 # Find the Laplace transforms of the following time functions. Solve problem 1(a) and 1 (b) using the Laplace transform definition i.e. integration. For Kye 2021-02-21 Answered
Find the Laplace transforms of the following time functions.
Solve problem 1(a) and 1 (b) using the Laplace transform definition i.e. integration. For problem 1(c) and 1(d) you can use the Laplace Transform Tables.
a)$f\left(t\right)=1+2t$ b)
c)$f\left(t\right)=\mathrm{sin}\left(2t\right)+2\mathrm{cos}\left(2t\right)+{e}^{-t}\mathrm{sin}\left(2t\right)$
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Step 1 To find the Laplace transform of the following functions.
Laplace transform of a function f(t) can be defined as $L\left(f\left(t\right)\right)={\int }_{0}^{\mathrm{\infty }}f\left(t\right){e}^{-st}dt$
Step 2
a) To find the Laplace transform of $f\left(t\right)=1+2t$
$L\left(f\left(t\right)\right)=L\left(1+2t\right)$
$={\int }_{0}^{\mathrm{\infty }}\left(1+2t\right){e}^{-st}dt$
$={\int }_{0}^{\mathrm{\infty }}{e}^{-st}dt+{\int }_{0}^{\mathrm{\infty }}\left(2t\right){e}^{-st}dt$
$={\left[\frac{{e}^{-st}}{-s}\right]}_{0}^{\mathrm{\infty }}+2\left[\left(t\right)\cdot \frac{{e}^{-st}}{-s}\right]-{\left[1\cdot \frac{{e}^{-st}}{{s}^{2}}\right]}_{0}^{\mathrm{\infty }}$
$=\left[\frac{0-1}{-s}\right]+2\left[\left(0-0\right)-\left(0-\frac{1}{{s}^{2}}\right]$
$=\frac{1}{s}+\frac{2}{{s}^{2}}$
$=\frac{s+2}{{s}^{2}}$ Thus, $L\left(f\left(t\right)\right)=\frac{s+2}{{s}^{2}}$
Step 3
b) To find the Laplace transform of $f\left(t\right)=\mathrm{sin}\omega t=\frac{{e}^{j\omega t}-{e}^{-j\omega t}}{2j}$
$L\left(f\left(t\right)\right)=L\left(\mathrm{sin}\omega t\right)$
$={\int }_{0}^{\mathrm{\infty }}\mathrm{sin}\omega t\cdot {e}^{-st}dt$
$={\int }_{0}^{\mathrm{\infty }}\frac{{e}^{j\omega t}-{e}^{-j\omega t}}{2j}\cdot {e}^{-st}dt$ $=\frac{1}{2j}{\int }_{0}^{\mathrm{\infty }}{e}^{\left(j\omega -s\right)t}-{e}^{-\left(j\omega +s\right)t}dt$
$=\frac{1}{2j}{\left[\left(\frac{{e}^{\left(j\omega -s\right)t\right)}}{j\omega -s}\right)-\left(\frac{{e}^{-\left(j\omega +s\right)t}}{-\left(j\omega +s\right)}\right)\right]}_{0}^{\mathrm{\infty }}$
$=\frac{1}{2j}{\left[\left(\frac{{e}^{-\left(s-j\omega \right)t}}{-\left(s-j\omega \right)}\right)-\left(\frac{{e}^{-\left(j\omega +s\right)t}}{-\left(j\omega +s\right)}\right)\right]}_{0}^{\mathrm{\infty }}$
$=\frac{1}{2j}\left[\frac{0-1}{-\left(s-j\omega \right)}-\frac{0-1}{-\left(j\omega +s\right)}\right]$
$=\frac{1}{2j}\left[\frac{1}{\left(s-j\omega \right)}-\frac{1}{\left(j\omega +s\right)}\right]$
$=\frac{1}{2j}\left[\frac{1}{\left(s-j\omega \right)}-\frac{1}{\left(s+j\omega \right)}\right]$ $=\frac{1}{2j}\left[\frac{\left(s+j\omega \right)-\left(s-j\omega \right)}{\left(s-j\omega \right)\left(s+j\omega \right)}\right]$
$=\frac{\omega }{{s}^{2}-\left(j\omega {\right)}^{2}}$
$=\frac{\omega }{{s}^{2}+{\omega }^{2}}$ Since ${\omega }^{2}=-1$