How Come the Low Pass Filter in Sobel Operator Isn't Normalized? I am relatively new to the field of computer vision and I have just learnt about the sobel operator. The sobel operator in the x direction is a convolution of the finite difference kernel $[1,0,-1]$ and the gaussian smoothing kernel $[1,2,1]$. Why is it the case that the smoothing kernel does not need to be normalised ?
For example, the vector below convolved with the image will result in pixel intensities that are higher than the original values. Eg, $[50,100,50]$ will result in the middle pixel getting a value of 300 which is not the intended effect of smoothing. If normalisation is applied, then the middle pixel would get a value 75, which blurs the image.
\begin{bmatrix}
1 \\
2 \\
1
\end{bmatrix}
I hope my question was clear in the sense that i don't see how applying $[1,2,1]$ filter results in blurring without normalisation.
EDIT
How the sobel operator is obtained.
The answer is simple, the Sobel Filter is a composition of Lows Pass Filter (LPF) and High Pass Filter (HPF). The composition is done by convolution.
Now, indeed the LPF presented above $ {\left[ 1, 2, 1 \right]}^{T} $ has amplification in the DC value (Its sum is 4 so the amplification is 4). Yet it is convolved with an HPF filter which rejects the DC component.
Convolution is multiplication in the Frequency Domain, since we multiply 4 by zero we essentially get zero.
Actually multiplication of LPF and HPF gives a Band Pass Filter (BPF) (In case they have some overlap in Frequency Domain). Hence in the case above, the Sobel Filter is actually a BPF.