1. ## Inverse Laplace transform

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?

2. ## Re: Inverse Laplace transform

Originally Posted by ajithmk91
Can i find the inverse laplace transform for s^2?
No, because $\mathcal{L}\{f(t)\}=F(s)\Rightarrow \lim_{s\to +\infty}F(s)=0$

3. ## Re: Inverse Laplace transform

I thought the inverse laplace of s^2 was $\delta ''(t)$

4. ## Re: Inverse Laplace transform

But the delta-dirac function seems to violate the formal definition. Is that so?

5. ## Re: Inverse Laplace transform

Okay I do a little signal processing, and knowing that s^2 is the LT of the double derivative of the Dirac delta is like Signals 101, very basic. You can google that anyay

6. ## Re: Inverse Laplace transform

Originally Posted by rebghb
Okay I do a little signal processing, and knowing that s^2 is the LT of the double derivative of the Dirac delta is like Signals 101, very basic. You can google that anyay
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 $\mathcal{L}\{f(t) \}=s^2$.

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? );

7. ## Re: Inverse Laplace transform

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