avissidep

## Answered question

2020-10-27

a) Find the Laplace transformation of the following function by using definition of Laplace transformation:

### Answer & Explanation

Nathanael Webber

Skilled2020-10-28Added 117 answers

Step 1
Given $f\left(t\right)=\mathrm{cos}\left(pt\right)$
Using the definition of Laplace Transform we have,
$L\left[f\left(t\right)\right]=F\left(s\right)={\int }_{0}^{\mathrm{\infty }}{e}^{-st}f\left(t\right)dt$
$⇒L\left[\mathrm{cos}pt\right]={\int }_{0}^{\mathrm{\infty }}{e}^{-st}\mathrm{cos}ptdt$
Now we will use Integration by parts to solve it

$I=\mathrm{cos}pt\int {e}^{-st}dt-\int \left(-p\mathrm{sin}pt\right)\frac{{e}^{-st}}{-s}dt$
$I=\mathrm{cos}pt\left[\frac{{e}^{-st}}{-s}\right]-\frac{p}{s}\int \mathrm{sin}pt{e}^{-st}dt$
$I=-\frac{1}{s}\mathrm{cos}pt{e}^{-st}-\frac{p}{s}\left[\mathrm{sin}pt\left(\frac{{e}^{-st}}{-s}\right)-\int p\mathrm{cos}pt\left(\frac{{e}^{-st}}{-s}\right)\right]$
$I={\left[-\frac{1}{s}\mathrm{cos}pt{e}^{-st}-\frac{p}{\left({s}^{2}\right)}\mathrm{sin}pt{e}^{-st}-\frac{\left({p}^{2}\right)}{\left({s}^{2}\right)}I\right]}_{0}^{\mathrm{\infty }}$
$⇒I\left[1+\frac{\left({p}^{2}\right)}{\left({s}^{2}\right)}\right]=-\frac{1}{s}{\left[\mathrm{cos}pt{e}^{-st}+\frac{p}{s}\mathrm{sin}pt{e}^{-st}\right]}_{0}^{\mathrm{\infty }}$
$⇒I\left[\frac{\left({s}^{2}+{p}^{2}\right)}{{s}^{2}}\right]=-\frac{1}{s}\left[\left(0+0\right)-\left(1+0\right)\right]$
$⇒I\left[\frac{\left({s}^{2}+{p}^{2}\right)}{{s}^{2}}\right]=\frac{1}{s}$
$⇒I=\frac{s}{\left({s}^{2}+{p}^{2}\right)}$
Step 2
Hence,
$I={\int }_{0}^{\mathrm{\infty }}{e}^{-st}\mathrm{cos}ptdt=L\left[\mathrm{cos}pt\right]=\frac{s}{\left({s}^{2}+{p}^{2}\right)}$

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