# Cálculo de la probabilidad de error para las diferentes modulaciones/Prob. de error de 4ASK

${\begin{aligned}&s_{4-ASK}(t)=A_{c}\sum \limits _{k=-\infty }^{\infty }{a_{I_{k}}p\left(t-kT_{s}\right)\cos \left(\omega _{c}t\right)}\\&a_{I_{k}}=\left\{0,A,+2A,+3A\right\}\\\end{aligned}}$
La constelación de una señal ASK de 4 niveles es:
${\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\{{\begin{aligned}&a_{'00'}=0\\&a_{'01'}=+A\\&a_{'10'}=+2A\\&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}}{0p^{*}\left(t\right)\partial t}=0\\&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}}{+2Ap(t)p^{*}\left(t\right)\partial t}=+2AT_{s}\\&m_{'11'}=\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}\\\end{aligned}}\right.\\&V_{T}=\left\{{\frac {AT_{s}}{2}},{\frac {3AT_{s}}{2}},{\frac {5AT_{s}}{2}}\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|{\frac {AT_{s}}{2}}-0\right|}{\sqrt {\sigma ^{2}}}}\right)+{\frac {1}{4}}\cdot 2Q\left({\frac {\left|{\frac {AT_{s}}{2}}-\left(AT_{s}\right)\right|}{\sqrt {\sigma ^{2}}}}\right)+{\frac {1}{4}}\cdot 2Q\left({\frac {\left|{\frac {3AT_{s}}{2}}-2AT_{s}\right|}{\sqrt {\sigma ^{2}}}}\right)+{\frac {1}{4}}\cdot Q\left({\frac {\left|{\frac {5AT_{s}}{2}}-3AT_{s}\right|}{\sqrt {\sigma ^{2}}}}\right)=\\&P_{e}\simeq {\frac {3}{2}}Q\left({\frac {\frac {AT_{s}}{2}}{\sqrt {\sigma ^{2}}}}\right)={\frac {3}{2}}Q\left({\frac {\frac {AT_{s}}{2}}{\sqrt {{\frac {\eta }{2}}T_{s}}}}\right)={\frac {3}{2}}Q\left({\sqrt {\frac {\frac {A^{2}T_{s}^{2}}{4}}{{\frac {\eta }{2}}T_{s}}}}\right)={\frac {3}{2}}Q\left({\sqrt {\frac {A^{2}T_{s}}{2\eta }}}\right)\\\end{aligned}}$
${\begin{aligned}&E_{sc}={\frac {1}{4}}\cdot E_{'00'}+{\frac {1}{4}}\cdot E_{'01'}+{\frac {1}{4}}\cdot E_{'10'}+{\frac {1}{4}}\cdot E_{'11'}\\&E_{'00'}=0\\&E_{'01'}=\int _{0}^{T_{s}}{s_{'01'}^{2}(t)\partial t}=\left(+A\right)^{2}T_{s}=A^{2}T_{s}\\&E_{'10'}=\int _{0}^{T_{s}}{s_{'10'}^{2}(t)\partial t}=\left(+2A\right)^{2}T_{s}=4A^{2}T_{s}\\&E_{'11'}=\int _{0}^{T_{s}}{s_{'11'}^{2}(t)\partial t}=\left(+3A\right)^{2}T_{s}=9A^{2}T_{s}\\&E_{sc}={\frac {14A^{2}T_{s}}{4}}={\frac {7A^{2}T_{s}}{2}}\\&E_{sc}={\frac {7A^{2}T_{s}}{2}}\leftrightarrow {\frac {A^{2}T_{s}}{2}}={\frac {E_{sc}}{7}}\\&P_{sc}\simeq {\frac {3}{2}}Q\left({\sqrt {\frac {A^{2}T_{s}}{2\eta }}}\right)={\frac {3}{2}}Q\left({\sqrt {\frac {E_{sc}}{7\eta }}}\right)\\&E_{s}=E_{sc_{1}}+\underbrace {E_{sc_{2}}} _{0}=E_{sc_{1}}\\&E_{s}=\log _{2}\left(M=4\right)E_{b}=2E_{b}\\&P_{s}\simeq {\frac {3}{2}}Q\left({\sqrt {\frac {E_{s}}{7\eta }}}\right)={\frac {3}{2}}Q\left({\sqrt {\frac {2E_{b}}{7\eta }}}\right)\\\end{aligned}}$