Reputation: 3517
Say I have this matrix in memory and I want to calculate the 3D FFT
T =
0 1 2 3
4 5 6 7
8 9 10 11
12 13 14 15
16 17 18 19
20 21 22 23
24 25 26 27
28 29 30 31
32 33 34 35
36 37 38 39
40 41 42 43
44 45 46 47
44 45 46 47
52 53 54 55
56 57 58 59
60 61 62 63
real(fft2(T))
ans =
2000 -32 -32 -32
-128 0 0 0
-112 0 0 0
-128 0 0 0
-144 0 0 0
-128 0 0 0
-112 0 0 0
-128 0 0 0
-144 0 0 0
-128 0 0 0
-112 0 0 0
-128 0 0 0
-144 0 0 0
-128 0 0 0
-112 0 0 0
-128 0 0 0
real(fftn(T))
ans =
2000 -32 -32 -32
-128 0 0 0
-112 0 0 0
-128 0 0 0
-144 0 0 0
-128 0 0 0
-112 0 0 0
-128 0 0 0
-144 0 0 0
-128 0 0 0
-112 0 0 0
-128 0 0 0
-144 0 0 0
-128 0 0 0
-112 0 0 0
-128 0 0 0
Why am I getting the same result? How 3D FFTs can be done in Matlab/Octave?
Upvotes: 0
Views: 710
Reputation: 112689
A 3D-FFT should be applied to a 3D-array. If you apply the 3D-FFT to a 2D-array you get the same result as a 2D-FFT, because there is no third dimension in the array.
Think about it this way: an N-dimensional FFT is just N 1-dimensional FFT's, one along each dimension. If there is no third dimension in the array, the FFT along that dimension does nothing.
Upvotes: 2