Image Processing Reference
In-Depth Information
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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