# Thread: How to solve this inverse Laplace transform?

1. ## How to solve this inverse Laplace transform?

L^−1 {1 − e^−s/s(1 − e^−2s)} = ?

2. 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

3. 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..?

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

5. 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

6. Originally Posted by mabel lizzy
Can you show me the Laplace inverse of -e^-s/s
You're expected to know that $LT^{-1}\left[ e^{-as} f(s) \right] = F(t - a)$ for $t > a$ and zero otherwise, where $F(t) = LT^{-1}[f(s)]$.