Question

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

$f\left(t\right)=\left\{\begin{array}{ll}0,& 0⩽t<3\pi /2\\ \sin t,& t⩾3\pi /2\end{array}\right.$

Short answer

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

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

is a function of time.

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,

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

$f\left(t\right)=\left\{\begin{array}{ll}0,& 0⩽t<3\pi /2\\ \sin t,& t⩾3\pi /2\end{array}\right.$

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,

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

So the given function can be expressed as

$f\left(t\right)=0+(\sin t-0)u\left(t-\frac{3\pi }{2}\right)=\sin tu\left(t-\frac{3\pi }{2}\right)$

Hence the function$f\left(t\right)=\left\{\begin{array}{ll}0,& 0⩽t<3\pi /2\\ \sin t,& t⩾3\pi /2\end{array}\right.$can be expressed as$\sin tu\left(t-\frac{3\pi }{2}\right)$ using unit step function.

Find the Laplace transform for the given function.

Take Laplace transform on both sides of the equation.

$\mathcal{L}\left\{f\right(t\left)\right\}=\mathcal{L}\left\{\sin tu\left(t-\frac{3\pi }{2}\right)\right\}$

Since

$$\begin{gathered} \mathcal{L}\left\{f\right(t\left)u\right(t-a\left)\right\}=e^{-as}\mathcal{L}\left\{f\right(t+a\left)\right\} \\ =e^{-\frac{3\pi }{2}s}\mathcal{L}\left\{\sin \left(t+\frac{3\pi }{2}\right)\right\} \end{gathered}$$

From trigonometry

$$\begin{gathered} \sin \left(x+\frac{3\pi }{2}\right)=-\cos x \\ =e^{-\frac{3\pi }{2}s}\mathcal{L}\{-\cos t\} \\ =-e^{-\frac{3\pi }{2}s}\frac{s}{s^2+1} \end{gathered}$$

Hence, for the function$f\left(t\right)=\left\{\begin{array}{ll}0,& 0⩽t<3\pi /2\\ \sin t,& t⩾3\pi /2\end{array}\right.$$theLaplacetransformis$$-e^{-\frac{3\pi }{2}s}\frac{s}{s^2+1}$