Find the inverse laplace trans.

$F\left(s\right)=\frac{10}{s({s}^{2}+9)}$

tricotasu
2021-02-19
Answered

Find the inverse laplace trans.

$F\left(s\right)=\frac{10}{s({s}^{2}+9)}$

You can still ask an expert for help

avortarF

Answered 2021-02-20
Author has **113** answers

Step 1

Given,

$F\left(s\right)=\frac{10}{s({s}^{2}+9)}$

Find the inverse Laplace transform of this function.

Step 2

$F\left(s\right)=\frac{10}{s({s}^{2}+9)}$

$=\frac{10}{9s}-\frac{10s}{9({s}^{2}+9)}$

Taking inverse Laplace transform of both sides,

${L}^{-1}\left[F\left(s\right)\right]={L}^{-1}[\frac{10}{9s}-\frac{10s}{9({s}^{2}+9)}]$

Then,$f\left(t\right)={L}^{-1}\left[\frac{10}{9s}\right]-{L}^{-1}\left[\frac{10s}{9({s}^{2}+9)}\right]$

$=\frac{10}{9}{L}^{-1}\left[\frac{1}{s}\right]-\frac{10}{9}{L}^{-1}\left[\frac{s}{{s}^{2}+9}\right]$

$=\frac{10}{9}{L}^{-1}\left[\frac{1}{s}\right]-\frac{10}{9}\cdot {L}^{-1}\left[\frac{s}{{s}^{2}+{3}^{2}}\right]$

Step 3

Use the formula such that

${L}^{-1}\left[\frac{1}{s}\right]=t$

${L}^{-1}\left[\frac{s}{{s}^{2}+{a}^{2}}\right]=\mathrm{cos}\left(at\right)$

$f\left(t\right)=\frac{10}{9}\cdot t-\frac{10}{9}\cdot \mathrm{cos}\left(3t\right)$

$=\frac{10}{9}[t-\mathrm{cos}\left(3t\right)]$

Step 4

Hence,

$f\left(t\right)=\frac{10}{9}[t-\mathrm{cos}\left(3t\right)]$

Given,

Find the inverse Laplace transform of this function.

Step 2

Taking inverse Laplace transform of both sides,

Then,

Step 3

Use the formula such that

Step 4

Hence,

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