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HW_Set_5_Solutions

# HW_Set_5_Solutions - Introduction to Signal Processing...

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Introduction to Signal Processing Hayder Radha - 1 - ECE 366 Homework Set #5 Solutions ECE 366 – Introduction to Signal Processing Spring 2009 Michigan State University Department of Electrical and Computer Engineering [1] Problem 6.13-10 (a) From Exercise E6.1a in the book, we have: ( ) ( ) 2 2 1 1 ( 1) 4 cos 3 n n x t n t n π π = = + . We also have, ( ) 2 x t t = for 1 1 t < < + with a period of 0 2 T = and hence a fundamental frequency 0 ω π = . We know that this is a power signal with a power x P that can be evaluated using the standard formula for average power. For a periodic signal, we have: ( ) 0 2 0 1 x T P x t dt T = . Parseval’s theorem states that the signal power x P can be evaluated using the Fourier series amplitudes: 2 2 0 1 1 2 x n n P C C = = + . Here, we need to show that both of the above two expressions for x P lead to the same value.

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Introduction to Signal Processing Hayder Radha - 2 - ECE 366 Now, using the standard power expression: ( ) ( ) 0 1 2 2 2 0 1 1 1 1 2 5 x T P x t dt t dt T + = = = . Now, using the second expression for x P : 2 2 2 2 2 0 2 2 1 1 1 ( 1) 1 1 4 2 3 2 n x n n n P C C n π = = = + = + . Hence, we have: 2 2 4 2 4 4 1 1 1 ( 1) 1 1 16 1 8 3 2 9 n x n n P n n π π = = = + = + . Now using the expression given by the problem statement regarding the summation of the infinite series: 4 4 1 1 90 n n π = = , leads to: 4 4 4 4 1 1 1 8 1 8 10 8 9 9 90 90 x n P n π π π = + = + = + = Hence, the power evaluation based on the Fourier series terms leads to: 18 1 90 5 x P = = .
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HW_Set_5_Solutions - Introduction to Signal Processing...

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