Question

The Laplace Transform of the following function is
f(t)=$\{\begin{array}{c}\sin tfor0\le t\le \pi \\ 0fort>\pi \end{array}$

• A. $\frac{1}{1+s^2}foralls>0$
• B. $\frac{1}{1+s^2}foralls<\pi$
• C. $\frac{1+e^{-\pi s}}{1+s^2}foralls>0$
• D. $\frac{e^{-\pi s}}{1+s^2}foralls>0$

Answer 1

The correct option is C $\frac{1+e^{-\pi s}}{1+s^2}foralls>0$

Let $F\left(s\right)=L\left[f\right(t\left)\right]={\int }_0^{\pi }{e}^{-st}\sin tdt$
Then $F\left(s\right)=\left[e^{-st}\right(-\cos ){]}_0^{\pi }-{\int }_0^{\pi }{e}^{-st}(-s)-(cost)dt$
$=(e^{-s\pi }+1)-s{\int }_0^{\pi }{e}^{-st}\cos tdt$

$=e^{-s\pi }+1-s\left\{\right[e^{-st}\sin {]}_0^{\pi }-{\int }_0^{\pi }{e}^{-st}(-s)\sin tdt\}$
$F\left(s\right)=1+e^{-s\pi }-s^{2}F\left(s\right)$
so, $F\left(s\right)+s^{2}F\left(s\right)=1+e^{-s\pi }$
F(s) $=\frac{1+e^{-s\pi }}{1+s^2}$