模拟计算机图：参数化的输入数量

Question

这是这个问题。

他修改了运算放大器，使得积分器的矩形出现，但我遇到了问题。

人们给了我一个解决方案，即删除加号和减号（这是理想的）。但是，代码没有考虑用户想要的模拟加法器和积分器的输入数量（在 CircuiTikz 包中，可以指定逻辑门的输入数量）。在之前的代码中，人们只是在积分器的中间画了一条线，这不是我想要的；我想要一个输入从那里伸出来，这取决于输入的数量。

以下是我所拥有的：

对应方程：

我查看了 CircuiTikz 中的逻辑门代码以及代码中提供的解决方案，但语法对我来说非常不友好。有人可以提供代码来执行此操作并详细解释代码中使用的术语吗？我不知道许多术语是什么意思。

Answer 1

这是实现此目的的一种方式；我使用组件的边框引脚和外部引脚之间的差异muxdemux来创建框。如果您保持和之间的比率不变NL，您将获得美观的对象，当然，将“矩形绘制”封装到宏中也相当容易。Lhw

\documentclass[border=10pt]{standalone}
\usepackage[siunitx, RPvoltages]{circuitikz}
\begin{document}

\begin{circuitikz}[
    ]
    \tikzset{small text/.style={
            font=\tiny,
            right,
            inner xsep=1pt,
        },
    }
    % four input plain thing. I am using two pins more to give space
    % for the text
    \node[muxdemux, no input leads, muxdemux def={Lh=4, NL=6, w=4, Rh=0, NR=1}]
         (A){adder};
    \draw (A.blpin 2) node[small text]{$1$} -- ++(-1,0) node[left]{$u$};
    \draw (A.blpin 3) node[small text]{$-1$} -- ++(-0.8,0);
    \draw (A.blpin 4) node[small text]{$-1.2$} -- ++(-0.6,0);
    \draw (A.blpin 5) node[small text]{$-1.2$} -- ++(-0.4,0);
    \draw (A.brpin 1) -- ++(1,0) coordinate(one);
    % integrator  --- I use three input pins to build the rectangle
    % I use the difference between border pin and exteral pin to build the
    % square block
    \node[muxdemux, no input leads, muxdemux def={Lh=2, NL=3, w=2, Rh=0, NR=1}, anchor=lpin 2]
         (B) at(one) {$\int\quad$};
    \draw[line width=0.8pt] (B.north west -| B.lpin 2) rectangle (B.south west -| B.blpin 2);
    % notice that there is a small overshoot of the triangle for the line bevel. You can adjust for that if you want.
    \draw (B.brpin 1) -- ++(1,0) coordinate(two);
    \node [small text] at (B.lpin 2) {$1$};
    % deriver ---
    \node[muxdemux, no input leads, muxdemux def={Lh=2, NL=3, w=2, Rh=0, NR=1}, anchor=lpin 2]
         (C) at(two) {$\frac{\mathrm{d}}{\mathrm{d}\,t}\quad$} ;
    \draw[line width=0.8pt] (C.north west -| C.lpin 2) rectangle (C.south west -| C.brpin 1);
    \node [small text] at (C.lpin 2) {$1$};
    \draw (C.brpin 1) -- ++(1,0) coordinate(three);
\end{circuitikz}
\end{document}

为了使事情更紧凑，你可以使用薛定谔的猫的绝招和pics；append after command在这种情况下，你需要添加锚点遗漏的部分muxdemux（我会在下一版本中添加它！）：

\documentclass[border=10pt]{standalone}
\usepackage[siunitx, RPvoltages]{circuitikz}
\makeatletter
\def\pgfaddtoshape#1#2{% https://tex.stackexchange.com/a/14772/38080
  \begingroup
  \def\pgf@sm@shape@name{#1}%
  \let\anchor\pgf@sh@anchor
  #2%
  \endgroup
}
% we need to add an anchor to muxdemux
\pgfaddtoshape{muxdemux}{
    \anchor{top left ext}{%
        \topleft\advance\pgf@x by -\extshift
}}
\makeatother
\begin{document}

\begin{circuitikz}[
    ]
    \tikzset{small text/.style={
            font=\tiny,
            right,
            inner xsep=1pt,
        },
    integral block/.style={append after command={%
        pic{inputrect}}
        },
    derivative block/.style={append after command={%
        pic{fullrect}}
        },
    pics/inputrect/.style={code={\let\mytikzlastnode\tikzlastnode
        % adjust thickness
        \draw[line width=0.8pt] (\mytikzlastnode.top left ext) rectangle (\mytikzlastnode.bottom left);}},
    pics/fullrect/.style={code={\let\mytikzlastnode\tikzlastnode
        % adjust thickness
        \draw[line width=0.8pt] (\mytikzlastnode.top left ext) rectangle (\mytikzlastnode.south east);}},
    }
    % four input plain thing. I am using two pins more to give space
    % for the text
    \node[muxdemux, no input leads, muxdemux def={Lh=4, NL=6, w=4, Rh=0, NR=1}](A){adder};
    \draw (A.blpin 2) node[small text]{$1$} -- ++(-1,0) node[left]{$u$};
    \draw (A.blpin 3) node[small text]{$-1$} -- ++(-0.8,0);
    \draw (A.blpin 4) node[small text]{$-1.2$} -- ++(-0.6,0);
    \draw (A.blpin 5) node[small text]{$-1.2$} -- ++(-0.4,0);
    \draw (A.brpin 1) -- ++(1,0) coordinate(one);
    % integrator  --- I use three input pins to build the rectangle
    % I use the difference between border pin and exteral pin to build the
    % square block
    \node[muxdemux, muxdemux def={Lh=2, NL=3, w=2, Rh=0, NR=1},
        no input leads, integral block,
        anchor=lpin 2](B) at(one) {$\int\quad$};
    % notice that there is a small overshoot of the triangle for the line bevel. You can adjust for that if you want.
    \draw (B.brpin 1) -- ++(1,0) coordinate(two);
    \node [small text] at (B.lpin 2) {$1$};
    % deriver ---
    \node[muxdemux,  muxdemux def={Lh=2, NL=3, w=2, Rh=0, NR=1},
        no input leads, derivative block,
        anchor=lpin 2](C) at(two) {$\frac{\mathrm{d}}{\mathrm{d}\,t}\quad$};
    \node [small text] at (C.lpin 2) {$1$};
    \draw (C.brpin 1) -- ++(1,0) coordinate(three);
    %
    % you can also rotate them...
    %
    \draw (B.brpin 1) ++(0.5,0) -- ++(0,1) coordinate(four);
    \node[muxdemux,  muxdemux def={Lh=2, NL=3, w=2, Rh=0, NR=1},
        no input leads, integral block,
        anchor=lpin 2, rotate=90](D) at(four) {$\int\quad$};
        \draw (B.brpin 1) ++(0.5,0) -- ++(0,-1) coordinate(five);
    \node[muxdemux,  muxdemux def={Lh=2, NL=3, w=2, Rh=0, NR=1},
        no input leads, derivative block,
        anchor=lpin 2, rotate=-90](D) at(five) {$\frac{\mathrm{d}}{\mathrm{d}\,t}\quad$};
\end{circuitikz}
\end{document}