Fringe spacing and frequency from image

Hi,

I am trying to calculate the fringe spacing and numbers from the attached image. However, the noise that appears in the dark spot of the fringe causing the error in the frequency estimation using fft function. Could someone please help me with this.

load matlab; whos

  Name         Size              Bytes  Class    Attributes

  ans          1x34                 68  char               
  x1        1280x720            921600  int8               

imshow(x1',[])

[Matt J]

Do the bands have a known orientation with respect to the heptagon? Can we assume they are perfectly vertical in the above image?

[MechenG]

Hi Matt,

Yes, one can assume they are vertical. Thanks!!

Answer 1

[Mathieu NOE]

hello

this is maybe a bit oversimplified but I assumed that I would not make a big error by considering that the fringes are parallel to the horizontal axis

my logic is just to take the mean of the x1 array , smooth a bit the result and pick the peaks (then you get a period in pixel units => up to you to convert in the freq units)

load('matlab.mat')
x1 = double(x1);
% x1(x1>1.5) = NaN; % not really needed , just to remove some large
% amplitude isolated spots

figure,
imagesc(x1)

s1 = mean(x1,2,'omitnan');
s1 = smoothdata(s1,'gaussian',25);

figure,
plot(s1)
hold on 
[PKS,LOCS] = findpeaks(s1,'MinPeakHeight',max(s1)/5);
plot(LOCS,PKS,'dr')

spatial_period_pixels = diff(LOCS)

spatial_period_pixels = 5×1

42 39 40 36 34

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[MechenG]

Hi Mathieu,

Thank you very much for the answer.

For the frequency this is perfect. I didn't think about findpeaks, as it correctly predicts the numbeer of fringes, either FFT or find peaks shoule be fine. However, there exist difference in fringe spacing, ideally it should be same. I think noise causing this issue?

[Mathieu NOE]

I tried to see what happens if I reduce the amount of smoothing but that does not really affect the results

I believe it has more to do with "noise" , I would rather call it some non uniform distribution of the fringes

but also is possible to get the same iamge before it gets binarized ? this would improve the results

lastly what you can do is simply take the mean value of those periods

[MechenG]

By default these images are binary. I will try to minimize the noise.

[Mathieu NOE]

fyi , I tried two methods , results are very similar

load('matlab.mat')
x1 = double(x1);
% x1(x1>1.5) = NaN;
figure,
imagesc(x1)

s1 = mean(x1,2,'omitnan');

%% findpeaks period counting
s1 = smoothdata(s1,'gaussian',25);
[PKS,LOCS] = findpeaks(s1,'MinPeakHeight',max(s1)/5);
figure,
plot(s1)
hold on 
plot(LOCS,PKS,'dr')
hold off

spatial_period_pixels1 = mean(diff(LOCS))

spatial_period_pixels1 = 38.2000

%% zero crossing period counting 
% first some high pass filtering 
[b,a] = butter(1,0.1,'high');
s1 = filtfilt(b,a,s1);

threshold = 0.1*max(s1); 
x = (1:numel(s1))';
t0_pos1 = find_zc(x,s1,threshold);
spatial_period_pixels2 = mean(diff(t0_pos1))

spatial_period_pixels2 = 38.3318

figure
plot(x,s1,'b',t0_pos1,threshold*ones(size(t0_pos1)),'*r','linewidth',2,'markersize',6);grid on
legend('signal','signal positive slope crossing points');

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function [Zx] = find_zc(x,y,threshold)

% positive slope "zero" crossing detection, using linear interpolation
y = y - threshold;
zci = @(data) find(diff(sign(data))>0);  %define function: returns indices of +ZCs
ix=zci(y);                      %find indices of + zero crossings of x
ZeroX = @(x0,y0,x1,y1) x0 - (y0.*(x0 - x1))./(y0 - y1); % Interpolated x value for Zero-Crossing 
Zx = ZeroX(x(ix),y(ix),x(ix+1),y(ix+1));

end

[MechenG]

Yes!!. I tried with some other better quality images, where the spacing is more uniform.