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Starting with some vector $x$, I am performing two operations: First, I convolve $x$ with another vector $g$ to compute $x*g$, where $~*~$ denotes convolution. Second, I pointwise multiply the result with a discretized rectangular function to compute $(x*g) \cdot r$, where $~\cdot~$ denotes point-wise product.

I know that point-wise multiplication with a rectangular function is equivalent to convolution with a sinc function (by the convolution theorem), so really we are finding $(x*g)*R$. And we also know that convolution commutes, so the answer should be equivalent to $(x*R)*g$, which could in turn be rewritten as $(x \cdot r)*g$. But it is obviously not the case:

For example, when $g$ is a discretized Gaussian, then you are smoothing $x$ with "Gaussian blur" and then clipping everything outside some window (specified by $r$ to be zero). Whereas, clipping $x$ and then smoothing does not include the influence that the clipped-off points can have on the smoothing (a single high-intensity peak that is clipped off would have multiplied with the tail of a Gaussian to influence a point in the actual result).

Who can figure out where my reasoning is (obviously) wrong?

user
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1 Answers1

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You're a little bit off with your interpretation of the convolution theorem. In particular, what we really have is as follows: $$ \newcommand{\F}{\mathscr{F}} \F [(x*g)\cdot r] = \F(x*g) * \F(r) = \F(x*g) * R $$ Where $\F$ here denotes the Fourier transform. The reason things work out so nicely when your vector is a Gaussian is that the Fourier transform of a Gaussian is another Gaussian.

So, on the one hand, we would have $$ \F[(x*g)\cdot r] = X G*R $$ On the other hand, $$ \F[(x*r)\cdot g] = XR * G $$ Ultimately, it is this successive multiplication and convolution that cannot be interchanged.

Ben Grossmann
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  • Thanks for the note-- I actually mean that it doesn't work for a Gaussian (I've tried massaging the Gaussian property that you're talking about to be able to reorder the operations). But good point about the frequency domain, I think that is the mistake. – user Jul 14 '14 at 16:32
  • @Oliver I mean that things should work nicely when you're just working with Gaussians. Maybe I can expand on my answer. – Ben Grossmann Jul 14 '14 at 16:36
  • Thanks a lot, I see what you mean. If only $x$ and $r$ were also Gaussian, things would be a lot better. I grew up in Philly, please enjoy omnoming a cheesesteak for me. – user Jul 14 '14 at 16:56
  • @Oliver Ha! will do :) – Ben Grossmann Jul 14 '14 at 17:14