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有人知道如何将两个 MIMO 块连接在一起吗？

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有人知道如何将两个 MIMO 块连接在一起吗？

有人知道如何绘制两个连接在一起的 MIMO 块吗？我正在做一个项目，需要使用 LATEX 绘制一个类似于下面的示意图。有人知道怎么做吗？ 在此处输入图片描述 具体来说，我想知道如何在块内放置文本。我还想知道如何创建这两个块以及如何正确连接它们。以下是我目前所拥有的。 在此处输入图片描述

\documentclass{article}
\usepackage{amsfonts}
\usepackage{siunitx}
\usepackage[table]{xcolor}
\usepackage{booktabs}
\usepackage[margin=0.5in]{geometry}
\usepackage{tikz}
\usetikzlibrary{arrows, automata, backgrounds, fit, matrix, arrows.meta, calc, positioning}
\begin{document}
\begin{tikzpicture}[>=stealth]
\matrix[matrix of math nodes, nodes in empty cells, draw] (blockA) at (0,2)
{
A & & & F \\
B & & & G \\
C & & & H \\
D & & & I  \\
& & & & E \\
};
\matrix[matrix of math nodes,nodes in empty cells,draw,minimum width=30pt,below=of blockA] (blockB) 
{
A & & & F \\
B & & & G \\
C & & & H \\
D & & & I \\
E & & & J \\
};
\node at (blockB) {$Block B$};

\foreach \i in {1,2,3,4,5}
{
\draw[->] let \p1 = (blockA.west), \p2 = (blockA-\i-\i.west), \p3 = (blockB.west), \p4 = (blockB-\i-1.west) in (\x3,\y4) -| 
++(-2.4+0.2*\i,1) |- (\x1,\y2);
\draw[->] let \p1 = (blockA.east), \p2 = (blockA-\i-\i.east), \p3 = (blockB.east), \p4 = (blockB-\i-1.east) in (\x1,\y2) -|
++(1.6-0.2*\i,-1) |- (\x3,\y4);
}
\end{tikzpicture}
\end{document}

答案1

为了使其对称，我使用绝对坐标作为中点。唯一棘手的是找到顶部块下方和底部块上方的 y 值。

由于行数为奇数，我只需将文本添加为​​矩阵的一部分即可。如果行数为偶数，则可以\hphantom{\text{Block A}}像以前一样使用并覆盖节点来保留空间。

\documentclass{article}
\usepackage{amsfonts}
\usepackage{siunitx}
\usepackage[table]{xcolor}
\usepackage{booktabs}
\usepackage[margin=0.5in]{geometry}
\usepackage{tikz}
\usetikzlibrary{arrows, automata, backgrounds, fit, matrix, arrows.meta, calc, positioning}
\begin{document}
\begin{tikzpicture}[>=stealth]
\matrix[matrix of math nodes, nodes in empty cells, draw] (blockA) at (0,2)
{
A & & F \\
B & & G \\
C & \text{Block A} & H \\
D & & I \\
E & & J \\
};
\matrix[matrix of math nodes,nodes in empty cells,draw,below=of blockA] (blockB) 
{
A & & F \\
B & & G \\
C & \text{Block B} & H \\
D & & I \\
E & & J \\
};

\foreach \i in {1,2,3,4,5}
{
\draw[->] let \p1 = (blockA.west), \p2 = (blockA-\i-1.west), \p3 = (blockB.west), \p4 = (blockB-\i-1.west) in (\x3,\y4) -| 
(-4+0.2*\i,0) |- (\x1,\y2);
\draw[->] let \p1 = (blockA.east), \p2 = (blockA-\i-1.east), \p3 = (blockB.east), \p4 = (blockB-\i-1.east) in (\x1,\y2) -|
(4-0.2*\i,0) |- (\x3,\y4);
}
\end{tikzpicture}
\end{document}

演示

答案2

这是一个解决方案，它定义了一个名为 的新形状block，可以与任意数量的输入和输出一起使用。我曾经为类似的问题做过一次。优点：

  • 任意数量的输入和输出
  • 输入/输出之间的可配置空间
  • Tikz 锚定在输入/输出处
  • rectangle形状与默认形状相同
  • flip可以选择右侧输入、左侧输出

如果您愿意，我可以详细说明代码，但您也可以直接使用它。

\documentclass[tikz]{standalone}

\usetikzlibrary{positioning}

\makeatletter
\newdimen\block@iospacing
\newdimen\block@height

\newif\if@block@flip

\tikzset{
    block/.code={
        \pgfkeys{
            /block/.cd,
            #1
        }
        \pgfmathsetlengthmacro{\block@height}{%
            max(10mm, int(max(\block@inputs,\block@outputs) * \block@iospacing))}
        \tikzset{
            draw,
            align=center,
            minimum width = 15mm,
            minimum height = \block@height,
            shape=block
        }
    }
}
\pgfkeys{
    /block/.is family,
    /block/.cd,
    inputs/.code={
        \pgfmathparse{int(#1)}
        \let\block@inputs=\pgfmathresult
    },
    inputs=1,
    outputs/.code={
        \pgfmathparse{int(#1)}
        \let\block@outputs=\pgfmathresult
    },
    outputs=1,
    io spacing/.code=\setlength\block@iospacing{#1},
    io spacing=5mm,
    flip/.is choice,
    flip/true/.code={\@block@fliptrue\def\@block@flipbool{1}},
    flip/false/.code={\@block@flipfalse\def\@block@flipbool{0}},
    flip/.default=true,
    flip=false,
}

\pgfdeclareshape{block}{
    \inheritsavedanchors[from={rectangle}]
    \savedanchor\centerpoint{%
        \pgf@x=.5\wd\pgfnodeparttextbox%
        \pgf@y=.5\ht\pgfnodeparttextbox%
        \advance\pgf@y by-.5\dp\pgfnodeparttextbox%
    }
    \inheritsavedanchors[from=rectangle]
    \inheritanchorborder[from=rectangle]
    \inheritanchor[from=rectangle]{north}
    \inheritanchor[from=rectangle]{north west}
    \inheritanchor[from=rectangle]{north east}
    \inheritanchor[from=rectangle]{center}
    \inheritanchor[from=rectangle]{text}
    \inheritanchor[from=rectangle]{west}
    \inheritanchor[from=rectangle]{east}
    \inheritanchor[from=rectangle]{mid}
    \inheritanchor[from=rectangle]{mid west}
    \inheritanchor[from=rectangle]{mid east}
    \inheritanchor[from=rectangle]{base}
    \inheritanchor[from=rectangle]{base west}
    \inheritanchor[from=rectangle]{base east}
    \inheritanchor[from=rectangle]{south}
    \inheritanchor[from=rectangle]{south west}
    \inheritanchor[from=rectangle]{south east}
    \savedmacro\blockinputs{%
        \pgfmathparse{int(\block@inputs)}%
        \let\blockinputs=\pgfmathresult}%
    \savedmacro\blockoutputs{%
        \pgfmathparse{int(\block@outputs)}%
        \let\blockoutputs=\pgfmathresult}%
    \savedmacro\blockmaxio{%
        \pgfmathparse{int(max(\block@inputs,\block@outputs))}%
        \let\blockmaxio=\pgfmathresult}%
    \savedmacro\blockflip{%
        \pgfmathparse{\@block@flipbool}%
        \let\blockflip=\pgfmathresult}%
    \saveddimen\halfwidth{\pgfmathsetlength\pgf@x{%
            \pgfkeysvalueof{/pgf/minimum width}/2}\pgfmathresult}
    \saveddimen\halfheight{\pgfmathsetlength\pgf@x{%
            \pgfkeysvalueof{/pgf/minimum height}/2}\pgfmathresult}
    \saveddimen\iospacing{\pgfmathsetlength\pgf@x{%
            \block@iospacing}\pgfmathresult}
    \inheritbackgroundpath[from={rectangle}]
    \pgfutil@g@addto@macro\pgf@sh@s@block{%
        \pgfmathloop%
        \ifnum\pgfmathcounter>\blockinputs\relax%
        \else%
        \pgfutil@ifundefined{pgf@anchor@block@input \pgfmathcounter}{%
            \expandafter\xdef\csname pgf@anchor@block@input %
            \pgfmathcounter\endcsname{\noexpand%
                \pgf@sh@lib@block@in@anchor{\pgfmathcounter}%
            }%
        }{}
        \repeatpgfmathloop%
        \pgfmathloop%
        \ifnum\pgfmathcounter>\blockoutputs\relax%
        \else%
        \pgfutil@ifundefined{pgf@anchor@block@output \pgfmathcounter}{%
            \expandafter\xdef\csname pgf@anchor@block@output %
            \pgfmathcounter\endcsname{\noexpand%
                \pgf@sh@lib@block@out@anchor{\pgfmathcounter}%
            }%
        }{}
        \repeatpgfmathloop%
    }%
}

\def\pgf@sh@lib@block@in@anchor#1{%
    \pgf@process{\centerpoint}%
    \pgf@ya=\pgf@y%
    \ifnum\blockflip=0\relax%
        \pgf@process{\southwest}%
    \else%
        \pgf@process{\northeast}%
    \fi%
    \pgfmathsetlength\pgf@y{\pgf@ya + (0.5*(\blockinputs+1)-#1)*\iospacing}
}
\def\pgf@sh@lib@block@out@anchor#1{%
    \pgf@process{\centerpoint}%
    \pgf@ya=\pgf@y%
    \ifnum\blockflip=0\relax%
        \pgf@process{\northeast}%
    \else%
        \pgf@process{\southwest}%
    \fi%
    \pgfmathsetlength\pgf@y{\pgf@ya + (0.5*(\blockoutputs+1)-#1)*\iospacing}
}
\makeatother

\begin{document}
    \begin{tikzpicture}[block/.append style={minimum width=3cm},iolabel/.style={text width=2.5mm,align=center}]
        \node[block={inputs=4,outputs=5}](blockA){Block A};
        \node[block={inputs=5,outputs=4,flip,io spacing=8mm},below=of blockA](blockB){Block B};

        \foreach[count = \i] \letter in {F,...,J}{
            \node[left,iolabel] at (blockA.output \i) {\letter};
            \node[left,iolabel] at (blockB.input \i) {\letter};
            \draw[->] (blockA.output \i) -- ++(2-\i*0.2,0) |- (blockB.input \i);
        }
        \foreach[count = \i] \letter in {A,...,D}{
            \node[right,iolabel] at (blockA.input \i) {\letter};
            \node[right,iolabel] at (blockB.output \i) {\letter};
            \draw[->] (blockB.output \i) -- ++(-2+\i*0.2,0) |- (blockA.input \i);
        }
    \end{tikzpicture}
\end{document}

该代码的结果是：

在此处输入图片描述

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