# Inverse laplace transform

Discussion in 'Homework Help' started by fdsa, Nov 6, 2013.

1. ### fdsa Thread Starter New Member

Aug 16, 2011
9
0
Hello I have the following function in the Laplace domain:

$\frac{1}{s \times (s+1)}$

For which I get the inverse Laplace transform: $1-e^{-t}$ but looking at the solution they claim it's $(1-e^{-t})\times u(t)$ where u(t) is the unit step function. I don't understand why that is?

Also for $\frac{1}{s^2 \times (s+1)}$ i use the method of partial fraction expansion and get $-u(t)+tu(t)+e^{-t}$ but they claim it's $(t-1+e^{-t})u(t)$.

2. ### WBahn Moderator

Mar 31, 2012
20,057
5,644
The Laplace Transform you are working with is the single-sided Laplace Transform, meaning that the function you are taking the transform of has to be identically zero for t<0. Thus, when you take the inverse transform, you have to get as a result a function that is identically zero for t<0.