inverse laplace transform

**hurz** · July 13th 2011, 08:39 AM

How to show that
$L^{-1} [\frac{6}{(s^2+2s+2)^2}]=3\exp(-t)sin(t)-3t\exp(-t)cos(t)$
where $L^{-1}[F(s)] = f(t)$ is the inverse laplace ?
Ty

**FernandoRevilla** · July 13th 2011, 09:36 AM

$\mathcal{L}\{3e^{-t}\sin t-3e^{-t}\cos t\}=3\mathcal{L}\{e^{-t}\sin t\}-3\mathcal{L}\{e^{-t}\cos t\}$ . Now use

$\mathcal{L}\{e^{-at}\sin bt\}=\frac{b}{(s+a)^2+b^2},\quad \mathcal{L}\{e^{-at}\cos bt\}=\frac{s+a}{(s+a)^2+b^2}$ .

**hurz** · July 13th 2011, 10:26 AM

Originally Posted by FernandoRevilla

$\mathcal{L}\{3e^{-t}\sin t-3e^{-t}\cos t\}=3\mathcal{L}\{e^{-t}\sin t\}-3\mathcal{L}\{e^{-t}\cos t\}$ . Now use

$\mathcal{L}\{e^{-at}\sin bt\}=\frac{b}{(s+a)^2+b^2},\quad \mathcal{L}\{e^{-at}\cos bt\}=\frac{s+a}{(s+a)^2+b^2}$ .

Done, thanks!
Now i have to find
$L^{-1} [\frac{6}{(s^2+2s+2)^2}]$
without knowing the result previously.
Thank

**waqarhaider** · July 13th 2011, 10:33 AM

use partial fractions

**General** · July 14th 2011, 12:50 PM

Partial Fraction Decomposition will not do anything here.

Use the Convolution Theorem.

**chisigma** · July 15th 2011, 10:11 AM

Originally Posted by General

Partial Fraction Decomposition will not do anything here.

Use the Convolution Theorem.

The way indicated by General is the 'winning key'!...

Setting $F(s)= \frac{1}{s^{2}+2\ s + 2}= \frac{1}{1+(s+1)^2}$ is $\mathcal{L}^{-1} \{F(s)\}= e^{-t}\ \sin t$ , so that is...

$\mathcal{L}^{-1} \{F^{2}(s)\} = \int_{0}^{t} e^{-\tau}\ e^{\tau - t}\ \sin \tau\ \sin (t-\tau)\ d \tau =$

$= e^{-t}\ (\sin t\ \int_{0}^{t} \sin \tau\ \cos \tau\ d \tau - \cos t\ \int_{0}^{t} \sin^{2} \tau\ d \tau) =$

$= \frac{e^{-t}}{2}\ (\sin t - \sin t\ \cos^{2} t - t\ \cos t + \sin t\ \cos^{2} t) =$

$= \frac{e^{-t}}{2}\ \{\sin t - t\ \cos t \}$

Kind regards

$\chi$ $\sigma$

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