2阶带复根的SISO线性连续系统的能达域边界计算

     已有研究指出在单位输入变量 $\left(\left\Vert u_{t}\right\Vert _{\infty}\leq1,\forall t\right)$ 约束下线性连续时间系统的能达域的定义为

$R_{x}=\left\{ z\left|z=\int_{0}^{T}e^{At}Bu_{t}\textrm{d}t,\left\Vert u_{t}\right\Vert _{\infty}\leq1\right.\right\}$

且该能达域的边界为

$\partial R_{x}=\left\{ z\left|z=\int_{0}^{T}e^{At}B\textrm{sgn(}c^{T}e^{At}B)\textrm{d}t,\forall c\in R^{n}\right.\right\}$

      对2阶带复根的SISO线性连续系统 $\Sigma(A,B)$ ,设其系统矩阵和输入矩阵为(或可转换为)

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

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

其中

        $\rho_{B}=\left(b_{1}^{2}+b_{2}^{2}\right)^{1/2},\quad\delta=\arctan\frac{b_{2}}{b_{1}}\in(-\pi,\pi]$

        $\rho_{A}=\textrm{sgn}(\sigma)\left(\sigma^{2}+\mu^{2}\right)^{1/2},\quad\theta=\arctan\frac{\mu}{\sigma}\in[0,\pi)$

则有

$\partial R_{x}=\left\{ z\left|z=\int_{0}^{T}e^{At}B\textrm{sgn}\left(e^{\rho_{A}t}\cos(\mathit{\theta t}-\delta-\textbfsymbol{\phi})\right)\textrm{d}t,\forall\textbfsymbol{\textbfsymbol{\phi}}\in(-\pi,\pi]\right.\right\}$   (1)

对给定的 $\textbfsymbol{\phi}\in[-\pi,\pi]$ ,上式中的 $\textrm{sgn}\text{()}$ 函数的值发生变号的时间点为如下方程的解

$e^{\rho_{A}t}\cos(\mathit{\theta t}-\delta-\textbfsymbol{\phi})=0\quad\quad0

上述方程的解可能有多个,这取决于复根的虚部 $\theta$ 代表的频率和积分时长 $T$ 。因此,若在 $[0,T]$ 时长内,对给定的 $\textbfsymbol{\phi}\in[-\pi,\pi]$ ,上式的解 $\tau_{i}$ 为

        $\mathit{\tau_{i}}=s_{i}+\frac{\hat{\delta}+\pi/2}{\theta}\qquad i=1,2,\ldots,N$

        $\mathit{s_{i}}=(i-1)\Delta\qquad i=1,2,\ldots,N$

        $N\mathrm{=int}\left(\left|\frac{\mathit{\theta T}-\delta-\textbfsymbol{\phi}}{\pi}\right|\right)$

        $\Delta=\frac{\pi}{\theta}$

        $\hat{\delta}=\delta+\textbfsymbol{\phi}-q\pi,\qquad\hat{\delta}\in(-\pi/2,\pi/2],q$ 为整数

且有

$\cos(\theta s\mathit{_{i}}-\delta-\textbfsymbol{\phi})=(-1)^{i}\cos(\delta+\textbfsymbol{\phi}),\qquad i=0,1,2,\ldots,N$

则式(1)所描述的边界为

$\partial R_{x}(\phi) =\left\{ \pm z\left|z=\frac{\rho_{b}}{\sqrt{\rho_{A}^{2}+\theta^{2}}}\left(\left[\begin{array}{c} \cos(-\delta+\psi)\\ -\sin(-\delta+\psi) \end{array}\right]+(-1)^{1+q}\frac{2\left(1-e^{N\rho_{A}\varDelta}\right)e^{\rho_{A}\tau_{1}}}{1-e^{\rho_{A}\varDelta}}\left[\begin{array}{c} \sin(\phi+\psi)\\ \cos(\phi+\psi) \end{array}\right]\right.\right.\right.$

          $\left.\left.+(-1)^{N}e^{\rho_{A}T}\left[\begin{array}{c} -\cos(\mathit{\theta T}-\delta+\psi)\\ \sin(\mathit{\theta T}-\delta+\psi) \end{array}\right]\right)\right\}$     (2)

where

        $\tau_{0}=0,\qquad\tau_{N+1}=T$

        $\psi=\arctan\frac{\rho_{A}}{\theta}$

     当 $T\rightarrow\infty$ 且 $\sigma<0$ 时,式(2)所描述的边界为

$\partial R_{x}(\phi) =\left\{ \pm z\left|z=\frac{\rho_{b}}{\sqrt{\rho_{A}^{2}+\theta^{2}}}\left[\left[\begin{array}{c} \cos(-\delta+\psi)\\ -\sin(-\delta+\psi) \end{array}\right]+(-1)^{1+q}\frac{2e^{\rho_{A}\tau_{1}}}{1-e^{\rho_{A}\varDelta}}\left[\begin{array}{c} \sin(\phi+\psi)\\ \cos(\phi+\psi) \end{array}\right]\right]\right.\right\}$

                        $\textbfsymbol{\phi}\in[0,\pi)$