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A BPSK scheme operating over an AWGN channel with noise power spectral density $N_0/2$ uses equiprobable signals $s_1(t)=\sqrt{2E/T}\,\sin(\omega_c t)$ and $s_2(t)=-\sqrt{2E/T}\,\sin(\omega_c t)$ over the symbol interval $(0,T)$. If the local oscillator in a coherent receiver is ahead in phase by $45^\circ$ with respect to the received signal, the probability of error in the resulting system is

1. $Q\!\left(\sqrt{\frac{2E}{N_0}}\right)$
2. $Q\!\left(\sqrt{\frac{E}{N_0}}\right)$
3. $Q\!\left(\sqrt{\frac{E}{2N_0}}\right)$
4. $Q\!\left(\sqrt{\frac{E}{4N_0}}\right)$

Correct answer: $Q\!\left(\sqrt{\frac{E}{N_0}}\right)$

Solution

With a phase error $\phi$, the effective signal amplitude in coherent BPSK becomes proportional to $\cos\phi$, so the SNR is reduced by a factor of $\cos^2\phi$. For $\phi=45^\circ$, $\cos^2 45^\circ=1/2$, giving $P_e=Q\!\left(\sqrt{2E/N_0\cdot 1/2}\right)=Q\!\left(\sqrt{E/N_0}\right)$.

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