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20082ee102_1_HW-4_SOL

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

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UCLA Department of Electrical Engineering EE102: SYSTEMS & SIGNALS 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 - τ ) cos( t - τ ) e - τ [cos( τ ) - sin( τ )] = Z t 0 e - t cos( t - τ )[cos( τ ) - sin( τ )] 3. # 3.7 s 2 Y ( s ) - y 0 (0) - sy (0) + 4 Y ( s ) = 1 s + 1 Y ( s ) = 1 ( s 2 + 4)( s + 1) + 1 s 2 + 4 = - s/ 5 s 2 + 4 + 6 / 5 s 2 + 4 + 1 / 5 s + 1 y ( t ) = - 1 5 cos(2 t ) U ( t ) + 3 5 sin(2 t ) U ( t ) + 1 5 e - t U ( t ) 4. (i) To find the IRF, the initial conditions are assumed to be zero. There- fore, H ( s ) = Y ( s ) X ( s ) = 1 s 2 + 3 s + 2 = 1 s + 1 + - 1 s + 2 h ( t ) = e - t U ( t ) - e - 2 t U ( t ) The system is LTI h ( t, σ ) = e - ( t - σ ) U ( t - σ ) - e - 2( t - σ ) U ( t - σ ) . If x ( t ) = U ( t - 2), X ( s ) = e - 2 s s 2
The system is at rest, therefore, Y ( s ) = H ( s ) X ( s ) = 1 s 2 + 3 s + 2 · 1 s e - 2 s = ( 1 / 2 s + - 1 s + 1 + 1 / 2 s + 2 ) e - 2 s y ( t ) = 1 2 U ( t - 2) - e - ( t - 2) U ( t - 2) + 1 2 e - 2( t - 2) U ( t - 2) (ii) X ( s ) = 1 ( s + 1) 2 . Therefore, Y ( s ) = y 0 (0) ( s + 1)( s + 2) + 1 ( s + 1)( s + 2) X ( s ) = s 2 + 2 s + 2 ( s + 1) 3 ( s + 2) = - 2 s + 2 - 1 ( s + 1) 3 + 1 ( s + 1) 2 + 2 s + 1 y ( t ) = - 2 e - 2 t U ( t ) - 1 2 t 2 e - t U ( t ) + t e - t U ( t ) + 2 e - t U ( t ) 5. # 3.22 (i) Y ( s ) = L{ δ ( t ) - 2 e - t U ( t ) } · L{ t 2 e - t U ( t ) } = (1 - 2 s

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

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