Question

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

1.$f\left(t\right)={\{}_{1,t⩾1}^{-1,0⩽t<1}$

Short answer

The Laplace transform of above function is,

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

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 $f\left(t\right)={\{}_{1,t⩾1}^{-1,0⩽t<1}$

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$[0,1)$and$[1,\infty )$Laplacian if f is$L\left\{f\left(t\right)\right\}$expressed as the sum of two integrals.

$$\begin{gathered} L\left\{f\left(t\right)\right\}={\int }_0^1{e}^{-st}f\left(t\right)dt+{\int }_1^{\infty }{e}^{-st}f\left(t\right)dt \\ ={\int }_0^1{e}^{-st}\left(-1\right)dt+{\int }_1^{\infty }{e}^{-st}\left(1\right)dt \\ =-{\int }_0^1{e}^{-st}dt+{\int }_1^{\infty }{e}^{-st}dt \\ =-{\left(-\frac{1}{s}{e}^{-st}\right)}_0^1+{\left(-\frac{1}{s}{e}^{-st}\right)}_1^{\infty } \end{gathered}$$

Simplification:

Continuation to above steps,

$$\begin{gathered} L\left\{f\left(t\right)\right\}=-{\left(-\frac{1}{s}{e}^{-st}\right)}_0^1+{\left(-\frac{1}{s}{e}^{-st}\right)}_1^{\infty } \\ =\frac{1}{s}{\left(e^{-st}\right)}_0^1-\frac{1}{s}{\left(e^{-st}\right)}_1^{\infty } \\ =\frac{1}{s}\left(e^{-s.1}-e^{-s.0}\right)-\frac{1}{s}\left(e^{-s.\infty }-e^{-s.1}\right) \\ =\frac{1}{s}\left(e^{-s}-e^0\right)-\frac{1}{s}\left(e^{-\infty }-e^{-s}\right) \\ =\frac{1}{s}\left(e^{-s}-1\right)-\frac{1}{s}\left(0-e^{-s}\right) \\ =\frac{1}{s}{e}^{-s}-\frac{1}{s}-0+\frac{1}{s}{e}^{-s} \\ =\frac{2}{s}{e}^{-s}-\frac{1}{s} \end{gathered}$$

Therefore the required Laplace transform of function is,

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