Question

Consider a binary channel transmitting bits independently. Each bit is demodulated with a 0 corresponding to 0 volts and a 1 corresponding to 5 volts. Thus, the received random variable v is normally distributed with variance σ² = 1 and mean μ = 0 or μ = 5. The demodulated voltage v is compared to a threshold τ to decide whether a bit is a 0 or 1, i.e., decide that a 1 was sent if v > τ and that a 0 was sent if v < τ. Bits are equally probable so that P ( μ = 0 ) = P ( μ = 5 ) = 1 / 2. Note that an error occurs either if a 0 was sent so that μ = 0 but v > τ and a 1 is decided or if a 1 was sent so that μ = 5 but v < τ and a 0 is decided. What are the conditional probabilities P ( v > τ | μ = 0 ) and P ( v < τ | μ = 5 )? What is the resulting probability of error? What value should one choose for τ to minimize the probability of error? If 1250 bytes (8 bits each) are sent, what is the expected number of errors?

Answer 1

P ( v > τ | μ = 0 ) = P[Z > (τ - 0)/1] = P[Z > τ] = 1 - P[Z τ] = 1 - (τ)

P ( v < τ | μ = 5 ) = P[Z > (τ - 5)/1] = P[Z < (τ-5)] = (τ-5)

By law of total probability,

The resulting probability of error,

P ( v > τ | μ = 0 ) P ( μ = 0) +  P ( v < τ | μ = 5 ) P ( μ = 5)

= (1 - (τ)) * 0.5 + (τ-5) * 0.5

= 0.5 - 0.5 (τ) + 0.5 * (τ-5)

To minimize the probability of error,  (τ) should be maximum and (τ-5) should be minimum. For this, we choose a mid-point between 0 and 5 which is 2.5

For τ = 2.5,

Probability of error =  0.5 - 0.5 (2.5) + 0.5 * (2.5-5) = 0.5 - 0.5 (2.5) + 0.5 * (-2.5)

= 0.5 - 0.5 * 0.9938 + 0.5 * 0.0062

= 0.0062

Expected number of errors = 1250 * 8 * 0.0062 = 62