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Inverse Laplace transform

Alexmahone

Active member
Jan 26, 2012
268
Find the inverse Laplace transform of $\displaystyle \frac{2s+7-e^{-2s}}{(s+1)^2}$.
 
Last edited:

dwsmith

Well-known member
Feb 1, 2012
1,673
Find the inverse Laplace transform of $\displaystyle\frac{2s+5-e^{-2s}}{s^2+s+1}$.
$s^2 + s + 1/4 + 1 - 1/4 = (s + 1/2)^2 + 3/4$

Then break up the numerator.
 

Alexmahone

Active member
Jan 26, 2012
268
$s^2 + s + 1/4 + 1 - 1/4 = (s + 1/2)^2 + 3/4$

Then break up the numerator.
I changed the question. (Sorry about that.)
 

dwsmith

Well-known member
Feb 1, 2012
1,673
I changed the question. (Sorry about that.)
Then look at

$$
\frac{2s+7}{(s+1)^2} - \frac{e^{-2s}}{(s+1)^2}
$$

The formula for the second piece is

$$
\frac{(t-\tau)^n}{n!}e^{-\alpha(t-\tau)}u(t-\tau) = \mathfrak{L}^{-1}\left[\frac{e^{-\tau s}}{(s+\alpha)^{n+1}}\right]
$$

The other one shouldn't be too bad. Just ask if you need help with that one.