problem 3b

# problem 3b - ECE 101 Linear Systems Winter 2009 Problem Set...

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ECE 101 - Linear Systems, Winter 2009 Problem Set # 3 Solutions (send comments/questions to [email protected]) Problem 1 (3.27 (modiﬁed) OW2) Using the Fourier series synthesis equation (3.94), x [ n ] = a 0 + a 1 e (2 π/N ) n + a - 1 e - (2 π/N ) n + a 2 e (4 π/N ) n + a - 2 e - (4 π/N ) n = 1 + e jπ/ 2 e (2 π/ 6) n + e - jπ/ 2 e - (2 π/ 6) n + e jπ/ 4 e (4 π/ 6) n + e - jπ/ 4 e - (4 π/ 6) n = 1 + 2 cos ± 2 n + 3 6 π + 2cos ± 8 n + 3 12 π = 1 - 2sin 3 · - 2sin ± 2 3 - π 4 where we have used cos( θ ) = - sin( θ - π/ 2). Problem 2 (3.28 (a)[Fig P3.28(c)],(c) OW2) 3.28 (a)[Fig P3.28(c)] We are given x[-2:3] = [-1 2 1 2 -1 0] and N=6 . Using the analysis equation (3.95) yields a k = 1 N 3 X n = - 2 x [ n ] e - jk (2 π/N ) n = 1 6 1 + 2 e - 2 πjk/ 6 + 2 e 2 πjk/ 6 - e - 4 πjk/ 6 - e 4 πjk/ 6 · = 1 6 (1 + 4 cos( πk/ 3) - 2cos(2 πk/ 3)) 3.28 (c) 1

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We are given x[n] = 1 - sin( π n/4) and N=4 . Thus, a k = 1 N 3 X n =0 x [ n ] e - jk (2 π/N ) n = 1 4 3 X n =0 (1 - sin(
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problem 3b - ECE 101 Linear Systems Winter 2009 Problem Set...

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