2D vs 3D FFT in Matlab/Octave -

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says that I have this matrix in memory and I want to calculate 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 Actual (FFT2 (T)) ans = 2000 -32 -32 -32 -128 0 0 -112 0 -128 0 0 - 144 0 0 -128 0 0 -112 0 -128 0 -144 0 -128 0 0 -112 0 -128 0 -144 0 0 -128 0 -112-000-2018 0 real (fftn (t) ans = 2000 -32 -32 -32 -128 0 -112 0 -128 0 -144 0 -128 0 0 -112 0 0 -128 0 0 0 - 144 0 0 -128 0 0 -112 0 0128 0 -144 0 0 -128 0 0112 0 -128 0 0 0

Why do I miss the same results Losing? How can 3D FFT be done in Matlab / Octave?

A 3D-FFT should be implemented in a 3D -array. If you apply 3D-FFT to 2D-Oraure, then you get the same results as 2D-FFT, because there is no third dimension in the array.

Think of it this way: An en-dimensional FFT is just N1-dimensional FFT, with each dimension. If there is no third dimension in the array, then FFT does nothing with that dimension.


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