PS-8-2009-Solutions

# PS-8-2009-Solutions - EE 261 The Fourier Transform and its...

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Unformatted text preview: EE 261 The Fourier Transform and its Applications Fall 2009 Problem Set Eight Due Wednesday, November 18 1. (10 points) Different definitions for the DFT This is an alternate version, in one respect, to Section 6.9 in the notes, on different definitions of the DFT. The purpose is to get a precise understanding of the statement: “The DFT can be defined over any set of N consecutive indices.” This statement is behind the different indexing conventions that are in use for the DFT. For periodic functions f ( t ) in the continuous setting, say functions of period 1, the n ’th Fourier coefficient is ˆ f ( n ) = integraldisplay 1 e- 2 πint f ( t ) dt. The periodicity of f ( t ) implies that ˆ f ( n ) can be obtained by integrating over any interval of length 1, meaning ˆ f ( n ) = integraldisplay a +1 a e- 2 πint f ( t ) dt for any a . Morally, we’re looking for the discrete version of this. Morally, we’ll give the same argument, which goes: Take the derivative with respect to a : d da integraldisplay a +1 a e- 2 πint f ( t ) dt = e- 2 πi ( a +1) n f ( a + 1) − e- 2 πian f ( a ) = e- 2 πian e- 2 πin f ( a + 1) − e- 2 πian f ( a ) = e- 2 πian f ( a ) − e- 2 πian f ( a ) , using e 2 πin = 1 and the periodicity of f ( t ). Since the derivative is identically zero the value of the integral is the same for any value of a . In particular, taking a = 0, integraldisplay a +1 a e- 2 πint f ( t ) dt = integraldisplay 1 e- 2 πint f ( t ) dt = ˆ f ( n ) . Back to the discrete case. We start over, so to speak, by giving a more general definition of the DFT, one that self-consciously depends on the indexing: 1 We consider discrete signals f : Z −→ C that are periodic of period N . Let P and Q be index sets of N consecutive integers, say P = [ p : p + N − 1] , Q = [ q : q + N − 1] . The DFT based on P and Q is defined by G f [ m ] = summationdisplay k ∈P f [ k ] e- 2 πimk/N = p + N- 1 summationdisplay k = p f [ k ] e- 2 πimk/N , m ∈ Q . We’ve called the transform G to distinguish it from F , which in the present setup corresponds to the special choice of index sets P = Q = [0 : N − 1]. Since f is periodic of period N , knowing f on any set of N consecutive numbers determines it everywhere. We hope the same will be true of a transform of f , but this is what must be shown. Thus one wants to establish that the definition of G is independent of P and Q . This is a sharper version of the informal statement in the first quote –The DFT can be defined over any set of N consecutive indices – but we have to say what “independent of P and Q ” means. First Q . Allowing for a general set Q of N consecutive integers to index the values G f [ m ] doesn’t really come up in applications and is included in the definition for the sake of gener- ality. We can dispose of the question quickly. To show that the transform is independent of Q we: (a) Extend G f to be periodic of period N ; thus G f [ m ] is defined, by periodicity, for all...
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PS-8-2009-Solutions - EE 261 The Fourier Transform and its...

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