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

Question

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

CoormaBak9 2021-01-31 Answered

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

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Expert Answer

Khribechy

Answered 2021-02-01 Author has 100 answers

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

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