Image Processing Reference
In-Depth Information
computed using either Cayley
Hamilton technique or z-transform. To use z-trans-
-
form, let
w(
k
) ¼
A
k
(
3
:
200
)
Then it is obvious that
) ¼
A
k
þ
1
w(
k
þ
1
¼
A
w(
k
)
(
3
:
201
)
w(
) ¼
I
0
Taking the z-transform from both sides of Equation 3.201, we have
z
F(
z
)
z
w(
0
) ¼
A
F(
z
)
(
3
:
202
)
Since
w(
0
) ¼
I, then we have
(
zI
A
)F(
z
) ¼
zI
(
3
:
203
)
Solving Equation 3.203 for
F(
z
)
,
F(
z
) ¼
z
(
zI
A
)
1
(
3
:
204
)
Therefore,
w(
k
) ¼
Z
1
[
z
(
zI
A
)
1
]
(
3
:
205
)
Example 3.50
Find A
k
,if
2
4
3
5
0
:
10
:
4
0
:
1
A ¼
0
:
15 0
:
6
0
:
05
0
:
20
:
40
:
2
S
OLUTION
Using z-transform, we have
2
4
3
5
1
z
0
:
1
0
:
4
0
:
1
¼ z(zI A)
1
F
(z)
¼ z
0
:
15
z
0
:
60
:
05
0
:
2
0
:
4
z
0
:
2
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