hw3slns

# hw3slns - Homework 3 Solutions Total: 55 points 8.19 (10...

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Homework 3 Solutions Total: 55 points 8.19 (10 points) S [ k ] = Nx [( - k ) N ] = Nx [0] , k = 0 Nx [ N - k ] , k = 1 ...N - 1 To get this, start with the deﬁnition: S [ k ] = N - 1 X n =0 X [ n ] e - j 2 π N kn Substitute the analysis expression for X [ n ]: S [ k ] = N - 1 X n =0 X [ n ] e - j 2 π N kn = N - 1 X n =0 N - 1 X m =0 x [ m ] e - j 2 π N mn ! e - j 2 π N kn = N - 1 X m =0 x [ m ] N - 1 X n =0 e - j 2 π N n ( m + k ) Here, we have the sum over one period of a complex exponential, which can be written as [( m + k ) N ]. This expression will be N when ( m + k ) mod N = 0 and 0 for all other values of m . The value for m for which the expression in the sum is non-0 is thus m = 0 when k = 0 and m = N - k when k = 1 ...N - 1. We then have S [ k ] = N - 1 X m =0 x [ m ] N - 1 X n =0 e - j 2 π N n ( m + k ) = N N - 1 X m =0 x [ m ] δ [( m + k ) N ] = Nx [( - k ) N ] = Nx [0] , k = 0 Nx [ N - k ] , k = 1 ...N - 1 8.20 (10 points) Start with the deﬁnition of Y [ k ]: Y [ k ] = 2 N - 1 X n =0 y [ n ] e - j 2 π 2 N kn Since only the even samples of y [ n ] have non-0 values, we can substitute m = n/ 2 n = 2 m as the index in the sum: Y [ k ] = N - 1 X m =0 y [2 m ] e - j π N k (2 m ) = N - 1 X m =0 x [ m ] e - j 2 π N km = X [ k ] ,k = 0 ...N - 1

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This resulting expression is the DFT of x [ n ] = X [ k ], except that X [ k ] is only deﬁned for k = 0 ...N - 1. However, if we input values of k > N into the sum, by periodicity: Y [ k ] = N - 1 X m =0 x [ m ] e - j 2 π N km = N - 1 X m =0 x [ m ] e - j 2 π N ( k - N ) m = X [ k - N ] ,k = N ... 2 N - 1 Thus, the ﬁnal expression for Y [ k ] is Y [ k ] = X [ k ] , k = 0 ...N
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hw3slns - Homework 3 Solutions Total: 55 points 8.19 (10...

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