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

Question
Laplace transform
asked 2021-01-31
Compute the Inverse Laplace Transform of $$F(s) =\frac{s}{R^2s^2+16\pi^2}$$
take R=70

## Answers (1)

2021-02-01
Step 1
Given that, R = 70.
The given function is, $$F(s) =\frac{s}{R^2s^2+16\pi^2}$$
$$F(s)=\frac{s}{70^2s^2+16\pi^2}$$
$$F(s)=\frac{s}{4900s^2+16\pi^2}$$
$$f(t)=L^{-1}\left(\frac{s}{4900s^2+16\pi^2}\right)$$
It is known that, $$L^{-1}(sF(s))=\frac{d}{(dt)}f(t)+f(0)$$
Step 2
Apply the above property and obtain the inverse Laplace transform of the given function as shown below.
$$f(t)=L^{-1}\left(\frac{s}{4900s^2+16\pi^2}\right)$$
$$f(t)=L^{-1}\left(\frac{1}{4900s^2+16\pi^2}\right)+L^{-1}\left(\frac{1}{4900s^2+16\pi^2}\right)(0)$$
$$f(t)=\frac{d}{(dt)}\left(\frac{1}{280} \sin \left(\frac{(2\pi t)}{35}\right)\right)+0$$
$$f(t)=\frac{1}{4900} \cdot \cos\left(\frac{2\pi t}{35}\right)$$
Thus, the required inverse Laplace transform is obtained.

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