a triangular (bartlett) window has a kink (discontinuity in first derivative). thinking in the continuous (not discrete) case, my intuition is that the fourier transform of any window with a kink must asymptote to a non-zero value (rather than exponentially decay), since it should take infinitely high frequencies to be able to sum to a precisely sharp "corner." treated as the kernel (impulse response) of a filter, this would correspond to passing arbitrarily high frequencies.

is this right? in general, how do i calculate the asymptotic value, and why isn't this a standard quoted feature of a window?


This actually isn't the case. Consider the continuous-time triangle function:

$$ \text{tri}(t) = \begin{cases} 1 - |t|, && |t| < 1 \\ 0, &&\text{otherwise} \end{cases} $$

This function has the Fourier transform:

$$ \text{Tri}(f) = \text{sinc}^2(f) $$

This follows readily from the fact that the triangle function can be constructed by convolving a rectangle function, which has a Fourier transform of $\text{sinc}(f)$, with itself. The $\text{sinc}$ function decays to zero as its argument approaches infinity (and its square decays even faster yet), so the triangle window does not have a highpass characteristic.

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  • $\begingroup$ thanks, can you help me intuit two troubles i have with this? 1) how do we make the kink in the triangle without summing (more than a vanishing amount) of infinitely high frequencies? 2) sinc is the fourier transform of a rectangular window, so even though it decays like 1/x, it seems to me to pass high frequencies because, as a filter kernel, a rectangle would not smooth or delay infinitely fast changes in an input signal. thanks! $\endgroup$ – user1441998 Sep 14 '15 at 13:05
  • $\begingroup$ 1) The triangle still has infinite bandwidth, precisely speaking. However, for practical purposes, it still has a lowpass characteristic, like the rectangle function. Note that the triangle isn't discontinuous in the time domain, only in its first derivative. 2) I think your intuition is just inaccurate in this case. A rectangle filter kernel does in fact smooth discontinuous changes in an input signal; it integrates over the width of the rectangle function, which can greatly attenuate high frequency content. $\endgroup$ – Jason R Sep 14 '15 at 14:06
  • $\begingroup$ 1) how do i use the fact that the triangle is only discontinuous in dx/dt to see why we don't need lots of high freq content to make a sharp corner? i agree if it were discontinuous in time, it would need even more... :) 2) i'm thinking of what a rectangle filter does to eg a unit impulse - turns it into a rectangle. still perfectly sharp onset and offset, so it seems like the highest freqs are preserved. as the simplest moving avg filter, surely it smoothes if you normalize gain, but couldn't it be adding low freqs onto something like a unity gain of the high freqs? how do i see that's wrong? $\endgroup$ – user1441998 Sep 14 '15 at 16:19
  • $\begingroup$ I'm not sure how to tell you to fix your intuitions; I see what you're going for, but they just aren't accurate. In my experience, the best way to understand the details is to look at the underlying mathematics. A triangle function isn't highpass because its spectrum does decay to zero. A rectangular filter kernel is lowpass because its spectrum decays to zero. High-frequency content is removed from the input, as a rectangle time-domain window is, practically speaking narrower in bandwidth than a time-domain impulse. $\endgroup$ – Jason R Sep 14 '15 at 17:19
  • $\begingroup$ thx for help and patience, glad u can see what i'm going for. maybe u help me rephrase the q to elicit answers that will clarify my intuition error? i hear you on math - but i don't consider something understood until i have both the symbol manipulation and "feel" why it is true. i have the math result and agree with your interp, i just don't get how to build a sharp corner from non-infinite freqs - you indicated it is relevant that the discontinuity is only in dx/dt, but can you say how that helps? and how we see the (unnormalized!) rectangle is removing high freqs rather than adding low? $\endgroup$ – user1441998 Sep 14 '15 at 19:56

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