Radial gradient using Convolution
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Hi, I have color images(here I), and considering its Green(G) channel and following is the part of code:
I=I(:,:,2);
dx=[-1 0 0;-1 8 -1;-1 0 0];
dy=[1 0 0;-1 8 -1;1 0 0];
s = [1 ; 1];
gx = conv2(conv2(double(I),dx,'same'),s,'same');
gy = conv2(conv2(double(I),dy,'same'),s','same');
now my question is: -- What gx and gy is giving here, is it a radial gradient??
-- And what will be there if convolved once only ?
Thanks..
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Image Analyst
el 25 de Abr. de 2013
Editada: Image Analyst
el 25 de Abr. de 2013
Why are you convolving twice? You're taking the output of conv() and convolving it again. And why not just use imgradent() or imgradentxy()? I don't know what radial gradient is. The gradient is the maximum slope and it has a specified direction. If you say, x or y, then it's the max slope in the x or y direction. But the radial direction? That's all directions and I think that would just be the same as the normal, regular gradient.
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Cedric
el 25 de Abr. de 2013
Editada: Cedric
el 25 de Abr. de 2013
Any function that depends on the norm of x-x0 only expresses what seems to be named "radial gradient" (around x0) in drawing applications. We might owe this terminology to Mac Paint 1986 or something like that ;-)
[X,Y] = meshgrid(-7:0.1:7, -5:0.1:5) ;
nrm = sqrt(X.^2 + Y.^2) ;
R = (sin(nrm)+1) ./ sqrt(nrm) ;
G = (sin(nrm+2*pi/3)+1) ./ sqrt((nrm+2*pi/3)) ;
B = (sin(nrm+4*pi/3)+1) ./ sqrt((nrm+4*pi/3)) ;
img = cat(3, R, G, B) ;
imshow(img)

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