# QAM
## QAM (Quaternary Amplitude Modulation)

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

## 16-QAM

${\begin{aligned}&a_{I_{k}}=\left\{-{\frac {3}{\sqrt {2}}},-{\frac {1}{\sqrt {2}}},{\frac {1}{\sqrt {2}}},{\frac {3}{\sqrt {2}}}\right\}\\&a_{Q_{k}}=\left\{-{\frac {3}{\sqrt {2}}},-{\frac {1}{\sqrt {2}}},{\frac {1}{\sqrt {2}}},{\frac {3}{\sqrt {2}}}\right\}\\\end{aligned}}$
${\begin{aligned}&s_{I}(t)=A_{c}\sum \limits _{k=-\infty }^{\infty }{a_{I_{k}}\cdot p\left(t-kT_{s}\right)}\\&s_{Q}(t)=A_{c}\sum \limits _{k=-\infty }^{\infty }{a_{Q_{k}}\cdot p\left(t-kT_{s}\right)}\to \\&{\bar {G}}_{I/Q}(f)=\sigma _{a_{k}}^{2}\cdot R_{s}\left|P(f)\right|^{2}+m_{a_{k}}^{2}\cdot R_{s}^{2}\sum \limits _{k=-\infty }^{\infty }{\left|P(kR_{s})\right|^{2}\delta \left(f-kR_{s}\right)}\\&\left|P(f)\right|^{2}=T_{s}^{2}\operatorname {sinc} ^{2}\left(T_{s}f\right)\\\end{aligned}}$
${\begin{aligned}&m_{a_{I_{k}}}=m_{a_{Q_{k}}}=0\\&P_{x}=\left(-{\frac {3}{\sqrt {2}}}\right)^{2}\cdot {\frac {4}{16}}+\left(-{\frac {1}{\sqrt {2}}}\right)^{2}\cdot {\frac {4}{16}}+\left({\frac {1}{\sqrt {2}}}\right)^{2}\cdot {\frac {4}{16}}+\left({\frac {3}{\sqrt {2}}}\right)^{2}\cdot {\frac {4}{16}}=\\&{\frac {1}{4}}\left({\frac {9}{2}}+{\frac {1}{2}}+{\frac {1}{2}}+{\frac {9}{2}}\right)={\frac {1}{4}}\cdot 10=2.5\\&\sigma _{a_{I_{k}}}^{2}=\sigma _{a_{Q_{k}}}^{2}=2.5\\&{\bar {G}}_{x}(f)=\sigma _{a_{k}}^{2}\cdot R_{s}\left|P(f)\right|^{2}+\underbrace {m_{a_{k}}^{2}} _{0}\cdot R_{s}^{2}\sum \limits _{k=-\infty }^{\infty }{\left|P(kR_{s})\right|^{2}\delta \left(f-kR_{s}\right)}=\sigma _{a_{k}}^{2}\cdot R_{s}\cdot T_{s}^{2}\operatorname {sinc} ^{2}\left(T_{s}f\right)=\\&{\bar {G}}_{I}(f)=G_{polar}(f)=A_{c}^{2}\sigma _{a_{k}}^{2}T_{s}\operatorname {sinc} ^{2}\left(T_{s}f\right)=A_{c}^{2}2.5\cdot T_{s}\operatorname {sinc} ^{2}\left(T_{s}f\right)\\\end{aligned}}$
${\begin{aligned}&m_{I_{k}}=m_{Q_{k}}=0\\&\sigma _{I_{k}}^{2}=\sigma _{Q_{k}}^{2}\\&{\bar {G}}_{I}(f)={\bar {G}}_{Q}(f)=A_{c}^{2}\sigma _{a_{k}}^{2}T_{s}\operatorname {sinc} ^{2}\left(T_{s}f\right)=A_{c}^{2}2.5\cdot T_{s}\operatorname {sinc} ^{2}\left(T_{s}f\right)\\&G_{x}(f)={\frac {G_{I}(f-f_{c})+G_{I}(f+f_{c})}{4}}+{\frac {G_{Q}(f-f_{c})+G_{Q}(f+f_{c})}{4}}\to \\&G_{16-QAM}(f)=2{\frac {G_{I/Q}(f-f_{c})+G_{I/Q}(f+f_{c})}{4}}={\frac {G_{I/Q}(f\pm f_{c})}{2}}={\frac {A_{c}^{2}}{2}}2.5\cdot T_{s}\operatorname {sinc} ^{2}\left(T_{s}f\right)\\\end{aligned}}$
Para la probabilidad de error BER:
(en proceso) BER de 16-QAM
* Proyecto: Departamento de Teoría de la Señal y Comunicaciones
* Anterior: FSK — QAM — Siguiente: MSK