Question

L{f(t)}.In Problems 1 and 2 use, the definition of the Laplace transform to find $\mathcal{L}\mathbf{\left\{}\mathit{f}\mathbf{\right(}\mathit{t}\mathbf{\left)}\mathbf{\right\}}\mathbf{.}$

$\mathit{f}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathbf{\{}\begin{array}{cc}\mathbf{0}& \mathbf{0}\mathbf{\le }\mathbf{t}\mathbf{<}\mathbf{2}\\ \mathbf{1}& \mathbf{2}\mathbf{\le }\mathbf{t}\mathbf{<}\mathbf{4}\\ \mathbf{0}& \mathbf{t}\mathbf{\ge }\mathbf{4}\end{array}$

Short answer

$\mathcal{L}\left(f\right(t\left)\right)=\left[\frac{e^{-4s}-e^{-2s}}{s}\right]\forall s>0$

Step-by-step solution

To Find the Laplace transfrom

Laplace transform of given piece wise continuous function is written as:

$f\left(t\right)=\{\begin{array}{cc}0& 0\le \mathrm{t}<2\\ 1& 2\le \mathrm{t}<4\\ 0& \mathrm{t}\ge 4\end{array}$

$\begin{array}{c}Find\mathcal{L}\left\{f\right(t\left)\right\}as:\\ \mathcal{L}\left(f\right(t\left)\right)={\int }_0^2\left(0\right)e^{-st}dt+{\int }_2^4\left(1\right)e^{-st}dt+{\int }_4^{\infty }\left(0\right)e^{-st}dt\\ ={\left.\left[-\frac{e^{-st}}{s}\right]\right|}_2^4\\ =\left[-\frac{e^{-2s}}{s}+\frac{e^{-4s}}{s}\right]\\ =\left[\frac{e^{-4s}-e^{-2s}}{s}\right]\forall s>0\\ Thus,\\ \mathcal{L}\left(f\right(t\left)\right)=\left[\frac{e^{-4s}-e^{-2s}}{s}\right]\forall s>0\end{array}$