5

I am trying to insert arrows between the terms of an equation explaining the partial results achieved. For example, in the following equation (I manually drew the arrows), V_i \frac{g_m}{1+g_mR_s} would give I_d, while V_i \frac{g_m}{1+g_mR_s} R_d would be equal to V_{R_d} and so on: enter image description here

Could anyone help me come up with a solution?

Thanks in advance.

2

A nested stack. I introduce \dnAr[<size>]{<label>} to do so, where <size> is stuff like \bigg (the default). In the 2nd example, I exercise the optional argument to \dnAr.

\documentclass{report}
\usepackage{stackengine,graphicx}
\newcommand\dnAr[2][\bigg]{\ensurestackMath{%
  \stackengine{-0.3pt}{#1\vert}{%
    \stackengine{1pt}{\rotatebox{55}{$\leftarrow\mkern-2mu$}}{%
      \mkern-5mu\scriptscriptstyle#2}{U}{l}{F}{T}{S}}{U}{r}{F}{T}{S}%
}}
\begin{document}
\[
V_u = V_i \frac{g_m}{1 + g_m R_s} \dnAr{I_d} R_d \dnAr{V_{R_d}}(-1)\dnAr{V_u}
\]
\[
V_u = V_i \frac{g_m}{1 + g_m R_s} \dnAr{I_d} R_d \dnAr[\big]{V_{R_d}}(-1)\dnAr[]{V_u}
\]
\end{document}

enter image description here

Here's a version with 45 degree angled arrow and a tighter under-label:

\documentclass{report}
\usepackage{stackengine,graphicx}
\newcommand\dnAr[2][\bigg]{\ensurestackMath{%
  \stackengine{-0.7pt}{#1\vert}{%
    \stackengine{-1pt}{\rotatebox{45}{$\leftarrow\mkern-1.5mu$}}{%
      \mkern-5mu\scriptscriptstyle#2}{U}{l}{F}{T}{S}}{U}{r}{F}{T}{S}%
}}
\begin{document}
\[
V_u = V_i \frac{g_m}{1 + g_m R_s} \dnAr{I_d} R_d \dnAr{V_{R_d}}(-1)\dnAr{V_u}
\]
\[
V_u = V_i \frac{g_m}{1 + g_m R_s} \dnAr{I_d} R_d \dnAr[\big]{V_{R_d}}(-1)\dnAr[]{V_u}
\]
\end{document}

enter image description here

  • @RenatoArino You are most welcome. I have added a variant that may be more to your taste. – Steven B. Segletes Apr 28 '16 at 18:57
2

A Tikz version.

Output

enter image description here

Code

\documentclass{article}
\usepackage{tikz}

\newcommand\arrd[1]{
    \tikz[baseline, inner xsep=-1cm]{
        \draw[->] (0,.4) -- (0,-.2) -- (-.1,-.3) node[below, font=\tiny] {#1};
    }\ % for spacing
}

\begin{document}

$V_u = V_i \frac{g_m}{1+g_mR_s} \arrd{$id$} R_d \arrd{$V_{R_d}$} (-1) \arrd{$V_u$}$

\end{document} 

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.