Cálculo de la probabilidad de error para las diferentes codificaciones

$\begin{array}{cc}& \underset{V_T}{\overset{\mathrm{\infty }}{\int }}\frac{1}{\sqrt{2\pi {\sigma }^2}}\cdot e^{\frac{-{(t-m)}^2}{2{\sigma }^2}}\partial t=Q\left(\frac{|V_T-m|}{\sqrt{{\sigma }^2}}\right)\\ & P_e=pr\left(\text{'}{0}^{\prime }sent\right)\cdot p\left({}^{error}╱{}_{\text{'}{0}^{\prime }sent}\right)+pr\left(\text{'}{1}^{\prime }sent\right)\cdot p\left({}^{error}╱{}_{\text{'}{1}^{\prime }sent}\right)\\ & m_{y_{PR}(t)}=\underset{0}{\overset{T_s}{\int }}{s}_T(t)k{p}^{*}\left(t\right)\partial t\\ & {\sigma }_{y_{PR}(t)}^2=\frac{\eta }{2}\cdot E_{p(t)}=\frac{\eta }{2}\underset{0}{\overset{T_s}{\int }}{p}^2(t)\partial t\\ & s_T(t)=\sum _{k=0}^{\mathrm{\infty }}{a}_{k}p(t-k{T}_s)\\ \end{array}$

Probabilidad de error para codificación unipolar NRZ

$\begin{array}{cc}& T_s=T_b\to p(t)=\prod \left(\frac{t-{}^{T_s}╱{}_2}{P}pp{T}_s\right)\\ & a_k=\{\begin{array}{cc}P& a_{\text{'}{1}^{\prime }}=A\\ Pml& a_{\text{'}{0}^{\prime }}=0\\ \end{array}\\ & m_{\text{'}{1}^{\prime }}=\underset{0}{\overset{T_s}{\int }}{s}_{\text{'}{1}^{\prime }}(t)k{p}^{*}\left(t\right)\partial t=\underset{0}{\overset{T_s}{\int }}Ap(t)p^{*}\left(t\right)\partial t=A{T}_s\\ & m_{\text{'}{0}^{\prime }}=\underset{0}{\overset{T_s}{\int }}{s}_{\text{'}{0}^{\prime }}(t)k{p}^{*}\left(t\right)\partial t=0\\ & V_T=\frac{m_{\text{'}{1}^{\prime }}+m_{\text{'}{0}^{\prime }}}{2}=\frac{A{T}_s}{2}\\ & {\sigma }_{y_{PR}(t)}^2=\frac{\eta }{2}\cdot E_{p(t)}=\frac{\eta }{2}\underset{0}{\overset{T_s}{\int }}{p}^2(t)\partial t=\frac{\eta }{2}{T}_s\\ & P_e=p_{\text{'}{1}^{\prime }}\cdot Q\left(\frac{|V_T-m|}{\sqrt{{\sigma }^2}}\right)+p_{\text{'}{0}^{\prime }}\cdot Q\left(\frac{|V_T-m|}{\sqrt{{\sigma }^2}}\right)=0.5\cdot Q\left(\frac{|\frac{A{T}_s}{2}-A{T}_s|}{\sqrt{\frac{\eta }{2}{T}_s}}\right)+0.5\cdot Q\left(\frac{|\frac{A{T}_s}{2}-0|}{\sqrt{\frac{\eta }{2}{T}_s}}\right)=\\ & P_e=Q\left(\frac{|\frac{A{T}_s}{2}-A{T}_s|}{\sqrt{\frac{\eta }{2}{T}_s}}\right)=Q\left(\frac{\frac{A{T}_s}{2}}{\sqrt{\frac{\eta }{2}{T}_s}}\right)=Q\left(\sqrt{\frac{\frac{A^2{T}_s^2}{4}}{\frac{\eta }{2}{T}_s}}\right)=Q\left(\sqrt{\frac{A^2{T}_s}{2\eta }}\right)\\ \end{array}$

$\begin{array}{cc}& E_b=0.5\cdot E_{\text{'}{1}^{\prime }}+0.5\cdot E_{\text{'}{0}^{\prime }}\\ & E_{\text{'}{1}^{\prime }}=E_p=\underset{0}{\overset{T_s}{\int }}{s}_{\text{'}{1}^{\prime }}^2(t)\partial t=A^2{T}_s\\ & E_{\text{'}{0}^{\prime }}=0\\ & E_b=\frac{A^2{T}_s}{2}\\ & Q\left(\sqrt{\frac{A^2{T}_s}{2\eta }}\right)=Q\left(\sqrt{\frac{E_b}{\eta }}\right)=Q\left(\sqrt{\frac{E_p}{2\eta }}\right)\\ \end{array}$

Probabilidad de error para codificación unipolar RZ

$\begin{array}{cc}& T_s=T_b\to p(t)=\prod \left(\frac{t-{}^{T_s}╱{}_4}{{}^{T_s}╱{}_2}\right)\\ & a_k=\{\begin{array}{cc}& a_{\text{'}{1}^{\prime }}=A\\ & a_{\text{'}{0}^{\prime }}=0\\ \end{array}\\ & m_{\text{'}{1}^{\prime }}=\underset{0}{\overset{T_s}{\int }}{s}_{\text{'}{1}^{\prime }}(t)k{p}^{*}\left(t\right)\partial t=\underset{0}{\overset{T_s}{\int }}Ap(t)p^{*}\left(t\right)\partial t=\frac{A{T}_s}{2}\\ & m_{\text{'}{0}^{\prime }}=\underset{0}{\overset{T_s}{\int }}{s}_{\text{'}{0}^{\prime }}(t)k{p}^{*}\left(t\right)\partial t=0\\ & V_T=\frac{m_{\text{'}{1}^{\prime }}+m_{\text{'}{0}^{\prime }}}{2}=\frac{A{T}_s}{4}\\ & {\sigma }_{y_{PR}(t)}^2=\frac{\eta }{2}\cdot E_{p(t)}=\frac{\eta }{2}\underset{0}{\overset{T_s}{\int }}{p}^2(t)\partial t=\frac{\eta }{2}\frac{T_s}{2}\\ & P_e=p_{\text{'}{1}^{\prime }}\cdot Q\left(\frac{|V_T-m|}{\sqrt{{\sigma }^2}}\right)+p_{\text{'}{0}^{\prime }}\cdot Q\left(\frac{|V_T-m|}{\sqrt{{\sigma }^2}}\right)=0.5\cdot Q\left(\frac{|\frac{A{T}_s}{4}-\frac{A{T}_s}{2}|}{\sqrt{\frac{\eta }{4}{T}_s}}\right)+0.5\cdot Q\left(\frac{|\frac{A{T}_s}{4}-0|}{\sqrt{\frac{\eta }{4}{T}_s}}\right)=\\ & P_e=Q\left(\frac{\frac{A{T}_s}{4}}{\sqrt{\frac{\eta }{4}{T}_s}}\right)=Q\left(\sqrt{\frac{\frac{A^2{T}_s^2}{16}}{\frac{\eta }{4}{T}_s}}\right)=Q\left(\sqrt{\frac{A^2{T}_s}{4\eta }}\right)\\ \end{array}$

$\begin{array}{cc}& E_b=0.5\cdot E_{\text{'}{1}^{\prime }}+0.5\cdot E_{\text{'}{0}^{\prime }}\\ & E_{\text{'}{1}^{\prime }}=\underset{0}{\overset{T_s}{\int }}{s}_{\text{'}{1}^{\prime }}^2(t)\partial t=\frac{A^2{T}_s}{2}\\ & E_{\text{'}{0}^{\prime }}=\underset{0}{\overset{T_s}{\int }}{s}_{\text{'}{0}^{\prime }}^2(t)\partial t=0\\ & E_b=\frac{A^2{T}_s}{4}\\ & Q\left(\sqrt{\frac{A^2{T}_s}{4\eta }}\right)=Q\left(\sqrt{\frac{E_b}{\eta }}\right)\\ \end{array}$

Probabilidad de error para codificación polar

$\begin{array}{cc}& T_s=T_b\to p(t)=\prod \left(\frac{t-{}^{T_s}╱{}_2}{T_s}\right)\\ & a_k=\{\begin{array}{cc}& a_{\text{'}{1}^{\prime }}=A\\ & a_{\text{'}{0}^{\prime }}=-A\\ \end{array}\\ & m_{\text{'}{1}^{\prime }}=\underset{0}{\overset{T_s}{\int }}{s}_{\text{'}{1}^{\prime }}(t)k{p}^{*}\left(t\right)\partial t=\underset{0}{\overset{T_s}{\int }}Ap(t)p^{*}\left(t\right)\partial t=A{T}_s\\ & m_{\text{'}{0}^{\prime }}=\underset{0}{\overset{T_s}{\int }}{s}_{\text{'}{0}^{\prime }}(t)k{p}^{*}\left(t\right)\partial t=\underset{0}{\overset{T_s}{\int }}-Ap(t)p^{*}\left(t\right)\partial t=-A{T}_s\\ & V_T=\frac{m_{\text{'}{1}^{\prime }}+m_{\text{'}{0}^{\prime }}}{2}=0\\ & {\sigma }_{y_{PR}(t)}^2=\frac{\eta }{2}\cdot E_{p(t)}=\frac{\eta }{2}\underset{0}{\overset{T_s}{\int }}{p}^2(t)\partial t=\frac{\eta }{2}{T}_s\\ & P_e=p_{\text{'}{1}^{\prime }}\cdot Q\left(\frac{|V_T-m|}{\sqrt{{\sigma }^2}}\right)+p_{\text{'}{0}^{\prime }}\cdot Q\left(\frac{|V_T-m|}{\sqrt{{\sigma }^2}}\right)=0.5\cdot Q\left(\frac{|0-A{T}_s|}{\sqrt{\frac{\eta }{2}{T}_s}}\right)+0.5\cdot Q\left(\frac{|0-(-A{T}_s)|}{\sqrt{\frac{\eta }{2}{T}_s}}\right)=\\ & P_e=Q\left(\frac{A{T}_s}{\sqrt{\frac{\eta }{2}{T}_s}}\right)=Q\left(\sqrt{\frac{A^2{T}_s^2}{\frac{\eta }{2}{T}_s}}\right)=Q\left(\sqrt{\frac{2A^2{T}_s}{\eta }}\right)\\ \end{array}$

$\begin{array}{cc}& E_b=0.5\cdot E_{\text{'}{1}^{\prime }}+0.5\cdot E_{\text{'}{0}^{\prime }}\\ & E_{\text{'}{1}^{\prime }}=\underset{0}{\overset{T_s}{\int }}{s}_{\text{'}{1}^{\prime }}^2(t)\partial t=A^2{T}_s\\ & E_{\text{'}{0}^{\prime }}=\underset{0}{\overset{T_s}{\int }}{s}_{\text{'}{0}^{\prime }}^2(t)\partial t=A^2{T}_s\\ & E_b=A^2{T}_s\\ & Q\left(\sqrt{\frac{2A^2{T}_s}{\eta }}\right)=Q\left(\sqrt{\frac{2E_b}{\eta }}\right)\\ \end{array}$

Probabilidad de error para codificación bipolar

$\begin{array}{cc}& T_s=T_b\to p(t)=\prod \left(\frac{t-{}^{T_s}╱{}_2}{T_s}\right)\\ & a_k=\{\begin{array}{cc}& a_{\text{'}{1}^{\prime }}=\pm A\\ & a_{\text{'}{0}^{\prime }}=0\\ \end{array}\\ & m_{\text{'}{1}^{\prime }}=\underset{0}{\overset{T_s}{\int }}{s}_{\text{'}{1}^{\prime }}(t)k{p}^{*}\left(t\right)\partial t=\underset{0}{\overset{T_s}{\int }}\pm Ap(t)p^{*}\left(t\right)\partial t=\pm A{T}_s\\ & m_{\text{'}{0}^{\prime }}=\underset{0}{\overset{T_s}{\int }}{s}_{\text{'}{0}^{\prime }}(t)k{p}^{*}\left(t\right)\partial t=0\\ & V_T=\frac{m_{\text{'}{1}^{\prime }}+m_{\text{'}{0}^{\prime }}}{2}=\pm \frac{A{T}_s}{2}\\ & {\sigma }_{y_{PR}(t)}^2=\frac{\eta }{2}\cdot E_{p(t)}=\frac{\eta }{2}\underset{0}{\overset{T_s}{\int }}{p}^2(t)\partial t=\frac{\eta }{2}{T}_s\\ & P_e=\frac{1}{4}\cdot Q\left(\frac{|V_T-m|}{\sqrt{{\sigma }^2}}\right)+\frac{1}{4}\cdot Q\left(\frac{|V_T-m|}{\sqrt{{\sigma }^2}}\right)+\frac{1}{2}\cdot 2Q\left(\frac{|V_T-m|}{\sqrt{{\sigma }^2}}\right)=\frac{3}{2}\cdot Q\left(\frac{|V_T-m|}{\sqrt{{\sigma }^2}}\right)\\ & P_e=\frac{3}{2}Q\left(\frac{\frac{A{T}_s}{2}}{\sqrt{\frac{\eta }{2}{T}_s}}\right)=\frac{3}{2}Q\left(\sqrt{\frac{\frac{A^2{T}_s^2}{4}}{\frac{\eta }{2}{T}_s}}\right)=\frac{3}{2}Q\left(\sqrt{\frac{A^2{T}_s}{2\eta }}\right)\\ \end{array}$

$\begin{array}{cc}& E_b=0.5\cdot E_{\text{'}{1}^{\prime }}+0.5\cdot E_{\text{'}{0}^{\prime }}\\ & E_{\text{'}{1}^{\prime }}=E_p=\underset{0}{\overset{T_s}{\int }}{s}_{\text{'}{1}^{\prime }}^2(t)\partial t=A^2{T}_s\\ & E_{\text{'}{0}^{\prime }}=0\\ & E_b=\frac{A^2{T}_s}{2}\\ & P_e\simeq \frac{3}{2}Q\left(\sqrt{\frac{A^2{T}_s}{2\eta }}\right)=\frac{3}{2}Q\left(\sqrt{\frac{E_b}{\eta }}\right)\\ \end{array}$

Probabilidad de error total sin aproximaciones

$\begin{array}{cc}& P_e=p\left({}^{error}╱{}_{+1sent}\right)+p\left({}^{error}╱{}_{-1sent}\right)+p\left({}^{error}╱{}_{0sent}\right)\simeq \\ & p\left({}^{0detected}╱{}_{+1sent}\right)+p\left({}^{0detected}╱{}_{-1sent}\right)+p\left({}^{+1detected}╱{}_{0sent}\right)+p\left({}^{-1detected}╱{}_{0sent}\right)\to \\ & P_e=p\left({}^{0detected}╱{}_{+1sent}\right)+p\left({}^{0detected}╱{}_{-1sent}\right)+p\left({}^{+1detected}╱{}_{0sent}\right)+p\left({}^{-1detected}╱{}_{0sent}\right)+\\ & p\left({}^{-1detected}╱{}_{+1sent}\right)+p\left({}^{+1detected}╱{}_{-1sent}\right)\\ & P_e=\frac{3}{2}Q\left(\sqrt{\frac{A^2{T}_s}{2\eta }}\right)+\frac{1}{4}\cdot Q\left(\frac{|V_T-m|}{\sqrt{{\sigma }^2}}\right)+\frac{1}{4}\cdot Q\left(\frac{|V_T-m|}{\sqrt{{\sigma }^2}}\right)=\\ & \frac{3}{2}Q\left(\sqrt{\frac{A^2{T}_s}{2\eta }}\right)+\frac{1}{4}\cdot Q\left(\frac{\frac{A{T}_s}{2}-(-A{T}_s)}{\sqrt{\frac{\eta }{2}{T}_s}}\right)+\frac{1}{4}\cdot Q\left(\frac{A{T}_s-(-\frac{A{T}_s}{2})}{\sqrt{\frac{\eta }{2}{T}_s}}\right)=\\ & \frac{3}{2}Q\left(\sqrt{\frac{A^2{T}_s}{2\eta }}\right)+\frac{1}{2}\cdot Q\left(\frac{\frac{3A{T}_s}{2}}{\sqrt{\frac{\eta }{2}{T}_s}}\right)=\frac{3}{2}Q\left(\sqrt{\frac{A^2{T}_s}{2\eta }}\right)+\frac{1}{2}\cdot Q\left(\sqrt{\frac{\frac{9A^2{T}_s^2}{4}}{\frac{\eta }{2}{T}_s}}\right)\\ & P_e=\frac{3}{2}Q\left(\sqrt{\frac{A^2{T}_s}{2\eta }}\right)+\frac{1}{2}\cdot Q\left(\sqrt{\frac{9A^2{T}_s}{2\eta }}\right)\\ & E_b=\frac{A^2{T}_s}{2}\to P_e=\frac{3}{2}Q\left(\sqrt{\frac{E_b}{\eta }}\right)+\frac{1}{2}\cdot Q\left(\sqrt{9\frac{E_b}{\eta }}\right)\\ \end{array}$

Probabilidad de error para codificación manchester

$\begin{array}{cc}& T_s=T_b\to p(t)=\prod \left(\frac{t-{}^{T_s}╱{}_4}{{}^{T_s}╱{}_2}\right)-\prod \left(\frac{t-{}^{3T_s}╱{}_4}{{}^{T_s}╱{}_2}\right)\\ & a_k=\{\begin{array}{cc}& a_{\text{'}{1}^{\prime }}=A\\ & a_{\text{'}{0}^{\prime }}=-A\\ \end{array}\\ & m_{\text{'}{1}^{\prime }}=\underset{0}{\overset{T_s}{\int }}{s}_{\text{'}{1}^{\prime }}(t)k{p}^{*}\left(t\right)\partial t=\underset{0}{\overset{T_s}{\int }}Ap(t)p^{*}\left(t\right)\partial t=A{T}_s\\ & m_{\text{'}{0}^{\prime }}=\underset{0}{\overset{T_s}{\int }}{s}_{\text{'}{0}^{\prime }}(t)k{p}^{*}\left(t\right)\partial t=\underset{0}{\overset{T_s}{\int }}-Ap(t)p^{*}\left(t\right)\partial t=-A{T}_s\\ & V_T=\frac{m_{\text{'}{1}^{\prime }}+m_{\text{'}{0}^{\prime }}}{2}=0\\ & {\sigma }_{y_{PR}(t)}^2=\frac{\eta }{2}\cdot E_{p(t)}=\frac{\eta }{2}\underset{0}{\overset{T_s}{\int }}{p}^2(t)\partial t=\frac{\eta }{2}{T}_s\\ & P_e=p_{\text{'}{1}^{\prime }}\cdot Q\left(\frac{|V_T-m|}{\sqrt{{\sigma }^2}}\right)+p_{\text{'}{0}^{\prime }}\cdot Q\left(\frac{|V_T-m|}{\sqrt{{\sigma }^2}}\right)=0.5\cdot Q\left(\frac{|0-A{T}_s|}{\sqrt{\frac{\eta }{2}{T}_s}}\right)+0.5\cdot Q\left(\frac{|0-(-A{T}_s)|}{\sqrt{\frac{\eta }{2}{T}_s}}\right)=\\ & P_e=Q\left(\frac{A{T}_s}{\sqrt{\frac{\eta }{2}{T}_s}}\right)=Q\left(\sqrt{\frac{A^2{T}_s^2}{\frac{\eta }{2}{T}_s}}\right)=Q\left(\sqrt{\frac{2A^2{T}_s}{\eta }}\right)\\ \end{array}$

$\begin{array}{cc}& E_b=0.5\cdot E_{\text{'}{1}^{\prime }}+0.5\cdot E_{\text{'}{0}^{\prime }}\\ & E_{\text{'}{1}^{\prime }}=\underset{0}{\overset{T_s}{\int }}{s}_{\text{'}{1}^{\prime }}^2(t)\partial t=A^2{T}_s\\ & E_{\text{'}{0}^{\prime }}=\underset{0}{\overset{T_s}{\int }}{s}_{\text{'}{0}^{\prime }}^2(t)\partial t=A^2{T}_s\\ & E_b=A^2{T}_s\\ & Q\left(\sqrt{\frac{2A^2{T}_s}{\eta }}\right)=Q\left(\sqrt{\frac{2E_b}{\eta }}\right)\\ \end{array}$

Probabilidad de error para codificación unipolar NRZ (TRIANGULAR)

$\begin{array}{cc}& T_s=T_b\to p(t)=\mathrm{\Lambda }\left(\frac{t-{}^{T_s}╱{}_4}{{}^{T_s}╱{}_2}\right)\\ & a_k=\{\begin{array}{cc}& a_{\text{'}{1}^{\prime }}=A\\ & a_{\text{'}{0}^{\prime }}=0\\ \end{array}\\ & m_{\text{'}{0}^{\prime }}=\underset{0}{\overset{T_s}{\int }}{s}_{\text{'}{0}^{\prime }}(t)k{p}^{*}\left(t\right)\partial t=0\\ & m_{\text{'}{1}^{\prime }}=\underset{0}{\overset{T_s}{\int }}{s}_{\text{'}{1}^{\prime }}(t)k{p}^{*}\left(t\right)\partial t=\underset{0}{\overset{T_s}{\int }}Ap(t)p^{*}\left(t\right)\partial t=A\cdot 2\underset{0}{\overset{{}^{T_s}╱{}_2}{\int }}{\left(\frac{t}{{}^{T_s}╱{}_2}\right)}^2\partial t=\\ & 2A\frac{4}{T_s^2}\underset{0}{\overset{{}^{T_s}╱{}_2}{\int }}{t}^2\partial t=2A\frac{4}{T_s^2}{\frac{t^3}{3}|}_0^{{}^{T_s}╱{}_2}=\frac{8A}{3T_s^2}{\left({}^{T_s}╱{}_2\right)}^3=\frac{8A{T}_s}{3\cdot 8}=\frac{A{T}_s}{3}\\ & V_T=\frac{m_{\text{'}{1}^{\prime }}+m_{\text{'}{0}^{\prime }}}{2}=\frac{A{T}_s}{6}\\ & {\sigma }_{y_{PR}(t)}^2=\frac{\eta }{2}\cdot E_{p(t)}=\frac{\eta }{2}\underset{0}{\overset{T_s}{\int }}{p}^2(t)\partial t=\frac{\eta }{2}\frac{T_s}{3}\\ & P_e=p_{\text{'}{1}^{\prime }}\cdot Q\left(\frac{|V_T-m|}{\sqrt{{\sigma }^2}}\right)+p_{\text{'}{0}^{\prime }}\cdot Q\left(\frac{|V_T-m|}{\sqrt{{\sigma }^2}}\right)=0.5\cdot Q\left(\frac{|\frac{A{T}_s}{6}-\frac{A{T}_s}{3}|}{\sqrt{\frac{\eta }{2}\frac{T_s}{3}}}\right)+0.5\cdot Q\left(\frac{|\frac{A{T}_s}{6}-0|}{\sqrt{\frac{\eta }{2}\frac{T_s}{3}}}\right)=\\ & P_e=Q\left(\frac{\frac{A{T}_s}{6}}{\sqrt{\frac{\eta T_s}{6}}}\right)=Q\left(\sqrt{\frac{\frac{A^2{T}_s^2}{6^2}}{\frac{\eta T_s}{6}}}\right)=Q\left(\sqrt{\frac{A^2{T}_s}{6\eta }}\right)\\ \end{array}$

$\begin{array}{cc}& E_b=0.5\cdot E_{\text{'}{1}^{\prime }}+0.5\cdot E_{\text{'}{0}^{\prime }}\\ & E_{\text{'}{1}^{\prime }}=E_p=\underset{0}{\overset{T_s}{\int }}{s}_{\text{'}{1}^{\prime }}^2(t)\partial t=\frac{A^2{T}_s}{3}\\ & E_{\text{'}{0}^{\prime }}=0\\ & E_b=\frac{A^2{T}_s}{6}\\ & Q\left(\sqrt{\frac{A^2{T}_s}{6\eta }}\right)=Q\left(\sqrt{\frac{E_b}{\eta }}\right)\\ \end{array}$

Probabilidad de error para triangular NRZ no óptima

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

$\begin{array}{cc}& T_s=T_b\to a_k\cdot p(t)=\mathrm{\Lambda }\left(\frac{t-{}^{T_s}╱{}_4}{{}^{T_s}╱{}_2}\right)\\ & p(t)=\prod \left(\frac{t}{T_s}\right)\\ & a_k=\{\begin{array}{cc}& a_{\text{'}{1}^{\prime }}=A\\ & a_{\text{'}{0}^{\prime }}=0\\ \end{array}\\ & m_{\text{'}{0}^{\prime }}=\underset{0}{\overset{T_s}{\int }}{s}_{\text{'}{0}^{\prime }}(t)k{p}^{*}\left(t\right)\partial t=0\\ & m_{\text{'}{1}^{\prime }}=\underset{0}{\overset{T_s}{\int }}{s}_{\text{'}{1}^{\prime }}(t)k{p}^{*}\left(t\right)\partial t=\underset{0}{\overset{T_s}{\int }}Ap(t)p^{*}\left(t\right)\partial t=\frac{A{T}_s}{2}\\ & V_T=\frac{m_{\text{'}{1}^{\prime }}+m_{\text{'}{0}^{\prime }}}{2}=\frac{A{T}_s}{4}\\ & {\sigma }_{y_{PR}(t)}^2=\frac{\eta }{2}\cdot E_{p(t)}=\frac{\eta }{2}\underset{0}{\overset{T_s}{\int }}{p}^2(t)\partial t=\frac{\eta }{2}{T}_s\\ & P_e=p_{\text{'}{1}^{\prime }}\cdot Q\left(\frac{|V_T-m|}{\sqrt{{\sigma }^2}}\right)+p_{\text{'}{0}^{\prime }}\cdot Q\left(\frac{|V_T-m|}{\sqrt{{\sigma }^2}}\right)=0.5\cdot Q\left(\frac{|\frac{A{T}_s}{4}-\frac{A{T}_s}{2}|}{\sqrt{\frac{\eta }{2}{T}_s}}\right)+0.5\cdot Q\left(\frac{|\frac{A{T}_s}{4}-0|}{\sqrt{\frac{\eta }{2}{T}_s}}\right)=\\ & P_e=Q\left(\frac{\frac{A{T}_s}{4}}{\sqrt{\frac{\eta }{2}{T}_s}}\right)=Q\left(\sqrt{\frac{\frac{A^2{T}_s^2}{16}}{\frac{\eta }{2}{T}_s}}\right)=Q\left(\sqrt{\frac{A^2{T}_s}{8\eta }}\right)\\ \end{array}$

$\begin{array}{cc}& E_b=0.5\cdot E_{\text{'}{1}^{\prime }}+0.5\cdot E_{\text{'}{0}^{\prime }}\\ & E_{\text{'}{1}^{\prime }}=E_p=\underset{0}{\overset{T_s}{\int }}{s}_{\text{'}{1}^{\prime }}^2(t)\partial t=\frac{A^2{T}_s}{3}\\ & E_{\text{'}{0}^{\prime }}=0\\ & E_b=\frac{A^2{T}_s}{6}\\ & Q\left(\sqrt{\frac{A^2{T}_s}{8\eta }}\right)=Q\left(\sqrt{\frac{3}{4}\frac{E_b}{\eta }}\right)\\ \end{array}$