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I am working on a university project in which I need to find a periodic net element in an image.

The net is a set of diagonal lines in a specific angle ( which I don't know in advance ) on a noisy image. They are hard to detect so I would like to use all the information I can get my hands on.

enter image description here

My current algorithm is based on using edge detection + hough transform to find lines at specific angle range.

I was wondering if there is any way to detect periodic signals in image? Something based on ff2 or something like that...

Thanks,

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  • $\begingroup$ can you show the picture? $\endgroup$ – endolith Jun 11 '13 at 16:11
  • $\begingroup$ now i can , addd the image to the question $\endgroup$ – Elad Walach Jun 11 '13 at 19:35
  • $\begingroup$ Ahh, the separation of the lines varies along their length. Autocorrelation won't pick up on that as well, since it's not really periodic in 2D. $\endgroup$ – endolith Jun 11 '13 at 20:17
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Another approach might be to perform the Radon / Hough transform first, then detect the points.

e.g. R = radon(I,0:179) in MATLAB.

It gives this image:

enter image description here

The x-axis is angle (0-180 deg) and the y-axis is distance from the centre. Each local minimum represents a line. It shows 6 lines ~75 degrees, 2 around 90 degrees, and 3 around 170 degrees. (This is MATLAB angles which go clockwise from x-axis because the y-coords are upside down)

Edit: Forgot Radon and Hough transforms were roughly the same.

Update:

I wrote some MATLAB code to locate the angles and mean separation between lines.


close all
I = imread('testt.jpg');
I = rgb2gray(I);
I = I(2:end-1,:);

% Radon transform
R = radon(I,0:179);
imagesc(R); colormap gray(256); pause;

% Radon transform of smoothed image
Rg = radon(imgaussian(I,2),0:179);
imagesc(Rg); colormap gray(256); pause;

% Take it away
Rf = R - Rg;
imagesc(Rf); colormap gray(256); pause;

% Chop off out of range parts
chop = size(Rf,1) - size(I,1);
chop = ceil(chop/2);
Rf = Rf(chop+1:end-chop,:);
imagesc(Rf); colormap gray(256); pause;

% Negative lines - threshold
Rf(Rf > 0) = 0; 
imagesc(Rf); colormap gray(256); pause;

% Plot sum - peaks are angles
Rp = sum(abs(Rf));
plot(Rp);

% Get the peaks sep by at least 15 deg
[p,a] = findpeaks(Rp,'minpeakdistance',15,'sortstr','descend');
hold on;
scatter(a,p,'r*');
hold off;
pause;

% Iterate through peaks and find fequencies
for j = 1:numel(a)
    % Get subsection of Radon transform around angle and transpose
    vstart = max([a(j)-10,1]);
    vend = min([a(j)+10,size(Rf,2)]);
    Rsub = Rf(:,vstart:vend).';
    imagesc(Rsub); colormap gray(256); pause;
    RsubP = sum(abs(Rsub));
    plot(RsubP); pause;

    % Find peak correlation with a bit of smoothing
    xp = xcorr(imgaussian(RsubP,2),imgaussian(RsubP,2));
    plot(xp); pause;
    [rp,rl] = findpeaks(xp,'sortstr','descend');
    wave(j) = abs(rl(1) - rl(2));
    disp(['Angle: ' num2str(a(j))]);
    disp(['Wavelength: ' num2str(wave(j))]);
    disp(['Strength: ' num2str(p(j))]);
    pause;
end

which results in: (red * are possible angles)

enter image description here

and

Angle: 73
Wavelength: 16
Strength: 12401.356

Angle: 92
Wavelength: 54
Strength: 9442.2545

Angle: 175
Wavelength: 33
Strength: 9030.1877
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  • $\begingroup$ Hough is to Radon as FFT is to DFT? $\endgroup$ – endolith Jun 12 '13 at 14:07
  • $\begingroup$ @endolith I think Hough is the discrete one? $\endgroup$ – geometrikal Jun 13 '13 at 1:33
  • $\begingroup$ I mean that Hough is a fast implementation of Radon, like FFT is a fast implementation of DFT. Not sure if that's exactly right, though. $\endgroup$ – endolith Jun 13 '13 at 16:09
  • $\begingroup$ @endolith I'll have to have a look. Had not properly connected Hough to Radon in my mind before a few days ago hehe. $\endgroup$ – geometrikal Jun 14 '13 at 11:15
  • $\begingroup$ dsp.stackexchange.com/q/470/29 :) $\endgroup$ – endolith Jun 14 '13 at 14:25
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You could apply a 2d FFT. You will get a frequency spectrum as z values on a 2d plane:

FFT of the example image ← That's log(abs(fftshift(fft2(image))))) where image is from the question, but with white borders trimmed

From the z values you can determine the highest peak and calculate its position on the 2d plane. With some geometric calculation, you can derive wavelength, direction and phaseshift from this.

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    $\begingroup$ I would multiply the spectrum by the radius from its centre. This compensates somewhat for lower frequency components being grouped into fewer bins and therefore having larger amplitudes, when finding the highest peak. $\endgroup$ – geometrikal Jun 11 '13 at 9:51
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    $\begingroup$ would autocorrelation be better than FFT alone? $\endgroup$ – endolith Jun 11 '13 at 16:14
  • $\begingroup$ @geometrikal: Is that equivalent to differentiating in the space domain? $\endgroup$ – endolith Jun 11 '13 at 20:22
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    $\begingroup$ @endolith I think differential needs the i as well. I have no mathematical basis for my comment actually.. Just something that seemed to work on some problems I had. Tried on this one and it didn't give anything useful. Though, that big line in the spectrum can be removed by getting rid of the first and last rows in the image, as they are white. $\endgroup$ – geometrikal Jun 11 '13 at 22:28
  • $\begingroup$ wow ,that looks good, exactly what i wanted to try..Thanks! $\endgroup$ – Elad Walach Jun 14 '13 at 18:30

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