MATLAB 程序:exe908.m
den1=conv([1 1],[1 2]);num1=[0 0 0;1 3 2;0 1 2;0 1 1];
[A,B,C,D] = tf2ss(num1,den1)
[Ac1,Bc1,Cc1,P1]=normal_control(A,B,C,D)
Ao1=Ac1';Bo1=Cc1';Co1=Bc1';Do1=D';
9-9 已知系统传递函数
(1) \(G(s) = \dfrac{s^2+s-6}{s^4+10s^3+35s^2+50s+24}\); (2) \(G(s) = \dfrac{4s^2+17s+16}{s^3+7s^2+16s+12}\);
(3) \(G(s) = \dfrac{2s^2+5s+1}{s^3+6s^2+12s+8}\); (4) \(G(s) = \dfrac{(s+1)^3}{s^3}\)。
试求系统的约当标准型或对角线标准型实现。
解 (1) 将系统传递函数分解为部分分式形式
\[G(s) = \frac{Y(s)}{U(s)} = \frac{(s+3)(s-2)}{(s+1)(s+2)(s+3)(s+4)}\]
\[= \frac{s-2}{(s+1)(s+2)(s+4)} = \frac{-1}{s+1} + \frac{2}{s+2} + \frac{-1}{s+4}\]
令
\[
\begin{cases}
X_1(s) = \dfrac{1}{s+1}U(s) \\[2mm]
X_2(s) = \dfrac{1}{s+2}U(s) \\[2mm]
X_3(s) = \dfrac{1}{s+4}U(s)
\end{cases}
\]
则
\[
\begin{cases}
\dot{x}_1 = -x_1 + u \\
\dot{x}_2 = -2x_2 + u \\
\dot{x}_3 = -4x_3 + u
\end{cases}
\]
写成向量-矩阵形式,可得系统的对角线标准型最小实现为
\[
\dot{\mathbf{x}} =
\begin{bmatrix}
-1 & 0 & 0 \\
0 & -2 & 0 \\
0 & 0 & -4
\end{bmatrix}
\mathbf{x} +
\begin{bmatrix}
1 \\ 1 \\ 1
\end{bmatrix}
u,\qquad
y = \begin{bmatrix} -1 & 2 & -1 \end{bmatrix}\mathbf{x}
\]
(2) 将系统的传递函数分解为部分分式形式
\[G(s) = \frac{Y(s)}{U(s)} = \frac{4s^2+17s+16}{(s+3)(s+2)^2} = \frac{-2}{(s+2)^2} + \frac{3}{s+2} + \frac{1}{s+3}\]
令
\[
\begin{cases}
X_1(s) = \dfrac{1}{(s+2)^2}U(s) = \dfrac{1}{s+2}X_2(s) \\[2mm]
X_2(s) = \dfrac{1}{s+2}U(s) \\[2mm]
X_3(s) = \dfrac{1}{s+3}U(s)
\end{cases}
\]
则
\[
\begin{cases}
\dot{x}_1 = -2x_1 + x_2 \\
\dot{x}_2 = -2x_2 + u \\
\dot{x}_3 = -3x_3 + u
\end{cases}
\]
写成向量-矩阵形式,可得系统的约当标准型实现为
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