EE567_Fall2009_HW3_Solutions

EE567_Fall2009_HW3_Solutions - Homework #3 Solutions EE 567...

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Homework #3 Solutions EE 567 Communication Systems Fall 2009
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EE 567 Communication Systems Fall 2009
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EE 567 Communication Systems Fall 2009
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From Given that the mean is zero and variance is 2: Var(n) = E{n 2 } – E{n} 2 = E{n 2 } = Hence, The pdf of the x can be easily written as Since y = x + n, EE 567 Communication Systems Fall 2009
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EE 567 Communication Systems Fall 2009
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EE 567 Communication Systems Fall 2009
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Computer Problem 1 Rx 0.06 0.07 0.08 0.09 001 0.02 0.03 0.04 0.05 0 10 20 30 40 50 60 -0.01 0 0.01 0.1 Sx 0.06 0.07 0.08 0.09 0.02 0.03 0.04 0.05 -0.5 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 0.5 0 0.01 EE 567 Communication Systems Fall 2009
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echo on ; clear all ; close all ; N = 1000; M =50; ave = 50; Rx_av = zeros (1,M+1); Sx_av = zeros (1, M+1); for j=1:ave % TAKE THE ENSEMBLE AVERAGE X = rand (1, N) - (1/2); Rx = Rx_est(X,M); Sx = fftshift(abs(fft(Rx))); Rx_av = Rx_av + Rx; Sx_av = Sx_av + Sx; end Rx_av = Rx_av/ave; Sx_av = Sx_av/ave; f = linspace(-0.5,0.5,M+1); figure(1) plot(Rx_av)
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This note was uploaded on 09/26/2011 for the course EE 567 taught by Professor Weber during the Fall '07 term at USC.

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EE567_Fall2009_HW3_Solutions - Homework #3 Solutions EE 567...

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