0
$\begingroup$

What's the inverse laplace transform of $$\frac{s}{(s-1/2)^2+1}\:?$$

$\endgroup$
1

1 Answer 1

1
$\begingroup$

Hint: $$\mathcal{L}(e^{-at}\cos\omega t) = \frac{s + a}{(s+a)^2 + \omega^2}$$ $$\mathcal{L}(e^{-at}\sin\omega t) = \frac{\omega}{(s+a)^2 + \omega^2}.$$

So maybe you can rewrite $$\frac{s}{(s-0.5)^2 + 1}$$ as a sum of two such functions and then take the inverse Laplace transform.

Alternatively: If you don't mind dealing with complex numbers, you might consider decomposing your function via partial fractions and using $\mathcal{L}(e^{at}) = 1/(s-a)$, though this might be annoying as a calculation (I haven't worked out the details).

$\endgroup$
2
  • 1
    $\begingroup$ Blah, seems so obvious now. Must be because it's 4 am. Thanks a bunch =) $\endgroup$
    – user9616
    Apr 15, 2011 at 7:34
  • $\begingroup$ @Anthony: Yeah, late night calculations are like that. No problem at all. :-) $\endgroup$ Apr 15, 2011 at 7:35