Image Processing Reference
In-Depth Information
or
A
1
z
z
p
1
þ
A
2
z
(
z
p
1
)
A
m
1
z
(
z
p
1
)
X
(
z
) ¼
A
0
þ
2
þþ
m
1
B
1
z
z
p
2
þ
B
2
z
(
z
p
2
)
B
m
2
z
(
z
p
2
)
þ
2
þþ
þ
(
3
:
76
)
m
2
where
z
¼
0
¼
X
(
A
0
¼
z
X
(
z
)
z
)
(
:
)
0
3
77
The residues A
1
, A
2
,
...
, A
m
1
corresponding to the pole p
1
are computed using
z
¼
p
1
k
m
i
d
z
k
m
i
(
z
p
1
)
A
k
¼
d
m
1
X
(
z
)
z
k
¼
m
i
, m
i
1
,
...
,2,1
(
3
:
78
)
Similarly, the other set of residues corresponding to other poles are com-
puted. Once the residues are computed, the time-domain function is
x
(
n
) ¼
A
0
d(
n
)
A
1
(
p
1
)
n
u
(
n
) þ
A
1
n
(
p
1
)
n
1
þ
u
(
n
)þ
þ
A
m
1
n
(
n
1
)(
n
2
) (
n
m
1
þ
1
)
n
m
1
u
(
n
m
1
þ
1
)
(
p
1
)
m
1
!
n
u
(
n
) þ
B
1
n
(
p
2
)
n
1
þ
B
1
(
p
2
)
u
(
n
)þ
þ
B
m
2
n
(
n
1
)(
n
2
) (
n
m
2
þ
1
)
n
m
2
u
(
n
m
2
þ
1
)
(
p
2
)
þ
m
2
!
(
3
:
79
)
Example 3.23
Find the inverse z-transform of the following function:
z
2
X(z)
¼
3)
, ROC
: jzj >
0
:
3
2)
2
(z
(z
0
:
0
:
S
OLUTION
Using partial-fraction expansion, we would have
¼
30z
z
2z
(z
30z
z
X(z)
2
2)
2
þ
0
:
0
:
3
0
:
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