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两输入两阶连续系统的能达丰富性
多输入连续系统的能达域定义
$R_{cx}=\left\{ z\left|z=\int_{0}^{T}e^{A(T-t)}Bu_{t}\textrm{d}t,\left\Vert u_{t}\right\Vert _{\infty}\leq1\right.\right\}$
$=\left\{ z\left|z=\int_{0}^{T}e^{As}Bu_{T-s}\textrm{d}t,\left\Vert u_{T-s}\right\Vert _{\infty}\leq1\right.\right\}$
其中 $u_{t}=[u_{1}(t),u_{2}(t),\cdots,u_{r}(t)]^{T}$ 和 $B=[B_{1},B_{2},\cdots,B_{r}]$ 分别为 $r$ 维输入变量向量及相应输入矩阵。
若将时间 $[0,T]$ 进行 $N$ 等分,则有如下离散时间点
$t_{i}=i\varDelta,\quad i=0,2,\cdots,N\noindent$
其中 $\varDelta=T/N$ 。则连续系统的能控域 $R_{cx}$ 可由下式近似
$R_{dx}=\left\{ z\left|z=\sum_{i=0}^{N-1}\int_{t_{i}}^{t_{i+1}}e^{At}\textrm{d}tBu_{T-t_{i}},\left\Vert u_{T-t_{i}}\right\Vert _{\infty}\leq1,i=0,1,\cdots,N-1\right.\right\}$
$=\left\{ z\left|z=\left[H,GH,\cdots,G^{N-1}H\right]F,\left\Vert F\right\Vert _{\infty}\leq1\right.\right\}$
其中 $F\in R^{\left(Nr\right)\times1}$ ,
$G=e^{A\Delta},\qquad H=[H_{1},H_{2},\cdots,H{}_{r}],$
$H_{i}=\int_{0}^{\varDelta}e^{At}\textrm{d}tB_{i},\qquad i=1,2,\cdots,r$
则可以利用离散系统的能控丰富性计算近似有限时间的连续系统的能控丰富性计算,即只要 $N$ 充分大(离散化步长 $\varDelta$ 充分小),则有
$\mathrm{Vol}\left(R_{cx}\right)\approx\mathrm{Vol}\left(R_{dx}\right)$
若两输入两阶系统的线性连续系统 $\varSigma(A,B)$ 的各矩阵可表示为(或可变换为)
$A=\left[\begin{array}{cc} \eta_{1} & 0\\ 0 & \eta_{2} \end{array}\right]\qquad B=\left[\begin{array}{cc} b_{11} & b_{12}\\ b_{21} & b_{22} \end{array}\right]$
不失一般性,可通过对输入变量进行变换,设
$b_{11},b_{22},\det\left(B\right)=b_{11}b_{22}-b_{12}b_{21}>0$
对连续系统 $\varSigma(A,B)$ ,离散化后的系统矩阵和输入矩阵的各向量为
$G=\left[\begin{array}{cc} \lambda_{1} & 0\\ 0 & \lambda_{2} \end{array}\right]=\left[\begin{array}{cc} e^{\eta_{1}\Delta} & 0\\ 0 & e^{\eta_{2}\Delta} \end{array}\right]$
$H=\left[\begin{array}{cc} h_{11} & h_{12}\\ h_{21} & h_{22} \end{array}\right]=\varLambda B$
$\varLambda=\int_{0}^{\varDelta}e^{At}\textrm{d}t=\left[\begin{array}{cc} \frac{e^{\eta_{1}\Delta}-1}{\eta_{1}} & 0\\ 0 & \frac{e^{\eta_{2}\Delta}-1}{\eta_{2}} \end{array}\right]$
若 $h_{11}h_{22}\lambda_{1}^{s}-h_{12}h_{21}\lambda_{2}^{s}$ 不变号,则无限时间能达丰富性为(当离散化步长 $\varDelta$ 充分小)
$\mathrm{Vol}\left(R_{cx}\right)=\frac{\left[\left|b_{11}b_{21}\right|+\left|b_{12}b_{22}\right|\right]\left(\eta_{2}-\eta_{1}\right)}{\eta_{1}\eta_{2}\left(\eta_{1}+\eta_{2}\right)}+\frac{\det\left(B\right)}{\eta_{1}\eta_{2}}$
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