Question

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

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

Short answer

Therefore, the solution is:

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

Step-by-step solution

Given Information.

The given equation is:

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

Determining the L{f(t)}

A piecewise function is one that is defined as

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

can be written as

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

as a result, the specified function

$$\begin{gathered} f\left(t\right)=\left\{t,0⩽t<1 \\ 2-t,1⩽t<2\right. \end{gathered}$$

can be written as

$f\left(t\right)=t-tU(t-1)+(2-t)U(t-1)-(2-t)U(t-2)$

both sides of the above equation's Laplace transform

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

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

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

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

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

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

Determining the L{etf(t)}

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

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

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

Final answer

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