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Calculating FFT for discrete values

  1. Jan 27, 2015 #1
    1. The problem statement, all variables and given/known data
    We have a set of values: f(n)=f(0,1,2)=(1,3,2)

    so f(0)=1, f(1)=3, f(2)=2, where n=0,1,2


    the number of values N=3

    The question is to calculate the FFT of this signal, the fourier spectrum the power spectrum and phase spectrum.

    I'm not sure concerning FFT. And also about the Fourier Spectrum what should I use to conduct it? the DFT or FFT?

    2. Relevant equations

    The calculation of DFT for these values.

    3. The attempt at a solution

    I personally calculated the Discrete Fourier Transform in this way:

    $$ F[n]=\sum_{k=0}^{N-1}f(k)e^{-j \frac{2\pi}{N}nk} $$

    so I had a set of values: F(0),F(1),F(2)
     
  2. jcsd
  3. Jan 29, 2015 #2

    RUber

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    Homework Helper

    For only 3 points, it seems unnecessary to try to do the FFT. However, the point of the FFT is to take advantage of symmetry.
    F(1) = f(0) +f(1) e^{-j 2pi/N * 1 }+f(2) e^{-j 2pi/N * 2 }
    F(2) = f(0) +f(1) e^{-j 2pi/N * 2 }+f(2) e^{-j 2pi/N * 4 }= f(0) +f(1) e^{-j 2pi/N * 2 }+f(2) e^{-j 2pi/N * 1 }
    So whay you really have is (for real-valued inputs):
    ## F(0) = \pmatrix{ f(0), f(1), f(2) } \pmatrix{ 1\\1\\1} , ##
    ## F(1) = \pmatrix{ f(0), f(1), f(2) } \pmatrix{ 1\\ e^{-j 2\pi/N }\\ e^{ j 2\pi/N }}##
    ## F(2) = \pmatrix{ f(0), f(1), f(2) } \pmatrix{ 1\\ e^{j 2\pi/N }\\ e^{- j 2\pi/N }} = F(1) ^* ##
    As I said, for 3 values, this is more a demonstration of the concept that F(N-k) = F(k)*, which should save you about 1 calculation by just replacing F(1) with its conjugate for F(2).
     
  4. Jan 29, 2015 #3
    That is a very good and summarized answer. Thank you RUber :)
     
  5. Jan 29, 2015 #4
    But I still don't understand how e^(-j*2*pi/N *2)= e^(j*2*pi/N). Are these two angles identical??
     
  6. Jan 29, 2015 #5

    RUber

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    Homework Helper

    This is the symmetry of the function. ##e^{ix}## is 2pi periodic, so ##e^{ix}=e^{ix+i 2\pi}##
    ##e^{-j2\pi/N 2+j 2\pi}= e^{-j4\pi/3 +j6\pi/3}=e^{j*2*pi/N}.##
     
  7. Jan 29, 2015 #6
    OK, thank you again.
     
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