Question

In Problems 29–32 express f in terms of unit step functions. Find$\mathit{L}\left\{f\left(t\right)\right\}$

and$\mathit{L}\left\{e^{t}f\left(t\right)\right\}$.

30.

Short answer

Therefore, the solution is

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

Step-by-step solution

Given Information.

The given equation is:

$$\begin{gathered} f\left(t\right)=\left\{0,0⩽t<a \\ g\left(t\right),a⩽t<b \\ 0,t⩾b\right. \end{gathered}$$

Determining the L{f(t)}

A piecewise function is one that is defined as

$$\begin{gathered} f\left(t\right)=\left\{0,0⩽t<a \\ g\left(t\right),a⩽t<b \\ 0,t⩾b\right. \end{gathered}$$

can be written as

$f\left(t\right)=g\left(t\right)\left[U\left(t-a\right)-U\left(t-b\right)\right]$

as a result, the specified function

$$\begin{gathered} f\left(t\right)=\left\{\backslash 0,0⩽t<\pi \\ sint,\pi ⩽t<3\pi \\ 0,t⩾3\pi \right. \end{gathered}$$

can be written as

$f\left(t\right)=sint\left[U\left(t-\pi \right)-U\left(t-3\pi \right)\right]$

both sides of the above equation's Laplace transform

$L\left\{f\left(t\right)\right\}=L\left\{sint\left[U\left(t-\pi \right)-U\left(t-3\pi \right)\right]\right\}$

$=L\left\{-\sin \left(t-\pi \right)U(t-\pi )\right\}-L\left\{-\sin \left(t-3\pi \right)U(t-3\pi )\right\}$

$=L\left\{\sin \left(t-\pi \right)U(t-\pi )\right\}+L\left\{\sin \left(t-3\pi \right)U(t-3\pi )\right\}$

$=\frac{e^{-\pi s}}{s^2+1}+\frac{e^{-3\pi s}}{s^2+1}$

$$\begin{gathered} \left\{IfF\left(s\right)=L\left\{f\left(t\right)\right\}anda>0,thenL\left\{f(t-a)U(t-a)=e^{-as}F\left(s\right)\right\} \\ L\left\{sinkt\right\}=\frac{k}{s^2+k^2}\right. \end{gathered}$$

Determining the L{etf(t)}

$L\left\{e^{t}f\left(t\right)\right\}={\left.L\left\{f\left(t\right)\right\}\right|}_{s\to s-1}$

$L\left\{e^{t}f\left(t\right)\right\}={\left.\left[\frac{e^{-\pi s}}{s^2+1}+\frac{e^{-3\pi s}}{s^2+1}\right]\right|}_{s\to s-1}$

$=\frac{e^{-\pi (s-1)}}{{\left(s-1\right)}^2+1}+\frac{e^{-3\pi s(s-1)}}{{\left(s-1\right)}^2+1}$

Final answer

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