这个等式该如何做？

Question

我想知道如何以这种风格来做这个等式，我试图把它变成一个矩阵，但却是灾难性的。

\begin{equation}
\begin{matrix} 
\frac{2}{3}\sinh{\frac{3}{2}t}&\longleftrightarrow&\frac{2}{3}\left[\frac{\frac{3}{2}}{s^{2}-\frac{9}{4}}\right]\\
        \downarrow e^{-3t}    &      \quad        &\downarrow s\rightarrow s-3\\
\frac{2}{3}e^{-3t}\sinh{\frac{3}{2}t}&\longleftrightarrow&\frac{1}{\left[s+\frac{1}{2}\right]^2-\frac{9}{4}}
\end{matrix}
\end{equation}

Answer 1

类似这样的蒂克兹。

\documentclass{article}
\usepackage{mathpazo}
\usepackage{amsmath}
\usepackage{tikz}
\usetikzlibrary{matrix,arrows.meta}
\begin{document}
\begin{tikzpicture}[semithick,>=Latex]
\matrix (m)[matrix of math nodes,column sep=2cm,row sep=1cm] 
{
\frac{2}{3}\sinh{\frac{3}{2}t}          &   \frac{2}{3}\left[\dfrac{\frac{3}{2}}{s^{2}-\frac{9}{4}}\right]\\
\frac{2}{3}e^{-3t}\sinh{\frac{3}{2}t}   &   \dfrac{1}{\left[s+\frac{1}{2}\right]^2-\frac{9}{4}}\\
};
\draw[<->] (m-1-1) -- (m-1-1-|m-1-2.west); 
\draw[<->] (m-2-1) -- (m-2-1-|m-2-2.west); 
\draw[->] (m-1-1) -- node[right] {$e^{-3t}$} (m-2-1);
\draw[->] (m-1-2) -- node[right] {$s\rightarrow s-3$} (m-2-2);
\end{tikzpicture}
\end{document}

Answer 2

很简单tikz-cd：

\documentclass{article}
\usepackage{mathpazo}
\usepackage{amsmath,mathtools}
\usepackage{tikz-cd}

\begin{document}
\[
\begin{tikzcd}
\frac{2}{3}\sinh{\frac{3}{2}t}
  \arrow[r,leftrightarrow]
  \arrow[d,"e^{-\frac{t}{2}}"]
&
\frac{2}{3}\left[\dfrac{\frac{3}{2}}{s^{2}-\frac{9}{4}}\right]
  \arrow[d,"s\to s+\frac{1}{2}"]
\\
\frac{2}{3}e^{-3t}\sinh{\frac{3}{2}t}
  \arrow[r,leftrightarrow]
&
\dfrac{1}{\left[s+\frac{1}{2}\right]^2-\frac{9}{4}}
\mathrlap{{}=\dfrac{1}{s^2+s+2}}
\end{tikzcd}
\hphantom{{}=\dfrac{1}{s^2+s+2}} % compensate for the rlapped fraction
\]
\end{document}