Inverse laplace transform

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

  1. fdsa

    Thread Starter New Member

    Aug 16, 2011
    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


    Mar 31, 2012
    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.