Bài giảng Digital signal processing - Chapter 5: Z-Transform - Nguyen Thanh Tuan

Click to edit Master subtitle style Nguyen Thanh Tuan, M.Eng. 
Department of Telecommunications (113B3) 
Ho Chi Minh City University of Technology 
Email: nttbk97@yahoo.com 
 z-Transform 
 Chapter 5 
Digital Signal Processing 2 
 The z-transform is a tool for analysis, design and implementation of 
discrete-time signals and LTI systems. 
 Convolution in time-domain multiplication in the z-domain 
z-Transform 
Digital Signal Processing 
Content 
3 z-Transform 
1. z-transform 
2. Properties of the z-transform 
3. Causality and Stability 
4. Inverse z-transform 
Digital Signal Processing 
1. The z-transform 
4 
 The z-transform of a discrete-time signal x(n) is defined as the power 
series: 
z-Transform 
2 1 2( ) ( ) ( 2) ( 1) (0) (1) (2)n
n
X z x n z x z x z x x z x z
 
 The region of convergence (ROC) of X(z) is the set of all values of 
z for which X(z) attains a finite value. 
})()(|C{ 
n
nznxzXzROC
 The z-transform of impulse response h(n) is called the transform 
function of the filter: 

n
nznhzH )()(
Digital Signal Processing 
Example 1 
5 
 Determine the z-transform of the following finite-duration signals 
z-Transform 
 a) x1(n)=[1, 2, 5, 7, 0, 1] 
 b) x2(n)=x1(n-2) 
 c) x3(n)=x1(n+2) 
 d) x4(n)=(n) 
 e) x5(n)=(n-k), k>0 
 f) x6(n)=(n+k), k>0 
Digital Signal Processing 
Example 2 
6 
 Determine the z-transform of the signal 
z-Transform 
 a) x(n)=(0.5)nu(n) 
 b) x(n)=-(0.5)nu(-n-1) 
Digital Signal Processing 
z-transform and ROC 
7 
 It is possible for two different signal x(n) to have the same z-
transform. Such signals can be distinguished in the z-domain by their 
region of convergence. 
z-Transform 
 z-transforms: 
 and their ROCs: 
 ROC of a causal signal is the 
exterior of a circle. 
 ROC of an anticausal signal 
is the interior of a circle. 
Digital Signal Processing 
Example 3 
8 
 Determine the z-transform of the signal 
z-Transform 
)1()()( nubnuanx nn
 The ROC of two-sided signal is a ring (annular region). 
Digital Signal Processing 
2. Properties of the z-transform 
9 
 Linearity: 
z-Transform 
111 ROCwith)()( zXnx
z 
222 ROCwith)()( zXnx
z 
 if 
 and 
 then 
212121 ROCROCROCwith)()()()()()(    zXzXzXnxnxnx
z
 Example: Determine the z-transform and ROC of the signals 
 a) x(n)=[3(2)n-4(3)n]u(n) 
 b) x(n)=cos(w0 n)u(n) 
 c) x(n)=sin(w0 n)u(n) 
Digital Signal Processing 
2. Properties of the z-transform 
10 
 Time shifting: 
z-Transform 
)()( zXnx z 
)()( zXzDnx Dz   
 if 
 then 
 The ROC of is the same as that of X(z) except for z=0 if 
D>0 and z= if D<0. 
)(zXz D 
Example: Determine the z-transform of the signal x(n)=2nu(n-1). 
 Convolution of two sequence: 
 if and 
)()()()()()( 2121 zXzXzXnxnxnx
z   then 
 the ROC is, at least, the intersection of that for X1(z) and X2(z). 
Example: Compute the convolution of x=[1 1 3 0 2 1] and h=[1, -2, 1] ? 
)()( 11 zXnx
z  )()( 22 zXnx
z 
Digital Signal Processing 
2. Properties of the z-transform 
11 
 Time reversal: 
z-Transform 
 if 
 then 
Example: Determine the z-transform of the signal x(n)=u(-n). 
21 || :ROC)()( rzrzXnx
z  
12
1 1||
r
1
 :ROC)()(
r
zzXnx z   
 Scaling in the z-domain: 
21 || :ROC)()( rzrzXnx
z   if 
21
1 |||||| :ROC)()( razrazaXnxa zn   then 
 for any constant a, real or complex 
Example: Determine the z-transform of the signal x(n)=ancos(w0n)u(n). 
Digital Signal Processing 
3. Causality and stability 
12 z-Transform 
 will have z-transform 
 A causal signal of the form 
 )()()( 2211 nupAnupAnx
nn
||max||ROC
11
)(
1
2
2
1
1
1
i
i
pz
zp
A
zp
A
zX 

 the ROC of causal signals are outside of the circle. 
 A anticausal signal of the form 
 )1()1()( 2211 nupAnupAnx
nn
||min||ROC
11
)(
1
2
2
1
1
1
i
i
pz
zp
A
zp
A
zX 

 the ROC of causal signals are inside of the circle. 
Digital Signal Processing 
3. Causality and stability 
13 z-Transform 
Mixed signals have ROCs that are the annular region between two 
circles. 
 It can be shown that a necessary and sufficient condition for the 
stability of a signal x(n) is that its ROC contains the unit circle. 
Digital Signal Processing 
4. Inverse z-transform 
14 z-Transform 
ROC ),()(
transformz zXnx   
)(ROC ),(
transform-z inverse nxzX  
 In inverting a z-transform, it is convenient to break it into its partial 
fraction (PF) expression form, i.e., into a sum of individual pole 
terms whose inverse z transforms are known. 
ROC),()( zXnx z 
 Note that with we have 
signals) l(anticausa |a||z| ROC if )1(
signals) (causal |a||z| ROC if)(
)(
nua
nua
nx
n
n
1-az-1
1
)( zX
Digital Signal Processing 
Partial fraction expression method 
15 z-Transform 
 In general, the z-transform is of the form 
 The poles are defined as the solutions of D(z)=0. There will be M 
poles, say at p1, p2,,pM . Then, we can write 
)1()1)(1()( 112
1
1
 zpzpzpzD M
 If N < M and all M poles are single poles. 
where 
M
M
N
N
zaza
zbzbb
zD
zN
zX


1
0
1
10
1)(
)(
)(
Digital Signal Processing 
Example 4d 
16 z-Transform 
 Compute all possible inverse z-transform of 
Solution: 
- Find the poles: 1-0.25z-2 =0 p1=0.5, p2=-0.5 
- We have N=1 and M=2, i.e., N < M. Thus, we can write 
 where 
Digital Signal Processing 
Example 5od 
17 z-Transform 
Digital Signal Processing 
Partial fraction expression method 
18 z-Transform 
 If N=M 
Where and for i=1,,M 
 If N> M 
Digital Signal Processing 
Example 6 
19 z-Transform 
 Compute all possible inverse z-transform of 
Solution: 
- Find the poles: 1-0.25z-2 =0 p1=0.5, p2=-0.5 
- We have N=2 and M=2, i.e., N = M. Thus, we can write 
 where 
Digital Signal Processing 
Example 6 (cont.) 
20 z-Transform 
Digital Signal Processing 
Example 7 (cont.) 
21 z-Transform 
 Determine the causal inverse z-transform of 
Solution: 
- We have N=5 and M=2, i.e., N > M. Thus, we have to divide the 
denominator into the numerator, giving 
Digital Signal Processing 
Partial fraction expression method 
22 z-Transform 
 Complex-valued poles: since D(z) have real-valued coefficients, the 
complex-valued poles of X(z) must come in complex-conjugate pairs 
Considering the causal case, we have 
Writing A1 and p1 in their polar form, say, 
with B1 and R1 > 0, and thus, we have 
As a result, the signal in time-domain is 
Digital Signal Processing 
Example 8 
23 z-Transform 
 Determine the causal inverse z-transform of 
Solution: 
Digital Signal Processing 
Example 8 (cont.) 
24 z-Transform 
Digital Signal Processing 
Some common z-transform pairs 
25 z-Transform 
Digital Signal Processing 
Review 
26 
 Định nghĩa biến đổi z 
 Ý nghĩa miền hội tụ của biến đổi z 
Mối liên hệ giữa miền hội tụ với đặc tính nhân quả và ổn định của 
tín hiệu/hệ thống-LTI rời rạc. 
 Biến đổi z của một số tín hiệu cơ bản: (n), anu(n), anu(-n-1) 
Một số tính chất cơ bản (tuyến tính, trễ, tích chập) của biến đổi z 
 Phân chia đa thức và biến đổi z ngược 
z-Transform 
Digital Signal Processing 
Homework 1 
27 z-Transform 
Digital Signal Processing 
Homework 2 
28 z-Transform 
Digital Signal Processing 
Homework 3 
29 z-Transform 
Digital Signal Processing 
Homework 4 
30 z-Transform 
Digital Signal Processing 
Homework 5 
31 z-Transform 
Digital Signal Processing 
Homework 6 
32 
 Tìm biến đổi z và miền hội tụ của các tín hiệu sau: 
1) (n + 2) – (n – 2) 
2) u(n – 2) 
3) u(n + 2) 
4) u(n + 2) – u(n – 2) 
5) u(–n) 
6) u(n) + u(–n) 
7) u(n) – u(–n) 
8) u(1–n) 
9) u(|n|) 
10) 2nu(–n) 
11) 2nu(n–1) 
12) 2nu(1–n) 
z-Transform 
Digital Signal Processing 
Homework 7 
33 
 Tìm biến đổi z và miền hội tụ của các tín hiệu sau: 
1) cos( n)u(n) 
2) cos( n/2)u(n) 
3) sin( n/2)u(n) 
4) cos( n/3)u(n) 
5) sin( n/3)u(n) 
6) cos( n)u(n-1) 
7) cos( n)u(1-n) 
8) cos( n)u(-n-1) 
9) 2ncos( n/2)u(n) 
10) 2nsin( n/2)u(n) 
11) 3ncos( n/3)u(n) 
12) 3nsin( n/3)u(n) 
z-Transform 
Digital Signal Processing 
Homework 8 
34 
 Liệt kê giá trị các mẫu (n=0, 1, 2, 3) của tín hiệu nhân quả có biến 
đổi z sau: 
1) 2z -1 /(1 – 2z -1) 
2) 2z -1 /(1 + 2z -1) 
3) 2/(1 – 4z -2) 
4) 2/(1 + 4z -2) 
5) 2z -1 /(1 – 4z -2) 
6) 2z -1 /(1 + 4z -2) 
7) 2z -2 /(1 – 4z -2) 
8) 2z -2 /(1 + 4z -2) 
9) 2z -1 /(1 – z -1 – 2z -2) 
10) 2z -2 /(1 – z -1 – 2z -2) 
11) 2z -1 /(1 – 3z -1 + 2z -2) 
12) 2z -2 /(1 – 3z -1 + 2z -2) 
z-Transform