第四章作业
1.
令
v
ˉ
k
+
1
=
v
ˉ
k
+
s
k
\bar{v}_{k+1} = \bar{v}_{k} + s_k
vˉk+1=vˉk+sk，则增广后的状态为
x
k
′
=
[
x
k
v
ˉ
k
+
1
]
x_k^{'} = \begin{bmatrix} x_k \\ \bar{v}_{k+1} \end{bmatrix}
xk′=[xkvˉk+1]
增广后的状态方程为
x
k
′
=
[
x
k
v
ˉ
k
+
1
]
=
[
1
1
0
1
]
[
x
k
−
1
v
ˉ
k
]
+
[
1
0
]
v
k
+
[
0
1
]
s
k
x_k^{'} = \begin{bmatrix} x_k \\ \bar{v}_{k+1} \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} x_{k-1} \\ \bar{v}_{k} \end{bmatrix} + \begin{bmatrix} 1 \\0 \end{bmatrix} v_k + \begin{bmatrix} 0 \\ 1 \end{bmatrix}s_k
xk′=[xkvˉk+1]=[1011][xk−1vˉk]+[10]vk+[01]sk
其中
A
′
=
[
1
1
0
1
]
,
B
′
=
[
1
0
]
A^{'} = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix},B^{'} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}
A′=[1011],B′=[10]
增广后的观测方程为
d
k
=
[
1
0
]
[
x
k
v
ˉ
k
+
1
]
d_k = \begin{bmatrix}1 & 0 \end{bmatrix} \begin{bmatrix} x_k \\ \bar{v}_{k+1} \end{bmatrix}
dk=[10][xkvˉk+1]
其中
C
′
=
[
1
0
]
C^{'} = \begin{bmatrix} 1 & 0\end{bmatrix}
C′=[10]
则可观性矩阵
O
′
=
[
C
′
C
′
A
′
]
=
[
1
0
1
1
]
O^{'} = \begin{bmatrix} C^{'} \\ C^{'}A^{'} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}
O′=[C′C′A′]=[1101]
r a n k ( O ′ ) = 2 = N + U rank(O^{'}) = 2 = N+U rank(O′)=2=N+U
所以系统是能观的。
2.
令
d
ˉ
k
=
d
ˉ
k
−
1
+
s
k
\bar{d}_{k} = \bar{d}_{k-1} + s_k
dˉk=dˉk−1+sk，则增广后的状态为
x
k
′
=
[
x
k
v
k
d
ˉ
k
]
x_k^{'} = \begin{bmatrix} x_k \\ v_k \\ \bar{d}_{k} \end{bmatrix}
xk′=⎣⎡xkvkdˉk⎦⎤
增广后的状态方程为
x
k
′
=
[
x
k
v
k
d
ˉ
k
]
=
[
1
1
0
0
1
0
0
0
1
]
[
x
k
−
1
v
k
−
1
d
ˉ
k
−
1
]
+
[
1
1
0
]
a
k
+
[
0
0
1
]
s
k
x_k^{'} = \begin{bmatrix} x_k \\ v_k \\ \bar{d}_{k} \end{bmatrix} = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x_{k-1} \\ v_{k-1}\\ \bar{d}_{k-1} \end{bmatrix} + \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} a_k + \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}s_k
xk′=⎣⎡xkvkdˉk⎦⎤=⎣⎡100110001⎦⎤⎣⎡xk−1vk−1dˉk−1⎦⎤+⎣⎡110⎦⎤ak+⎣⎡001⎦⎤sk
其中
A
′
=
[
1
1
0
0
1
0
0
0
1
]
,
B
′
=
[
1
1
0
]
A^{'} = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix},B^{'} = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}
A′=⎣⎡100110001⎦⎤,B′=⎣⎡110⎦⎤
增广后的观测方程为
d
k
=
[
d
1
,
k
d
2
,
k
]
=
[
1
0
0
1
0
1
]
[
x
k
v
k
d
ˉ
k
]
d_k = \begin{bmatrix} d_{1,k} \\ d_{2,k} \end{bmatrix} = \begin{bmatrix}1 & 0 & 0 \\ 1 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} x_k \\ v_k \\ \bar{d}_{k} \end{bmatrix}
dk=[d1,kd2,k]=[110001]⎣⎡xkvkdˉk⎦⎤
其中
C
′
=
[
1
0
0
1
0
1
]
C^{'} = \begin{bmatrix}1 & 0 & 0 \\ 1 & 0 & 1 \\ \end{bmatrix}
C′=[110001]
则可观性矩阵
O
′
=
[
C
′
C
′
A
′
C
′
A
′
2
]
=
[
1
0
0
1
0
1
1
1
0
1
1
1
1
2
0
1
2
1
]
O^{'} = \begin{bmatrix} C^{'} \\ C^{'}A^{'} \\ C^{'}{A^{'}}^2\end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \\ 1 & 2 & 0 \\ 1 & 2 & 1\end{bmatrix}
O′=⎣⎢⎡C′C′A′C′A′2⎦⎥⎤=⎣⎢⎢⎢⎢⎢⎢⎡111111001122010101⎦⎥⎥⎥⎥⎥⎥⎤
r a n k ( O ′ ) = 3 = N + U rank(O^{'}) = 3 = N+U rank(O′)=3=N+U
所以系统是能观的。
3.
将参数
w
=
0.1
,
n
=
3
,
p
=
0.999
w=0.1,n=3,p = 0.999
w=0.1,n=3,p=0.999代入公式(5.37)中可得
k
=
l
n
(
1
−
0.999
)
l
n
(
1
−
0.
1
3
)
=
6904.30
k = \frac{ln(1-0.999)}{ln(1 - 0.1 ^3)} = 6904.30
k=ln(1−0.13)ln(1−0.999)=6904.30
所以理论情况下最少需要6905次迭代