1

I would like to represent the following integral equation using latex. The square pulse in the integral denotes a square pulse of a clock. Can anyone suggest me how I can use the square pulse here? That is the only problem I am having while writing the equation.

enter image description here

Thanks in advance.

Arash Esbati
  • 7,416

3 Answers3

1

If you don't mind using TikZ, here is an ad-hoc solution. Otherwise, you can use the ifsym package as pointed out by @egreg, but, the documentation is in German.

\documentclass{article}
\usepackage{tikz,amsmath}

\begin{document}

\newcommand{\pulse}{%
\begin{tikzpicture}[line width=.7pt,scale=.3]
\draw (0,0)--++(.3,0)--++(0,.5)--++(.5,0)--++(0,-.5)--++(.3,0);
\end{tikzpicture}
}

\begin{equation}
E_\text{dyn}=V_\text{DD}\int\limits_{\pulse} I_\text{tot}(t)-P_\text{stat}\cdot T
\end{equation}

\end{document}

enter image description here

AboAmmar
  • 46,352
  • 4
  • 58
  • 127
1

With ifsym:

\documentclass{article}
\usepackage{amsmath, amssymb}
\usepackage{moresize} 
\usepackage[electronic]{ifsym} 

\begin{document}

\[ E_{\text{dyn}} = V_{\text{DD}} \underset{\textbf{\ssmall\PulseHigh}}{\int} I_{\text{tot}(t)\,\mathrm dt-P_{\text{stat}\cdot T} )} \]%

\end{document} 

enter image description here

Bernard
  • 271,350
1

You could use the ifsym package (but it only provides bitmap fonts) or draw yourself the symbol. Equation (1) below shows the hand-made symbol, equation (2) uses ifsym. If you go for method (1), loading ifsym is unnecessary, but pict2e is.

One should note that the symbols provided by ifsym are for text, so if used in math they have to be surrounded by \text.

\documentclass{article}
\usepackage{amsmath}

\usepackage{pict2e}
\usepackage[electronic]{ifsym}

\DeclareRobustCommand{\pulse}{%
  \text{%
    \settoheight{\unitlength}{T}%
    \begin{picture}(2.2,1)
    \roundcap
    \polyline(0.1,0)(0.6,0)(0.6,1)(1.6,1)(1.6,0)(2.1,0)
    \end{picture}%
  }%
}

\begin{document}

\begin{equation}
E_{\mathrm{dyn}} =
V_{\mathrm{DD}}\int\limits_{\pulse} I_{\mathrm{tot}}(t)\,dt
- P_{\mathrm{stat}}\cdot T
\end{equation}

\begin{equation}
E_{\mathrm{dyn}} =
V_{\mathrm{DD}}\int\limits_{\text{\PulseHigh}} I_{\mathrm{tot}}(t)\,dt
- P_{\mathrm{stat}}\cdot T
\end{equation}

\end{document}

enter image description here

egreg
  • 1,121,712