lab_solution4

lab_solution4 - ECE 101 Lab #4 Solutions Introduction: The...

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ECE 101 – Lab #4 Solutions Introduction: The assignment is to complete the Basic Problems (a)-(c) and the Intermediate Problems (d)-(h) of the textbook. Results: Problem 3.5 (a) – Based on the DTFS coefficients and Table 3.2 (page 221) in [OW2], which states that conjugate symmetry for real signals gives a k = a* -k , x[n] has to be a purely real signal. Problem 3.5 (b) – a 0 = 1 a -1 = a -4 = 2 a 2 = a 3 = a -2 = a 4 = 2 Problem 3.5 (c) – The prediction in part (a) was accurate. Problem 3.5 (e) 1 1 0 0 [] N N n ax n = = Invoking the above formula, we get: a1(1) = 1/64 x (64) = 1 a2(1) = 1/64 x (8x4) = 0.5 a3(1) = 1/64 x (8x2) = 0.25 Hence, my predicted values tally with those obtained with MATLAB.
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Problem 3.5 (g) – By conjugate symmetry for x3_all[n], a k = a* -k for k = 1:15. a 0 and a 16 which are the only 2 remaining terms are real valued terms as well, hence proving that x3_all[n] must be a real signal. M-files: Lab4_P3_5_b.m % Problem 3.5 (b): a = ones(1,5); a(2) = 2*exp((pi/3)*(-j)); a(3) = exp((pi/4)*(j)); a(4) = exp((pi/4)*(-j)); a(5) = 2*exp((pi/3)*(j)); Lab4_P3_5_c.m % Problem 3.5 (c): for n = 1:5 x(n) = 0; for k = 1:5 x(n) = x(n) +
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lab_solution4 - ECE 101 Lab #4 Solutions Introduction: The...

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