How to solve this inverse Laplace transform?

**essedra** · December 9th 2009, 04:36 AM

L^−1 {1 − e^−s/s(1 − e^−2s)} = ?

**CaptainBlack** · December 9th 2009, 04:43 AM

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

**essedra** · December 11th 2009, 05:32 AM

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..?

**Danny** · December 11th 2009, 06:28 AM

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.

**mabel lizzy** · January 15th 2010, 07:51 AM

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

**mr fantastic** · January 15th 2010, 02:55 PM

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 $F(t) = LT^{-1}[f(s)]$ .

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