102_1_midterm_solution - Systems and Signals EE102 Lee...

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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
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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
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Problem 2. Properties of convolution (20 Points) Determine whether the assertions are true or false, and provide a supporting argument.
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