传递函数变换到状态空间

1. 分子为1的传递函数

例：

G (s) = \frac{1}{s^{3} + a_{2} s^{2} + a_{1} s + a_{0}}

$G(s)=\frac{1}{s^3+a_2s^2+a_1s+a_0}$

首先写成输入输出关系：

(s^{3} + a_{2} s^{2} + a_{1} s + a_{0}) Y (s) = U (s)

$(s^3+a_2s^2+a_1s+a_0)Y(s)=U(s)$

对应的微分方程：

\overset{⃛}{y} (t) + a_{2} \ddot{y} (t) + a_{1} \dot{y} (t) + a_{0} y (t) = u (t) \overset{⃛}{y} (t) = - a_{2} \ddot{y} (t) - a_{1} \dot{y} (t) - a_{0} y (t) + u (t)

$\dddot{y}(t)+a_2\ddot y(t)+a_1\dot y(t)+a_0y(t)=u(t)\\ \dddot{y}(t)=-a_2\ddot y(t)-a_1\dot y(t)-a_0y(t)+u(t)\\$

令：

x_{1} = y, x_{2} = \dot{y}, x_{3} = \ddot{y}

$x_1=y,\ x_2=\dot y,\ x_3=\ddot y$

得到状态空间模型：

\begin{array}{l} {\dot{x}}_{1} = x_{2} \\ {\dot{x}}_{2} = x_{3} \\ {\dot{x}}_{3} = - a_{0} x_{1} - a_{1} x_{2} - a_{2} x_{3} + u \end{array} \Rightarrow {\begin{cases} [\begin{matrix} {\dot{x}}_{1} \\ {\dot{x}}_{2} \\ {\dot{x}}_{3} \end{matrix}] = [\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ - a_{0} & - a_{1} & - a_{2} \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] + [\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}] u \\ y = [\begin{matrix} 1 & 0 & 0 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] \end{cases}

$\begin{array}{l} \dot x_1=x_2\\ \dot x_2 = x_3\\ \dot x_3 = -a_0x_1-a_1x_2-a_2x_3+u \end{array} \Rightarrow \begin{cases} \begin{bmatrix} \dot x_1\\\dot x_2\\\dot x_3 \end{bmatrix}= \begin{bmatrix} 0&1&0\\0&0&1\\ -a_0&-a_1&-a_2 \end{bmatrix} \begin{bmatrix} x_1\\x_2\\x_3 \end{bmatrix}+ \begin{bmatrix} 0\\0\\1 \end{bmatrix}u\\ y= \begin{bmatrix} 1&0&0 \end{bmatrix} \begin{bmatrix} x_1\\x_2\\x_3 \end{bmatrix} \end{cases}$

模拟图：

2. 分子不为1的传递函数

例：

G (s) = \frac{b_{2} s^{2} + b_{1} s + b_{0}}{s^{3} + a_{2} s^{2} + a_{1} s + a_{0}} = \frac{b (s)}{a (s)}

$G(s)=\frac{b_2s^2+b_1s+b_0}{s^3+a_2s^2+a_1s+a_0}=\frac{b(s)}{a(s)}$

可以看成：

Y (s) = \frac{b (s)}{a (s)} U (s) = b (s) [\frac{1}{a (s)} U (s)] = b (s) Y_{1} (s)

$Y(s)=\frac{b(s)}{a(s)}U(s)=b(s)\left [\frac{1}{a(s)}U(s) \right ]=b(s)Y_1(s)$

对于

Y_{1} (s) = \frac{1}{a (s)} U (s)

$Y_1(s)=\frac{1}{a(s)}U(s)$

可直接写出

G_{1} (s) = \frac{1}{a (s)} = \frac{1}{s^{3} + a_{2} s^{2} + a_{1} s + a_{0}} ⇓ {\begin{cases} [\begin{matrix} {\dot{x}}_{1} \\ {\dot{x}}_{2} \\ {\dot{x}}_{3} \end{matrix}] = [\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ - a_{0} & - a_{1} & - a_{2} \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] + [\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}] u \\ y_{1} = x_{1} \end{cases}

$G_1(s)=\frac{1}{a(s)}=\frac{1}{s^3+a_2s^2+a_1s+a_0}\\ \Downarrow\\ \begin{cases} \begin{bmatrix} \dot x_1\\ \dot x_2\\ \dot x_3 \end{bmatrix}= \begin{bmatrix} 0&1&0\\0&0&1\\ -a_0&-a_1&-a_2 \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}+ \begin{bmatrix} 0\\0\\ 1 \end{bmatrix}u\\ y_1=x_1 \end{cases}$

对于

Y (s) = b (s) Y_{1} (s) = b (s) X_{1} (s) = b_{2} s^{2} X_{1} (s) + b_{1} s X_{1} (s) + b_{0} X_{1} (s)

$Y(s)=b(s)Y_1(s)=b(s)X_1(s)=b_2s^2X_1(s)+b_1sX_1(s)+b_0X_1(s)$

可得

y = b_{2} {\ddot{x}}_{1} + b_{1} {\dot{x}}_{1} + b_{0} x_{1} y = b_{2} x_{3} + b_{1} x_{2} + b_{0} x_{1}

$y = b_2\ddot x_1+b_1 \dot x_1 +b_0 x_1\\ y=b_2x_3+b_1x_2+b_0x_1$

状态空间模型：

{\begin{cases} [\begin{matrix} {\dot{x}}_{1} \\ {\dot{x}}_{2} \\ {\dot{x}}_{3} \end{matrix}] = [\begin{matrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ - a_{0} & - a_{1} & - a_{2} \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] + [\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}] u \\ y = [\begin{matrix} b_{0} & b_{1} & b_{2} \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] \end{cases}

$\begin{cases} \begin{bmatrix} \dot x_1\\ \dot x_2\\ \dot x_3 \end{bmatrix}= \begin{bmatrix} 0&1&0\\0&0&1\\ -a_0&-a_1&-a_2 \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}+ \begin{bmatrix} 0\\0\\ 1 \end{bmatrix}u\\ y=\begin{bmatrix} b_0&b_1&b_2 \end{bmatrix} \begin{bmatrix} x_1\\x_2\\x_3 \end{bmatrix} \end{cases}$

模拟图：