07hw03sol - EE544 Homework Assignment # 3 Solution 1. (a)...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: EE544 Homework Assignment # 3 Solution 1. (a) Let s ( t ) be the output of the matched filter h ( t ) = s ( T- t ) due to s ( t ). Then s ( t ) = s ( t ) * h ( t ) = Z - s ( ) h ( t- ) d = Z - s ( ) s ( T- ( t- )) d = A 2 T 2 Z T ( T- t + ) P T ( T- t + ) d = A 2 T 2 R t ( T- t + ) P T ( T- t + ) d, t < T A 2 T 2 R T t- T ( T- t + ) P T ( T- t + ) d, T t < 2 T , t < 0 or t 2 T = A 2 6 T 2 t 2 (3 T- t ) , t < T A 2 6 T 2 ( t + T )( t- 2 T ) 2 , T t < 2 T , t < 0 or t 2 T The MF output due to s ( t ) is shown in Figure 1. Since s ( t ) and s 1 ( t ) are antipodal signals, s ( T ) = A 2 T 3 , s 1 ( T ) =- A 2 T 3 Noise variance, 2 n = N 2 k h k 2 = N 2 k s k 2 = A 2 TN 6 Hence, P e, = 1- s ( T ) n = 1- A r 2 T 3 N ! = Q A r 2 T 3 N ! Similarly, P e, 1 = 1- A r 2 T 3 N ! = Q A r 2 T 3 N ! (b) Let s ( t ) be the correlator output due to s ( t ). s ( t ) = Z t- s ( ) s ( ) d = A 2 T 2 Z t- 2 P T ( ) d = , t < A 2 T 2 R t 2 d, t < T A 2 T 2 R T 2 d, t T = , t < A 2 t 3 3 T 2 , t < T A 2 T 3 , t T The correlator output due to s ( t ) is shown in Figure 1. 1 Correlator Matched filter T 2T A T 3 2 Figure 1: Matched filter output vs correlator output in part (b) and (c) of Prob. 1 Since the two signals s ( t ) and s 1 ( t ) are antipodal, s ( T ) = A 2 T 3 , s 1 ( T ) =- A 2 T 3 Let n ( t ) be the AWGN process. Then the correlator output due to the noise process at t = T is n ( T ) = Z T n ( ) A T d E[ n ( T )] = Z T A T E[ n ( )] d = 0 Var( n ( T )) = A 2 T 2 E Z T Z T N 2 ( - ) dd = A 2 T 2 Z T 2 d = A 2 TN 6 Hence, P e, = P e, 1 = 1- A r 2 T 3 N ! = Q A r 2 T 3 N ! (c) In (a) and (b), the matched filter receiver and the correlator receiver show the same error performance in the Gaussian channel. However, as is evident in Figure 1, if the sampling time is not optimal the signal component of each receiver is not the same. In other words,time is not optimal the signal component of each receiver is not the same....
View Full Document

This note was uploaded on 02/13/2012 for the course ECE 544 taught by Professor Jameslehnert during the Spring '12 term at Purdue University-West Lafayette.

Page1 / 8

07hw03sol - EE544 Homework Assignment # 3 Solution 1. (a)...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online