Digital Signal Processing - 2018 Faculty of Engineering Helwan University

1. ‫ر‬َ‫ـد‬ْ‫ق‬‫ِـ‬‫ن‬،،،‫لما‬‫اننا‬ ‫نصدق‬ْْ‫ق‬ِ‫ن‬‫ر‬َ‫د‬
LECTURE (4)
The Z-Transform
Amr E. Mohamed
Faculty of Engineering - Helwan University

2. Agenda
 Introduction
 The z-Transform
 Properties of the z-Transform
 Inverse z-Transform
 Direct Division Method
 Partial Fraction-expansion Method
 Inversion Integral Method (Residue Theorem)
 System Representation in the z-Domain
 The Transfer Function of a Discrete-Time System
 The Frequency Response of a Discrete-Time System
 System Properties in the z-Domain
 Causality
 Stability
 Relationships Between System Representations
2

3. Introduction
 A LTI discrete system, as discussed in Chapter 2, represented in time domain
using the impulse response, h(n), frequency domain using frequency response,
H(ej) or in the Z domain using transfer function, H(z) as shown in Fig. 3.1.
 Advantages of using frequency response:
1) The computation of steady state response is greatly facilitated.
2) The response to any arbitrary signal x(n) can be easily computed.
 Disadvantages of using frequency response:
1) There are some useful signals for which DTFT does not exist.
2) The transient response of a system due to initial conditions can not be
computed.
3Fig. 3.1 LTI discrete system representation.
 nh
 j
eH
 nx
 j
eX
  )(*)( nhnxny 
      jjj
eHeXeY 
 zX  zH      zHzXzY 

4. The z-Transform
 Counterpart of the Laplace transform for discrete-time signals
 Generalization of the Fourier Transform
 Fourier Transform does not exist for all signals
 The z-Transform is often time more convenient to use
 Definition:
 Compare to DTFT definition:
 z is a complex variable that can be represented as z=r ej
 Substituting z=ej will reduce the z-transform to DTFT
4
   




n
n
znxzX
    nj
n
j
enxeX 





5. z-transform of the Unit impulse
 Let x[n] = δ(n)
 Then
 And the region of convergence is all z, where z is a complex number.
5


 

otherwise0
0nfor1
(n)
    1(n)  






n
n
n
n
zznxzX 

6. z-transform of the a Shifted Unit impulse
 Let x[n] = δ(n-q)
 Then
 And the region of convergence is all z except for z=0.
6


 

otherwise0
qnfor1
q)-(n
    q
n
n
n
n
zzznxzX 






  q)-(n

7. z-transform of a Unit-Step Function
 Let x[n] = u[n]
 Then
 And the region of convergence will be |z|>1.
7


 

0nfor0
0nfor1
(n)

u
   
11
1
(n) 1
0 


 









 z
z
z
zzuznxzX
n
n
n
n
n
n

8. z-transform of anu[n]
 Let x[n] = anu[n]
 Then
 With the region of convergence |z|>|a| and a is any real or complex
number.
8


 

0nfor0
0nfor
(n)

n
n a
ua
    










0
(n)
n
nn
n
nn
n
n
zazuaznxzX
 
az
z
az
azzX
n
n



 



 1
0
1
1
1
)(

9. z-transform of nanu[n]
 Let x[n] = nanu[n]
 Then
 With the region of convergence |z|>|a| and a is any real or complex
number.
9


 

0nfor0
0nfor
(n)

n
n na
una
    










0
n(n)
n
nn
n
nn
n
n
zazunaznxzX
  221
0
1
)()1(
1
)(n
az
z
az
azzX
n
n



 





10. z-transform of Sinusoids
 Let x[n] = (cos Ωn) u[n]
 Then
 Similarly it can be shown:
10


 

0nfor0
0nfor)cos(
(n))cos(

n
un
2
n)(cos
njnj
ee 


)}()({
2
1
u(n)}n)({cos nuenueZ njnj 








  jj
ez
z
ez
z
2
1
u(n)}n)({cos
1)(cos2
)(cos
)(cos21
)(cos1
u(n)}n)({cos 2
2
21
1





 

zz
zz
zz
z
1)(cos2
)(s
)(cos21
)(s
u(n)}n)({sin 221
1





 

zz
inz
zz
inz

11. The z-transform and the DTFT
 The z-transform is a function of the complex z variable
 Convenient to describe on the complex z-plane
 If we plot z=ej for =0 to 2 we get the unit circle
11
Re
Im
Unit Circle

r=1
0
2 0 2

 j
eX

12. Right-Sided Exponential Sequence Example
 For Convergence we require
 Hence the ROC is defined as
 Inside the ROC series converges to
12
         







0
1
n
n
n
nnn
azznuazXnuanx





0n
n
1
az
az1az
n
1

    az
z
az
azzX
n
n



 




0
1
1
1
1
Re
Im
a 1
o x
• Region outside the circle of
radius a is the ROC
• Right-sided sequence ROCs
extend outside a circle

13. Left-Sided Exponential Sequence Example
 Hence the ROC is defined as
• Region inside the circle of radius a is the ROC
• Left-sided sequence ROCs extend inside a circle
13
          





1
1
11
n
n
n
nnn
azznuazXnuanx
azza 
11
Re
Im
a 1
o x
  ...]1[... 33221133221
 
zazazazazazazazX
 
az
z
za
zazX



 

1
1
1
1

14. Two-Sided Exponential Sequence Example
14
     1-n-u
2
1
-nu
3
1
nx
nn













11
1
0
1
0n
n
1
z
3
1
1
1
z
3
1
1
z
3
1
z
3
1
z
3
1





























11
0
11
1
n
n
1
z
2
1
1
1
z
2
1
1
z
2
1
z
2
1
z
2
1






























z
3
1
1z
3
1
:ROC 1

 
z
2
1
1z
2
1
:ROC 1


 


























2
1
z
3
1
z
12
1
zz2
z
2
1
1
1
z
3
1
1
1
zX
11
Re
Im
2
1
oo
12
1
xx3
1


15. 15
Finite Length Sequence
 


 

otherwise0
1Nn0a
nx
n
     
az
az
z
1
az1
az1
azzazX
NN
1N1
N11N
0n
n1
1N
0n
nn





 







 Geometric series formula:





2
1
21N
Nn
1NN
n
a1
aa
a

16. Stability, Causality, and the ROC
 Consider a system with impulse response h[n]
 The z-transform H(z) and the pole-zero plot
shown
 Without any other information h[n] is not
uniquely determined
 |z|>2 or |z|<½ or ½<|z|<2
 If system stable ROC must include unit-circle:
½<|z|<2
 If system is causal must be right sided: |z|>2
16

17. 17
Signal, 𝐱(𝐧) Z-Transform, 𝐗(𝐳) ROC
1 𝛅(𝐧) 1 all Z
2 𝛅(𝐧 − 𝐧 𝟎) 𝐳−𝐧 𝟎 𝐳 ≠ 𝟎
3 𝐮(𝐧)
𝐳
𝐳 − 𝟏
𝐳 > 𝟏
4 −𝐮(−𝐧 − 𝟏)
𝐳
𝐳 − 𝟏
𝐳 < 𝟏
5 𝐚 𝐧
𝐮(𝐧)
𝐳
𝐳 − 𝐚
𝐳 > 𝐚
6 −𝐚 𝐧
𝐮(−𝐧 − 𝟏)
𝐳
𝐳 − 𝐚
𝐳 < 𝐚
7 𝐧 𝐮(𝐧)
𝐳
(𝐳 − 𝟏) 𝟐
𝐳 > 𝟏
8 𝐧 𝐚 𝐧
𝐮(𝐧)
𝐚𝐳
(𝐳 − 𝐚) 𝟐
𝐳 > 𝐚
9 𝐜𝐨𝐬(𝛚 𝟎 𝐧)
𝐳 𝟐
− 𝐳𝐜𝐨𝐬(𝛚 𝟎)
𝐳 𝟐 − 𝟐𝐳 𝐜𝐨𝐬 𝛚 𝟎 + 𝟏
𝐳 > 𝟏
10 𝐬𝐢𝐧(𝛚 𝟎 𝐧)
𝐳 𝐬𝐢𝐧(𝛚 𝟎)
𝐳 𝟐 − 𝟐𝐳 𝐜𝐨𝐬 𝛚 𝟎 + 𝟏
𝐳 > 𝟏
11 𝐫 𝐧
𝐜𝐨𝐬(𝛚 𝟎 𝐧)
𝐳 𝟐
− 𝐳 𝐫 𝐜𝐨𝐬(𝛚 𝟎)
𝐳 𝟐 − 𝟐𝐳 𝐫 𝐜𝐨𝐬 𝛚 𝟎 + 𝐫 𝟐
𝐳 > 𝐫
12 𝐫 𝐧
𝐬𝐢𝐧(𝛚 𝟎 𝐧)
𝐳 𝐫 𝐬𝐢𝐧(𝛚 𝟎)
𝐳 𝟐 − 𝟐𝐳 𝐫 𝐜𝐨𝐬 𝛚 𝟎 + 𝐫 𝟐
𝐳 > 𝐫

18. Z-Transform Properties
18

19. 19
Z-Transform Properties: Linearity
 Notation
 Linearity
 Note that the ROC of combined sequence may be larger than either ROC
 This would happen if some pole/zero cancellation occurs
 Example:
• Both sequences are right-sided
• Both sequences have a pole z=a
• Both have a ROC defined as |z|>|a|
• In the combined sequence the pole at z=a cancels with a zero at z=a
• The combined ROC is the entire z plane except z=0
 We did make use of this property already, where?
    x
Z
RROCzXnx  
        21 xx21
Z
21 RRROCzbXzaXnbxnax  
     N-nua-nuanx nn


20. Z-Transform Properties: Time Shifting
 Here no is an integer
 If positive the sequence is shifted right
 If negative the sequence is shifted left
 Add or remove poles at z=0 or z=
 Example:
20
    x
nZ
o RROCzXznnx o
 
 
4
1
z
z
4
1
1
1
zzX
1
1

















   1-nu
4
1
nx
1-n







    x
n
nnZ
o RROCzzXznnx
o
o






 

x(n)-
0n

21. Z-Transform Properties: Multiplication by Exponential
 ROC is scaled by |zo|
 All pole/zero locations are scaled
 If zo is a positive real number: z-plane shrinks or expands
 If zo is a complex number with unit magnitude it rotates
 Example: We know the z-transform pair
 Let’s find the z-transform of
21
    x
Zn
RROCzXnx   /
  1z:ROC
z-1
1
nu 1-
Z
 
             nurenurenunrnx
njnj
o
n oo 
 

2
1
2
1
cos
  rz
zre1
2/1
zre1
2/1
zX 1j1j oo




 

22. Z-Transform Properties: Differentiation
 Example: We want the inverse z-transform of
 Let’s differentiate to obtain rational expression
 Making use of z-transform properties and ROC
22
   
x
Z
RROC
dz
zdX
znnx  
    azaz1logzX 1
 
   
1
1
1
2
az1
1
az
dz
zdX
z
az1
az
dz
zdX









     1nuaannx
1n


     1nu
n
a
1nx
n
1n



23. Z-Transform Properties: Time Reversal
 ROC is inverted
 Example:
 Time reversed version of
23
   
x
Z
R
1
ROCz/1Xnx  
   nuanx n
 
 nuan
  1
11-
1-1
az
za-1
za-
az1
1
zX 






24. Z-Transform Properties: Convolution
 Convolution in time domain is multiplication in z-domain
 Example: Let’s calculate the convolution of
 Multiplications of z-transforms is
 ROC: if |a|<1 ROC is |z|>1 if |a|>1 ROC is |z|>|a|
 Partial fractional expansion of Y(z)
24
        2x1x21
Z
21 RR:ROCzXzXnxnx  
       nunxandnuanx 2
n
1 
  az:ROC
az1
1
zX 11 

 
  1z:ROC
z1
1
zX 12 

 
     
  1121
z1az1
1
zXzXzY 


  1z:ROCassume
1
1
1
1
1
1
11









 
azza
zY
      nuanu
a1
1
ny 1n




25. Z-Transform Properties: Initial and Final Value Theorems
 Initial Value Theorem:
 The value of x(n) as n → 0 is given by:
 Final Value Theorem:
 The value of x(n) as n →  is given by:
25
)(lim)(lim)0(
0
zXnxx
zn 

)]()1[(lim)(lim)(
1
zXznxx
zn



26. Z-Transform Properties: Complex Conjugation
 If X(z) = Z[x(n)], then:
26
XRzzXnx  *)(*)](*[Z

27. 27

28. The inverse Z-Transform
28

29. 29
The Inverse Z-Transform
 Formal inverse z-transform is based on a Cauchy integral
 Less formal ways sufficient most of the time
1) Direct or Long Division Method
2) Partial fraction expansion
3) Inversion Integral Method (Residue-theorem)
       dzzzX
j
zXZnx n
C
11
2
1 



30. 30
Inverse Z-Transform: Power Series Expansion
 The z-transform is power series
 In expanded form
 Z-transforms of this form can generally be inversed easily
 Especially useful for finite-length series
 Example
   




n
n
znxzX
            ...21012... 2112
 
zxzxxzxzxzX
    
12
1112
2
1
1
2
1
z
11
2
1
1










zz
zzzzzX
         1n
2
1
n1n
2
1
2nnx 
 














Otherwise
n
n
n
n
nx
0
1
2
1
01
1
2
1
21

31. Inverse Z-Transform: Power Series Expansion
 Example: Find x(n) for n = 0, 1, 2, 3, 4, when X(z) is given by:
 Solution:
 First, rewrite X(z) as a ratio of polynomial in 𝑧−1
, as follows:
 Dividing the numerator by the denominator, we have:
31
 
  2.01
510



zz
z
zX
  21
21
2.02.11
510





zz
zz
zX
4321
68.184.181710 
 zzzz
21
2.02.11 
 zz 21
510 
 zz
32
217 
 zz
432
4.34.2017 
 zzz
43
4.34.18 
 zz
543
68.308.224.18 
 zzz
54
68.368.18 
 zz
654
736.3416.2268.18 
 zzz
321
21210 
 zzz Therefore,
• x(0) = 0,
• x(1) = 10,
• x(2) = 17,
• x(3) = 18.4
• x(4) = 18.68

32. Inverse Z-Transform: Partial Fraction Expansion
 Assume that a given z-transform can be expressed as
 Apply partial fractional expansion
 First term exist only if M>N
 Br is obtained by long division
 Second term represents all first order poles
 Third term represents an order s pole
 There will be a similar term for every high-order pole
 Each term can be inverse transformed by inspection
32
 






 N
0k
k
k
M
0k
k
k
za
zb
zX
 
  










s
1m
m1
i
m
N
ik,1k
1
k
k
NM
0r
r
r
zd1
C
zd1
A
zBzX

33. Inverse Z-Transform: Partial Fraction Expansion
 Coefficients are given as
 Easier to understand with examples
33
 
  










s
1m
m1
i
m
N
ik,1k
1
k
k
NM
0r
r
r
zd1
C
zd1
A
zBzX
    kdzkk zXzdA 

 1
1
   
     1
1
1
!
1















idw
s
ims
ms
ms
i
m wXwd
dw
d
dms
C

34. Example:
 Find x(n) if X(z) is given by:
 Solution: We first expand X(z)/z into partial fractions as follows :
 Then we obtain
 then
34
 
  2.01
10


zz
z
zX
 
      2.0
5.12
1
5.12
2.01
10






zzzzz
zX
  









2.01
5.12
z
z
z
z
zX
  )()2.0(5.12)(5.12 nununx n

  )(11
nZ 
 nu
z
z
Z 







1
1
 nua
az
z
Z n







1

35. Another Solution
 In this example, if X(z), rather than X(z)/z, is expanded into partial
fractions, then we obtain:
 However, by use of the shifting theorem we find :
 Then
35
 
   2.0
5.2
1
5.12
2.01
10






zzzz
z
zX
  








 
2.0
5.2
1
5.12 11
z
z
z
z
z
zzX
  )1(2.015.12)1()2.0(
2.0
5.2
)1(5.12)(
)1()2.0(5.2)1(5.12)( 1

 
nunununx
nununx
nn
n
 
 
 
 
 
..
..
48.124
4.123
122
101
00





x
x
x
x
x
     o
n
nnxzXzZ o
1

36. Example:
 Find x(n) if X(z) is given by:
 Solution: Expand X(z)/z into partial fraction as follows:
 Then:
 By referring to tables we find that x(n) is given by:
 and therefore,
 x(0) = 0 -1 +3 = 2
 x(1) = 18 – 2 + 3= 19
36
 
   12
2
2
3



zz
zz
zX
 
      1
3
2
1
2
9
12
12
22
2









zzzzz
z
z
zX
 
     1
3
22
9
2






z
z
z
z
z
z
zX
  )(3)()2()()2(5.4 nununrnx nn

 nnu
n
nn
nr 


 

00
0
)(

 
 nr
z
z
Z 







2
1
1
 
 nra
az
az
Z n








2
1

37. Solution of Difference Equations
37

38. Transfer Function Representation
 First – Order Case
 Let y[n] + ay[n-1] = bx[n]
 Then take the z-transform to get:
• Y(z) + a z-1Y(z) = b X(z)
 Simplifying:
• Y(z) (1 + az-1) = b X(z)
• Y(z) = (b X(z))/ (1 + az-1)
 And we have the transfer function H(z):
• Y(z) = H(z) X(z)
• so  H(z) = b z /(z + a)
38

39. Step Response of a First Order System
 If a discrete-Time system has the following difference equation y[n] + ay[n-1]
= bx[n]
 Find the impulse response
 Solution:
 In time Domain:
 y[n] = bx[n] - ay[n-1]
 For n<0  y[n] = 0
 at n=0  y[0] = b δ[0] – ay[-1] = b
 at n=1  y[1] = b δ[1] – ay[0] = -ab
 at n=2  y[2] = b δ[2] – ay[1] = a2b
 at n=3  y[3] = b δ[3] – ay[2] = -a3b
 For n>0  y[n] = (-a)n b
39

40. Step Response of a First Order System
 If a discrete-Time system has the following difference equation
y[n] + ay[n-1] = bx[n]
 Find the impulse response
 Solution:
 In z-Domain:
 By z-transform  Y[z] + az-1 y[z] = bX[z]
Y[z][1 + az-1] = bX[z]
Y[z] = b X[z] /[1 + az-1]
 Since, x[n] = δ[n]  X[z] = 1
Y[z] = b /[1 + az-1]
 By inverse z-transform
y[n] = (-a)nb u(n)
40

41. Second-Order Case
 Consider the following difference equation:
 Y[n] +a1y[n-1] +a2y[n-2] = b0x[n] + b1x[n-1]
• H(z) = (b0z2 + b1z)/(z2 + a1z + a2)
41

42. Nth-Order System
 y[n] +a1y[n-1] +…+aNy[n-N] =b0x[n]+…+bMx[n-M]
 And H(z) = B(z)/A(z)
 Where H(z) =(b0+…+bMz-M)/(1 + a1z-1+ …+aNz-N)
 Now multiply by (zN/zN)
 H(z) = (b0zN +…+bMzN-M)/ (zN + a1zN-1 + …+ aN)
 So we have
 B(z) = (b0zN +…+bMzN-M)
 A(z) = (zN + a1zN-1 + …+ aN)
42

43. System Representation in the z-
Domain
43

44. The Transfer Function of a Discrete-Time System
 The transfer function of a discrete-time system is defined as the ratio of the z
transform of the response to the z transform of the excitation as shown in Fig. 3.6.
 Consider a linear, time-invariant, discrete-time system, and let x(n), y(n), and h(n)
be the excitation, response, and impulse response, respectively. From the
convolution summation in Eq. (2.7), we have:
 and, therefore, from the real-convolution theorem, we get:
 or
44
     



k
knhkxny
      





 

k
knhkxZnyZ
Fig. 3.6 The transfer function of a discrete-time system.
H(z)
X(z) Y(z)
Input
Excitation
Output
Response
System Function
Transfer Function         
 zX
zY
zHzHzXzY 

45. Derivation of the Transfer Function from Difference Equation
 A causal, linear, time-invariant, recursive discrete-time system can be
represented by the difference equation Eq. (2.10) as:
 Apply the z transform for both sides, we get:
45
      





  
 
N
i
M
i
ii inybinxaZnyZ
0 1
 
 


N
i
M
i
i
i
i
i zbzYzazXzY
0 1
)()()(
   
 
 
 zD
zN
zb
za
zX
zY
zH M
i
i
i
N
i
i
i









1
0
1
      
 

N
i
M
i
ii inyainxany
0 1

46. Derivation of the Transfer Function from the System Network
 The z-domain characterizations of the unit delay, the adder, and the multiplier
are obtained from Fig. 2.9 and shown in Fig. 3.7.
46Fig. 3.7 Elements of discrete-time systems (a) Unit delay, (b) Adder, (c) Multiplier.

47. The Transfer Function of a Discrete-Time System
 Example 3.14 Find the transfer function of the system shown in Fig. 3.8.
 Solution From Fig. 3.8, we can write:
 Therefore,
47Fig. 3.8 Second-order recursive system.
  )(
4
1
)(
2
1
)( 21
zWzzWzzXzW 

    )()1()()( 11
zWzzYzWzzWzY 

 
21
4
1
2
1
1
)(



zz
zX
zW
 
4
1
2
1
)1(
)(
)(
2



zz
zz
zX
zY
zH

48. The Frequency Response of a Discrete-Time System
 If x(n) is absolutely sumable, that is:
 Then the discrete-time Fourier transform is given by:
 The operator [.] transforms a discrete time signal x(n) into a complex valued
continuous function X(ej) of real variable  = 𝟐𝝅𝒇 .
 X(ej) is a complex valued continuous function.
  = 𝟐𝝅𝒇 [rad/sec] and f is the digital frequency measured in [C/S or Hz].
 To find the frequency response 𝑯(𝒆𝒋) from the transfer function, H(z), replace z
by 𝒆𝒋 as follow:
48
  

0n
nx
      




0n
njj
enxnxeX 
   


j
ezn
njj
zHenheH




  )(
 nx      nhnxny 
 j
eX  j
eH
 nh
      jjj
eHeXeY 
    j
eYny 1


49. Example
 For the system described by its impulse response do:
a) Derive the frequency response, 𝐻(𝑒𝑗) .
b) Evaluate and draw its magnitude and phase at 𝜔 = 0, 0.25𝜋, 0.5𝜋, 0.75𝜋 𝑎𝑛𝑑 𝜋.
 Solution
a) Using DTFT Eq.(3.17 and 3.18), we find :
Hence:
Or
and
49
     nunh
n
9.0
     
         

sin9.0cos9.01
1
9.01
1
9.0
0 je
eenheH j
n n
njnnjj



 





 
 
                  22222
sincos9.0cos9.021
1
sin9.0cos9.01
1





j
eH
   

cos8.181.1
1

j
eH
   
 









cos9.01
sin9.0
arctanj
eH

50. Example (Cont.)
b) The magnitude and phase plots are shown in Fig. 3.9.
50
Fig. 3.9 The frequency response of a discrete-time system:
(a) The magnitude plot and (b) The phase plot.
(a) (b)

51. Causality
 H(z) is described as:
 H(z) is causal If the order of the denominator, D(z) ≥ the order of numerator N(z).
 Example 3.16: Check the system causality if its transfer function, H(z) is:
a) b)
 Solution
a) The system is not causal as the O[N(z)] > O[D(z)].
b) H(z) can be rewritten as:
Therefore, he system is causal as the O[N(z)] = O[D(z)].
51
   
 zD
zN
zH 
 
8
1
4
1
2
2
23



zz
zzz
zH   1
1 21
1
3
1
1
1

 



zz
zH
 
1)3/5(
)3/5(2
21
1
)3/1(1
1
2
2
11






 
zz
zz
zz
zH

52. Stability
 A LTI system is said to be stable If the poles of H(z) lies inside the unit
circle (z=1)
 Example 3.16 Check the system stability if its transfer function, H(z) is:
a) b)
 Solution
a) H(z) can be rewritten as:
This system is unstable as it has one pole at z = 4.
a) H(z) can be rewritten as:
This system is stable as it has one pole at z = 0.5.
52
  1
41
1



z
zH   


 

1
1
)2/1(1
)2/1(1
z
z
zH
 
)2/1(
)2/1(
)2/1(1
)2/1(1
1
1





 

z
z
z
z
zH
 
441
1
1



 
z
z
z
zH

53. 53