Laplace transform.

Question

This is a past exam question. I'm having a bit of trouble at finding the inverse laplace transform of the following function. Any help would be great. $$\frac{s^2+1}{(s^2+4s+5)^2}$$

Thanks for your help.

Answer 1

From the standard Laplace transforms $$L(\sin at)=\frac{a}{s^2+a^2}\quad\hbox{and}\quad L(\cos at)=\frac{s}{s^2+a^2}$$ and the general rule $$L(tf(t))=-\frac{d}{ds}L(f(t))$$ we obtain $$L(t\sin at)=\frac{2as}{(s^2+a^2)^2}$$ and a similar rule for $L(t\cos at)$. Your function can be written $$\eqalign{\frac{s^2+1}{(s^2+4s+5)^2} &=\frac{(s^2+4s+5)-(4s+4)}{(s^2+4s+5)^2}\cr &=\frac{1}{s^2+4s+5}-\frac{4s+4}{(s^2+4s+5)^2}\cr &=\frac{1}{(s+2)^2+1}-\frac{4(s+2)}{((s+2)^2+1)^2}+\frac{4}{((s+2)^2+1)^2} \ ,\cr}$$ and the result follows from the above and shifting on the $s$ axis.

Answer 2

$$\dfrac{s^2+1}{(s^2+4s+5)^2}=\dfrac{s^2+4s+5-4s-4}{(s^2+4s+5)^2}=$$

$$\dfrac{1}{s^2+4s+5}-\dfrac{4s+4}{(s^2+4s+5)^2}=(\text{1st part is easy})$$

$\dfrac{1}{s^2+4s+5}=\dfrac{1}{(s+2)^2+1^2}=e^{-2t}sin(t)$