# Compute the Inverse Laplace Transform of F(s) =frac{s}{R^2s^2+16pi^2}take R=70

Compute the Inverse Laplace Transform of $F\left(s\right)=\frac{s}{{R}^{2}{s}^{2}+16{\pi }^{2}}$
take $R=70$

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Step 1
Given that, $R=70$.
The given function is, $F\left(s\right)=\frac{s}{{R}^{2}{s}^{2}+16{\pi }^{2}}$
$F\left(s\right)=\frac{s}{{70}^{2}{s}^{2}+16{\pi }^{2}}$
$F\left(s\right)=\frac{s}{4900{s}^{2}+16{\pi }^{2}}$
$f\left(t\right)={L}^{-1}\left(\frac{s}{4900{s}^{2}+16{\pi }^{2}}\right)$
It is known that, ${L}^{-1}\left(sF\left(s\right)\right)=\frac{d}{\left(dt\right)}f\left(t\right)+f\left(0\right)$
Step 2
Apply the above property and obtain the inverse Laplace transform of the given function as shown below.
$f\left(t\right)={L}^{-1}\left(\frac{s}{4900{s}^{2}+16{\pi }^{2}}\right)$
$f\left(t\right)={L}^{-1}\left(\frac{1}{4900{s}^{2}+16{\pi }^{2}}\right)+{L}^{-1}\left(\frac{1}{4900{s}^{2}+16{\pi }^{2}}\right)\left(0\right)$
$f\left(t\right)=\frac{d}{\left(dt\right)}\left(\frac{1}{280}\mathrm{sin}\left(\frac{\left(2\pi t\right)}{35}\right)\right)+0$
$f\left(t\right)=\frac{1}{4900}\cdot \mathrm{cos}\left(\frac{2\pi t}{35}\right)$
Thus, the required inverse Laplace transform is obtained.