Cálculo de la probabilidad de error para las diferentes modulaciones/Prob. de error de M-QAM rectangular

Para una señal 4-QAM, los posibles niveles eran 2 por portadora:

${\begin{aligned}&a_{I_{k}}=\left\{\pm {\frac {1}{\sqrt {2}}}\right\}\\&a_{Q_{k}}=\left\{\pm {\frac {1}{\sqrt {2}}}\right\}\\\end{aligned}}$

Para una señal 16-QAM, los posibles niveles eran 4 por portadora:

${\begin{aligned}&a_{I_{k}}=\left\{\pm {\frac {1}{\sqrt {2}}},\pm {\frac {3}{\sqrt {2}}}\right\}\\&a_{Q_{k}}=\left\{\pm {\frac {1}{\sqrt {2}}},\pm {\frac {3}{\sqrt {2}}}\right\}\\\end{aligned}}$

Generalizando:

${\begin{aligned}&a_{I_{k}}=a_{Q_{k}}=\left\{\pm {\frac {1}{\sqrt {2}}},\pm {\frac {3}{\sqrt {2}}},\pm {\frac {2n+1}{\sqrt {2}}},...\right\},n=0,1,2...+{\frac {\sqrt {M}}{2}}-1\\&A={\frac {1}{\sqrt {2}}}\\\end{aligned}}$

Por lo que al demodular, tendremos en total ${\sqrt {M}}$ símbolos. La probabilidad de error será:

Para los símbolos de las esquinas:

$p\left(simbolo\right)\cdot Q\left({\frac {AT_{s}}{\sqrt {\sigma ^{2}}}}\right)$

Para todos los demás:

$p\left(simbolo\right)\cdot 2Q\left({\frac {AT_{s}}{\sqrt {\sigma ^{2}}}}\right)$

El 2 es debido a que el error puede producirse tanto por demodular incorrectamente por la izquierda, como por demodular incorrectamente por la derecha.

Entonces, de todos los símbolos, 2 de ellos tendrán la primera probabilidad de error, mientras que todos los demás tendrán la última. Si tenemos M niveles, tenemos $\log _{2}\left(M\right)$ bits. Si la señal es rectangular, ${\sqrt {M}}$ símbolos.

La probabilidad de símbolo es sencilla:

${\begin{aligned}&4QAM\to M=4\to {\sqrt {M}}={\sqrt {4}}=2\to pr={\frac {1}{2}}\\&16QAM\to M=16\to {\sqrt {M}}={\sqrt {16}}=4\to pr={\frac {1}{4}}\\&M-QAM\to pr={\frac {1}{\sqrt {M}}}\\\end{aligned}}$

${\begin{aligned}&P_{sc}=2\times {\frac {1}{\sqrt {M}}}\cdot Q\left({\frac {AT_{s}}{\sqrt {{\frac {\eta }{2}}T_{s}}}}\right)+\sum \limits _{i=1}^{{\sqrt {M}}-2}{{\frac {1}{\sqrt {M}}}\cdot 2Q\left({\frac {AT_{s}}{\sqrt {{\frac {\eta }{2}}T_{s}}}}\right)}=\\&2\times {\frac {1}{\sqrt {M}}}\cdot Q\left({\frac {AT_{s}}{\sqrt {{\frac {\eta }{2}}T_{s}}}}\right)+\left({\sqrt {M}}-2\right){\frac {1}{\sqrt {M}}}\cdot 2Q\left({\frac {AT_{s}}{\sqrt {{\frac {\eta }{2}}T_{s}}}}\right)=\\&\left(2+\left({\sqrt {M}}-2\right)\cdot 2\right){\frac {1}{\sqrt {M}}}\cdot Q\left({\frac {AT_{s}}{\sqrt {\frac {\eta }{2}}}}\right)=\left(2+2{\sqrt {M}}-4\right){\frac {1}{\sqrt {M}}}\cdot Q\left({\frac {AT_{s}}{\sqrt {\frac {\eta }{2}}}}\right)=\\&\left(2{\sqrt {M}}-2\right){\frac {1}{\sqrt {M}}}\cdot Q\left({\sqrt {\frac {2A^{2}T_{s}}{\eta }}}\right)=2{\frac {\left({\sqrt {M}}-1\right)}{\sqrt {M}}}\cdot Q\left({\sqrt {\frac {2A^{2}T_{s}}{\eta }}}\right)=\\&2\left(1-{\frac {1}{\sqrt {M}}}\right)\cdot Q\left({\sqrt {\frac {2A^{2}T_{s}}{\eta }}}\right)\\\end{aligned}}$

Ahora, la energía por portadora será ($E_{sc}$ ):

${\begin{aligned}&a_{I_{k}}=a_{Q_{k}}=\left\{\pm A,\pm 3A,\pm \left(2n+1\right)A,...\right\},n=0,1,2...{\frac {\sqrt {M}}{2}}-1;A={\frac {1}{\sqrt {2}}}\\&4-QAM\to E_{sc}=A^{2}T_{s}\\&16-QAM\to E_{sc}={\frac {1}{2}}A^{2}T_{s}+{\frac {1}{2}}\left(\pm 3A\right)^{2}T_{s}={\frac {1}{2}}A^{2}T_{s}+{\frac {1}{2}}9A^{2}T_{s}=5A^{2}T_{s}\\&4-QAM\to {\frac {\sqrt {M=4}}{2}}-1=0\\&16-QAM\to {\frac {\sqrt {M=16}}{2}}-1=1\\&M-QAM\to E_{sc}={\frac {1}{\sqrt {M}}}\sum \limits _{i=0}^{+{\frac {\sqrt {M}}{2}}-1}{2\left(2i+1\right)^{2}\cdot A^{2}T_{s}=}{\frac {1}{\sqrt {M}}}A^{2}T_{s}\sum \limits _{i=0}^{+{\frac {\sqrt {M}}{2}}-1}{2\left(2i+1\right)^{2}}=\\&{\frac {1}{\sqrt {M}}}A^{2}T_{s}\sum \limits _{i=0}^{+{\frac {\sqrt {M}}{2}}-1}{2\left(2i+1\right)^{2}}\\\end{aligned}}$

${\begin{aligned}&{\frac {1}{\sqrt {M}}}A^{2}T_{s}2\sum \limits _{i=0}^{+{\frac {\sqrt {M}}{2}}-1}{4i^{2}+1+4i}\to \sum \limits _{i=0}^{+{\frac {\sqrt {M}}{2}}-1}{4i^{2}+1+4i}=\sum \limits _{i=0}^{+{\frac {\sqrt {M}}{2}}-1}{1}+4\sum \limits _{i=0}^{+{\frac {\sqrt {M}}{2}}-1}{i}+4\sum \limits _{i=0}^{+{\frac {\sqrt {M}}{2}}-1}{i^{2}}=\\&\sum \limits _{i=1}^{n}{a}=na\to \sum \limits _{i=0}^{+{\frac {\sqrt {M}}{2}}-1}{1}={\frac {\sqrt {M}}{2}}\\&\sum \limits _{i=1}^{n}{i}={\frac {n\left(n+1\right)}{2}}\to 4\sum \limits _{i=0}^{+{\frac {\sqrt {M}}{2}}-1}{i}=4\times {\frac {\left({\frac {\sqrt {M}}{2}}-1\right){\frac {\sqrt {M}}{2}}}{2}}=4\times {\frac {{\frac {M}{4}}-{\frac {\sqrt {M}}{2}}}{2}}={\frac {M-2{\sqrt {M}}}{2}}\\&\sum \limits _{i=1}^{n}{i^{2}}={\frac {n\left(n+1\right)\left(2n+1\right)}{6}}\to 4\sum \limits _{i=0}^{+{\frac {\sqrt {M}}{2}}-1}{i^{2}}=4\times {\frac {\left({\frac {\sqrt {M}}{2}}-1\right){\frac {\sqrt {M}}{2}}\left({\sqrt {M}}-1\right)}{6}}=\\&4\times {\frac {\left({\frac {M}{4}}-{\frac {\sqrt {M}}{2}}\right)\left({\sqrt {M}}-1\right)}{6}}=4\times {\frac {\left(-{\frac {M}{4}}+{\frac {\sqrt {M}}{2}}+{\frac {{\sqrt {M}}M}{4}}-{\frac {M}{2}}\right)}{6}}=4\times {\frac {\left(-{\frac {3M}{4}}+{\frac {{\sqrt {M}}\left(M+2\right)}{4}}\right)}{6}}=\\&{\frac {-3M+{\sqrt {M}}\left(M+2\right)}{6}}\\&{\frac {1}{\sqrt {M}}}A^{2}T_{s}2\sum \limits _{i=0}^{+{\frac {\sqrt {M}}{2}}-1}{4i^{2}+1+4i}={\frac {1}{\sqrt {M}}}A^{2}T_{s}2\left({\frac {\sqrt {M}}{2}}+{\frac {M-2{\sqrt {M}}}{2}}+{\frac {-3M+{\sqrt {M}}\left(M+2\right)}{6}}\right)=\\&{\frac {1}{\sqrt {M}}}A^{2}T_{s}2\left({\frac {+{\sqrt {M}}\left(M+2+3-6\right)}{6}}\right)={\frac {1}{\sqrt {M}}}A^{2}T_{s}2\left({\frac {+{\sqrt {M}}\left(M-1\right)}{6}}\right)=A^{2}T_{s}2\left({\frac {M-1}{6}}\right)\\&E_{sc}=A^{2}T_{s}2\left({\frac {M-1}{6}}\right)=A^{2}T_{s}\left({\frac {M-1}{3}}\right)\\&E_{s}=\log _{2}\left(M\right)\cdot E_{b}\\&E_{s}=2\cdot E_{sc}\to \\&E_{sc}=A^{2}T_{s}\left({\frac {M-1}{3}}\right)\to E_{s}=2A^{2}T_{s}\left({\frac {M-1}{3}}\right)\\\end{aligned}}$

${\begin{aligned}&P_{sc}=2\left(1-{\frac {1}{\sqrt {M}}}\right)\cdot Q\left({\sqrt {\frac {2A^{2}T_{s}}{\eta }}}\right)\\&E_{s}=2A^{2}T_{s}\left({\frac {M-1}{3}}\right)\leftrightarrow 2A^{2}T_{s}={\frac {E_{s}3}{M-1}}\\&P_{sc}=2\left(1-{\frac {1}{\sqrt {M}}}\right)\cdot Q\left({\sqrt {\frac {2A^{2}T_{s}}{\eta }}}\right)=2\left(1-{\frac {1}{\sqrt {M}}}\right)\cdot Q\left({\sqrt {{\frac {3}{M-1}}{\frac {E_{s}}{\eta }}}}\right)\\&P_{sc}=2\left(1-{\frac {1}{\sqrt {M}}}\right)\cdot Q\left({\sqrt {{\frac {3}{M-1}}{\frac {E_{s}}{\eta }}}}\right)=2\left(1-{\frac {1}{\sqrt {M}}}\right)\cdot Q\left({\sqrt {{\frac {3}{M-1}}{\frac {\log _{2}\left(M\right)E_{b}}{\eta }}}}\right)\\\end{aligned}}$