有人能给我一个初步的想法，如何像相邻的图像一样将平面每个象限中的 4 个图分组吗？

Question

我已经有时间使用 pgfplots 和 tikz，但我不知道如何实现这样的目标。

Answer 1

这个可能只需要 Ti钾Z 和图片。

\documentclass[tikz,border=3mm]{standalone}
\usetikzlibrary{backgrounds,shapes.misc}
\usepackage{amsmath}
\DeclareMathOperator{\re}{Re}
\DeclareMathOperator{\im}{Im}
\begin{document}
\begin{tikzpicture}[pics/graph/.style={code={
    \begin{scope}[local bounding box=temp,xshift=-1cm]
    \draw[thick,solid]      plot[variable=\t,domain=0:2,samples=101,smooth] 
     ({\t},{#1});
     \edef\temp{\noexpand\path \pgfkeysvalueof{/tikz/graph/extra points};}
     \temp
    \end{scope}
    \begin{scope}[on background layer]
    \draw[fill=gray!30,rounded corners,local bounding box=graph] ([xshift=-2mm,yshift=-2mm]temp.south west)
    rectangle ([xshift=4mm,yshift=2mm]temp.north east);
    \draw[-stealth] ([yshift=-1mm]temp.south west) -- 
    ([yshift=1mm]temp.north west);
    \draw[-stealth] (-1,0) -- 
    (1.2,0);
    \end{scope}
    }},/tikz/graph/extra points/.initial={(0,0)},
    xmark/.style={cross out,minimum size=1ex,solid,draw},
    Graph/.style={matrix,node contents={\pic{graph=#1};\\}}]
  \draw[-stealth] (-5,0) -- (5,0) node[below]{$\re(s)$};
  \draw[-stealth] (0,-5) -- (0,5) node[right] {$\im(s)$};
  \draw[densely dotted,every edge/.append style={-stealth,solid}] 
   (2,0) |-  (0,3)  node[pos=0.5,xmark] (x1) {}
   (4,1.2) pic{graph={cos(1440*\t)*exp(\t-2)/2}}
   (x1) edge[bend left] (graph.north)
   (0,3.4) -| (-2,0) node[pos=0.5,xmark] (x2) {}
   (-3.7,2) pic{graph={cos(1440*\t)*exp(-\t)/2}}
   (x2) edge[bend right] (graph.north)
   (0,-3.4) -| (-2,0) node[pos=0.5,xmark]  {}
   (-2.5,0) node[xmark] (x3) {}
   (-3.7,-1.4) pic{graph={exp(-\t)}}
   (x3) edge[bend right] (graph.north)
   (0,-3) -| (2,0) node[pos=0.5,xmark]  {}
   (0.5,0) node[xmark] (x4) {}
   (4,-1.8) pic{graph={exp(\t-2)}}
   (x4) edge[bend right] (graph.west);
\end{tikzpicture}
\end{document}