两输入两阶系统的复根时的能达丰富性

         若两输入两阶系统的特征根为复根，即线性离散系统 $\varSigma(A,B)$ 的各矩阵可表示为（或可变换为）

            $A_{N}=[B,AB,...,A^{N-1}B]$

            $A=\left[\begin{array}{cc} \sigma & -\mu\\ \mu & \sigma \end{array}\right]=\lambda\left[\begin{array}{cc} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{array}\right]$

            $B=\left[\begin{array}{cc} b_{11} & b_{12}\\ b_{21} & b_{22} \end{array}\right]=\left[\begin{array}{cc} \rho_{1}\left[\begin{array}{c} \cos\delta_{1}\\ \sin\delta_{1} \end{array}\right] & \rho_{2}\left[\begin{array}{c} \cos\delta_{2}\\ \sin\delta_{2} \end{array}\right]\end{array}\right]$

其中

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

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

故，

$A_{N}=\left[\rho_{1}\left[\begin{array}{c} \cos\delta_{1}\\ \sin\delta_{1} \end{array}\right],\rho_{2}\left[\begin{array}{c} \cos\delta_{2}\\ \sin\delta_{2} \end{array}\right],\lambda\rho_{1}\left[\begin{array}{c} \cos(\theta+\delta_{1})\\ \sin(\theta+\delta_{1}) \end{array}\right],\lambda\rho_{2}\left[\begin{array}{c} \cos(\theta+\delta_{2})\\ \sin(\theta+\delta_{2}) \end{array}\right],...,\lambda^{N-1}\rho_{1}\left[\begin{array}{c} \cos[(N-1)\theta+\delta_{1}]\\ \sin[(N-1)\theta+\delta_{1}] \end{array}\right]\lambda^{N-1}\rho_{2}\left[\begin{array}{c} \cos[(N-1)\theta+\delta_{2}]\\ \sin[(N-1)\theta+\delta_{2}] \end{array}\right]\right]$

因此，有 $N$ 步能达丰富性为

$V_{2}(C_{2}(A_{N})) =V_{2}(C_{2}(A_{N}^{(1)}))+V_{2}(C_{2}(A_{N}^{(2)}))+\sum_{i=0}^{N-1}\sum_{j=0}^{N-1}\lambda^{i+j}\rho_{1}\rho_{2}\left|\sin[(j-i)\theta+\delta_{2}-\delta_{1}]\right|$

相应的无限时间能达丰富性为

$V_{2}(C_{2}(A_{\infty}))= \frac{\rho_{1}^{2}+\rho_{2}^{2}}{\left(1-\lambda^{2}\right)\left(1-\lambda^{K}\right)}\delta_{1,K}^{\lambda}(\theta,0)+\frac{\rho_{1}\rho_{2}}{\left(1-\lambda^{2}\right)}\left|\sin[\delta_{2}-\delta_{1}]\right|+\frac{\rho_{1}\rho_{2}}{\left(1-\lambda^{2}\right)\left(1-\lambda^{K}\right)}\left[\delta_{1,K}^{\lambda}(\theta,\delta_{2}-\delta_{1})+\delta_{1,K}^{\lambda}(\theta,\delta_{1}-\delta_{2})\right]$

其中 $K$ 为满足 $K\theta$ 为 $\pi$ 的最小整倍数的 $K$ ，

$\delta_{1,K}^{\lambda}(\theta,\delta)=\sum_{k=1}^{K}\left|\lambda^{k}\sin[k\theta+\delta]\right|$