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

Step 1 To find the Laplace transform of the following functions.
Laplace transform of a function f(t) can be defined as $L(f(t))={\int }_0^{\mathrm{\infty }}f(t)e^{-st}dt$
Step 2
a) To find the Laplace transform of $f(t)=1+2t$
$L(f(t))=L(1+2t)$
$={\int }_0^{\mathrm{\infty }}(1+2t)e^{-st}dt$
$={\int }_0^{\mathrm{\infty }}{e}^{-st}dt+{\int }_0^{\mathrm{\infty }}(2t)e^{-st}dt$
$={\left[\frac{e^{-st}}{-s}\right]}_0^{\mathrm{\infty }}+2[(t)\cdot \frac{e^{-st}}{-s}]-{[1\cdot \frac{e^{-st}}{s^2}]}_0^{\mathrm{\infty }}$
$=\left[\frac{0-1}{-s}\right]+2[(0-0)-(0-\frac{1}{s^2}]$
$=\frac{1}{s}+\frac{2}{s^2}$
$=\frac{s+2}{s^2}$ Thus, $L(f(t))=\frac{s+2}{s^2}$
Step 3
b) To find the Laplace transform of $f(t)=\sin \omega t=\frac{e^{j\omega t}-e^{-j\omega t}}{2j}$
$L(f(t))=L(\sin \omega t)$
$={\int }_0^{\mathrm{\infty }}\sin \omega t\cdot e^{-st}dt$
$={\int }_0^{\mathrm{\infty }}\frac{e^{j\omega t}-e^{-j\omega t}}{2j}\cdot e^{-st}dt$ $=\frac{1}{2j}{\int }_0^{\mathrm{\infty }}{e}^{(j\omega -s)t}-e^{-(j\omega +s)t}dt$
$=\frac{1}{2j}{[\left(\frac{e^{(j\omega -s)t)}}{j\omega -s}\right)-\left(\frac{e^{-(j\omega +s)t}}{-(j\omega +s)}\right)]}_0^{\mathrm{\infty }}$
$=\frac{1}{2j}{[\left(\frac{e^{-(s-j\omega )t}}{-(s-j\omega )}\right)-\left(\frac{e^{-(j\omega +s)t}}{-(j\omega +s)}\right)]}_0^{\mathrm{\infty }}$
$=\frac{1}{2j}[\frac{0-1}{-(s-j\omega )}-\frac{0-1}{-(j\omega +s)}]$
$=\frac{1}{2j}[\frac{1}{(s-j\omega )}-\frac{1}{(j\omega +s)}]$
$=\frac{1}{2j}[\frac{1}{(s-j\omega )}-\frac{1}{(s+j\omega )}]$ $=\frac{1}{2j}\left[\frac{(s+j\omega )-(s-j\omega )}{(s-j\omega )(s+j\omega )}\right]$
$=\frac{\omega }{s^2-(j\omega {)}^2}$
$=\frac{\omega }{s^2+{\omega }^2}$ Since ${\omega }^2=-1$

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