# Cálculo de la probabilidad de error para las diferentes codificaciones

${\begin{aligned}&\int \limits _{V_{T}}^{\infty }{{\frac {1}{\sqrt {2\pi \sigma ^{2}}}}\cdot e^{\frac {-\left(t-m\right)^{2}}{2\sigma ^{2}}}\partial t}=Q\left({\frac {\left|V_{T}-m\right|}{\sqrt {\sigma ^{2}}}}\right)\\&P_{e}=pr\left('0'sent\right)\cdot p\left(^{error}\!\!\diagup \!\!_{'0'sent}\;\right)+pr\left('1'sent\right)\cdot p\left(^{error}\!\!\diagup \!\!_{'1'sent}\;\right)\\&m_{y_{PR}(t)}=\int _{0}^{T_{s}}{s_{T}(t)kp^{*}\left(t\right)\partial t}\\&\sigma _{y_{PR}(t)}^{2}={\frac {\eta }{2}}\cdot E_{p(t)}={\frac {\eta }{2}}\int _{0}^{T_{s}}{p^{2}(t)\partial t}\\&s_{T}(t)=\sum \limits _{k=0}^{\infty }{a_{k}p\left(t-kT_{s}\right)}\\\end{aligned}}$

## Probabilidad de error para codificación unipolar NRZ

${\begin{aligned}&T_{s}=T_{b}\to p(t)=\prod {\left({\frac {t-^{T_{s}}\!\!\diagup \!\!_{2}\;}{P}}pp{T_{s}}\right)}\\&a_{k}=\left\{{\begin{aligned}P&a_{'1'}=A\\Pml&a_{'0'}=0\\\end{aligned}}\right.\\&m_{'1'}=\int _{0}^{T_{s}}{s_{'1'}(t)kp^{*}\left(t\right)\partial t}=\int _{0}^{T_{s}}{Ap(t)p^{*}\left(t\right)\partial t}=AT_{s}\\&m_{'0'}=\int _{0}^{T_{s}}{s_{'0'}(t)kp^{*}\left(t\right)\partial t}=0\\&V_{T}={\frac {m_{'1'}+m_{'0'}}{2}}={\frac {AT_{s}}{2}}\\&\sigma _{y_{PR}(t)}^{2}={\frac {\eta }{2}}\cdot E_{p(t)}={\frac {\eta }{2}}\int _{0}^{T_{s}}{p^{2}(t)\partial t}={\frac {\eta }{2}}T_{s}\\&P_{e}=p_{'1'}\cdot Q\left({\frac {\left|V_{T}-m\right|}{\sqrt {\sigma ^{2}}}}\right)+p_{'0'}\cdot Q\left({\frac {\left|V_{T}-m\right|}{\sqrt {\sigma ^{2}}}}\right)=0.5\cdot Q\left({\frac {\left|{\frac {AT_{s}}{2}}-AT_{s}\right|}{\sqrt {{\frac {\eta }{2}}T_{s}}}}\right)+0.5\cdot Q\left({\frac {\left|{\frac {AT_{s}}{2}}-0\right|}{\sqrt {{\frac {\eta }{2}}T_{s}}}}\right)=\\&P_{e}=Q\left({\frac {\left|{\frac {AT_{s}}{2}}-AT_{s}\right|}{\sqrt {{\frac {\eta }{2}}T_{s}}}}\right)=Q\left({\frac {\frac {AT_{s}}{2}}{\sqrt {{\frac {\eta }{2}}T_{s}}}}\right)=Q\left({\sqrt {\frac {\frac {A^{2}T_{s}^{2}}{4}}{{\frac {\eta }{2}}T_{s}}}}\right)=Q\left({\sqrt {\frac {A^{2}T_{s}}{2\eta }}}\right)\\\end{aligned}}$
${\begin{aligned}&E_{b}=0.5\cdot E_{'1'}+0.5\cdot E_{'0'}\\&E_{'1'}=E_{p}=\int _{0}^{T_{s}}{s_{'1'}^{2}(t)\partial t}=A^{2}T_{s}\\&E_{'0'}=0\\&E_{b}={\frac {A^{2}T_{s}}{2}}\\&Q\left({\sqrt {\frac {A^{2}T_{s}}{2\eta }}}\right)=Q\left({\sqrt {\frac {E_{b}}{\eta }}}\right)=Q\left({\sqrt {\frac {E_{p}}{2\eta }}}\right)\\\end{aligned}}$

## Probabilidad de error para codificación unipolar RZ

${\begin{aligned}&T_{s}=T_{b}\to p(t)=\prod {\left({\frac {t-^{T_{s}}\!\!\diagup \!\!_{4}\;}{^{T_{s}}\!\!\diagup \!\!_{2}\;}}\right)}\\&a_{k}=\left\{{\begin{aligned}&a_{'1'}=A\\&a_{'0'}=0\\\end{aligned}}\right.\\&m_{'1'}=\int _{0}^{T_{s}}{s_{'1'}(t)kp^{*}\left(t\right)\partial t}=\int _{0}^{T_{s}}{Ap(t)p^{*}\left(t\right)\partial t}={\frac {AT_{s}}{2}}\\&m_{'0'}=\int _{0}^{T_{s}}{s_{'0'}(t)kp^{*}\left(t\right)\partial t}=0\\&V_{T}={\frac {m_{'1'}+m_{'0'}}{2}}={\frac {AT_{s}}{4}}\\&\sigma _{y_{PR}(t)}^{2}={\frac {\eta }{2}}\cdot E_{p(t)}={\frac {\eta }{2}}\int _{0}^{T_{s}}{p^{2}(t)\partial t}={\frac {\eta }{2}}{\frac {T_{s}}{2}}\\&P_{e}=p_{'1'}\cdot Q\left({\frac {\left|V_{T}-m\right|}{\sqrt {\sigma ^{2}}}}\right)+p_{'0'}\cdot Q\left({\frac {\left|V_{T}-m\right|}{\sqrt {\sigma ^{2}}}}\right)=0.5\cdot Q\left({\frac {\left|{\frac {AT_{s}}{4}}-{\frac {AT_{s}}{2}}\right|}{\sqrt {{\frac {\eta }{4}}T_{s}}}}\right)+0.5\cdot Q\left({\frac {\left|{\frac {AT_{s}}{4}}-0\right|}{\sqrt {{\frac {\eta }{4}}T_{s}}}}\right)=\\&P_{e}=Q\left({\frac {\frac {AT_{s}}{4}}{\sqrt {{\frac {\eta }{4}}T_{s}}}}\right)=Q\left({\sqrt {\frac {\frac {A^{2}T_{s}^{2}}{16}}{{\frac {\eta }{4}}T_{s}}}}\right)=Q\left({\sqrt {\frac {A^{2}T_{s}}{4\eta }}}\right)\\\end{aligned}}$
${\begin{aligned}&E_{b}=0.5\cdot E_{'1'}+0.5\cdot E_{'0'}\\&E_{'1'}=\int _{0}^{T_{s}}{s_{'1'}^{2}(t)\partial t}={\frac {A^{2}T_{s}}{2}}\\&E_{'0'}=\int _{0}^{T_{s}}{s_{'0'}^{2}(t)\partial t}=0\\&E_{b}={\frac {A^{2}T_{s}}{4}}\\&Q\left({\sqrt {\frac {A^{2}T_{s}}{4\eta }}}\right)=Q\left({\sqrt {\frac {E_{b}}{\eta }}}\right)\\\end{aligned}}$

## Probabilidad de error para codificación polar

${\begin{aligned}&T_{s}=T_{b}\to p(t)=\prod {\left({\frac {t-^{T_{s}}\!\!\diagup \!\!_{2}\;}{T_{s}}}\right)}\\&a_{k}=\left\{{\begin{aligned}&a_{'1'}=A\\&a_{'0'}=-A\\\end{aligned}}\right.\\&m_{'1'}=\int _{0}^{T_{s}}{s_{'1'}(t)kp^{*}\left(t\right)\partial t}=\int _{0}^{T_{s}}{Ap(t)p^{*}\left(t\right)\partial t}=AT_{s}\\&m_{'0'}=\int _{0}^{T_{s}}{s_{'0'}(t)kp^{*}\left(t\right)\partial t}=\int _{0}^{T_{s}}{-Ap(t)p^{*}\left(t\right)\partial t}=-AT_{s}\\&V_{T}={\frac {m_{'1'}+m_{'0'}}{2}}=0\\&\sigma _{y_{PR}(t)}^{2}={\frac {\eta }{2}}\cdot E_{p(t)}={\frac {\eta }{2}}\int _{0}^{T_{s}}{p^{2}(t)\partial t}={\frac {\eta }{2}}T_{s}\\&P_{e}=p_{'1'}\cdot Q\left({\frac {\left|V_{T}-m\right|}{\sqrt {\sigma ^{2}}}}\right)+p_{'0'}\cdot Q\left({\frac {\left|V_{T}-m\right|}{\sqrt {\sigma ^{2}}}}\right)=0.5\cdot Q\left({\frac {\left|0-AT_{s}\right|}{\sqrt {{\frac {\eta }{2}}T_{s}}}}\right)+0.5\cdot Q\left({\frac {\left|0-\left(-AT_{s}\right)\right|}{\sqrt {{\frac {\eta }{2}}T_{s}}}}\right)=\\&P_{e}=Q\left({\frac {AT_{s}}{\sqrt {{\frac {\eta }{2}}T_{s}}}}\right)=Q\left({\sqrt {\frac {A^{2}T_{s}^{2}}{{\frac {\eta }{2}}T_{s}}}}\right)=Q\left({\sqrt {\frac {2A^{2}T_{s}}{\eta }}}\right)\\\end{aligned}}$
${\begin{aligned}&E_{b}=0.5\cdot E_{'1'}+0.5\cdot E_{'0'}\\&E_{'1'}=\int _{0}^{T_{s}}{s_{'1'}^{2}(t)\partial t}=A^{2}T_{s}\\&E_{'0'}=\int _{0}^{T_{s}}{s_{'0'}^{2}(t)\partial t}=A^{2}T_{s}\\&E_{b}=A^{2}T_{s}\\&Q\left({\sqrt {\frac {2A^{2}T_{s}}{\eta }}}\right)=Q\left({\sqrt {\frac {2E_{b}}{\eta }}}\right)\\\end{aligned}}$

## Probabilidad de error para codificación bipolar

${\begin{aligned}&T_{s}=T_{b}\to p(t)=\prod {\left({\frac {t-^{T_{s}}\!\!\diagup \!\!_{2}\;}{T_{s}}}\right)}\\&a_{k}=\left\{{\begin{aligned}&a_{'1'}=\pm A\\&a_{'0'}=0\\\end{aligned}}\right.\\&m_{'1'}=\int _{0}^{T_{s}}{s_{'1'}(t)kp^{*}\left(t\right)\partial t}=\int _{0}^{T_{s}}{\pm Ap(t)p^{*}\left(t\right)\partial t}=\pm AT_{s}\\&m_{'0'}=\int _{0}^{T_{s}}{s_{'0'}(t)kp^{*}\left(t\right)\partial t}=0\\&V_{T}={\frac {m_{'1'}+m_{'0'}}{2}}=\pm {\frac {AT_{s}}{2}}\\&\sigma _{y_{PR}(t)}^{2}={\frac {\eta }{2}}\cdot E_{p(t)}={\frac {\eta }{2}}\int _{0}^{T_{s}}{p^{2}(t)\partial t}={\frac {\eta }{2}}T_{s}\\&P_{e}={\frac {1}{4}}\cdot Q\left({\frac {\left|V_{T}-m\right|}{\sqrt {\sigma ^{2}}}}\right)+{\frac {1}{4}}\cdot Q\left({\frac {\left|V_{T}-m\right|}{\sqrt {\sigma ^{2}}}}\right)+{\frac {1}{2}}\cdot 2Q\left({\frac {\left|V_{T}-m\right|}{\sqrt {\sigma ^{2}}}}\right)={\frac {3}{2}}\cdot Q\left({\frac {\left|V_{T}-m\right|}{\sqrt {\sigma ^{2}}}}\right)\\&P_{e}={\frac {3}{2}}Q\left({\frac {\frac {AT_{s}}{2}}{\sqrt {{\frac {\eta }{2}}T_{s}}}}\right)={\frac {3}{2}}Q\left({\sqrt {\frac {\frac {A^{2}T_{s}^{2}}{4}}{{\frac {\eta }{2}}T_{s}}}}\right)={\frac {3}{2}}Q\left({\sqrt {\frac {A^{2}T_{s}}{2\eta }}}\right)\\\end{aligned}}$
${\begin{aligned}&E_{b}=0.5\cdot E_{'1'}+0.5\cdot E_{'0'}\\&E_{'1'}=E_{p}=\int _{0}^{T_{s}}{s_{'1'}^{2}(t)\partial t}=A^{2}T_{s}\\&E_{'0'}=0\\&E_{b}={\frac {A^{2}T_{s}}{2}}\\&P_{e}\simeq {\frac {3}{2}}Q\left({\sqrt {\frac {A^{2}T_{s}}{2\eta }}}\right)={\frac {3}{2}}Q\left({\sqrt {\frac {E_{b}}{\eta }}}\right)\\\end{aligned}}$

### Probabilidad de error total sin aproximaciones

${\begin{aligned}&P_{e}=p\left(^{error}\!\!\diagup \!\!_{+1sent}\;\right)+p\left(^{error}\!\!\diagup \!\!_{-1sent}\;\right)+p\left(^{error}\!\!\diagup \!\!_{0sent}\;\right)\simeq \\&p\left(^{0detected}\!\!\diagup \!\!_{+1sent}\;\right)+p\left(^{0detected}\!\!\diagup \!\!_{-1sent}\;\right)+p\left(^{+1detected}\!\!\diagup \!\!_{0sent}\;\right)+p\left(^{-1detected}\!\!\diagup \!\!_{0sent}\;\right)\to \\&P_{e}=p\left(^{0detected}\!\!\diagup \!\!_{+1sent}\;\right)+p\left(^{0detected}\!\!\diagup \!\!_{-1sent}\;\right)+p\left(^{+1detected}\!\!\diagup \!\!_{0sent}\;\right)+p\left(^{-1detected}\!\!\diagup \!\!_{0sent}\;\right)+\\&p\left(^{-1detected}\!\!\diagup \!\!_{+1sent}\;\right)+p\left(^{+1detected}\!\!\diagup \!\!_{-1sent}\;\right)\\&P_{e}={\frac {3}{2}}Q\left({\sqrt {\frac {A^{2}T_{s}}{2\eta }}}\right)+{\frac {1}{4}}\cdot Q\left({\frac {\left|V_{T}-m\right|}{\sqrt {\sigma ^{2}}}}\right)+{\frac {1}{4}}\cdot Q\left({\frac {\left|V_{T}-m\right|}{\sqrt {\sigma ^{2}}}}\right)=\\&{\frac {3}{2}}Q\left({\sqrt {\frac {A^{2}T_{s}}{2\eta }}}\right)+{\frac {1}{4}}\cdot Q\left({\frac {{\frac {AT_{s}}{2}}-\left(-AT_{s}\right)}{\sqrt {{\frac {\eta }{2}}T_{s}}}}\right)+{\frac {1}{4}}\cdot Q\left({\frac {AT_{s}-\left(-{\frac {AT_{s}}{2}}\right)}{\sqrt {{\frac {\eta }{2}}T_{s}}}}\right)=\\&{\frac {3}{2}}Q\left({\sqrt {\frac {A^{2}T_{s}}{2\eta }}}\right)+{\frac {1}{2}}\cdot Q\left({\frac {\frac {3AT_{s}}{2}}{\sqrt {{\frac {\eta }{2}}T_{s}}}}\right)={\frac {3}{2}}Q\left({\sqrt {\frac {A^{2}T_{s}}{2\eta }}}\right)+{\frac {1}{2}}\cdot Q\left({\sqrt {\frac {\frac {9A^{2}T_{s}^{2}}{4}}{{\frac {\eta }{2}}T_{s}}}}\right)\\&P_{e}={\frac {3}{2}}Q\left({\sqrt {\frac {A^{2}T_{s}}{2\eta }}}\right)+{\frac {1}{2}}\cdot Q\left({\sqrt {\frac {9A^{2}T_{s}}{2\eta }}}\right)\\&E_{b}={\frac {A^{2}T_{s}}{2}}\to P_{e}={\frac {3}{2}}Q\left({\sqrt {\frac {E_{b}}{\eta }}}\right)+{\frac {1}{2}}\cdot Q\left({\sqrt {9{\frac {E_{b}}{\eta }}}}\right)\\\end{aligned}}$

## Probabilidad de error para codificación manchester

${\begin{aligned}&T_{s}=T_{b}\to p(t)=\prod {\left({\frac {t-^{T_{s}}\!\!\diagup \!\!_{4}\;}{^{T_{s}}\!\!\diagup \!\!_{2}\;}}\right)}-\prod {\left({\frac {t-^{3T_{s}}\!\!\diagup \!\!_{4}\;}{^{T_{s}}\!\!\diagup \!\!_{2}\;}}\right)}\\&a_{k}=\left\{{\begin{aligned}&a_{'1'}=A\\&a_{'0'}=-A\\\end{aligned}}\right.\\&m_{'1'}=\int _{0}^{T_{s}}{s_{'1'}(t)kp^{*}\left(t\right)\partial t}=\int _{0}^{T_{s}}{Ap(t)p^{*}\left(t\right)\partial t}=AT_{s}\\&m_{'0'}=\int _{0}^{T_{s}}{s_{'0'}(t)kp^{*}\left(t\right)\partial t}=\int _{0}^{T_{s}}{-Ap(t)p^{*}\left(t\right)\partial t}=-AT_{s}\\&V_{T}={\frac {m_{'1'}+m_{'0'}}{2}}=0\\&\sigma _{y_{PR}(t)}^{2}={\frac {\eta }{2}}\cdot E_{p(t)}={\frac {\eta }{2}}\int _{0}^{T_{s}}{p^{2}(t)\partial t}={\frac {\eta }{2}}T_{s}\\&P_{e}=p_{'1'}\cdot Q\left({\frac {\left|V_{T}-m\right|}{\sqrt {\sigma ^{2}}}}\right)+p_{'0'}\cdot Q\left({\frac {\left|V_{T}-m\right|}{\sqrt {\sigma ^{2}}}}\right)=0.5\cdot Q\left({\frac {\left|0-AT_{s}\right|}{\sqrt {{\frac {\eta }{2}}T_{s}}}}\right)+0.5\cdot Q\left({\frac {\left|0-\left(-AT_{s}\right)\right|}{\sqrt {{\frac {\eta }{2}}T_{s}}}}\right)=\\&P_{e}=Q\left({\frac {AT_{s}}{\sqrt {{\frac {\eta }{2}}T_{s}}}}\right)=Q\left({\sqrt {\frac {A^{2}T_{s}^{2}}{{\frac {\eta }{2}}T_{s}}}}\right)=Q\left({\sqrt {\frac {2A^{2}T_{s}}{\eta }}}\right)\\\end{aligned}}$
${\begin{aligned}&E_{b}=0.5\cdot E_{'1'}+0.5\cdot E_{'0'}\\&E_{'1'}=\int _{0}^{T_{s}}{s_{'1'}^{2}(t)\partial t}=A^{2}T_{s}\\&E_{'0'}=\int _{0}^{T_{s}}{s_{'0'}^{2}(t)\partial t}=A^{2}T_{s}\\&E_{b}=A^{2}T_{s}\\&Q\left({\sqrt {\frac {2A^{2}T_{s}}{\eta }}}\right)=Q\left({\sqrt {\frac {2E_{b}}{\eta }}}\right)\\\end{aligned}}$

## Probabilidad de error para codificación unipolar NRZ (TRIANGULAR)

${\begin{aligned}&T_{s}=T_{b}\to p(t)=\Lambda \left({\frac {t-^{T_{s}}\!\!\diagup \!\!_{4}\;}{^{T_{s}}\!\!\diagup \!\!_{2}\;}}\right)\\&a_{k}=\left\{{\begin{aligned}&a_{'1'}=A\\&a_{'0'}=0\\\end{aligned}}\right.\\&m_{'0'}=\int _{0}^{T_{s}}{s_{'0'}(t)kp^{*}\left(t\right)\partial t}=0\\&m_{'1'}=\int _{0}^{T_{s}}{s_{'1'}(t)kp^{*}\left(t\right)\partial t}=\int _{0}^{T_{s}}{Ap(t)p^{*}\left(t\right)\partial t}=A\cdot 2\int _{0}^{^{T_{s}}\!\!\diagup \!\!_{2}\;}{\left({\frac {t}{^{T_{s}}\!\!\diagup \!\!_{2}\;}}\right)^{2}\partial t}=\\&2A{\frac {4}{T_{s}^{2}}}\int _{0}^{^{T_{s}}\!\!\diagup \!\!_{2}\;}{t^{2}\partial t}=2A{\frac {4}{T_{s}^{2}}}\left.{\frac {t^{3}}{3}}\right|_{0}^{^{T_{s}}\!\!\diagup \!\!_{2}\;}={\frac {8A}{3T_{s}^{2}}}\left(^{T_{s}}\!\!\diagup \!\!_{2}\;\right)^{3}={\frac {8AT_{s}}{3\cdot 8}}={\frac {AT_{s}}{3}}\\&V_{T}={\frac {m_{'1'}+m_{'0'}}{2}}={\frac {AT_{s}}{6}}\\&\sigma _{y_{PR}(t)}^{2}={\frac {\eta }{2}}\cdot E_{p(t)}={\frac {\eta }{2}}\int _{0}^{T_{s}}{p^{2}(t)\partial t}={\frac {\eta }{2}}{\frac {T_{s}}{3}}\\&P_{e}=p_{'1'}\cdot Q\left({\frac {\left|V_{T}-m\right|}{\sqrt {\sigma ^{2}}}}\right)+p_{'0'}\cdot Q\left({\frac {\left|V_{T}-m\right|}{\sqrt {\sigma ^{2}}}}\right)=0.5\cdot Q\left({\frac {\left|{\frac {AT_{s}}{6}}-{\frac {AT_{s}}{3}}\right|}{\sqrt {{\frac {\eta }{2}}{\frac {T_{s}}{3}}}}}\right)+0.5\cdot Q\left({\frac {\left|{\frac {AT_{s}}{6}}-0\right|}{\sqrt {{\frac {\eta }{2}}{\frac {T_{s}}{3}}}}}\right)=\\&P_{e}=Q\left({\frac {\frac {AT_{s}}{6}}{\sqrt {\frac {\eta T_{s}}{6}}}}\right)=Q\left({\sqrt {\frac {\frac {A^{2}T_{s}^{2}}{6^{2}}}{\frac {\eta T_{s}}{6}}}}\right)=Q\left({\sqrt {\frac {A^{2}T_{s}}{6\eta }}}\right)\\\end{aligned}}$
${\begin{aligned}&E_{b}=0.5\cdot E_{'1'}+0.5\cdot E_{'0'}\\&E_{'1'}=E_{p}=\int _{0}^{T_{s}}{s_{'1'}^{2}(t)\partial t}={\frac {A^{2}T_{s}}{3}}\\&E_{'0'}=0\\&E_{b}={\frac {A^{2}T_{s}}{6}}\\&Q\left({\sqrt {\frac {A^{2}T_{s}}{6\eta }}}\right)=Q\left({\sqrt {\frac {E_{b}}{\eta }}}\right)\\\end{aligned}}$

## Probabilidad de error para triangular NRZ no óptima

(Señal original-> triangular, filtro de recepcion = cuadrado)
${\begin{aligned}&T_{s}=T_{b}\to a_{k}\cdot p(t)=\Lambda \left({\frac {t-^{T_{s}}\!\!\diagup \!\!_{4}\;}{^{T_{s}}\!\!\diagup \!\!_{2}\;}}\right)\\&p(t)=\prod {\left({\frac {t}{T_{s}}}\right)}\\&a_{k}=\left\{{\begin{aligned}&a_{'1'}=A\\&a_{'0'}=0\\\end{aligned}}\right.\\&m_{'0'}=\int _{0}^{T_{s}}{s_{'0'}(t)kp^{*}\left(t\right)\partial t}=0\\&m_{'1'}=\int _{0}^{T_{s}}{s_{'1'}(t)kp^{*}\left(t\right)\partial t}=\int _{0}^{T_{s}}{Ap(t)p^{*}\left(t\right)\partial t}={\frac {AT_{s}}{2}}\\&V_{T}={\frac {m_{'1'}+m_{'0'}}{2}}={\frac {AT_{s}}{4}}\\&\sigma _{y_{PR}(t)}^{2}={\frac {\eta }{2}}\cdot E_{p(t)}={\frac {\eta }{2}}\int _{0}^{T_{s}}{p^{2}(t)\partial t}={\frac {\eta }{2}}T_{s}\\&P_{e}=p_{'1'}\cdot Q\left({\frac {\left|V_{T}-m\right|}{\sqrt {\sigma ^{2}}}}\right)+p_{'0'}\cdot Q\left({\frac {\left|V_{T}-m\right|}{\sqrt {\sigma ^{2}}}}\right)=0.5\cdot Q\left({\frac {\left|{\frac {AT_{s}}{4}}-{\frac {AT_{s}}{2}}\right|}{\sqrt {{\frac {\eta }{2}}T_{s}}}}\right)+0.5\cdot Q\left({\frac {\left|{\frac {AT_{s}}{4}}-0\right|}{\sqrt {{\frac {\eta }{2}}T_{s}}}}\right)=\\&P_{e}=Q\left({\frac {\frac {AT_{s}}{4}}{\sqrt {{\frac {\eta }{2}}T_{s}}}}\right)=Q\left({\sqrt {\frac {\frac {A^{2}T_{s}^{2}}{16}}{{\frac {\eta }{2}}T_{s}}}}\right)=Q\left({\sqrt {\frac {A^{2}T_{s}}{8\eta }}}\right)\\\end{aligned}}$
${\begin{aligned}&E_{b}=0.5\cdot E_{'1'}+0.5\cdot E_{'0'}\\&E_{'1'}=E_{p}=\int _{0}^{T_{s}}{s_{'1'}^{2}(t)\partial t}={\frac {A^{2}T_{s}}{3}}\\&E_{'0'}=0\\&E_{b}={\frac {A^{2}T_{s}}{6}}\\&Q\left({\sqrt {\frac {A^{2}T_{s}}{8\eta }}}\right)=Q\left({\sqrt {{\frac {3}{4}}{\frac {E_{b}}{\eta }}}}\right)\\\end{aligned}}$
* Proyecto: Departamento de Teoría de la Señal y Comunicaciones
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