Codificaciones digitales

De Wikiversidad

Sabemos que:

s_{T}(t)=\sum\limits_{k=0}^{\infty }{a_{k}p\left( t-kT_{s} \right)}

Ahora que sabemos la DEP de una señal digital ...

Contenido

[editar] Codificación Unipolar NRZ

P(t).png


CodificacionUnipolarNRZ.png


\begin{align} & a_{'1'}=A \\ & a_{'0'}=0 \\ & m_{a_{k}}=a_{'1'}\cdot \frac{1}{2}+a_{'0'}\cdot \frac{1}{2}=A\cdot \frac{1}{2}+0\cdot \frac{1}{2}=\frac{A}{2} \\ & P_{a_{k}}=a_{'1'}^{2}\cdot \frac{1}{2}+a_{'0'}^{2}\cdot \frac{1}{2}=A^{2}\cdot \frac{1}{2}+0\cdot \frac{1}{2}=\frac{A^{2}}{2} \\ & \sigma _{a_{k}}^{2}=P_{a_{k}}-m_{a_{k}}^{2}=\frac{A^{2}}{2}-\left( \frac{A}{2} \right)^{2}=\frac{2A^{2}}{4}-\frac{A^{2}}{4}=\frac{A^{2}}{4} \\ & T_{s}=T_{b} \\ & p(t)=\prod{\left( \frac{t-{}^{T_{s}}\!\!\diagup\!\!{}_{2}\;}{T_{s}} \right)}\to P(f)=T_{s}\operatorname{sinc}\left( T_{s}f \right)e^{-j2\pi f\frac{T_{s}}{2}}\to \left| P(f) \right|^{2}=T_{s}^{2}\operatorname{sinc}^{2}\left( T_{s}f \right) \\ & R_{s}=\frac{1}{T_{s}} \\ & \bar{G}_{x}(f)=\sigma _{a_{k}}^{2}\cdot R_{s}\left| P(f) \right|^{2}+m_{a_{k}}^{2}\cdot R_{s}^{2}\sum\limits_{k=-\infty }^{\infty }{\left| P(kR_{s}) \right|^{2}\delta \left( f-kR_{s} \right)} \\ & \bar{G}_{x}(f)=\frac{A^{2}}{4}R_{s}\cdot T_{s}^{2}\operatorname{sinc}^{2}\left( T_{s}f \right)+\frac{A^{2}}{4}R_{s}^{2}\sum\limits_{k=-\infty }^{\infty }{T_{s}^{2}\operatorname{sinc}^{2}\left( T_{s}\left( kR_{s} \right) \right)\delta \left( f-kR_{s} \right)}= \\ & \bar{G}_{x}(f)=\frac{A^{2}}{4}T_{s}\operatorname{sinc}^{2}\left( T_{s}f \right)+\frac{A^{2}}{4}\sum\limits_{k=-\infty }^{\infty }{\operatorname{sinc}^{2}\left( k \right)\delta \left( f-kR_{s} \right)}\to \\ & \operatorname{sinc}(k)=\left\{ \begin{align} & 1,k=0 \\ & 0,k\ne 0 \\ \end{align} \right. \\ & \bar{G}_{x}(f)=\frac{A^{2}}{4}T_{s}\operatorname{sinc}^{2}\left( T_{s}f \right)+\frac{A^{2}}{4}\delta \left( f-R_{s} \right) \\ \end{align}


Para la probabilidad de error (BER):

Unipolar NRZ

[editar] Codificación Unipolar RZ

P(t)RZ.png


CodificacionUnipolarRZ.png


\begin{align} & a_{'1'}=A \\ & a_{'0'}=0 \\ & m_{a_{k}}=a_{'1'}\cdot \frac{1}{2}+a_{'0'}\cdot \frac{1}{2}=A\cdot \frac{1}{2}+0\cdot \frac{1}{2}=\frac{A}{2} \\ & P_{a_{k}}=a_{'1'}^{2}\cdot \frac{1}{2}+a_{'0'}^{2}\cdot \frac{1}{2}=A^{2}\cdot \frac{1}{2}+0\cdot \frac{1}{2}=\frac{A^{2}}{2} \\ & \sigma _{a_{k}}^{2}=P_{a_{k}}-m_{a_{k}}^{2}=\frac{A^{2}}{2}-\left( \frac{A}{2} \right)^{2}=\frac{2A^{2}}{4}-\frac{A^{2}}{4}=\frac{A^{2}}{4} \\ & T_{s}=T_{b} \\ & p(t)=\prod{\left( \frac{t-{}^{T_{s}}\!\!\diagup\!\!{}_{4}\;}{{}^{T_{s}}\!\!\diagup\!\!{}_{2}\;} \right)}\to P(f)=\frac{T_{s}}{2}\operatorname{sinc}\left( \frac{T_{s}}{2}f \right)e^{-j2\pi f\frac{T_{b}}{4}}\to \\ & \left| P(f) \right|^{2}=\left( \frac{T_{s}}{2} \right)^{2}\operatorname{sinc}^{2}\left( \frac{T_{s}}{2}f \right)=\frac{T_{s}^{2}}{4}\operatorname{sinc}^{2}\left( \frac{T_{s}}{2}f \right) \\ & R_{s}=\frac{1}{T_{s}} \\ & \bar{G}_{x}(f)=\sigma _{a_{k}}^{2}\cdot R_{s}\left| P(f) \right|^{2}+m_{a_{k}}^{2}\cdot R_{s}^{2}\sum\limits_{k=-\infty }^{\infty }{\left| P(kR_{s}) \right|^{2}\delta \left( f-kR_{s} \right)} \\ & \bar{G}_{x}(f)=\frac{A^{2}}{4}R_{s}\cdot \frac{T_{s}^{2}}{4}\operatorname{sinc}^{2}\left( \frac{T_{s}}{2}f \right)+\frac{A^{2}}{4}R_{s}^{2}\sum\limits_{k=-\infty }^{\infty }{\frac{T_{s}^{2}}{4}\operatorname{sinc}^{2}\left( \frac{T_{s}}{2}\left( kR_{s} \right) \right)\delta \left( f-kR_{s} \right)}= \\ & \bar{G}_{x}(f)=\frac{A^{2}}{16}T_{s}\operatorname{sinc}^{2}\left( \frac{T_{s}}{2}f \right)+\frac{A^{2}}{16}\sum\limits_{k=-\infty }^{\infty }{\operatorname{sinc}^{2}\left( \frac{k}{2} \right)\delta \left( f-kR_{s} \right)} \\ \end{align}

Para la probabilidad de error (BER):

Unipolar RZ

[editar] Codificación Polar

P(t).png


CodificacionPolar.png


\begin{align} & a_{'1'}=A \\ & a_{'0'}=-A \\ & m_{a_{k}}=a_{'1'}\cdot \frac{1}{2}+a_{'0'}\cdot \frac{1}{2}=A\frac{1}{2}-A\frac{1}{2}=0 \\ & P_{a_{k}}=a_{'1'}^{2}\cdot \frac{1}{2}+a_{'0'}^{2}\cdot \frac{1}{2}=A^{2}\cdot \frac{1}{2}+\left( -A \right)^{2}\cdot \frac{1}{2}=A^{2} \\ & \sigma _{a_{k}}^{2}=P_{a_{k}}-m_{a_{k}}^{2}=A^{2}-0=A^{2} \\ & T_{s}=T_{b} \\ & p(t)=\prod{\left( \frac{t-{}^{T_{s}}\!\!\diagup\!\!{}_{2}\;}{T_{s}} \right)}\to P(f)=T_{s}\operatorname{sinc}\left( T_{s}f \right)e^{-j2\pi f\frac{T_{s}}{2}}\to \left| P(f) \right|^{2}=T_{s}^{2}\operatorname{sinc}^{2}\left( T_{s}f \right) \\ & R_{s}=\frac{1}{T_{s}} \\ & \bar{G}_{x}(f)=\underbrace{\sigma _{a_{k}}^{2}}_{A^{2}}\cdot R_{s}\left| P(f) \right|^{2}+\underbrace{m_{a_{k}}^{2}}_{0}\cdot R_{s}^{2}\sum\limits_{k=-\infty }^{\infty }{\left| P(kR_{s}) \right|^{2}\delta \left( f-kR_{s} \right)} \\ & \bar{G}_{x}(f)=A^{2}R_{s}\cdot T_{s}^{2}\operatorname{sinc}^{2}\left( T_{s}f \right)=A^{2}T_{s}\operatorname{sinc}^{2}\left( T_{s}f \right) \\ \end{align}

Para la probabilidad de error (BER):

Polar

[editar] Codificación Bipolar

P(t).png


CodificacionBipolar.png


\begin{align} & a_{'1'}=\pm A \\ & a_{'0'}=0 \\ \end{align}

\begin{align} & m_{a_{k}}=a_{'1+'}\cdot \frac{1}{4}+a_{'0'}\cdot \frac{1}{4}+a_{'1-'}\cdot \frac{1}{4}+a_{'0'}\cdot \frac{1}{4}=A\cdot \frac{1}{4}+0\cdot \frac{1}{4}+\left( -A \right)\cdot \frac{1}{4}+0\cdot \frac{1}{4}=0 \\ & P_{a_{k}}=a_{'1+'}^{2}\cdot \frac{1}{4}+a_{'0'}^{2}\cdot \frac{1}{4}+a_{'1-'}^{2}\cdot \frac{1}{4}+a_{'0'}^{2}\cdot \frac{1}{4}=A^{2}\cdot \frac{1}{4}+0\cdot \frac{1}{4}+\left( -A \right)^{2}\cdot \frac{1}{4}+0\cdot \frac{1}{4}=\frac{A^{2}}{2} \\ \end{align}

\begin{align} & \sigma _{a_{k}}^{2}=P_{a_{k}}-m_{a_{k}}^{2}=\frac{A^{2}}{2} \\ & T_{s}=T_{b} \\ & p(t)=\prod{\left( \frac{t-{}^{T_{s}}\!\!\diagup\!\!{}_{2}\;}{T_{s}} \right)}\to P(f)=T_{s}\operatorname{sinc}\left( T_{s}f \right)e^{-j2\pi f\frac{T_{s}}{2}}\to \left| P(f) \right|^{2}=T_{s}^{2}\operatorname{sinc}^{2}\left( T_{s}f \right) \\ & R_{s}=\frac{1}{T_{s}} \\ & \bar{G}_{x}(f)=\underbrace{\sigma _{a_{k}}^{2}}_{\frac{A^{2}}{2}}\cdot R_{s}\left| P(f) \right|^{2}+\underbrace{m_{a_{k}}^{2}}_{0}\cdot R_{s}^{2}\sum\limits_{k=-\infty }^{\infty }{\left| P(kR_{s}) \right|^{2}\delta \left( f-kR_{s} \right)} \\ & \bar{G}_{x}(f)=\frac{A^{2}}{2}\cdot R_{s}T_{s}^{2}\operatorname{sinc}^{2}\left( T_{s}f \right)=\frac{A^{2}}{2}T_{s}\operatorname{sinc}^{2}\left( T_{s}f \right) \\ \end{align}

Para la probabilidad de error (BER):

Bipolar

[editar] Codificación Manchester

P(t)Manchester.png


CodificacionManchester.png


\begin{align} & a_{'1'}=A \\ & a_{'0'}=-A \\ & m_{a_{k}}=a_{'1'}\cdot \frac{1}{2}+a_{'0'}\cdot \frac{1}{2}=A\cdot \frac{1}{2}+\left( -A \right)\cdot \frac{1}{2}=0 \\ & P_{a_{k}}=a_{'1'}^{2}\cdot \frac{1}{2}+a_{'0'}^{2}\cdot \frac{1}{2}=A^{2}\cdot \frac{1}{2}+\left( -A \right)^{2}\cdot \frac{1}{2}=A^{2} \\ & \sigma _{a_{k}}^{2}=P_{a_{k}}-m_{a_{k}}^{2}=A^{2} \\ & T_{s}=T_{b} \\ & p(t)=\prod{\left( \frac{t-{}^{T_{s}}\!\!\diagup\!\!{}_{4}\;}{{}^{T_{s}}\!\!\diagup\!\!{}_{2}\;} \right)}-\prod{\left( \frac{t-{}^{3\cdot T_{s}}\!\!\diagup\!\!{}_{4}\;}{{}^{T_{s}}\!\!\diagup\!\!{}_{2}\;} \right)}\to \\ & P(f)=\frac{T_{s}}{2}\operatorname{sinc}\left( \frac{T_{s}}{2}f \right)e^{-j2\pi f\frac{T_{s}}{4}}-\frac{T_{s}}{2}\operatorname{sinc}\left( \frac{T_{s}}{2}f \right)e^{-j2\pi f\frac{3T_{s}}{4}}\Rightarrow \left\{ \sin x=\frac{e^{jx}-e^{-jx}}{2j}\to 2j\sin x=e^{jx}-e^{-jx} \right\} \\ & =\frac{T_{s}}{2}\operatorname{sinc}\left( \frac{T_{s}}{2}f \right)e^{-j2\pi f\frac{T_{s}}{2}}\left( e^{+j2\pi f\frac{T_{s}}{4}}-e^{-j2\pi f\frac{T_{s}}{4}} \right)=\frac{T_{s}}{2}\operatorname{sinc}\left( \frac{T_{s}}{2}f \right)e^{-j2\pi f\frac{T_{s}}{2}}2j\sin \left( 2\pi f\frac{T_{s}}{4} \right) \\ & \left| P(f) \right|^{2}=\left( \frac{T_{s}}{2} \right)^{2}\operatorname{sinc}^{2}\left( \left( \frac{T_{s}}{2} \right)f \right)4\sin ^{2}\left( 2\pi f\frac{T_{s}}{4} \right)=T_{s}^{2}\operatorname{sinc}^{2}\left( \frac{T_{s}f}{2} \right)\sin ^{2}\left( \pi f\frac{T_{s}}{2} \right) \\ & R_{s}=\frac{1}{T_{s}} \\ & \bar{G}_{x}(f)=\underbrace{\sigma _{a_{k}}^{2}}_{A^{2}}\cdot R_{s}\left| P(f) \right|^{2}+\underbrace{m_{a_{k}}^{2}}_{0}\cdot R_{s}^{2}\sum\limits_{k=-\infty }^{\infty }{\left| P(kR_{s}) \right|^{2}\delta \left( f-kR_{s} \right)} \\ & \bar{G}_{x}(f)=A^{2}R_{s}T_{s}^{2}\operatorname{sinc}^{2}\left( \frac{T_{s}f}{2} \right)\sin ^{2}\left( \pi f\frac{T_{s}}{2} \right)=A^{2}T_{s}\operatorname{sinc}^{2}\left( \frac{T_{s}f}{2} \right)\sin ^{2}\left( \pi f\frac{T_{s}}{2} \right) \\ \end{align}

Para la probabilidad de error (BER):

Manchester



Proyecto: Departamento de Teoría de la Señal y Comunicaciones
Anterior: ISI (Inter Symbol Interference) — Codificaciones digitales — Siguiente: Modulaciones digitales


Obtenido de "http://es.wikiversity.org/wiki/Codificaciones_digitales"
Herramientas personales