Codificaciones digitales

Sabemos que:

$s_T(t)=\sum _{k=0}^{\mathrm{\infty }}{a}_{k}p(t-k{T}_s)$

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

Codificación Unipolar NRZ

$\begin{array}{cc}& a_{\text{'}{1}^{\prime }}=A\\ & a_{\text{'}{0}^{\prime }}=0\\ & m_{a_k}=a_{\text{'}{1}^{\prime }}\cdot \frac{1}{2}+a_{\text{'}{0}^{\prime }}\cdot \frac{1}{2}=A\cdot \frac{1}{2}+0\cdot \frac{1}{2}=\frac{A}{2}\\ & P_{a_k}=a_{\text{'}{1}^{\prime }}^2\cdot \frac{1}{2}+a_{\text{'}{0}^{\prime }}^2\cdot \frac{1}{2}=A^2\cdot \frac{1}{2}+0\cdot \frac{1}{2}=\frac{A^2}{2}\\ & {\sigma }_{a_k}^2=P_{a_k}-m_{a_k}^2=\frac{A^2}{2}-{\left(\frac{A}{2}\right)}^2=\frac{2A^2}{4}-\frac{A^2}{4}=\frac{A^2}{4}\\ & T_s=T_b\\ & p(t)=\prod \left(\frac{t-{}^{T_s}╱{}_2}{T_s}\right)\to P(f)=T_s\mathrm{sinc}\left(T_{s}f\right)e^{-j2\pi f\frac{T_s}{2}}\to {|P(f)|}^2=T_s^2{\mathrm{sinc}}^2\left(T_{s}f\right)\\ & R_s=\frac{1}{T_s}\\ & {\overline{G}}_x(f)={\sigma }_{a_k}^2\cdot R_s{|P(f)|}^2+m_{a_k}^2\cdot R_s^2\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{|P(k{R}_s)|}^2\delta (f-k{R}_s)\\ & {\overline{G}}_x(f)=\frac{A^2}{4}{R}_s\cdot T_s^2{\mathrm{sinc}}^2\left(T_{s}f\right)+\frac{A^2}{4}{R}_s^2\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{T}_s^2{\mathrm{sinc}}^2\left(T_s\left(k{R}_s\right)\right)\delta (f-k{R}_s)=\\ & {\overline{G}}_x(f)=\frac{A^2}{4}{T}_s{\mathrm{sinc}}^2\left(T_{s}f\right)+\frac{A^2}{4}\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{\mathrm{sinc}}^2\left(k\right)\delta (f-k{R}_s)\to \\ & \mathrm{sinc}(k)=\{\begin{array}{cc}& 1,k=0\\ & 0,k\ne 0\\ \end{array}\\ & {\overline{G}}_x(f)=\frac{A^2}{4}{T}_s{\mathrm{sinc}}^2\left(T_{s}f\right)+\frac{A^2}{4}\delta \left(f\right)\\ \end{array}$

Para la probabilidad de error (BER):

Unipolar NRZ

Codificación Unipolar RZ

$\begin{array}{cc}& a_{\text{'}{1}^{\prime }}=A\\ & a_{\text{'}{0}^{\prime }}=0\\ & m_{a_k}=a_{\text{'}{1}^{\prime }}\cdot \frac{1}{2}+a_{\text{'}{0}^{\prime }}\cdot \frac{1}{2}=A\cdot \frac{1}{2}+0\cdot \frac{1}{2}=\frac{A}{2}\\ & P_{a_k}=a_{\text{'}{1}^{\prime }}^2\cdot \frac{1}{2}+a_{\text{'}{0}^{\prime }}^2\cdot \frac{1}{2}=A^2\cdot \frac{1}{2}+0\cdot \frac{1}{2}=\frac{A^2}{2}\\ & {\sigma }_{a_k}^2=P_{a_k}-m_{a_k}^2=\frac{A^2}{2}-{\left(\frac{A}{2}\right)}^2=\frac{2A^2}{4}-\frac{A^2}{4}=\frac{A^2}{4}\\ & T_s=T_b\\ & p(t)=\prod \left(\frac{t-{}^{T_s}╱{}_4}{{}^{T_s}╱{}_2}\right)\to P(f)=\frac{T_s}{2}\mathrm{sinc}\left(\frac{T_s}{2}f\right)e^{-j2\pi f\frac{T_b}{4}}\to \\ & {|P(f)|}^2={\left(\frac{T_s}{2}\right)}^2{\mathrm{sinc}}^2\left(\frac{T_s}{2}f\right)=\frac{T_s^2}{4}{\mathrm{sinc}}^2\left(\frac{T_s}{2}f\right)\\ & R_s=\frac{1}{T_s}\\ & {\overline{G}}_x(f)={\sigma }_{a_k}^2\cdot R_s{|P(f)|}^2+m_{a_k}^2\cdot R_s^2\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{|P(k{R}_s)|}^2\delta (f-k{R}_s)\\ & {\overline{G}}_x(f)=\frac{A^2}{4}{R}_s\cdot \frac{T_s^2}{4}{\mathrm{sinc}}^2\left(\frac{T_s}{2}f\right)+\frac{A^2}{4}{R}_s^2\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}\frac{T_s^2}{4}{\mathrm{sinc}}^2\left(\frac{T_s}{2}\left(k{R}_s\right)\right)\delta (f-k{R}_s)=\\ & {\overline{G}}_x(f)=\frac{A^2}{16}{T}_s{\mathrm{sinc}}^2\left(\frac{T_s}{2}f\right)+\frac{A^2}{16}\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{\mathrm{sinc}}^2\left(\frac{k}{2}\right)\delta (f-k{R}_s)\\ \end{array}$

Para la probabilidad de error (BER):

Unipolar RZ

Codificación Polar

$\begin{array}{cc}& a_{\text{'}{1}^{\prime }}=A\\ & a_{\text{'}{0}^{\prime }}=-A\\ & m_{a_k}=a_{\text{'}{1}^{\prime }}\cdot \frac{1}{2}+a_{\text{'}{0}^{\prime }}\cdot \frac{1}{2}=A\frac{1}{2}-A\frac{1}{2}=0\\ & P_{a_k}=a_{\text{'}{1}^{\prime }}^2\cdot \frac{1}{2}+a_{\text{'}{0}^{\prime }}^2\cdot \frac{1}{2}=A^2\cdot \frac{1}{2}+{(-A)}^2\cdot \frac{1}{2}=A^2\\ & {\sigma }_{a_k}^2=P_{a_k}-m_{a_k}^2=A^2-0=A^2\\ & T_s=T_b\\ & p(t)=\prod \left(\frac{t-{}^{T_s}╱{}_2}{T_s}\right)\to P(f)=T_s\mathrm{sinc}\left(T_{s}f\right)e^{-j2\pi f\frac{T_s}{2}}\to {|P(f)|}^2=T_s^2{\mathrm{sinc}}^2\left(T_{s}f\right)\\ & R_s=\frac{1}{T_s}\\ & {\overline{G}}_x(f)=\underset{A^2}{\underbrace{{\sigma }_{a_k}^2}}\cdot R_s{|P(f)|}^2+\underset{0}{\underbrace{m_{a_k}^2}}\cdot R_s^2\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{|P(k{R}_s)|}^2\delta (f-k{R}_s)\\ & {\overline{G}}_x(f)=A^2{R}_s\cdot T_s^2{\mathrm{sinc}}^2\left(T_{s}f\right)=A^2{T}_s{\mathrm{sinc}}^2\left(T_{s}f\right)\\ \end{array}$

Para la probabilidad de error (BER):

Polar

Codificación Bipolar

$\begin{array}{cc}& a_{\text{'}{1}^{\prime }}=\pm A\\ & a_{\text{'}{0}^{\prime }}=0\\ \end{array}$

$\begin{array}{cc}& m_{a_k}=a_{\text{'}1{+}^{\prime }}\cdot \frac{1}{4}+a_{\text{'}{0}^{\prime }}\cdot \frac{1}{4}+a_{\text{'}1{-}^{\prime }}\cdot \frac{1}{4}+a_{\text{'}{0}^{\prime }}\cdot \frac{1}{4}=A\cdot \frac{1}{4}+0\cdot \frac{1}{4}+(-A)\cdot \frac{1}{4}+0\cdot \frac{1}{4}=0\\ & P_{a_k}=a_{\text{'}1{+}^{\prime }}^2\cdot \frac{1}{4}+a_{\text{'}{0}^{\prime }}^2\cdot \frac{1}{4}+a_{\text{'}1{-}^{\prime }}^2\cdot \frac{1}{4}+a_{\text{'}{0}^{\prime }}^2\cdot \frac{1}{4}=A^2\cdot \frac{1}{4}+0\cdot \frac{1}{4}+{(-A)}^2\cdot \frac{1}{4}+0\cdot \frac{1}{4}=\frac{A^2}{2}\\ \end{array}$

$\begin{array}{cc}& {\sigma }_{a_k}^2=P_{a_k}-m_{a_k}^2=\frac{A^2}{2}\\ & T_s=T_b\\ & p(t)=\prod \left(\frac{t-{}^{T_s}╱{}_2}{T_s}\right)\to P(f)=T_s\mathrm{sinc}\left(T_{s}f\right)e^{-j2\pi f\frac{T_s}{2}}\to {|P(f)|}^2=T_s^2{\mathrm{sinc}}^2\left(T_{s}f\right)\\ & R_s=\frac{1}{T_s}\\ & {\overline{G}}_x(f)=\underset{\frac{A^2}{2}}{\underbrace{{\sigma }_{a_k}^2}}\cdot R_s{|P(f)|}^2+\underset{0}{\underbrace{m_{a_k}^2}}\cdot R_s^2\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{|P(k{R}_s)|}^2\delta (f-k{R}_s)\\ & {\overline{G}}_x(f)=\frac{A^2}{2}\cdot R_s{T}_s^2{\mathrm{sinc}}^2\left(T_{s}f\right)=\frac{A^2}{2}{T}_s{\mathrm{sinc}}^2\left(T_{s}f\right)\\ \end{array}$

Para la probabilidad de error (BER):

Bipolar

Codificación Manchester

$\begin{array}{cc}& a_{\text{'}{1}^{\prime }}=A\\ & a_{\text{'}{0}^{\prime }}=-A\\ & m_{a_k}=a_{\text{'}{1}^{\prime }}\cdot \frac{1}{2}+a_{\text{'}{0}^{\prime }}\cdot \frac{1}{2}=A\cdot \frac{1}{2}+(-A)\cdot \frac{1}{2}=0\\ & P_{a_k}=a_{\text{'}{1}^{\prime }}^2\cdot \frac{1}{2}+a_{\text{'}{0}^{\prime }}^2\cdot \frac{1}{2}=A^2\cdot \frac{1}{2}+{(-A)}^2\cdot \frac{1}{2}=A^2\\ & {\sigma }_{a_k}^2=P_{a_k}-m_{a_k}^2=A^2\\ & T_s=T_b\\ & p(t)=\prod \left(\frac{t-{}^{T_s}╱{}_4}{{}^{T_s}╱{}_2}\right)-\prod \left(\frac{t-{}^{3\cdot T_s}╱{}_4}{{}^{T_s}╱{}_2}\right)\to \\ & P(f)=\frac{T_s}{2}\mathrm{sinc}\left(\frac{T_s}{2}f\right)e^{-j2\pi f\frac{T_s}{4}}-\frac{T_s}{2}\mathrm{sinc}\left(\frac{T_s}{2}f\right)e^{-j2\pi f\frac{3T_s}{4}}\Rightarrow \{\sin x=\frac{e^{jx}-e^{-jx}}{2j}\to 2j\sin x=e^{jx}-e^{-jx}\}\\ & =\frac{T_s}{2}\mathrm{sinc}\left(\frac{T_s}{2}f\right)e^{-j2\pi f\frac{T_s}{2}}(e^{+j2\pi f\frac{T_s}{4}}-e^{-j2\pi f\frac{T_s}{4}})=\frac{T_s}{2}\mathrm{sinc}\left(\frac{T_s}{2}f\right)e^{-j2\pi f\frac{T_s}{2}}2j\sin \left(2\pi f\frac{T_s}{4}\right)\\ & {|P(f)|}^2={\left(\frac{T_s}{2}\right)}^2{\mathrm{sinc}}^2\left(\left(\frac{T_s}{2}\right)f\right)4{\sin }^2\left(2\pi f\frac{T_s}{4}\right)=T_s^2{\mathrm{sinc}}^2\left(\frac{T_{s}f}{2}\right){\sin }^2\left(\pi f\frac{T_s}{2}\right)\\ & R_s=\frac{1}{T_s}\\ & {\overline{G}}_x(f)=\underset{A^2}{\underbrace{{\sigma }_{a_k}^2}}\cdot R_s{|P(f)|}^2+\underset{0}{\underbrace{m_{a_k}^2}}\cdot R_s^2\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{|P(k{R}_s)|}^2\delta (f-k{R}_s)\\ & {\overline{G}}_x(f)=A^2{R}_s{T}_s^2{\mathrm{sinc}}^2\left(\frac{T_{s}f}{2}\right){\sin }^2\left(\pi f\frac{T_s}{2}\right)=A^2{T}_s{\mathrm{sinc}}^2\left(\frac{T_{s}f}{2}\right){\sin }^2\left(\pi f\frac{T_s}{2}\right)\\ \end{array}$

Para la probabilidad de error (BER):

Manchester

Plantilla:Navegación