4阶有2对复根的线性离散系统的能控丰富性计算

     本人的文章arXiv1705.08064(On Controllable Abundance Of Saturated-input Linear Discrete Systems) 里定义了线性离散系统的controllable abundance(能控丰富性、能控充裕性)及其计算。当系统状态空间维数 $n=4$ 且矩阵 $A$ 的特征根为两对复根,即系统矩阵可表示为(或经变换可表示为)

      $A=\left[\begin{array}{cccc} \sigma_{1} & \mu_{1} & 0 & 0\\ -\mu_{1} & \sigma_{1} & 0 & 0\\ 0 & 0 & \sigma_{2} & \mu_{2}\\ 0 & 0 & -\mu_{2} & \sigma_{2} \end{array}\right]=\left[\begin{array}{cccc} \rho_{A_{1}}\cos\theta_{1} & \rho_{A_{1}}\sin\theta_{1} & 0 & 0\\ -\rho_{A_{1}}\sin\theta_{1} & \rho_{A_{1}}\cos\theta_{1} & 0 & 0\\ 0 & 0 & \rho_{A_{2}}\cos\theta_{2} & \rho_{A_{2}}\sin\theta_{2}\\ 0 & 0 & -\rho_{A_{2}}\sin\theta_{2} & \rho_{A_{2}}\cos\theta_{2} \end{array}\right]$

      $B=\left[\begin{array}{c} b_{1}\\ b_{2}\\ b_{3}\\ b_{4} \end{array}\right]=\left[\begin{array}{c} \rho_{B_{1}}\cos\delta_{1}\\ \rho_{B_{1}}\sin\delta_{1}\\ \rho_{B_{2}}\cos\delta_{2}\\ \rho_{B_{2}}\sin\delta_{2} \end{array}\right]$

其中

          $\rho_{A_{i}}=\left(\sigma_{2i-1}^{2}+\mu_{2i-1}^{2}\right)^{1/2},\quad\theta_{i}=\arctan\frac{\mu_{2i-1}}{\sigma_{2i-1}}$

          $\rho_{B_{i}}=\left(b_{2i-1}^{2}+b_{2i}^{2}\right)^{1/2},\quad\delta_{i}=\arctan\frac{b_{2i}}{b_{2i-1}}$

则有能控丰富性计算结果如下

  $V_{4}(C_{4}(A_{4}))=\rho_{B_{1}}^{2}\rho_{B_{2}}^{2}\rho_{A_{1}}\rho_{A_{2}}\left|\rho_{A_{2}}^{4}\sin\theta_{1}\sin\theta_{2}-\rho_{A_{1}}\rho_{A_{2}}^{3}\sin2\theta_{1}\sin2\theta_{2}+\rho_{A_{1}}^{2}\rho_{A_{2}}^{2}\sin3\theta_{1}\sin\theta_{2}\right.$

              $\quad\quad\left.+\rho_{A_{1}}^{4}\sin\theta_{2}\sin\theta_{1}-\rho_{A_{2}}\rho_{A_{1}}^{3}\sin2\theta_{2}\sin2\theta_{1}+\rho_{A_{2}}^{2}\rho_{A_{1}}^{2}\sin3\theta_{2}\sin\theta_{1}\right|\noindent$

当 $\rho_{A_{1}}$ 和 $\rho_{A_{2}}$ 有上界 $\rho$ 时，上式最大值为

        $V_{4}(C_{4}(A_{4}))=\rho_{B_{1}}^{2}\rho_{B_{2}}^{2}\rho^{6}\left|\sin\theta_{1}\sin\theta_{2}\right|(\cos\theta_{1}-\cos\theta_{2})^{2}$

对于矩阵 $A$ 的两对复根，其角度满足 $\theta_{1}=45\textdegree,\quad\theta_{2}=135\textdegree$ 时(即系统的4个极点(特征根)分布最均匀),该线性离散系统的 $N=4$ 时的能控丰富性达到如下极值

$V_{4}(C_{4}(A_{4}))=\frac{1}{4}\rho_{B_{1}}^{2}\rho_{B_{2}}^{2}\rho^{6}$

上述结果于2017年7月推导并给出.