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I Extracting sinusoids from FFT

  1. Oct 26, 2017 #1

    Thank you for taking time to read my post.

    Background: I have a accelerometer project that I am playing with that gives me the acceleration of the object. I can plot this data and it looks very nice. I want to integrate this to get the velocity and then integrate it again to get the position over time.

    Question: is there a way to extract the individual sinusoids out of a fast Fourier transform of a discrete set of data points?
    I understand that the FFT of the set of data points gives me the amplitude of each sinusoids for its respective frequency. But how can I extract the Fourier series representation of my set of data points?
  2. jcsd
  3. Oct 26, 2017 #2


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    A FFT will give you a complex value for each harmonic in your set of discrete data points.

    Say that the FFT value of the 4. harmonic is ( 0.3 + 0.5i ), you may interprete it as 0.3*cos(4ωt) + 0.5*sin(4ωt)

    . . . if I understand you correct.
  4. Oct 26, 2017 #3
    Works perfectly! I can't believe that it was that straight forward. You have been very helpful. Here is my python code as a contribution to the community (this is obviously for a dummy signal):

    Code (Python):

    import numpy as n
    import matplotlib.pyplot as plt
    from scipy.fftpack import fft, ifft, fftfreq

    #create our dummy signal
    xaxis = n.linspace(0,2*n.pi,num=1000)
    for i in range(0,xaxis.size):
        x.append( n.exp(-xaxis[i])*n.sin((xaxis[i])))

    #take the fft of the dummy signal
    xfft = fft(x)

    #find out what the 10 main frequencies are and add them to a list
    main =[]
    tempMain = 0;
    for i in range(0,999):
        for j in range(-10,10):
            a_ = n.real(xfft[j])
            b_ = n.imag(xfft[j])
            w_ = 1 #really w=2pi*f, f=1/T, T=2pi therefore w=1
            n_ = 1000
            tempMain += (a_*n.cos(j*w_*xaxis[i])+b_*n.sin(j*w_*xaxis[i]))/n_
        tempMain = 0

    #this will reverse the order of the list. FFT will cause the data to be mirrored.

    #plot the beauties. blue is the original, main is the fit
    plt.plot(main, 'r')


    Thanks again!
  5. Oct 26, 2017 #4


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    Well, I have recently done some experiments, transforming some shapes like the letters 'E' and 'F'.

    Then I calculate the transfer function from 'F' to 'E':

    FFT(H) = FFT('E') / FFT('F').

    Now, if I transform the letter 'O' and calculate:

    IFFT( FFT('O') * FFT(H) ), will I get a 'Q' ??
  6. Oct 27, 2017 #5
    will iFFT( FFT('O') * FFT(H) ), result in a backwards 9?
    I haven't ever find anything like that before. It seems interesting though.
  7. Oct 28, 2017 #6


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    You can do a lot with these transforms.

    Say you have a sattelite photo of a milititary airport and you want to know how many jet fighters of which type is parked in this airport, you can employ a lot of people with magnifying glasses to count these planes. But also you could stuff the photo into a computer, that will place silhouttes of the planes into a complex plane. Now the computer can convert edge pixels of the silhouette to complex numbers, that can be FFT-transformed.

    The 0. harmonic tells about the mean illumination in the photo. That's not interesting, so all harmonics in the transform is divided by the 0. harmonic to "standardize" the illumination.

    The 1. harmonic tells about in which direction the plane is parked. Different directions could be confusing to the computer, but if you divide all the harmonics by the 1. harmonic, so that the 1. harmonic becomes ( 1 + 0i ), all the planes will be parked in the same direction, and with the same size. In this way you have made a "stadardized" FFT of the jet plane, which you could regard as a "finger print" of the plane.

    Now the computer just have to find a matching finger print in a look up table to identify the type of plane.

    Another example is to remove blur in a photo due to linear motion of an object ( passing car, where the registration number cannot be read due to blur ).
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