Is it possible to correctly perform a multi-dimensional FFT on a 1D linearised version of a 3D array?
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I have a 3D array indexed by A_3D(ind_y,ind_x,ind_z). This has then been converted to a single 1D linear vector B_1D(ind) such that the coordinates cycle through the z coorindinates fastest, followed by y, and finally by x which is the slowest to increment.
I am able to take the FFT of the 3D matrix A_3D no problem, using fftn(). I would like to know if it is possible in Matlab to successfully take a FFT of B_1D directly. In other words, is it possible to do a multi-dimensional FFT when working with a linearised 3D array?
My attempt below fails, because when calling fft() it simply takes a 1D FFT of the entire vector, not caring that there are blocks of points which correspond to different dimensions:
Nx = 3;
Ny = 2;
Nz = 4;
N = Nx*Ny*Nz;
[X , Y , Z] = meshgrid(1:Nx,1:Ny,1:Nz);
%%% Do the FFT on a 3D matrix, and reshape the output to 1D vector
A_3D = rand(size(X)); % Create random 3D matrix, indexed by A_3D(indy,indx,indz)
A_3D_FFT = fftn(A_3D); % 3D FFT
A_1D_FFT = reshape( permute(A_3D_FFT,[3,1,2]),[1 N] ); % Reshape output of FFT so cycle through z fastest, then y, then x
%%% Do FFT correctly on a reshaped, 1D version of the original 3D array?
B_1D = reshape( permute(A_3D,[3,1,2]),[1 N] ); % Instead of reshaping output of FFT, reshape original matrix instead
B_1D_FFT = fftn(B_1D, [Ny,Nx,Nz]); % Is there a way to do this, even though B_1D has been flattened?
max( abs(A_1D_FFT - B_1D_FFT) ) % Check if equivalent
I know that this is possible in C++, because in the FFTW library help docs you can specify a "FFT plan" for fftwnd using
my_plan = fftw3d_create_plan_specific(Nx,Ny,Nz,...
which gives information about how the initial 3D matrix was unwrapped to 1D. Then you can do something like
B_1D_FFT = fftwnd(my_plan , 1 , B_1D , 1 , Nx*Ny*Nz , ... );
and in this case the transform is done correctly. I would like to do the same thing in Matlab (which I believe is based on the same fftw library)
5 Comments
Matt J
on 27 Jul 2021
Edited: Matt J
on 27 Jul 2021
That's almost what I mean. It's not the reshape() that is costing you time. It is the permute(). There is no need to permute that I can see. You could just as easily do,
Nx = 256;
Ny = 256;
Nz = 128;
N = Nx*Ny*Nz;
A0 = rand(N,1);
tic
for k = 1:20
B = reshape( A0, [Nz,Ny,Nx] ) ;
A_3D = fftn(B);
A_1D = reshape( A_3D, N,1);
end
toc
and we can see below that the time spent is essentially the same time as if only the fftn() operation is included:
B= reshape( A0, [Nz,Ny,Nx] ) ;
tic
for k = 1:20
A_3D = fftn(B);
end
toc
Accepted Answer
Matt J
on 27 Jul 2021
Edited: Matt J
on 27 Jul 2021
Reshaping to and from 3D format should not add significant cost:
Nx = 256;
Ny = 256;
Nz = 128;
N = Nx*Ny*Nz;
A0 = rand(N,1);
tic
for k = 1:20
B = reshape( A0, [Nz,Ny,Nx] ) ;
A_3D = fftn(B);
A_1D = reshape( A_3D, N,1); %Strangely, A_1D=A_3D(:) is much slower ????
end
toc
When compared to processing without reshaping, it is almost the same:
B= reshape( A0, [Nz,Ny,Nx] ) ;
tic
for k = 1:20
A_3D = fftn(B);
end
toc
0 Comments
More Answers (1)
rajmouli jujjavarapu
on 23 Jul 2021
F_d = fft(t_d, [], n);
At the n term if you could specify the dimension, in your case it is 3. I guess this computes each layer individually.
4 Comments
rajmouli jujjavarapu
on 23 Jul 2021
I understand. I didnot find any function for this, but lets wait and see. somebody will have a good idea and will reply.
Thank you.
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