过满和欠满错误。无法更正，请帮忙

Question

\documentclass[11pt]{article}
\usepackage[utf8]{inputenc}
\usepackage{listings}
\usepackage{amsmath}
\usepackage{amssymb}
\begin{document}
Similarly, the state-space averaging is used to described the AC small signal dynamics:
\begin{multline}
\begin{bmatrix}
L_1 & 0 & 0 & 0\\
0 & L_2 & 0 & 0\\
0 & 0 & C_1 & 0\\
0 & 0 & 0 & C_2
\end{bmatrix} \frac{d}{dt} \begin{bmatrix}
i_{L1}(t)\\
i_{L2}(t)\\
v_{C1}(t)\\
v_{C2}(t)
\end{bmatrix} =\\ \begin{bmatrix}
0 & 0 & D & -D'\\
0 & 0 & -D' & D\\
-D & D' & -\frac{D'}{R_l} & -\frac{D'}{R_l}\\
D' & -D & -\frac{D'}{R_l} & -\frac{D'}{R_l}
\end{bmatrix} \cdot \begin{bmatrix}
i_{L1}(t)\\
i_{L2}(t)\\
v_{C1}(t)\\
v_{C2}(t)
\end{bmatrix} + \begin{bmatrix}
D'\\
D'\\
\frac{D'}{R_l}\\
\frac{D'}{R_l}
\end{bmatrix} \cdot \hat{v}_g(t) + \\ \begin{bmatrix}
V_{C1} + V_{C2} -(V_g - V_D)\\
V_{C1} + V_{C2} -(V_g - V_D)\\
-I_{L1} -I_{L2} + \frac{V_{C1}}{R_l} + \frac{V_{C2}}{R_l} - \frac{V_g - V_D}{R_l}\\
-I_{L1} -I_{L2} + \frac{V_{C1}}{R_l} + \frac{V_{C2}}{R_l} - \frac{V_g - V_D}{R_l}
\end{bmatrix} \cdot \hat{d}(t)
\end{multline}
Also, $\hat{v}_{C1}(t) = \hat{v}_{C2}(t) = \hat{v}_C(t) \ and \ \hat{i}_{L1}(t) = \hat{i}_{L2}(t) = \hat{i}_L(t)$. While DC state equations are:
\begin{equation}
\begin{bmatrix}
0\\
0\\
0\\
0
\end{bmatrix} = \begin{bmatrix}
0 & 0 & D & -D'\\
0 & 0 & -D' & D\\
-D & D' & -\displaystyle\frac{D'}{R_l} & -\displaystyle\frac{D'}{R_l}\\
D' & -D' & -\displaystyle\frac{D'}{R_lZ_l} & -\displaystyle\frac{D'}{R_l}
\end{bmatrix}\cdot \begin{bmatrix}
I_{L1}\\
I_{L2}\\
V_{C1}\\
V_{C2}
\end{bmatrix} + \begin{bmatrix}
D'\\
D'\\
\displaystyle\frac{D'}{R_l}\\
\displaystyle\frac{D'}{R_l}
\end{bmatrix} \cdot (V_g - V_D)
\end{equation}
The state space equations hence become,
\begin{multline}
sL\cdot \hat{i}_L(s) = (D - D')\cdot \hat{v}_C(s) + D'\cdot \hat{v}_g(s) + (2V_C-V_g+V_D)\cdot \hat{d}(s) \\
sC\cdot \hat{v}_C(s) = (D'-D)\cdot \hat{i}_L(s) - \displaystyle\frac{2D'}{R_l} \hat{v}_C(s) + \displaystyle\frac{D'}{R_l} \hat{v}+g(s) + (-2I_L + 2\displaystyle\frac{2V_C}{r_l} - \displaystyle\frac{V_g-V_D}{R_l})\cdot \hat{d}(s)
\end{multline}
Steady state values are:
\begin{equation}
V_C = \displaystyle\frac{1-D}{1-2D} \cdot (V_g-V_D)
\end{equation}
and \\
\begin{equation}
I_L=\displaystyle\frac{1-D}{(1-2D)^2} \cdot \displaystyle\frac{V_g}{R_l}
\end{equation}
\end{document}

警告：过满或不足。

Answer 1

如果我编译你的代码，我会收到以下警告：

在第 69 行检测到 \hbox 过满（宽 33.51332pt）

段落第 73-75 行的 \hbox 未满（badness 10000）

我假设您收到相同的错误消息。现在，有了这个，消除错误真的很容易，您只需查看这些行即可。

第 68 行：

sC\cdot \hat{v}_C(s) = (D'-D)\cdot \hat{i}_L(s) - \displaystyle\frac{2D'}{R_l} \hat{v}_C(s) + \displaystyle\frac{D'}{R_l} \hat{v}+g(s) + (-2I_L + 2\displaystyle\frac{2V_C}{r_l} - \displaystyle\frac{V_g-V_D}{R_l})\cdot \hat{d}(s)

这是一个很长的公式，如果将其分成两行或缩短，超额警告就会消失。

第 74 行：

and \\

正如@David Carlisle 在他的评论中提到的那样，去掉那些 \ ，警告就会消失。

遇到此类警告时，最简单的解决方法是查看相应的行（因为警告会给出行号），看看删除该行是否能使警告消失。如果能，那么你就知道错误在哪里，可以尝试处理该行，直到找到可接受的解决方案。

Answer 2

拆分长方程的一个解决方案如下（使用对齐代替多线）：

\documentclass[11pt]{article}
\usepackage[utf8]{inputenc}
\usepackage{listings}
\usepackage{amsmath}
\usepackage{amssymb}
\begin{document}
Similarly, the state-space averaging is used to described the AC small signal dynamics:
\begin{multline}
\begin{bmatrix}
L_1 & 0 & 0 & 0\\
0 & L_2 & 0 & 0\\
0 & 0 & C_1 & 0\\
0 & 0 & 0 & C_2
\end{bmatrix} \frac{d}{dt} \begin{bmatrix}
i_{L1}(t)\\
i_{L2}(t)\\
v_{C1}(t)\\
v_{C2}(t)
\end{bmatrix} =\\ \begin{bmatrix}
0 & 0 & D & -D'\\
0 & 0 & -D' & D\\
-D & D' & -\frac{D'}{R_l} & -\frac{D'}{R_l}\\
D' & -D & -\frac{D'}{R_l} & -\frac{D'}{R_l}
\end{bmatrix} \cdot \begin{bmatrix}
i_{L1}(t)\\
i_{L2}(t)\\
v_{C1}(t)\\
v_{C2}(t)
\end{bmatrix} + \begin{bmatrix}
D'\\
D'\\
\frac{D'}{R_l}\\
\frac{D'}{R_l}
\end{bmatrix} \cdot \hat{v}_g(t) + \\ \begin{bmatrix}
V_{C1} + V_{C2} -(V_g - V_D)\\
V_{C1} + V_{C2} -(V_g - V_D)\\
-I_{L1} -I_{L2} + \frac{V_{C1}}{R_l} + \frac{V_{C2}}{R_l} - \frac{V_g - V_D}{R_l}\\
-I_{L1} -I_{L2} + \frac{V_{C1}}{R_l} + \frac{V_{C2}}{R_l} - \frac{V_g - V_D}{R_l}
\end{bmatrix} \cdot \hat{d}(t)
\end{multline}
Also, $\hat{v}_{C1}(t) = \hat{v}_{C2}(t) = \hat{v}_C(t) \ and \ \hat{i}_{L1}(t) = \hat{i}_{L2}(t) = \hat{i}_L(t)$. While DC state equations are:
\begin{equation}
\begin{bmatrix}
0\\
0\\
0\\
0
\end{bmatrix} = \begin{bmatrix}
0 & 0 & D & -D'\\
0 & 0 & -D' & D\\
-D & D' & -\displaystyle\frac{D'}{R_l} & -\displaystyle\frac{D'}{R_l}\\
D' & -D' & -\displaystyle\frac{D'}{R_lZ_l} & -\displaystyle\frac{D'}{R_l}
\end{bmatrix}\cdot \begin{bmatrix}
I_{L1}\\
I_{L2}\\
V_{C1}\\
V_{C2}
\end{bmatrix} + \begin{bmatrix}
D'\\
D'\\
\displaystyle\frac{D'}{R_l}\\
\displaystyle\frac{D'}{R_l}
\end{bmatrix} \cdot (V_g - V_D)
\end{equation}
The state space equations hence become,
\begin{align}
sL\cdot \hat{i}_L(s) &= (D - D')\cdot \hat{v}_C(s) + D'\cdot \hat{v}_g(s) + (2V_C-V_g+V_D)\cdot \hat{d}(s)\nonumber \\
sC\cdot \hat{v}_C(s) &= (D'-D)\cdot \hat{i}_L(s) - \displaystyle\frac{2D'}{R_l} \hat{v}_C(s)\nonumber\\
&\phantom{=} + \displaystyle\frac{D'}{R_l} \hat{v}+g(s) + (-2I_L + 2\displaystyle\frac{2V_C}{r_l} - \displaystyle\frac{V_g-V_D}{R_l})\cdot \hat{d}(s)
\end{align}
Steady state values are:
\begin{equation}
V_C = \displaystyle\frac{1-D}{1-2D} \cdot (V_g-V_D)
\end{equation}
and 
\begin{equation}
I_L=\displaystyle\frac{1-D}{(1-2D)^2} \cdot \displaystyle\frac{V_g}{R_l}
\end{equation}
\end{document}

Answer 3

这是重复的，但由于我有这个：

\documentclass[11pt]{article}
\usepackage[utf8]{inputenc}
\usepackage{listings}
\usepackage{amsmath}
\usepackage{amssymb}
\begin{document}
Similarly, the state-space averaging is used to described the AC small signal dynamics:
\begin{multline}
\begin{bmatrix}
L_1 & 0 & 0 & 0\\
0 & L_2 & 0 & 0\\
0 & 0 & C_1 & 0\\
0 & 0 & 0 & C_2
\end{bmatrix} \frac{d}{dt} \begin{bmatrix}
i_{L1}(t)\\
i_{L2}(t)\\
v_{C1}(t)\\
v_{C2}(t)
\end{bmatrix} =\\ \begin{bmatrix}
0 & 0 & D & -D'\\
0 & 0 & -D' & D\\
-D & D' & -\frac{D'}{R_l} & -\frac{D'}{R_l}\\
D' & -D & -\frac{D'}{R_l} & -\frac{D'}{R_l}
\end{bmatrix} \cdot \begin{bmatrix}
i_{L1}(t)\\
i_{L2}(t)\\
v_{C1}(t)\\
v_{C2}(t)
\end{bmatrix} + \begin{bmatrix}
D'\\
D'\\
\frac{D'}{R_l}\\
\frac{D'}{R_l}
\end{bmatrix} \cdot \hat{v}_g(t) + \\ \begin{bmatrix}
V_{C1} + V_{C2} -(V_g - V_D)\\
V_{C1} + V_{C2} -(V_g - V_D)\\
-I_{L1} -I_{L2} + \frac{V_{C1}}{R_l} + \frac{V_{C2}}{R_l} - \frac{V_g - V_D}{R_l}\\
-I_{L1} -I_{L2} + \frac{V_{C1}}{R_l} + \frac{V_{C2}}{R_l} - \frac{V_g - V_D}{R_l}
\end{bmatrix} \cdot \hat{d}(t)
\end{multline}
Also, $\hat{v}_{C1}(t) = \hat{v}_{C2}(t) = \hat{v}_C(t) \ and \ \hat{i}_{L1}(t) = \hat{i}_{L2}(t) = \hat{i}_L(t)$. While DC state equations are:
\begin{equation}
\begin{bmatrix}
0\\
0\\
0\\
0
\end{bmatrix} = \begin{bmatrix}
0 & 0 & D & -D'\\
0 & 0 & -D' & D\\
-D & D' & -\frac{D'}{R_l} & -\frac{D'}{R_l}\\
D' & -D' & -\frac{D'}{R_lZ_l} & -\frac{D'}{R_l}
\end{bmatrix}\cdot \begin{bmatrix}
I_{L1}\\
I_{L2}\\
V_{C1}\\
V_{C2}
\end{bmatrix} + \begin{bmatrix}
D'\\
D'\\
\frac{D'}{R_l}\\
\frac{D'}{R_l}
\end{bmatrix} \cdot (V_g - V_D)
\end{equation}
The state space equations hence become,
\begin{align}
sL\cdot \hat{i}_L(s) &= (D - D')\cdot \hat{v}_C(s) + D'\cdot \hat{v}_g(s) + (2V_C-V_g+V_D)\cdot \hat{d}(s) \\
sC\cdot \hat{v}_C(s) &=
\!\begin{aligned}[t]&(D'-D)\cdot \hat{i}_L(s) - \frac{2D'}{R_l} \hat{v}_C(s)\\
& + \frac{D'}{R_l} \hat{v}+g(s) + (-2I_L + 2\frac{2V_C}{r_l} -
 \frac{V_g-V_D}{R_l})\cdot \hat{d}(s)
\end{aligned}
\end{align}
Steady state values are:
\begin{equation}
V_C = \frac{1-D}{1-2D} \cdot (V_g-V_D)
\end{equation}
and% never do this \\
\begin{equation}
I_L=\frac{1-D}{(1-2D)^2} \cdot \frac{V_g}{R_l}
\end{equation}
\end{document}