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561_hmk3_soln

# 561_hmk3_soln - ECE 561 Digital Communications Systems...

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ECE 561 – Digital Communications Systems Solutions for Homework #3 1. a. μ = x 2 e 2 x dx 0 = ... = 1 / 2 b. F X ( x ) = 2 e 2 x dx 0 x = ... = 1 e 2 x c. P (0 x 1) = F X (1) = 1 e 2 = 0.865 d. P (1 x 2) = F X (2) F X (1) = 1 e 4 1 e 2 ( ) = 0.117 2. From the given form of a Gaussian pdf: a. 4 b. 9 c. P ( X 2) = P z 2 4 3 = P ( z ≤ − 2 / 3) = 1 Q 2 3 = 1 1 Q 2 3 = Q 2 3 = 0.253 d. P ( X > 7) = P z > 7 4 3 = P ( z > 1) = Q (1) = 0.1587 e. P ( 2 X 10) = P 2 4 3 z 10 4 3 = P ( 2 z 2) = P ( z 2) P ( z ≤ − 2) = 1 Q (2) 1 Q ( 2) [ ] = 1 Q (2) 1 + 1 Q (2) = 1 2 Q (2) = 0.9544

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561_hmk3_soln - ECE 561 Digital Communications Systems...

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