This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Systems and Signals Lee, Spring 200910 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....
View
Full
Document
This note was uploaded on 01/19/2011 for the course EE 102 taught by Professor Levan during the Spring '08 term at UCLA.
 Spring '08
 Levan

Click to edit the document details