Explained: "laplace inverse of s/(s^2+16)^2"

faheem khan

Answered question

2022-07-04

laplace inverse of s/(s^2+16)^2

Answer & Explanation

nick1337

Expert2023-05-28Added 777 answers

To find the Laplace inverse of

\frac{s}{(s^{2} + 16)^{2}}

, we can use the Laplace transform table and the properties of Laplace transforms.
Let's start by writing the given function in a more convenient form:

\frac{s}{(s^{2} + 16)^{2}} = \frac{s}{(s^{2} + 4^{2})^{2}}

Using the Laplace transform property

ℒ {t^{n}} = \frac{n!}{s^{n + 1}}

, we can rewrite the given function as:

\frac{s}{(s^{2} + 4^{2})^{2}} = \frac{d}{d s} (\frac{- 1}{s^{2} + 4^{2}})

Now, we can apply the inverse Laplace transform to find the original function. Let's denote the inverse Laplace transform operator as

ℒ^{- 1}

ℒ^{- 1} {\frac{d}{d s} (\frac{- 1}{s^{2} + 4^{2}})} = - ℒ^{- 1} {\frac{1}{s^{2} + 4^{2}}}^{'}

Using the Laplace transform property

ℒ^{- 1} {\frac{F^{'} (s)}{s}} = \int_{0}^{t} f (τ) d τ

, where

F (s) = ℒ {f (t)}

, we have:

- ℒ^{- 1} {\frac{1}{s^{2} + 4^{2}}}^{'} = - \int_{0}^{t} f (τ) d τ

where

f (τ)

is the inverse Laplace transform of

\frac{1}{s^{2} + 4^{2}}

.
The inverse Laplace transform of

\frac{1}{s^{2} + 4^{2}}

can be found using the Laplace transform table:

ℒ^{- 1} {\frac{1}{s^{2} + 4^{2}}} = \sin (4 t)

Substituting this result back into our expression, we get:

- \int_{0}^{t} \sin (4 τ) d τ

Integrating this expression, we have:

- {[- \frac{1}{4} \cos (4 τ)]}_{0}^{t} = \frac{1}{4} (1 - \cos (4 t))

Therefore, the Laplace inverse of

\frac{s}{(s^{2} + 16)^{2}}

\frac{1}{4} (1 - \cos (4 t))

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