I'm stuck with a question about Inverse Laplace Transform here, but the use of inverse laplace transform equation is forbidden.

Magdalena Norton
2022-04-21
Answered

Inverse Laplace Transform without using equations

I'm stuck with a question about Inverse Laplace Transform here, but the use of inverse laplace transform equation is forbidden.

$Y\left(s\right)=\frac{{e}^{-\pi s}}{s[{(s+1)}^{2}+1]}$

I'm stuck with a question about Inverse Laplace Transform here, but the use of inverse laplace transform equation is forbidden.

You can still ask an expert for help

dagars5nx

Answered 2022-04-22
Author has **12** answers

The inverse Laplace transform of $\frac{1}{{s}^{2}+1}$ is $\mathrm{sin}t$. Therefore we have

${\mathcal{L}}^{-1}\left\{\frac{1}{(s+1{)}^{2}+1}\right\}={e}^{-t}\mathrm{sin}\left(t\right)=\frac{1}{2i}({e}^{(-1+i)t}-{e}^{(-1-i)t})$

Therefore

${\mathcal{L}}^{-1}\left\{\frac{1}{s}\frac{1}{(s+1{)}^{2}+1}\right\}={\int}_{0}^{t}{e}^{-u}\mathrm{sin}\left(u\right)du=\frac{1}{2i}(\frac{{e}^{(-1+i)t}-1}{-1+i}-\frac{{e}^{(-1-i)t}-1}{-1-i})$

$=\frac{1}{2}-\frac{1}{2}{e}^{-t}(\mathrm{sin}\left(t\right)+\mathrm{cos}\left(t\right))$

The last step is to use the shift theorem ${\mathcal{L}}^{-1}\left\{{e}^{-as}F\left(s\right)\right\}=f(t-a)\mathrm{\Theta}(t-a)$ and we get

${\mathcal{L}}^{-1}\left\{{\mathrm{e}}^{-\mathrm{\pi s}}\frac{1}{\mathrm{s}}\frac{1}{(\mathrm{s}+1{)}^{2}+1}\right\}=[\frac{1}{2}-\frac{1}{2}{\mathrm{e}}^{-(\mathrm{t}-\mathrm{\pi})}(\mathrm{sin}(\mathrm{t}-\mathrm{\pi})+\mathrm{cos}(\mathrm{t}-\mathrm{\pi})\left)\right]\mathrm{\Theta}(\mathrm{t}-\mathrm{\pi})$

$=(\frac{1}{2}+\frac{1}{2}{\mathrm{e}}^{\mathrm{\pi}-\mathrm{t}}(\mathrm{cos}\left(\mathrm{t}\right)+\mathrm{sin}\left(\mathrm{t}\right)\left)\right)\mathrm{\Theta}(\mathrm{t}-\mathrm{\pi})$

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