Question

The constellation diagram of a binary modulation scheme, with equiprobable symbols, is shown below. The two equiprobable symbols shown in the diagram are transmitted through an $AWGN$ channel with two sided nose $PSD0.50W/Hz$. If a correlator receiver with optimum threshold detection is used at the receiver end, then the bit error rate $\left(BER\right)$ of the system will be

• A. $Q\left(1\right)$
• B. $Q\left(3\right)$
• C. $Q\left(4\right)$
• D. $Q\left(2\right)$

Answer 1

The correct option is D $Q\left(2\right)$

$BER=Q\left(\sqrt{\frac{d_{min}^2}{2N_0}}\right)$ $\because$ Equiprobable symbols with optimum threshold
where $d_{min}=$ Euclidean distance between ${\overline{S}}_0$ and ${\overline{S}}_1$ and $\frac{N_0}{2}=\frac{1}{2}W/Hz\Rightarrow N_0=1W/Hz$
$d_{min}=|{\overline{S}}_1-{\overline{S}}_0|=\sqrt{4+4}=\sqrt{8}$
$BER=Q\left(\sqrt{\frac{8}{2\left(1\right)}}\right)=Q\left(2\right)$