# Is this true: inverse Laplace transform of exp(−s)F(s)=(t−1)u(t−1)u(t−1)

Iris Vaughn 2022-10-12 Answered
$y\left(s\right)=1/s-1/{s}^{2}+\mathrm{exp}\left(-s\right)\cdot \left(1/{s}^{2}\right)$
I'm struggling with
$\mathrm{exp}\left(-s\right)\cdot \left(1/{s}^{2}\right)$
formulas:
$f\left(t-1\right)u\left(t-1\right)\to \mathrm{exp}\left(-s\right)F\left(s\right)$
But
$f\left(t\right)=t\cdot u\left(t\right)$
is this true: inverse Laplace transform of
$\mathrm{exp}\left(-s\right)F\left(s\right)=\left(t-1\right)u\left(t-1\right)u\left(t-1\right)$
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## Answers (1)

flasheadita237m
Answered 2022-10-13 Author has 17 answers
It's simply:
${\mathcal{L}}^{\mathcal{-}\mathcal{1}}\left({e}^{-s}\frac{1}{{s}^{2}}\right)=u\left(t-1\right)\left(t-1\right)$
Since you have that:
${\mathcal{L}}^{\mathcal{-}\mathcal{1}}\left({e}^{-cs}F\left(s\right)\right)=u\left(t-c\right)f\left(t-c\right)$
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