Using Inverse Laplace to find the frequency response of a transfer function - Help needed!

The frequency response is the inverse Laplace transform of a transfer function. I am tasked to apply the inverse Laplace on the transfer function below in order to convert it into the time domain.

$$H(s) =\frac{20\cdot10^3s}{{s^2} +{20\cdot10^3s}+{500\cdot10^6}}$$

So far, I have had no success in doing so. Any help in solving this would be greatly appreciated. Note that I have to use Laplace transforms to solve this.

Solution 1:

The 'trick' is to write the transfer function in zero-pole form $$ H(s) = \frac{\alpha s}{(s-p)(s-p^*)} $$ where $p = a+ib$.

Then use partial fraction decomposition to discover that $$ H(s) = \frac{\alpha}{p-p^*} \left[ \frac{p}{s-p}-\frac{p^*}{s-p^*} \right] $$

Then you can use inverse Laplace transform \begin{eqnarray*} h(t) &=& \frac{\alpha}{p-p^*} \left[ p e^{p t} - p^* e^{p^* t} \right]\\ &=& \frac{\alpha}{2ib} e^{at} \left[ p e^{ibt} - p^* e^{-ibt} \right] \\ &=& \frac{\alpha}{2ib} e^{at} \left[ (a+ib) e^{ibt} - (a-ib) e^{-ibt} \right] \\ &=& \alpha e^{at} \left[ \cos(bt)+\frac{a}{b} \sin(bt) \right] \end{eqnarray*}

This seems reasonable since $$ H(s) = \frac{\alpha s}{(s-a)^2+b^2} = \alpha \frac{(s-a)+(a/b)b}{(s-a)^2+b^2} $$