Question

Write each function in terms of unit step functions. Find the Laplace transform of the given function.

$f\left(t\right)=\left\{\begin{array}{ll}sint,& 0⩽t<2\pi \\ 0,& t⩾2\pi \end{array}\right.$

Short answer

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

Step-by-step solution

Definition of Laplace Transform

A Laplace transform of a function $f\left(t\right)$in a time domain, where is the real number greater than or equal to zero, is given as where there

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

S is the complex frequency domain.

Find the Laplace transform of the above function;

The unit step “switch” representation is :

$\text{sin}\left(t\right)-\text{sin}\left(t\right)U\left(t-2\pi \right)$

$f\left(t\right)=\left\{\begin{array}{ll}sint,& 0⩽t<2\pi \\ 0,& t⩾2\pi \end{array}\right.$

Again, we need to replace the "t" in the sine function that is multiplied by the unit step function by $\left(t-2\pi \right)$

So,

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

Since were adding a full revolution to it, don’t forget, we’re treating "t" as $\text{(}t-2\pi )$

So,

$$\begin{gathered} \text{sin}\left(t\right)=\text{sin}\left(t+2\pi \right)\text{.} \\ y\left(t\right)=\text{sin}\left(t\right)-\text{sin}\left(t-2\pi \right)U\left(t-2\pi \right) \end{gathered}$$

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

$$\begin{gathered} 9(t-a)=\left\{\begin{array}{ll}0,& 0⩽t<a\\ 1& t\ge a\end{array}\right. \\ Y\left(s\right)=L\left\{\sin \right(t\left)\right\}-L\left\{\sin \right(t-2\pi \left)U\right(t-2\pi \left)\right\} \\ Y\left(s\right)=\frac{1}{s^2+1}-\frac{1}{s^2+1}{e}^{-2\pi s} \end{gathered}$$

Laplace transform of the above function is $Y\left(s\right)=\frac{1}{s^2+1}-\frac{1}{s^2+1}{e}^{-2\pi s}$