Cálculo de la probabilidad de error para las diferentes codificaciones

De Wikiversidad

\begin{align} & \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{align}

Contenido

[editar] Probabilidad de error para codificación unipolar NRZ

\begin{align} & T_{s}=T_{b}\to p(t)=\prod{\left( \frac{t-{}^{T_{s}}\!\!\diagup\!\!{}_{2}\;}{T_{s}} \right)} \\ & a_{k}=\left\{ \begin{align} & a_{'1'}=A \\ & a_{'0'}=0 \\ \end{align} \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{align}

\begin{align} & 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{align}

[editar] Probabilidad de error para codificación unipolar RZ

\begin{align} & T_{s}=T_{b}\to p(t)=\prod{\left( \frac{t-{}^{T_{s}}\!\!\diagup\!\!{}_{4}\;}{{}^{T_{s}}\!\!\diagup\!\!{}_{2}\;} \right)} \\ & a_{k}=\left\{ \begin{align} & a_{'1'}=A \\ & a_{'0'}=0 \\ \end{align} \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{align}

\begin{align} & 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{align}

[editar] Probabilidad de error para codificación polar

\begin{align} & T_{s}=T_{b}\to p(t)=\prod{\left( \frac{t-{}^{T_{s}}\!\!\diagup\!\!{}_{2}\;}{T_{s}} \right)} \\ & a_{k}=\left\{ \begin{align} & a_{'1'}=A \\ & a_{'0'}=-A \\ \end{align} \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{align}

\begin{align} & 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{align}

[editar] Probabilidad de error para codificación bipolar

\begin{align} & T_{s}=T_{b}\to p(t)=\prod{\left( \frac{t-{}^{T_{s}}\!\!\diagup\!\!{}_{2}\;}{T_{s}} \right)} \\ & a_{k}=\left\{ \begin{align} & a_{'1'}=\pm A \\ & a_{'0'}=0 \\ \end{align} \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{align}

\begin{align} & 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{align}


[editar] Probabilidad de error total sin aproximaciones

\begin{align} & 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{align}

[editar] Probabilidad de error para codificación manchester

\begin{align} & 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{align} & a_{'1'}=A \\ & a_{'0'}=-A \\ \end{align} \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{align}

\begin{align} & 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{align}


[editar] Probabilidad de error para codificación unipolar NRZ (TRIANGULAR)

\begin{align} & T_{s}=T_{b}\to p(t)=\Lambda \left( \frac{t-{}^{T_{s}}\!\!\diagup\!\!{}_{4}\;}{{}^{T_{s}}\!\!\diagup\!\!{}_{2}\;} \right) \\ & a_{k}=\left\{ \begin{align} & a_{'1'}=A \\ & a_{'0'}=0 \\ \end{align} \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{align}

\begin{align} & 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{align}

[editar] Probabilidad de error para triangular NRZ no optima

(Señal original-> triangular, filtro de recepcion = cuadrado)


\begin{align} & 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{align} & a_{'1'}=A \\ & a_{'0'}=0 \\ \end{align} \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{align}

\begin{align} & 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{align}


Proyecto: Departamento de Teoría de la Señal y Comunicaciones
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