How to calculate a gradient by fft ???

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

Answer 1

[David Young]

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?

[3omyer 3omayer]

thanks for your response ;

I would like to calculate gradient (function [GradX,GradY] = gradient(I) in MatLab) not to use this usual method, but calculate with fft.

my professeur wants to see the results in two ways.

This code calculates the gradient of the fft along the x axis, in y axis is the same?

[David Young]

OK I understand now. The code I've given you above computes a result which is close to (but not quite the same as) GradX from the gradient function. As it's an exercise, you should understand what my code does, so that you can adjust it to compute the y gradient - the principle is the same.

[3omyer 3omayer]

thank you;

[3omyer 3omayer]

I read your code, but i still not understood how can i calculate the gradient in y direction ???

you may give me an idea or steps that i should follows ??

[Walter Roberson]

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

[3omyer 3omayer]

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, []);

[David Young]

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.

Answer 2

[Walter Roberson]

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.)

[3omyer 3omayer]

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