我正在输入一个较长的等式,需要在左括号后断开,并在左括号同一行的末尾放置三个点。这里有一个示例,其中,我需要在第三个左括号(即加号之前的列向量)后面跟三个点,同时将换行符中的等式的其余部分与等号对齐。
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{center}
\begin{equation}\left[
\begin{array}{c}
\dot{\omega}_m \\
\dot{\theta}_m \\
\dot{\omega}_v \\
\dot{I}_m \\
\end{array}
\right]
=
\left[
\begin{array}{cccc}
-\frac{C_m}{J_m} & -\frac{P_s K_s \psi}{2 \pi r_s^2 J_m} & 0 & \frac{K_{tm}}{J_m} \\
1 & 0 & 0 & 0 \\
0 & -\frac{P_s K_s \mu_p R_{mp}}{2 \pi r_s J_v} & -\frac{C_v}{J_v} & 0 \\
-\frac{K_{em}}{L_m} & 0 & 0 & -\frac{R_m}{L_m} \\
\end{array}
\right]
\left[
\begin{array}{c}
\omega_m \\
\theta_m \\
\omega_v \\
I_m \\
\end{array}
\right]
+
\left[
\begin{array}{cc}
0 & 0 \\
0 & 0 \\
\frac{1}{J_v} & 0 \\
0 & \frac{1}{L_m} \\
\end{array}
\right]
\left[
\begin{array}{c}
F_v \\
V_m(t) \\
\end{array}
\right]
\end{equation}
\end{center}
\end{document}
提前致谢
答案1
完全不清楚你想做什么;我会这样拆分方程
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{equation}
\begin{aligned}
\begin{bmatrix}
\dot{\omega}_m \\
\dot{\theta}_m \\
\dot{\omega}_v \\
\dot{I}_m \\
\end{bmatrix}
&=
\begin{bmatrix}
-\frac{C_m}{J_m} & -\frac{P_s K_s \psi}{2 \pi r_s^2 J_m} & 0 & \frac{K_{tm}}{J_m} \\
1 & 0 & 0 & 0 \\
0 & -\frac{P_s K_s \mu_p R_{mp}}{2 \pi r_s J_v} & -\frac{C_v}{J_v} & 0 \\
-\frac{K_{em}}{L_m} & 0 & 0 & -\frac{R_m}{L_m} \\
\end{bmatrix}
\begin{bmatrix}
\omega_m \\
\theta_m \\
\omega_v \\
I_m \\
\end{bmatrix}
\\
&\qquad+
\begin{bmatrix}
0 & 0 \\
0 & 0 \\
\frac{1}{J_v} & 0 \\
0 & \frac{1}{L_m} \\
\end{bmatrix}
\begin{bmatrix}
F_v \\
V_m(t) \\
\end{bmatrix}
\end{aligned}
\end{equation}
\end{document}
环境center
不合适;bmatrix
环境给出了比更好的矩阵\left[\begin{array}{..}...\end{array}\right]
。
减小列宽后,可以尝试挤压矩阵列并在产品处折断。以下是列宽为 229.5pt(3.2 英寸或 8 厘米)的示例:
\documentclass[twocolumn]{article}
\usepackage{amsmath}
\begin{document}
\pagestyle{empty}
\begin{equation}
\setlength{\arraycolsep}{1pt}
\begin{aligned}
\begin{bmatrix}
\dot{\omega}_m \\
\dot{\theta}_m \\
\dot{\omega}_v \\
\dot{I}_m \\
\end{bmatrix}
&=
\begin{bmatrix}
-\frac{C_m}{J_m} & -\frac{P_s K_s \psi}{2 \pi r_s^2 J_m} & 0 & \frac{K_{tm}}{J_m} \\
1 & 0 & 0 & 0 \\
0 & -\frac{P_s K_s \mu_p R_{mp}}{2 \pi r_s J_v} & -\frac{C_v}{J_v} & 0 \\
-\frac{K_{em}}{L_m} & 0 & 0 & -\frac{R_m}{L_m} \\
\end{bmatrix}
\\
&\qquad\cdot
\begin{bmatrix}
\omega_m \\
\theta_m \\
\omega_v \\
I_m \\
\end{bmatrix}
+
\begin{bmatrix}
0 & 0 \\
0 & 0 \\
\frac{1}{J_v} & 0 \\
0 & \frac{1}{L_m} \\
\end{bmatrix}
\begin{bmatrix}
F_v \\
V_m(t) \\
\end{bmatrix}
\end{aligned}
\end{equation}
\end{document}
答案2
你的意思不是很清楚对齐到底是什么,但对于任何类型的数学对齐,通常最好使用该amsmath
包。我思考你的意思是:
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\newcommand\arrayb{
\begin{array}{c}
\omega_m \\
\theta_m \\
\omega_v \\
I_m
\end{array}}
\begin{align}
\left[
\begin{array}{c}
\dot{\omega}_m \\
\dot{\theta}_m \\
\dot{\omega}_v \\
\dot{I}_m
\end{array}
\right]
&=
\left[
\begin{array}{cccc}
-\frac{C_m}{J_m} & -\frac{P_s K_s \psi}{2 \pi r_s^2 J_m} & 0 & \frac{K_{tm}}{J_m} \\
1 & 0 & 0 & 0 \\
0 & -\frac{P_s K_s \mu_p R_{mp}}{2 \pi r_s J_v} & -\frac{C_v}{J_v} & 0 \\
-\frac{K_{em}}{L_m} & 0 & 0 & -\frac{R_m}{L_m}
\end{array}
\right]
\left[\vphantom{\arrayb}\right.\cdots
\\\notag
&
\left.\arrayb\right]
+
\left[
\begin{array}{cc}
0 & 0 \\
0 & 0 \\
\frac{1}{J_v} & 0 \\
0 & \frac{1}{L_m}
\end{array}
\right]
\left[
\begin{array}{c}
F_v \\
V_m(t)
\end{array}
\right]
\end{align}
\end{document}