Question

Find the transform of

$\mathit{f}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\{\begin{array}{ll}sin(x-vt),& t>x/v\\ 0,& t<x/v\end{array}$

Where xand vare constants.

Short answer

The transform of L$\left[f\left(t\right)\right]=e^{\frac{xp}{v}}\left(\frac{-v}{p^2+v^2}\right)$ is .

Step-by-step solution

Given information

The given function is$f\left(t\right)=\{\begin{array}{ll}\sin (\mathrm{x}-\mathrm{vt}),& \mathrm{t}>\mathrm{x}/\mathrm{v}\\ 0,& \mathrm{t}<\mathrm{x}/\mathrm{v}\end{array}$

Definition of Laplace Transformation

A transformation of a function f(x) into the function g(t) that is useful especially in reducing the solution of an ordinary linear differential equation with constant coefficients to the solution of a polynomial equation.

The inverse Laplace transform of a function F(s) is the piecewise-continuous and exponentially-restricted real function f(t)

Transform of the given function

The given expression is,

$$\begin{gathered} f\left(t\right)=\left\{\begin{array}{ll}\sin (\mathrm{x}-\mathrm{vt}),& \mathrm{t}>\mathrm{x}/\mathrm{v}\\ 0,& \mathrm{t}<\mathrm{x}/\mathrm{v}\end{array}\right\} \\ f\left(t\right)=\left\{\begin{array}{cc}\sin \left[-v\left(t-\frac{x}{v}\right)\right],& t>x/v\\ 0,& t<x/v\end{array}\right\} \end{gathered}$$

Use the expression

$L_{28}L\left[f\left(t\right)\right]=e^{-pa}.G\left(p\right)$

Let

$a=\underset{=}{v}$

And,

$$\begin{gathered} g(t-a)=\sin \left[-v(t-a)\right] \\ g\left(t\right)=\sin (-vt) \\ =-\sin \left(vt\right) \end{gathered}$$

Calculate the value of G (p).

$$\begin{gathered} G\left(p\right)\mathit{=}L\left[g\left(t\right)\right] \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}L\left[-\sin \mathrm{\left(}\mathrm{vt}\mathrm{\right)}\right] \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}\frac{\mathit{-}\mathit{v}}{{\mathit{p}}^{\mathit{2}}\mathit{+}{\mathit{v}}^{\mathit{2}}} \end{gathered}$$

Calculate the Laplace transform of the function.

$$\begin{gathered} L\left[f\left(t\right)\right]=e^{-pa}.G\left(p\right) \\ =e^{\frac{p}{v}}\left(\frac{-v}{p^2+v^2}\right) \end{gathered}$$

Thus, the Laplace transform of$\left\{\begin{array}{c}\sin (x-vt),t>x/v\\ \Omega \end{array}\right\}$is localid="1659248824919" $L\left[f\left(t\right)\right]=e^{\frac{px}{v}}\left(\frac{-v}{n^2+v^2}\right)$.