Question

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 Laplace transformation of the given function is$Y\left(s\right)=\frac{1}{s^2}-\frac{1}{s^2}{e}^{-2s}-2\frac{e^{-2s}}{s}$

Step-by-step solution

Definition of Laplace transform

The Laplace transform formula is used to define the Laplace transform of$f\left(t\right)$which is denoted by$Lf\left(t\right)$ or F$\left(s\right)$

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

Find the Laplace transform of the given function.

The representation of the given piecewise function as a step "switch"

Function is: $t-tU\left(t-2\right)$

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

We need to replace “t” that is multiplied by the step function by $(t-2)$without effecting the function.

So,

$t-\left[\left(t-2\right)+2\right]U\left(t-2\right)=\text{t}-\left(\text{t}-2\right)U\left(\text{t}-2\right)+2\text{U}\left(\text{t}-2\right)$

The unit step function $J\left(t-a\right)$is defined to be,

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

$L\left\{t\right\}-L\left\{\left(t-2\right)U\left(t-2\right)\right\}+L\left\{2U\left(t-2\right)\right\}$

$Y\left(s\right)=\frac{1}{s^2}-\frac{1}{s^2}{e}^{-2s}-2\frac{e^{-2s}}{s}$

Hence, the Laplace transform of the given function is

$Y\left(s\right)=\frac{1}{s^2}-\frac{1}{s^2}{e}^{-2s}-2\frac{e^{-2s}}{s}$