Probabilidad de error de una señal aleatoria gaussiana

BER: Bit Error Rate

$\begin{array}{cc}& \frac{1}{\sqrt{2\pi {\sigma }^2}}\cdot e^{\frac{-{(t-m)}^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(\mathrm{\infty })=1\\ & \underset{-\mathrm{\infty }}{\overset{V_T}{\int }}\frac{1}{\sqrt{2\pi {\sigma }^2}}\cdot e^{\frac{-{(t-m)}^2}{2{\sigma }^2}}\partial t=?,\underset{V_T}{\overset{\mathrm{\infty }}{\int }}\frac{1}{\sqrt{2\pi {\sigma }^2}}\cdot e^{\frac{-{(t-m)}^2}{2{\sigma }^2}}\partial t=?,\\ & \{\begin{array}{cc}& -{\lambda }^2=\frac{-{(t-m)}^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 \{\begin{array}{cc}& t=\mathrm{\infty }\to \lambda =\mathrm{\infty }\\ & t=V_T\to \lambda =\frac{V_T-m}{\sqrt{2{\sigma }^2}}\end{array}\end{array}\\ & \underset{t=V_T}{\overset{t=\mathrm{\infty }}{\int }}\frac{1}{\sqrt{2\pi {\sigma }^2}}\cdot e^{\frac{-{(t-m)}^2}{2{\sigma }^2}}\partial t=\underset{t=V_T}{\overset{t=\mathrm{\infty }}{\int }}\frac{1}{\sqrt{2\pi {\sigma }^2}}\cdot e^{-{\lambda }^2}\sqrt{2{\sigma }^2}\partial \lambda =\frac{1}{\sqrt{\pi }}\underset{t=V_T}{\overset{t=\mathrm{\infty }}{\int }}{e}^{-{\lambda }^2}\partial \lambda =\frac{1}{\sqrt{\pi }}\underset{\lambda =\frac{V_T-m}{\sqrt{2{\sigma }^2}}}{\overset{\lambda =\mathrm{\infty }}{\int }}{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 }}\underset{\lambda =\frac{V_T-m}{\sqrt{2{\sigma }^2}}}{\overset{\lambda =\mathrm{\infty }}{\int }}{e}^{-{\lambda }^2}\partial \lambda =\frac{1}{\sqrt{\pi }}\underset{\lambda =0}{\overset{\lambda =\mathrm{\infty }}{\int }}{e}^{-{\lambda }^2}\partial \lambda -\frac{1}{\sqrt{\pi }}\underset{\lambda =0}{\overset{\lambda =\frac{V_T-m}{\sqrt{2{\sigma }^2}}}{\int }}{e}^{-{\lambda }^2}\partial \lambda =\frac{erf(\mathrm{\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)\\ & \underset{V_T}{\overset{\mathrm{\infty }}{\int }}\frac{1}{\sqrt{2\pi {\sigma }^2}}\cdot e^{\frac{-{(t-m)}^2}{2{\sigma }^2}}\partial t=Q\left(\frac{V_T-m}{\sqrt{{\sigma }^2}}\right)\\ & \underset{-\mathrm{\infty }}{\overset{V_T}{\int }}\frac{1}{\sqrt{2\pi {\sigma }^2}}\cdot e^{\frac{-{(t-m)}^2}{2{\sigma }^2}}\partial t=1-\underset{V_T}{\overset{\mathrm{\infty }}{\int }}\frac{1}{\sqrt{2\pi {\sigma }^2}}\cdot e^{\frac{-{(t-m)}^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{|V_T-m|}{\sqrt{{\sigma }^2}}\right)\end{array}$

Cálculo de la probabilidad de error

$\begin{array}{cc}& \underset{V_T}{\overset{\mathrm{\infty }}{\int }}\frac{1}{\sqrt{2\pi {\sigma }^2}}\cdot e^{\frac{-{(t-m)}^2}{2{\sigma }^2}}\partial t=Q\left(\frac{|V_T-m|}{\sqrt{{\sigma }^2}}\right)\\ & P_e=pr\left(\text{'}{0}^{\prime }sent\right)\cdot p\left({}^{error}╱{}_{\text{'}{0}^{\prime }sent}\right)+pr\left(\text{'}{1}^{\prime }sent\right)\cdot p\left({}^{error}╱{}_{\text{'}{1}^{\prime }sent}\right)\end{array}$

Proyecto: Departamento de Teoría de la Señal y Comunicaciones

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