FP ridge orientation code only returning 45 or 135 degrees. Please help
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Hi all, I am trying calculate fp orientation using equation given the image (Jain 1995). I am only getting two values 2.3562 and 0.7875 rad(45 deg and 135 deg) in the output matrix. Please help me to know where i am going wrong.
clc;clear all;close all;
img1=imread('1_1.png');
img1=normalise(img1);
[m n]=size(img1);
[gx gy]=imgradientxy(img1);
for i=9:m-9
for j=9:n-9
gxy=0;gxx=0;gyy=0;
for i1=-8:+8
for j1=-8:+8
gxy=gxy+((gx(i+i1,j+j1)).*(gy(i+i1,j+j1)));
gxx=gxx+power((gx(i+i1,j+j1)),2);
gyy=gyy+power((gx(i+i1,j+j1)),2);
end
end
Q(i,j)=pi/2+0.5*(atan2(double(2*gxy),double((gxx-gyy))));
end
end
1 件のコメント
GOPEE Ajit Kumar
2017 年 2 月 2 日
Hello Sam I am new to Fingerprint image processing. I wish to know the steps to reach FG orientation. I am using FVC 2002 fingerprint images. First hw do you thin the images? Then how you locate the singular points? Any help/advice/suggestions would be highly appreciated Thanks a lot in advance rgds ajit
回答 (1 件)
Image Analyst
2013 年 9 月 23 日
Are gx and gy integers or doubles before you start multiplying them together. If things are being quantized, the first thing I look at is if it's doing integer math instead of floating point math. By the way, you can use ^ to raise to a power instead of calling the power() function.
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