Use L28 to find the Laplace transform of

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

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

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)

From (L28).

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

Then,

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

Compare the given function with the defined general form of f (t) an obtain.

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

FromL 28)

Result for the given function and substitute the above values.

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

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