2阶有复根的离散系统的能控丰富性极值计算(整理中)

     本人的博文“2阶有复根的离散系统的能控丰富性计算”(http://blog.sciencenet.cn/home.php?mod=space&uid=3343777&do=blog&id=1065494) 中，对系统状态空间维数 $n=2$ 且矩阵 $A$ 的特征根为一对复根的线性离散系统，计算其controllable abundance(能控丰富性、能控充裕性)为  

  $\rho_{A}=1:\quad V_{2}(C_{2}(A_{N})) =\rho_{B}^{2}\times\sum_{k=1}^{N-1}(N-k)\left|\sin(k\theta)\right|" original="http://latex.codecogs.com/gif.latex?\rho_{A}=1:\quad V_{2}(C_{2}(A_{N})) =\rho_{B}^{2}\times\sum_{k=1}^{N-1}(N-k)\left|\sin(k\theta)\right|" style="margin:0px;padding:0px;word-wrap:break-word;max-width:620px;display:inline;$

    $\rho_{A}<1:\quad V_{2}(C_{2}(A_{N})) =\rho_{B}^{2}\times\sum_{k=1}^{N-1}\left|\sin(k\theta)\right|\frac{\rho_{A}^{k}-\rho_{A}^{2N-k}}{1-\rho_{A}^{2}}" original="http://latex.codecogs.com/gif.latex?\rho_{A}<1:\quad V_{2}(C_{2}(A_{N})) =\rho_{B}^{2}\times\sum_{k=1}^{N-1}\left|\sin(k\theta)\right|\frac{\rho_{A}^{k}-\rho_{A}^{2N-k}}{1-\rho_{A}^{2}}" style="margin:0px;padding:0px;word-wrap:break-word;max-width:620px;display:inline;$

    $\rho_{A}>1:\quad V_{2}(C_{2}(A_{N}))=\rho_{B}^{2}\times\sum_{k=1}^{N-1}\left|\sin(k\theta)\right|\frac{\rho_{A}^{2(N-k-1)+k}-\rho_{A}^{k-2}}{1-\rho_{A}^{-2}}" original="http://latex.codecogs.com/gif.latex?\rho_{A}>1:\quad V_{2}(C_{2}(A_{N}))=\rho_{B}^{2}\times\sum_{k=1}^{N-1}\left|\sin(k\theta)\right|\frac{\rho_{A}^{2(N-k-1)+k}-\rho_{A}^{k-2}}{1-\rho_{A}^{-2}}" style="margin:0px;padding:0px;word-wrap:break-word;max-width:620px;display:inline;$

    当 $\rho_{A}=1$ 时,若 $\theta=\frac{\pi}{N}+\delta$ 且 $(N-1)(\frac{\pi}{N}+\delta)\leq\pi$ ,有

$V_{2}(C_{2}(A_{N}))=\rho_{B}^{2}\times\left[\sum_{k=1}^{N-1}(N-k)\sin\frac{k\pi}{N}+\delta\sum_{k=1}^{N-1}(N-k)k\cos\frac{k\pi}{N}-\frac{\delta^{2}}{2}\sum_{k=1}^{N-1}(N-k)k^{2}\sin\frac{k\pi}{N}\right]+o(\delta^{2})$

考虑到

$\sum_{k=1}^{N-1}(N-k)k\cos\frac{k\pi}{N}=0,\qquad\sum_{k=1}^{N-1}(N-k)k^{2}\sin\frac{k\pi}{N}>0$

故有，函数 $V_{2}(C_{2}(A_{N}))$ 的极值点为 $\theta^{*}=\frac{\pi}{N}$ ,函数 $V_{2}(C_{2}(A_{N}))$ 的极值(最大能控丰富性)为 $\max_{\theta\in[0,\pi]}V_{2}(C_{2}(A_{N}))=\rho_{B}^{2}\sum_{k=1}^{N-1}(N-k)\sin\frac{k\pi}{N}$

即，当 $N=2,3,4$ 时， $V_{2}(C_{2}(A_{N}))$ 的极值和极值点分别为：

    $\max_{\theta\in[0,\pi]}V_{2}(C_{2}(A_{2}))=\rho_{B}^{2},\qquad\theta^{*}=\frac{\pi}{2}$

    $\max_{\theta\in[0,\pi]}V_{2}(C_{2}(A_{3}))=\frac{3\sqrt{3}}{2}\rho_{B}^{2},\qquad\theta^{*}=\frac{\pi}{3}$  

    $\max_{\theta\in[0,\pi]}V_{2}(C_{2}(A_{4}))=\left(2+2\sqrt{2}\right)\rho_{B}^{2},\qquad\theta^{*}=\frac{\pi}{4}$

     由上述结果并可推知，当 $\rho_{A}<1$ 时，函数 $V_{2}(C_{2}(A_{N}))$ 的极值点为 $\theta^{*}\geq\frac{\pi}{N}$ ;当 $\rho_{A}>1$ 时，函数 $V_{2}(C_{2}(A_{N}))$ 的极值点为 $\theta^{*}\leq\frac{\pi}{N}$ .

上述递推式于2017年7月推导并给出.