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

${\begin{aligned}&s_{PSK}(t)=A_{c}\sum \limits _{k=-\infty }^{\infty }{p\left(t-kT_{s}\right)\cos \left(\omega _{c}t+\varphi _{k}\right)=}\\&A_{c}\sum \limits _{k=-\infty }^{\infty }{I_{k}p\left(t-kT_{s}\right)\cos \left(\omega _{c}t\right)-A_{c}\sum \limits _{k=-\infty }^{\infty }{Q_{k}p\left(t-kT_{s}\right)\sin \left(\omega _{c}t\right)}}\\&\\&y_{T}(t)=s_{T}(t)+n(t)\\&y_{T}(t)\cdot \cos \left(\omega _{c}t\right)=s_{T}(t)\cos \left(\omega _{c}t\right)+n(t)\cos \left(\omega _{c}t\right)\to H_{LPF}(f)?\to \\&y_{R}(t)=A_{R}\sum \limits _{k=-\infty }^{\infty }{I_{k}p\left(t-kT_{s}\right)+n_{I}(t)}\\&y_{R}^{'}(t)=A_{R}\sum \limits _{k=-\infty }^{\infty }{Q_{k}p\left(t-kT_{s}\right)+n_{Q}(t)}\\\end{aligned}}$

Constellation diagram for QPSK with Gray coding. Each adjacent symbol only differs by one bit.

${\begin{aligned}&\cos ^{2}x+\sin ^{2}x=1\\&I_{k}^{2}+Q_{k}^{2}=1\\\end{aligned}}$

Igual que antes, la señal QPSK demodulada es equivalente a una señal polar.

${\begin{aligned}&I_{k}=\left\{\pm {\frac {1}{\sqrt {2}}}\right\}\\&Q_{k}=\left\{\pm {\frac {1}{\sqrt {2}}}\right\}\\\end{aligned}}$

Demodulación de la señal QPSK In-Phase (Proyección horizontal de la constelación QPSK)

$a_{I_{k}}=a_{k}\cdot \cos(\varphi _{k})$

${\begin{aligned}&y_{R}(t)=A_{R}\sum \limits _{k=-\infty }^{\infty }{I_{k}p\left(t-kT_{s}\right)+n_{I}(t)}\\&I_{k}=\left\{-{\frac {1}{\sqrt {2}}},{\frac {1}{\sqrt {2}}}\right\}\\&A={\frac {1}{\sqrt {2}}}\\&a_{k}=\left\{{\begin{aligned}&a_{'1'}=A\\&a_{'0'}=-A\\\end{aligned}}\right.\\&m_{'1'}=\int _{0}^{T_{s}}{s_{'1'}(t)kp^{*}\left(t\right)\partial t}=\int _{0}^{T_{s}}{Ap(t)p^{*}\left(t\right)\partial t}=AT_{s}\\&m_{'0'}=\int _{0}^{T_{s}}{s_{'0'}(t)kp^{*}\left(t\right)\partial t}=\int _{0}^{T_{s}}{-Ap(t)p^{*}\left(t\right)\partial t}=-AT_{s}\\&V_{T}=0\\&P_{sc}=p_{'1'}\cdot Q\left({\frac {\left|V_{T}-m\right|}{\sqrt {\sigma ^{2}}}}\right)+p_{'0'}\cdot Q\left({\frac {\left|V_{T}-m\right|}{\sqrt {\sigma ^{2}}}}\right)=Q\left({\frac {AT_{s}}{\sqrt {{\frac {\eta }{2}}T_{s}}}}\right)=Q\left({\sqrt {\frac {2A^{2}T_{s}}{\eta }}}\right)\\\end{aligned}}$

Ahora, calcularemos la $E_{sc}$ :

${\begin{aligned}&A={\frac {1}{\sqrt {2}}}\\&E_{sc}=p_{'1'}\cdot E_{'1'}+p_{'0'}\cdot E_{'0'}\\&E_{'11'}=\int _{0}^{T_{s}}{s_{'1'}^{2}(t)\partial t}=A^{2}T_{s}\\&E_{'10'}=\int _{0}^{T_{s}}{s_{'0'}^{2}(t)\partial t}=A^{2}T_{s}\\&E_{sc}=A^{2}T_{s}\\&P_{sc}=Q\left({\sqrt {\frac {2A^{2}T_{s}}{\eta }}}\right)=Q\left({\sqrt {\frac {2E_{sc}}{\eta }}}\right)\\\end{aligned}}$

${\begin{aligned}&E_{s}=\log _{2}\left(M=4\right)\cdot E_{b}=2\cdot E_{b}\\&E_{s}=2\cdot E_{sc}\to E_{sc}=E_{b}\\\end{aligned}}$

$P_{b}=P_{sc}=Q\left({\sqrt {\frac {2E_{sc}}{\eta }}}\right)=Q\left({\sqrt {\frac {2E_{b}}{\eta }}}\right)$

Demodulación de la señal QPSK Quadrature (Proyección vertical de la constelación QPSK)

$a_{Q_{k}}=a_{k}\cdot \sin(\varphi _{k})$

Es totalmente análogo, por lo que:

${\begin{aligned}&E_{sc}=p_{'1'}\cdot E_{'1'}+p_{'0'}\cdot E_{'0'}=A^{2}T_{s}\\&P_{b}=P_{sc}=Q\left({\sqrt {\frac {2E_{sc}}{\eta }}}\right)=Q\left({\sqrt {\frac {2E_{b}}{\eta }}}\right)\\\end{aligned}}$

Finalmente:

$QPSK\to P_{b}=Q\left({\sqrt {\frac {2E_{b}}{\eta }}}\right)$