dalllc

2022-10-06

Calculate inverse Laplace transform of $\frac{{e}^{-as}}{{s}^{2}}$
$F\left(s\right)=\frac{{e}^{-as}}{{s}^{2}}\phantom{\rule{2em}{0ex}}a\in \mathbb{R}$
${\mathcal{L}}^{-1}\left\{\frac{1}{{s}^{2}}\right\}\left(t\right)=t$
So (for the time delay property):
$f\left(t\right)=\left(t-a\right)u\left(t-a\right)$
Is it correct?

dheasca8d

We can use, when $\mathrm{\Re }\left(\text{s}\right)>0$:
${\mathcal{L}}_{\text{s}}^{-1}{\left[\frac{{e}^{-\text{a}\text{s}}}{{\text{s}}^{\text{n}}}\right]}_{\left(t\right)}=\frac{{\left(t-\text{a}\right)}^{\text{n}-1}\theta \left(t-\text{a}\right)}{\mathrm{\Gamma }\left(\text{n}\right)}$
Where $\theta \left(t\right)$ is he Heaviside theta function.
To solve it a harder way we can use the convolution theorem, when $\mathrm{\Re }\left(\text{s}\right)>0$:
${\mathcal{L}}_{\text{s}}^{-1}{\left[\frac{{e}^{-\text{a}\text{s}}}{{\text{s}}^{\text{n}}}\right]}_{\left(t\right)}={\mathcal{L}}_{\text{s}}^{-1}{\left[{e}^{-\text{a}\text{s}}\right]}_{\left(t\right)}\ast {\mathcal{L}}_{\text{s}}^{-1}{\left[\frac{1}{{\text{s}}^{\text{n}}}\right]}_{\left(t\right)}=\delta \left(t-\text{a}\right)\ast \frac{{t}^{\text{n}-1}}{\mathrm{\Gamma }\left(\text{n}\right)}$

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