3阶有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=3且矩阵A的特征根为有一对复根,即系统矩阵可表示为(或经变换可表示为)

      $A=\left[\begin{array}{ccc} \lambda & 0 & 0\\ 0 & \sigma & \mu\\ 0 & -\mu & \sigma \end{array}\right]=\left[\begin{array}{ccc} \lambda & 0 & 0\\ 0 & \lambda_{A}\cos\theta & -\lambda_{A}\sin\theta\\ 0 & \lambda_{A}\sin\theta & \lambda_{A}\cos\theta \end{array}\right]=\lambda\left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & r\cos\theta & -r\sin\theta\\ 0 & r\sin\theta & r\cos\theta \end{array}\right]$

      $B=\left[\begin{array}{c} b_{1}\\ b_{2}\\ b_{3} \end{array}\right]=\left[\begin{array}{c} b_{1}\\ \rho\cos\delta\\ \rho\sin\delta \end{array}\right]$

其中

        $\rho=\textrm{sgn}(b_{2})\left(b_{2}^{2}+b_{3}^{2}\right)^{1/2},\quad\delta=\arctan\frac{b_{3}}{b_{2}}\in\left[\frac{-\pi}{2},\frac{\pi}{2}\right)$  

        $\lambda_{A}=\textrm{sgn}(\sigma)\left(\sigma^{2}+\mu^{2}\right)^{1/2},\quad\theta=\arctan\frac{\mu}{\sigma}\in\left[\frac{-\pi}{2},\frac{\pi}{2}\right)$

        $\lambda_{A}=r\lambda$

则有能达富性计算如下

        $V_{3}(C_{3}(A_{N}))=\left|b_{1}\rho^{2}\right|\sum_{(k_{1},k_{2},k_{3})\in\Omega_{0,N-1}^{3}}\left|\lambda^{k_{1}+k_{2}+k_{3}}\right|\left|r^{k_{2}+k_{3}}\sin[(k_{3}-k_{2})\theta]$

                        $-r^{k_{1}+k_{3}}\sin[(k_{3}-k_{1})\theta]\right. +r^{k_{1}+k_{2}}\sin[(k_{2}-k_{1})\theta]\right|$

当 $r=1$ 时,对上述能达丰富性的上界可估计如下

$V_{3}(C_{3}(A_{N}))\leq\left|\frac{3\sqrt{3}\left|b_{1}\rho^{2}\right|}{2}\left[\frac{\lambda^{3}-\lambda^{3N-3}}{\left(1-\lambda\right)^{2}\left(1+\lambda\right)\left(1-\lambda^{3}\right)}+\frac{\lambda^{2N}-\lambda^{3N-2}-\lambda^{N+1}+\lambda^{3N-3}}{\left(1-\lambda\right)^{3}\left(1+\lambda\right)}\right]\right|$

当 $\lambda=\lambda_{A}<1$ 时,无限时间能达丰富性的上界可估计如下

$\lim_{N\rightarrow\infty}V_{3}(C_{3}(A_{N}))\leq\left|\frac{3\sqrt{3}\left|b_{1}\rho^{2}\right|\lambda^{3}}{2\left(1-\lambda\right)\left(1-\lambda^{2}\right)\left(1-\lambda^{3}\right)}\right|$

       对 $\lambda=\lambda_{A}<1$ ，有更精准的无限时间能达丰富性估计为

$\lim_{N\rightarrow\infty}V_{3}(C_{3}(A_{N}))\leq\left|\frac{3\theta^{3}\left|b_{1}\rho^{2}\right|\left(1+\lambda^{2\pi/\theta}\right)}{2\ln\lambda\left(1-\lambda^{3}\right)\left(1-\lambda^{2\pi/\theta}\right)\left[\theta^{2}+\ln^{2}\lambda\right]\left[\theta^{2}+4\ln^{2}\lambda\right]}\right|$

当 $\theta=90\textdegree,120\textdegree$ 度时，无限时间能达丰富性的精确值为

        $\left.V_{3}(C_{3}(A_{\infty}))\right|_{\theta=120}=\frac{3^{3/2}\left|b_{1}\rho^{2}\right|\left(\lambda^{3}+\lambda^{6}\right)}{2\left(1-\lambda^{3}\right)^{2}\left(1-\lambda^{6}\right)}$

        $\left.V_{3}(C_{3}(A_{\infty}))\right|_{\theta=90}=\frac{2\left|b_{1}\rho^{2}\right|\left(\lambda^{3}+\lambda^{4}+\lambda^{5}+\lambda^{7}+\lambda^{8}+\lambda^{9}\right)}{\left(1-\lambda^{3}\right)\left(1-\lambda^{4}\right)\left(1-\lambda^{8}\right)}$