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Fourier transform

  1. Jul 29, 2015 #1
    In making Fourier Transform of a function, why is it said that large space (r) corresponds to low wavenumber(k)?
  2. jcsd
  3. Jul 29, 2015 #2


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    Actually it's more about the feature size in the object space rather than the absolute position in that space. You can roughly picture it with the help of the fact that Fourier transform is an infinite sum of weighted sinusoidal disturbance. When there is a large feature in the object space, a large contribution to it must come from sinusoidal terms which have large wavelength (slow variation), or in term of frequency or wavenumber the lower ones.
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