Sampling a signal and do the discrete Fourier transform

  1. When I sample a certain digital signal with increasing sampling frequency, the fast Fourier transform of the sampled signal becomes finer and finer. (the image follows) Previously I thought higher sampling frequency makes the sampled signal more similar to the original one, so the Fourier transform of a signal sampled at very high frequency would be the same as the FT of the original signal. But in fact the FT of the sampled signal is much narrower.

    How to explain this phenomenon? As the FT is different in different sampling conditions, why the original signal can still be correctly reconstructed?

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  2. jcsd
  3. AlephZero

    AlephZero 7,300
    Science Advisor
    Homework Helper

    If you doubled the sampling frequency, the frequency range of the DTFT doubles (because the Nyquist frequency is double).

    The X axis of your CTFT plot is in rad/sec. The X axis of your DTFT plot is labelled in rad, which is wrong, because "radians" are not a unit for frequency. If you convert the DTFT scales into rad/sec, the plots will look the same - except that the second one covers twice the frequency range of the first one.

    You can extend the CTFT plot to cover any frequency range you like, of course.
    Last edited: Jul 26, 2014
  4. DrGreg

    DrGreg 1,915
    Science Advisor
    Gold Member

    I wouldn't say this is "wrong", but it is what's causing the confusion. The DTFTs are plotted against normalised (angular) frequency instead of (angular) frequency. When plotted against (angular) frequency, the confusion should be removed.
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