Question

The Inverse laplace transform of \[ \fract{2 e^{- \pi s} }{s^2 + 2s +2} \]

Is the following but I cant figure out how to get there, can someone help me? Thanks

\[ -2 e^{\pi-t} sin(t) \]

Answer 1

\[ \frac{2e^{−πs}}{s2+2s+2} \] Sorry

Answer 2

\[ \frac{2 e^{-\pi s}}{s^2 + 2s+ 2} \]

Answer 3

First complete the square in the denominator. Doesn't look like it factors nicely.

Answer 4

use convolution ... not sure if I used correct formula.
\[ \frac{2 e^{-\pi s}}{s^2 + 2s+ 2} = \frac{2 e^{-\pi s}}{(s+1)^2 + 1} = \int_0^t u(\tau - \pi ) \sin (\tau) e^{-\tau} d\tau \]

Answer 5

woops!! looks like o

Answer 6

I'm not so sure convolution is necessary here.

Given some Laplace transform \(e^{-as}F(s)\), its inverse transform would have the form \(f(t-a)\).

Given some transform \(F(s+c)\), its inverse would have the form \(e^{-ct}f(t)\).

Use these two properties for this problem.