# Cálculo de la probabilidad de error para las diferentes modulaciones/Prob. de error de 16-QAM

${\begin{aligned}&s_{QAM}(t)=A_{c}\sum \limits _{k=-\infty }^{\infty }{a_{I_{k}}p\left(t-kT_{s}\right)\cos \left(\omega _{c}t\right)-A_{c}\sum \limits _{k=-\infty }^{\infty }{a_{Q_{k}}p\left(t-kT_{s}\right)\sin \left(\omega _{c}t\right)}}\\&a_{I_{k}}=\left\{-{\frac {3}{\sqrt {2}}},-{\frac {1}{\sqrt {2}}},{\frac {1}{\sqrt {2}}},{\frac {3}{\sqrt {2}}}\right\}\\&a_{Q_{k}}=\left\{-{\frac {3}{\sqrt {2}}},-{\frac {1}{\sqrt {2}}},{\frac {1}{\sqrt {2}}},{\frac {3}{\sqrt {2}}}\right\}\\\end{aligned}}$

### Demodulación de la señal QAM In-Phase (Proyección horizontal de la constelación 16-QAM)

$a_{I_{k}}=a_{k}\cdot \cos(\varphi _{k})$
Una demodulación sin errores nos proporcionará 2 bits; podemos verlo matemáticamente: $\log _{2}\left(M=4\right)=2bit$
${\begin{aligned}&y_{R}(t)=A_{R}\sum \limits _{k=-\infty }^{\infty }{a_{I_{k}}p\left(t-kT_{s}\right)+n_{I}(t)}\\&a_{I_{k}}=\left\{-{\frac {3}{\sqrt {2}}},-{\frac {1}{\sqrt {2}}},{\frac {1}{\sqrt {2}}},{\frac {3}{\sqrt {2}}}\right\}\\&A={\frac {1}{\sqrt {2}}}\\&a_{I_{k}}=\left\{{\begin{aligned}&a_{'00'}=-3A\\&a_{'01'}=-A\\&a_{'10'}=+A\\&a_{'11'}=+3A\\\end{aligned}}\right\}\to \left\{{\begin{aligned}&m_{'00'}=\int _{0}^{T_{s}}{s_{'00'}(t)kp^{*}\left(t\right)\partial t}=\int _{0}^{T_{s}}{-3Ap(t)p^{*}\left(t\right)\partial t}=-3AT_{s}\\&m_{'01'}=\int _{0}^{T_{s}}{s_{'01'}(t)kp^{*}\left(t\right)\partial t}=\int _{0}^{T_{s}}{-Ap(t)p^{*}\left(t\right)\partial t}=-AT_{s}\\&m_{'10'}=\int _{0}^{T_{s}}{s_{'10'}(t)kp^{*}\left(t\right)\partial t}=\int _{0}^{T_{s}}{+Ap(t)p^{*}\left(t\right)\partial t}=+AT_{s}\\&m_{'11'}=\int _{0}^{T_{s}}{s_{'11'}(t)kp^{*}\left(t\right)\partial t}=\int _{0}^{T_{s}}{+3Ap(t)p^{*}\left(t\right)\partial t}=+3AT_{s}\\\end{aligned}}\right.\\&V_{T}=\left\{0,\pm 2AT_{s}\right\}\\&{\sqrt {\sigma ^{2}}}={\sqrt {{\frac {\eta }{2}}T_{s}}}\\&P_{e}\simeq {\frac {1}{4}}\cdot Q\left({\frac {\left|V_{T}-m\right|}{\sqrt {\sigma ^{2}}}}\right)+{\frac {1}{4}}\cdot 2Q\left({\frac {\left|V_{T}-m\right|}{\sqrt {\sigma ^{2}}}}\right)+{\frac {1}{4}}\cdot 2Q\left({\frac {\left|V_{T}-m\right|}{\sqrt {\sigma ^{2}}}}\right)+{\frac {1}{4}}\cdot Q\left({\frac {\left|V_{T}-m\right|}{\sqrt {\sigma ^{2}}}}\right)=\\&P_{e}\simeq {\frac {1}{4}}\cdot Q\left({\frac {\left|-2AT_{s}-\left(-3AT_{s}\right)\right|}{\sqrt {\sigma ^{2}}}}\right)+{\frac {1}{4}}\cdot 2Q\left({\frac {\left|-2AT_{s}-\left(-AT_{s}\right)\right|}{\sqrt {\sigma ^{2}}}}\right)+{\frac {1}{4}}\cdot 2Q\left({\frac {\left|0-AT_{s}\right|}{\sqrt {\sigma ^{2}}}}\right)+{\frac {1}{4}}\cdot Q\left({\frac {\left|2AT_{s}-3AT_{s}\right|}{\sqrt {\sigma ^{2}}}}\right)=\\&P_{e}\simeq {\frac {3}{2}}Q\left({\frac {AT_{s}}{\sqrt {\sigma ^{2}}}}\right)={\frac {3}{2}}Q\left({\frac {AT_{s}}{\sqrt {{\frac {\eta }{2}}T_{s}}}}\right)={\frac {3}{2}}Q\left({\sqrt {\frac {A^{2}T_{s}^{2}}{{\frac {\eta }{2}}T_{s}}}}\right)={\frac {3}{2}}Q\left({\sqrt {\frac {2A^{2}T_{s}}{\eta }}}\right)\\\end{aligned}}$
${\begin{aligned}&E_{sc}={\frac {1}{4}}\cdot E_{'00'}+{\frac {1}{4}}\cdot E_{'01'}+{\frac {1}{4}}\cdot E_{'10'}+{\frac {1}{4}}\cdot E_{'11'}\\&E_{'00'}=E_{'11'}=\int _{0}^{T_{s}}{s_{'00'}^{2}(t)\partial t}=\left(\pm 3A\right)^{2}T_{s}=9A^{2}T_{s}\\&E_{'01'}=E_{'10'}=\int _{0}^{T_{s}}{s_{'01'}^{2}(t)\partial t}=\left(\pm A\right)^{2}T_{s}=A^{2}T_{s}\\&E_{sc}=5A^{2}T_{s}\\&P_{e}\simeq {\frac {3}{2}}Q\left({\sqrt {\frac {2A^{2}T_{s}}{\eta }}}\right)={\frac {3}{2}}Q\left({\sqrt {{\frac {2}{5}}{\frac {E_{sc}}{\eta }}}}\right)\\\end{aligned}}$
Sabiendo que:
${\begin{aligned}&E_{s}=\log _{2}\left(M\right)E_{b}=\log _{2}\left(16\right)E_{b}=4E_{b}\\&E_{s}=2E_{sc}\to \\&P_{sc}={\frac {3}{2}}Q\left({\sqrt {{\frac {1}{5}}{\frac {E_{s}}{\eta }}}}\right)={\frac {3}{2}}Q\left({\sqrt {{\frac {4}{5}}{\frac {E_{b}}{\eta }}}}\right)\\\end{aligned}}$

### Demodulación de la señal QAM Quadrature (Proyección vertical de la constelación 16-QAM)

$a_{Q_{k}}=a_{k}\cdot \sin(\varphi _{k})$
$a_{Q_{k}}=\left\{-{\frac {3}{\sqrt {2}}},-{\frac {1}{\sqrt {2}}},{\frac {1}{\sqrt {2}}},{\frac {3}{\sqrt {2}}}\right\}$
Como se ve, tenemos los mismos símbolos, por lo que la probabilidad de error será análoga.
$P_{sc}={\frac {3}{2}}Q\left({\sqrt {{\frac {1}{5}}{\frac {E_{s}}{\eta }}}}\right)={\frac {3}{2}}Q\left({\sqrt {{\frac {4}{5}}{\frac {E_{b}}{\eta }}}}\right)$
Finalmente, la probabilidad de error total es:
Teniendo en cuenta que la componente en fase y la componente en cuadratura son independientes, y que el error podría darse por, un error en la componente en fase, pero no en la componente en cuadratura; un error en la componente en cuadratura, pero no en la componente en fase; o un error en ambas componentes:
${\begin{aligned}&P_{e}=P_{errorI}P_{noerrorQ}+P_{noerrorI}P_{errorQ}+P_{errorI}P_{errorQ}\\&P_{e}=2{\frac {3}{2}}Q\left({\sqrt {{\frac {4}{5}}{\frac {E_{b}}{\eta }}}}\right)\left(1-{\frac {3}{2}}Q\left({\sqrt {{\frac {4}{5}}{\frac {E_{b}}{\eta }}}}\right)\right)+\left({\frac {3}{2}}Q\left({\sqrt {{\frac {4}{5}}{\frac {E_{b}}{\eta }}}}\right)\right)^{2}\\&P_{e}=3Q\left({\sqrt {{\frac {4}{5}}{\frac {E_{b}}{\eta }}}}\right)-{\frac {9}{4}}\left(Q\left({\sqrt {{\frac {4}{5}}{\frac {E_{b}}{\eta }}}}\right)\right)^{2}\\&P_{e}\simeq 3Q\left({\sqrt {{\frac {4}{5}}{\frac {E_{b}}{\eta }}}}\right)\end{aligned}}$