Chapter 2
The z Transform
Introduction
• A mathematical tool used to the analysis and synthesis
of discrete-time control systems: the z transform.
– The Laplace transform in continuous-time systems.
– With the z transform method, the solutions to linear difference
equations become algebraic in nature.
• Discrete-Time Signals
– The sampled signal: x(0), x(T), x(2T),…, where T is the sampling
period.
– If the system involves an iterative process carried out by a digital
computer, the signal involved is a number sequence: x(0), x(1),
x(2),…. – it can be considered as a sampled signal of x(t) when
the sampling period T is 1 sec.
41-142
Chap. 2 The z Transform
2
The z Transform
• One-sided z Transform

X ( z )  Z x(t )  Z x(kT )   x(kT ) z  k
k 0

X ( z )  Z x(k )   x(k ) z  k
k 0
– For most engineering applications the one-sided z transform will
have a convenient close-form solution in its region of
convergence.
X ( z )  x(0)  x(T ) z 1  x(2T ) z 2    x(kT ) z  k  
– The z transform of any continuous-time function may be written
in the series form by inspection.
– The inverse transform can be obtained by inspection if X(z) is
given in the series form.
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Chap. 2 The z Transform
3
z Transforms of Elementary Functions
• Unit-Step Function
1(t ) 0  t
x(t )  
 0 t0

X ( z )  Z 1(t )  1z
k 0
k

  z k
k 0
 1  z 1  z  2  z 3  
1
1  z 1
z

z 1

– The series converges if z  1
– Unit step sequence
1 0  t
1(k )  
0 t  0
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Chap. 2 The z Transform
4
z Transforms of Elementary Functions
• Unit-Ramp Function
t 0  t
x(t )  
0 t  0
x(kT )  kT , k  0,1,2,

X ( z )  Z t    x(kT ) z
k 0

  kTz
k 0
k

 T  kz k
k 0
 T ( z 1  2 z  2  3 z 3  )
T

41-142
k
z 1
1  z 
1 2
Tz
z  12
Chap. 2 The z Transform
5
z Transforms of Elementary Functions
• Polynomial Function ak
a k
x(t )  
0
k  0,1,2,
k 0
 

X ( z )  Z a   x(k ) z
k
k
k 0

  a k z k
k 0
 1  az 1  a 2 z  2  a 3 z 3  
1
1  az 1
z

za

• Exponential Function
e  at
x(t )  
 0
0t
t0
x(kT )  e  akT ,
41-142
k  0,1,2,
    x(kT ) z
X ( z)  Z e
 at

k 0
k

  e  akT z  k
k 0
 1  e  aT z 1  e  2 aT z  2  e 3aT z 3  

1
1  e  aT z 1
z

z  e  aT
Chap. 2 The z Transform
6
z Transforms of Elementary Functions
• Sinusoidal Function
sin t 0  t
x(t )  
t0
 0
 1 j t

X ( z )  Z sin t   Z 
e  e  jt 
2 j

1 
1
1





2 j  1  e jT z 1 1  e  jT z 1 

e jt  cos t  j sin t
e  jt  cos t  j sin t
sin t 
 

1 jt  jt
e e
2j
Z e  at 
1
1  e  aT z 1


1
(e jT  e  jT ) z 1

2 j 1  (e jT  e  jT ) z 1  z  2
z 1 sin T

1  2 z 1 cos T  z  2
z sin T
 2
z  2 z cos T  1
See Table 2-1 on pages 29-30
41-142
Chap. 2 The z Transform
7
Important Properties and Theorems of the z Transform
• Multiplication by a Constant

Z ax(t )   ax(kT ) z
k
k 0

 a  x(kT ) z  k  aX ( z )
k 0
• Linearity of the z Transform
x(k )  f (k )  g (k )
X ( z )  F ( z )  G ( z )
X ( z )  Z x(k )  Z f (k )  g (k )

  f (k )  g (k )z  k
k 0

   f (k ) z
k 0
k

  g (k )z  k
k 0
 Z  f (k )  Z g (k )
 F ( z )  G ( z )
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Chap. 2 The z Transform
8
Important Properties and Theorems of the z Transform
• Multiplication by ak



Z a x(k )   a x(k ) z
k
k
k 0
k

  x(k )( a 1 z )  k
k 0
 X (a 1 z )
• Shifting Theorem
Z x(t  nT )  z X ( z )
n

Z x(t  nT )   x(kT  nT ) z  k
k 0
z
n

 x(kT  nT ) z
( k  n )
k 0
z
n

 x(mT ) z
m
m n
z
n

 x(mT ) z
m
m 0
 z n X ( z )
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Chap. 2 The z Transform
9
Important Properties and Theorems of the z Transform
• Shifting Theorem (cont.)
n 1


Z x(t  nT )  z  X ( z )   x(kT ) z k 
k 0


n

Z x(t  nT )   x(kT  nT ) z  k
k 0
z

n
 x(kT  nT ) z
( k  n )
k 0




n 1
n 1


k
k
( k  n )
n
  x(kT ) z   x(kT ) z 
 z  x(kT  nT ) z


0
k 0
 k 0
k 



n1
x ( mT ) z m   x ( kT ) z k



k 0
mn
n 1


k
 z  x(kT ) z   x(kT ) z  k 
k 0

 k 0
n
n 1


 z  X ( z )   x(kT ) z  k 
k 0


n
41-142
Chap. 2 The z Transform
10
Important Properties and Theorems of the z Transform
• Example 2-3
1
z 1
Z 1(t  T )  z Z 1(t )  z

1
1 z
1  z 1
1
1
1
z 4
Z 1(t  4T )  z Z 1(t )  z

1
1 z
1  z 1
4
4
– Note that z- 1 represents a delay of 1 sampling period T,
regardless of the value of T.
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Chap. 2 The z Transform
11
Important Properties and Theorems of the z Transform
• Complex Translation Theorem



Z e x(t )   x(kT )e
 at
k 0
 akT
z
k

  x(kT )( ze aT )  k  X ( ze aT )
k 0
• Initial Value Theorem
x(0)  lim X ( z )
z 

X ( z )   x(k ) z  k  x(0)  x(1) z 1  x(2) z  2  
k 0
41-142
Chap. 2 The z Transform
12
Important Properties and Theorems of the z Transform
• Final Value Theorem

lim x(k )  lim (1  z 1 ) X ( z )
k 
z 1


Z x(k )  X ( z )   x(k ) z  k
k 0

Z x(k  1)  z X ( z )   x(k  1) z  k
1
k 0
Hence

 x(k ) z
k 0
k

  x(k  1) z  k  X ( z )  z 1 X ( z )
k 0




k
lim  x(k ) z   x(k  1) z  k   lim X ( z )  z 1 X ( z )
z 1
k 0
 k 0
 z 1


 x(k )  x(k  1)  x(0)  x(1)  x(1)  x(0)  x(2)  x(1)    x()  lim x(k )
k 
k 0
41-142
Chap. 2 The z Transform
13
Important Properties and Theorems of the z Transform
• Example 2-9
Determine the final value x() of
1
1
X ( z) 

,
1
 aT 1
1 z
1 e z

x()  lim (1  z 1 ) X ( z )
z 1
a0


1
 1

 lim (1  z 1 )

1
 aT 1  
z 1
1

z
1

e
z 



1  z 1 

 lim 1 
 aT 1 
z 1
1

e
z


1
x(t )  1  e  at


x()  lim 1  e  at  1
t 
41-142
Chap. 2 The z Transform
14
The Inverse z Transform
• The notation for the inverse z transform is Z-1. The
inverse z transform of X(z) yields the corresponding time
sequence x(k).
– Only the time sequence at the sampling instants is obtained from
the inverse z transform.
– The inverse z transform yields a time sequence that specifies the
values of x(t) only at discrete instants of time.
• Many different time functions x(t) can have the same x(kT).
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Chap. 2 The z Transform
15
The Inverse z Transform
• A z transform table: an obvious method for finding the
inverse z transform.
– Table 2-2
• Four methods for obtaining the inverse z transform:
–
–
–
–
Direct division method
Computational method
Partial-fraction-expansion method
Inversion integral method
41-142
Chap. 2 The z Transform
16
The Inverse z Transform
• Poles and Zeros in the z Plane
b0 z m  b1 z m 1    am
X ( z)  n
z  a1 z n 1    an
b0 ( z  z1 )( z  z 2 )  ( z  z m )

( z  p1 )( z  p2 )  ( z  pn )
– The locations of the poles and zeros of X(z) determine the
characteristics of x(k), the sequence of values or numbers.
– We often use a graphical display in the z plane of the locations of
the poles and zeros of X(z).
z 2  0.5z
z ( z  0.5)
X ( z)  2

z  3z  2 ( z  1)( z  2)
poles at z=-1, z=-2
zeros at z=0, z=-0.5
1  0.5 z 1
1  0.5z 1
X ( z) 

1
2
1  3z  2 z
(1  z 1 )( z  2 z 1 )
41-142
Chap. 2 The z Transform
17
The Inverse z Transform
• Direct Division Method

X ( z )   x(k ) z  k
k 0
 x(0)  x(1) z 1  x(2) z  2    x(k ) z k  
– Example 2-10
10 z  5
X ( z) 
( z  1)( z  0.2)
10 z 1  5 z 2
X ( z) 
1  1.2 z 1  0.2 z 2
X ( z )  10 z 1  17 z 2  18.4 z 3  18.68z 4  
x ( 0)  0

X ( z )   x(k ) z
k 0
k
x (1)  10
x ( 2)  17
x (3)  18.4
x ( 4)  18.68
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Chap. 2 The z Transform
18
The Inverse z Transform
• Computational Method
– Consider a system
0.4673z 1  0.3393z 2
G( z ) 
1  1.5327 z 1  0.6607 z 2
– For the Kronecker delta input, X(z) =1
Y ( z)
0.4673z 1  0.3393z 2
0.4673z  0.3393
G( z ) 


X ( z ) 1  1.5327 z 1  0.6607 z 2 z 2  1.5327 z  0.6607
– The inverse z transform of G(z) is given by y(0),y(1),y(2),…
– Two approaches to obtain the inverse transform:
• MATLAB approach
• Difference equation approach
41-142
Chap. 2 The z Transform
19
The Inverse z Transform
• Computational Method (cont.)
– MATLAB approach
0.4673z 1  0.3393z 2
0.4673z  0.3393
Y ( z )  G( z ) 

1  1.5327 z 1  0.6607 z 2 z 2  1.5327 z  0.6607
num=[0 0.4673 -0.3393]
den=[1 -1.5327 0.6607]
x=[1 zeros(1,40)]
y=filter(num,den,x)
y =
0 0.4673 0.3769 0.2690 0.1632 0.0725 0.0032 -0.0429
-0.0679 -0.0758 -0.0712 -0.0591 -0.0436 -0.0277 -0.0137 -0.0027
0.0050 0.0094 0.0111 0.0108 0.0092 0.0070 0.0046 0.0025
0.0007 -0.0005 -0.0013 -0.0016 -0.0016 -0.0014 -0.0011 -0.0008
-0.0004 -0.0002 0.0000 0.0002 0.0002 0.0002 0.0002 0.0002
0.0001
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Chap. 2 The z Transform
20
The Inverse z Transform
• Computational Method (cont.)
– MATLAB approach
41-142
Chap. 2 The z Transform
21
The Inverse z Transform
• Computational Method (cont.)
– Difference Equation Approach
( z 2  1.5327 z  0.6607)Y ( z )  (0.4673z  0.3393) X ( z )
y (k  2)  1.5327 y (k  1)  0.6607 y (k )  0.4673x(k  1)  0.3393x(k )
where x(0)  1 and x(k )  0 for k  0, and y (k )  0 for k  0.
• for k = -2
y (0)  1.5327 y (1)  0.6607 y (2)  0.4673x(1)  0.3393x(2)
y (0)  0
• for k = -1
y (1)  1.5327 y (0)  0.6607 y (1)  0.4673x(0)  0.3393x(1)
y (1)  0.4673
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Chap. 2 The z Transform
22
The Inverse z Transform
• Partial-Fraction-Expansion Method
– Each expanded term has a form that may easily be found from
commonly available z transform tables.
b0 z m  b1 z m1    bm
b0 z m  b1 z m1    bm
X ( z) 
X ( z)  n
, mn
n 1
( z  p1 )( z  p2 )  ( z  pn )
z  a1 z    an
– for simple pole
an
X ( z)
a1
a2



z
z  p1 z  p2
z  pn
– for multiple pole
X ( z)
c1
c2


z
z  p1 2 z  p1
41-142
X ( z) 

ai   z  pi 
z  z  pi


2 X ( z) 
c1   z  p1 
z  z  p1

d 
2 X ( z) 
c2    z  p1 

dz
z

  z  p1

Chap. 2 The z Transform
23
The Inverse z Transform
• Partial-Fraction-Expansion Method (cont.)
– Example 2-14
1  e z
X ( z) 
z  1z  e 
 aT
 aT
X ( z)
1
1


z
z  1 z  e  aT
 1 
Z 1 
1
1 
1  z 
X ( z) 
1
1

1  z 1 1  e  aT z 1
x(kT )  1  e  akT ,
k  0,1,2,...
1


Z 1 
 e  akT
 aT 1 
1  e z 
41-142
Chap. 2 The z Transform
24
The Inverse z Transform
• Inversion Integral Method
Z 1 X ( z )  x(kT )  x(k ) 
1
k 1
X
(
z
)
z
dz

2j C
x(kT )  x(k )  K1  K 2    K m
m

  residue of X ( z ) z k 1 at pole z  zi of X ( z ) z k 1

i 1
– for a simple pole

K  lim z  zi X ( z ) z k 1
z  zi

– for a multiple pole
1
d q 1
q
K
lim q 1 z  z j  X ( z ) z k 1
q  1! zz j dz

41-142
Chap. 2 The z Transform

25
The Inverse z Transform
• Inversion Integral Method (cont.)
– Example 2-16



1  e  aT z k
z 1  e  aT
k 1
X ( z) z 
X ( z) 
z  1z  eaT 
z  1 z  eaT
• Two simple poles: z=z1=1 and z=z2=e-aT




(1  e  aT ) z k
x(k )   residue of
at
pole
z

z
i
( z  1)( z  e aT )
i 1 

 K1  K 2

(1  e  aT ) z k 
K1  lim ( z  1)
1
 aT 
z 1
( z  1)( z  e ) 

2

(1  e  aT ) z k 
 aT
K 2  limaT ( z  e )
 e  akT
 aT 
z e
( z  1)( z  e ) 

x(kT )  K1  K 2  1  e  akT
41-142
Chap. 2 The z Transform
26
z Transform Method for Solving Difference Equations
• The linear time-invariant discrete-time system
characterized by the following linear difference equation:
x(k )  a1 x(k  1)    an x(k  n)  b0u(k )  b1u(k  1)    bnu(k  n)
Discrete function
41-142
z Transform
x ( k  3)
z 3 X ( z )  z 3 x(0)  z 2 x(1)  zx(2)
x ( k  2)
z 2 X ( z )  z 2 x(0)  zx(1)
x(k  1)
zX ( z )  zx (0)
x(k )
X (z )
x(k  1)
x ( k  2)
z 1 X ( z )
z 2 X ( z )
x(k  3)
z 3 X ( z )
Chap. 2 The z Transform
27
z Transform Method for Solving Difference Equations
• Example 2-18
x(k  2)  3x(k  1)  2 x(k )  0,
x(0)  0,
x(1)  1
Z x(k  2)  z 2 X ( z )  z 2 x(0)  zx (1)
Z x(k  1)  zX ( z )  zx (0)
Z x(k )  X ( z )
z 2 X ( z )  z 2 x(0)  zx(1)  3zX ( z )  3zx(0)  2 X ( z )  0
z
z
z
z



z 2  3z  2 ( z  1)( z  2) z  1 z  2
1
1


1  z 1 1  2 z 1
X ( z) 
x(k )  (1) k  (2) k ,
41-142
k  0,1,2,
Chap. 2 The z Transform
28
Concluding Comments
• The basic theory of the z transform method has been
presented.
– z transform : linear time invariant discrete-time systems
– Laplace transform: linear time-invariant continuous-time systems
• With the z transform method, linear time-invariant
difference equations can be transformed into algebraic
equations.
• The z transform method allows us to use conventional
analysis and design techniques available to analog
control systems.
41-142
Chap. 2 The z Transform
29