FFT of 3D array in MATLAB

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Janee 2024-8-6

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https://ww2.mathworks.cn/matlabcentral/answers/2143236-fft-of-3d-array-in-matlab

评论： Janee 2024-8-7

采纳的回答： Matt J

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I am trying to understand how the FFT of different directions in MATLAB works to reproduce in C/C++ instead.

So far I have the following simple example in MATLAB:

clearvars; clc; close all;
%3D FFT test
Nx = 8;  
Ny = 4; 
Nz= 6;
Lx =16;
Ly = 6;
dx = Lx/Nx; 
dy = Ly/Ny;
%-----------
xi_x =  (2*pi)/Lx;
yi_y =  (2*pi)/Ly;
xi  = ((0:Nx-1)/Nx)*(2*pi);
yi  = ((0:Ny-1)/Ny)*(2*pi);
x =  xi/xi_x;
y =  yi/yi_y;
zlow = 0; %a
zupp =6; %b
Lz = (zupp-zlow);
eta_zgl  = 2/Lz;
[D,zgl] = cheb(Nz);
zgl = (1/2)*(((zupp-zlow)*zgl) + (zupp+zlow));
[X,Z,Y] = meshgrid(x,zgl,y); %this gives 3d grid with z-by-x-by-y size (i.e. ZXY)
%ICs
A = 2*pi / Lx;
B = 2*pi / Ly;
u = (Z-zlow) .* (Z-zupp) .* sin(A*X).* sin(B*Y);
uh1 =(fft(u,[],3));%ZXY
uh2 =(fft(u,[],1));%ZXY
uh3 =(fft(u,[],2));%ZXY

So, in C/C++ I have a 3D tensor with (Nz+1) rows and Nx coumns and Ny matrices and taking the 1D FFT along each "row" of u returns the same results as the following in MATLAB:

uh3 =(fft(u,[],2));%ZXY

While taking the 1D FFT of u along each column of u in C/C++ returns the same result as the following in MATLAB:

uh2 =(fft(u,[],1));%ZXY

Then my question is what does this 1D FFT represent? and how should I represent it in C/C++?

uh1 =(fft(u,[],3));%ZXY

The cheb(N) function is:

function [ D, x ] = cheb ( N )
  if ( N == 0 )
    D = 0.0;
    x = 1.0;
    return
  end
  x = cos ( pi * ( 0 : N ) / N )';
  c = [ 2.0; ones(N-1,1); 2.0 ] .* (-1.0).^(0:N)';
  X = repmat ( x, 1, N + 1 );
  dX = X - X';
  
%  Set the off diagonal entries.
  D =( c * (1.0 ./ c )' ) ./ ( dX + ( eye ( N + 1 ) ) );
%  Diagonal entries.
  D = D - diag ( sum ( D' ) );
 
  return
 end