# Inverse Laplace transform of Second Order System underdamped

I was wondering how to perform this

to

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

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

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This is straight forward (just plugging the formulas):

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

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

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

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

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

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

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If you take $\frac{\zeta}{\sqrt{1 + \zeta^2}}$ out as a constant, it works. But unfortunately my inverse tables to not have those formulas on them. Also one would hope it works because it is the basis of much of control system theory. –  Supernovah Nov 8 '11 at 7:19
Yes, thats correct. Please note that Laplace transform (and its inverse) is defined in terms of integral, which by the rules of CALCULUS is LINEAR. So, it is correct that $L(af+bg)=aF(s)+bG(s)$ as well as its inverse, $L^{-1}(aF+bG)=af(t)+bg(t)$. –  Tapu Nov 8 '11 at 8:02