I have the following image .I want to find the frequency or time interval of the periodic pattern.

I mean i want to find the periodic frequency peak.I know may be i have to use fft2 of matlab.But i could not ablle to see the frequency peak after applying the fft2.

</matlabcentral/answers/uploaded_files/45290/Capture.PNG> I want to find that kind of response from my image by using fft.Please help me.I have no idea how to do it. Thanks in advance

[sheli whitson]

Image Analyst please give me some idea.How can i do that for both regular and near regular texture image.

Answer 1

[Image Analyst]

I would not use fft2() because the frequency is so low that its spike will be located very close to the main peak at DC and you might not get good resolution. What I'd probably try is to use normalized cross correlation to find template match locations. Then threshold the correlation image and find the location of the weighted centroids with regionprops() and then inspect each centroid to find the average distance to the closest 6 neighbors. I attach a normxcorr2 demo. Using regionprops to get centroids and finding distances from each centroid to all the others is trivial by using a for loop, or even pdist2() if you have the Stats toolbox. Give it a shot and let us know how it went.

[sheli whitson]

mean first i will find the repeating structures then find the center of the repeating structure ,then calculate the distances between each image center?right?

[sheli whitson]

If i want to do it by fft how can i do that?I actually want to do it by fft.

[Image Analyst]

Didn't you try fft2()??? I'm sure you must have, but what happened when you did it??? Please attach your code or error message. I can't know what went wrong with your code if you don't tell me what it is.

See attached demos.

[sheli whitson]

Sorry image analyst for the late reply.Here is my code.

grayImage=imread('Capture12.png')  ;
[rows columns numberOfColorChannels] = size(grayImage)
if numberOfColorChannels > 1
  f = rgb2gray(grayImage);
end

fsx=50;
    fsy=50;
    dx=1/fsx;
    dy=1/fsy;
    [m,n]=size(f);
    x=dx*(0:n-1);
    y=dy*(0:m-1);
    dfx=fsx/n;
    dfy=fsy/m;
    fx=(-fsx/2:dfx:fsx/2-dfx);
    fy=(-fsy/2:dfy:fsy/2-dfy);
    fftOriginal = fft2((f));

imshow(log(fftshift(fftOriginal)),[]);

Here is my result

 When i tried to plot the fft image with respect to fx it is showing that result.

I do not know what exactly this image is showing. I actually want to find out the peak points that represent the repeated structures so that i can know the interval of repeated objects. How can i get that by using fft?

[Image Analyst]

Well there you go. Now you know why I said to do it in the spatial domain, not the frequency domain.

[sheli whitson]

How can i solve this problem by using fft2?

[Image Analyst]

It's harder because, as you can see, the pattern is more complicated. I'd start with flattening the pattern with adapthisteq(), then I'd assume the pattern is aligned with the rows and columns so I'd sum the image vertically and horizontally. Then use findpeaks to try to find the major peaks. Attached is an fft filtering demo (not sure if it uses that algorithm though).

[sheli whitson]

Thank you.I will try your idea.The attached algorithms did not use this concept.

[sheli whitson]

Hello Image Analyst , How can i sum the image vertically and horizontally?

[Image Analyst]

verticalProfile = sum(grayImage, 2);
horizontalProfile = sum(grayImage, 1);

[sheli whitson]

Here is my code .I actually did not under stand what to do with vertical and horizontal profile and how to use find peaks and what to plot.I read matlab findpeaks demo but did not understand how to use it in my case. grayImage=imread('Capture12.png'); figure; imshow(grayImage)

[rows columns numberOfColorChannels] = size(grayImage) if numberOfColorChannels > 1 f = rgb2gray(grayImage); end

J = adapthisteq(f); [m,n]=size(f) figure imshow(J) figure, imhist(J)

vprofile = sum(J, 2); hprofile = sum(J, 1);

fsx=50; fsy=50; dx=1/fsx; dy=1/fsy; [m,n]=size(f); x=dx*(0:n-1); y=dy*(0:m-1); dfx=fsx/n; dfy=fsy/m; fx=(-fsx/2:dfx:fsx/2-dfx); fy=(-fsy/2:dfy:fsy/2-dfy); r=fftshift(fft2(J)); dBspots = 20*log10(abs(r(m/2+1:m)));

     [pks,locs] = findpeaks(dBspots,'MinPeakDistance',6);

Please help me.Thanks in advance.

[Image Analyst]

You were supposed to call adapthisteq() on the spectrum, not the gray scale spatial domain image. And you were supposed to pass the profiles, not dBspots into findpeaks().

[sheli whitson]

grayImage=imread('Capture12.png');

figure;

imshow(grayImage)

[rows columns numberOfColorChannels] = size(grayImage)

if numberOfColorChannels > 1 f = rgb2gray(grayImage);

end

fsx=size(f,1);

    fsy=size(f,2);
    dx=1/fsx;
    dy=1/fsy;
    [m,n]=size(f);
    x=dx*(0:n-1);
    y=dy*(0:m-1);
    dfx=fsx/n;
    dfy=fsy/m;
    fx=(-fsx/2:dfx:fsx/2-dfx);

    fy=(-fsy/2:dfy:fsy/2-dfy);

    r=log10(fftshift(fft2(f)));

    plot(fx,r)

     J=imread('untitled.tif');

     [rows columns numberOfColorChannels] = size(grayImage)

if numberOfColorChannels > 1

g = rgb2gray(J);

end

     j=adapthisteq(g);

     vprofile=sum(j,2);

     hprofile=sum(j,1);
     [pk,MaxFreq] = findpeaks(vprofile,'NPeaks',1,'SortStr','descend');

Than what to plot? I plot(pk) and get an empty image.

[Image Analyst]

I don't have time tonight to look at your code. In the meantime, read this and fix the formatting: http://www.mathworks.com/matlabcentral/answers/13205-tutorial-how-to-format-your-question-with-markup so I can read it.

[sheli whitson]

ok.Thank you.

[sheli whitson]

Here is my updated code

grayImage = imread('capture.png');

[rows columns numberOfColorChannels] = size(grayImage)

if numberOfColorChannels > 1 f = rgb2gray(grayImage); end

h = size(f,2);

w = size(f,1);

g = zeros(h, w);

for x = 1:w

    for y = 1:h

        g(y,x) = g(y,x) +f(x,y);%image with zero frequency 

      end
end

% dft 2d NFFTY = 2^nextpow2(h); NFFTX = 2^nextpow2(w); % eliminate zero frequency component

av = sum(g(:)) / length(g(:));

g = g - av;

% sampling rate of 1

samplingfreq = 1000;

spatialfreqsx = samplingfreq/2*linspace(0,1,NFFTX/2+1);

spatialfreqsy = samplingfreq/2*linspace(0,1,NFFTY/2+1);

spectrum2D = fft2(g, NFFTY,NFFTX);

figure;

plot(spatialfreqsy, abs(spectrum2D(1:NFFTY/2+1, 1:NFFTX/2+1)));

xlabel('fy HZ')

ylabel('abs(y)')

figure;

plot(spatialfreqsx,abs(spectrum2D(1:NFFTY/2+1, 1:NFFTX/2+1)))

xlabel('fx HZ')

ylabel('abs(y)')

figure;

plot(spatialfreqsy,10*log10(

abs(spectrum2D(1:NFFTY/2+1, 1:NFFTX/2+1))));

xlabel('fy db')

ylabel('abs(y)')

figure;

plot(spatialfreqsx,10*log(abs(spectrum2D(1:NFFTY/2+1, 1:NFFTX/2+1))))

xlabel('fx db')

ylabel('abs(y)')

<<

>>

[sheli whitson]

Here is my new image.

I want to know why I am getting the blue curve.It seems that blue curve is representing the peak points .Am I right?

[sheli whitson]

Hello Image Analyst, I need your help.I am still trying to do this thing. Here is my code and i am getting some peaks.But I am not sure what exactly these peaks are showing .Can you please explain that?

clear all; close all; grayImage = imread('image1.png');

[rows , colums , numberOfColorChannels] =size(grayImage);

if numberOfColorChannels > 1

f = im2double(rgb2gray(grayImage));

end

figure; imshow(f)

[n,m] = size(f); % sampling freq Fs = 1; % zero padding

m = 2^nextpow2(m);

n = 2^nextpow2(n);

% Frequency spacing in pixels per pixel

fx = Fs*((-m/2:m/2-1)/m);

fy = Fs*((-n/2:n/2-1)/n);

% fftshift centers the zero frequency

spectrum2D = fftshift(fft2(f, n,m));

%spectrum2D = fftshift(fft2(f, n,m))

spectrum2D_abs = abs(spectrum2D);

% show fft2 results

figure;

imagesc(1+log10(spectrum2D_abs))

colormap gray

% frequency peaks

figure;

plot(fx(m/2+1:end),log10(spectrum2D_abs(n/2+1,m/2+1:end)))

xlabel('fx-Hz'); ylabel('Power/frequency/')

figure;

plot((fy(n/2+1:end)),100*log10(spectrum2D_abs(n/2+1 :end,m/2+1)),'r')

xlabel('fyHz');

ylabel('Power/frequency')

Another thing I want to know ,can i find the peak of repeating things by doing 1 dimensional fft (first taking row then column).I would be grateful if you could give me a demo code of applying 1d fft in 2 dimensional image.

[Image Analyst]

Like I said, I don't think FFT is the way to go. Believe it or not we actually did this problem this week with some crystalline "balls" in a lattice structure in SEM images. We used delaunay triangles and found the mean and standard deviation of the separations between each ball and the neighbors defined by the delaunay triangulation. From the spacing, you can invert it to get the frequency.

[sheli whitson]

Hello Image Analyst,

Now i am thinking to do it like that paper mentioned(fig-2b) if there are repeating structures i will get one peak for those structures and if there is another type of repeating structures i will get another peak so that i can decide how many different repeating structures are present in my image.But they did not mention how they plot the result.So how can get similar types of plots ? Is that possible to do that by using fft like they did or by using your technique?Finally I will apply similar techniques on other images .

Thank you very much .If fft technique is not working i have to show someone the prove.That is why I am trying to do it by using fft.I hope you understand .Thanks again for your help. I am looking forward to hear from you.

Answer 2

[sheli whitson]

Hello Image Analyst,

Do you have any suggestion how can i get similar peaks by using the attached paper's technique?I actually could not under stand what they plot(fig-2b). Thanks in advance