# Window functions with rippleless spectra

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?

• 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. – robert bristow-johnson May 1 '14 at 0:54
• 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. – MackTuesday May 1 '14 at 1:04
• Oh, I was supposed to call it compact support. – MackTuesday May 1 '14 at 1:20
• The only infinitely differentiable function which is zero over any non-zero width interval is a constant. – hotpaw2 May 1 '14 at 1:45
• 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. – MackTuesday May 1 '14 at 1:56

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