PSK

PSK (Phase Shift Keying)

$\begin{align} & s_{PSK}(t)=A_{c}\sum\limits_{k=-\infty }^{\infty }{p\left( t-kT_{s} \right)\cos \left( \omega _{c}t+\varphi _{k} \right)=} \\ & A_{c}\sum\limits_{k=-\infty }^{\infty }{p\left( t-kT_{s} \right)\cos \left( \omega _{c}t \right)\underbrace{\cos \left( \varphi _{k} \right)}_{I_{k}}-A_{c}\sum\limits_{k=-\infty }^{\infty }{p\left( t-kT_{s} \right)\sin \left( \omega _{c}t \right)\underbrace{\sin \left( \varphi _{k} \right)}_{Q_{k}}}} \\ & \cos ^{2}x+\sin ^{2}x=1\to \\ & I_{k}^{2}+Q_{k}^{2}=1 \\ \end{align}$

Dependiendo del numero de niveles tenemos diferentes tipos de PSK

BPSK (Binary Phase Shift Keying)

Tambien llamada PRK (Phase Reversal Keying).

Constellation diagram for BPSK.

$\begin{align} & s_{BPSK}(t)=A_{c}\sum\limits_{k=-\infty }^{\infty }{p\left( t-kT_{s} \right)\cos \left( \omega _{c}t+\varphi _{k} \right)=} \\ & \varphi _{k}=\left\{ 0,\pi \right\}\to s_{BPSK}(t)=A_{c}\sum\limits_{k=-\infty }^{\infty }{p\left( t-kT_{s} \right)\cos \left( \omega _{c}t \right)\underbrace{\cos \left( \varphi _{k} \right)}_{I_{k}}} \\ \end{align}$

$\varphi _{k}$	$I_{k}$
$0$	$+1$
$\pi$	$-1$

Por lo que vemos, para este caso particular de PSK, la señal puede ser modelada como una codificacion polar modulada por un coseno

$\begin{align} & x_{BPSK}(t)=A_{c}\sum\limits_{k=-\infty }^{\infty }{I_{k}\cdot p\left( t-kT_{s} \right)\cos \left( \omega _{c}t \right)} \\ & x_{I}(t)=A_{c}\sum\limits_{k=-\infty }^{\infty }{I_{k}\cdot p\left( t-kT_{s} \right)},x_{Q}(t)=0 \\ & \bar{G}_{I}(f)=G_{polar}(f)=A^{2}T_{s}\operatorname{sinc}^{2}\left( T_{s}f \right) \\ & G_{x}(f)=\frac{G_{I}(f-f_{c})+G_{I}(f+f_{c})}{4}+\frac{G_{Q}(f-f_{c})+G_{Q}(f+f_{c})}{4}\to \\ & G_{x}(f)=\frac{G_{I}(f\pm f_{c})}{4}\to \\ & G_{x_{BPSK}}(f)=\frac{A^{2}T_{s}\operatorname{sinc}^{2}\left( T_{s}\left( f\pm f_{c} \right) \right)}{4} \\ \end{align}$

Para la probabilidad de error (BER):

BER de BPSK

QPSK (Quadrature Phase Shift Keying)

$\begin{align} & s_{PSK}(t)=A_{c}\sum\limits_{k=-\infty }^{\infty }{p\left( t-kT_{s} \right)\cos \left( \omega _{c}t+\varphi _{k} \right)=} \\ & A_{c}\sum\limits_{k=-\infty }^{\infty }{p\left( t-kT_{s} \right)\cos \left( \omega _{c}t \right)\underbrace{\cos \left( \varphi _{k} \right)}_{I_{k}}-A_{c}\sum\limits_{k=-\infty }^{\infty }{p\left( t-kT_{s} \right)\sin \left( \omega _{c}t \right)\underbrace{\sin \left( \varphi _{k} \right)}_{Q_{k}}}} \\ \end{align}$

Existe más de un tipo de QPSK, la que se utiliza en la práctica:

Constellation diagram for QPSK with Gray coding. Each adjacent symbol only differs by one bit.

$\varphi _{k}$	$I_{k}$	$Q_{k}$
$+{}^{\pi }\!\!\diagup\!\!{}_{4}\;$	$\frac{1}{\sqrt{2}}$	$\frac{1}{\sqrt{2}}$
$+{}^{3\pi }\!\!\diagup\!\!{}_{4}\;$	$-\frac{1}{\sqrt{2}}$	$\frac{1}{\sqrt{2}}$
$-{}^{3\pi }\!\!\diagup\!\!{}_{4}\;$	$-\frac{1}{\sqrt{2}}$	$-\frac{1}{\sqrt{2}}$
$-{}^{\pi }\!\!\diagup\!\!{}_{4}\;$	$\frac{1}{\sqrt{2}}$	$-\frac{1}{\sqrt{2}}$

$\begin{align} & s_{QPSK}(t)=A_{c}\sum\limits_{k=-\infty }^{\infty }{I_{k}\cdot p\left( t-kT_{s} \right)\cos \left( \omega _{c}t \right)-A_{c}\sum\limits_{k=-\infty }^{\infty }{Q_{k}\cdot p\left( t-kT_{s} \right)\sin \left( \omega _{c}t \right)}} \\ & s_{I}(t)=A_{c}\sum\limits_{k=-\infty }^{\infty }{I_{k}\cdot p\left( t-kT_{s} \right)} \\ & s_{Q}(t)=A_{c}\sum\limits_{k=-\infty }^{\infty }{Q_{k}\cdot p\left( t-kT_{s} \right)} \\ \end{align}$

Ahora, para sacar la densidad espectral de potencia:

$\begin{align} & s_{I}(t)=A_{c}\sum\limits_{k=-\infty }^{\infty }{I_{k}\cdot p\left( t-kT_{s} \right)}\to \\ & I_{k}=\left\{ \frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}} \right\} \\ & \bar{G}_{x}(f)=\sigma _{a_{k}}^{2}\cdot R_{s}\left| P(f) \right|^{2}+m_{a_{k}}^{2}\cdot R_{s}^{2}\sum\limits_{k=-\infty }^{\infty }{\left| P(kR_{s}) \right|^{2}\delta \left( f-kR_{s} \right)} \\ & \left| P(f) \right|^{2}=T_{s}^{2}\operatorname{sinc}^{2}\left( T_{s}f \right) \\ & m_{I_{k}}=\frac{1}{\sqrt{2}}\cdot \frac{1}{4}+\left( -\frac{1}{\sqrt{2}} \right)\cdot \frac{1}{4}+\left( -\frac{1}{\sqrt{2}} \right)\cdot \frac{1}{4}+\frac{1}{\sqrt{2}}\cdot \frac{1}{4}=0 \\ & P_{I_{k}}=\left( \frac{1}{\sqrt{2}} \right)^{2}\cdot \frac{1}{4}+\left( -\frac{1}{\sqrt{2}} \right)^{2}\cdot \frac{1}{4}+\left( -\frac{1}{\sqrt{2}} \right)^{2}\cdot \frac{1}{4}+\left( \frac{1}{\sqrt{2}} \right)^{2}\cdot \frac{1}{4}=\frac{1}{2} \\ & \sigma _{I_{k}}^{2}=P_{I_{k}}-m_{I_{k}}^{2}=\frac{1}{2} \\ & \bar{G}_{x}(f)=\sigma _{a_{k}}^{2}\cdot R_{s}\left| P(f) \right|^{2}+\underbrace{m_{a_{k}}^{2}}_{0}\cdot R_{s}^{2}\sum\limits_{k=-\infty }^{\infty }{\left| P(kR_{s}) \right|^{2}\delta \left( f-kR_{s} \right)}=\underbrace{\sigma _{a_{k}}^{2}}_{\left( {}^{1}\!\!\diagup\!\!{}_{2}\; \right)^{2}}\cdot R_{s}\cdot T_{s}^{2}\operatorname{sinc}^{2}\left( T_{s}f \right)= \\ & \bar{G}_{I}(f)=A_{c}^{2}\sigma _{a_{k}}^{2}T_{s}\operatorname{sinc}^{2}\left( T_{s}f \right)=\frac{A_{c}^{2}}{2}T_{s}\operatorname{sinc}^{2}\left( T_{s}f \right) \\ \end{align}$

$\begin{align} & s_{Q}(t)=A_{c}\sum\limits_{k=-\infty }^{\infty }{Q_{k}\cdot p\left( t-kT_{s} \right)}\to \\ & Q_{k}=\left\{ \frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}} \right\}\to m_{a_{k}}=0,\sigma _{a_{k}}^{2}=\frac{1}{2} \\ & m_{Q_{k}}=\frac{1}{\sqrt{2}}\cdot \frac{1}{4}+\frac{1}{\sqrt{2}}\cdot \frac{1}{4}+\left( -\frac{1}{\sqrt{2}} \right)\cdot \frac{1}{4}+\left( -\frac{1}{\sqrt{2}} \right)\cdot \frac{1}{4}=0 \\ & P_{Q_{k}}=\left( \frac{1}{\sqrt{2}} \right)^{2}\cdot \frac{1}{4}+\left( \frac{1}{\sqrt{2}} \right)^{2}\cdot \frac{1}{4}+\left( -\frac{1}{\sqrt{2}} \right)^{2}\cdot \frac{1}{4}+\left( -\frac{1}{\sqrt{2}} \right)^{2}\cdot \frac{1}{4}=\frac{1}{2} \\ & \sigma _{Q_{k}}^{2}=P_{Q_{k}}-m_{Q_{k}}^{2}=\frac{1}{2} \\ & \bar{G}_{Q}(f)=A_{c}^{2}\sigma _{a_{k}}^{2}T_{s}\operatorname{sinc}^{2}\left( T_{s}f \right)=\frac{A_{c}^{2}}{2}T_{s}\operatorname{sinc}^{2}\left( T_{s}f \right) \\ \end{align}$

$\begin{align} & m_{I_{k}}=m_{Q_{k}}=0 \\ & \sigma _{I_{k}}^{2}=\sigma _{Q_{k}}^{2}=\frac{1}{2} \\ & \bar{G}_{I}(f)=\bar{G}_{Q}(f)=A_{c}^{2}\sigma _{a_{k}}^{2}T_{s}\operatorname{sinc}^{2}\left( T_{s}f \right)=\frac{A_{c}^{2}}{2}T_{s}\operatorname{sinc}^{2}\left( T_{s}f \right) \\ & G_{x}(f)=\frac{G_{I}(f-f_{c})+G_{I}(f+f_{c})}{4}+\frac{G_{Q}(f-f_{c})+G_{Q}(f+f_{c})}{4}\to \\ & G_{QPSK}(f)=2\frac{G_{I/Q}(f-f_{c})+G_{I/Q}(f+f_{c})}{4}=\frac{G_{I/Q}(f\pm f_{c})}{2}=\frac{A_{c}^{2}}{4}T_{s}\operatorname{sinc}^{2}\left( T_{s}\left( f\pm f_{c} \right) \right) \\ \end{align}$

Para la probabilidad de error (BER):

BER de QPSK

Existe otro tipo de QPSK:

4PSK

$\varphi _{k}$	$I_{k}$	$Q_{k}$
$0$	$+1$	$0$
${}^{\pi }\!\!\diagup\!\!{}_{2}\;$	$0$	$+1$
$\pi$	$-1$	$0$
$-{}^{\pi }\!\!\diagup\!\!{}_{2}\;$	$0$	$-1$

$\begin{align} & s_{Q}(t)=A_{c}\sum\limits_{k=-\infty }^{\infty }{I_{k}\cdot p\left( t-kT_{s} \right)}\to \\ & I_{k}=\left\{ +1,0,-1,0 \right\} \\ & m_{a_{k}}=+1\cdot \frac{1}{4}+0\cdot \frac{1}{4}+\left( -1 \right)\cdot \frac{1}{4}+0\cdot \frac{1}{4}=0 \\ & P_{a_{k}}=+1^{2}\cdot \frac{1}{4}+0\cdot \frac{1}{4}+\left( -1 \right)^{2}\cdot \frac{1}{4}+0\cdot \frac{1}{4}=\frac{1}{2} \\ & \sigma _{a_{k}}^{2}=P_{a_{k}}-m_{a_{k}}^{2}=\frac{1}{2} \\ & s_{Q}(t)=A_{c}\sum\limits_{k=-\infty }^{\infty }{Q_{k}\cdot p\left( t-kT_{s} \right)}\to \\ & Q_{k}=\left\{ 0,+1,0,-1 \right\} \\ & m_{a_{k}}=0\cdot \frac{1}{4}+1\cdot \frac{1}{4}+0\cdot \frac{1}{4}+\left( -1 \right)\cdot \frac{1}{4}=0 \\ & P_{a_{k}}=0\cdot \frac{1}{4}+1^{2}\frac{1}{4}+0\cdot \frac{1}{4}+\left( -1 \right)^{2}\cdot \frac{1}{4}=\frac{1}{2} \\ & \sigma _{a_{k}}^{2}=P_{a_{k}}-m_{a_{k}}^{2}=\frac{1}{2} \\ \end{align}$

$G_{4PSK}(f)=\frac{A_{c}^{2}}{4}T_{s}\operatorname{sinc}^{2}\left( T_{s}\left( f\pm f_{c} \right) \right)$

Para la probabilidad de error (BER):

BER de 4PSK

Como consecuencia final, vemos que la media y varianza de una señal PSK es constante:

$\begin{align} & m_{I_{k}}=m_{Q_{k}}=0 \\ & \sigma _{I_{k}}^{2}=\sigma _{Q_{k}}^{2}=\frac{1}{2} \\ \end{align}$

8-PSK,16-PSK…

Como se ha visto, la media y varianza de la señal no cambia, por lo que la densidad espectral de potencia será siempre igual independientemente del número de símbolos usados:

$G_{PSK}(f)=\frac{A_{c}^{2}}{4}T_{s}\operatorname{sinc}^{2}\left( T_{s}\left( f\pm f_{c} \right) \right)$

Constellation diagram for 8-PSK with Gray coding.

Para la probabilidad de error (BER):

BER de M-PSK

OQPSK

Signal doesn't cross zero, because only one bit of the symbol is changed at a time

π/4–QPSK

Dual constellation diagram for π/4-QPSK. This shows the two separate constellations with identical Gray coding but rotated by 45° with respect to each other.

Proyecto: Departamento de Teoría de la Señal y Comunicaciones

Anterior: ASK — PSK — Siguiente: FSK