Fourier transform

  • Thread starter shirin
  • Start date
  • #1
46
0
Hi
In making Fourier Transform of a function, why is it said that large space (r) corresponds to low wavenumber(k)?
 

Answers and Replies

  • #2
blue_leaf77
Science Advisor
Homework Helper
2,629
784
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.
 

Related Threads on Fourier transform

Replies
3
Views
2K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
7
Views
778
  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
2
Views
2K
Top