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HW05f08solutions

# HW05f08solutions - Georgia Institute of Technology School...

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Georgia Institute of Technology School of Electrical and Computer Engineering ECE 2025 Fall 2008 Solutions for Problem Set #5 5.1 A signal is defined by the equation x ( t ) = cos 3 (100 πt + 0 . 25 π ) (a) Determine the fundamental frequency of this signal. x ( t ) = e jπ/ 4 e j 100 πt + e - jπ/ 4 e - j 100 πt 2 3 = 1 8 e jπ/ 4 e j 100 πt + e - jπ/ 4 e - j 100 πt e jπ/ 2 e j 200 πt + e - jπ/ 2 e - j 200 πt + 2 = 1 8 e j 3 π/ 4 e j 300 πt + e jπ/ 4 e j 100 πt + e - jπ/ 4 e - j 100 πt + e - j 3 π/ 4 e - j 300 πt + 2 e jπ/ 4 e j 100 πt + 2 e - jπ/ 4 e - j 100 πt = 1 8 3 e jπ/ 4 e j 100 πt + 3 e - jπ/ 4 e - j 100 πt + e j 3 π/ 4 e j 300 πt + e - j 3 π/ 4 e - j 300 πt = 3 4 cos(100 πt + π/ 4) + 1 4 cos(300 πt + π/ 4) f 0 = GCD(50 , 150) = 50Hz (b) Determine the Fourier series coefficients of this signal. a k = 0 k = 0 (3 / 8) e ± jπ/ 4 k = ± 1 0 k = ± 2 (1 / 8) e ± j 3 π/ 4 k = ± 3 0 | k | ≥ 4 5.2 A signal x ( t ) is periodic with period T 0 = 4. Therefore, it can be represented as a Fourier series of the form x ( t ) = X k = -∞ a k e j (2 π/ 4) kt Suppose that the Fourier series coefficients for x ( t ) are given by the integral a k = 1 4 Z 2 0 e - 3 t e - j (2 π/ 4) kt dt (a) Determine a general formula for the Fourier coefficients a k for x ( t ) above. Express your answer as complex numbers in polar form. 1

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a k = 1 4 Z 2 0 e - 3 t e - j (2 π/ 4) kt dt = 1 4 Z 2 0 e - (6+ jπk ) t/ 2 dt = - 2 4 (6 + jπk ) e - (6+ jπk ) t/ 2 2 t =0 = - 1 (12 + j 2 πk ) e - (6+ jπk ) - 1 = h p 144 + 4 π 2 k 2 e j tan - 1 (2 πk/ 12) i - 1 1 - e - (6+ jπk ) = 1 - e - 6 e - jπk 2 36 + π 2 k 2 e - j tan - 1 ( πk/ 6) = ( 1 - e - 6 2 36+ π 2 k 2 e - j tan - 1 ( πk/ 6) for k even 1+ e - 6 2 36+ π 2 k 2 e - j tan - 1 ( πk/ 6) for k odd (b) Plot the spectrum for this signal versus frequency. Label the frequency axis in Hz. Show the first
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