Question

In Problems 29–32 express f in terms of unit step functions. Find

and.

31.

Short answer

Therefore, the solution is:

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

Step-by-step solution

Given Information.

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

Determining the L{f(t)}

A piecewise function is one that is defined as

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

can be written as

$\mathrm{f}\left(\mathrm{t}\right)=\mathrm{g}\left(\mathrm{t}\right)-\mathrm{g}\left(\mathrm{t}\right)\mathrm{U}(\mathrm{t}-\mathrm{a})+\mathrm{h}\left(\mathrm{t}\right)\mathrm{U}(\mathrm{t}-\mathrm{a})$

as a result, the specified function

$$\begin{gathered} f\left(t\right)=\{2,0⩽t<2 \\ t,t⩾2 \end{gathered}$$

can be written as

$f\left(t\right)=2-2U(t-2)+tU(t-2)$

both sides of the above equation's Laplace transform

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

$=2L\left\{1\right\}+L\left\{(t-2)U(t-2)\right\}$

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

$$\begin{gathered} \{IfF\left(s\right)=L\left\{f\left(t\right)\right\}anda>0,thenL\left\{f\right(t-a\left)\right\}=e^{-as}F\left(s\right) \\ L\left\{1\right\}=\frac{1}{s},L\left\{t^n\right\}=\frac{n!}{s^{n+1}} \end{gathered}$$

Determining the L{etf(t)}

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

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

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

Determining the Result

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