AnsClassEx2 - EE3118 Linear System and Signal Analysis Q1...

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Unformatted text preview: EE3118 Linear System and Signal Analysis Q1 (a) ω0 = Answer of Class Exercise 2 π 2 (b) Consider the period from 0 to 4s, 2 − t 0 ≤ t < 2 x(t ) = 2≤t <4 0 24 an = ∫ x(t ) cos(nω0t )dt 40 12 = ∫ (2 − t ) cos(0.5nπt )dt 20 12 sin(0.5nπt ) = − 2 ∫0 t cos(0.5nπt )dt 0.5nπ 0 1 t sin(0.5nπt ) cos(0.5nπt ) =0− + 2 0.5nπ (0.5nπ ) 2 0 2 2 [1 − cos(nπ )] (nπ ) 2 2 1 − (−1) n = (nπ ) 2 = 2 [ (c) If the frequency component is larger than 0.8π, H (ω ) = 0. We only need to consider the d.c. term and the term with n=1. For n=1, or ω = 0.5π , H (ω ) = 3.75, ∠H (ω ) = 0.5π . Therefore, 1 4 2 y (t ) = × 10 + 2 × 3.75 × cos(0.5πt + 0.5π ) + × 3.75 × sin(0.5πt + 0.5π ) 2 π π 15 7.5 = 5 + 2 cos(0.5πt + 0.5π ) + sin(0.5πt + 0.5π ) π π Q2. F [sgn (t )] = 2 jω ⇒ F [− j sgn (t )] = − j2 − 2 = jω ω By duality, F [Z (t )] = F [− j sgn (t )] = 2π z (− ω ) , Hence, 2π z (− ω ) = −2 ω ⇒ z (t ) = 1 πt (replacing symbol ω by t) ...
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This note was uploaded on 01/11/2011 for the course EE 3118 taught by Professor Kitsangtsang during the Spring '08 term at City University of Hong Kong.

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