# How to solve this inverse Laplace transform?

• December 9th 2009, 04:36 AM
essedra
How to solve this inverse Laplace transform?
L^−1 {1 − e^−s/s(1 − e^−2s)} = ?
• December 9th 2009, 04:43 AM
CaptainBlack
Quote:

Originally Posted by essedra
L^−1 {1 − e^−s/s(1 − e^−2s)} = ?

Do you mean:

$
\mathcal{L}^{-1}\left[
\frac{1-e^{-s}}{s(1-e^{-2s})}
\right]=
\mathcal{L}^{-1}\left[
\frac{1}{s(1+e^{-s})}
\right]
$
?

If so please more brackets in future to make your meaning clear.

CB
• December 11th 2009, 05:32 AM
essedra
Quote:

Originally Posted by CaptainBlack
Do you mean:

$
\mathcal{L}^{-1}\left[
\frac{1-e^{-s}}{s(1-e^{-2s})}
\right]=
\mathcal{L}^{-1}\left[
\frac{1}{s(1+e^{-s})}
\right]
$
?

If so please more brackets in future to make your meaning clear.

CB

Yes, this was the expression that I've tried to write... How can I solve this..?
• December 11th 2009, 06:28 AM
Jester
Quote:

Originally Posted by essedra
Yes, this was the expression that I've tried to write... How can I solve this..?

Quote:

Originally Posted by CaptainBlack
Do you mean:

$
\mathcal{L}^{-1}\left[
\frac{1-e^{-s}}{s(1-e^{-2s})}
\right]=
\mathcal{L}^{-1}\left[
\frac{1}{s(1+e^{-s})}
\right]
$
?

If so please more brackets in future to make your meaning clear.

CB

With CB's simplification re-write as

$
\mathcal{L}^{-1}\left[
\frac{1}{s(1+e^{-s})}
\right] =
\mathcal{L}^{-1}\left[
\frac{1}{s} - \frac{e^{-s}}{s} + \frac{e^{-2s}}{s} - \frac{e^{-3s}}{s} \pm \cdots
\right]
$

The inverse Laplace transform of each gives a series of step functions and together a square wave.
• January 15th 2010, 07:51 AM
mabel lizzy
Quote:

Originally Posted by Danny
With CB's simplification re-write as

$
\mathcal{L}^{-1}\left[
\frac{1}{s(1+e^{-s})}
\right] =
\mathcal{L}^{-1}\left[
\frac{1}{s} - \frac{e^{-s}}{s} + \frac{e^{-2s}}{s} - \frac{e^{-3s}}{s} \pm \cdots
\right]
$

The inverse Laplace transform of each gives a series of step functions and together a square wave.

Can you show me the Laplace inverse of -e^-s/s(Wait)
• January 15th 2010, 02:55 PM
mr fantastic
Quote:

Originally Posted by mabel lizzy
Can you show me the Laplace inverse of -e^-s/s(Wait)

You're expected to know that $LT^{-1}\left[ e^{-as} f(s) \right] = F(t - a)$ for $t > a$ and zero otherwise, where $F(t) = LT^{-1}[f(s)]$.