CHAPTER 5
Z-Transform.
School of Electrical System Engineering, UniMAP
NORSHAFINASH BT SAUDIN
[email protected]
EKT 230
1
5.0 Z-Transform.
5.1 Introduction.
5.2 The z-Transform.
5.2.1 Convergence.
5.2.2 z-Plane.
5.2.3 Poles and Zeros.
5.3 Properties of Region of Converges (ROC).
5.4 Properties of z-Transform.
5.5 Inverse of z-Transform.
5.6 Transfer Function.
5.7 Causality and Stability.
5.8 Discrete and Continuous Time Transformation.
5.9 Unilateral z-Transformation.
2
5.1 Introduction.
 In Laplace Transform we evaluate the complex sinusoidal
representation of a continuous signal.
 In the z-Transform, it is on the complex sinusoidal
representation of a discrete-time signal.
3
5.2 The z-Transform.
 Let z = rejW be a complex number with magnitude r and angle W.
The signal x[n]=zn is a complex exponential signal.
 We may write,
x[n]  r cosWn   jr sin Wn .
n
n
 The real part of x[n] is an exponential damped cosine.
 And the imaginary part is an exponential damped sine as shown
in Figure 5.1.
 The positive number of r determine the damping factor and W is the
sinusoidal frequency.
4
Cont’d…
Figure 5.1: Real and imaginary parts of the signal zn.
5
Cont’d…
 The transfer function,
H ( z) 

k
h
[
k
]
z

k  
 The z-transform of an arbitrary signal x[n] is,
X ( z) 

n
x
[
n
]
z

n  
 The inverse z-transform is,
1
n 1
x[n] 
X
(
z
)
z
dz

2j
6
5.2.1 Convergence.
 The region of converges (ROC) is the range of r for which the
below equation is satisfied:


x[n]r  n  .
n  
z
Xz  A
, z 
z
A
Az1
1
Xz 
 1
, z
1  z z  

7
5.2.2 z-Plane.
 It is convenience to represent the complex frequency z as a location
in z-plane as shown in Figure 5.2.
Figure 5.2: The z-plane. A point z = rejW is located at a distance r– from the origin and
an angle W relative to the real axis.
 The point z=rej W is located at a distance r from the origin and the
angle Wfrom the positive real axis.
Figure 5.3: The unit circle,
z = ejW, in the z-plane.
 Figure 5.3 is a unit circle in the z-plane. z=rej W describes a circle
8
of unit radius centered on the origin in the z-plane.
5.2.3 Poles and Zeros.
 The z-transform form, a ratio of two polynomial in z-1,
b0  b1 z 1  ...  bM z  M
X ( z) 
a0  a1 z 1  ...  a N z  N
 The X(z) can be rewrite as a product of terms involving the roots
of the numerator and denominator polynomial,
~ M
b k 1 (1  ck z 1 )
X ( z) 
N
1
(
1

d
z
 Where,
k 1 k )
~
b  b0 / a0
ck= the roots of the numerator polynomial and the zeros (o) of
X(z).
dk= the roots of the denominator polynomial and the poles (x) of
X(z)
9
5.3 The Properties of ROC.
Figure 5.4: The relationship between the ROC and the time extent of a signal.
(a) A right-sided signal has an ROC of the form |z| > r+.
(b) A left-sided signal has an ROC of the form |z| < r–.
(c) A two-sided signal has an ROC of the form r+ < |z| < r–.
10
Cont’d…
Pg 565 Text
Figure 5.5: ROCs for Example 7.5 (text).
(a) Two-sided signal x[n] has ROC in between the poles.
(b) Right-sided signal y[n] has ROC outside of the circle containing the pole of
largest magnitude.
11
(c) Left-sided signal w[n] has ROC inside the circle containing the pole of smallest
magnitude.
Cont’d…
 Taking a path analogous to that used the development of the
 Laplace transform, the z transform of the causal DT signal is
A un  ,   0
n



n
n0
n0
Xz  A   n un z n  A  n z n  A 
n

 
 z 
and the series converges if |z| > |a|. This
defines the ROC as the exterior of a circle in
the z plane centered at the origin, of radius|a|.
 The z transform is
z
Xz  A
, z 
z
Causal
12
Cont’d…
 By similar reasoning, the z transform and region of convergence of
the anti-causal signal below, are
A un  ,   0
n
A
Az1
1
Xz 
 1
, z
1  z z  

Anti-Causal
13
Example 5.0: Two-Sided Exponential Sequence
Example
 1
1
xn     un -   u- n - 1
n

n
3
2
ROC : 
0
 1 1 
 1 1 

z



 z 
n

1
3



 3

1 
Solution:

z




1
3

n0 
1  z 1
3
n
 1 1 
 z  


n   2
1
Xz  
 1 1 
 z 
2


1

  z 1 
2

1
1  z 1
2
1
1

1
1
1  z  1 1  z 1
3
2
Time Domain -> s Domain


1
1
1  z 1
3
0

1
1
1  z 1
2
1 

2z z 

12 


1 
1

 z   z  
3 
2

1 1
z
1
3
1
 z
3
ROC :
1 1
z
1
2
1
 z
2
Im
1

3x
oo1
1
2
x
12
14
Re
5.4 Properties of z-Transform.
 Most properties of z-transform are similar to the DTFT. We
assumed that,
z
x[n] 

X  z , with
y[n] 
 Y  z , with
z
ROC
Rx
ROC
Ry
 The effect of an operation on the ROC is described by a change in
the radii of the ROC boundaries.
(1) Linearity,
zu
ax[n]  by[n] 
aX z   bY z , with ROC
atleast
Rx  R y
15
Cont’d…
(2) Time Reversal.
1
x[n] 
 X  , with ROC
z
z
1
.
Rx
(3) Time Shift.
x[n  n0 ] 
 z
z
 n0
X z 
with ROC Rx, except possibly z=0 or |z|= infinity.
16
Cont’d…
(4) Multiplication by an Exponential Sequence.
z
 x[n] 
 X  , with ROC |  | R x
a
z
n
(5) Convolution.
z
x[n] * y[n] 

X z Y z , with ROC
atleast
Rx  R y
(6) Differentiation in the z-Domain.
d
nx[n] 
  z
X  z, , with
dz
z
ROC
Rx
17
5.5 The Inverse Z-Transform.
 There are two common methods;
5.5.1 Partial-Fraction Expression.
5.5.2 Power-Series Expansion.
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5.5.1 Partial-Fraction Expansion.
Example 5.1: Inversion by Partial-Fraction Expansion.
Find the inverse z-transform of,
with ROC 1<|z|<2.
1  z 1  z 2
X ( z) 
(1 
1 1
z )(1  2 z 1 )(1  z 1 )
2
Figure 5.6: Locations of poles and ROC.
Solution:
Step 1: Use the partial fraction expansion of Z(s) to write
X ( z) 
Solving the A, B and C will give
A
B
C


1 1
(1  2 z 1 ) (1  z 1 )
(1  z )
2
X ( z) 
1
2
2


1 1
(1  2 z 1 ) (1  z 1 )
(1  z )
2
19
Cont’d…
Step 2: Find the Inverse z-Transform for each Terms.
- The ROC has a radius greater than the pole at z=1/2, it is the rightsided inverse z-transform.
n
1
1
z

  u[n] 
1 1
2
(1  z )
2
- The ROC has a radius less than the pole at z=2, it is the left-sided
inverse z-transform.
2
z
n
 2(2) u[n  1] 

(1  2 z 1 )
- Finally the ROC has a radius greater than the pole at z=1, it is the
right-sided inverse z-transform.
2
 2u[n] 

(1  z 1 )
z
20
Cont’d…
Step 3: Combining the Terms.
1
2
2
X ( z) 


1
1 1
(1  2 z ) (1  z 1 )
(1  z )
2
n
1
n
x[n]    u[n]  2(2) u[n  1]  2u[n].
2
 .
21
Example 5.2: Inversion of Improper Rational Function.
Find the inverse z-transform of,
z 3  10 z 2  4 z  4
with ROC |z|<1.
X ( z) 
2
2z  2z  4
Figure 5.7: Locations of poles and ROC.
Solution:
Step 1: Convert X(z) into Ratio of Polynomial in z-1.
Factor z3 from numerator and 2z2 from denominator.
z3
X ( z)  2
2z
 1  10 z 1  4 z 2  4 z 3 


1
2
1  z  2z


z  1  10 z 1  4 z 2  4 z 3 

X ( z )  
1
2
2
1  z  2z

22
Step 2: Use long division to reduce order of numerator
polynomial.
Factor z3 from numerator and 2z2 from denominator.
 2 z  2  z 1
 2 z 1  3 __________
 1 4 z 3  4 z  2  10 z 1  1
4 z 3  2 z  2  2 z 1
 6 z  2  8 z 1  1
 6 z  2  3 z 1  3
 5 z 1  2
1
1  10 z 1  4 z 2  4 z 3

5
z
2
1
 2 z  3 
1
2
1  z  2z
1  z 1  2 z  2
1

5
z
2
1
  2z  3 
(1  z 1 )(1  2 z 1 )
23
Factor z3 from numerator and 2z2 from denominator.
 5 z 1  2
1
3


1
2
1
1  z  2z
(1  z ) (1  2 z 1 )
1
X ( z )  zW ( z )
2
We define,
W ( z )  2 z 1  3 
Where,
1
3

1  z 1 (1  2 z 1 )
With ROC|z|<1
Step 3: Find the Inverse z-Transform for each Terms.
w[n]  2 [n  1]  3 [n]  (1) n u[n  1]  3(2) n u[n  1]
1
w[n  1]
2
3
1
x[n]   [n]   [n  1]  (1) n 1 u[n  2]  3(2) n u[n  2]
2
2
24
x[n] 
 .
5.6 Transfer Function.
 The transfer function is defined as the z-transform of the impulse
response. y[n]= h[n]*x[n]
 Take the z-transform of both sides of the equation and use the
convolution properties result in,
Y ( z)  H ( z) X ( z)
 Rearrange the above equation result in the ratio of the z-transform
of the output signal to the z-transform of the input signal.
Y ( z)
H ( z) 
X ( z)
 The definition applies at all z in the ROC of X(z) and Y(z) for
which X(z) is nonzero.
25
Example 5.3: Find the Transfer Function.
Find the transfer function and the impulse response of a causal LTI
system if the input to the system is
n
 1
x[n]     u[n] and
 3
output,
n
1
y[n]  3(1) u[n]    u[n].
3
n
Solution:
Step 1: Find the z-Transform of the input X(z) and output Y(z).
1
X ( z) 
  1  1 
1    z 
 3 
3
Y ( z) 

(1  z 1 )
1
1
(1  z 1 )
3
4

1
(1  z 1 )(1  z 1 )
3
With ROC |z|>1/3
With ROC |z|>1.
26
Step 2: Solve for H(z).
  1  1 
41    z 
 3 

H ( z) 
 1  1  , with ROC |z|>1.
1 
(1  z )1    z 
  3 
Solve for impulse response h[n],
H ( z) 
2
2

(1  z 1 )   1  1 
1    z 
  3   , with ROC |z|>1.
So the impulse response h[n] is,
h[n]  2(1) u[n]  2(1 / 3) u[n].
n
n
 .
27
5.7 Causality and Stability.
 The impulse response of a causal system is zero for n<0.
 The impulse response of a casual LTI system is determined from the
transfer function by using right-sided inverse transform.
 The poles inside the unit circle, contributes an exponentially
decaying term to the impulse response.
 The poles outside the unit circle, contributes an exponentially
increasing term.
Figure 5.8: Pole and impulse response
characteristic of a causal system.
(a) A pole inside the unit circle contributes
an exponentially decaying term to the
impulse response.
(b) A pole outside the unit circle
contributes an exponentially increasing
term to the impulse response.
28
Cont’d…
 Stable system; the impulse response is absolute summable and the
DTFT of impulse response exist.
 The impulse response of a casual LTI system is determined from
the transfer function by using right-sided inverse transform.
 The poles inside the unit circle, contributes a right-sided decaying
exponential term to the impulse response.
 The poles outside the unit circle, contributes a left-sided decaying
exponentially term to the impulse response.
 Refer to Figure 5.9.
Figure 5.9: Pole and impulse response
characteristics for a stable system.
(a) A pole inside the unit circle contributes a
right-sided term to the impulse response.
(b) A pole outside the unit circle contributes
a left-sided term to the impulse response.
29
Cont’d…
Stable/Causal ?
 From the ROC below the system is stable, because all the poles
within the unit circle and causal because the right-sided
decaying exponential in terms of impulse response.
Figure 7.16: A system that is both stable and causal must have all its poles
inside the unit circle in the z-plane, as illustrated here.
30
5.8 Implementing Discrete-Time
LTI System.
 The system is represented by the differential equation.
y[n]  a1 y[n  1]  a2 y[n  2]  b0 x[n]  b1 x[n  1]  b2 x[n  2]
 Taking the z-transform of difference equation gives,
(1  a1 z 1  a2 z 2 )Y ( z )  (b0  b1 z 1  b2 z 2 ) X ( z )
 The transfer function of the system,
Y ( z)
H ( z) 
X ( z)
b0  b1 z 1  b2 z 2
H ( z) 
1  a1 z 1  a2 z 2
31
Example 5.4 : Causality and Stability.
Can this system be both stable and causal?
Solution:
Step 1: Find the characteristic equation of the system.
From the plot, the system is unstable
because the pole at z = 1.2 is outside
the unit circle.
 .
32
Cont’d…
(1  a1 z 1  a2 z 2 )Y ( z )  (b0  b1 z 1  b2 z 2 ) X ( z )
Figure 7.26: Block diagram of the transfer function.
33
Cont’d…
Figure 7.27: Development of the direct form II representation of an LTI system. (a)
Representation of the transfer function H(z) as H2(z)H1(z).
(b) Direct form II implementation of the transfer function H(z) obtained from (a) by
34
collapsing the two sets of z–1 blocks.
5.9 The Unilateral z-Transform.
 The unilateral z-Transform of a signal x[n] is defined as,

Xz   xnz
n
n0
Properties.
 If two causal DT signals form these transform pairs,
gn 
Gz and hn 
Hz
Z
Z
(1) Linearity.
 gn   hn
 Gz   Hz
Z
(2) Time Shifting.
gn  n0 
z n0 Gz , n0  0
Z
n 0 1

Z
n0
 m 
gn  n0 
z Gz   gm z  , n0  0


m 0
35