Image Processing Reference
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Therefore,
4) 1 A
4) 2 A þ
A ¼
0
:
75(0
:
0
:
125(0
:
2
(3
:
32)
Solving for A yields
2
64
3
A ¼
4) 2 ¼
(3
:
33)
4) 1
1
0
:
75(0
:
þ
0
:
125(0
:
Therefore,
64
3 (0
4) n
y h (n)
¼
:
n
0
(3
:
34)
The total solution is the sum of the homogenous and the zero-input response.
Therefore,
64
3 (0
5) n
25) n
5) n
25) n
4) n
y(n)
¼ C 1 (0
:
þ C 2 (0
:
þ y h (n)
¼ C 1 (0
:
þ C 2 (0
:
:
(3
:
35)
The constants C 1 and C 2 are found by applying the initial conditions
160
3
160
3
5) 1
25) 1
y(
1)
¼ C 1 (0
:
þ C 2 (0
:
¼
2C 1 þ
4C 2
¼
0
(3
:
36)
400
3
400
3
5) 2
25) 2
y(
2)
¼ C 1 (0
:
þ C 2 (0
:
¼
4C 1 þ
16C 2
¼
0
(3
:
37)
10
3 . Therefore,
Solving Equations 3.36 and 3.37, we have C 1 ¼
20 and C 2 ¼
10
3 (0
64
3 (0
5) n
25) n
4) n
y(n)
¼
20(0
:
þ
:
:
n
2
(3
:
38)
ne
z-transform, inverse z-transform, and properties associated with the transform. We
will then use z-transform to solve discrete-time linear systems.
DE can also be solved using z-transform. In the next section, we de
3.5 z-TRANSFORM
The z-transform of discrete-time signal x ( n )
is de
ned as
1
x ( n ) z n
X ( z ) ¼
(
3
:
39
)
n ¼1
where z is a complex variable. The in
nite sum in Equation 3.39 may not converge
for all values of z. The region where the complex function X ( z )
converges is known
as the region of convergence. This is referred to as double-sided z-transform. If the
signal x ( n ) ¼
0 for n <
0, then we have one-sided z-transform, which is de
ned as
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