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Fourier transform frequency resolution

  1. Mar 5, 2013 #1

    Wox

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    If I have a signal, sampled at [itex]N[/itex] data points with a time-interval of [itex]T[/itex], does this restrict the frequency resolution I can obtain in Fourier space?

    I understand that from the Nyquist-Shannon sampling theorem it follows that all information on the Fourier transform of a [itex]T[/itex]-sampled signal is mapped in the frequency interval [itex][-\frac{1}{2T},\frac{1}{2T}][/itex]. Hence the sampling rate restricts the maximal frequency that can be measured (if there are higher frequencies, you get aliasing effects).

    However it is not clear to me how the frequency resolution is limited by [itex]N[/itex] and/or [itex]T[/itex]. Typical FFT algorithms return as many data points in Fourier space as there are data points in real space. Taking into account the [itex][-\frac{1}{2T},\frac{1}{2T}][/itex] limits it follows that the frequency resolution is in this case [itex]\frac{1}{NT}[/itex]. But is there a fundamental reason for having as many data points in Fourier space as there are data points in real space, or is this just convenient to implement?
     
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  3. Mar 5, 2013 #2

    olivermsun

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    The Fourier basis functions are just sinusoids which are periodic on time T; thus you can have exactly 1 period in T, 2 periods in T, etc. (up to the Nyquist frequency).
     
  4. Mar 6, 2013 #3

    Wox

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    So if I understand correctly, if I define the following:
    • [itex]N[/itex] are the number of sampling points
    • [itex]T[/itex] the time between two sampling points
    • [itex]NT[/itex] the total time sampled
    then
    • A signal with one cycle in the sampled time span [itex]NT[/itex] has frequency [itex]\nu_{1}=\frac{1}{NT}[/itex]
    • A signal with two cycles in the sampled time span [itex]NT[/itex] has frequency [itex]\nu_{2}=\frac{2}{NT}[/itex]
    • ...
    • A signal with [itex]\frac{N}{2}[/itex] cycles in the sampled time span [itex]NT[/itex] has frequency [itex]\nu_{\frac{N}{2}}=\frac{1}{2T}[/itex] (which is the maximal frequency).
    So far so good, but I fail to see why one should need to consider only these frequencies. Why not any other frequency between [itex]0[/itex] and [itex]\frac{1}{2T}[/itex]? So why not for example [itex]\nu_{1.5}=\frac{1.5}{NT}[/itex]?
     
  5. Mar 6, 2013 #4

    AlephZero

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    Because the signal values of any other frequency, at the ##NT## sampling points, can be represented exactly as a linear combination of the the Fourier basis functions.

    The important point is when you sample a continous signal, first you are only looking at a finite time-segment ##NT## of the signal, and the Fourier transform assumes the rest of the signal is periodic with period ##NT## (which "in real life" it would not be, for a signal with frequency ##\frac{1.5}{NT}##). And second, you are assuming the signal is bandwith limited to the Nyquist frequency.

    Those two assumptions mean the sampled signal usually contains MUCH less "information" than the continuous signal before sampling. That's why the sampled signal can be represented exactly by the "small" amount of information in a finite number of Fourier coefficients.
     
  6. Mar 6, 2013 #5

    olivermsun

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    In general you will need at least N degrees of freedom to represent N data points (think of the example of fitting N data points with polynomials of degree N-1). Sines and cosines with frequencies 1, 2, ..., N/2 cycles per period are convenient because they are orthogonal to one another, so you know that N of them they are "enough" to represent arbitrary data at N points. On the other hand, if you have "too many" degrees of freedom (e.g., too many frequencies) then the fits are no longer unique.
     
  7. Mar 7, 2013 #6

    Wox

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    Thanks! Great answers!
     
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