Question

The constellation diagram of a binary modulation scheme is shown below. The two in the diagram are transmitted through an AWGN channel with two sides noise $PSD\frac{1}{2}W/Hz$. If a correlator receiver with optimum threshold detection is used at the receiver end, then the bit error rate (BER) of the system will be

Answer 1

$BERor{P}_n=Q\left(\frac{d_{min}}{2{\sigma }_{nQ}}\right)\because$ equiprobable messages with optimum threshold

where $d_{min}=$ Euclidean distane between ${\overline{S}}_{0}and\overline{S_1},{\sigma }_{n0}^2=\frac{N_0}{2}=\frac{1}{2}W/Hz$

$S_1(1,1)and{S}_0(-1,-1)$

$d_{min}=|\overline{S_1}-\overline{S_0}|=\sqrt{4+4}$

$d_{min}=\sqrt{8}$

${\sigma }_{nQ}^2=\frac{1}{2}\Rightarrow {\sigma }_{nQ}=\sqrt{\frac{1}{2}}$

$P_n=Q\left(\frac{d_{min}}{2{\sigma }_{nQ}}\right)$

$P_n=Q\left(\frac{\sqrt{8}}{2.\sqrt{\frac{1}{2}}}\right)=Q\left(\sqrt{\frac{8}{2}}\right)$

= Q(2)