状态转移矩阵计算方法及其离散化转换（含举例）

1. 何为状态转移矩阵

一般地，对于一个线性定常系统，可以写成如下的柯西标准型形式
$$\{\begin{array}{l}\dot{x}(t)=A(t)x(t)+B(t)u(t)\\ y(t)=C(t)x(t)+D(t)u(t)\end{array}$$其中$x(t),u(t),y(t)$分别为系统状态和、系统输入控制量和输出量，为向量。该方程组的解为
$$x(t)=e^{At}{x}_0=\left(I+At+\frac{A^2{t}^2}{2!}+\cdots \right)x_0=\Phi (t)x_0\text{\hspace{1em}(1)}$$其中$I$为单位矩阵。
式(1)中的$\Phi (t)$即为状态转移矩阵。

2. 状态转移矩阵计算举例

从式(1)可以看出，状态转移矩阵$\Phi (t)=e^{At}$虽然具有指数形式，但由于$A$是矩阵，因而其计算并不是简单地将$A$中所有元素变为$e$的指数就行的。相反，其计算需要具有式(1)的形式：
$$\Phi (t)=e^{At}=I+At+\frac{A^2{t}^2}{2!}+\frac{A^3{t}^3}{3!}+\cdots +\frac{A^i{t}^i}{i!}+\cdots \text{\hspace{1em}(2)}$$以下列传递函数为例，进行状态转移矩阵的计算。
例：$$W(s)=\frac{s+1}{s^2+12s+32}$$可以写出其对应的状态空间形式：$$A=\left[\begin{array}{cc}-12& -32\\ 1& 0\end{array}\right],$$$$B=\left[\begin{array}{c}1\\ 0\end{array}\right],$$$$C=\left[\begin{array}{cc}1& 1\end{array}\right],$$$$D=0,$$则
$$A^2=\left[\begin{array}{cc}112& 384\\ -12& -32\end{array}\right],A^3=\left[\begin{array}{cc}-960& -3584\\ 112& 384\end{array}\right],\cdots$$代入式(2)有
$$\begin{array}{rl}\Phi (t)& =e^{At}=I+At+\frac{A^2{t}^2}{2!}+\frac{A^3{t}^3}{3!}+\cdots +\frac{A^i{t}^i}{i!}+\cdots \\ & =\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]+\left[\begin{array}{cc}-12& -32\\ 1& 0\end{array}\right]t+\frac{1}{2}\left[\begin{array}{cc}112& 384\\ -12& -32\end{array}\right]t^2+\frac{1}{6}\left[\begin{array}{cc}-960& -3584\\ 112& 384\end{array}\right]t^3+\cdots \\ & =\left[\begin{array}{cc}1-12t+56t^2-160t^3+\cdots & -32t+192t^2-\frac{1792}{3}{t}^3+\cdots \\ t-6t^2+\frac{56}{3}{t}^3+\cdots & 1-16t^2+64t^3+\cdots \end{array}\right]\end{array}$$

3. 离散系统的状态方程

对于离散系统
$$\{\begin{array}{l}x(k+1)=A_d(k)x(k)+B_d(k)u(k)\\ y(k)=C_d(k)x(k)+D_d(k)u(k)\end{array}$$其中$k$指量化后的第$k$时刻，即：若离散化的每个时间单位为$T_0$，则$x(k+1)$即指在$(k+1)T_0$时刻下的状态量。
在离散化的系统下，矩阵$A_d,B_d,C_d,D_d$与原连续系统下的$A,B,C,D$具有如下转换关系：
$$A_d=\Phi \left(T_0\right)=e^{A{T}_0},B_d={\int }_0^{T_0}\Phi (q)Bdq,C_d=C,D_d=D\text{\hspace{1em}(3)}$$

4. 离散系统的状态方程计算举例

仍以如下连续系统传递函数为例：
$$W(s)=\frac{s+1}{s^2+12s+32}$$设量化时间单位为$T_0=0.01s$。连续系统下的矩阵$A,B,C,D$为$$A=\left[\begin{array}{cc}-12& -32\\ 1& 0\end{array}\right],B=\left[\begin{array}{c}1\\ 0\end{array}\right],C=\left[\begin{array}{cc}1& 1\end{array}\right],D=0,$$则$$\begin{array}{rl}{A}_d& =\Phi \left(T_0\right)=e^{A{T}_0}\\ & =I+A{T}_0+\frac{A^2{T}_0^2}{2!}+\frac{A^3{T}_0^3}{3!}+\cdots \\ & =\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]+\left[\begin{array}{cc}-12& -32\\ 1& 0\end{array}\right]T_0+\frac{1}{2}\left[\begin{array}{cc}112& 384\\ -12& -32\end{array}\right]T_0^2+\frac{1}{6}\left[\begin{array}{cc}-960& -3584\\ 112& 384\end{array}\right]T_0^3+\cdots \\ & =\left[\begin{array}{cc}1-12T_0+56T_0^2-160T_0^3+\cdots & -32T_0+192T_0^2-\frac{1792}{2}{T}_0^3+\cdots \\ T_0-6T_0^2+\frac{56}{3}{T}_0^3+\cdots & 1-16T_0^2+64T_0^3+\cdots \end{array}\right]{|}_{T_0=0.01}\\ & \approx \left[\begin{array}{cc}0.8854& -0.3014\\ 0.0094& 0.9985\end{array}\right]\end{array}$$同时$$\begin{array}{rl}{B}_d& ={\int }_0^{T_0}\Phi (q)Bdq={\int }_0^{T_0}{e}^{Aq}Bdq\\ & ={\int }_0^{T_0}\left(I+Aq+\frac{A^2{q}^2}{2!}+\frac{A^3{q}^3}{3!}+\cdots \right)Bdq\\ & ={\int }_0^{T_0}Bdq+{\int }_0^{T_0}ABqdq+\frac{1}{2}{\int }_0^{T_0}{A}^{2}B{q}^{2}dq+\frac{1}{6}{\int }_0^{T_0}{A}^{3}B{q}^{3}dq+\cdots \\ & ={\int }_0^{T_0}Bdq+{\int }_0^{T_0}ABqdq+\frac{1}{2}{\int }_0^{T_0}{A}^{2}B{q}^{2}dq+\frac{1}{6}{\int }_0^{T_0}{A}^{3}B{q}^{3}dq+\cdots \\ & =Bq{|}_0^{T_0}+AB\cdot \frac{1}{2}{q}^2{|}_0^{T_0}+\frac{1}{2}{A}^{2}B\cdot \frac{1}{3}{q}^3{|}_0^{T_0}+\frac{1}{6}{A}^{3}B\cdot \frac{1}{4}{q}^4{|}_0^{T_0}+\cdots \\ & =B{T}_0+\frac{1}{2}AB{T}_0^2+\frac{1}{6}{A}^{2}B{T}_0^3+\frac{1}{24}{A}^{3}B{T}_0^4+\cdots \end{array}\text{\hspace{1em}(4)}$$下面计算$AB,A^{2}B,A^{3}B$。由于$$A=\left[\begin{array}{cc}-12& -32\\ 1& 0\end{array}\right],B=\left[\begin{array}{c}1\\ 0\end{array}\right]$$故$$AB=\left[\begin{array}{c}-12\\ 1\end{array}\right],A^{2}B=\left[\begin{array}{c}112\\ -12\end{array}\right],A^{3}B=\left[\begin{array}{c}-960\\ 112\end{array}\right]$$则式(4)为
$$\begin{array}{rl}{B}_d& =B{T}_0+\frac{1}{2}AB{T}_0^2+\frac{1}{6}{A}^{2}B{T}_0^3+\frac{1}{24}{A}^{3}B{T}_0^4+\cdots \\ & =\left[\begin{array}{c}1\\ 0\end{array}\right]T_0+\frac{1}{2}\left[\begin{array}{c}-12\\ 1\end{array}\right]T_0^2+\frac{1}{6}\left[\begin{array}{c}112\\ -12\end{array}\right]T_0^3+\frac{1}{24}\left[\begin{array}{c}-960\\ 112\end{array}\right]T_0^4{|}_{T-0=0.01}\\ & =\left[\begin{array}{c}0.00942\\ 4.8047\times 10^{-5}\end{array}\right]\end{array}$$进一步地，$$C_d=C=\left[\begin{array}{cc}1& 1\end{array}\right],D_d=D=0$$