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GATE 2014 EC (EC2) – Question 60

Communications · Digital Communications · 2 marks · Multiple choice

Coherent orthogonal binary FSK sends equiprobable $s_1(t)=\alpha\cos(2\pi f_1t)$ and $s_2(t)=\alpha\cos(2\pi f_2t)$, with $\alpha=4$ mV, through AWGN having two-sided PSD $N_0/2=0.5\times10^{-12}$ W/Hz. For optimal reception at 500 kbps, find bit-error probability. Here $Q(v)=\frac1{\sqrt{2\pi}}\int_v^\infty e^{-u^2/2}\,du$.

  1. $Q(2)$
  2. $Q(2\sqrt2)$
  3. $Q(4)$
  4. $Q(4\sqrt2)$

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Correct answer: (C) $Q(4)$

Explanation

The bit duration is1/(500kbps)=2µs. For amplitude4mV, Eb=a²Tb/2=(4×10⁻³)²×2×10⁻⁶/2=16×10⁻¹²J in the implied unit resistance. Since N0/2=0.5×10⁻¹², N0=10⁻¹². Coherent orthogonal BFSK has Pb=Q(√(Eb/N0))=**Q(4),C**.