How Come the Low Pass Filter in Sobel Operator Isn't Normalized?

Question

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.

Answer 1

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.

Answer 2

For numerical details, you can check What's Logic Behind the Construction of Sobel's Filter in Image Processing?. Hereafter is an explanation.

For early image preprocessing tasks, normalization is not mandatory, as long as they only add a common multiplication factor for all images. Indeed, one is often mostly interested in the relative "importance" of features, in detection or localization. As much as "integer" pixel values are somewhat arbitrary, multiplying them a constant often does not matter a lot.

Here is an analogy: given an elevation map of some landscape, finding the flattest road, or the two tallest mountains is not very sensitive to measure given in meters or kilometers.

Here, the normalization could be applied to the derivative part as well: to get a correct derivative/slope estimation, one should divide by 2 : $[1,\,0,\,-1]/2$, just like you would want to divide the smoothing part by $4$. But...

But starting with integer pixel values, normalizing yields floating point. Sobel filters were designed when every operation mattered. Here, you only have integer dyadic values in the filter ($0$, $1$, $2$), the most complicated 2-fold product can be implemented by a left-shift. So, unnormalization is a cheap price of the one of "simplest"-to-implement edge detectors.