Inverse Laplace Transform

**raed** · August 18th 2010, 04:38 AM

Hi All;
I need the solution of the following question :
Find the inverse Laplace Transform of $\frac{1}{s^n-1} ; n$ is postive integer

**yeKciM** · August 18th 2010, 05:47 AM

Originally Posted by raed

Hi All;
I need the solution of the following question :
Find the inverse Laplace Transform of $\frac{1}{s^n-1} ; n$ is postive integer

use formula for inverse Laplace transformation

just solve integral...

$\displaystyle f(t)= \frac {1}{2\pi i} \lim_{T \to \infty } \int _{\sigma - iT} ^{\sigma + iT } F(S) e^{st} ds$

**chisigma** · August 18th 2010, 07:23 AM

A confortable alternative is to develop $F(s)$ in partial fractions...

$\displaystyle \frac{1}{s^{n}-1} = \sum_{k=0}^{n-1}\frac {r_{k}}{s-e^{i 2 \pi \frac{k}{n}}} = \sum_{k=0}^{n-1}\frac {e^{i 2 \pi \frac{k}{n}}}{n\ (s-e^{i 2 \pi \frac{k}{n}})}$ (1)

... so that is...

$\displaystyle \mathcal {L}^{-1} \{\frac{1}{s^{n}-1} \} = \sum_{k=0}^{n-1}\frac {e^{i 2 \pi \frac{k}{n}}}{n}\ e^{i 2 \pi \frac{k}{n}\ t}$ (2)

Kind regards

$\chi$ $\sigma$

**Ackbeet** · August 18th 2010, 07:37 AM

Which, considering the contour you'd have had to integrate over, and the residue calculus you'd have had to do in order to compute the integral, probably amounts to the same underlying math.

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