I found the Hann–Poisson window on Wikipedia. It has no sidelobe for $\alpha \ge 2$.

I am interested to know more examples of such window function with no sidelobes and what they are useful for.

I found the function $\frac{1}{|\frac{2n}{N-1}-1|^2}$, $\ln^2|\frac{2n}{N-1}-1|$ and $\ln^2(1-\ln|\frac{2n}{N-1}-1|)$ seems to work (looks fine visually not verified with math), though they have poles at $n=\frac{N-1}{2}$, so they can only make type-II filter. This post mentions the Gaussian filter which almost works but still has some very small ripple.

So far, the patterns are functions that have a large peak at the center in the time domain.

*edit: The square of a Bartlett window also seems to work, and delta function, for obvious reasons.

  • $\begingroup$ Welcome! I wasn't clear on my first reading of your question what you were asking, so I've reformatted and slightly reworded it to be a bit clearer (for me). Feel free to update if I've got the wrong end of the stick. $\endgroup$
    – Peter K.
    Jan 9 at 21:43
  • $\begingroup$ Thanks for the edit. $\endgroup$ Jan 10 at 9:33
  • $\begingroup$ If I'm not mistaken, a Gaussian window doesn't have sidelobes for a certain range of standard deviation values. For example, if there's too much standard deviation, the window starts to be come sufficiently flat to where its Fourier transform approximates a delta. But the Fourier transform of a Gaussian is a Gaussian, and there's a range of standard deviations I believe where the no sidelobe condition is met. $\endgroup$
    – Baddioes
    Jan 10 at 17:34
  • $\begingroup$ More generally, if you have any function that is monotonically decreasing outward from the origin in both directions, and then you inverse Fourier transform this function, I believe you would end up with a window that has no sidelobes. Whether or not this is a good window probably depends on more factors, though. $\endgroup$
    – Baddioes
    Jan 10 at 17:38

1 Answer 1


A Gaussian window, theoretically, has no sidelobes because the Fourier Transform of a Gaussian in the time domain is another Gaussian in the frequency domain. That function is smooth and ripple-free.

But a Gaussian function goes on forever and never quite gets to zero, but it gets close to zero very quickly, so we can truncate the Gaussian to zero at some point where it is close enough (perhaps $10^{-7}$) and then you have a window function that is finite in length. But that truncation will cause tiny ripples.

In the comments to this answer regarding monotonic soft clipping polynomials there is some discussion of filterbanks and the maxflat half-band filter. If a cosine goes into $f(x)$, then a finite number of cosines will come out and you can derive an FIR low-pass filter which will have this smooth frequency response with no ripples. It's possible that you might create an FIR LPF with lower cutoff than half Nyquist. But the repeated images of the original half band will become tiny ripples or sidelobes. So it's pretty smooth but not completely ripple free.

  • $\begingroup$ Thanks for your answer. Polynomials are a lot easier to handle compared to directly analysing Fourier transforms. Changing the polynomial to $f(x) = K \int_{-1}^{x} {(u+1)}^N \,du$ can makes the mainlobe slimmer. $\endgroup$ Jan 11 at 17:28

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