Image Processing Reference
In-Depth Information
Therefore,
5) k
25) k
5) k
25) k
3(0
:
2(0
:
(0
:
þ
(0
:
w
(k)
¼
5) k
25) k
5) k
25) k
6(0
:
6(0
:
2(0
:
þ
3(0
:
(b)
z -Transform Approach
: Using z-transform, we have
1
z
z
10
:
25
zþ
0
:
25
0
:
25
¼ z(zIA) 1
F
(z)
¼ z
¼
1
:
5 zþ
0
:
25
1
:
5
z
1
z 2
0
:
75zþ
0
:
125
2
4
3
5
z þ
0
:
25
0
:
25
(z
0
:
5)(z
0
:
25)
(z
0
:
5)(z
0
:
25)
F
(z)
¼ z
1
:
5
z
1
(z
0
:
5)(z
0
:
25)
(z
0
:
5)(z
0
:
25)
Partial fraction expansion yields
2
4
3
5 ¼
2
4
3
5
3
z
2
1
1
3z
z
2z
z
z
z
5 þ
5 þ
5 þ
5 þ
0
:
z
0
:
25
z
0
:
z
0
:
25
0
:
z
0
:
25
0
:
z
0
:
25
F
(z)
¼z
6
z
6
2
3
6z
z
6z
25
2z
3z
z
5 þ
5 þ
5 þ
5 þ
0
:
z
0
:
25
z
0
:
z
0
:
25
0
:
z
0
:
z
0
:
0
:
25
Hence,
5) k
25) k
5) k
25) k
3(0
:
2(0
:
(0
:
þ
(0
:
¼ Z 1 [
w
(k)
F
(z)]
¼
5) k
25) k
5) k
25) k
6(0
:
6(0
:
2(0
:
þ
3(0
:
4.8.3 C OMPLETE S OLUTION OF S TATE E QUATIONS
Assuming that the initial state is x (
0
)
, we have
x (
) ¼ Ax (
) þ Bu (
)
1
0
0
) ¼ A 2 x (
x (
2
) ¼ Ax (
1
) þ Bu (
1
0
) þ ABu (
0
) þ Bu (
1
)
) ¼ A 3 x (
) þ A 2 Bu (
x (
3
) ¼ Ax (
2
) þ Bu (
2
0
0
) þ ABu (
1
) þ Bu (
2
)
.
x ( k ) ¼ Ax ( k
) ¼ A k x (
) þ A k 1 Bu (
) þ A k 2 Bu (
1
) þ Bu ( k
1
0
0
1
)þþ Bu ( k
1
)
(
4
:
117
)
Therefore,
X
k 1
x ( k ) ¼ A k x (
A k 1 n Bu ( n )
) þ
(
:
)
0
4
118
n ¼ 0
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