4阶有2对复根的线性离散系统的能达丰富性计算

       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2对复根,即系统矩阵和输入矩阵可表示为(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

          $\rho_{B_{i}}=\textrm{sgn}(b{}_{2i-1})\left(b_{2i-1}^{2}+b_{2i}^{2}\right)^{1/2},\quad\delta_{i}=\arctan\frac{b_{2i}}{b_{2i-1}}\in[0,\pi)$

          $\rho_{A_{i}}=\textrm{sgn}(\sigma_{i})\left(\sigma_{2i-1}^{2}+\mu_{2i-1}^{2}\right)^{1/2},\quad\theta_{i}=\arctan\frac{\mu_{2i-1}}{\sigma_{2i-1}}\in[0,\pi)$

则有能达富性计算如下

$V_{4}(C_{4}(A_{N}))=\sum_{(i_{1},i_{2},i_{3},i_{4})\in\Omega_{0,N-1}^{4}}\rho_{B_{1}}^{2}\rho_{B_{2}}^{2}\rho_{A_{1}}^{i_{1}+i_{2}+i_{3}+i_{4}}\left|k^{i_{3}+i_{4}}\sin(i_{2}-i_{1})\theta_{1}\sin(i_{4}-i_{3})\theta_{2}-k^{i_{2}+i_{4}}\sin(i_{3}-i_{1})\theta_{1}\sin(i_{4}-i_{2})\theta_{2}\right.$

                $\quad\quad+k^{i_{2}+i_{3}}\sin(i_{4}-i_{1})\theta_{1}\sin(i_{3}-i_{2})\theta_{2}+k^{i_{1}+i_{2}}\sin(i_{2}-i_{1})\theta_{2}\sin(i_{4}-i_{3})\theta_{1}$

              $\left.-k^{i_{1}+i_{3}}\sin(i_{3}-i_{1})\theta_{2}\sin(i_{4}-i_{2})\theta_{1}+k^{i_{1}+i_{4}}\sin(i_{4}-i_{1})\theta_{2}\sin(i_{3}-i_{2})\theta_{1}\right|$

其中 $\rho_{A_{2}}=k\rho_{A_{1}}$ 。

     当 $k=1$ ,即 $\rho_{A_{2}}=\rho_{A_{1}}=\rho_{A}$ ，且有 $N\rightarrow\infty$ 时

$V_{4}(C_{4}(A_{\infty}))=\sum_{k_{1}=0}^{\infty}\sum_{k_{2}=1}^{\infty}\sum_{k_{3}=1}^{\infty}\sum_{k_{4}=1}^{\infty}\rho_{B_{1}}^{2}\rho_{B_{2}}^{2}\rho_{A}^{4k_{1}+3k_{2}+2k_{3}+k_{4}}\left|\sin(k_{2}\theta_{1})\sin(k_{4}\theta_{2})-\sin(k_{2}+k_{3})\theta_{1}\sin(k{}_{3}+k{}_{4})\theta_{2}\right.$

              $+\sin(k_{2}+k_{3}+k_{4})\theta_{1}\sin(k_{3}\theta_{2})+\sin(k_{2}\theta_{2})\sin(k_{4}\theta_{1})$

              $\quad\quad\left.-\sin(k_{2}+k_{3})\theta_{2}\sin(k_{3}+k{}_{4})\theta_{1}+\sin(k_{2}+k_{3}+k_{4})\theta_{2}\sin(k_{3}\theta_{1})\right|$

当 $\rho_{A}<1$ ,有如下估计

            $V_{4}(C_{4}(A_{\infty}))\leq\sum_{k_{1}=0}^{\infty}\sum_{k_{2}=1}^{\infty}\sum_{k_{3}=1}^{\infty}\sum_{k_{4}=1}^{\infty}\rho_{B_{1}}^{2}\rho_{B_{2}}^{2}\rho_{A}^{4k_{1}+3k_{2}+2k_{3}+k_{4}}\widetilde{V}$

                        $=\frac{\rho_{B_{1}}^{2}\rho_{B_{2}}^{2}\rho_{A}^{9}\widetilde{V}}{\left(1-\rho_{A}\right)\left(1-\rho_{A}^{2}\right)\left(1-\rho_{A}^{3}\right)\left(1-\rho_{A}^{4}\right)}$

其中 $\widetilde{V}$ 为如下上界估计值

  $\widetilde{V}=\max_{\theta_{i}\in(-\pi,\pi]}\left|\sin(\theta_{2})\sin(\theta_{8})-\sin(\theta_{2}+\theta_{3})\sin(\theta{}_{7}+\theta{}_{8})+\sin(\theta_{2}+\theta_{3}+\theta_{4})\sin(\theta_{7})\right.$

          $\qquad\left.+\sin(\theta_{6})\sin(\theta_{4})-\sin(\theta_{6}+\theta_{7})\sin(\theta_{3}+\theta{}_{4})+\sin(\theta_{6}+\theta_{7}+\theta_{8})\sin(\theta_{3})\right|$

上述上界估计值目前还未得到好的结果。