20082ee102_1_HW-2_SOL

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

This preview shows pages 1–3. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
3. (i)
This is the end of the preview. Sign up to access the rest of the document.

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online