20082ee102_1_HW-4_SOL

# 20082ee102_1_HW-4_SOL - UCLA Department of Electrical...

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UCLA Department of Electrical Engineering Spring 2008 Homework # 4 Solutions 1. f ( t ) = Z t 0 1 2 sin[2( t - τ )] = 1 2 Z -∞ sin[2( t - τ )] U ( t - τ ) U ( τ ) = 1 2 sin(2 t ) U ( t ) * U ( t ) F ( s ) = 1 2 L{ sin(2 t ) U ( t ) } · L{ U ( t ) } = 1 2 · 2 s 2 + 4 · 1 s = 1 s ( s 2 + 4) 2. (i) ( i ) L{ tf ( t ) } = ( - 1) d ds F ( s ) where F ( s ) , L{ f ( t ) } ⇒ L{ Z t 0 τf ( τ ) } = 1 s · [ - d ds F ( s )] ( ii ) L{ f ( t ) } = ( - 1) 2 d 2 ds 2 F ( s ) = d 2 ds 2 F ( s ) ⇒ L{ Z t 0 τ 2 f ( τ ) } = 1 s · [ d 2 ds 2 F ( s )] (ii) F ( s ) = , F 1 ( s ) z }| { s + 1 ( s + 1) 2 + 1 · , F 2 ( s ) z }| { s ( s + 1) 2 + 1 = F 1 ( s ) · F 2 ( s ) f ( t ) = f 1 ( t ) * f 2 ( t ) = Z -∞ f 1 ( t - τ ) f 2 ( τ ) 1

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Where, f 1 ( t ) = L - 1 { F 1 ( s ) } = e - t cos( t ) U ( t ) And f 2 ( t ) = L - 1 { F 2 ( s ) } = L - 1 { s + 1 ( s + 1) 2 + 1 - 1 ( s + 1) 2 + 1 } = e - t [cos( t ) - sin( t )] U ( t ) Therefore, f ( t ) = Z t 0 e - ( t
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## This note was uploaded on 04/13/2010 for the course EE ee102 taught by Professor Levan during the Spring '09 term at UCLA.

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20082ee102_1_HW-4_SOL - UCLA Department of Electrical...

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