# inverse laplace transform

• Jul 13th 2011, 08:39 AM
hurz
inverse laplace transform
How to show that
$\displaystyle L^{-1} [\frac{6}{(s^2+2s+2)^2}]=3\exp(-t)sin(t)-3t\exp(-t)cos(t)$
where $\displaystyle L^{-1}[F(s)] = f(t)$ is the inverse laplace ?
Ty
• Jul 13th 2011, 09:36 AM
FernandoRevilla
Re: inverse laplace transform
$\displaystyle \mathcal{L}\{3e^{-t}\sin t-3e^{-t}\cos t\}=3\mathcal{L}\{e^{-t}\sin t\}-3\mathcal{L}\{e^{-t}\cos t\}$ . Now use

$\displaystyle \mathcal{L}\{e^{-at}\sin bt\}=\frac{b}{(s+a)^2+b^2},\quad \mathcal{L}\{e^{-at}\cos bt\}=\frac{s+a}{(s+a)^2+b^2}$ .
• Jul 13th 2011, 10:26 AM
hurz
Re: inverse laplace transform
Quote:

Originally Posted by FernandoRevilla
$\displaystyle \mathcal{L}\{3e^{-t}\sin t-3e^{-t}\cos t\}=3\mathcal{L}\{e^{-t}\sin t\}-3\mathcal{L}\{e^{-t}\cos t\}$ . Now use

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

Done, thanks!
Now i have to find
$\displaystyle L^{-1} [\frac{6}{(s^2+2s+2)^2}]$
without knowing the result previously.
Thank
• Jul 13th 2011, 10:33 AM
waqarhaider
Re: inverse laplace transform
use partial fractions
• Jul 14th 2011, 12:50 PM
General
Re: inverse laplace transform
Partial Fraction Decomposition will not do anything here.

Use the Convolution Theorem.
• Jul 15th 2011, 10:11 AM
chisigma
Re: inverse laplace transform
Quote:

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

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

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$