Question

In Problems 1–18 use Definition 7.1.1 to find$L\left\{f\left(t\right)\right\}.$

2.$\mathit{f}\left(\mathit{t}\right)=\left\{\begin{array}{c}4,0\le t<2\\ 0,t\ge 2\end{array}\right.$

Short answer

The Laplace transform of above function is,

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

Step-by-step solution

Definition 7.1.1 Laplace transform

Let$f$be a function define for$t⩾0$.Then the integral

$L\left\{f\left(t\right)\right\}={\int }_0^{\infty }{e}^{-st}f\left(t\right)dt$

is said to be Laplace transform of$f,$provide that integral converges.

Applying the definition

Consider the function$f\left(t\right)={\{}_{0,t⩾2}^{4,0⩽t<2}$

The objective is to find $L\left\{f\left(t\right)\right\}$using the definition.

Note that, the function$f$is defined for$t⩾0$

From the definition,

$L\left\{f\left(t\right)\right\}={\int }_0^{\infty }{e}^{-st}f\left(t\right)dt$

Sinceis defined in two pieces $[0,2)$and $[2,\infty ),$Laplacian if$f$is $L\left\{f\left(t\right)\right\}$expressed as the sum of two integrals.

$$\begin{gathered} L\left\{f\left(t\right)\right\}={\int }_0^2{e}^{-st}f\left(t\right)dt+{\int }_2^{\infty }{e}^{-st}f\left(t\right)dt \\ ={\int }_0^2{e}^{-st}\left(4\right)dt+{\int }_2^{\infty }{e}^{-st}\left(0\right)dt \\ =4{\int }_0^2{e}^{-st}dt+0 \\ =4{\left(-\frac{1}{s}{e}^{-st}\right)}_0^2 \end{gathered}$$

Simplification:

Continuation to above steps,

$$\begin{gathered} L\left\{f\left(t\right)\right\}=4{\left(-\frac{1}{s}{e}^{-st}\right)}_0^2 \\ =\frac{4}{s}\left(e^{-s.2}-e^{-s.0}\right) \\ =\frac{4}{s}\left(e^{-2s}-e^0\right) \\ =\frac{4}{s}\left(e^{-2s}-1\right) \end{gathered}$$

Therefore the required Laplace transform of function is,

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