Why is it that one finds the degree of numerator always less than that of the denominator when applying inverse laplace transform? For example, Can i find the inverse laplace transform for s^2?
Just to thow in my 2 cents....
First the questions you have asked has multiple different answers and Both correct answers have been given.
First there is no FUNCTION that satisfies $\displaystyle \mathcal{L}\{f(t) \}=s^2$.
Now if you want to use some measure theorey and think of the problem in terms of distributions. http://en.wikipedia.org/wiki/Distribution_(mathematics) and use weak differentation then the solution is correct that it is the 2nd derivative of the delta distribution.
@rebghb why do you know that it is true? How can it be proved? );
Well the s-domain equivelent of the derivitave operator in time is the multiplication by "s". Since the LT of a Dirac delta is 1, then s^2 = s^2*1 is the LT of the double derivaive of the Dirac delta.
Unfortunately, I do not know how to get this answer from the definition of the transform, I know how to get the LT of a delta but not its derivatives, I would be grateful if you would point that out.
Thank you