# Probabilidad de error de una señal aleatoria gaussiana

BER: Bit Error Rate
${\begin{aligned}&{\frac {1}{\sqrt {2\pi \sigma ^{2}}}}\cdot e^{\frac {-\left(t-m\right)^{2}}{2\sigma ^{2}}}\\&erf(x)={\frac {2}{\sqrt {\pi }}}\int _{\lambda =0}^{\lambda =x}{e^{-\lambda ^{2}}\partial \lambda }\to erfc(x)=erf(-x)=1-erf(x),erf(\infty )=1\\&\int \limits _{-\infty }^{V_{T}}{{\frac {1}{\sqrt {2\pi \sigma ^{2}}}}\cdot e^{\frac {-\left(t-m\right)^{2}}{2\sigma ^{2}}}\partial t}=?,\int \limits _{V_{T}}^{\infty }{{\frac {1}{\sqrt {2\pi \sigma ^{2}}}}\cdot e^{\frac {-\left(t-m\right)^{2}}{2\sigma ^{2}}}\partial t}=?,\\&\left\{{\begin{aligned}&-\lambda ^{2}={\frac {-\left(t-m\right)^{2}}{2\sigma ^{2}}}\Rightarrow \lambda ={\frac {t-m}{\sqrt {2\sigma ^{2}}}}\leftrightarrow t={\sqrt {2\sigma ^{2}}}\lambda +m\Rightarrow \partial t={\sqrt {2\sigma ^{2}}}\partial \lambda \\&t={\sqrt {2\sigma ^{2}}}\lambda +m\to \left\{{\begin{aligned}&t=\infty \to \lambda =\infty \\&t=V_{T}\to \lambda ={\frac {V_{T}-m}{\sqrt {2\sigma ^{2}}}}\\\end{aligned}}\right.\\\end{aligned}}\right.\\&\int \limits _{t=V_{T}}^{t=\infty }{{\frac {1}{\sqrt {2\pi \sigma ^{2}}}}\cdot e^{\frac {-\left(t-m\right)^{2}}{2\sigma ^{2}}}\partial t}=\int \limits _{t=V_{T}}^{t=\infty }{{\frac {1}{\sqrt {2\pi \sigma ^{2}}}}\cdot e^{-\lambda ^{2}}{\sqrt {2\sigma ^{2}}}\partial \lambda =}{\frac {1}{\sqrt {\pi }}}\int \limits _{t=V_{T}}^{t=\infty }{e^{-\lambda ^{2}}\partial \lambda }={\frac {1}{\sqrt {\pi }}}\int \limits _{\lambda ={\frac {V_{T}-m}{\sqrt {2\sigma ^{2}}}}}^{\lambda =\infty }{e^{-\lambda ^{2}}\partial \lambda }\\&{\frac {erf(x)}{2}}={\frac {1}{\sqrt {\pi }}}\int _{\lambda =0}^{\lambda =x}{e^{-\lambda ^{2}}\partial \lambda }\\&{\frac {1}{\sqrt {\pi }}}\int \limits _{\lambda ={\frac {V_{T}-m}{\sqrt {2\sigma ^{2}}}}}^{\lambda =\infty }{e^{-\lambda ^{2}}\partial \lambda }={\frac {1}{\sqrt {\pi }}}\int \limits _{\lambda =0}^{\lambda =\infty }{e^{-\lambda ^{2}}\partial \lambda }-{\frac {1}{\sqrt {\pi }}}\int \limits _{\lambda =0}^{\lambda ={\frac {V_{T}-m}{\sqrt {2\sigma ^{2}}}}}{e^{-\lambda ^{2}}\partial \lambda }={\frac {erf(\infty )=1}{2}}-{\frac {erf\left({\frac {V_{T}-m}{\sqrt {2\sigma ^{2}}}}\right)}{2}}=\\&{\frac {1-erf\left({\frac {V_{T}-m}{\sqrt {2\sigma ^{2}}}}\right)}{2}}={\frac {erfc\left({\frac {V_{T}-m}{\sqrt {2\sigma ^{2}}}}\right)}{2}}\to Q(x)={\frac {erfc\left({\frac {x}{\sqrt {2}}}\right)}{2}}\to Q\left({\frac {V_{T}-m}{\sqrt {\sigma ^{2}}}}\right)\\&\int \limits _{V_{T}}^{\infty }{{\frac {1}{\sqrt {2\pi \sigma ^{2}}}}\cdot e^{\frac {-\left(t-m\right)^{2}}{2\sigma ^{2}}}\partial t}=Q\left({\frac {V_{T}-m}{\sqrt {\sigma ^{2}}}}\right)\\&\int \limits _{-\infty }^{V_{T}}{{\frac {1}{\sqrt {2\pi \sigma ^{2}}}}\cdot e^{\frac {-\left(t-m\right)^{2}}{2\sigma ^{2}}}\partial t}=1-\int \limits _{V_{T}}^{\infty }{{\frac {1}{\sqrt {2\pi \sigma ^{2}}}}\cdot e^{\frac {-\left(t-m\right)^{2}}{2\sigma ^{2}}}\partial t}=1-Q\left({\frac {V_{T}-m}{\sqrt {\sigma ^{2}}}}\right)=Q\left({\frac {m-V_{T}}{\sqrt {\sigma ^{2}}}}\right)\\&Q\left({\frac {\left|V_{T}-m\right|}{\sqrt {\sigma ^{2}}}}\right)\\\end{aligned}}$

## Cálculo de la probabilidad de error

${\begin{aligned}&\int \limits _{V_{T}}^{\infty }{{\frac {1}{\sqrt {2\pi \sigma ^{2}}}}\cdot e^{\frac {-\left(t-m\right)^{2}}{2\sigma ^{2}}}\partial t}=Q\left({\frac {\left|V_{T}-m\right|}{\sqrt {\sigma ^{2}}}}\right)\\&P_{e}=pr\left('0'sent\right)\cdot p\left(^{error}\!\!\diagup \!\!_{'0'sent}\;\right)+pr\left('1'sent\right)\cdot p\left(^{error}\!\!\diagup \!\!_{'1'sent}\;\right)\\\end{aligned}}$
* Proyecto: Departamento de Teoría de la Señal y Comunicaciones
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