# Q2S - ECE 402 Communications Engineering Quiz 2 Fall 2009...

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ECE 402 Communications Engineering Quiz 2 Fall 2009 KEY NAME: STUDENT NUMBER: INSTRUCTIONS Write your name and student number in the space provided above. This quiz is closed book and closed notes. Two handwritten 8.5” × 11” reference sheets may be used (both sides). Conventional calculators may be used (but no computers, PDAs or other wireless devices) Give step-by-step solutions that include all details. Assume the graders know only the basics. Box your ﬁnal answer . GRADES 1 2 3 4 Total 1

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Problem 1 (25 pts): The heights of American women are well modeled as a Gaussian distribution with a mean of 64” (5’ 4”) and a standard deviation of 3”. What fraction of American women are over 5’ 10” (=70”)? Solution: If X represents the height of a person chosen at random from the population, then X is Gaussian with μ = 64 and σ = 3. Hence Pr { X 70 } = Q (distance from mean in std) = Q ± 70 - μ σ = Q (2) 0 . 02275 . 2
Problem 2 (25 pts): One of two equally likely messages is sent over an AWGN channel with noise PSD N 0 / 2 using pulse-width modulation (PWM) , where 0 τ T : 0 T 0 ( ) x t D 1 ( ) x t 0 T D τ (a) Find the optimal receiver in its simplest form. In what sense is it optimal? (b) Find the bit-error-rate of the optimal receiver in terms of D , T , τ and N 0 . What value of τ gives the best bit-error-rate? Solution: (a) The optimal receiver is the correlation receiver. It is optimal in the sense of minimizing the error probability. This receiver chooses ˆ b = 1 if R T 0 Y ( t ) x 1 ( t ) dt - E 1 / 2 > R T 0 Y ( t ) x 0 ( t ) dt - E 0 / 2, or = D Z T 0 Y ( t ) dt - D 2 T/ 2 > D Z τ 0 Y ( t ) dt - D 2 τ/ 2 = Z T τ Y ( t ) dt > D ( T - τ ) / 2 which yields the receiver below: U ( ) Y t T dt 1, ( )/ 2 ˆ 0, ( ) / 2 U D T b U D T - = - ˆ b The bit-error-rate is given by P b = Q s R T 0 ( x 1 ( t ) - x 0 ( t )) 2 dt 2 N 0 = Q s R T τ D 2 dt 2 N 0 =

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Q2S - ECE 402 Communications Engineering Quiz 2 Fall 2009...

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