# Differential equations - inverse laplace transforms

1. Mar 7, 2008

### sugarplum31

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

We are given the following LaPlace Transformations and we are to find g(t) and h(t).

L(g)(t) = 1/[(1-e^(-2s))(s-1)]

L(h)(t) = (e^-s)/s(1-e^-s)

2. Relevant equations

By looking at the Table of LaPlace Transformations, the only one I found that could possibly work for both of these is

f(t+T) = f(t) ---> (integral from 0 to T of e^(-st)f(t) dt)/ 1-e^(-sT)

3. The attempt at a solution

For the first one, I broke it into two separate equations, 1/(1-e^-2s) * 1/(s-1) and for the second part it would be e^t. However, I am very confused on the equation I found in the Table and I'm not sure how to use it on either of these.

2. Mar 8, 2008

### Tom Mattson

Staff Emeritus
That formula makes it easy to compute the Laplace transform of a periodic function, but I don't see how it's going to make it easy to do the inverse transform. I think the following is a better approach.

* Use the summation formula for convergent geometric series:

$$\sum_{n=0}^\infty x^n = \frac{1}{1-x}$$, $$|x|<1$$.

Here, $x=e^-Ts$. Since $T>0$, you can apply the above sum formula.

Once you have done that, then...

* Make repeated use the time shift property of Laplace transforms:

$$e^{-Ts}F(s) \longrightarrow f(t-T)U(t-T)$$

I think you'll have more success taking that route.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook