\[
\dot{x} = \begin{bmatrix} -2 & 1 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -3 \end{bmatrix} x + \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} u, \quad y = \begin{bmatrix} -2 & 3 & 1 \end{bmatrix} x
\]
(3) 将系统的传递函数分解为部分分式形式
\[
G(s) = \frac{Y(s)}{U(s)} = \frac{2s^2+5s+1}{(s+2)^3} = \frac{-1}{(s+2)^3} + \frac{-3}{(s+2)^2} + \frac{2}{s+2}
\]
令
\[
\begin{cases}
X_1(s) = \dfrac{1}{(s+2)^3}U(s) = \dfrac{1}{s+2}X_2(s) \\[2mm]
X_2(s) = \dfrac{1}{(s+2)^2}U(s) = \dfrac{1}{s+2}X_3(s) \\[2mm]
X_3(s) = \dfrac{1}{s+2}U(s)
\end{cases}
\]
则
\[
\begin{cases}
\dot{x}_1 = -2x_1 + x_2 \\
\dot{x}_2 = -2x_2 + x_3 \\
\dot{x}_3 = -2x_3 + u
\end{cases}
\]
写成向量-矩阵形式,可得系统的约当标准型实现为
\[
\dot{x} = \begin{bmatrix} -2 & 1 & 0 \\ 0 & -2 & 1 \\ 0 & 0 & -2 \end{bmatrix} x + \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} u, \quad y = \begin{bmatrix} -1 & -3 & 2 \end{bmatrix} x
\]
(4) 将系统的传递函数分解为部分分式形式
\[
G(s) = \frac{Y(s)}{U(s)} = \frac{(s+1)^3}{s^3} = 1 + \frac{3s^2+3s+1}{s^3} = 1 + \frac{1}{s^3} + \frac{3}{s^2} + \frac{3}{s}
\]
令
\[
\begin{cases}
X_1(s) = \dfrac{1}{s^3}U(s) = \dfrac{1}{s}X_2(s) \\[2mm]
X_2(s) = \dfrac{1}{s^2}U(s) = \dfrac{1}{s}X_3(s) \\[2mm]
X_3(s) = \dfrac{1}{s}U(s)
\end{cases}
\]
则
\[
\begin{cases}
\dot{x}_1 = x_2 \\
\dot{x}_2 = x_3 \\
\dot{x}_3 = u
\end{cases}
\]
写成向量-矩阵形式,可得系统的约当标准型实现为
\[
\dot{x} = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} x + \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} u, \quad y = \begin{bmatrix} 1 & 3 & 3 \end{bmatrix} x + u
\]
(5) MATLAB验证。最后,利用MATLAB的分式分解函数residue不难验证上述计算结果的正确性。
MATLAB程序:exe909.m
num1=[1 1 -6];den1=[1 10 35 50 24];[r1,p1,k1]=residue(num1,den1)
num2=[4 17 16];den2=[1 7 16 12];[r2,p2,k2]=residue(num2,den2)
num3=[2 5 1];den3=[1 6 12 8];[r3,p3,k3]=residue(num3,den3)
num4=conv(conv([1 1],[1 1]),[1 1]);den4=[1 0 0 0];[r4,p4,k4]=residue(num4,den4)
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