Question

In Problems 55 – 62 write each function in terms of unit stepfunctions. Find the Laplace transform of the given function62.

Short answer

The Laplace transform of the given function is$Lf\left(t\right)=\frac{e^{(-s)}}{s}+\frac{e^{(-2s)}}{s}+\frac{e^{(-3s)}}{s}.$

Step-by-step solution

Definition of Laplace Transform"

• The integral transform of a given derivative function with real variable t into a complex function with variable s is known as the Laplace transforms.
• Let f(t) be supplied for t(0), and assume that the function meets certain constraints that will be presented subsequently.
• The Laplace transform formula defines the Laplace transform of f(t), which is indicated by Lf(t) or F(s):

$F\left(s\right)={\int }_0^{+\infty }\left(t\right)·e^{-s·t}·dt$

Find the Laplace transform

From the given graph, we can write a function as a piecewise function.

$f\left(t\right)=\left\{\begin{array}{l}00⩽t<1\\ 11⩽t<2\\ 22⩽t<3\\ 33⩽t<4\end{array}\right.$

As we know, the piecewise-defined function

$f\left(t\right)=\left\{\begin{array}{l}g\left(t\right)0⩽t<a\\ h\left(t\right)t⩾a\end{array}\right.$

The equation can be written as:

role="math" localid="1663921902372" $f\left(t\right)=g\left(t\right)+\left(h\left(t\right)-g\left(t\right)\right)u\left(t-a\right)$

Thus given function we can write as

$f\left(t\right)=0+u\left(t-1\right)+u\left(t-2\right)+u\left(t-3\right)$

Now, Laplace transform will be

$$\begin{gathered} Lf\left(t\right)=L\left\{u\left(t-1\right)+u\left(t-2\right)+u\left(t-3\right)\right\} \\ =\frac{e^{-s}}{s}+\frac{e^{-2s}}{s}+\frac{e^{-3s}}{s} \end{gathered}$$

The Laplace transform for the given function is$Lf\left(t\right)=\frac{e^{(-s)}}{s}+\frac{e^{(-2s)}}{s}+\frac{e^{(-3s)}}{s}.$