PS-7-2007-Solutions

PS-7-2007-Solutions - EE 261 The Fourier Transform and its...

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EE 261 The Fourier Transform and its Applications Fall 2007 Problem Set Seven Solutions 1. (15 points) DFT basics. (a) Prove the shift theorem for the discrete Fourier transform: F ( τ p f )= ω p F f . where τ p f [ n ]=f [ n p ] . (b) Replication. Suppose that the signal f =( f [0] , f [1] ,..., f [ N 1]), has discrete Fourier transform F . W ecreateanews igna lg [ n ], n =0 , 1 2 N 1w ith twice the number of points defined by, g [ n ]= = ± f [ n ] ,n , 1 ,...N 1 f [ n N ] = N,N +1 ,... 2 N 1 Find the DFT of g in terms of F . (c) Zero-Padding. Consider a vector of N samples, x =(x [0] , x [1] x [ N 1]). We augment this vector by appending M zeros to the end of it to form a new signal ˜x of length N + M . Express ˜ X = F ˜x and X = F x in terms of samples of a continuous Fourier transform and compare the two. Why might we want to ‘zero-pad’? Solution: (a) For the DFT of the shift τ p f we have F ( τ p f N 1 ² n =0 τ p f [ n ] ω n , = N 1 ² n =0 f [ n p ] ω n . 1
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Now change the index, letting k = n p : F ( τ p f )= N 1 p ± k = p f [ k ] ω ( k + p ) = N 1 p ± k = p f [ k ] ω k ω p = ω p N 1 p ± k = p f [ k ] ω k = ω p F f because when calculating the DFT we can sum over any complete period of the input. (b) Remember that to take the N -po intDFToff forces us to think of f as periodic of period N . Forming the replicated signal g isthesamea slook inga tf as a periodic signal of period 2 N : ... g ² ³´ µ f [0] , f [1] ,..., f [ N 1] ´ µ² ³ f , f [ N ]=f [0] , f [ N +1]=f [1] f [2 N 1] = f [ N 1] ´ µ² ³ f Thus to compute the DFT of g we regard f as periodic of period 2 N and find its 2 N - point DFT. We indicate the orders of the DFT’s involved with a subscript N or 2 N .W e have, for m =0 , 1 2 N 1, F 2 N g [ m ]= 2 N 1 ± k =0 f [ k ] e 2 πimk/ 2 N . We first break this up into sums over the two halves, from 0 to N 1andfrom N 1 to 2 N 1: F 2 N g [ m N 1 ± k =0 f [ k ] e 2 πimk/ 2 N + 2 N 1 ± k = N f [ k ] e 2 πimk/ 2 N . To bring in the N -point DFT of f we have to get an N in the denominator of the complex exponential, not a 2 N . For this we distinguish the cases when m is even or odd. Suppose first that m is even, say m =2 ± .H e r ew ecanl e t ± range from 0 to N 1 and we’ll get all the even indices from 0 to 2 N 1. We have F 2 N g [2 ± N 1 ± k =0 f [ k ] e 2 πi 2 ±k/ 2 N + 2 N 1 ± k = N f [ k ] e 2 2 ±k/ 2 N = N 1 ± k =0 f [ k ] e 2 πi±k/N + 2 N 1 ± k = N f [ k ] e 2 πi±k/N = F N f [ ± ]+ F N f [ ± ] (the second sum also gives F f [ ± ] because any N -consecutive indices are OK) F N f [ ± ] 2
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Next suppose that m is odd. This time we can write m =2 ± + 1 where again ± ranges from 0 to N 1. For the DFT of g we have F 2 N g [2 ± +1]= N 1 ± k =0 f [ k ] e 2 πi (2 ± +1) k/ 2 N + 2 N 1 ± k = N f [ k ] e 2 (2 ± +1) 2 N = N 1 ± k =0 f [ k ] e 2 πi±k/N e 2 πik/ 2 N + 2 N 1 ± k = N f [ k ] e 2 πi±k/N e 2 πik/ 2 N = N 1 ± k =0 f [ k ] e 2 πi±k/N e πik/N + 2 N 1 ± k = N f [ k ] e 2 πi±k/N e πik/N But if we rewrite the second sum (using periodicity of f ), 2 N 1 ± k = N f [ k ] e 2
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PS-7-2007-Solutions - EE 261 The Fourier Transform and its...

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