Image Processing Reference
In-Depth Information
Example 5.8
Consider the dynamic system given by
x(k)
u(k)
0
1
0
1
x(k þ
1)
¼
þ
0
:
72 1
:
7
y(k)
¼
½
10
x(k)
Design a full-state observer for the system. Place the poles of the observer at
l
1
¼ l
2
¼
0
:
25
S
OLUTION
The characteristic polynomial of the observer is
2
P(
l
)
¼
(
l l
1
)(
l l
2
)
¼
(
l
0
:
25)(
l
0
:
25)
¼ l
0
:
5
l þ
0
:
0625
The observability matrix is
¼
C
CA
10
01
Since the observability matrix is full rank, the system is completely state observ-
able. The observer gain vector is given by
2
4
3
5
2
4
3
5
1
C
CA
CA
2
.
CA
N1
0
0
.
0
1
1
C
CA
0
1
[A
2
K ¼ P(A)
¼
0
:
5A þ
0
:
0625I]
10
01
¼
0
:
6575
0
:
33
0
1
0
:
33
¼
0
:
2376
0
:
7136
0
:
7136
The state observer is given by
x(k þ
1)
¼ Ax(k)
þ Bu(k)
þ K[y(k)
Cx(k)]
x(k)
u(k)
(y(k)
0
1
0
1
0
:
33
x(k þ
1)
¼
þ
þ
½
10
x(k))
0
:
72 1
:
7
0
:
7136
or
x(k)
u(k)
y(k)
0
:
33
1
0
1
0
:
33
x(k þ
1)
¼
þ
þ
0
:
0064 0
:
17
0
:
7136
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