对齐环境中的双重方程对齐问题

Question

改为并改为。\begin{align}\begin{alignat}{2}\end{align}\end{alignat}

我还添加了split前两个等式，以便更清楚地表明该数字指的是它们两个。

\documentclass{article}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage[swedish]{babel}
\usepackage{amsmath}

\DeclareMathOperator{\laplace}{\mathcal{L}} %Laplace symbol

\begin{document}

\begin{alignat}{2}
  \begin{split}
  u_{\mathrm{in}}(t) &= u_{\mathrm{ut}} + u(t)\\
  %
  i(t) &= i_L(t) + i_R(t)
  \end{split}
  %
\intertext{Laplace-transformering av ovanstående ekvationer ger:}
  %
  U_{\mathrm{ut}}(s) &= \laplace\{ u_{\mathrm{ut}}(t) \}(s) &&= \frac{1}{Cs}I(s) \label{eq:u_ut}\\
  %
  U(s) &= \laplace\{ u(t) \}(s) =  L[sI_L(s)-i(0)]          &&= sLI_L(s) \label{eq:u}\\
  %
  U(s) &= \laplace\{ \widetilde{u}(t) \}(s)                 &&= RI_R(s) \label{eq:omhs}\\
  %
  U_{\mathrm{in}}(s) &= \laplace\{ u_{\mathrm{in}}(t) \}(s) &&= U_{\mathrm{ut}}(s) + U(s) \label{eq:u_in}\\
  %
  I(s) &= \laplace\{ i(t) \}(s)                             &&= I_L(s) + I_R(s) \label{eq:i}\\
  %
  \intertext{Överföringsfunktionen blir då:}   
  %
  H(s) &= \frac{U_{\mathrm{ut}}(s)}{U_{\mathrm{in}}(s)}
\end{alignat}

\end{document}

对于更长的显示，您可以做同样的事情，但您必须“隐藏”更长的方程式（它们可能会在您不知情的情况下超出边界）。

（我修复了最常见的错误，例如无用的\left和\right，不直立的下标和不必要的大分数）。

\documentclass{article}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage[swedish]{babel}
\usepackage{amsmath,mathtools}

\DeclareMathOperator{\laplace}{\mathcal{L}} %Laplace symbol

\begin{document}
\begin{alignat}{2}
  \nonumber
  u_{\mathrm{ut}}(t) &= \frac{1}{C}\int_0^ti(t)dt\\ 
  %           
  \nonumber
  u(t) &= L\frac{di_L(t)}{dt}\\
  %
\intertext{Ohm's lag:}
  %
  \nonumber
  u(t) &= \widetilde{u}(t) = i_R(t)R\\
  %
  \nonumber
\intertext{Kirchoff's lagar:}    
  %
  \nonumber
  u_{\mathrm{in}}(t) &= u_{\mathrm{ut}} + u(t)\\
  %
  \nonumber
  i(t) &= i_L(t) + i_R(t)
  %
\intertext{Laplace-transformering av ovanstående ekvationer ger:}
  %
  U_{\mathrm{ut}}(s) &= \laplace\{ u_{\mathrm{ut}}(t) \}(s)          &&= \frac{1}{Cs}I(s) \label{eq:u_ut}\\
  %
  U(s) &= \laplace\{ u(t) \}(s) =  L[sI_L(s)-i(0)] &&= sLI_L(s) \label{eq:u}\\
  %
  U(s) &= \laplace\{ \widetilde{u}(t) \}(s)        &&= RI_R(s) \label{eq:omhs}\\
  %
  U_{\mathrm{in}}(s) &= \laplace\{ u_{\mathrm{in}}(t) \}(s)          &&= U_{\mathrm{ut}}(s) + U(s) \label{eq:u_in}\\
  %
  I(s) &= \laplace\{ i(t) \}(s)                    &&= I_L(s) + I_R(s) \label{eq:i}\\
  %
\intertext{Överföringsfunktionen blir då:}   
  %
  \nonumber
  H(s) &= \frac{U_{\mathrm{ut}}(s)}{U_{\mathrm{in}}(s)} = 
  %
\intertext{Insättning av ekvationerna \ref{eq:u_ut} \& \ref{eq:u} ger:}
  %
       &=\mathrlap{\frac{I(s)}{sC(\frac{1}{sC}I(s)+U(s))} = \frac{I(s)}{I(s)+sCU(s)}} \label{eq:h1}\\
  %
\intertext{Insättning av ekvation \ref{eq:i} i ekvation \ref{eq:omhs} ger:}
  %
  \nonumber
  U(s) &= RI_R(s) = R(I(s)-I_L(s)) =
  %
\intertext{Insättning av ekvation \ref{eq:u} ger:}
  %
  \nonumber       
       &= R\left(I(s)-\frac{U(s)}{sL}\right)\\
  %
\intertext{Alltså:} 
  %
  \nonumber
  U(s) &=     R\left(I(s)-\frac{U(s)}{sL}\right)
  %
\intertext{Bryt ut $U(s)$:}
  %
  U(s) &= \frac{R}{(1+\frac{R}{sL})}I(s) = \frac{sRL}{R+sL}I(s) \label{eq:u2}
  %
\intertext{Insättning av ekvation \ref{eq:u2} i ekvation \ref{eq:h1} ger:}
  %
  \nonumber
  H(s) &= \mathrlap{\frac{I(s)}{I(s)+\dfrac{s^2RLC}{R+sL}I(s)}
        = \frac{1}{1+\dfrac{s^2RLC}{R+sL}}
        = \frac{R+sL}{R+sL+s^2RLC}} \\
  %
\intertext{Alltså:}
  %
  H(s) &= \frac{R+sL}{R+sL+s^2RLC} \label{eq:h_laplace}
\end{alignat}

\end{document}

很遗憾，结果很糟糕。我宁愿只在真正需要时才使用对齐。前两个等式中的等号没有理由与下一页最后一个等式中的等号对齐。

顺便说一句，前两个方程最好在同一条线上，并以宽大的（\qquad）分隔，就像基尔霍夫定律一样（据我所知，名称中有两个 h）。

这是我建议的版本：

\documentclass{article}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage[swedish]{babel}
\usepackage{amsmath,mathtools}

\DeclareMathOperator{\laplace}{\mathcal{L}} %Laplace symbol

\begin{document}
\begin{equation*}
  u_{\mathrm{ut}}(t) = \frac{1}{C}\int_0^ti(t)dt
  \qquad
  u(t) = L\frac{di_L(t)}{dt}
\end{equation*}
Ohm's lag:
\begin{equation*}
  u(t) = \widetilde{u}(t) = i_R(t)R
\end{equation*}
Kirchhoff's lagar:
\begin{equation*}
  u_{\mathrm{in}}(t) = u_{\mathrm{ut}} + u(t)
  \qquad
  i(t) = i_L(t) + i_R(t)
\end{equation*}
Laplace-transformering av ovanstående ekvationer ger:
\begin{alignat}{2}
  U_{\mathrm{ut}}(s) &= \laplace\{ u_{\mathrm{ut}}(t) \}(s)
    &&= \frac{1}{Cs}I(s) \label{eq:u_ut}\\
  U(s) &= \laplace\{ u(t) \}(s) =  L[sI_L(s)-i(0)]
    &&= sLI_L(s) \label{eq:u}\\
  U(s) &= \laplace\{ \widetilde{u}(t) \}(s)
    &&= RI_R(s) \label{eq:omhs}\\
  U_{\mathrm{in}}(s) &= \laplace\{ u_{\mathrm{in}}(t) \}(s)
    &&= U_{\mathrm{ut}}(s) + U(s) \label{eq:u_in}\\
  I(s) &= \laplace\{ i(t) \}(s)
    &&= I_L(s) + I_R(s) \label{eq:i}
\end{alignat}
Överföringsfunktionen blir då:
\begin{align}
  \nonumber
  H(s) &= \frac{U_{\mathrm{ut}}(s)}{U_{\mathrm{in}}(s)} = 
\intertext{Insättning av ekvationerna \ref{eq:u_ut} \& \ref{eq:u} ger:}
       &=\frac{I(s)}{sC(\frac{1}{sC}I(s)+U(s))} = \frac{I(s)}{I(s)+sCU(s)} \label{eq:h1}
\end{align}
Insättning av ekvation \ref{eq:i} i ekvation \ref{eq:omhs} ger:
\begin{align*}
  U(s) &= RI_R(s) = R(I(s)-I_L(s)) =
\intertext{Insättning av ekvation \ref{eq:u} ger:}
       &= R\left(I(s)-\frac{U(s)}{sL}\right)
\end{align*}
Alltså:
\begin{equation*}
  U(s) = R\left(I(s)-\frac{U(s)}{sL}\right)
\end{equation*}
Bryt ut $U(s)$:
\begin{equation}
  U(s) = \frac{R}{(1+\frac{R}{sL})}I(s) = \frac{sRL}{R+sL}I(s) \label{eq:u2}
\end{equation}
Insättning av ekvation \ref{eq:u2} i ekvation \ref{eq:h1} ger:
\begin{equation*}
  H(s) = \frac{I(s)}{I(s)+\dfrac{s^2RLC}{R+sL}I(s)}
       = \frac{1}{1+\dfrac{s^2RLC}{R+sL}}
       = \frac{R+sL}{R+sL+s^2RLC}
\end{equation*}
Alltså:
\begin{equation}
  H(s) = \frac{R+sL}{R+sL+s^2RLC} \label{eq:h_laplace}
\end{equation}

\end{document}

Answer 1