20082ee102_1_HW_3_solution1

20082ee102_1_HW_3_solution1 - HW3 solution SPRING 2008 EE...

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Unformatted text preview: HW3 solution, SPRING 2008, EE 102 1. Reformulate y(t) into y ( t ) = Z ∞-∞ U ( t- τ ) x ( τ ) dτ + Z ∞-∞ U ( τ- t ) e t e- τ x ( τ ) dτ, = Z ∞-∞ [ U ( t- τ ) + U ( τ- t ) e t e- τ ] x ( τ ) dτ, Therefore by BT h ( t,τ ) = U ( t- τ ) + U (- ( t- τ )) e t- τ We obtain h ( t ) = U ( t ) + U (- t ) e t . By BT, and we introduce the notation max(a,b)= the larger of a or b. For example, max(2,5)=5. We have g ( t ) = Z ∞-∞ h ( t- τ ) U ( τ ) dτ = Z ∞-∞ [ U ( t- τ ) + U (- ( t- τ )) e t- τ ] U ( τ ) dτ, = Z ∞-∞ U ( t- τ ) U ( τ ) dτ + Z ∞-∞ U (- t + τ ) e t- τ U ( τ ) dτ, = U ( t ) Z t 1 dτ + e t Z ∞ max ( t, 0) e- τ dτ, = tU ( t ) + e t- max ( t, 0) , If t < 0, the above formula simplifies to g ( t ) = e t . If t ≥ 0, the above simplifies to g ( t ) = t + 1 . 2. IRF: h ( t,σ ) = Z t-∞ e- ( t- τ ) δ ( τ- σ ) dτ, = U ( t- σ ) e- ( t- σ ) , ⇒ h ( t ) = U ( t ) e- t 1 Let x 1 ( t ) = e- ( t- 1) U ( t- 1) , and x 2 ( 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_3_solution1 - HW3 solution SPRING 2008 EE...

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