The derivative of a 1D signal can be obtained from convolution with the kernel [1 -1]. To get the derivative of a 2D image, can someone help me with what the kernel is - something like [1 -1] [-1 1]? I guess. And for the second derivative, in 1D it's [-1 2 -1] and for 2D [ 1 -2 1] [-2 4 -2] [ 1 -2 1]? Thanks for helping!
Convolution kernel for derivative in 2D
Started by ●June 16, 2004
Reply by ●June 17, 20042004-06-17
trouble wrote:> > The derivative of a 1D signal can be obtained from convolution with > the kernel [1 -1].No it can't. By definition the derivative applies to a continuous function. What you are describing is very much like the derivative in some respects, but not exactly the same.>To get the derivative of a 2D image, can someone > help me with what the kernel is - something like > > [1 -1] > [-1 1]? >Yes, this would be analogous to the 1d case. That is, it is equivalent to applying the kernel [1 -1] separately in x and y. But, you may find that the half pixel shift in the result is a nuisance. To avoid that you might use: 1 0 -1 0 0 0 1 0 -1 or maybe: 0 1 0 1 0 -1 0 -1 0 or any number of other more complex symmetrical operators have been used.> I guess. And for the second derivative, in 1D it's [-1 2 -1] and for > 2D > > [ 1 -2 1] > [-2 4 -2] > [ 1 -2 1]? >That would be equivalent to applying your original kernel twice. -jim -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =-----