HW4%20Solution

# HW4%20Solution - φ t = 1 2 φ(2 t φ(2 t − 1 1 2 φ(2 t...

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H.W. 6 Problem 1 Let p ( t ) = h φ ( τ ) ( τ + t ) i . Then P ( j Ω) = Z -∞ h φ ( τ ) ( τ + t ) i e - j Ω t dt = Z -∞ Z -∞ φ ( τ ) φ ( τ + t ) dτe - j Ω t dt Making the substitution α = τ + t , we get t = α - τ and dt = . So P ( j Ω) = Z -∞ Z -∞ φ ( τ ) φ ( α ) e - j Ω( α - τ ) dαdτ = Z -∞ φ ( τ ) e j Ω τ Z -∞ φ ( α ) e - j Ω α = Φ * ( j Ω)Φ( j Ω) = | Φ( j Ω) | 2 Next deﬁne p d [ n ] = p ( n ) i.e., p ( t ) sampled with a sampling period of T = 1. Then P d ( e ) = X k P ( + 2 πk ) . Substituting for P ( ), we get P d ( e ) = X k | Φ( + 2 πk ) | 2 But since φ ( t ) and φ ( t + n ) are orthonormal for n Z , p d [ n ] = δ [ n ], so P d ( e ) = 1. Therefore, X k | Φ( + 2 πk ) | 2 = 1 . 1

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H.W. 6 Problem 2 (a) k φ ( t ) k 2 =2 Z 1 0 A 2 t 2 dt =2 [ 1 3 A 2 t 3 ] 1 0 = 2 A 2 3 So k φ ( t ) k =1 A = q 3 2 .

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Unformatted text preview: φ ( t ) = 1 2 φ (2 t ) + φ (2 t − 1) + 1 2 φ (2 t − 2) . So, φ ( t ) = √ 2 2 X n =0 h [ n ] φ (2 t − n ) with h [0] = 1 2 √ 2 h [1] = 1 √ 2 h [2] = 1 2 √ 2 1 (c) With N = 3, g [ n ] = ( − 1) n h [2 − n ], and g [0] = 1 2 √ 2 g [1] = − 1 √ 2 g [2] = 1 2 √ 2 See fgure below For ψ ( t ). 2 A/2 \ ± t ) 1 2 2 Ͳ A 3...
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## This note was uploaded on 04/21/2009 for the course ECE 700 taught by Professor Un during the Winter '09 term at Ohio State.

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HW4%20Solution - φ t = 1 2 φ(2 t φ(2 t − 1 1 2 φ(2 t...

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