I'm having trouble finding the inverse Laplace transform of: $$G(s) = \frac{1}{2s^2}e^{-Cs}$$ where C is a constant.
I know that the inverse Laplace transform of $s^{-2}$ on its own is $t$, and that the inverse transform of $e^{-Cs}$ on its own is $\delta{(t-C)}$, but I'm unable to combine these results.
Could someone please help?
Well, we are trying to find the following inverse Laplace transform:
$$\text{y}\left(x\right):=\mathscr{L}_\text{s}^{-1}\left[\frac{\exp\left(-\alpha\text{s}\right)}{\text{n}\text{s}^\beta}\right]_{\left(x\right)}=\frac{1}{\text{n}}\cdot\mathscr{L}_\text{s}^{-1}\left[\frac{\exp\left(-\alpha\text{s}\right)}{\text{s}^\beta}\right]_{\left(x\right)}\tag1$$
Using the 'time shifting' property of the Laplace transform, we can write:
$$\text{y}\left(x\right)=\frac{\theta\left(x-\alpha\right)}{\text{n}}\cdot\mathscr{L}_\text{s}^{-1}\left[\frac{1}{\text{s}^\beta}\right]_{\left(x-\alpha\right)}=\frac{\theta\left(x-\alpha\right)}{\text{n}}\cdot\frac{\left(x-\alpha\right)^{\beta-1}}{\Gamma\left(\beta\right)}\tag2$$