Inverse Laplace Transform Involving a Step (Heaviside) Function

1. Nov 12, 2007

cswall

Hello,

I have a fairly straight forward question regarding an inverse laplace transform. The function follows along with the solution. The solution (provided in my text) has been confirmed by Maple; however, i cannot account for the (1/3) in the solution. Any help would be greatly appreciated; thanks in advance!

function to transform:
$$F(s)=\frac{e^{-2s}}{(s+2)*(s-1)}$$
for Maple: exp(-2*s)/((s+2)*(s-1))

solution:
$$f(t)=\frac{1}{3}*Heaviside(t-2)*(-e^{-2t+4}+e^{t-2})$$
for Maple: (1/3)*Heaviside(t-2)*(-exp(-2*t+4)+exp(t-2))

On a side note, this is my first post, and any suggestions regarding my formatting, etiquette, or topic selection would also be appreciated.

Last edited: Nov 12, 2007
2. Nov 12, 2007

rbj

you need to think of it as

$$F(s) = e^{-2s} G(s)$$

where

$$G(s)=\frac{1}{(s+2)(s-1)} = \frac{-1/3}{s+2}+\frac{1/3}{s-1}$$

(i did the partial fraction expansion for you since it was so easy.)

now, find out what the inverse L.T. of G(s) is (gonna be unstable or unbounded, but who cares?), then, knowing that intermediate result, tell us what multiplying e-2s does to it.