Calculate inverse Laplace transform of $\frac{e^{-as}}{s^2}$

Question

$$F(s)=\frac{e^{-as}}{s^2} \qquad a \in \mathbb{R}$$

$$\mathscr{L}^{-1}\Big\{\frac{1}{s^2} \Big\}(t)=t$$

So (for the time delay property):

$$f(t)=(t-a) u(t-a)$$

Is it correct?

Thanks!

Answer 1

We can use, when $\Re\left(\text{s}\right)>0$:

$$\mathcal{L}_\text{s}^{-1}\left[\frac{e^{-\text{a}\text{s}}}{\text{s}^\text{n}}\right]_{\left(t\right)}=\frac{\left(t-\text{a}\right)^{\text{n}-1}\theta\left(t-\text{a}\right)}{\Gamma\left(\text{n}\right)}$$

Where $\theta\left(t\right)$ is he Heaviside theta function.

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To solve it a harder way we can use the convolution theorem, when $\Re\left(\text{s}\right)>0$:

$$\mathcal{L}_\text{s}^{-1}\left[\frac{e^{-\text{a}\text{s}}}{\text{s}^\text{n}}\right]_{\left(t\right)}=\mathcal{L}_\text{s}^{-1}\left[e^{-\text{a}\text{s}}\right]_{\left(t\right)}*\mathcal{L}_\text{s}^{-1}\left[\frac{1}{\text{s}^\text{n}}\right]_{\left(t\right)}=\delta\left(t-\text{a}\right)*\frac{t^{\text{n}-1}}{\Gamma\left(\text{n}\right)}$$