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2018南京航天航空大学820自动控制原理第十题

分类: 文章 • 2024-05-19 20:41:58

2018南京航天航空大学820自动控制原理第十题
1.系统状态方程的齐次解
x(t)=Φ(t)x(0) \boldsymbol {x} (t) = \boldsymbol {\varPhi} (t) \boldsymbol {x} (0)
可得
[cos⁡4t−sin⁡4t2cos⁡4t−sin⁡4t−cos⁡4t−sin⁡4t−cos⁡4t−2sin⁡4t]=Φ(t)[12−1−1] \begin{bmatrix} \cos4t - \sin4t & 2\cos4t - \sin4t \\ -\cos4t - \sin4t & -\cos4t - 2\sin4t \end{bmatrix} = \boldsymbol {\varPhi} (t) \begin{bmatrix} 1& 2 \\ -1 & -1 \end{bmatrix}
解得
Φ(t)=[cos⁡4t−sin⁡4t2cos⁡4t−sin⁡4t−cos⁡4t−sin⁡4t−cos⁡4t−2sin⁡4t][12−1−1]−1=[cos⁡4tsin⁡4t−sin⁡4tcos⁡4t] \begin{aligned} \boldsymbol {\varPhi} (t) &= \begin{bmatrix} \cos4t - \sin4t & 2\cos4t - \sin4t \\ -\cos4t - \sin4t & -\cos4t - 2\sin4t \end{bmatrix} \begin{bmatrix} 1& 2 \\ -1 & -1 \end{bmatrix} ^{-1} \\ & = \begin{bmatrix} \cos4t & \sin4t \\ -\sin4t & \cos4t \end{bmatrix} \end{aligned}
矩阵A\boldsymbol A为
A=Φ˙(0)=[−4sin⁡4t4cos⁡4t−4cos⁡4t−4sin⁡4t]t=0=[04−40] \begin{aligned} \boldsymbol A &= \boldsymbol {\dot{\varPhi}} (0) \\ & = \begin{bmatrix} -4\sin4t & 4\cos4t \\ -4\cos4t & -4\sin4t \end{bmatrix}_{t=0} \\ & = \begin{bmatrix} 0 & 4 \\ -4 & 0 \end{bmatrix} \end{aligned}
2.由1可知
Φ(t)=[cos⁡4tsin⁡4t−sin⁡4tcos⁡4t] \boldsymbol {\varPhi} (t) = \begin{bmatrix} \cos4t & \sin4t \\ -\sin4t & \cos4t \end{bmatrix}
可得
G(T)=Φ(t)t=T=[cos⁡4Tsin⁡4T−sin⁡4Tcos⁡4T]h(T)=∫0TΦ(t)Bdt=∫0T[cos⁡4tsin⁡4t−sin⁡4tcos⁡4t][04]dt=[−cos⁡4Tsin⁡4T] \begin{aligned} \boldsymbol G(T) & = \boldsymbol {\varPhi} (t) _{t=T} \\ & = \begin{bmatrix} \cos4T & \sin4T \\ -\sin4T & \cos4T \end{bmatrix} \end{aligned} \\ \begin{aligned} \boldsymbol h(T) &= \int_{0}^{T} \boldsymbol {\varPhi} (t) \boldsymbol B dt \\ &= \int_{0}^{T} \begin{bmatrix} \cos4t & \sin4t \\ -\sin4t & \cos4t \end{bmatrix} \begin{bmatrix} 0 \\ 4 \end{bmatrix} dt \\ &= \begin{bmatrix} -\cos4T \\ \sin4T \end{bmatrix} \end{aligned}
离散化后系统的状态空间表达式为
{x˙[(k+1)T]=[cos⁡4Tsin⁡4T−sin⁡4Tcos⁡4T]x(kT)+[−cos⁡4tsin⁡4t]u(kT)y=[20]x(kT) \left\{\begin{array}{l} \boldsymbol {\dot x} [(k+1)T] = \begin{bmatrix} \cos4T & \sin4T \\ -\sin4T & \cos4T \end{bmatrix} \boldsymbol {x} (kT) + \begin{bmatrix} -\cos4t \\ \sin4t \end{bmatrix} u(kT) \\ y = \begin{bmatrix} 2 & 0 \end{bmatrix} \boldsymbol {x} (kT) \end{array}\right.
系统能观性矩阵
Nc=[202cos⁡4T2sin⁡4T] \boldsymbol {N_c} = \begin{bmatrix} 2 & 0 \\ 2\cos4T & 2\sin4T \end{bmatrix}
当离散化系统能观时
T≠kπ(k=0,±1,±2⋯ ) T\not=k\pi(k=0,\pm1,\pm2\dotsb)

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