Figure showing the tradeoff between aliasing and bluring on a gaussian-like filter in the frequency domain

I've seen figures in various books about the the tradeoff between aliasing and blurring when using a gaussian-like filter: the narrower it is, the more it cuts off low frequencies and thus blurs it, and the wider it is the more it leaves high frequencies and aliases. Often this is accompanied by a helpful arrows, which point to the fourier-domain box-filter (an ideal filter), overlaid with the gaussian-like filter of interest, and shades the delta them. The arrows mark the Box - Gaussian areas as "blurring" and the Gaussian - Box filter as "aliasing". (Figure shown above).

The surrounding text also warns that the more box-like the gaussian-like is made, the more ringing will occur, (so there is this kind of three way tradeoff, {ringing <=> {aliasing <=> blur}).

  • My question concerns the fact that there is no nice arrow pointing to how much ringing will occur; is there a way to get an intuition for how much ringing will occur?
    • Can one predict the amount of ringing by integrating the negative lobes in the spatial domain?
    • What effect does the large support have on ringing?

1 Answer 1


Intuitively, for a linear-phase filter the sharper the transition band of a given filter is, the more that filter will ring. Do a search on "Gibbs phenomenon".

I'm editing this. I had said "One can probably assess the likely amount of ringing by integrating the negative lobes, or by comparing the width of the response at the first zero-crossing vs. the width where the response is below some percentage of the maximum." Then I realized that there are filters that ring, but whose impulse response never goes negative. A readily accessible example of this is a piecewise triangular (in frequency) filter, i.e. $$H(\omega) = \begin{cases} 1 + \omega & -1 < \omega \le 0 \\ 1 - \omega & 0 < \omega \le 1 \\ 0 & \mathrm {otherwise} \end{cases}.$$ The time-domain impulse of such a filter is $h(t) = (\mathrm {sinc}\ \omega t)^2$ (with some scaling). Because $\mathrm {sinc}\ t$ is real-valued, it's square must be non-negative everywhere.

Large support by itself doesn't cut down on ringing. The reason that filters that go to zero quickly tend to ring is because they -- of necessity -- have sharp skirts, and because even if the filter shape itself is continuous-ish*, at some higher derivative it'll become discontinuous -- which creates ringing.

Probably the best way to assess for ringing is to just look at the filter's time-domain impulse response. If it rings -- with or without going negative -- then you'll see ringing at the edges of a filtered image. Or just filter some images that you know to be challenging, and see what the results look like.

* I'm not sure what the right meaning of "continuous" is in a discrete-time filter.

  • 1
    $\begingroup$ Is that metric continuous-ish or imperial continuous-ish? :D $\endgroup$
    – Peter K.
    Nov 11, 2022 at 2:15
  • 1
    $\begingroup$ I believe it's in cubits. So it's quantum? $\endgroup$
    – TimWescott
    Nov 11, 2022 at 2:22

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