代码:
$$x(t) =
\left[ \begin{array}{c}
i_{L1}(t)\\
i_{L2}(t)\\
v_{C1}(t)\\
v_{C2}(t)\\
i_l(t)
\end{array}\right]
$$
Recall that the input voltage $V_g$ is an independent voltage source and the load impedence is $Z_l = R_l + sL_l$. Apparently, the load is short-circuited by $S_2$ in Mode 1.\\
The circuit equations in Mode 1 can be written in the state space form $K\cdot x = A_1\cdot x + B_1\cdot u$, that is,
$$\left[ \begin{array}{ccccc}
L_1 & 0 & 0 & 0 & 0\\
0 & L_2 & 0 & 0 & 0\\
0 & 0 & C_1 & 0 & 0\\
0 & 0 & 0 & C_2 & 0\\
0 & 0 & 0 & 0 & L_l
end{array}\right] \frac{d}{dt} \left[ \begin{array}{c}
i_{L1}(t)\\
i_{L2}(t)\\
v_{C1}(t)\\
v_{C2}(t)\\
i_l(t)
\end{array}\right] = \left[ \begin{array}{ccccc}
0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 1 & 0\\
-1 & 0 & 0 & 0 & 0\\
0 & -1 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & -R_l
\end{array}\right] \left[ \begin{array}{c}
i_{L1}(t)\\
i_{L2}(t)\\
v_{C1}(t)\\
v_{C2}(t)\\
i_l(t)
\end{array}\right]$$
错误:
! Extra \right.
l.25 end{array}\right]
\frac{d}{dt} \left[ \begin{array}{c}
答案1
使用包设置矩阵更容易。这还可以改善水平间距。问题评论中已经涵盖了amsmath
缺失\
或不使用等问题。$$
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\[
x(t) =
\begin{bmatrix}
i_{L1}(t)\\
i_{L2}(t)\\
v_{C1}(t)\\
v_{C2}(t)\\
i_l(t)
\end{bmatrix}
\]
Recall that the input voltage $V_g$ is an independent voltage source and the
load impedence is $Z_l = R_l + sL_l$. Apparently, the load is
short-circuited by $S_2$ in Mode~1.\\
The circuit equations in Mode~1 can be written in the state space form
$K\cdot x = A_1\cdot x + B_1\cdot u$, that is,
%
\[
\begin{bmatrix}
L_1 & 0 & 0 & 0 & 0\\
0 & L_2 & 0 & 0 & 0\\
0 & 0 & C_1 & 0 & 0\\
0 & 0 & 0 & C_2 & 0\\
0 & 0 & 0 & 0 & L_l
\end{bmatrix}
\frac{d}{dt}
\begin{bmatrix}
i_{L1}(t)\\
i_{L2}(t)\\
v_{C1}(t)\\
v_{C2}(t)\\
i_l(t)
\end{bmatrix}
=
\begin{bmatrix}
0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 1 & 0\\
-1 & 0 & 0 & 0 & 0\\
0 & -1 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & -R_l
\end{bmatrix}
\begin{bmatrix}
i_{L1}(t)\\
i_{L2}(t)\\
v_{C1}(t)\\
v_{C2}(t)\\
i_l(t)
\end{bmatrix}
\]
\end{document}