questions in orientation of fingerprint recognition!!!!

[kejunconsult]

Hi, everyone:
in the following code, Gx,Gy already get the gradient of the image. so we can get
the orient by Gy/Gx. Why we need to calculate the Gxx, Gyy, Gxy?? what did they mean?? why sin2theta = Gxy./denom?? why cos2theta = (Gxx-Gyy)./denom??
is there anybody can help me???
Thank you very much!

*************************************************
function [orientim, reliability] = ...
ridgeorient(im, gradientsigma, blocksigma, orientsmoothsigma)

[rows,cols] = size(im);

% Calculate image gradients.
sze = fix(6*gradientsigma); if ~mod(sze,2); sze = sze+1; end
f = fspecial('gaussian', sze, gradientsigma); % Generate Gaussian filter.
[fx,fy] = gradient(f); % Gradient of Gausian.

Gx = filter2(fx, im); % Gradient of the image in x
Gy = filter2(fy, im); % ... and y

% Estimate the local ridge orientation at each point by finding the
% principal axis of variation in the image gradients.

Gxx = Gx.^2; % Covariance data for the image gradients
Gxy = Gx.*Gy;
Gyy = Gy.^2;

% Now smooth the covariance data to perform a weighted summation of the
% data.
sze = fix(6*blocksigma); if ~mod(sze,2); sze = sze+1; end
f = fspecial('gaussian', sze, blocksigma);
Gxx = filter2(f, Gxx);
Gxy = 2*filter2(f, Gxy);
Gyy = filter2(f, Gyy);

% Analytic solution of principal direction
denom = sqrt(Gxy.^2 + (Gxx - Gyy).^2) + eps;
sin2theta = Gxy./denom; % Sine and cosine of doubled angles
cos2theta = (Gxx-Gyy)./denom;

sze = fix(6*orientsmoothsigma); if ~mod(sze,2); sze = sze+1; end
f = fspecial('gaussian', sze, orientsmoothsigma);
cos2theta = filter2(f, cos2theta); % Smoothed sine and cosine of
sin2theta = filter2(f, sin2theta); % doubled angles

orientim = pi/2 + atan2(sin2theta,cos2theta)/2;

 

[kejunconsult]

nobody to deal with such problem?

 

[leekk8]

This is the algorithm in Raymond Thai's paper. Try to refer to his equations.

 

[kejunconsult]

hi leekk8:
I know the algorithm of this code. but I don't know why he didn't calculate the orient by Gy/Gx?
Thank you for reply!

 

[leekk8]

From the algorithm of the computation, the equation is (Gx)^2 and (Gy)^2, so the code has Gxx and Gyy. About the details of this, I'm also not sure.

 

[kejunconsult]

hi leekk8;
Thank you all the same!