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How to calculate a gradient by fft ???

 Hi;
    someone can help me, how to calculate Gradient using fft ???

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2 Answers

Answer by David Young
on 20 Sep 2011
 Accepted Answer

It's still not clear what you mean by 'simple "gradient"' - that could refer to a variety of things.

One possibility is that you mean just the differences between adjacent values - what you'd get by computing

diff(im, 1, 2) % x differences - x component of gradient

and

diff(im, 1, 1) % y differences - y component of gradient

for example. You can do a similar operation in the frequency domain using the fact that the differentiation operator transforms to multiplication with ik (where k is the transform variable). The code looks like this for differentiating with respect to x:

im = imread('pout.tif');        % data
% compute differencing operator in the frequency domain
nx = size(im, 2);
hx = ceil(nx/2)-1;
ftdiff = (2i*pi/nx)*(0:hx);     % ik 
ftdiff(nx:-1:nx-hx+1) = -ftdiff(2:hx+1);  % correct conjugate symmetry
% compute "gradient" in x using fft
g = ifft2( bsxfun(@times, fft2(im), ftdiff) );
imshow(g, []);      % see result

As you can see, it's simpler to do it in the space domain. Why use the FFT?

  7 Comments

step 1: copy the portion of the code after the imread.
step 2: in the copy, change all of the x to y
step 3: in the copy, change size(im,2) to something appropriate for calculating the size in the other direction

I actually changed size(im,2) by size(im,1), but it gives me the same results ??
this is the following code

im = imread('pout.tif');
ny = size(im, 1);
hy = ceil(ny/2)-1;
ftdiff2 = (2i*pi/ny)*(0:hy);
ftdiff2(ny:-1:ny-hy+1) = -ftdiff2(2:hy+1);
% compute "gradient" in x using fft
g2 = ifft2( bsxfun(@times, fft2(im), ftdiff2) );
figure(2),
imshow(g2, []);

Well, you have to think about the shapes of the arrays. Differentiating with respect to x means multiplying each row of the the Fourier transform by ftdiff. If you want to differentiate with respect to y, you have to multiply each column by ftdiff. Changing the size of ftdiff is a start, but you also have to change it from a row vector to a column vector.

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Answer by Walter Roberson
on 19 Sep 2011

Is that plain "gradient", or is it "conjugate gradient" ?

(I notice you posted the same question to some other locations, all of which seem to have replied asking for clarification.)

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I want to calculate simple "gradient" for images by using "fft"

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