hw09su09_soln

# hw09su09_soln - GEORGIA INSTITUTE OF TECHNOLOGY ECE 2025...

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GEORGIA INSTITUTE OF TECHNOLOGY ECE 2025 SOLUTIONS Problem Set # 9 Summer 2009 Problem 9.1 (b) From: sin(4 πt ) ←→ jπδ ( ω + 4 π ) - jπδ ( ω - 4 π ) sin(50 πt ) ←→ jπδ ( ω + 50 π ) - jπδ ( ω - 50 π ) sin(4 πt ) sin(50 πt ) ←→ 1 2 π ( ) 2 [ δ ( ω + 4 π ) - δ ( ω - 4 π )] * [ δ ( ω + 50 π ) - δ ( ω - 50 π )] = - π 2 [ δ ( ω + 54 π ) - δ ( ω + 46 π ) - δ ( ω ( - 46 π ) + δ ( ω - 54 π )] . (c) From: sin(4 πt ) πt ←→ u ( ω + 4 π ) - u ( ω - 4 π )] sin(50 πt ) ←→ jπδ ( ω + 25 π ) - jπδ ( ω - 25 π ) sin(4 πt ) πt sin(50 πt ) ←→ 1 2 π ( )[ u ( ω + 4 π ) - u ( ω - 4 π )] * [ δ ( ω + 50 π ) - δ ( ω - 50 π )] = j 2 [ u ( ω + 54 π ) - δ ( ω + 46 π ) - δ ( ω ( - 24 π ) + δ ( ω - 54 π )] . Problem 9.2 (a) e - 2 t u ( t ), delayed by 3. Thus e - 2 t +6 u ( t - 3) . (b) e - 2 t u ( t ), differentiated. Thus - 2 e - 2 t u ( t ) + e - 2 t δ ( t ) = - 2 e - 2 t u ( t ) . 1

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(c) e - 2 t u ( t ), delayed and differentiated. Thus - 2 e - 2 t +6 u ( t - 3) . (d) 2 sin ω ω ←→ u ( t + 1) - u ( t - 1), and k = -∞ π 5 δ ( ω - 2 πk/ 10) ←→ k = -∞ δ ( t - 10 k ). Thus X ( ) ←→ X k = -∞ [ u ( t + 1) - u ( t - 1)] * X k = -∞ δ ( t - 10 k ) = X k = -∞ [ u ( t + 1 - 10 k ) - u ( t - 1 - 10 k )] . Problem 9.3 (a) H bp ( jw ) = [ u ( ω + ω co 2 ) - u ( ω - ω co 2 )] - [ u ( ω + ω co 1 ) - u ( ω - ω co 1 )]. Hence, h bp ( t ) = sin( ω co 2 t ) πt - sin( ω co 1 t ) πt = 2 cos( ω + ω co 1 + ω co 2 2 ) sin ω co 2 - ω co 1 2 πt Clearly, ω 0 = ω co 1 + ω co 2 2 ω co = ω co 2 - ω co 1 2 (b) see (a).
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