Inverse Laplace transform of Second Order System underdamped

Question

I was wondering how to perform this

$C(s) = \frac{1}{s} + \frac{(s + \zeta\omega_n) + \frac{\zeta}{\sqrt{1 + \zeta^2}}\omega_n\sqrt{1 - \zeta^2}}{(s + \zeta\omega_n)^2 + \omega^2_n(1 - \zeta^2)}$

to

$C(t) = 1 - e^{-\zeta\omega_nt}\left(\cos\omega_n\sqrt{1 - \zeta^2}t + \frac{\zeta}{\sqrt{1 - \zeta^2}}\sin\omega_n\sqrt{1 - \zeta^2}t\right)$

The inverse Laplace transform $C(t)$ of that s-domain $C(s)$ response to a $u(t)$ step

Answer 1

This is straight forward (just plugging the formulas):

1. $L^{-1}\left(\frac{1}{s}\right)=1$

2. $L^{-1}\left(\frac{b}{(s+a)^2+b^2}\right)=e^{-at}\sin bt$

3. $L^{-1}\left(\frac{s+a}{(s+a)^2+b^2}\right)=e^{-at}\cos bt$

Note that in 2. and 3. shifting property have been used. So, your inverse transform

$=L^{-1}\left(\frac{1}{s}\right)+L^{-1}\left(\frac{s+a}{(s+a)^2+b^2}\right)+\frac{\xi}{\sqrt{1+\xi^2}}L^{-1}\left(\frac{b}{(s+a)^2+b^2}\right)$

with appropriate $a$ and $b$. I hope you can do it now, without difficulty. Note that the answer you provided is wrong.