Question

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

$\mathit{f}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathbf{\{}\begin{array}{l}\mathbf{3}\mathbf{,}\mathbf{0}\mathbf{\le }\mathbf{t}\mathbf{\le }\mathbf{2}\\ \mathbf{6}\mathbf{-}\mathbf{t}\mathbf{,}\mathbf{2}\mathbf{<}\mathbf{t}\end{array}$

Short answer

$L\left\{f\right(t\left)\right\}\left(s\right)=\frac{3}{s}+\frac{e^{-2s}}{s}-\frac{e^{-2s}}{s^2}$

Step-by-step solution

Step 1:Given Information

The given function is$f\left(t\right)=\{\begin{array}{l}3,0\le t\le 2\\ 6-t,2<t\end{array}$

Determining the L{f}

Using the Laplace transform definition, we get

$\begin{array}{c}L\left\{f\right(t\left)\right\}={\int }_0^{\infty }{e}^{-st}f\left(t\right)dt\\ ={\int }_0^{2}3e^{-st}dt+{\int }_2^{\infty }(6-t)e^{-st}dt\\ =3{[-\frac{e^{-st}}{s}]}_0^2+\underset{N\to \infty }{\lim }{\int }_2^N(6-t)e^{-st}dt\\ =\frac{3}{s}(1-e^{-2s})+\underset{N\to \infty }{\lim }{\int }_2^N(6-t)e^{-st}dt\end{array}$

Letrole="math" localid="1664044470656" $\left|\begin{array}{c}\begin{array}{ll}6-t=u& e^{-st}dt=dv\\ -dt=du& v=-\frac{e^{-st}}{s}\end{array}\\ \end{array}\right|$in second integral, then we can write as:

$\begin{array}{c}L\left\{f\right(t\left)\right\}=\frac{3}{s}(1-e^{-2s})+\underset{N\to \infty }{\lim }({\left[-(6-t)\frac{e^{-st}}{s}\right]}_2^N-{\int }_2^N\frac{e^{-st}}{s}dt)\\ =\frac{3}{s}(1-e^{-2s})+\underset{N\to \infty }{\lim }\left(\frac{4e^{-2s}}{s}-\frac{(6-N)e^{-sN}}{s}+{\left[\frac{e^{-st}}{s^2}\right]}_2^N\right)\\ =\frac{3}{s}(1-e^{-2s})+\underset{N\to \infty }{\lim }\left(\frac{4e^{-2s}}{s}-\frac{(6-N)e^{-sN}}{s}+\frac{e^{-sN}}{s^2}-\frac{e^{-2s}}{s^2}\right)\\ =\frac{3}{s}(1-e^{-2s})+\underset{N\to \infty }{\lim }\frac{4e^{-2s}}{s}-\underset{N\to \infty }{\lim }\frac{(6-N)e^{-sN}}{s}+\underset{N\to \infty }{\lim }\frac{e^{-sN}}{s^2}-\underset{N\to \infty }{\lim }\frac{4e^{-2s}}{s^2}\end{array}$

Simplify further as:

$\begin{array}{c}L\left\{f\right(t\left)\right\}=\frac{3}{s}(1-e^{-2s})+\frac{4e^{-2s}}{s}-0+0-\frac{4e^{-2s}}{s^2}\\ =\frac{3}{s}(1-e^{-2s})+\frac{4e^{-2s}}{s}-\frac{4e^{-2s}}{s^2}\\ =\frac{3}{s}+\frac{e^{-2s}}{s}-\frac{e^{-2s}}{s^2}\end{array}$

Determining the Result

Thus, the required Laplace transform is$L\left\{f\right(t\left)\right\}\left(s\right)=\frac{3}{s}+\frac{e^{-2s}}{s}-\frac{e^{-2s}}{s^2}$