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Unformatted text preview: Homework 1 Solutions 1. 2. (a) Let c (1) k be the Fourier series coefficients of x * [ n ]. Then c (1) k = 1 N N 1 X n =0 x * [ n ] e j 2 π N kn Using the properties ( ab ) * = a * b * and ∑ a * = ( ∑ a ) * gives c (1) k = 1 N N 1 X n =0 x [ n ] e j 2 π N kn ! * Now, substitute m = n , which gives c (1) k = 1 N X m = ( N 1) x [ m ] e j 2 π N km * Remember that both x [ n ] and e j 2 π N kn are periodic with period N , so x [ n ] = x [ n + N ] and e j 2 π N kn = e j 2 π N k ( n + N ) . Then, if we define l = m + N , we get c (1) k = 1 N N 1 X l =0 x [ l ] e j 2 π N kl ! * = c * k (b) Let c (2) k be the Fourier series coefficients of x R [ n ]. Then, use the property that x R [ n ] = x [ n ] + x * [ n ] 2 Let c (3) k be the Fourier series coefficients of x * [ n ]. It can be shown using a very similar proof to the one given above in part (a) that c (3) k = c * k . We want to focus on the Fourier series coefficients between 0 and N 1, so using the periodicity of c k , we know that c * k = c * N k (the conjugate operator has no effect on periodicity).(the conjugate operator has no effect on periodicity)....
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This document was uploaded on 05/26/2010.
 Spring '09

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