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A single bit, equally likely to be 0 and 1, is to be sent across an additive white Gaussian noise (AWGN) channel with power spectral density N0/2. Binary signaling, with 0 → p(t) and 1 → q(t), is used for the transmission, along with an optimal receiver that minimizes the bit-error probability. Let φ1(t), φ2(t) form an orthonormal signal set. If we choose p(t) = φ1(t) and q(t) = -φ1(t), we would obtain a certain bit-error probability P_b. If we keep p(t) = φ1(t), but take q(t) = √E φ2(t), for what value of E would we obtain the same bit-error probability P_b?

1. 0
2. 1
3. 2
4. 3

Correct answer: 3

Solution

BER depends only on the squared distance between the two signal points. Antipodal p=phi1, q=-phi1 gives distance^2 = |2 phi1|^2 = 4. For orthogonal p=phi1, q=sqrt(E) phi2 the distance^2 = 1+E. Setting 1+E=4 gives E=3, not the stored E=1.

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