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


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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!