The Inverse
z-Transform
The inverse z transformation can be performed by
using a few results from complex variable theory.
Suppose that we integrate one (1) from z=0 to z=1:
1

0
1
(1)dz  z  1  0  1.
0
The integrals become more involved when the path
of integration is in the complex plane.
For example, suppose that we wish to integrate
around a unit circle in the (complex) z-plane. To do
the integration we make a substitution of variables:
z=ejq. We have dz = j ejq dq.
Im{z}
q  p/2
qp
ze
q
radius = 1
q  3p/2
q0
jq
Re{z}
Suppose, first, that we wish integrate one (1) around
a semicircle C around the origin from 0 to p.
Im{z}
z  e jq
C
qp
q
q0
Re{z}
p
(
1
)
dz

je
d
q


jq
0
C
p
 j  cosq  j sin q dq
0
 j 0  j 2  2.
Suppose that we integrate one (1) around a closed
unit circle C around the origin (0 to 2p).
Im{z}
C
z  e jq
q
q0
q  2p
Re{z}
We have,
(
1
)
dz



2p
0
jq
je dq  0.
C
(The areas under the sine and cosine sum to zero
over a full period.)
Suppose that we integrate z around a closed unit
circle C around the origin (0 to 2p).
2p
z
dz

e
je
d
q


jq
jq
0
C
2p
 j e
0
2 jq
dq  0.
Similarly, if we integrate z2 , z3 or zk (k  0) around
the unit circle we get zero.
What about integrating z-1?
2p
z
dz

e


1
0
 jq
jq
je dq
C
 j
2p
0
dq  2p j.
We do not get zero when we integrate z-1 around
the unit circle.
What about integrating z-2?
2p
z
dz

e


2
 j 2q
0
jq
je dq
C
2p
 j e
0
 jq
dq  0.
Similarly, if we integrate z-3 , z-4 or z-k (k>1) around
the unit circle we get zero.
So integrating z-1 is unique, and
z
dz

2
p
j
.

1
C
So if we were to integrate a z-transform
X ( z) 

n
x
[
n
]
z
,

n  
we get
 X ( z)dz 
C


n  
x[n] z n dz  2p jx[1].
C
(Only z-1 does not integrate to zero.)
In other words the integration “picks-out” the sample
x1. We can “pick-out” another sample by adjusting
the power of z:
 X ( z) zdz 
C


n  
x[n] z n1  2p jx[2].
C
(The integral is zero except when –n+1 = -1, or n=2.)
Similarly,
2
X
(
z
)
z
dz 

C


n  
xk  z n2  2p jx[3],
C
and
m1
X
(
z
)
z
dz 

C


n  
xk  z nm1  2p jx[m].
C
Thus, we have found an expression for the inverse
z-transform:
x[n] 
1
X
(
z
)
z
dz
.

n 1
2p j C
The path of the integral must be carried-out over a
region of convergence of X(z). [CROC.]
Example: Find the inverse z-transform of X(z)=1
using the complex inversion integral.
Solution: Plugging-in X(z)=1, we have
x[n] 
1
z

2p j
C
n 1
dz.
We know that the integral is equal to zero except
when n=0. When n=0, the integral is equal to 2pj.
Thus, the final inverse transform is equal to
x[n] 
1
2p j
2p j  1
when n=0. Hence we have x[n]=d[n].
Example: Find the inverse z-transform of X(z)=z-k
using the complex inversion integral.
Solution: Plugging-in X(z)=z-k, we have
x[n] 
1
z

2p j
C
k
n 1
z dz.
We know that the integral is equal to zero except
when n=0. When n-k=0, or when n=k. Hence we
have x[n]=d[n-k].
Example: Find the inverse z-transform of
1
X ( z) 
1
1 z
using the complex inversion integral.
Solution: Let us expand 1/(1-z-1) using the
geometric series:
1
1
2
 1 z  z 
1
1 z
Solution: Plugging this into the complex inversion
integral we have
x[n] 

1 z

2p j
1
1
2

n 1
 z   z dz.
C
For n=0, the only term in 1+z-1+z-2+… which
“survives” a non-zero integration is 1.
For n=1, the only term in 1+z-1+z-2 which “survives”
is z-1.
We see that there is exactly one term that “survives”
when n0. The resultant integral in each case is 2pj,
and the corresponding value for x[n] in each case is
one. Thus,
x[n]  u[n].
Example: Find the inverse z-transform of
1
X ( z) 
1 2
(1  z )
using the complex inversion integral.
Solution: Using the expansion for 1/(1-z-1), we have
2


2
 1 
1
2

1

z

z




1
1 z 
1
2
 1  2 z  3z  
Solution: Plugging this into the complex inversion
integral we have
x[n] 

1 2z

2p j
1
1
2

n 1
 3z   z dz.
C
For n=0, the only term in 1+z-1+z-2+… which
“survives” a non-zero integration is 1.
For n=1, the only term in 1+z-1+z-2 which “survives”
is 2z-1.
The resultant inverse z-transform is
x[n]  (n  1)u[n  1]  (n  1)u[n].
Example: Find the inverse z-transform of
1
X ( z) 
1
1  az
using the complex inversion integral.
Solution: Expanding 1/(1-az-1) as a geometric
series:
1
1
2 2
 1  az  a z  
1
1  az
Plugging this into the complex inversion integral we
have
x[n] 

1  az

2p j
1
1
2 2

n 1
 a z   z dz.
C
For n=0, the only term in 1+az-1+a2z-2+… which
“survives” a non-zero integration is 1.
For n=1, the only term in 1+az-1+a2z-2 which
“survives” is az-1.
The resultant integral in the first case is 2pj, the
resultant integral in the second case is 2pja, etc.
The inverse z-transform is
x[n]  a nu[n].
Example: Find the inverse z-transform of
1
1
X ( z) 
1
1
1  az 1  bz
using the complex inversion integral.
Solution: We expand both 1/(1-az-1) and 1/(1-bz-1):
1
1
2 2
 1  az  a z  
1
1  az
1
1
2 2
 1  bz  b z  
1
1  bz
The product is
1
1
1
1
1  az 1  bz
1
2 2
1
2 2
 1  az  a z   1  bz  b z  


1

2
 1  (a  b) z  (a  ab  b ) z  
2
2
Plugging this into the complex inversion integral, we
wind-up with
1
a  b

x[n]   2
2
a

ab

b

 ...
(n  0),
( n  1),
(n  2),
This same problem can be solved in a slightly
different manner:
1
X ( z) 
1
1
(1  az )(1  bz )
2
z

( z  a )(z  b)


z
 z

 ( z  a )(z  b) 
B 
 A
 z


z

a
z

b


A
B


,
1
1
1  az
1  bz
where
z
1
A

1
z  b z a 1  bz
,
z a
and
z
B
za
z b
1

1
1  az
.
z b
The inverse z-transform is


x[n]  Aa  Bb u[n].
n
n
It can be shown algebraically (exercise) that this
expression is equivalent to the previous expression
for x[n].
In general, to find the inverse z-transform of
1
1
1
X ( z) 
,
1
1
1
1  az 1  bz 1  cz
we perform a partial fraction expansion:
A
B
C
X ( z) 


,
1
1
1
1  az
1  bz
1  cz
where
1
A  (1  az ) X ( z )
z a
1
B  (1  bz ) X ( z )
z b
1
C  (1  cz ) X ( z )
etc.
z c
,
,
,
The resultant inverse z-transform is


x[n]  Aan  Bbn  Cc n   u[n].
Example: Find the inverse z-transform of
z 1
X ( z)  e .
using the complex inversion integral.
Solution: We take the MacLaurin series for ez-1
e
z 1
1
2
3
 1  z  z  z  .
1
2
1
3!
The inverse z-transform follows rather naturally:
1
x[n]  .
n!
We could obtain this result through the complex
inversion integral, or we could recognize that
x[n]=1/n! corresponds to the coefficients of the ztransform.