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′​=[xk​vˉk+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′​=[xk​vˉk+1​​]=[10​11​][xk−1​vˉk​​]+[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′=[10​11​],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​=[1​0​][xk​vˉk+1​​]
其中
C ′ = [ 1 0 ] C^{'} = \begin{bmatrix} 1 & 0\end{bmatrix} C′=[1​0​]
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′​]=[11​01​]

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′​=⎣⎡​xk​vk​dˉk​​⎦⎤​
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′​=⎣⎡​xk​vk​dˉk​​⎦⎤​=⎣⎡​100​110​001​⎦⎤​⎣⎡​xk−1​vk−1​dˉk−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′=⎣⎡​100​110​001​⎦⎤​,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,k​d2,k​​]=[11​00​01​]⎣⎡​xk​vk​dˉk​​⎦⎤​
其中
C ′ = [ 1 0 0 1 0 1 ] C^{'} = \begin{bmatrix}1 & 0 & 0 \\ 1 & 0 & 1 \\ \end{bmatrix} C′=[11​00​01​]
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​⎦⎥⎤​=⎣⎢⎢⎢⎢⎢⎢⎡​111111​001122​010101​⎦⎥⎥⎥⎥⎥⎥⎤​

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