Inverse of Laplace

22 Views Asked by At

I am studying engineering and I am stuck on a Laplace inverse transformation, could you please help me out?

I am supposed to solve this inverse transformation: $$F(s) = \dfrac {(1 - 2s) }{ s^2+4s+5}$$ So far, I managed to reorganize like that:

$$F(s) = \dfrac {(1 - 2s) }{ (s + 2)^2 + 1}$$

Any ideas about how I could continue?

Thanks in advance

1

There are 1 best solutions below

0
On

$$F(s) = \dfrac {(1 - 2s) }{ (s + 2)^2 + 1}$$ $$F(s) = \dfrac {(5- 2(s+2)) }{ (s + 2)^2 + 1}$$ $$F(s) = \dfrac 5{ (s + 2)^2 + 1}- \dfrac {2(s+2) }{ (s + 2)^2 + 1}$$ Apply the formula: $$\mathscr{L^{-1}} \left( \dfrac b {(s-a)^2+b^2} \right)=e^{at}\sin (bt)$$ $$\mathscr{L^{-1}} \left( \dfrac {s-a} {(s-a)^2+b^2} \right)=e^{at}\cos (b t)$$ You find that: $$f(t)=e^{-2t} (5\sin t-2\cos (t))$$