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erfc table and formulas
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huettern committed Jan 7, 2020
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25 changes: 14 additions & 11 deletions Communication Systems/communication_systems.tex
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% @Author: Noah Huetter
% @Date: 2019-09-24 17:26:28
% @Last Modified by: noah
% @Last Modified time: 2020-01-07 11:46:45
% @Last Modified time: 2020-01-07 12:21:57
% ---------------------------------------------------------------------------

\documentclass[a4paper, fontsize=8pt, landscape, DIV=1]{scrartcl}
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$f^{(n)}$ $n^\text{th}$ derivation \\
$f^*$ complex conjugate

\subsection{Probability}
\begin{empheq}{align*}
&\P(X > x) = \frac{1}{2}\erfc\left(\frac{x}{\sqrt{2}\sigma_X}\right) \\
&\erf(x) = \frac{1}{\sqrt{\pi}}\int_{-x}^x \e^{-t^2}\dt = \frac{2}{\sqrt{\pi}}\int_{0}^x \e^{-t^2}\dt \\
&\erfc(x) = 1 - \erf{x} = \frac{2}{\sqrt{\pi}}\int_{x}^\infty \e^{-t^2}\dt \\
&Q(x) = \frac{1}{\sqrt{2\pi}}\int_x^\infty\exp(-y^2/2)\dy\leq\exp(-x^2/2)
\end{empheq}

\vfill\null
\columnbreak
\subsection{Sums}
\begin{empheq}{align*}
& \sum_{k=0}^n q^k k &&=&& \frac{nq^{n+2}-(n+1)q^{n+1}+q}{(q-1)^2} \\
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& \sum_{k=0}^\infty q^k &&=&& \frac{1}{1-q} \\
\end{empheq}

\subsection{Probability}
\begin{empheq}{align*}
&\P(X > x) = \frac{1}{2}\erfc\left(\frac{x}{\sqrt{2}\sigma_X}\right) \\
&\erf(x) = \frac{1}{\sqrt{\pi}}\int_{-x}^x \e^{-t^2}\dt = \frac{2}{\sqrt{\pi}}\int_{0}^x \e^{-t^2}\dt \\
&\erfc(x) = 1 - \erf{x} = 2Q(\sqrt{2}x) = \frac{2}{\sqrt{\pi}}\int_{x}^\infty \e^{-t^2}\dt \\
&Q(x) = \frac{1}{\sqrt{2\pi}}\int_x^\infty\exp(-y^2/2)\dy\leq\exp(-x^2/2) \\
&Q(x) = \frac{1}{2}\erfc\left(\frac{x}{\sqrt{2}}\right)
\end{empheq}
\cgraphic{0.999}{img/erfcq.png}

% \vfill\null
% \columnbreak


\end{multicols*}

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Binary file added Communication Systems/img/erfcq.png
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