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如果矩阵方程太长，如何将其一部分对齐到另一行而不改变方程的标签

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如果矩阵方程太长，如何将其一部分对齐到另一行而不改变方程的标签 
\documentclass[11pt]{article}
\usepackage[utf8]{inputenc}
\usepackage{listings}
\usepackage{amsmath}
\usepackage{amssymb}
\begin{document}

The load current is assumed continuous because of an inductive load. We define the state variables (i.e., capacitor volatges, inductor currents and load current) as a vector, 

\begin{align}
x(t) = \left[ \begin{array}{c}
i_{L1}(t)\\
i_{L2}(t)\\
v_{C1}(t)\\
v_{C2}(t)\\
i_l(t)
\end{array}\right]
\end{align}
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 $K\dot{x} = A_1\cdot x + B_1\cdot u$, that is,
\begin{align}
\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]
\end{align}
\\where
$$ K = \left[ \begin{array}{ccccc}
L_1 & 0 & 0 & 0 & 0\\
0 & L_2 & 0 & 0 & 0\\
0 & 0 & C_1 & 0 & 0\\
0 & 0 & 0 & C_2 & 0\\
0 & 0 & 0 & 0 & L_l \end{array} \right], A_1= \left[ \begin{array}{ccccc}
0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 1 & 0\\
-1 & 0 & 0 & 0 & 0\\
0 & -1 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & -R_l
\end{array} \right] and\ B_1 = \left[\begin{array}{c}
0\\
0\\
0\\
0\\
0 \end{array} \right]$$ 
\\
The equations in Mode 2 can be expresses in the state space form \\ $K\dot{x} = A_2\cdot x c+ B_2\cdot u$ , that is 
\begin{align*}
\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 & 0 & -1 & 0\\
0 & 0 & -1 & 0 & 0\\
0 & 1 & 0 & 0 & -1\\
1 & 0 & 0 & 0 & -1\\
0 & 0 & 1 & 1 & -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] + \\
&\left[\begin{array}{c}
1\\
1\\
0\\
0\\
-1 \end{array} \right] (v_g(t) - V_D)
\end{align*}

错误：公式的编号没有连续

答案1

如果您希望方程式被标记，请不要使用align*星号。我建议输入以下内容。（您可能还想看看split，\notag以及此主页上显示长方程式拆分的所有数百篇帖子。

请注意，我已将所有数组更改为bmatrix。这可使间距更好，括号大小更稳定。您应该将它们用于此类目的。

显示数学形式$$...$$仅应在纯 TeX 中使用。

% arara: pdflatex

\documentclass[11pt]{article}
\usepackage[utf8]{inputenc}
\usepackage{mathtools}

\begin{document}    
The load current is assumed continuous because of an inductive load. We define the state variables (i.e., capacitor voltages, inductor currents and load current) as a vector,  
\begin{equation}
    x(t) = 
    \begin{bmatrix}
    i_{L1}(t)\\
    i_{L2}(t)\\
    v_{C1}(t)\\
    v_{C2}(t)\\
    i_l(t)
    \end{bmatrix}
\end{equation}
Recall that the input voltage $V_g$ is an independent voltage source and the load impedance 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 $K\dot{x} = A_1\cdot x + B_1\cdot u$, that is,
\begin{equation}
    K \frac{\mathrm{d}}{\mathrm{d}t} 
    \begin{bmatrix}
    i_{L1}(t)\\
    i_{L2}(t)\\
    v_{C1}(t)\\
    v_{C2}(t)\\
    i_l(t)
    \end{bmatrix} 
    = A_1  
    \begin{bmatrix}
    i_{L1}(t)\\
    i_{L2}(t)\\
    v_{C1}(t)\\
    v_{C2}(t)\\
    i_l(t) 
    \end{bmatrix}
\end{equation}
where
\begin{align*} K &= 
    \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},\\
    A_1 &= 
    \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},\\
    \shortintertext{and}
    B_1 &= 
    \begin{bmatrix}
    0\\
    0\\
    0\\
    0\\
    0 
    \end{bmatrix}.
\end{align*} 

The equations in Mode 2 can be expressed in the state space form $K\dot{x} = A_2\cdot x c+ B_2\cdot u$ , that is 
\begin{multline}
    \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{\mathrm{d}}{\mathrm{d}t} 
    \begin{bmatrix}
        i_{L1}(t) \\
        i_{L2}(t) \\
        v_{C1}(t) \\
        v_{C2}(t) \\
        i_l(t)    \\
    \end{bmatrix}\\
    = 
    \begin{bmatrix}
        0 & 0 & 0  & -1 & 0    \\
        0 & 0 & -1 & 0  & 0    \\
        0 & 1 & 0  & 0  & -1   \\
        1 & 0 & 0  & 0  & -1   \\
        0 & 0 & 1  & 1  & -R_l \\
    \end{bmatrix}
    \begin{bmatrix}
        i_{L1}(t) \\
        i_{L2}(t) \\
        v_{C1}(t) \\
        v_{C2}(t) \\
        i_l(t)    \\
    \end{bmatrix}  
    + \begin{bmatrix}
    1\\
    1\\
    0\\
    0\\
    -1 \end{bmatrix}  (v_g(t) - V_D)
\end{multline}
\end{document}

在此处输入图片描述

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