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vchen

I'm still having a little trouble wrapping my head around going from spatial to frequency. Why would a larger Gaussian in space convert to a smaller signal in the frequency domain?

hal

hmm let's say you take a gaussian (just in the 2D) and give it a standard deviation sigma. In the time domain, it'll look something like ~ e^{(t/sigma)^2}. When you take the fourier transform of the gaussian, the sigma^2 ends up actually being in the numerator and it'll look something like ~ e^{(f sigma)^2}. So the standard deviation part is inverted. If in the time domain, you have a really narrow gaussian, the Fourier transform flips it and you end up with a very broad gaussian in frequency domain. Now going the spatial domain, if you spin that 2D gaussian about the y axis into a 3D gaussian, we see the same reasoning as before applies when taking it to the frequency domain.

hal

I hope that was helpful...Maybe a more intuitive picture would be to think about how one would go about getting a sharper and sharper pulse in the time domain. Like if you wanted to make the dirac delta function, for example, you'd need to add infinitely many frequencies to make that sharp pulse. In fact, the fourier transform of the dirac delta is just 1 across all frequencies. So maybe if you imagine slowly morphing those "into" a gaussian, it provides more of an intuitive picture? maybe useful gif

jchh

A simpler explanation might be that in the spatial domain, the Gaussian is bigger. That means its "period" must be bigger and thus its "frequency" is thus smaller. Smaller frequency = values closer to the origin in the frequency domain.