EE567_Fall2009_HW3_Solutions

# EE567_Fall2009_HW3_Solutions - Homework #3 Solutions EE 567...

This preview shows pages 1–9. Sign up to view the full content.

Homework #3 Solutions EE 567 Communication Systems Fall 2009

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
EE 567 Communication Systems Fall 2009
EE 567 Communication Systems Fall 2009

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
EE 567 Communication Systems Fall 2009

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
EE 567 Communication Systems Fall 2009
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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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)
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 09/26/2011 for the course EE 567 taught by Professor Weber during the Fall '07 term at USC.

### Page1 / 10

EE567_Fall2009_HW3_Solutions - Homework #3 Solutions EE 567...

This preview shows document pages 1 - 9. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online