Question

In problems, 55-62 write each function in terms of unit step functions. Find the Laplace transform of the given function.

$f\left(t\right)=\left\{\begin{array}{l}0,\\ t^2,\end{array}\right.\left.\begin{array}{r}0⩽t<1\\ t\ge 1\end{array}\right\}$

Short answer

The given equation $f\left(t\right)=\left\{\begin{array}{l}0,\\ t^2,\end{array}\right.\left.\begin{array}{r}0⩽t<1\\ t\ge 1\end{array}\right\}$can be expressed in terms of unit step function as $f\left(t\right)=t^{2}U(t-1)$. The Laplace transform of the function$f\left(t\right)=\left\{\begin{array}{l}0,\\ t^2,\end{array}\right.\left.\begin{array}{r}0⩽t<1\\ t\ge 1\end{array}\right\}is{e}^{-s}\left(\frac{2}{s^3}+\frac{2}{s^2}+\frac{1}{s}\right)$ is .

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,

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

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

$f\left(t\right)=t^{2}U(t-1)$

Hence the function$f\left(t\right)=\left\{\begin{array}{l}0,\\ t^2,\end{array}\right.\left.\begin{array}{r}0⩽t<1\\ t\ge 1\end{array}\right\}$can be expressed as using $f\left(t\right)=t^{2}U(t-1)$aunit step function.

Find the Laplace transform for the given function.

Take Laplace transform on both sides of the equation.

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

Hence, for the function$f\left(t\right)=\left\{\begin{array}{l}0,\\ t^2,\end{array}\right.\left.\begin{array}{r}0⩽t<1\\ t\ge 1\end{array}\right\}$the Laplace transform is$e^{-s}\left(\frac{2}{s^3}+\frac{2}{s^2}+\frac{1}{s}\right)$