We know that the square window gives the lowest main lobe width possible, and that other windows after that trade main lobe width for side lobe height. I also understand that the main lobe width is inversely proportional to the length of the DFT aperature.

Given this, is there then a physical limit to how narrow a main-lobe can be, for a fixed aperture length? For example, can we ever hope to have main lobes that are narrower than that given by a square window, (for a fixed aperture length again), via windowing or any other method?

A second related question I have here has to do with energy conservation - is the interpretation that windows trade side lobe height for main lobe width, because energy must be conserved? Thus any reduction in side lobe height means the energy "has to go somewhere" and ends up in the main lobe? Would this be a correct interpretation?

  • $\begingroup$ I would use the term "mathematics" instead of "physics". There isn't necessarily any physical relationship implied in the Fourier transform process. $\endgroup$
    – Jason R
    Sep 24, 2013 at 16:54
  • $\begingroup$ @JasonR Yes, but in terms of the physics involves, (conservation of energy), there might be. That is what I would like to ascertain. (Second paragraph). $\endgroup$ Sep 24, 2013 at 17:42
  • $\begingroup$ @TheGrapeBeyond: You seem to be referring to Parseval's theorem, which specifies that the unitary Fourier transform is energy-preserving. The properties of the Fourier transform are independent of any physical phenomena, so there will be no required tie-in to any conservation laws in physics. $\endgroup$
    – Jason R
    Sep 24, 2013 at 18:14
  • $\begingroup$ @JasonR Thus I take it that by extension, the trade where sidelobes are minimized at the expense of an expansion of the main lobe has absolutely nothing to do with preserving energy? $\endgroup$ Sep 24, 2013 at 18:23
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    $\begingroup$ The time-bandwidth product of signals (product of the rms time duration and the rms or Gabor bandwidth) has a lower bound which all signals must satisfy. Only Gaussian pulses e.g. $x(t) =e^{-\pi t^2}, -\infty < t < \infty$ meet this bound exactly; all others have larger time-bandwidth products. This is akin to the Heisenberg uncertainty principle in quantum mechanics (physics again, JasonR!) and in fact, the calculations are mathematically identical in the two cases. $\endgroup$ Sep 24, 2013 at 19:29

1 Answer 1


A non-rectangular bell-shaped window reduces information magnitude at the edges of the window. This either is, or acts like, information loss. Thus the main lobe gets fatter to indicate a greater uncertainty with respect to its peak due to information loss. (relative to the noise floor, etc. etc.)

The concepts of information theory seem to be used in both physics and signal analysis.

The side lobes get lower because the high frequency energy related to producing any discontinuities between the FFT's 2 sides using circular basis vectors is reduced by a bell shaped window.

The two effects are similarly opposite in term of energy trade-off, but not exactly opposite, as different windows have differing areas under the curve, thus producing another degree-of-freedom in the trade-off between main lobe and side lobe energy, even given Parseval's theorem.

The width of the main lobe is inversely related to the length of the information and S/N of the information fed to the FFT. In absolutely zero noise, just 3 or 4 non-aliased points can be used to calculate the exact frequency of a single pure sinusoid in the time domain (e.g. equivalent to an interpolated main lobe width of zero in the frequency domain). Additional information outside the DFT data can also be used to estimate a narrower main lobe: For instance, many FFT post-processing frequency estimation methods assume a-priori that the data is stationary outside the widow and thus make use of the results from other offset windows, etc.

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    $\begingroup$ What do you mean by 'information magnitude'? Also what do you mean by 'either is, or acts like, information loss'? What information is lost? $\endgroup$
    – niaren
    Sep 25, 2013 at 8:08
  • $\begingroup$ Multiplying a sample by zero (or by a number small enough in relation to the numerical quantization or stationary noise floor) renders results that do not distinguish different inputs. $\endgroup$
    – hotpaw2
    Sep 25, 2013 at 13:31

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