Signals 11 XỬ LÝ TÍN HIỆU SỐ

8. INVERSE Z-TRANSFORM

8. INVERSE Z-TRANSFORM
The process by which a Z-transform of a time –series xk , namely X(z), is returned to
the time domain is called the inverse Z-transform. The inverse Z-transform is defined by:
xk   Z 1 Xz 

Computer study
M-file iztrans.m is used to find inverse Z-transform.
Example 8.1
» syms z k
z
X( z )  2
» x=z/(z^2+5*z+6);
z  5z  6
» iztrans(x)

xk    2 U s k    3 U s k 
k

k

ans =
(-2)^k-(-3)^k

Four practican techniques can be used to implement an inverse transform. They are:
1.
2.
3.
4.

Long divisions
Partial fractions
Residue theorem
Difference equations

8.1 Inverse Z-transforms via long division
For causal sequences, the z-transform X(z) can be expended into a power serises in z 1 .
For a rational X(z), a convenient way to determine the power serise is an expansion by
long division.
b z n  b n 1z n 1  .......  b1 z  b 0
X( z )  n n
 c 0  c1z 1  c 2 z 2  .......
z  a n 1z n 1  .......  a 1z  a 0
where c 0 , c1 , c 2 ..... are power serise coefficients.
z
Example 8.2
X( z )  2
z  1.414z 1  1
z
z-1.414+z-1
1.414-z-1
1.414-2z-1+1.414z-2
z-1-1.414z-2
z-1-1.414z-2+z-3
-z-3
-z-3+1.414z-4-z-5

z2-1.414z+1
z-1+1.414z-2+z-3-z-5. . .

X[k]=[k-1]+1.414[k-2]+  [k-3]+0-[k-5]. . .

113


8. INVERSE Z-TRANSFORM

Computer study
The inverse of a rational z-transform can also be readıly calculated using MATLAB. The
function impz can be utilized for this purpose. Three versions of this function are as
follows:
 [h,t]=impz(num,den)
 [h,t]=impz(num,den, L)
 [h,t]=impz(num,den, L, FT)
Where the input data consists of the vector num and den containing the coefficients of the
numerator and the denominator polynomials of the z-transform given in the descending
powers of z, the output impulse response vector h, and the time index vector t. The first
form, the length L of h is determined automatically by the computer with t=0:L-1,
whereas in the remaining two forms it is supplied by the user through the input data L. In
1
the last form, the sampling interval is
. The default value of FT is 1. The following two
FT
examples show application h, t   impznum, den file to and plot power.

X ( z) 

Example 8.3

» num=[1 0];

» den=[1 -1.414 +1];
» L=8;
» [x,k]=impz(num,den,L)
x=
1.0000
1.4140
0.9994
-0.0009
-1.0006
-1.4140
-0.9988
0.0017
k=
0
1
2
3
4
5
6
7
» stem(k,x,’fill’,’k’)

z
z  1.414 z  1
2

Power series coefficiens for

X ( z) 

z
z  1.414 z  1
2

1.5

1

0.5

0

-0.5

-1

-1.5

0

1

2

3

4

5

k

114

6

7


8. INVERSE Z-TRANSFORM

X ( z) 

z 2  z 1
2 z 2  3z  1

Example 8.4
» num=[1 1 -1];
» den= [2 3 1];
» L=10;
» [x,t]=impz(num,den,L)
x=
0.5000
-0.2500
-0.3750
0.6875
-0.8438
0.9219
-0.9609
0.9805

-0.9902
0.9951
» stem(k,x,’fill’,’k’)

z 2  z 1
Power series coefficients for X ( z )  2
2 z  3z  1
1
0.8
0.6
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1

1

2

3

4

5

6

7

8

8.2 The Inverse Z-Transform Using Partial Fractions
We now derive the expression for the inverse z-transform and outline the two
methods for its computation.
Recall that, for z  re j , the z-transform G(z) given by the equation is merely the Fourier
transform of the modified sequence gk r  k . Accordingly, by the inverse Fourier
transform, we have:

1
k
gk  r 
G (re j )e jk d
(8.2)

2  
By making the change of variable z  re j , the above equation can be converted into a
contour integral given by :
gk 
residues of G(z)z k 1at the poles inside C
(8.3)
Note that theequation mentioned above needs to be equated at all values of k which can
be quite complicated in most cases.
A rational G(z) can be expressed as:
P( z)
G (z) 
D( z )





115

9


8. INVERSE Z-TRANSFORM

where P(z) and D(z) are the polynomials in z 1 . If the degree M of the numerator
polynomial P(z) is grester than or equal to the degree N of the denominator polynomial
D(z), we can divide P(z) by D(z) and re-express G(z) as:
M N
P ( z)
G ( z )   ai z  k  1
D( z )
i 0
where the degree of the polynomial P1 (z) is less that that of D(z). The rational function
P1(z)
is called a proper fraction.
D( z )
The expression of Eq (8.2) can be computed in a number of ways. Consider the following
cases:
Case 1: G(z) is a proper fraction with simple poles. Let the poles of G(z) be at z  p k ,
k=1,2,3,……,N, where pk are distinct. A partial-fraction expansion of G(z) then is of the
form :
N

az
(8.4)
G (z)   i
k 1 z  p i
where the constants  l in the above expression, called the residues, are given by:
G (z)
(8.5)
a i  (z  p i )
z z  pi
Each term of the sum on the right-hand side of Eq.(8.4) has an ROC given by z  pk,
therefore, the inverse transform g[k] of G(z) is given by
N

gk    a i p i U s k 
k

k 1

Note that the above approach with slight modifications can also be used to determine the
inverse z-transform of a noncausal sequence with a rational z-transform.
Example 8.5
Let the z-transform of a causal sequence g[k] be given by :

z(z  2.0)
(z  0.2)(z  0.6)
az
a 2z
G (z)  1 
z  0.2 z  0.6
z  2.0

2.2
a 1  (z  0.2)

 2.75
(z  0.2)(z  0.6) z 0.2 0.8
G (z) 

a 2  (z  0.6)



(z  2.0)
(z  0.2)(z  0.6)

z  0.6



1.4
 1.75
 0.8



gk   2.75(0.2) k  1.75(0.6) k U s k 

Example 8.6
Using MATLAB determine the partial fraction expansion of X(z):

X( z ) 

3z 3  12
2z 3  3.5z 2  1.5

116


8. INVERSE Z-TRANSFORM

num=[3 0 0 12];den=[2 -3.5 0 -1.5];
» [r,p,k]=residuez(num,den)
r=
1.9219
3.7891 - 0.3013i
3.7891 + 0.3013i
p=
1.9477
-0.0989 - 0.6126i
3.6
2.625
-0.0989 + 0.6126i 2.47
X( z) 


k=
z  1.5 z  1 z  0.5
-8
» r=[ 1.9219 3.7891 - 0.3013i
+ 0.3013i];
1.5z 3  3.7891

6z 2
(z)  -3 0.6126i 2 -0.0989 +
» p=[ 1.9477 X
-0.0989
z  1.75z  0.75
0.6126i];
» [num,den]=residuez(r,p,k)
num =
1.5001 -0.0000 0.0008 5.9999
den =
1.0000 -1.7499 -0.0000 -0.7500
Multiplying the numerator and the denominator by 2.

G (z) 

3z 3  12
2z 3  3.5z 2  1.5

Case 2. G(z) has multiple poles, for example, if the pole at z   is of multiplicity r and
the remaining N-r poles are simple and at z  p k , k  1,2,3,.....N  r, then the general
partial-fraction expansion of G(z) takes the form
G (z) 

MN

a
k 0

Nr

k

z k 
k i

r
akz
z
  a ri
z  p k i 1
(1  z 1 ) i

where the constant ari (no longer called the residues for i1) are computed using the
formula:
1
d r i 
G (z) 
a ri 
( z  ) r
, i=1,2,3,…..,r
r i 
(r  i)! d(z) 
z  z 
Example 8.7
a z
a z
z2
X(z)  21  22 2
X(z) 
;

2
(z  1) (z  1)
(z  1)
a
a 22
X( z )
 21 
;
z
(z  1) (z  1) 2

a0 

zX(z)
 0;
z

d z  1 X(z)
d
 z 1
dz
z
dz z 1
z 1
2

a 22 

a 21 

117

(z  1) 2 X(z)
1
z
z 1


8. INVERSE Z-TRANSFORM

Which results in the following time-series
xk   1  k u s k 
z
Example 8.8
G (z) 
(z  0.5)(z  1) 2
G ( z)
G (z)
a 1  (z  0.5)
 4;
a 22  (z  1) 2
 2;
z z 0.5
z z 1

a 21 

d 
2 G (z) 
(

z

1
)
 4
dz 
z  z 1
0.5

z
z
z
4
2
z  0.5
z 1
(z  1) 2
Consider the following three cases:
1)
z1
g[k]  4(0.5) k Uk  4Uk   2kUk 
1
2)
z
2
U s k   U s  k  1
G ( z)  4

1.0

z1
0.5

g[k]  4(0.5) k U s  k  1  4U s  k  1  2kU s  k  1
1
3) z1
2
g[k]  4(0.5) k U s k   4U s k  1  2kU s  k  1

1.0

z
0.5

1.0

Example 8.9

3z 3  5z 2  3z
(z  1) 2 (z  0.5)
a z
a z
a z
X(z)  11  21  22 2
z  0.5 z  1 (z  1)
X( z) 

1
z1
2

where a 11z /( z  0.5) an exponential, a 21z /( z  1) a step function, and a 22z /( z  1) 2 a ramp
function. What is desired, however, is the partial fraction expansion of X(z)/z, where:

a 11
a
a 22
X( z)

 21 
z
z  0.5 z  1 (z  1) 2
where

a1 

(z  0.5)X(z)
3z 3  5z 2  3z

z
z(z  1) 2
z 0.5

a 22 

(z  1) X(z)
3z  5z  3z

z
z(z  1 / 2)

z 1
2

3

5
z 0.5

2

2
z 1

 9z  10  3 2(3z 3  5z 2  3)(z  1 / 4) 
d (z  1) 2 X(z)


  2
2
dz
z
z
(
z

1
/
2
)
(

z
(
z

1
/
2
))

 z 1
z 1
which results in
a 21 

118

1
2


8. INVERSE Z-TRANSFORM





xk   5(0.5) k  2  2k u s k 

Example 8.10

18z 3
Solve using Matlab:
H( z ) 
18z 3  3z 2  4z  1
0.24
0.4
0.36
H( z ) 


1
1 2
z  0.5
1  0.33z
(1  0.33z )
» num=[18]; den=[18 3 -4 -1];
» [r,p,k]=residuez(num,den)
r=
0.2400
0.4000
0.3600
p=
-0.3333
-0.3333
0.5000
k =[]
» [num,den]=residuez(r,p,k)
num =
1.0000 0.0000 0.0000
den =

1.0000 0.1667 -0.2222 -0.0556
Using the numerator and the denominator coefficients we have:

X( z ) 

z3
z 3  0.1667z 2  0.2222z  0.0556

It can be seen that the coefficients will be same as in the equation of the question if we
multiply each coefficient by 18.
Example 8.11. Find the inverse Z-transform of

X( z) 

(z  1) 3
(z  0.5)(z  0.5) 2

a1
a 21
a 22
X( z) a 0




z
z (z  0.5) (z  0.5) (z  0.5) 2
a0 

zX(z)

z 0  8
z

119


8. INVERSE Z-TRANSFORM

(z  0.5)X(z)
a1 
z

a 22 

(z  0.5)(z  1) 3
z  0.5 
(z  0.5) 2

(z  0.5) 2 X(z)
z

z  0.5



(z  1) 3
(z  0.5)

z  0.5



z  0.5



1
4

27
4

d (z  0.5) 2 X(z)
d (z  1) 3
27

z  0.5
z  0.5  
dz
z
dz (z  0.5)
4
X( z )
1
z
27
z
27
z
8



z
4 (z  0.5) 4 (z  0.5) 4 (z  0.5) 2
a 21 

27
27
1

xk   8k    (0.5) k  (0.5) k 
k (0.5) k  U s k 
4
4
4


» num=[1 3 3 1]; den=poly([0 -0.5 0.5 0.5])
den =
1.0000 -0.5000 -0.2500 0.1250
0
» [r,p,k]=residue(num,den)
r=
-0.2500
-6.7500
6.7500
8.0000
p=
-0.5000
0.5000

0.5000
0
k = []

Case 3. X(z) has a complex pole


Example 8.12. The second-order X(z)  (3z 2  1.5z) /( z 2  cos( )z  1 / 4) has non
6
repeated complex roots. The partial expansion of X(z) is defined by:

z
z
 a2
( z  )
(z   * )
a1
a2
X( z)


z
( z  ) ( z   * )

X( z )  a 1

where =0.433  j0.25 and

a1 

( z  ) X( z)
(3z 2  1.5z)

z
z(z   * ) z 
z 

120


8. INVERSE Z-TRANSFORM

a2 

( z   * ) X( z )
(3z  1.5z)
*

 a1
z
z(z  ) z *
z  *

Also note that
 1 
x 1 k   Z 1 
  k uk 

z  
 1 

and x 2 k   Z 1 
 ( * ) k uk 
*
z   

Where =0.433013-j0.25=0.5exp(-j/6). Therefore,
z
z
X( z )  a 1
 a 1*
z  0.5 exp( j / 6)
z  0.5 exp( j / 6)
which corresponds to a time-series, for k0
k

k

1
1
xk   1.55 exp(  j / 12)  exp(  jk / 6)  1.55 exp( j / 12)  exp( jk / 6)
2
2
k

1
 1.55  (exp(  jk / 6  j / 12)  exp( jk / 6   / 12))
2
k

1

 3.1  cos(k / 6   / 12)
2
which is seen to be a causal phase-shifted cosine wave with an exponentially descending
envelope. Also observe that x0  3.1cos( / 12)  3 , which can be verified using the
initial value theorem.

8.3 Difference Equations
Long division can be intensive and tedious computational process. If a computer-based
signal processing is desired, the use of difference equation is generally more efficient.
Assume that the Z-transform of a time series is xk  is X(z), where
M

X(z) 

 bi z

i

 aiz

i

i 0
N

i 0

Recall that the Zk   1 and Zk  n   z  n . Therefore, it follows that:

a 0 xk   a 1 xk  1  ....  a M 1 xk  (M  1)  a M xk  M 

b 0 k   b1k  1  ....  b N 1k  ( N  1)  b N k  N

121


8. INVERSE Z-TRANSFORM

The response xk  can be simulated by implementing the difference equation.

Example 8.13
Consider causal

3z 3  5z 2  3z
X( z ) 
(z  1)(z  0.5)

from example 11.

X( z ) 

3z 3  5z 2  3z
3  5z 1  3z 2
3  5z 1  3z 2


z 3  2.5z 2  2z  0.5 (1  z 1 ) 2 (1  0.5z 1 ) 1  2.5z 1  2z 2  0.5z 3

Which produces a time-serises

xk   (5(0.5) k  2  2k)u s k 

Then xk  , for k  0 , can be simulated using

xk  2.5xk  1  2xk  2  0.5xk  3  3k  5k  1  3k  2

122