All input options for fft so I can correctly do fft with respect to a single dimension in multidimensional array.

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So I have a array A with dimensions [x y z t] of size [64 64 64 19] and I want to take the fft for every [x y z] position with respect to t. For this I thought I could evaluate B=fft(A, [], 4); taking the fft with respect to the fourth dimension. However if i then take for instance B(32,32,32,:) it is not the same as fft(A(32,32,32,:)). What am I doing wrong? Has it got to do with the [] input for fft? Nowhere in the documentation can I find how to use those brackets.

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Matt J
Matt J 2013년 9월 23일
편집: Matt J 2013년 9월 23일
However if i then take for instance B(32,32,32,:) it is not the same as fft(A(32,32,32,:)). What am I doing wrong?
I can't reproduce this observation, I'm afraid. As an experiment, I have done
A=rand(64,64,64,19);
B=fft(A,[],4);
C=fft(A(32,32,32,:));
Then at the command line, I find the agreement expected
>> isequal(B(32,32,32,:),C)
ans =
1
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Lennart
Lennart 2013년 9월 23일
편집: Lennart 2013년 9월 23일
Hmm then my code must have an error somewhere else... strange. Ill check again. thanks anyways

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