For a signal sampled at $f_s$, the frequency resolution (or "bin width") for an $N$ point FFT is $f_s/N$. Does this mean that the $k$th bin will contain energy from sinusoids within $0.5f_s/N$ on either side of the center frequency of that bin (ignoring spectral leakage for now)? Is this energy averaged? Is it the same as taking a $2N$ point FFT and then averaging every two bins?

  • 1
    $\begingroup$ You can't "ignore spectral leakage for now", because that is exactly the point here. $\endgroup$
    – chirlu
    Sep 9 '13 at 2:21
  • $\begingroup$ @chirlu Does a longer FFT length imply lesser leakage from $f_1$ into $f_2$? $\endgroup$
    – user5400
    Sep 9 '13 at 4:02
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    $\begingroup$ Not necessarily. See hotpaw2 answer below: the "leakage" is given by a Sinc function which in turn is determined by the FFT length. Since the Sinc function isn't monotonous, leakage at certain frequencies can actually increase with FFT length $\endgroup$
    – Hilmar
    Sep 9 '13 at 14:47
  • $\begingroup$ is N the number of samples in time series? $\endgroup$
    – user16307
    Oct 20 '15 at 14:02

The energy bin "width" of an FFT is the width of the transform of the window used (a rectangular window by default).

The transform of a rectangle the width of the FFT is a Sinc function (sin(x)/x) with a main lobe width equal to twice the DFT/FFT bin spacing, plus smaller ripples extending the full width of the FFT result (and actually circling around forever making it technically a Dirichlet function). Note that the frequency response of each bin filter overlaps with others except at exact bin centers (representing exactly periodic-in-aperture pure sinusoids).

For the default rectangular window, only about 87% of a bin's passband is within 0.5 of a bin width from the bin center frequency, which is why ignoring so-called leakage might not be a good idea. A longer FFT has a narrower main lobe bin response, with the ripples ("leakage") dying out faster. But 2 half width Sinc functions do not sum to a single Sinc, so averaging a longer FFT will not produce identical results (except for sinusoids that are exactly periodic in both FFT lengths.)

A window other than rectangular (von Hann, etc.) will have a main lobe wider that 2 bin spacings, but, in return, far smaller ripples away from the main lobe.

The DFT/FFT resolution depends on whether you are trying to resolve closely spaced peaks as clearly separate peaks (requiring around double or more the bin spacing), or estimating the frequency of a single isolated peak with a relatively high local S/N ratio (sometimes in low noise allowing a fraction of a bin spacing in resolution after high quality interpolation).


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