Question

Use L28 to find the Laplace transform of

$\mathit{f}\left(t\right)\mathbf{=}\{\begin{array}{ll}sin\left(t-\pi /2\right),& t>\pi /2.\\ 0,& t<\pi /2.\end{array}$

Short answer

Answer

The values of $f\left(t\right)$is $L\left\{f\left(t\right)\right\}=e^{-\frac{5}{2}}\left(\frac{1}{p^2+1}\right)$.

Step-by-step solution

Given information

The given function is $f\left(t\right)=\left\{\begin{array}{c}\sin \left(t-\mathrm{\pi }/2\right),t>\mathrm{\pi }/2\\ \\ 0,t<\mathrm{\pi }/2\end{array}\right\}$

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)

Differentiate the given function

From $\left(L28\right)$.

$f\left(t\right)=\left\{\begin{array}{c}g\left(t-a\right),t>a0\\ \\ 0,t<a\end{array}\right\}$

Then,

$$\begin{gathered} L\left\{f\left(t\right)\right\}=e^{px}G\left(p\right) \\ G\left(p\right)=L\left\{g\left(t\right)\right\} \end{gathered}$$

Compare the given function with the defined general form ofan obtain.

$$\begin{gathered} a-\mathrm{\pi }/2 \\ \mathrm{g}\left(\mathrm{t}\right)=\sin \mathrm{t} \end{gathered}$$

L 28)

Result for the given function and substitute the above values.

$$\begin{gathered} L\left[f\left(t\right)\right]=e^{-px}G\left(p\right) \\ L\left\{f\left(t\right)\right\}=e^{pw/2}L\left(\sin t\right) \\ L\left\{f\left(t\right)\right\}=e^{-\frac{\mathrm{\pi }}{2}}\left(\frac{1}{p^2+1}\right) \\ L\left\{\sin ar\right\}=\frac{a}{p^2+{\alpha }^2} \end{gathered}$$

Thus, the values of $f\left(f\right)$is$L\left[f\left(t\right)\right]=e^{\frac{m}{2}}\left(\frac{1}{p^2+1}\right)$.