Chapter 7.RP, Problem 1RP

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

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

To determine

The Laplace transform of the function using definition.

Answer to Problem 1RP

Solution:

The Laplace transform of $f\left(t\right)$ is $\frac{3}{s}+e^{-2s}\left(\frac{1}{s}-\frac{1}{s^2}\right)$.

Explanation of Solution

Given:

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

Approach:

$\mathcal{L}\left\{f\right\}\left(s\right)={\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-st}}f\left(t\right)dt$

Calculation:

The Laplace transform of $f\left(t\right)$ is,

$\begin{array}{l}\mathcal{L}\left\{f\right\}\left(s\right)={\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-st}}f\left(t\right)dt\\ \mathcal{L}\left\{f\right\}\left(s\right)=3{\displaystyle \underset{0}{\overset{2}{\int }}{e}^{-st}dt}+{\displaystyle \underset{2}{\overset{\infty }{\int }}\left(6-t\right)e^{-st}dt}\\ \mathcal{L}\left\{f\right\}\left(s\right)=3{\displaystyle \underset{0}{\overset{2}{\int }}{e}^{-st}dt}+\underset{n\to \infty }{\lim }\left(6{\displaystyle \underset{2}{\overset{n}{\int }}{e}^{-st}dt-{\displaystyle \underset{2}{\overset{n}{\int }}t{e}^{-st}dt}}\right)\end{array}$

$\begin{array}{l}\mathcal{L}\left\{f\right\}\left(s\right)={{\displaystyle \left[\frac{-3e^{-st}}{s}\right]}}_0^2+\underset{n\to \infty }{\lim }\left(\frac{-6e^{-sn}}{s}+\frac{n{e}^{-sn}}{s}+\frac{e^{-sn}}{s^2}+\frac{6e^{-2s}}{s}-\frac{2e^{-2s}}{s}-\frac{e^{-2s}}{s^2}\right)\\ \mathcal{L}\left\{f\right\}\left(s\right)=\frac{3}{s}-\frac{3e^{-2s}}{s}+\frac{4e^{-2s}}{s}-\frac{e^{-2s}}{s^2}\\ \mathcal{L}\left\{f\right\}\left(s\right)=\frac{3}{s}+e^{-2s}\left(\frac{1}{s}-\frac{1}{s^2}\right)\end{array}$

Therefore, the Laplace transform of $f\left(t\right)$ is $\frac{3}{s}+e^{-2s}\left(\frac{1}{s}-\frac{1}{s^2}\right)$.

Conclusion:

Hence, the Laplace transform of $f\left(t\right)$ is $\frac{3}{s}+e^{-2s}\left(\frac{1}{s}-\frac{1}{s^2}\right)$.