A ternary communication channel is shown in the following figure. Let X denotes

Question

A ternary communication channel is shown in the following figure. Let X denotes the input and Y the output. Suppose that X can take a symbol of 0, 1, and 2 with probability 1/2, 1/4, and 1/4, respectively. At the receiver side, observed Y depends on the input X. (a) What are the probabilities of output symbols, i.e., P(Y=0), P(Y=1), and P(Y=2)? (b) Find the total error probability of this communication system? (c)Suppose that a symbol 1 is observed as an output, what is the probability that input was symbol 0? 1? 2?

Explanation / Answer

Part (a)

P(Y = 0) = P(X = 0, Y = 0) + P(X = 2, Y = 0) = ½ (1 – ) + ¼ = ½ - ¼ = (2 - )/4.

P(Y = 1) = P(X = 1, Y = 1) + P(X = 0, Y = 1) = ¼ (1 – ) + ½ = ¼ + ¼ = (1 + )/4.

P(Y = 2) = P(X = 1, Y = 2) + P(X = 2, Y = 2) = ¼ +¼ (1 – ) = 1/4.

Part (b)

P(system error) = P(X = 0, Y = 1) + P(X = 1, Y = 2) + P(X = 2, Y = 0) = ½ + ¼ +¼ = .

Part (c)

Y = 1 is possible only in 2 ways – X = 0 and Y = 1 or X = 1 and Y = 1. So, if Y = 1, P(X = 0) = ,

P(X = 1) = 1 – and P(X = 2) =0.

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