Image Processing Reference
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and the residue A
2
corresponding to the second pole is
3)
n1
z
2
3
2
(0
0
:
:
3)
z
n1
3)
n
A
2
¼
(z
0
:
3)
j
z¼0:3
¼
2)
¼
3(0
:
(z
0
:
2)(z
0
:
(0
:
3
0
:
Therefore,
2)
n
u(n)
3)
n
u(n)
x(n)
¼ A
1
þ A
2
¼
2(0
:
þ
3(0
:
(b)
Inversion by Power Series Expansion
: In this case, we expand X
(
z
)
as a
power series of z
1
, that is, X
(
z
) ¼
P
n
¼1
a
n
z
n
. The expansion coeffi-
-
cients are x
(
n
)
is a rational function,
the power series can be obtained using long division as illustrated by the
following example.
, that is, x
(
n
) ¼
a
n
. If the function X
(
z
)
Example 3.20
z
z
Find the inverse z-transform of the function X(z)
¼
2
using power series
0
:
expansion.
S
OLUTION
The power series expansion of X(z)is
z
z
1
2z
1
2)
2
z
2
2)
3
z
3
X(z)
¼
2
¼
2z
1
¼
1
þ
0
:
þ
(0
:
þ
(0
:
þ
0
:
1
0
:
Therefore,
¼
2)
n
u(n)
2)
2
2)
3
2)
4
x(n)
¼
10
:
2(0
:
(0
:
(0
:
(0
:
Example 3.21
Find the inverse z-transform of
z
3
2
:
3z
2
þ
0
:
84z
X(z)
¼
z
3
1
:
4z
2
þ
0
:
63z
0
:
09
S
OLUTION
Expanding X(z), we have
z
3
3z
2
2
:
þ
0
:
84z
X(z)
¼
z
3
1
:
4z
2
þ
0
:
63z
0
:
09
9z
1
05z
2
813z
3
5577z
4
¼
1
0
:
þ
1
:
0
:
0
:
þ
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