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Unformatted text preview: EE 7615 Solutions to Problem Set #1 Sept. 2011 Problem 2.17 Problem 2.18 1 Problem 2.16 1) Let T and R denote input and output respectively. Having received A we calculate P ( T = 0) P ( R = A  T = 0) = . 4 (1 / 4) = . 1 and P ( T = 1) P ( R = A  T = 1) = . 6 (1 / 3) = . 2 . Since the latter is larger we choose 1 as the transmitted message. For B we have P ( T = 0) P ( R = B  T = 0) = . 4 (1 / 2) = . 2 , P ( T = 1) P ( R = B  T = 1) = . 6 (1 / 3) = . 2 Since we have equality we can choose either or 1 . The probability of error will be the same. For C , P ( T = 0) P ( R = C  T = 0) = . 4 (1 / 4) = . 1 , P ( T = 1) P ( R = C  T = 1) = . 6 (1 / 3) = . 2 In this case again we choose 1 as our decision. 2) The answer depends on what we do for B . Suppose we always choose when B is received. Then P ( E  T = 0) = P ( R = A  T = 0) + P ( T = C  R = 0) = 1 / 4 + 1 / 4 = 1 / 2 3) We also need P ( E  T = 1) ....
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 Fall '11
 Naragipour

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