20082ee102_1_HW-2_SOL

20082ee102_1_HW-2_SOL - UCLA Department of Electrical...

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UCLA Department of Electrical Engineering Spring 2008 Homework # 2 Solutions 1. (i) y ( t ) = T [ x ( t )] = x (4 t ) , t R ± T [ k 1 x 1 ( t ) + k 2 x 2 ( t )] = k 1 T [ x 1 ( t )] + k 2 T [ x 2 ( t )] T is L . ± T is NC for t > 0 and C for t < 0. ± T is TV since: y ( t - a ) = x (4 t - 4 a ) T [ x ( t - a )] = x (4 t - a ) ± y ( t - a ) 6 = T [ x ( t - a )] . (ii) y ( t ) = T [ x ( t )] = Z -∞ e - t ( t - τ ) U ( τ - t ) x ( τ ) dτ, t R y ( t ) = Z t e - t ( t - τ ) x ( τ ) ± T [ k 1 x 1 ( t ) + k 2 x 2 ( t )] = k 1 T [ x 1 ( t )] + k 2 T [ x 2 ( t )] T is L . ± T is NC . ± T is TV since: y ( t - a ) = Z t - a e - ( t - a ) ( t - a - τ ) x ( τ ) T [ x ( t - a )] = Z t e - t ( t - τ ) x ( τ - a ) = Z t - a e - t ( t - a - τ ) x ( τ ) y ( t - a ) 6 = T [ x ( t - a )] . 2. y ( t ) = Z -∞ e - ( t - σ ) U ( σ - t ) U (1 - σ ) y ( t ) = ² e - t R 1 t e σ = e 1 - t - 1 t 1 0 t > 1 1
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3. (i)
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20082ee102_1_HW-2_SOL - UCLA Department of Electrical...

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