what is the result of phase correlation mentioned in wikipedia

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Any one tried phase correlation for images mentioned in wikipedia? https://en.wikipedia.org/wiki/Phase_correlation
At the site it is given translation of (30,33). I am getting a translation of (24,21). Anybody got same results?

Answers (1)

Walter Roberson
Walter Roberson on 23 Sep 2016
Difficult to say without seeing your code.
I speculate, though, that you might not know that in MATLAB, the x coordinate corresponds to columns (second index) and the y coordinate corresponds to rows (first index), so you might be indexing into the wrong vectors to determine the values.
But that is just speculation. Maybe you used / where you needed ./ . Maybe you are suffering from integer arithmetic saturation due to the way calculations on uint8 work. Maybe something else. The Magic 8 Ball says "Situation is hazy, try again later"
  2 Comments
aravind S
aravind S on 23 Sep 2016
Edited: aravind S on 23 Sep 2016
Thank you for noticing. Yes,i used./ and converted unit8 to double No,I cant do it later. Have you tried it? Can u try and get me the answer. Here's my code
clear all
clc
imtool close all
close all
im1=imread('E:\Mtech internship\Image processing\image processing 3\frames\4 new 0th.jpg');
im2=imread('E:\Mtech internship\Image processing\image processing 3\frames\4 new 120th.jpg');
im1=rgb2gray(im1);
im2=rgb2gray(im2);
i1=imcrop(im1,[296,264,17,17]);
i2=imcrop(im2,[296,264,17,17]);
ima1=double(i1);
ima2=double(i2);
[Nx, Ny]=size(ima1);
Mx=(Nx-1)/2;
My=(Ny-1)/2;
nx=-Mx:1:Mx;
ny=-My:1:My;
w=hanning(Nx);% applying hanning window
w=w*w';
image1=ima1*w;
image2=ima2*w;
f1=fft2(ima1);
f2=fft2(ima2);
n=f1.*conj(f2);
n1=n./abs(n);
n1(isnan(n1))=1;
h=fspecial('gaussian',[Nx,Ny],1.5);
n1=imfilter(n1,h);
result=ifft2(n1);
result=fftshift(abs(result));
figure;
mesh((-Mx:Mx),(-My:My),real(result));
figure;
plot((-Mx:Mx),(result'));
title('Plot of X- axis');
figure;
plot((-Mx:Mx),(result));
title('Plot of Y- axis');
rmax = max(max(result));
[x , y] = find(result == rmax);
x=x-ceil(Nx/2)
y=y-ceil(Nx/2)
Thanks
Deming Peng
Deming Peng on 29 Jul 2020
Edited: Walter Roberson on 6 Aug 2020
May I ask - what does this part mean after you've got the 'x' and 'y' :
x = x-ceilNx / 2
y = y-ceilNx / 2
And I noticed that the operation of these two lines of codes will always result in 'x=1,y=1'.
I suppose it is because the 'rmax' seems always at the centre of the 'result' matrix, just one pixel bigger than
Nx/2 and Ny/2. So I'm a bit confused about the shift it shows here and hope that we can discuss more on it soon. Thank u!

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