102_1_midterm_solution

# 102_1_midterm_solution - Systems and Signals Lee, Spring...

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Unformatted text preview: Systems and Signals Lee, Spring 2009-10 EE102 Midterm Exam Solutions Problem 1. Computing Fourier Transforms (30 points) The signal f ( t ) is plotted below: 2- 2 t f ( t ) 1- 4 4 6- 6 It consists of the signal u ( t ) e- t/ 2 that has been shifted, and a shifted and reversed version of the same signal. a) Find an expression for f ( t ) Solution: The signal is u ( t ) e- t/ 2 delayed by 2, which is u ( t- 2) e- ( t- 2) / 2 and a time reversed version of the same signal u ((- t )- 2) e- ((- t )- 2) / 2 = u (- ( t + 2)) e ( t +2) / 2 The result is f ( t ) = u ( t- 2) e- ( t- 2) / 2 + u (- ( t + 2)) e ( t +2) / 2 f ( t ) = u ( t- 2) e- ( t- 2) / 2 + u (- ( t + 2)) e ( t +2) / 2 1 b) Find F ( jω ), the Fourier transform of f ( t ). Make sure to simplify your answer. Solution From the Lecture notes we know that F h u ( t ) e- at i = 1 a + jω and F h u (- t ) e at i = 1 a- jω Using these, plus the shift theorem, and simplifying F h u ( t- 2) e- ( t- 2) / 2 + u (- ( t + 2)) e ( t +2) / 2 i = e- j 2 ω 1 2 + jω + e j 2 ω 1 2- jω = e- j 2 ω 1 2- jω + e j 2 ω 1 2 + jω 1 4 + ω 2 = 1 2 ( e- j 2 ω + e j 2 ω ) + jω ( e j 2 ω- e- 2 jω ) 1 4 + ω 2 = cos(2 ωt ) + jω ( j 2 sin(2 ωt )) 1 4 + ω 2 = cos(2 ωt )- 2 ω sin(2 ωt ) 1 4 + ω 2 As a check, note that this is a real and even function of ω , which is what we’d expect given that f ( t ) is real and even....
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## This note was uploaded on 01/19/2011 for the course EE 102 taught by Professor Levan during the Spring '08 term at UCLA.

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102_1_midterm_solution - Systems and Signals Lee, Spring...

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