Chapter 7, Problem 1RE

In Problems 1 and 2 use the definition of the Laplace transform to find $ℒ\{f(t)\}$.

1. $f(t)=\{\begin{array}{ll}t,\hfill & 0\le t<1\hfill \\ 2-t,\hfill & t\ge 1\hfill \end{array}$

To determine

To find: The expression for the Laplace transform of the given function

Answer to Problem 1RE

The Laplace transform of the given function is $\frac{-2e^{-s}}{s^2}+\frac{1}{s^2}$.

Explanation of Solution

Given:

The given expression is $f\left(t\right)=\{\begin{array}{l}t,0\le t<1\\ 2-t,t\ge 1\end{array}$ .

Concept:

If $f\left(t\right)$ is a piecewise continuous function and of exponential of order $t>0$ such that $f\left(t\right)$ is defined in two pieces; then Laplace transform of $f\left(t\right)$ is expressed as the sum of two integrals.

 $\begin{array}{c}L\left\{f\left(t\right)\right\}\text{=}{\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-st}f\left(t\right)}dt\\ ={\displaystyle \underset{0}{\overset{1}{\int }}{e}^{-st}}\left(t\right)dt+{\displaystyle \underset{1}{\overset{\infty }{\int }}{e}^{-st}}\left(2-t\right)dt\end{array}$

Calculation:

Split the domain of $f\left(t\right)$ in two parts, 0 to 1 and 1 to $\infty$.

Use the definition above to write the Laplace transform of the given function.

$\begin{array}{c}L\left\{f\left(t\right)\right\}={\displaystyle \underset{0}{\overset{1}{\int }}{e}^{-st}}\left(t\right)dt+{\displaystyle \underset{1}{\overset{\infty }{\int }}{e}^{-st}}\left(2-t\right)dt\\ ={\left[t\left\{\frac{e^{-st}}{-s}\right\}-{\displaystyle \int \left(1\right)}\left\{\frac{e^{-st}}{-s}\right\}dt\right]}_0^1+{\left[\left(2-t\right)\left\{\frac{e^{-st}}{-s}\right\}-{\displaystyle \int \left(-1\right)}\left\{\frac{e^{-st}}{-s}\right\}dt\right]}_1^{\infty }\\ ={\left[t\left\{\frac{e^{-st}}{-s}\right\}-\left\{\frac{e^{-st}}{s^2}\right\}\right]}_0^1+{\left[\left(2-t\right)\left\{\frac{e^{-st}}{-s}\right\}+\left\{\frac{e^{-st}}{s^2}\right\}\right]}_1^{\infty }\\ =\left[\left\{\frac{e^{-s}}{-s}\right\}-\left\{\frac{e^{-s}}{s^2}\right\}+\frac{1}{s^2}\right]+\left[-\left\{\frac{e^{-s}}{-s}\right\}-\left\{\frac{e^{-s}}{s^2}\right\}\right]\end{array}$

Further solving the above expression as,

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

Therefore, the Laplace transform of the given function is $\frac{-2e^{-s}}{s^2}+\frac{1}{s^2}$.