Question

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

10.

Short answer

The Laplace transform of above function is,

$L\left\{f\left(t\right)\right\}=\frac{c}{s}\left(e^{-s.a}-e^{-s.b}\right)$

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$f\left(t\right)=\left\{\begin{array}{c}0,a<t<a\\ c,a<t<b\\ 0,t>b\end{array}\right.$

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,

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

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

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

$={\int }_0^a{e}^{-st}\left(0\right)dt+{\int }_a^b{e}^{-st}\left(c\right)dt+{\int }_b^{\infty }{e}^{-st}\left(0\right)dt$

$=0+{\int }_a^b{e}^{-st}\left(c\right)dt+0$

$=c{\int }_a^b{e}^{-st}dt$

Simplification

Continuation to above steps,

$L\left\{f\left(t\right)\right\}=c{\left(-\frac{1}{s}{e}^{-st}\right)}_a^b$

$L\left\{f\left(t\right)\right\}=c\left(-\frac{1}{s}{e}^{-s.b}+\frac{1}{s}{e}^{-s.a}\right)$

$L\left\{f\left(t\right)\right\}=\frac{c}{s}\left(e^{-s.a}-e^{-s.b}\right)$

Therefore the required Laplace transform of function is,

$L\left\{f\left(t\right)\right\}=\frac{c}{s}\left(e^{-s.a}-e^{-s.b}\right)$