Radial gradient using Convolution

2 vues (au cours des 30 derniers jours)
RUPESH
RUPESH le 25 Avr 2013
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..

Réponses (1)

Image Analyst
Image Analyst le 25 Avr 2013
Modifié(e) : Image Analyst le 25 Avr 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.
  1 commentaire
Cedric
Cedric le 25 Avr 2013
Modifié(e) : Cedric le 25 Avr 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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