Image Processing Reference
In-Depth Information
x
(
k
+1)
y
(
k
)
x
(
k
)
Unit
delay
C
u
(
k
)
B
+
+
A
FIGURE 5.5
Block diagram of the system.
The state observer is given by
x
(
k
þ
1
) ¼
Ax
(
k
) þ
Bu
(
k
) þ
Ky
(
k
)
Cx
(
k
)
½
(
5
:
78
)
where
x
2
R
N
N
is an estimate of the state of the system x
(
k
)
K is a gain matrix controlling the dynamic behavior of the observer
Let the estimation error e
(
k
)
be
e
(
k
) ¼
x
(
k
)
x
(
k
)
(
5
:
79
)
Then
e
(
k
þ
1
) ¼
x
(
k
þ
) ¼
Ax
(
k
) þ
Bu
(
k
)
Ax
(
k
)
Bu
(
k
)
K
[
y
(
k
)
Cx
(
k
)]
¼
A
(
x
(
k
)
x
(
k
))
K
[
Cx
(
k
)
Cx
(
k
)] ¼ (
A
KC
)(
x
(
k
)
x
(
k
))
¼ (
A
KC
)
e
(
k
)
1
)
x
(
k
þ
1
(
5
:
80
)
Equation 5.80 has a solution that is given by
k
e
(
e
(
k
) ¼ (
A
KC
)
0
)
(
5
:
81
)
Therefore if A
KC is a stable matrix (i.e., eigenvalues inside unit circle in complex
z-plane), the error signal e
(
k
)
. Hence we need to choose
gain matrix K by assigning eigenvalues to matrix A
KC.Thisissimilartothepole-
placement algorithm discussed before, except that in pole placement we assign eigen-
values to A
BK. Since A
KC
¼ (
A
T
C
T
K
T
)
T
, and eigenvalues are invariant
under transpose operation, the problem is same as assigning eigenvalues to
A
T
C
T
K
T
. Comparing the pole placement with this problem it is obvious that we
can use any pole-placement algorithm with the following simple substitutions:
will approach zero as k
!1
A
!
A
T
B
!
C
T
K
!
K
T
(
5
:
82
)
For example the Ackermann
s formula becomes
'
n
o
T
1
P
(
A
T
)
2
(
A
)
T
(
N
1
)
C
T
K
¼
½
00
01
C
T
A
T
C
T
A
T
C
T
(
5
:
83
)
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