Zane Decker
2022-04-14
Answered

Laplace transform of $f\left(t\right)=t{e}^{-t}\mathrm{sin}\left(2t\right)$

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maggionmoo

Answered 2022-04-15
Author has **16** answers

Here is an approach.

$\mathcal{L}\left(\mathrm{sin}2t\right)=\frac{2}{{s}^{2}+{2}^{2}}=\frac{2}{{s}^{2}+4}$ , using the table.

$\mathcal{L}\left({e}^{-t}\mathrm{sin}2t\right)=\frac{2}{{(s+1)}^{2}+4}$ , using frequency shifting.

$\mathcal{L}\left(t{e}^{-t}\mathrm{sin}2t\right)=-\frac{d}{ds}\left(\frac{2}{{(s+1)}^{2}+4}\right)=\frac{4(s+1)}{{({(s+1)}^{2}+4)}^{2}}$ , using frequency differentiation.

veselrompoikm

Answered 2022-04-16
Author has **12** answers

Let $f\left(t\right)=t{e}^{-t}\mathrm{sin}\left(2t\right)=tg\left(t\right)$ and $g\left(t\right)={e}^{-t}\mathrm{sin}\left(2t\right)={e}^{-t}h\left(t\right)$ with $h\left(t\right)=\mathrm{sin}\left(2t\right)$ . So $F\left(s\right)=-{G}^{\prime}\left(s\right)$ and $G\left(s\right)=H(s+1)$ with $H\left(s\right)=\frac{2}{{s}^{2}+4}$

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