what is the drawback in Laplacian of Gaussian filter? why are we going for Difference of gaussian?

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1 Answers
There is no drawback in Laplace of Gaussian. I use it all the time. Difference of Gaussians is an approximation, but both need the same amount of computation:
LoG: convolution with the second derivative along x of a Gaussian + convolution with the second derivative along y of a Gaussian
DoG: convolution with a Gaussian - convolution with another Gaussian
Each of those convolutions is a separable operation, so both require 4 1D convolutions, and 1 intermediate image to store one of the two results.
Many people implement these operations differently, for example the LoG as a convolution with a Gaussian and then with a discrete Laplace operator. This is, again, an approximation, and could be slightly faster.
There are also separable approximations to the DoG (which require thus only 2 1D convolutions), but these are much less isotropic (which means not invariant to rotations of the image).
Little known fact: as the two sigmas in the Difference of Gaussians approach each other, the approximation becomes more similar to the Laplace of Gaussian.
EDIT: I have just posted a more elaborate answer over at Signal Processing.

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Thank you! Could you tell me some applications of DoG where as LoG doesn't have? – van Mar 19 '18 at 05:31
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@van4: The DoG appears naturally in scale spaces. You build a scale space by filtering the image with a Gaussian at a series of different sigmas. The difference between subsequent scales is a DoG. These differences then form a sort of "band-pass" space, each one contains image details of different sizes. – Cris Luengo Mar 19 '18 at 12:51
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And obviously, didn't think about mentioning this tastier, a Gaussian is a low-pass filter, the difference between two low-pass filters is a band-pass filter. With a DoG you can tune both limits of the band-pass filter. – Cris Luengo Mar 19 '18 at 12:54