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Fourier Transform

  1. Oct 17, 2015 #1
    In a book the Fourier transform is defined like this. Let g(t) be a nonperiodic deterministic signal... and then the integrals are presented.

    So, I understand that the signal must be deterministic and not random. But why it has to be nonperiodic (aperiodic).
    The sin function is periodic and we can calculate its Fourier transform.

    Is it because a nonperiodic signal is absolutely integrable?

    And with the sin function. Yes, I can calculate. But deltas appear.

    This is the answer?
  2. jcsd
  3. Oct 17, 2015 #2


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    Fourier transform is a continuous form of Fourier series. In computing Fourier series, the signal in time domain must be periodic, i.e.it has finite period, and that you find that the spectrum contains combs separated by a fixed value which is reciprocal to the signal's period. So, the longer the period, the closer the frequency combs are to its neighbors. When the signal is not periodic, we can suppose that its period is infinitely long, therefore the corresponding frequency combs is separated by infinitesimal distance, which leads to a continuous spectrum.
  4. Oct 17, 2015 #3

    But in computing the Fourier transform of a signal, that signal must be absolutely necessary nonperiodic?
  5. Oct 17, 2015 #4


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    The use of the term non-periodic generalize the applicability of Fourier transformation to any integrable functions, be it periodic or non-periodic.
  6. Oct 17, 2015 #5
    Aa, ok. Thanks for the answer.

    I think this is a problem with today's books. They are not written in a more Euclidean way.
  7. Oct 21, 2015 #6
    Also. When deriving the Fourier transform from the Fourier series, we have a finite-length signal and repeat it multiple times over the time axis. And then expand it into a Fourier series. And then calculations. And then we get the Fourier transform.

    So the Fourier transform is for finite-length signals.

    The fact that we can calculate Fourier transforms for periodic signals or signals like the unit step is because we involve deltas functions there.

    But what about the decaying exponential? Its Fourier transform does not involve deltas and it is not of finite length.
    How can this decaying exponential be viewed as a finite-length signal that gets repeated multiple times over the time axis. Its period is infinite.

    So my question. How does one attach a Fourier transform to such decaying exponential? It could be the fact that for such signals we actually have other derivation of the Fourier transform but we haven't found it yet?
    We derived the Fourier transform for finite-length signals and with it we just calculated the Fourier transform for decaying exponential?
    Or one can think of having multiple infinities into one infinite. So we have that notion that some infinities are bigger than others?
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