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Solution - EE 261 The Fourier Transform and its...

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Unformatted text preview: EE 261 The Fourier Transform and its Applications Fall 2011 Problem Set Eight Solutions 1. (20 points) A True Story : Professor Osgood and a graduate student were working on a discrete form of the sampling theorem. This included looking at the DFT of the discrete rect function f [ n ] = ( 1 , | n | N 6 ,- N 2 + 1 n <- N 6 , N 6 < n N 2 The grad student, ever eager, said Let me work this out. A short time later the student came back saying I took a particular value of N and I plotted the DFT using MATLAB (their FFT routine). Here are plots of the real part and the imaginary part. 10 8 6 4 2 2 4 6 8 10 2 2 4 6 8 k real part of F(k) 10 8 6 4 2 2 4 6 8 10 2 1 1 2 k imaginary part of F(k) 1 (a) Produce these figures. Professor Osgood said, That cant be correct. (b) Is Professor Osgood right to object? If so, what is the basis of his objection, and produce the correct plot. If not, explain why the student is correct. Solution The student executed the following MATLAB command. f = [0 0 0 0 0 0 0 1 1 1 1 1 1 1 0 0 0 0 0 0 0]; F = fftshift(fft(fftshift(f))); DFT in MATLAB is done assuming DC at the first component of the array. Therefore the command fftshift is used to alter the center of the input array. There is a fine detail here that is worth noting. When the array size is N, fftshift shifts the ceil(N/2)+1 component to be the first component of the array. Therefore, to obtain the desired result, the rect function should be centered around the ceil(N/2)+1 component. The following command will give F without any imaginary compo- nent. f = [0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 0 0 0 0 0 0 0]; F = fftshift(fft(fftshift(f))); Heres the correct plot: 2 10 8 6 4 2 2 4 6 8 10 2 2 4 6 8 k real part of F(k) 10 8 6 4 2 2 4 6 8 10 2 1 1 2 k imaginary part of F(k) 3 2. (15 points) DFTs frequency response and spectral leakage Compute the DFT of a discrete signal of frequency x/N : f [ k ] = e 2 ixk/N i.e., find a closed form expression for F f . This is the frequency response of the DFT. Qual- itatively, how is the spectrum different when x is an integer and when x is not an integer? Plot the magnitude of the frequency response for x = 2 . 5 and N = 8. Does the plot agree with what you expect? Solutions: The DFT is calculated by F f [ n ] = N- 1 X k =0 f [ k ] - kn = N- 1 X k =0 e 2 ixk/N e- 2 ink/N = N- 1 X k =0 e 2 i ( x- n ) k/N = 1- e 2 i ( x- n ) 1- e 2 i ( x- n ) /N = e 2 i ( x- n ) / 2 e 2 i ( x- n ) / 2 N e- 2 i ( x- n ) / 2- e 2 i ( x- n ) / 2 e- 2 i ( x- n ) / 2 N- e 2 i ( x- n ) / 2 N = e 2 i ( x- n ) / 2(1- 1 /N ) sin(2 ( x- n ) / 2) sin(2 ( x- n ) / 2 N ) The magnitude of the frequency response is |F f [ n ] | = sin(2 ( x- n ) / 2) sin(2 ( x- n ) / 2 N ) When x is an integer, the DFT is a delta function, agreeing with what we saw in class. But if x is not an integer, we see something like: 1 2 3 4 5 6 7 1...
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Solution - EE 261 The Fourier Transform and its...

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