Laplace Transforms

1. Jun 15, 2013

nobodyuknow

1. The problem statement, all variables and given/known data

Laplace transform: te^t*H(t-1)

2. Relevant equations

L{te^t} = 1/(s-1)^2
L{H(t-1)} = e-s/s

3. The attempt at a solution

Using e^(at)f(t) -> F(s-a)
with..e^(1t)H(t-1)
We can get, e^(-1s-1)/(s-1) = e^(-s-1)/(s-1) [I think?]
Thus applying tf(t) -> -dF/ds
Derive[e^(-s-1)/(s-1)] w.r.t. s, you get:
-se^(-s-1)/(s-1)^2

Which is close, but, not exact.

I know the answer is se^s-1/(s-1)^2

But how do they come to that?
I don't think the order of my working would affect the final outcome.

All help appreciated

Last edited: Jun 15, 2013
2. Jun 15, 2013

vela

Staff Emeritus
You need to show some attempt at solving the problem on your own.

3. Jun 15, 2013

Ray Vickson

No. L{H(t-1)} is not e-s/s = e-1; it is e^(-s)/s. You must learn to use parentheses when writing things in plain text.

If I were doing the problem I would not bother trying to combine LTs using some "rules"; I would just write out the actual integration and go on from there.

4. Jun 15, 2013

nobodyuknow

Thanks, that was a mistake sorry

So basically use the integration rule and integrate te^t and then apply the Heaviside function?
I'm sorry, I'm not too confident with the Integral method, but..

integral{te^t} = (t-1)e^(t-st)

I'm not sure how to work from there.

[Googling Heaviside Integration as of now]

5. Jun 16, 2013

Ray Vickson

What do you mean by the "integral method"? What do you think the Laplace transform of f(t) actually IS? Go back and look at the basic definitions---it is, by definition, an integral of a certain type. All the so-called "rules" you want to use are just properties of integrals, applied to some special cases.

And no, you do NOT do the integral and then apply the Heaviside function; the Heaviside function is part of the definition of f(t) in this case, and comes before any integrations.

It is not at all clear to me that you really understand what you are doing; that is why I suggest you go right back to square one and start with basic definitions.