Question

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

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

29.

Short answer

Therefore, the solution is

$$\begin{gathered} L\left\{f\left(t\right)\right\}=\frac{1}{s^2}-\frac{1}{s^2}{e}^{-s}-\frac{1}{s}{e}^{-4s} \\ L\left\{e^{t}f\left(t\right)\right\}=\frac{1}{{\left(s-1\right)}^2}-\frac{1}{{\left(s-1\right)}^2}{e}^{-\left(s-1\right)}-\frac{1}{s-1}{e}^{-4\left(s-1\right)} \end{gathered}$$

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

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

can be written as

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

both sides of the above equation's Laplace transform

$L\left\{f\left(t\right)\right\}=L\left\{t\right\}-L\left\{\left(t-1+1\right)U\left(t-1\right)\right\}+L\left\{U\left(t-1\right)\right\}-L\left\{U\left(t-4\right)\right\}$

$=L\left\{t\right\}-L\left\{\left(t-1\right)U\left(-1\right)\right\}-L\left\{U\left(t-1\right)\right\}+L\left\{U\left(t-1\right)\right\}-L\left\{U\left(t-4\right)\right\}$

$\frac{1}{s^2}-\frac{1}{s^2}{e}^{-s}-\frac{1}{s}{e}^{-4s}$

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

Determining the L{etf(t)}

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

$L\left\{e^{t}f\left(t\right)\right\}={\left.\left[\frac{1}{s^2}-\frac{1}{s^2}{e}^{-s}-\frac{1}{s}{e}^{-4s}\right]\right|}_{s\to s-1}$

$=\frac{1}{{\left(s-1\right)}^2}-\frac{1}{{\left(s-1\right)}^2}{e}^{-\left(s-1\right)}-\frac{1}{s-1}{e}^{-4\left(s-1\right)}$

Final answer

$$\begin{gathered} L\left\{f\left(t\right)\right\}=\frac{1}{s^2}-\frac{1}{s^2}{e}^{-s}-\frac{1}{s}{e}^{-4s} \\ L\left\{e^{t}f\left(t\right)\right\}=\frac{1}{{\left(s-1\right)}^2}-\frac{1}{{\left(s-1\right)}^2}{e}^{-\left(s-1\right)}-\frac{1}{s-1}{e}^{-4\left(s-1\right)} \end{gathered}$$