102_1_midterm_solution - Systems and Signals EE102 Lee...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
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: 0 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
Image of page 1

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

View Full Document Right Arrow Icon
b) Find F ( j ω ), the Fourier transform of f ( t ). Make sure to simplify your answer. Solution From the Lecture notes we know that F u ( t ) e - at = 1 a + j ω and F u ( - t ) e at = 1 a - j ω Using these, plus the shift theorem, and simplifying F u ( t - 2) e - ( t - 2) / 2 + u ( - ( t + 2)) e ( t +2) / 2 = e - j 2 ω 1 2 + j ω + e j 2 ω 1 2 - j ω = e - j 2 ω 1 2 - j ω + e j 2 ω 1 2 + j ω 1 4 + ω 2 = 1 2 ( e - j 2 ω + e j 2 ω ) + j ω ( e j 2 ω - e - 2 j ω ) 1 4 + ω 2 = cos(2 ω t ) + j ω ( j 2 sin(2 ω t )) 1 4 + ω 2 = cos(2 ω t ) - 2 ω sin(2 ω t ) 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. F ( j ω ) = cos(2 ω t ) - 2 ω sin(2 ω t ) 1 4 + ω 2 2
Image of page 2
Problem 2. Properties of convolution (20 Points) Determine whether the assertions are true or false, and provide a supporting argument.
Image of page 3

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

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern