# BER for BPSK over AWGN Channel

bpsk

A BPSK scheme operating over an AWGN channel with noise power spectral density of \$N/2\$, uses equiprobable signals \$S_1(t) = \sqrt{\frac{2E}{T} sin\ t}\$ and \$S_2(t) = -\sqrt{\frac{2E}{T}sin\ t}\$ over the symbol interval \$(0, T)\$.

If the local oscillator in a coherent receiver is ahead in phase by 45 degrees with respect to the received signal, the probability of error in the resulting system is (Answer key given is \$Q(\sqrt{\frac{E}{N}})\$**

The AWGN channel with noise power spectral density $$\frac{N}{2}$$ And the signals: $$s_1(t)=\sqrt{2\frac{E}{T}}\sin(t)\qquad t\in[0,T]$$ $$s_2(t)=-\sqrt{2\frac{E}{T}}\sin(t)=-s_1(t)$$

In a normal coherent oscillator, you would receive something like the following in the constellation diagram (noise not represented): (the picture may be wrong because of the usage of sine or cosine, but you get the idea).

The red dashed line is the decision boundary, and the distance from the signal to this boundary represents the actual energy that the receiver can use to make the decision. So, in this case, it is obvious that the BER will be $$Q\bigg(\sqrt{\frac{2E}{N}}\bigg)$$

If the local oscillator in a coherent receiver is ahead in phase by 45°, it means that the receiver has is decision boundary rotated in the constellation diagram: So the distance from the signal to the boundary is actually less than before, therefore the BER increases. How much? It depends on the distance to the boundary, so the new BER is: $$Q\bigg(\sqrt{\frac{E}{N}}\bigg)$$