Is the Sobel Filter meant to be normalized?

Question

The x-derivative Sobel looks that way:

-1 0 +1
-2 0 +2
-1 0 +1

Lets say there are two samples of my image which look like that (0=black, 1=white):

0 0 1            1 0 0
0 0 1      &     1 0 0
0 0 1            1 0 0

If I perform convolution I'll end up with 4 and -4 respectively.

So my natural response would be to normalize the result by 8 and translate it by 0.5 - is that correct? (I am wondering as can't find Wikipedia etc. mentioning any normalization)

EDIT: I use the Sobel Filter to create a 2D Structure Tensor (with the derivatives dX and dY):

                   A B 
Structure Tensor = C D

with  A = dx^2 
      B = dx*dy
      C = dx*dy 
      D = dy^2

Ultimately I want to store the result in [0,1], but right now I'm just wondering if I have to normalize the Sobel result (by default, not just in order to store it) or not, i.e.:

A = dx*dx 
//OR
A = (dx/8.0)*(dx/8.0)
//OR
A = (dx/8.0+0.5)*(dx/8.0+0.5)

Answer 1

The Sobel filter is the composition of a finite difference filter in one dimension:

[ 1  0  -1 ] / 2

and a smoothing filter in the other dimension:

[ 1  2  1 ] / 4

Therefore, the proper normalization to the kernel as typically defined is 1/8.

This normalization is required when a correct estimate of the derivative is needed. When computing the gradient magnitude for detecting edges, the scaling is irrelevant.

The 1/4 in the smoothing filter is to normalize it to 1. The 1/2 in the finite difference filter comes from the distance between the two pixels compared. The derivative is defined as the limit of h to zero of [f(x+h)-f(x)]/h. For the finite difference approximation, we can choose h=1, leading to a filter [1,-1], or h=2, leading to the filter above. The advantage with h=2 is that the filter is symmetric, with h=1 you end up computing the derivative in the middle between the two pixels, thus the result is shifted by half a pixel.