ringearV

2020-12-24

Find the laplace transform of the following:
MULTIPLICATION BY POWER OF t
$g\left(t\right)=\left({t}^{2}-3t+2\right)\mathrm{sin}\left(3t\right)$

aprovard

Step 1
To find Laplace transform of $g\left(t\right)=\left({t}^{2}-3t+2\right)\mathrm{sin}\left(3t\right)$, we simplify the expression and we get
$g\left(t\right)={t}^{2}\mathrm{sin}\left(3t\right)-3t\mathrm{sin}\left(3t\right)+2\mathrm{sin}\left(3t\right)$
Using the following results:
$1.L\left(f\left(t\right)\right)=F\left(s\right)$
$2.L\left({t}^{n}f\left(t\right)\right)=\left(-1{\right)}^{n}\frac{{d}^{n}F\left(s\right)}{d{s}^{n}}$
$3.L\left(\mathrm{sin}\left(at\right)\right)=\frac{a}{{s}^{2}+{a}^{2}}$
Step 2
Using the above results and finding the required transforms,
$L\left(\mathrm{sin}\left(3t\right)\right)=\frac{3}{{s}^{2}+{3}^{2}}=\frac{3}{{s}^{2}+9}$
$L\left({t}^{2}\mathrm{sin}\left(3t\right)\right)=\left(-1\right)\frac{2}{d}$
Now,
$L\left(-3t\mathrm{sin}\left(3t\right)\right)=-3\left(-1\right)\frac{d}{ds}\left(L\left(\mathrm{sin}\left(3t\right)\right)\right)$
$=3\frac{d}{ds}\left(\frac{3}{{s}^{2}+9}\right)$
$=3\left(\frac{-3\cdot 2s}{\left({s}^{2}+9{\right)}^{2}}\right)$
$=\frac{-18s}{\left({s}^{2}+9{\right)}^{2}}$
Now,
$L\left(2\mathrm{sin}\left(3t\right)\right)=2\frac{3}{\left({s}^{2}+9\right)}$
$=\frac{6}{\left({s}^{2}+9\right)}$
Step 3
Therefore,
$L\left(g\left(t\right)\right)=L\left({t}^{2}\mathrm{sin}\left(3t\right)-3t\mathrm{sin}\left(3t\right)+2\mathrm{sin}\left(3t\right)\right)$
$=L\left[{t}^{2}\mathrm{sin}\left(3t\right)\right]+L\left[-3t\mathrm{sin}\left(3t\right)\right]+L\left[2\mathrm{sin}\left(3t\right)\right]$
Substituting the values of Laplace transforms from above step, we get
$G\left(s\right)=\frac{6\left[3{s}^{2}-9\right]}{\left({s}^{2}+9{\right)}^{3}}-\frac{18s}{\left({s}^{2}+9{\right)}^{2}}+\frac{6}{\left({s}^{2}+9\right)}$

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