Ver el código de «QAM»

== QAM (Quaternary Amplitude Modulation) ==

QAM = ASK + PSK

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

== 16-QAM ==

[[Image:16QAM Gray Coded.svg|200px|left|thumb|Constellation diagram for rectangular 16-QAM with Gray coding. Each adjacent symbol only differs by one bit.]]
<br style="clear:both;">

<math>\begin{align}
  & a_{I_{k}}=\left\{ -\frac{3}{\sqrt{2}},-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},\frac{3}{\sqrt{2}} \right\} \\ 
 & a_{Q_{k}}=\left\{ -\frac{3}{\sqrt{2}},-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},\frac{3}{\sqrt{2}} \right\} \\ 
\end{align}</math>

<math>\begin{align}
  & s_{I}(t)=A_{c}\sum\limits_{k=-\infty }^{\infty }{a_{I_{k}}\cdot p\left( t-kT_{s} \right)} \\ 
 & s_{Q}(t)=A_{c}\sum\limits_{k=-\infty }^{\infty }{a_{Q_{k}}\cdot p\left( t-kT_{s} \right)}\to  \\ 
 & \bar{G}_{I/Q}(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) \\ 
\end{align}</math>


<math>\begin{align}
  & m_{a_{I_{k}}}=m_{a_{Q_{k}}}=0 \\ 
 & P_{x}=\left( -\frac{3}{\sqrt{2}} \right)^{2}\cdot \frac{4}{16}+\left( -\frac{1}{\sqrt{2}} \right)^{2}\cdot \frac{4}{16}+\left( \frac{1}{\sqrt{2}} \right)^{2}\cdot \frac{4}{16}+\left( \frac{3}{\sqrt{2}} \right)^{2}\cdot \frac{4}{16}= \\ 
 & \frac{1}{4}\left( \frac{9}{2}+\frac{1}{2}+\frac{1}{2}+\frac{9}{2} \right)=\frac{1}{4}\cdot 10=2.5 \\ 
 & \sigma _{a_{I_{k}}}^{2}=\sigma _{a_{Q_{k}}}^{2}=2.5 \\ 
 & \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)}=\sigma _{a_{k}}^{2}\cdot R_{s}\cdot T_{s}^{2}\operatorname{sinc}^{2}\left( T_{s}f \right)= \\ 
 & \bar{G}_{I}(f)=G_{polar}(f)=A_{c}^{2}\sigma _{a_{k}}^{2}T_{s}\operatorname{sinc}^{2}\left( T_{s}f \right)=A_{c}^{2}2.5\cdot T_{s}\operatorname{sinc}^{2}\left( T_{s}f \right) \\ 
\end{align}</math>


<math>\begin{align}
  & m_{I_{k}}=m_{Q_{k}}=0 \\ 
 & \sigma _{I_{k}}^{2}=\sigma _{Q_{k}}^{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)=A_{c}^{2}2.5\cdot 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_{16-QAM}(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}}{2}2.5\cdot T_{s}\operatorname{sinc}^{2}\left( T_{s}f \right) \\ 
\end{align}</math>


Para la probabilidad de error BER:

(en proceso)
[[Calculo de la probabilidad de error para las diferentes modulaciones#16-QAM|BER de 16-QAM]]

{{navegación|[[Departamento de Teoría de la Señal y Comunicaciones]]|[[FSK]]|[[MSK]]}}

[[Categoría:Electrónica]]