title: Is the Sobel Filter meant to be normalized?
body: <p>The x-derivative Sobel looks that way: </p>
  
  <pre><code>-1 0 +1
  -2 0 +2
  -1 0 +1
  </code></pre>
  
  <p>Lets say there are two samples of my image which look like that (0=black, 1=white):</p>
  
  <pre><code>0 0 1            1 0 0
  0 0 1      &amp;     1 0 0
  0 0 1            1 0 0
  </code></pre>
  
  <p>If I perform convolution I'll end up with 4 and -4 respectively.</p>
  
  <p>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)</p>
  
  <p><strong>EDIT:</strong>
  I use the Sobel Filter to create a 2D Structure Tensor (with the derivatives dX and dY):</p>
  
  <pre><code>                   A B 
  Structure Tensor = C D
  
  with  A = dx^2 
        B = dx*dy
        C = dx*dy 
        D = dy^2
  </code></pre>
  
  <p>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.: </p>
  
  <pre><code>A = dx*dx 
  //OR
  A = (dx/8.0)*(dx/8.0)
  //OR
  A = (dx/8.0+0.5)*(dx/8.0+0.5)
  </code></pre>