Fringe spacing and frequency from image

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MechenG
MechenG 2025-4-18
评论： MechenG 2025-4-18
Ran in:
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',[])
  2 个评论
Matt J
Matt J 2025-4-18
编辑：Matt J 2025-4-18
Do the bands have a known orientation with respect to the heptagon? Can we assume they are perfectly vertical in the above image?
MechenG
MechenG 2025-4-18
Hi Matt,
Yes, one can assume they are vertical. Thanks!!

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Mathieu NOE
Mathieu NOE 2025-4-18
编辑：Mathieu NOE 2025-4-18
Ran in:
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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Mathieu NOE
Mathieu NOE 2025-4-18
Ran in:
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
MechenG 2025-4-18
Yes!!. I tried with some other better quality images, where the spacing is more uniform.

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