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Coherent orthogonal binary FSK modulation is used to transmit two equiprobable symbol waveforms s1(t) = a cos 2πf1t and s2(t) = a cos 2πf2t, where a = 4 mV. Assume an AWGN channel with two-sided noise power spectral density N0/2 = 0.5 × 10⁻¹² W/Hz. Using an optimal receiver and the relation Q(v) = 1/√(2π) ∫_v^∞ e^(-u²/2) du, the bit error probability for a data rate of 500 kbps is

1. Q(2)
2. Q(2√2)
3. Q(4)
4. Q(4√2)

Correct answer: Q(4)

Solution

Eb = (a^2/2)*Tb with a=4mV, Tb=1/500k=2us gives Eb=1.6e-11 J; N0=1e-12, so Eb/N0=16. For coherent orthogonal binary FSK, Pe = Q(sqrt(Eb/N0)) = Q(sqrt 16) = Q(4). Stored Q(2sqrt2) is wrong.

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