opatovaL

2021-02-21

Find the Laplace transforms of the given functions.
$g\left(t\right)=4\mathrm{cos}\left(4t\right)-9\mathrm{sin}\left(4t\right)+2\mathrm{cos}\left(10t\right)$

### Answer & Explanation

Dora

Step 1 : Introduction
Given function is,
$g\left(t\right)=4\mathrm{cos}\left(4t\right)-9\mathrm{sin}\left(4t\right)+2\mathrm{cos}\left(10t\right)$
We have to evaluate the Laplace transform of given function g(t)
step 2
$g\left(t\right)=4\mathrm{cos}\left(4t\right)-9\mathrm{sin}\left(4t\right)+2\mathrm{cos}\left(10t\right)$
using Laplace transform properties
$L\left\{\mathrm{sin}at\right\}=\frac{a}{{s}^{2}+{a}^{2}}$
$L\left\{\mathrm{cos}at\right\}=\frac{s}{{s}^{2}+{a}^{2}}$
Applying Laplace transform on g(t) function
$\therefore L\left\{g\left(t\right)\right\}=L\left\{4\mathrm{cos}\left(4t\right)-9\mathrm{sin}\left(4t\right)+2\mathrm{cos}\left(10t\right)\right\}$
$L\left\{g\left(t\right)\right\}=4L\left\{\mathrm{cos}\left(4t\right)\right\}-9L\left\{\mathrm{sin}\left(4t\right)\right\}+2L\left\{\mathrm{cos}\left(10t\right)\right\}$
$=4\left(\frac{s}{{s}^{2}+{4}^{2}}\right)-9\left(\frac{4}{{s}^{2}+{4}^{2}}\right)+2\left(\frac{s}{{s}^{2}+{10}^{2}}\right)$
$=\frac{4s}{{s}^{2}+16}-\frac{36}{{s}^{2}+16}+\frac{2s}{{s}^{2}+100}$
Thus , Laplace transform of given function g(t) is
$L\left\{g\left(t\right)\right\}=\frac{4s}{{s}^{2}+16}-\frac{36}{{s}^{2}+16}+\frac{2s}{{s}^{2}+100}$

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