-TRANSFORM

The z-transform of a sequence is defined as

This equation is, in general ,an infinite sum or infinite power

series ,with z being a complex variable. And refer to the z-

transform operator ,defined as

n

n

X z x n z

( )n

n

Z x n x n z X z

x n

Z

With this interpretation, the z-transform operator is seen

to transform the sequence into the function

where z is a continuous complex variable. The

correspondence between a sequence and its z-transform is

indicates by the notation

zx n X z

X z x n

The z-transform ,as we have defined it in Eq.(3.2) is often referred to as t

he two –sided or bilateral-transform ,in contrast to the one-sided or unilat

eral-transform, which is defined as

Clearly ,the bilateral and unilateral transform are equivalent only if

for

0

( ) n

n

X z x n z

0x n 0n

There is a close relationship between the Fourier transform and

the z-transform. In particular, if we replace the complex variable z in

Eq.(32). with the complex, variable ,then the z-transform reduces

to the Fourier transform.

More generally ,we can express the complex variable z in polar form as

je

jz re

With z expressed in this form ,Eq.(3.2)becomes

or

(3.6)

Equation (3.6)can be interpreted as the Fourier transform of the product of the original sequence and the exponential sequence

nj j

n

X re x n re

j n j n

n

X re x n r e

x nnr

Since the z-transform is a function of a complex variable, it is conveni

ent to describe and interpret it using the complex z-transform. In the

z-plane, the contour corresponding to is a circle of unit radius ,a

s illustrated in Figure 3.1. Note that is the angle between the vector

to a point z on the unit circle and the real axis of the complex z-plane.

1z

If we evaluate at points on the unit circle in the z-plane beginning at

through to we obtain the Fourier transform for

Figure3.1

jz e

X z1( . ., 0)z i e

( . ., / 2)z j i e 1( . ., )z i e 0

REGION OF CONVERGENCE （ ROC 收敛域）

As we stated in sec.2.7, if the sequence is absolutely summable ,the Fourier transform converges to a continuous function of ω . Applying this criterion to Eq.(3.6) leads to the condition

(3.7)

for convergence of the z-transform .

n

n

x n r

It should be clear from Eq.(3.7)that ,because of the multiplicati

on of the sequence by the real exponential it is possible for t

he z-transform to converge even if the Fourier transform does n

ot. For example ,the sequence is not absolutely summ

able and therefore ,the Fourier transform does not converge abs

olutely. However , is absolutely summable if

x n u n

nr u n

1r

nr

Example 3.1 Right-sided Exponential Sequence（右边指数序列）

Consider the signal . Because it is nonzero only for , this is an example of a right-sided sequence . From Eq.(3.2),

For convergence of ,we require that

nx n a u n0n

1

0

nn n

n n

X z a u n z az

1

0

n

n

az

X z

Thus the region of convergence is the range of values of z for which or equivalently, .inside the region of convergence, the infinite series converges to

For is the unit step sequence with z-transform

11

0

1, .

1

n

n

zX z az z a

az z a

1 1az z a

1a

1

1, 1.

1X z z

z

Example 3.2 Left-Sided Exponential Sequence （左边指数序列）

Now let since the sequence is nonzero

only for ,this is a left –sided sequence . Then

1nx n a u n

1n

1

1 0

1 (3.12)nn n

n n

a z a z

1n n n n

n n

X z a u n z a z

If or, equivalently, the sum in Eq.(3.12) conver

ges ,and

The pole-zero plot and region of convergence for this exampl

e are shown in Figure 3.4. Note that for ,the sequence

grows exponentially as and thus ,the Fouri

er transform does not exist.

1 1

1 11

1 1

zX z

a z az z a

1 1a z z a

1a

1na u n n

Figure3.3 Figure3.4

单位圆Z 平面 Z 平面

单位圆

Example3.3 Sum of two Exponential Sequences

Consider a signal that is the sum of two real exponentials;

The z-transform is then 1 1

2 3

n n

x n u n u n

1 1

2 3

1 1

2 3

1 1

0 0

( )

1 1

2 3

n nn

n

n nn n

n n

n n

X z u n u n z

u n z u n z

z z

1 1

1

1 1

1 11 1

1 12 3

12 1

121 1

1 12 3

12

12 ( 1/ 2)

1 12 3

Z Z

Z

Z Z

z zz

z z

Example 3.4 Sum of Two Exponentials (Again)

Again ,let be given by Eq.(3.14), then using the general result of Example 3.1 with and ,the z-transforms of the two individual terms are easily seen to be

1

2a

1

3a

1

1

1

2

1

3

1 1

2 2

1 1

3 3

1,

1

1,

1

nZ

nZ

u n zz

u n zz

x n

and ,consequently,

As we had determined in Example 3.3. the pole-zero plot and

ROC for the z-transform of each of the individual terms and for

the combined signal are shown in Figure3.5

1 11 1

2 3

1 1 1

2 3 2

1 1,

1 1

n nZu n u n z

z z

Example 3.5 Two-sided Exponential Sequence （双边指数序列）

Consider the sequence

Note that this sequence grows exponentially as

Using the general result of Example 3.1 with ,we obtain

and using the result of Example 3.2 with yields

1 1

3 21

n n

x n u n u n

1

1

13

3

1

3

1, ,

1

nzu n z

z

1

1

12

2

1

2

11 , ,

1

nzu n z

z

1

3a

n

1

2a

Thus , by the linearity of the z-transform,

In this case , the ROC is the annular region

1 1

1

1 1

1 1 1 1

1 1 3 21 1

3 2

1 12 1 2

12 12

1 1 1 11 1

2 3 2 3

, ,z z

z z z

z z z z

X z z z

1 1

3 2z

Note that the rational function in this example is identical to rational function in Example 3.3, but the ROC is different in the three cases. The pole-zero plot and the ROC for this example is shown in Figure 3.6

1

2

z plane

Re

Lm

1

121

3

Example 3.6 Finite-Length Sequence （有限长序列）

Consider the signal

Then

where we have used the general formula in Eq.(2.56) to sum the finite series. The ROC is determined by the set of values of z for which

,0 1

0,

n

xa n N

notherwise

1 11

0 0

1

1 1

1 1

1

nN Nz n

n n

NN N

N

X z a z az

az z a

az z z a

11

0

.nN

n

az

The N roots of the numerator polynomial are at

(note that these values satisfy the equation and

when ,these complex values are the Nth roots of unity ) the

zero at k=0 cancels the pole at z=a.

2 / , 0,1,..... 1.j k Nkz ae k N

N Nz a1a

Figure 3.7 Pole-zero for Example 3.6 with N=16 and a real such that 0<a<1. the region of convergence in this example consists of all values of z except z=0

Table 3.1 SOME COMMON z-TRANSFORM PAIRS( see book on the page of 104)

3.2 PROPERTIES OF THE REGION OF

CONVERGENCE FOR THE Z-TRANSFORM

PROPERTY 1: The ROC is a ring or disk in the z-plane centered at the origin: i.e.,

PROPERTY 2; The Fourier transform of converges absolutes if and only if the ROC of the z-transform of includes the unit circle

PROPERTY 3; The ROC cannot contain any poles. PROPERTY 4; If is a finite-duration sequence ,i.e., a sequence that is zero

except in a finite interval then the ROC is the entire z-plane, except possibly or

0 R Lr z r

1 2N n N 0z z

x n

x n x n

PROPERTY 5 ; If is a right-sided sequence ,i.e., a sequence that is zero for ,the ROC extends outward from the outermost (i.e., largest magnitude) finite pole in to ( and possibly including)

PROPERTY 6 ;If is a left-sided sequence ,i.e., a sequence that is zero for ,the ROC extends inward from the innermost (smallest magnitude) for pole in to ( and possibly including)

PROPERTY 7; A two-sided sequence is an infinite-duration sequence that is neither right sided nor left sided . If 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 ,not containing any poles.

PROPERTY; the ROC must be a connected region.

x n

1n N X z

z x n

2n N X z

0z

x n

As discussed in Section 3.1 ,property 1 results from the fact that convergence of Eq.(3.2) for a given is depended only on ,and property 2 is a consequence of the fact that Eq.(3.2) reduces to the Fourier transform when . Property 3 follows from the recognition that is infinite at a pole and therefore , by definition . does not converge.

Properties 4 through 7 can all be developed more or less directly from the interpretation of the z-transform as the Fourier transform of the original sequence, modified by an exponential weighting.

Property 8 is somewhat more difficult to develop formally , but at least intuitively, it is strongly suggested by our discussion of properties 4 through 7.

x n

X z1z

z

3.3 THE INVERSE z-TRANSFORM

One of the important roles of the z-transform is in the analysis of discrete-time linear systems . Often ,this analysis involves finding the z-transform of sequence and, after some

manipulation of the algebraic expressions, finding the

inverse z-transform.

In Sections 3.3.1-3.3.3 we consider some of these procedures , specifically the inspection method, partial fraction expansion , and power series expansion.

3.3.1 Inspection Method （观察法）

Evaluate the z-transform for sequence of the form where a can be either real or complex.

If we need to find the inverse z-transform of

and we recall the z-transform pair of Eq.(3.35), we shall recognize

“by inspection” the associated sequence as

nx n a u n

1

1,

1Zna u n z a

az

1

1

1 1 / 2

1

2zX z z

1( )

2

n

x n u n

If the ROC associated with in Eq(3.36) had been

we can recall transform pair 6 in Table 3.1 to find by inspection that

1

2z X z

11

2

n

x n u n

3.3.2 Partial Fraction Expansion部分分式展开法

To see how to obtain a partial fraction expansion ,let us assume that is expressed as a ratio of polynomials in ; i.e.,

an equivalent expression is

1z X z

0

0

Mk

kN

k

k

k

k

b zX z

za

0

0

MN M k

kN

M N k

k

k

k

z b zX z

z za

(3.38)

(3.37)

Equation (3.38 )explicitly shows that for such functions there will be M zeros and N poles at nonzero locations in the z-plane. In addition ,

there will be either M-N poles at z=0 if M>N

or N-M zeros at z=0 if N>MNote that could be expressed in the form

1

1

1

1

0

0

1

1

M

kN

k

k

k

c zb

X za

d z

X z

where the ‘s are the nonzero zeros of and the ‘s are nonzero poles of . If M<N and the poles are all first order , then can be expressed as

Obviously , the common denominator of the fractions in Eq.(3.40)is the same as the denominator in Eq.(3.39). Multiplying both sides of Eq.(3.40). by and evaluating for shows that the coefficients, can be found from

kc X z kd

X z X z

11 1

Nk

k k

AX z

d z

11 kd z

kz d

11kk k z dA d z X z

(3.40)

kA

(3.41)

Example 3.8 Second-Order z-Transform

Consider a sequence with z-transform

The pole-zero plot for is shown in Figure 3.12. from the region of

convergence and property 5 ,Section 3.2 ,we see that is a right-sided sequence .

x n

1 1

1

1 1 2

4 2

1,

1 1X z z

z z

X z

3.42

Since the poles are both first order , can be expresse

d in the form of Eq.(3.40);i.e.,

From Eq.(3.41)

1 2

1 11 1

4 21 1

A AX z

z z

11 1

4

12 1

2

1

4

1

2

1

1

1

2

z

z

A z X z

A z X z

X z

Therefore

Since is right sided , the ROC for each term extends

outward from the outermost pole . From Table 3.1 and the

linearity of the z-transform , it then follows that

1 11 1

4 2

1 2

1 1X z

z z

x n

1 12

2 4

n n

x n u n u n

Figure 3.12 pole-zero and ROC for Example 3.8

Re

z plane

1

4

1

2

Lm

1 2

1 1 1 11 2

0

1

1 10 1 1

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

( )

( ) [ ( )] [ ( )] [ ( )] ... [ ( )]

( )( )

( ) 1

( )

( )1 [1 ]

K

K

Mi

iiN

km

k

M N M r rn k k

n kn k kk i

B zX z X z X z X z

A z

x n X z X z X z X z

b zB z

X zA z a z

X z

A CX z B z

z z z z

z

则 可以展开成以下的部分分式形式

其中 ( ) ( )

( )i k nX z r z X z B

X z

为 的一个 阶极点，各个 为 的单阶极点， 为

的整式部分的系数。

1

1

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

1, 2,... -

1 1{ [(1 ) )]} , 1, 2,...

( )

k k

i

k

k k z z k z z

k

r k

k i z zr k r ki

A

X z X zA z z X z z z RES

z zk M r

C

dC z z X z k r

z r k dz

根据留数定理，系数 可用下式求得：（ （

（

系数 可以用以下关系取得：

＝ （（ ） ！

z展开式诸项确定后，再分别求右边各项的 反变换，以求得各个相加序列，即可得原序列。

Example 3.9 Inverse by partial Fractions

Consider a sequence with z-transform

The pole-zero plot for is shown in Figure3.13 .From the region of conve

rgence and property 5. Section 3.2 it is clear that is a right-sided sequen

ce . Since M=N=2 and the poles are all first order can be represented as

x n

x n

211 2

1 2 1 13 1 1

2 2 2

11 2(3.46)

1 (1 ) 1

zz zX z

z z z z

X z

X z

111

2

1 20

11

A AX z B

zz

(3.47)

The constant B0 can be found by long division

Since the remainder after one step of long division is o

f degree 1 in the variable it is not necessary to conti

nue to divide . Thus , can be expressed as

2 1

2 1

1

2 11 3

2 2

2

1 2 1

3 2

5 1

z z

z z

z

z z

1

1 11

2

1 52

(1 ) 1

zX z

z z

1z

X z

Now the coefficient A1 and A2 can be found by applying Eq.(3.41) to Eq.(3.46) or equivalently ,Eq.(3.47). Using Eq.(3.47), we obtain

Therefore

11

9 82

1 112

X zzz

1

1 1

11 5

(1 ) 1

1 12

12 1

29

12

z

z z

zzA

1

1 1

11 5

(1 ) 1

2 12 1 812

z

z z

zzA

From Table3.1 ,we see that since the ROC is

Thus ,from the linearity of the z-transform

1

1

1

1 2

2

2 2

1( )

1

1

1

Z

Z n

Z

n

u nz

u nz

12 9( ) 8

2nx n n u n u n

Figure 3.13 Pole-zero plot for the z-transform in Example 3.9

z plane

Re1

2

1

Lm

2.2 Inverse z-Transform

求 z 反变换的方法通常有三种：围线积分法（留数法）、部分分式展开法和长除法。一、 Residue Theorem （留数法）

-

-

-

1

1 1

0

1, k=0 1 {

0, k 0 2

( ) ( ) ,

1 1( ) ( )

2 2

x x

x x

k

n

x xn

k n k

n

X z R z R

R R

z dzj

X z x n z R z R

X z z dz x n z z dzj j

＋

＋

＋

根据柯西积分理论，若函数（）在环状区域

（ ， ）内是解析的，则在此区域内：

而

1 1

1

1

1 1

2 2

1

2

10 1 2

2-

( ) ( )

( ) ( ) ,

( ) ( ) , , , , ... ( , )

( )

k n k

n n

k

n

x x

X z z dz x n z dzj j

x k X z z dzj

x n X z z dz n C R Rj

z C X z

＋即：

这就是用围线积分的 反变换公式。其中围线 是在 的环状解析域内环绕原点的一条反时针的闭合单围线。直接计算围线积分比较麻烦，一般采用留数定理来求解。按留

1

1

1

1

2

( ( )

, ( , ),

[ ] ( )

[ ( ) ]

k

n

k m

n

C

nz z

k

F z X z z c c K

z c M z M K

x n X z z dzj

RES X z z

数定理，若函数 ）＝ 在围线 上连续，在 内有 个极点 而在 以外有 个极点 为有限值 则有

1 1

1 1

1 1

1

1 ( ) [ ( ) ]

2

1 1 ( ) ( )

2 2

1 [ ] ( ) [ ( ) ]

2

( ( )

m

m

n nz zC

m

n n

C C

n nz zC

m

n

X z z dz RES X z zj

X z z dz X z z dzj j

x n X z z dz RES X z zj

X z z

或

采用此公式必须满足

1

1

1 1

1

( )

( )

[ ( ) ] [( ) ( ) ]

( )

r r

nr

nr

n nz z r z z

nr

z z

X z z z

z X z z

RES X z z z z X z z

z X z z l

的分母多项式 的阶次比分子多项式 的阶次 高两阶或两阶以上）

在任一极点 处的留数的计算

若 是 的单极点，则有

若 是 的多重（阶）极点，则有1

1 11

1[ ( ) ] [( ) ( ) ]

1)!

r r

ln l n

z z r z zl

dRES X z z z z X z z

l dz

（

2

21 1

z2.5 X(z)= , 1/4< z 4,[(4-z)(z-1/4)]

1 1 z [ ] ( )

2 2 [(4-z)(z-1/4)]

X(z)

n n

C C

z

x n X z z dz z dzj j

c

例 已知 求 反变换。

=解：

为 的收敛域那的闭合围线。如图所示：

2 11

1 1

1

1/ 4

1[(4 - )( -1/ 4)] [(4 - )( -1/ 4)]

1/ 4

1[ ] ( ) [ ( ) ]

2

[ ][(4 - )( -1/ 4)]

k

nn

n nz zC

k

n

z

c

z zn z

z z z z

c z c

x n X z z dz RES X z zj

zRES

z z

＋

＋

现在来看极点在围线 内部及外部分布情况及极点阶数：

当 －时，函数

在围线 内只有 处的一个一阶极点，因此采用围线 内部的极点求留数较方便。

1

1/ 4

( -1/ 4)[(4 - )( -1/ 4)]

1 1 ( ) , 1

15 4

n

z

n

zz

z z

n

＋

－

2 11

1

1 1

1

2[(4 - )( -1/ 4)] [(4 - )( -1/ 4)]

4 ( )

1[ ] ( ) [ ( ) ]

2

[ ][(4 - )( -1/ 4)]

m

nn

n

n nz zC

M

n

z zn z

z z z z

c z X z z

c

x n X z z dz RES X z zj

zRES

z z

＋

＋

当 － 时，函数

在围线 外部只有 处的一个一阶极点，（且满足 分母的阶次比分子的阶次高两阶以上）因此采用围线 外部的极点求留数较方便。

1

4

4

2

2

( - 4)[(4 - )( -1/ 4)]

1 (4) , 2

15

1 1( ) , 1

15 4 ( )1

(4) , 215

n

z

z

n

n

n

zz

z z

n

nx n

n

＋

－

综合以上，可得

－

－

1

11

1

11 1

2 1

1

2

0

1

2

0

0

k

n

C

n

C

k

n n

z zC

nn

z a

n

X z z a x naz

x n z dzj az

zdz

j z a

n z a z a

z zx n dz RES

j z a z a

zz a a n

z a

zn

z a

( ) , , ( ).

[ ]

,

[ ] [ ]

( - )

例：已知： 求

解：

时，有单阶极点 已知

（ ）

当 时，函数 在围线 1

0

0 0

n

n

c X z z

c

x n n

x n a u n z a x n

( )

( ) ( )

( ) ( ) ( ( )

外部无极点，（且满足 分母的阶次

比分子的阶次高两阶以上）因此采用围线 外部的极点求留数较方便。因为围线外部无极点，所以留数围 。

为一因果序列）

1

11

1

1

11 1

2 1

1

2

0

0

0 0

0 0

n

C

n

C

k

n

n

X z z a x naz

x n z dzj az

zdz

j z a

n z a z a

x n n

zn c z a c z n

z a

X z z

( ) , , ( ).

[ ]

,

( ) ( )

,

( )

例：已知： 求

解：

时，有单阶极点 已知

在围线内无极点，留数为

当 时，函数 在围线 外部有一极点 在围线 内部 处有 阶

极点，（且满足 分母的阶

1 11

2

0

0 0

0

m

n nz zC

M

n nn

z a

z a

n

c

x n X z z dz RES X z zj

z zRES z a a n

z a z a

nx n x n

a n

[ ] ( ) [ ( ) ]

[ ] ( - ) ,

,( ) ( )

,

次比分子的阶次高两阶以上）因此采用围线 外部的极点求留数较方便。

即 1 na u n z a x n ( ) ( ( )为一左边序列）

1

2 1 2

1 2 3

( )

( ) ( ) [ 2] [ 1] [0] [1] [2]

( ) [ ].

1 11. ( ) , 0. ( ) ( ) 1

2 3!

( )

z

H z

a H z x z x z x x z x z

b x n

X z e z a H z z z z

b

本思想：将 表示为指数序列：

再用观察法求

例

再用

1

1 1

11 2

1 2

1[ ] [ ].

!

1 22. ( ) , [ ] .

1(1 )(1 )

2

2 1 1: ( ) 1 ,

2 3 2 41

[ ] ( ) [ ]2

n

x n u nn

zX z x n

z z

zX z z z

z z

x n u n

观察法求

例 因果序列

有理函数的长除法

二、 Power Series 幂级数数展开法

3.3.3 Power Series Expansion 幂级数数展开法

The defining expression for the z-transform is a Laurent series where the

sequence values are the coefficients of . Thus ,if the z-transfor

m is given as a power series in the form

We can determine any particular value of the sequence by finding the co

efficient of the appropriate power of 1z

nz x n

2 1 2.... 2 1 0 1 2 ...,

n

n

X z x n z

x z x z x x z x z

Example 3.10 Finite-length Sequence

Suppose is given in the form

We can express as

2 1 1 11

21 1 1X z z z z z

2 11 1

2 21X z z z z

X z

X z

Therefore ,by inspection , is seen to be

Equivalently

x n

0,

, 1

1, 0

1, 2

1

2

1, 1

2otherwise

n

n

n

x n

n

1 1

2 22 1 1x n n n n n

Example 3.11 Inverse transform by Power Series Expansion

Consider the z-transform

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

Therefore

1

1

1n n n

n

a zX z

n

11

, 1

0. 0

nn a

x n nnn

1log 1 ,X z az z a

1x

Example 3.12 Power Series Expansion by Long Division

Consider the z-transform

Since the region of convergence is the exterior of a circle ,the

sequence is a right-sided one. Furthermore, since approaches a

finite constant as z approaches infinity ,the sequence is causal.

Thus , we divide, so as to obtain a series in power of .

1

1

1X z z a

az

X z

1z

Carrying out the long division, we obtain1 2 2

1

1

1

1 2 2

2 2

1 ...1 1

1

...

az a zaz

az

az

az a z

a z

1 2 21

11 ...

1az a z

az

nx n a u n

Or

hence,

Example 3.13 Power Series Expansion for a Left-Sided Sequence

As another example, we can consider the same ratio of polynomials as in Eq.

(3.53), but with a different region of convergence

Because of the region of convergence , the sequence is a left-sided one, and

since at z=0 is finite , the sequence is zero for n>0.

1

1

1X z z a

az

X z

thus , we divide, so as to obtain a sequence in powers of

z as follows:

Therefore

1 2 2

1 2

1 2

...a z a za z z

z a z

a z

1nx n a u n

3.4 z-TRANSFORM PROPERTIES(Z 变换性质 )

Many of the properties of the z-transform are particularly useful

in studying discrete-time signals and systems.

In the following discussion , denotes the z-transform of

, and the ROC of is indicates by Rx: i.e.,

X z X z x n

,ZROC Rxx n X z

As we have seen ,Rx represents a set of values of z such that

For properties that involve two sequence and associated z-

transform, the transform pairs will be denoted as

R Lr z r

1

2

1 1

2 2

,

,

Z

Z

ROC Rx

ROC Rx

x n X z

x n X z

The linearity property states that

For example

1 2

1 2 1 2Zax n bx n ax z bx z

ROCcontainsRx Rx

1 1

1 11 ,1

1 1nx n u n a u n z a

z az

3.4.1 Linearity 线性

According to the time-shifting property, ROC=Rx( except for the possibleaddition or deletion of z=0 or z= )The quantity is an integer. If is positive, the original sequence is shifted right ,and if is negative, is shifted left. The derivation of this property follows directly from the z-transform is

With the substitution of variables

00 Z nx n n z X z

0n x n x n

0 n

n

Y z x n n z

0m n n

0 0m n n m

m m

Y z x m z z x m z

0nY z z X z

0n

0n

0n

3.4.2 Time Shifting 序列的移位

Example 3.14 Shifted Exponential Sequence

Consider the z-transform

From the ROC, we identify this as corresponding to a right-sided se

quence. We can first rewrite in the form

1 1,

1 44

X z zz

X z

1

1

1,

1 414

zX z z

z

This z-transform is of the form of Eq.(3.39) with M=N=

1, and its expansion in the form of Eq.(3.43) is

From Eq.(3.56), it follows that can be expresses as

1

44

11

4

X zz

x n

14 4

4

n

x n n u n

An expression for can be obtained more directly by applying the time-shifting property . first can be written as

From the time-shifting property

It is easily verified that Eqs.(3.57) and (3.59) are the same for all values of n; i.e.

1

11

4

n

x n u n

X z x n

1 1

1

4

1,

4z

X z z z

3.4.3 multiplication by an Exponential Sequence乘以指数序列

The exponential multiplication property is expressed mathematically as

The notation denotes that the ROC is Rx scaled by ; i.e., if Rx is the set of values of z such that

then is the set of values of z such that

0 0 0/ ,Znz x n X z z ROC z Rx

0ROC z Rx

0z Rx

0z

R Lr z r 0 0R Lz r z z r

Example 3.15 Exponential Multiplication

Staring with the transform pair

we can use the exponential multiplication property to determine the

z-transform of

First , is expressed as

1

11, (3.60)

1Zu n z

z

0cosnx n r n u n x n

0 01 1

2 2

n nj ju n re u n re u n

Then ,using Eq.(3.60) and the exponential multiplication prope

rty ,we see that

From the linearity property, it follows that

00 1

1

1 2

2,

1

n Zjj

re u n z rre z

0 1 0 1

1 12 2 ,

1 1j jX z z r

re z re z

1

1 2 2

0

0

1 cos,

1 2 cos

r zz r

r z r z

00 1

1

1 2

2,

1

n Zjj

re u n z rre z

3.4.4 Differentiation of X(z) X(z) 的微分

The differentiation property states that

This property is verified by differentiating the z-transform expression of Eq.(3.2); i.e.,

,Z dX z

nx n z ROC Rxdz

1

,n

n

n

n

n

n

X z x n z

dX zz z n x n z

dz

nx n z Z nx n

Example 3.16 Inverse of Non-Rational z-Transform

In this use the differentiation property together with the time-

shifting property to determine the inverse z-transform considered in

Example3.11.with

We first differentiation to obtain a rational expression:

1log 1 ,X z az z a

2

11

dX z az

dz az

From the differentiation property

The inverse transform of Eq.(3.63) can be obtained by the combined use of the z-transform pair of Example3.1 , the linearity property and the time-shifting property. Specifically , we can express as

Therefore

1

1,

1Z dX z az

nx n z adz az

11

nnx n a a u n

1 11 1 log 1 ,n

n Zax n u n az z a

n

nx n

Example 3.17 Second-Order pole

As another example of the use of the differentiation property, let us determine the z-transform of the sequence

From the z-transform pair of Example 3.1 and the differentiation property, it follows that

therefore

nx n na u n

1

1 2,

(1 )Zn az

na u n z aaz

1

1

1 2

1,

1

,(1 )

dX z z z a

dz az

azz a

az

3.4.5 conjugation of a Complex Sequence共轭序列

The conjugation property is expressed as

This property follows in a straightforward manner from

the definition of the z-transform, the details of which are

left as an exercise( Problem 3.51)

* * * ,Zx n X z ROC Rx

: ) ( ),

[ )] ) [ )( ) ]

( )

n n

n n

x x

x n X z ROC

x n x n z x n z

X z R z R

(共轭 不变

( ( (证明：

【即证】

3.4.6 Time Reversal 翻褶序列

The notation ROC=1/Rx implies that Rx is inverted ; i.e., if the set of values of z such that , then the ROC is the set of values of z such that . Thus, if is in the ROC for , then is the ROC for the z-transform of . If the sequence is real or we do not conjugate a complex sequence, the result becomes

* * * 11/ ,Zx n X z ROC

Rx

R Lr z r 1/ 1/ RLr z r 0z

x n

*01/ z

*x n x n

1

1 1 1

: [ ] ( ), 1/

[ )] ) )

)( ) ( )

n n

n n

nx x

n

x n X z ROC ROC

x n x n z x n z

x n z X z R z R

反序

(- (- (证明：

( 【即证】

Example 3.18 Time-Reversal Exponential Sequence

As an example of the use of the property of time reversal , consider the sequence

Which is a time-reversed version of . From the time-reversal property, it follows that

nx n a u n

na u n

1 1

11 1

1,

1 1

a zX z z a

az a z

3.4.7 Convolution of Sequences 序列的卷积

According to the convolution property contains

To derive this property formally, we consider

So that

1 * 2 1 2 ,Zx n x n X z X z ROC 1 2Rx Rx

1 2k

y n x k x n k

1 2n n

k k k

Y z y n z x k x n k z

If we interchange the order of summation

Changing the index of summation in the second sum from n

to m=n-k, we obtain

Thus, for values of z inside the regions of convergence of

both

and ,we can write

Where the region of convergence includes the intersection of

the regions of convergence of and

1 2 n

k k

Y z x k x n k z

1 2 m k

k m

Y z x k x m z z

1X z 2X z

1 2Y z X z X z

2X z 1X z

Example 3.19 Evaluating a Convolution Using the z-Transform

Let and . The corresponding are

and

If , the z-transform of the convolution of with

is then

1nx n a u n 2x n u n

10

11

,1

n n

n

X z a z z aaz

10

21

, 11

n

n

X z z zz

1a

2x n 1x n

2

1 1

1, 1

11 1

zY z z

z a zaz z

The poles and zeros of are plotted in Figure 3.14 , and

the region of convergence is seen to be the overlap region. T

he sequence can be obtained by determining the inverse

z-transform . Expanding in Eq.(3.64) in a partial fraction

expansion, we get

Therefore ,

Y z

y n

Y z

1 1

1 1 1, 1

1 1 1Y z z

a z az

11

1ny n u n a u n

a

Figure 3.14Pole-zero plot for the z-transform of the convolution of the sequence u [n] and an u [n]

Re

z plane

Lm

1a

Input z-Transform X(z) and system function H(z), determinethe ROC for output z-Transform Y(z)

-1

-1

-1 -1

-1

-1

1( ) ( ) 2

1- 21

( ) 31-1/ 3

1 1( ) ( ) 3

1 5(1- )(1 3 )5

1 3 ( ) 1/ 3

1 1/ 3

A X Z ZZ

H Z ZZ

B X Z ZZ Z

ZH Z Z

Z

If is zero for ( i.e., if is causal), then x n 0n

0 limz

x X z

3.4.8 Initial-Value Theorem 初值定理

x n

1 2

( ) 0, 0 lim ( ) 0

) ) (0) (1) (2) ...

lim ( ) 0

z

n

n

z

x n x n n X z x .

X z x n z x x z x z

X z x

对于因果序列（ ）即 则有： （）

( (

所以 （）【即证】

, (0)0.3

z x

-1

-1 -2

1+0. 2z1 X(z)= >1,例 ：已知 求

1-0. 7z z

(0) lim ( ) 1z

x X z

解：

, 2, (0))( )

z x-1 -2

-1 -1

1+z +z2 X(z)=例 ：已知 求

(1-z 1-2z

(0) lim ( ) 1z

x X z

解：

2.4 System Function

: /

[ ] [ ] [ ]

( )

[ ] [ ] [ ],

( ) ( ) ( ) ( ) ( ) ( )

[ ] [ ]

k

I O

h nx n y n

H z

y n h n x n

Y zY z H z X z H z X z

y n x n

a

出发点有理的系统函数与差分方程的 相对应。

若 和 的关系如下：

0 0

0

0

[ ] [ ]

( )

N M

mk m

Mm

mm

Nk

kk

y n k b x n m

z

b zH z

a z

则，系统冲激响应对应的 变换为下述多项式：

已知线性移不变系统稳定的充要条件：

当 |z|=1 时，上式变成

这就是系统稳定的充要条件。

因此，若系统函数在单位圆上收敛，则系统是稳定的。这也意味着，如果系统函数 H(z) 的收敛域包括单位圆，则系统是稳定的。反之，如果系统稳定，则系统函数 H(z) 的收敛域一定也包括单位圆。

一、因果稳定系统

显然，一个因果系统的系统函数的收敛域应该是

( )

1

H z

z

z

综上两点看出：一个因果稳定的系统函数 必须从单位圆到 的整个 域内收敛，即：

也就是说系统函数的全部极点必须在单位圆内。

线性移不变系统可以用线性常系数差分方程描述：

对上式两边求 Z 变换，利用线性性质和时不变性质，得

因此

可见系统函数的系数也正是其差分方程的系数。

二、系统函数和差分方程的关系

系统函数还可以进一步分解成：

式中， {dk) 和 {cr} 分别表示 H(z) 在 z 平面上的极点和零点。这样，系统函数可以用 z 平面上的极点、零点和常数 A 来确定。例 根据系统函数求差分方程

求该系统的差分方程。

为了求满足该系统输入输出的差分方程，可以将 H(z) 的分子和分母各因式乘开，而得到如下的形式：

于是，

其差分方程就是

三、系统的频率响应的意义( ) ( )

( ) ( ) |

[ ( )] [ ( )* ( ]

( ) ( ) ( )

( ) [ ( )]

( )

j

j

j

z e

j j j

j

j

H e H z

H e H z

DTFT y n DTFT x n h n

Y e X e H e

H e DTFT h n

H e

系统的频率响应 正是系统函数 在单位圆上的值，

即

有了系统频率响应的概念，现在对线性移不变系统，建立任意输入情况下，输入输出两者之间的傅立叶关系 ）

即 ＝

其中 ＝

是系统的

( )

1( ) ( ) ( )

2j j j n

y n

y n X e H e e d

－

频率响应，对于线性移不变系统，其输出序列的傅立叶变换等于输入序列的傅立叶变换与系统频率响应的乘积。利用发变换，可求得输出序列 ：

＝

四、频率响应的几何确定法

1

( )1 1

1

1 1

( )1 1

1 1

)

(1 ) ( )( )

(1 ) ( )

(1 ) ( )( )

(1 ) ( )

M M

m mN Mm m

N N

k kk k

j

M Mj j

m mj j N Mm m

Nj j

k kk k

H z z

c z z cH z K Kz

d z z d

K z e

c e e cH e K Ke

d e e d

－

－

可以利用（ 在 平面上零点、极点分布，通过几何法直观求出系统频率响应。

其中 为实数，用 代入，即得系统得频率响应为：

arg[ ( )] = ( )j

N

j j H eH e e

1

1

1 1

( ) ( )

( )

arg[ ( )] arg[ ] arg( ) arg( ) ( )

) )

Mj

mj m

Nj

kk

M Nj j j

m km k

m k

e cH e K

e d

H e K e c e d N M

z c H z d H z

其模等于：

其相角为：

=

z= z=在 平面上 表示（ 的零点， 表示（ 的极点。

k

k k k

1

1

1 1

,

,

( )

arg[ ( )] arg[ ] (

m

k

j jm m m

jjm m m m m

j jk k

jjk k

M

mj m

N

kk

M Nj

m km k

e c c e C

e c C C C e

e d d e D

e d D D D e

H e K

H e K

用一根由零点 指向单位圆上 点的矢向量 表示，

即 设矢向量 为

用一根由极点 指向单位圆上 点的矢向量 表示，

即 设矢向量 为

其模等于：

其相角为：

= )N M

) ( 1) ( )

( ), ( ); ( )

arg[ ( )]

j j

j

y n by n x n

h n H e H e

H e

:例：已知一稳定的因果系统可表示为 （

求：系统的脉冲响应 频率响应 幅频特性 ；

相频特性 。

1

n

2

1 ( ) ( , 1)

1

h(n)=b ( )

1 1( ) ( ) |

1 1 cos sin

1( )

1 2 cossin

arg[ ( )] ( )1 cos

j

jjz e

j

j

z

H z z b bbz

u n

H e H zbe b jb

H eb b

bH e arctg

b

-1Y(Z)=bz Y(Z)+X(z)解： 变换得：

Y(Z)X(z)

频率响应

四、 IIR 系统与 FIR 系统

1.IIR 系统：若系统的单位抽样（冲激）响应延伸到无穷长，称之为“无限长单位冲激响应系统”，简写成 IIR 系统。其对应的差分方程有输出端递归到到输入断，又称递归系统。

2.IIR 系统：若系统的单位抽样（冲激）响应是一个有限长序列，称之为“有限长单位冲激响应系统”，简写成 FIR 系统。其对应的差分方程无输出端递归到到输入断，又称非递归系统。

We have presented and discussed a number of the theorems and

properties of z-transforms , many of which are useful in

manipulating z-transform . These properties and a number of others

are summarized for convenient reference in Table3.2( see the book

on page.126)

3.4.9 Summary of Some z-Transform Properties

3.5 SUMMARY

In this chapter , we have defined the z-transform of a

sequence and shown that it is a generalization of the Fourier

transform. The discussion focused on the properties of the z-

transform and techniques for obtaining the z-transform of a

sequence and vice versa.

An important part of the chapter was a discussion of some

of the many properties of the z-transform that make it useful

in analyzing discrete-time signals and systems.