Codificaciones digitales

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 ...

Codificación Unipolar NRZ

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

Para la probabilidad de error (BER):

Unipolar NRZ

Codificación Unipolar RZ

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

Para la probabilidad de error (BER):

Unipolar RZ

Codificación Polar

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

Para la probabilidad de error (BER):

Polar

Codificación Bipolar

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

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

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

Para la probabilidad de error (BER):

Bipolar

Codificación Manchester

${\begin{aligned}&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{aligned}}$

Para la probabilidad de error (BER):

Manchester

Proyecto: Departamento de Teoría de la Señal y Comunicaciones

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