Stumped by Laplace Transform

[Steven O.]

I am studying the Laplace Transform, and I was rather puzzled to find
that the standard tables in the two textbooks we have -- and even the
tables I found at the Wolfram Research site -- do not provide an
inverse transform for the function F(s) = s.

In fact, the Wolfram site does not even give the inverse transform for
F(s) = 1, although my text gives that as the Dirac Delta function of
t. However, judging from one of the homework problems, the text I
have seems to clearly imply that the inverse of F(s) = s is the
derivative with respect to time of the Dirac Delta function of t.
However (i) I have no idea how they figure that out, and (ii) I can't
even imagine what the derivative of the Dirac Delta function would be.

Can anyone help me out on either score?

Thanks in advance for all replies.

Steve O.




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[John O'Flaherty]

I am studying the Laplace Transform, and I was rather puzzled to find
that the standard tables in the two textbooks we have -- and even the
tables I found at the Wolfram Research site -- do not provide an
inverse transform for the function F(s) = s.

In fact, the Wolfram site does not even give the inverse transform for
F(s) = 1, although my text gives that as the Dirac Delta function of
t. However, judging from one of the homework problems, the text I
have seems to clearly imply that the inverse of F(s) = s is the
derivative with respect to time of the Dirac Delta function of t.
However (i) I have no idea how they figure that out, and (ii) I can't
even imagine what the derivative of the Dirac Delta function would be.

The inverse transform for 1 is dirac(t),as you said. For F(s) = s,
Mupad gives dirac(t,1), which is the first derivative of dirac(t), so
that concurs with what you said. This makes sense, since multiplication
by s is differentiation in the frequency domain. The derivative of
dirac(t) is called a unit doublet, consisting of a spike to plus
infinity at 0-, and a spike to minus infinity at 0+. This site confirms
that its Laplace transform is just s:

http://www2.latech.edu/~sajones/BIEN 225 Web Pages/Special Functions.htm
Special Functions

 

[John O'Flaherty]

The inverse transform for 1 is dirac(t),as you said. For F(s) = s,
Mupad gives dirac(t,1), which is the first derivative of dirac(t), so
that concurs with what you said. This makes sense, since multiplication
by s is differentiation in the frequency domain.

Sorry, I should have said that multiplication by s in the frequency
domain corresponds to differentiation in the time domain.