Z Transform, Causal, Anti-Causal and Two sided sequence, Region of Convergence, Properties, Inverse Z Transform -by long division -Partial fraction

Z - Transform
•Basic Definition
•Causal, Anti-Causal and Two sided sequence
•Region of Convergence
•Properties
•Calculations
•Inverse Z Transform
-by long division
-Partial fraction
1

z-Transform
• The DTFT provides a frequency-domain
representation of discrete-time signals and LTI
discrete-time systems
• Because of the convergence condition, in
many cases, the DTFT of a sequence may not
exist
• As a result, it is not possible to make use of
such frequency-domain characterization in
these cases

z-Transform
• A generalization of the DTFT defined by
leads to the z-transform
• z-transform may exist for many sequences
for which the DTFT does not exist
• Moreover, use of z-transform techniques
permits simple algebraic manipulations








n
n
j
j
e
n
x
e
X ]
[
)
(

z-Transform
• Consequently, z-transform has become an
important tool in the analysis and design of
digital filters
• For a given sequence g[n], its z-transform
G(z) is defined as
where z = Re(z) + jIm(z) is a complex
variable






n
n
z
n
g
z
G ]
[
)
(

z-Transform
• If we let , then the z-transform
reduces to
• The above can be interpreted as the DTFT of
the modified sequence
• For r = 1 (i.e., |z| = 1), z-transform reduces to
its DTFT, provided the latter exists

 j
e
r
z









n
n
j
n
j
e
r
n
g
e
r
G ]
[
)
(
}
]
[
{ n
r
n
g 

z-Transform
• The contour |z| = 1 is a circle in the z-plane
of unity radius and is called the unit circle
• Like the DTFT, there are conditions on the
convergence of the infinite series
• For a given sequence, the set R of values of
z for which its z-transform converges is
called the region of convergence (ROC)





n
n
z
n
g ]
[

Causal Sequence
• Example - Determine the z-transform X(z)
of the causal sequence and its
ROC
• Now
• The above power series converges to
• ROC is the annular region |z| > |a|
]
[
]
[ n
n
x n

a

a

 
a








0
]
[
)
(
n
n
n
n
n
n
z
z
n
z
X
1
for
,
1
1
)
( 1
1

a
a

 

z
z
z
X

Causal Sequence
• Example - The z-transform (z) of the unit
step sequence [n] can be obtained from
by setting a = 1:
• ROC is the annular region
1
for
,
1
1
)
( 1
1

a
a

 

z
z
z
X


 z
1
1
for
1
1 1
1


 

z
z
z ,
)
(


Example of z-transform
n n1 0 1 2 3 4 5 N>5
x[n] 0 2 4 6 4 2 1 0
  5
4
3
2
1
2
4
6
4
2 









 z
z
z
z
z
z
X

Anti-Causal Sequence
• Note: The unit step sequence [n] is not
absolutely summable, and hence its DTFT
does not converge uniformly
• Example - Consider the anti-causal
sequence
]
1
[
]
[ 


a

 n
n
y n

Anti-Causal Sequence
• Its z-transform is given by
• ROC is the annular region
a


 a









1
1
)
(
m
m
m
n
n
n
z
z
z
Y
1
for
,
1
1 1
1

a
a

 

z
z
z
z
z
z
m
m
m
1
1
0
1
1 





a

a


a
a


a

z

Two-sided sequence
• Example - Consider the two-sided sequence
• Let and
with X(z) and Y(z) denoting, respectively,
their z-transforms
• Now
and
]
1
[
]
[
]
[ 





a
 n
n
n
v n
n
]
[
]
[ n
n
x n

a
 ]
1
[
]
[ 




 n
n
y n
a

a

 
z
z
z
X ,
1
1
)
( 1




 
z
z
z
Y ,
1
1
)
( 1

Two-sided sequence
• Using the linearity property we arrive at
• The ROC of V(z) is given by the overlap
regions of and
• If , then there is an overlap and the
ROC is an annular region
• If , then there is no overlap and V(z)
does not exist
1
1
1
1
1
1








z
z
z
Y
z
X
z
V

a
)
(
)
(
)
(
a

z 

z


a



a z


a

Two-sided sequence
• Example - Consider the two-sided sequence
where a can be either real or complex
• Its z-transform is given by
• The first term on the RHS converges for
, whereas the second term
converges for
n
n
u a

]
[













 a

a

a

1
0
)
(
n
n
n
n
n
n
n
n
n z
z
z
z
U
a

z
a

z

Two-sided sequence
• There is no overlap between these two
regions
• Hence, the z-transform of does not
exist
n
n
u a

]
[

z-Transform Properties
• Example - Determine the z-transform and
its ROC of the causal sequence
• We can express x[n] = v[n] + v*[n] where
• The z-transform of v[n] is given by
]
[
)
(cos
]
[ n
n
r
n
x o
n 


]
[
]
[
]
[
2
1
2
1 n
n
e
r
n
v n
n
j
n o 
a


 
r
z
z
e
r
z
z
V
o
j

a




a


 


,
1
1
1
1
)
( 1
2
1
1
2
1

• Using the conjugation property we obtain
the z-transform of v*[n] as
• Finally, using the linearity property we get
,
1
1
*
1
1
*)
(
* 1
2
1
1
2
1



 


a



z
e
r
z
z
V
o
j
*)
(
*
)
(
)
( z
V
z
V
z
X 












 



 1
1
2
1
1
1
1
1
z
e
r
z
e
r o
o j
j
a

z

• or,
• Example - Determine the z-transform Y(z)
and the ROC of the sequence
• We can write where
r
z
z
r
z
r
z
r
z
X
o
o 





 


,
)
cos
2
(
1
)
cos
(
1
)
( 2
2
1
1
]
[
)
1
(
]
[ n
n
n
y n
a


]
[
]
[
]
[ n
x
n
x
n
n
y 

]
[
]
[ n
n
x n
a


• Now, the z-transform X(z) of is
given by
• Using the differentiation property, we arrive at
the z-transform of as
]
[
]
[ n
n
x n
a

]
[n
x
n
a

a

 
z
z
z
X ,
1
1
)
( 1
a

a

a

 

z
z
z
dz
z
X
d
z ,
)
1
(
)
(
1
1

• Using the linearity property we finally obtain
2
1
1
1 )
1
(
1
1
)
( 

 a

a

a


z
z
z
z
Y
a

a

 
z
z
,
)
1
(
1
2
1

Table: Commonly Used z-
Transform Pairs

Inverse z-Transform
• By making a change of variable , the
previous equation can be converted into a
contour integral given by
where is a counterclockwise contour of
integration defined by |z| = r

 j
e
r
z
C
dz
z
z
G
j
n
g
C
n




1
)
(
2
1
]
[

Inverse Transform by
Partial-Fraction Expansion
• Example - Let the z-transform H(z) of a
causal sequence h[n] be given by
• A partial-fraction expansion of H(z) is then
of the form
)
.
)(
.
(
)
.
)(
.
(
)
(
)
( 1
1
1
6
0
1
2
0
1
2
1
6
0
2
0
2











z
z
z
z
z
z
z
z
H
1
2
1
1
6
.
0
1
2
.
0
1
)
( 







z
z
z
H

• Now
and
75
.
2
6
.
0
1
2
1
)
(
)
2
.
0
1
(
2
.
0
1
1
2
.
0
1
1 











z
z
z
z
z
H
z
75
.
1
2
.
0
1
2
1
)
(
)
6
.
0
1
(
6
.
0
1
1
6
.
0
1
2 














z
z
z
z
z
H
z

• Hence
• The inverse transform of the above is
therefore given by
1
1
6
0
1
75
1
2
0
1
75
2






z
z
z
H
.
.
.
.
)
(
]
[
)
6
.
0
(
75
.
1
]
[
)
2
.
0
(
75
.
2
]
[ n
n
n
h n
n






Inverse z-Transform via Long
Division
• Example - Consider
• Long division of the numerator by the
denominator yields
• As a result
2
1
1
12
.
0
4
.
0
1
2
1
)
( 






z
z
z
z
H
.
.
.
.
.
.
.
.
)
( 




 


 4
3
2
1
2224
0
4
0
52
0
6
1
1 z
z
z
z
z
H
0
2224
0
4
0
52
0
6
1
1 


 n
n
h },
.
.
.
.
.
.
.
.
{
]}
[
{


Inverse z-Transform Using
MATLAB
• The function impz can be used to find the
inverse of a rational z-transform G(z)
• The function computes the coefficients of the
power series expansion of G(z)
• The number of coefficients can either be user
specified or determined automatically

Z Transform, Causal, Anti-Causal and Two sided sequence, Region of Convergence, Properties, Inverse Z Transform -by long division -Partial fraction

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