# Cálculo de la probabilidad de error para las diferentes modulaciones/Prob. de error de M-ASK

${\begin{aligned}&s_{M-ASK}(t)=A_{c}\sum \limits _{k=-\infty }^{\infty }{a_{I_{k}}p\left(t-kT_{s}\right)\cos \left(\omega _{c}t\right)}\\&a_{I_{k}}=\left\{0,A,+2A,...\right\}\\\end{aligned}}$
${\begin{aligned}&y_{R}(t)=A_{R}\sum \limits _{k=-\infty }^{\infty }{a_{I_{k}}p\left(t-kT_{s}\right)+n_{I}(t)}\\&a_{I_{k}}=\left\{0,A,2A,...M-1\right\}\\&m_{I_{k}}=0,AT_{s},2AT_{s},...\\&E_{s}=E_{sc}\\&2\to {\frac {1}{M}}\cdot Q\left({\frac {\left|V_{T}-m\right|}{\sqrt {\sigma ^{2}}}}\right)\\&M-2\to {\frac {1}{M}}\cdot 2Q\left({\frac {\left|V_{T}-m\right|}{\sqrt {\sigma ^{2}}}}\right)\\\end{aligned}}$
${\begin{aligned}&\left|V_{T}-m\right|={\frac {AT_{s}}{2}}\\&{\sqrt {\sigma ^{2}}}={\sqrt {{\frac {\eta }{2}}T_{s}}}\\&P_{s}\simeq 2\times {\frac {1}{M}}\cdot Q\left({\frac {\left|V_{T}-m\right|}{\sqrt {\sigma ^{2}}}}\right)+\sum \limits _{i=1}^{M-2}{{\frac {1}{M}}\cdot 2Q\left({\frac {\left|V_{T}-m\right|}{\sqrt {\sigma ^{2}}}}\right)}=\\&2\times {\frac {1}{M}}\cdot Q\left({\frac {\left|V_{T}-m\right|}{\sqrt {\sigma ^{2}}}}\right)+\left(M-2\right){\frac {1}{M}}\cdot 2Q\left({\frac {\left|V_{T}-m\right|}{\sqrt {\sigma ^{2}}}}\right)=2\cdot {\frac {M-1}{M}}Q\left({\frac {\left|V_{T}-m\right|}{\sqrt {\sigma ^{2}}}}\right)=\\&2\cdot {\frac {M-1}{M}}Q\left({\frac {\frac {AT_{s}}{2}}{\sqrt {{\frac {\eta }{2}}T_{s}}}}\right)=2\cdot {\frac {M-1}{M}}Q\left({\sqrt {\frac {\frac {A^{2}T_{s}^{2}}{4}}{{\frac {\eta }{2}}T_{s}}}}\right)=2\cdot {\frac {M-1}{M}}Q\left({\sqrt {\frac {A^{2}T_{s}}{2\eta }}}\right)\\\end{aligned}}$
${\begin{aligned}&E_{s}=\sum \limits _{i=0}^{M-1}{pr(M)\cdot E_{i}}=\sum \limits _{i=0}^{M-1}{{\frac {1}{M}}\cdot E_{i}}={\frac {1}{M}}\sum \limits _{i=0}^{M-1}{E_{i}}\\&E_{i}=\left\{0,A^{2}T_{s},4A^{2}T_{s},9A^{2}T_{s},..\right\}\\&E_{s}={\frac {1}{M}}\sum \limits _{i=0}^{M-1}{E_{i}}={\frac {1}{M}}\sum \limits _{i=0}^{M-1}{i^{2}A^{2}T_{s}}={\frac {1}{M}}A^{2}T_{s}\sum \limits _{i=0}^{M-1}{i^{2}}\\&\sum \limits _{i=1}^{n}{i^{2}}={\frac {n\left(n+1\right)\left(2n+1\right)}{6}}\to \sum \limits _{i=0}^{M-1}{i^{2}}={\frac {\left(M-1\right)M\left(2\left(M-1\right)+1\right)}{6}}={\frac {\left(M-1\right)M\left(2M-1\right)}{6}}\\\end{aligned}}$
${\begin{aligned}&\sum \limits _{i=0}^{M-1}{i^{2}}={\frac {\left(M-1\right)M\left(2M-1\right)}{6}}\\&E_{s}={\frac {1}{M}}A^{2}T_{s}\sum \limits _{i=0}^{M-1}{i^{2}}={\frac {1}{M}}A^{2}T_{s}\cdot {\frac {\left(M-1\right)M\left(2M-1\right)}{6}}={\frac {A^{2}T_{s}}{6}}\left(M-1\right)\left(2M-1\right)\\&E_{s}={\frac {A^{2}T_{s}}{2\cdot 3}}\left(M-1\right)\left(2M-1\right)\leftrightarrow {\frac {A^{2}T_{s}}{2}}={\frac {3E_{s}}{\left(M-1\right)\left(2M-1\right)}}\\&P_{s}=2\cdot {\frac {M-1}{M}}Q\left({\sqrt {\frac {A^{2}T_{s}}{2\eta }}}\right)\Rightarrow \\&P_{s}=2\cdot {\frac {M-1}{M}}Q\left({\sqrt {{\frac {3}{\left(M-1\right)\left(2M-1\right)}}{\frac {E_{s}}{\eta }}}}\right)\\\end{aligned}}$
Comprobación:
${\begin{aligned}&4-ASK\to P_{s}\simeq {\frac {3}{2}}Q\left({\sqrt {\frac {E_{s}}{7\eta }}}\right)\\&P_{s}=\left.2\cdot {\frac {M-1}{M}}\right|_{M=4}\left.Q\left({\sqrt {{\frac {3}{\left(M-1\right)\left(2M-1\right)}}{\frac {E_{s}}{\eta }}}}\right)\right|_{M=4}={\frac {3}{2}}Q\left({\sqrt {{\frac {3}{3\cdot 7}}{\frac {E_{s}}{\eta }}}}\right)={\frac {3}{2}}Q\left({\sqrt {{\frac {1}{7}}{\frac {E_{s}}{\eta }}}}\right)\to c.q.d.\\\end{aligned}}$