即
\[
\begin{cases}
\dot{x}_1=-\dfrac{R_1+R_2}{L}x_1+\dfrac{1}{L}x_2+\dfrac{R_2}{L}u \\
\dot{x}_2=-\dfrac{1}{C}x_1+\dfrac{1}{C}u
\end{cases}
\]
写成向量-矩阵形式可得系统的状态方程为
\[
\dot{\boldsymbol{x}}=
\begin{bmatrix}
-\dfrac{R_1+R_2}{L} & \dfrac{1}{L} \\[2mm]
-\dfrac{1}{C} & 0
\end{bmatrix}
\boldsymbol{x}+
\begin{bmatrix}
\dfrac{R_2}{L} \\[2mm]
\dfrac{1}{C}
\end{bmatrix}
u
\]
(2) 取状态变量 \(x_1=v_C,x_2=i_1,x_3=i_2\),根据电路定律可列写如下方程:
\[
\begin{cases}
C\dfrac{\mathrm{d}v_C}{\mathrm{d}t}+i_1+i_2=0 \\[2mm]
L_2\dfrac{\mathrm{d}i_2}{\mathrm{d}t}+R_2 i_2=v_C \\[2mm]
L_1\dfrac{\mathrm{d}i_1}{\mathrm{d}t}+R_1 i_1+u=v_C
\end{cases}
\]
经整理后得
\[
\begin{cases}
\dfrac{\mathrm{d}v_C}{\mathrm{d}t}=-\dfrac{1}{C}i_1-\dfrac{1}{C}i_2 \\[2mm]
\dfrac{\mathrm{d}i_1}{\mathrm{d}t}=\dfrac{1}{L_1}v_C-\dfrac{R_1}{L_1}i_1-\dfrac{1}{L_1}u \\[2mm]
\dfrac{\mathrm{d}i_2}{\mathrm{d}t}=\dfrac{1}{L_2}v_C-\dfrac{R_2}{L_2}i_2
\end{cases}
\]
即
\[
\begin{cases}
\dot{x}_1=-\dfrac{1}{C}x_2-\dfrac{1}{C}x_3 \\
\dot{x}_2=\dfrac{1}{L_1}x_1-\dfrac{R_1}{L_1}x_2-\dfrac{1}{L_1}u \\
\dot{x}_3=\dfrac{1}{L_2}x_1-\dfrac{R_2}{L_2}x_3
\end{cases}
\]
写成向量-矩阵形式可得系统的状态方程式为
\[
\dot{\boldsymbol{x}}=
\begin{bmatrix}
0 & -\dfrac{1}{C} & -\dfrac{1}{C} \\[2mm]
\dfrac{1}{L_1} & -\dfrac{R_1}{L_1} & 0 \\[2mm]
\dfrac{1}{L_2} & 0 & -\dfrac{R_2}{L_2}
\end{bmatrix}
\boldsymbol{x}+
\begin{bmatrix}
0 \\[2mm]
-\dfrac{1}{L_1} \\[2mm]
0
\end{bmatrix}
u
\]
(3) 选取各电容器两端的电压(自左至右)为状态变量,根据电路定律可列写如下方程:
· 472 ·