20082ee102_1_HW-2_SOL

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

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UCLA Department of Electrical Engineering EE102: SYSTEMS & SIGNALS 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 ) = e - t R 1 t e σ = e 1 - t - 1 t 1 0 t > 1 1

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3. (i) S is LTI and h ( t ) = δ ( t ) - 2 e - t U ( t ) , t R y ( t ) = Z -∞ x ( τ )[ δ ( t - τ ) - 2 e - ( t - τ ) U ( t - τ )] = Z t -∞ x ( τ )[ δ ( t - τ ) - 2 e - ( t - τ ) ] (ii) x ( t ) = U ( t - 1) U (4 - t ) x ( t ) = 0 if t < 1 1 if 1 t 4 0 if t > 4 (a) t < 1: y ( t ) = 0 (b) 1 t 4: y ( t ) = Z t 1 [ δ ( t - τ ) - 2 e - ( t - τ ) ] = 1 - 2 e - t ( e t - e ) = 2 e 1 - t - 1 (c) t > 4: y ( t ) = Z 4 1 [ δ (
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