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

$\begin{array}{cc}& s_{4-ASK}(t)=A_c\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_{I_k}p(t-k{T}_s)\cos \left({\omega }_{c}t\right)\\ & a_{I_k}=\{0,A,+2A,+3A\}\\ \end{array}$

La constelación de una señal ASK de 4 niveles es:

4-ASK

$\begin{array}{cc}& y_R(t)=A_R\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_{I_k}p(t-k{T}_s)+n_I(t)\\ & a_{I_k}=\left\{\begin{array}{cc}& a_{\text{'}{00}^{\prime }}=0\\ & a_{\text{'}{01}^{\prime }}=+A\\ & a_{\text{'}{10}^{\prime }}=+2A\\ & a_{\text{'}{11}^{\prime }}=+3A\end{array}\right\}\to \{\begin{array}{cc}& m_{\text{'}{00}^{\prime }}={\int }_0^{T_s}{s}_{\text{'}{00}^{\prime }}(t)k{p}^{\ast }\left(t\right)\partial t={\int }_0^{T_s}0p^{\ast }\left(t\right)\partial t=0\\ & m_{\text{'}{01}^{\prime }}={\int }_0^{T_s}{s}_{\text{'}{01}^{\prime }}(t)k{p}^{\ast }\left(t\right)\partial t={\int }_0^{T_s}+Ap(t)p^{\ast }\left(t\right)\partial t=+A{T}_s\\ & m_{\text{'}{10}^{\prime }}={\int }_0^{T_s}{s}_{\text{'}{10}^{\prime }}(t)k{p}^{\ast }\left(t\right)\partial t={\int }_0^{T_s}+2Ap(t)p^{\ast }\left(t\right)\partial t=+2A{T}_s\\ & m_{\text{'}{11}^{\prime }}={\int }_0^{T_s}{s}_{\text{'}{00}^{\prime }}(t)k{p}^{\ast }\left(t\right)\partial t={\int }_0^{T_s}+3Ap(t)p^{\ast }\left(t\right)\partial t=+3A{T}_s\end{array}\\ & V_T=\{\frac{A{T}_s}{2},\frac{3A{T}_s}{2},\frac{5A{T}_s}{2}\}\\ & \sqrt{{\sigma }^2}=\sqrt{\frac{\eta }{2}{T}_s}\\ & P_e\simeq \frac{1}{4}\cdot Q\left(\frac{|V_T-m|}{\sqrt{{\sigma }^2}}\right)+\frac{1}{4}\cdot 2Q\left(\frac{|V_T-m|}{\sqrt{{\sigma }^2}}\right)+\frac{1}{4}\cdot 2Q\left(\frac{|V_T-m|}{\sqrt{{\sigma }^2}}\right)+\frac{1}{4}\cdot Q\left(\frac{|V_T-m|}{\sqrt{{\sigma }^2}}\right)=\\ & P_e\simeq \frac{1}{4}\cdot Q\left(\frac{|\frac{A{T}_s}{2}-0|}{\sqrt{{\sigma }^2}}\right)+\frac{1}{4}\cdot 2Q\left(\frac{|\frac{A{T}_s}{2}-\left(A{T}_s\right)|}{\sqrt{{\sigma }^2}}\right)+\frac{1}{4}\cdot 2Q\left(\frac{|\frac{3A{T}_s}{2}-2A{T}_s|}{\sqrt{{\sigma }^2}}\right)+\frac{1}{4}\cdot Q\left(\frac{|\frac{5A{T}_s}{2}-3A{T}_s|}{\sqrt{{\sigma }^2}}\right)=\\ & P_e\simeq \frac{3}{2}Q\left(\frac{\frac{A{T}_s}{2}}{\sqrt{{\sigma }^2}}\right)=\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}$

$\begin{array}{cc}& E_{sc}=\frac{1}{4}\cdot E_{\text{'}{00}^{\prime }}+\frac{1}{4}\cdot E_{\text{'}{01}^{\prime }}+\frac{1}{4}\cdot E_{\text{'}{10}^{\prime }}+\frac{1}{4}\cdot E_{\text{'}{11}^{\prime }}\\ & E_{\text{'}{00}^{\prime }}=0\\ & E_{\text{'}{01}^{\prime }}={\int }_0^{T_s}{s}_{\text{'}{01}^{\prime }}^2(t)\partial t={(+A)}^2{T}_s=A^2{T}_s\\ & E_{\text{'}{10}^{\prime }}={\int }_0^{T_s}{s}_{\text{'}{10}^{\prime }}^2(t)\partial t={(+2A)}^2{T}_s=4A^2{T}_s\\ & E_{\text{'}{11}^{\prime }}={\int }_0^{T_s}{s}_{\text{'}{11}^{\prime }}^2(t)\partial t={(+3A)}^2{T}_s=9A^2{T}_s\\ & E_{sc}=\frac{14A^2{T}_s}{4}=\frac{7A^2{T}_s}{2}\\ & E_{sc}=\frac{7A^2{T}_s}{2}\leftrightarrow \frac{A^2{T}_s}{2}=\frac{E_{sc}}{7}\\ & P_{sc}\simeq \frac{3}{2}Q\left(\sqrt{\frac{A^2{T}_s}{2\eta }}\right)=\frac{3}{2}Q\left(\sqrt{\frac{E_{sc}}{7\eta }}\right)\\ & E_s=E_{s{c}_1}+\underset{0}{\underbrace{E_{s{c}_2}}}=E_{s{c}_1}\\ & E_s={\log }_2(M=4)E_b=2E_b\\ & P_s\simeq \frac{3}{2}Q\left(\sqrt{\frac{E_s}{7\eta }}\right)=\frac{3}{2}Q\left(\sqrt{\frac{2E_b}{7\eta }}\right)\end{array}$