Question

The Laplace transform of the function $f\left(t\right)=k,0<t<c=0,c<t<\infty$, is

• A. $\left(\frac{k}{s}\right)e^{-sc}$
• B. $\left(\frac{k}{s}\right)e^{sc}$
• C. $k{e}^{-sc}$
• D. $\left(\frac{k}{s}\right)(1-e^{-sc})$

Answer 1

The correct option is D $\left(\frac{k}{s}\right)(1-e^{-sc})$

$f\left(t\right)=\{\begin{array}{c}k,0<t<c\\ 0,c<t<\infty \end{array}$

$L\left[f\right(t\left)\right]={\int }_0^{\infty }{e}^{-st}f\left(t\right)dt$

$={\int }_0^c{e}^{-st}kdt$

$=k{\left[\frac{e^{-st}}{-s}\right]}_0^c$

$=\frac{k}{(-s)}[e^{-sc}-1]$

$=\frac{k}{s}(1-e^{-sc})$