Can we rotate sobel operator and still get the same results for gradient image

5 views (last 30 days)
I have a 5*5 image and I applied imgradient() on it and got the directions(just focussing on directions). I also tried to calculate the directions for the same image on paper using the following sobel operators.
gx=[-1 0 1; -2 0 2; -1 0 1] gy=[-1 -2 -1; 0 0 0; 1 2 1]
I found theta using
Gdir = atan2(-gy,gx)*180/pi %Note: gy is negative as y moves from top to bottom
and got same answers as were obtained using imgradient().
But when I used
gx=[1 0 -1; 2 0 -2; 1 0 -1] gy=[1 2 1; 0 0 0; -1 -2 -1]
I got different answers. Why?
Can't we rotate sobel operator and still get the same results. Does MATLAB uses a fixed set of sobel operators to find gradient and never uses the rotated version? What is the problem. Please explain.

Answers (1)

Anand
Anand on 9 Feb 2016
The imgradient function uses the following kernels for the 'sobel':
hx = -fspecial('sobel')'
hx =
-1 0 1
-2 0 2
-1 0 1
hy = -fspecial('sobel')
hy =
-1 -2 -1
0 0 0
1 2 1
You get different results with the rotated kernels because the Sobel kernel is not symmetric.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!