Image Processing Reference
In-Depth Information
Since
w(
0
) ¼
I, then
(
zI
A
)F(
z
) ¼
zI
(
4
:
114
)
and
F(
z
) ¼
z
(
zI
A
)
1
(
4
:
115
)
Therefore,
F(
k
) ¼
Z
1
[
z
(
zI
A
)
1
]
(
4
:
116
)
Example 4.12
Find the state-transition matrix of the following dynamic system using the two
methods presented previously:
x(k)
1
0
:
25
x(k þ
1)
¼
1
:
5
0
:
25
S
OLUTION
(a)
Modal Matrix Approach
: The characteristic polynomial of matrix A is
¼l
jlIAj¼
l
10
:
25
2
0
:
75
l20:75lþ0:125¼
0
:
125
¼
(
l
0
:
5)(
l
0
:
25)
¼
0
1
:
5
l20:75lþ0:125¼
0
:
25
Therefore, the eigenvalues of A are
l
1
¼
0
:
5
l
2
¼
0
:
25
The corresponding eigenvectors are
T
AV
1
¼ l
1
V
1
,
therefore V
1
¼
½
12
T
AV
2
¼ l
2
V
2
,
therefore V
2
¼
½
13
Using the modal matrix approach,
"
#
11
23
1
11
23
5)
k
0
0(0
(0
:
(k)M
1
w
(k)
¼ ML
¼
25)
k
:
"
#
3
5)
k
25)
k
1
(0
:
(0
:
¼
5)
k
3(0
25)
k
21
2(0
:
:
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