GATE 2014 EC (EC1) – Question 61
A BPSK system on AWGN with two-sided noise PSD $N_0/2$ has BER $Q(\sqrt\gamma)$, where $\gamma$ depends on bit energy and noise PSD. In the shown system the same modulated signal goes through two independent identical AWGN channels, each PSD $N_0/2$, and their outputs are summed before BPSK demodulation. If BER is $Q(b\sqrt\gamma)$, find b.
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Correct answer: 1.4 to 1.42
Explanation
Each branch contains the same BPSK waveform plus independent noise. Summation doubles the signal amplitude, hence quadruples its energy, while the independent noise variances add and only double. The effective signal-to-noise ratio is therefore doubled. In Q(b√γ), this gives **b=√2=1.414214**.