图 9-8 离散系统的零输入状态响应(MATLAB)
其中 \(\boldsymbol{\Phi}(T)\) 与连续系统状态转移矩阵 \(\boldsymbol{\Phi}(t)\) 的关系为
\[\boldsymbol{\Phi}(T) = \boldsymbol{\Phi}(t)\mid_{t=T}, \qquad \boldsymbol{G}(T) = \int_0^T \boldsymbol{\Phi}(\tau)\boldsymbol{b}\,\mathrm{d}\tau\]
(1) 系统(1)的离散时间状态方程。先求连续系统的状态转移矩阵
\[\boldsymbol{\Phi}(t) = \mathscr{L}^{-1}[(s\boldsymbol{I}-\boldsymbol{A})^{-1}] = \mathscr{L}^{-1}\begin{bmatrix} \dfrac{3}{s-2}-\dfrac{2}{s-3} & \dfrac{1}{s-3}-\dfrac{1}{s-2} \\[2mm] \dfrac{6}{s-2}-\dfrac{6}{s-3} & \dfrac{3}{s-3}-\dfrac{2}{s-2} \end{bmatrix}\]
\[=\begin{bmatrix} 3\mathrm{e}^{2t}-2\mathrm{e}^{3t} & \mathrm{e}^{3t}-\mathrm{e}^{2t} \\ 6\mathrm{e}^{2t}-6\mathrm{e}^{3t} & 3\mathrm{e}^{3t}-2\mathrm{e}^{2t} \end{bmatrix}\]
\[\boldsymbol{\Phi}(T) = \boldsymbol{\Phi}(t)\mid_{t=T} = \begin{bmatrix} 3\mathrm{e}^{2T}-2\mathrm{e}^{3T} & \mathrm{e}^{3T}-\mathrm{e}^{2T} \\ 6\mathrm{e}^{2T}-6\mathrm{e}^{3T} & 3\mathrm{e}^{3T}-2\mathrm{e}^{2T} \end{bmatrix}\]
\[\boldsymbol{G}(T) = \int_0^T \boldsymbol{\Phi}(\tau)\boldsymbol{b}\,\mathrm{d}\tau = \int_0^T \begin{bmatrix} 3\mathrm{e}^{2\tau}-2\mathrm{e}^{3\tau} & \mathrm{e}^{3\tau}-\mathrm{e}^{2\tau} \\ 6\mathrm{e}^{2\tau}-6\mathrm{e}^{3\tau} & 3\mathrm{e}^{3\tau}-2\mathrm{e}^{2\tau} \end{bmatrix}\begin{bmatrix} 1 \\ -2 \end{bmatrix}\mathrm{d}\tau = \begin{bmatrix} \dfrac{5}{2}\mathrm{e}^{2T}-\dfrac{4}{3}\mathrm{e}^{3T}-\dfrac{7}{6} \\[2mm] -4\mathrm{e}^{3T}+5\mathrm{e}^{2T}-1 \end{bmatrix}\]
相应的离散时间状态方程为
\[\boldsymbol{x}(k+1) = \begin{bmatrix} 3\mathrm{e}^{2T}-2\mathrm{e}^{3T} & \mathrm{e}^{3T}-\mathrm{e}^{2T} \\ 6\mathrm{e}^{2T}-6\mathrm{e}^{3T} & 3\mathrm{e}^{3T}-2\mathrm{e}^{2T} \end{bmatrix}\boldsymbol{x}(k) + \begin{bmatrix} \dfrac{5}{2}\mathrm{e}^{2T}-\dfrac{4}{3}\mathrm{e}^{3T}-\dfrac{7}{6} \\[2mm] -4\mathrm{e}^{3T}+5\mathrm{e}^{2T}-1 \end{bmatrix}u(k)\]
(2) 系统(2)的离散时间状态方程。先求连续系统的状态转移矩阵
\[\boldsymbol{\Phi}(t) = \mathscr{L}^{-1}[(s\boldsymbol{I}-\boldsymbol{A})^{-1}] = \mathscr{L}^{-1}\begin{bmatrix} \dfrac{1}{s} & \dfrac{1}{2}\left(\dfrac{1}{s}-\dfrac{1}{s+2}\right) \\[2mm] 0 & \dfrac{1}{s+2} \end{bmatrix} = \begin{bmatrix} 1 & \dfrac{1}{2}(1-\mathrm{e}^{-2t}) \\ 0 & \mathrm{e}^{-2t} \end{bmatrix}\]
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