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On Wikipedia I found the Hann-Poisson window, and the article claims the spectrum is smooth, but it isn't theoretically smooth, as it turns out. In practice you achieve partial smoothness by jacking up its $\alpha$ parameter.

Can I find what I'm looking for in a function with finite support? Is it mathematically possible?

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    $\begingroup$ i won't put this in an answer, because i am not sure of every mathematical detail. i think that the answer is "no". any window function with "finite support" means that it is something multiplied by a rectangular function, which means in the other domain the spectrum is the transform of the something convolved with a $\mathrm{sinc}()$ function. convolving with the $\mathrm{sinc}()$ will cause bumps because the the $\mathrm{sinc}$ is bumpy. $\endgroup$
    – robert bristow-johnson
    Commented May 1, 2014 at 0:54
  • $\begingroup$ What if that something is infinitely differentiable at the endpoints? Then there's no discontinuity to speak of, so perhaps there's no rectangular function to speak of. $\endgroup$
    – MackTuesday
    Commented May 1, 2014 at 1:04
  • $\begingroup$ Oh, I was supposed to call it compact support. $\endgroup$
    – MackTuesday
    Commented May 1, 2014 at 1:20
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    $\begingroup$ The only infinitely differentiable function which is zero over any non-zero width interval is a constant. $\endgroup$
    – hotpaw2
    Commented May 1, 2014 at 1:45
  • $\begingroup$ I was thinking of gluing together two of them, one reversed with respect to x, so I'd have two such endpoints at zero. Anyway, I just tried that with $y = \text{e}^{-\frac{1}{x^2}}$, continuing with $y = 0$ for $x < 0$. This doesn't work perfectly either. $\endgroup$
    – MackTuesday
    Commented May 1, 2014 at 1:56

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If you take an infinitely smooth bump, for example $e^{-\frac{1}{1-x^2}}$, its Fourier transform will be analytical - not just infinitely smooth - but, of course, it won't be with a finite support like the nonanalytical bump itself. At least it decays faster than most other windowing functions you can take. I'm not sure if this answers your question.

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That's an interesting question, I personally don't know if, in theory, that's possible or not (would be interesting to investigate). But as a practical matter perhaps you could design a Chebyshev Type II filter, take the IFFT, and use a sufficiently large number of taps. Chebyshev Type II filters have ripple in the stopband but not the passband (although that is monotonically decreasing). I'm not sure how much the truncation would influence the ripple, or if you can handle that large of a FIR. But it might get you close enough depending on your application.

This page details some of the results. http://iowahills.com/B2PolynomialFIRFilters.html

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$y = 1 - \sqrt{1 - x^2}\space\space\text{for}\space x \in [-1,1]$ under a rectangular window is rippleless. The derivative of the transform has ripple, but the transform itself does not.

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