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07hw03sol

07hw03sol - EE544 Homework Assignment 3 Solution 1(a Let ˆ...

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Unformatted text preview: EE544 Homework Assignment # 3 Solution 1. (a) Let ˆ s ( t ) be the output of the matched filter h ( t ) = s ( T- t ) due to s ( t ). Then ˆ s ( t ) = s ( t ) * h ( t ) = Z ∞-∞ s ( τ ) h ( t- τ ) dτ = Z ∞-∞ s ( τ ) s ( T- ( t- τ )) dτ = A 2 T 2 Z T τ ( T- t + τ ) P T ( T- t + τ ) dτ = A 2 T 2 R t τ ( T- t + τ ) P T ( T- t + τ ) dτ, ≤ t < T A 2 T 2 R T t- T τ ( T- t + τ ) P T ( T- t + τ ) dτ, 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, ˆ s ( T ) = A 2 T 3 , ˆ s 1 ( T ) =- A 2 T 3 Noise variance, σ 2 n = N 2 k h k 2 = N 2 k s k 2 = A 2 TN 6 Hence, P e, = 1- Φ µ ˆ s ( T ) σ n ¶ = 1- Φ ˆ A r 2 T 3 N ! = Q ˆ A r 2 T 3 N ! Similarly, P e, 1 = 1- Φ ˆ A r 2 T 3 N ! = Q ˆ A r 2 T 3 N ! (b) Let ˜ s ( t ) be the correlator output due to s ( t ). ˜ s ( t ) = Z t-∞ s ( τ ) s ( τ ) dτ = A 2 T 2 Z t-∞ τ 2 P T ( τ ) dτ = , t < A 2 T 2 R t τ 2 dτ, ≤ t < T A 2 T 2 R T τ 2 dτ, 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 ( τ ) A T dτ E[˜ n ( T )] = Z T A T E[ n ( τ )] dτ = 0 Var(˜ n ( T )) = A 2 T 2 E •Z T Z T αβ N 2 δ ( α- β ) dαdβ ‚ = A 2 T 2 Z T α 2 dα = A 2 TN 6 Hence, P e, = P e, 1 = 1- Φ ˆ A r 2 T 3 N ! = Q ˆ 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....
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07hw03sol - EE544 Homework Assignment 3 Solution 1(a Let ˆ...

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