Question

In problems, write each function in terms of unit step functions. Find the Laplace transform of the given function.

$f\left(t\right)=\left\{\begin{array}{l}1,\\ 0,\\ 1,\end{array}\right.\left.\begin{array}{r}0⩽t<4\\ 4⩽t<5\\ ⩽t\ge 5\end{array}\right\}$

Short answer

The given equation$f\left(t\right)=\left\{\begin{array}{l}1,\\ 0,\\ 1,\end{array}\right.\left.\begin{array}{r}0⩽t<4\\ 4⩽t<5\\ ⩽t\ge 5\end{array}\right\}$can be expressed in terms of unit step function as$1-u(t-4)+u(t-5)$ . The Laplace transform of the function$f\left(t\right)=\left\{\begin{array}{l}1,\\ 0,\\ 1,\end{array}\right.\left.\begin{array}{r}0⩽t<4\\ 4⩽t<5\\ ⩽t\ge 5\end{array}\right\}$is$\frac{1}{s}-\frac{e^{-4s}}{s}+\frac{e^{-5s}}{s}$

Step-by-step solution

Define the unit step function.

The unit step function is defined as

$u\left(t\right)=\left\{\begin{array}{ll}0& t<0\\ 1& t>0\end{array}\right.$

Where

u is a function of time t.

The unit step function is an elementary function with only a positive side and a negative side of zero.

Find the unit step function of the given function.

A piecewise function is defined as,

$ft=\left\{\begin{array}{ll}g\left(t\right),& 0⩽t<a\\ h\left(t\right),& t\ge a\end{array}\right.$

It can be expressed as

$f\left(t\right)=g\left(t\right)+\left(h\right(t)-g(t\left)\right)u(t-a)$

Where$U(t-a)$is the unit function defined as,

$U(t-a)=\left\{\begin{array}{ll}0,& 0⩽t<a\\ 1,& t\ge a\end{array}\right.$

The given function can be expressed as

$f\left(t\right)=1+(0-1)u(t-4)+(1-0)u(t-5)=1-u(t-4)+u(t-5)$

Hence the function$f\left(t\right)=\left\{\begin{array}{l}1,\\ 0,\\ 1,\end{array}\right.\left.\begin{array}{r}0⩽t<4\\ 4⩽t<5\\ ⩽t\ge 5\end{array}\right\}$can be expressed as $1-u(t-4)+u(t-5)$using a unit step function.

Find the Laplace transform for the given function.

Take Laplace transform on both sides of the equation $f\left(t\right)=1-u(t-4)+u(t-5)$.

. $$\begin{gathered} \mathcal{L}\left\{f\right(t\left)\right\}=\mathcal{L}\{1-u(t-4)+u(t-5\left)\right\} \\ =\frac{1}{s}-\frac{e^{-4s}}{s}+\frac{e^{-5s}}{s} \end{gathered}$$

Hence, for the function $f\left(t\right)=\left\{\begin{array}{l}1,\\ 0,\\ 1,\end{array}\right.\left.\begin{array}{r}0⩽t<4\\ 4⩽t<5\\ ⩽t\ge 5\end{array}\right\}$the Laplace transform is $\frac{1}{s}-\frac{e^{-4s}}{s}+\frac{e^{-5s}}{s}$.