91Ó°ÊÓ

The equation is$\frac{\mathrm{p}}{({\mathrm{p}}^2+{\mathrm{b}}^2)({\mathrm{p}}^2+{\mathrm{b}}^2)}$ .

The piecewise-continuous and exponentially-restricted real function f(t) is the inverse Laplace transform of a function $\mathit{F}\left(s\right)$, and it has the property:

$$\begin{gathered} \mathbf{L}\left\{f\right\}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{=}\mathbf{L}\mathbf{\{}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{\left\}}\mathbf{\right(}\mathbf{s}\mathbf{)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{F}\left(s\right) \end{gathered}$$

where L is the Laplace transform.

As per Convolution theorem, if we have two functions, taking their convolution and then Laplace is the same as taking the Laplace first (of the two functions separately) and then multiplying the two Laplace Transforms.

Consider the equation.

$\frac{\mathrm{p}}{({\mathrm{p}}^2+{\mathrm{b}}^2)({\mathrm{p}}^2+{\mathrm{b}}^2)}=\frac{1}{({\mathrm{p}}^2+{\mathrm{b}}^2)}â‹\dots \frac{\mathrm{p}}{({\mathrm{p}}^2+{\mathrm{b}}^2)}$

As per the convolution theorem.

role="math" localid="1659271235261" $$\begin{gathered} L^{-1}\left[\frac{1}{\left(p^2+b^2\right)}·\frac{p}{\left(p^2+b^2\right)}\right]={∫}_0^t{g}_1·g_{2}dx \\ ={∫}_0^t{g}_1\left(t\right)·g_2\left(t-x\right)dt \\ ={∫}_0^t\frac{1}{a}\sin ax\cos b\left(t-x\right)dx \\ =\frac{1}{a}{∫}_0^t\sin ax\cos b\left(t-x\right)dx \end{gathered}$$

$$\begin{gathered} \left[\frac{1}{\left(p^2+b^2\right)}·\frac{p}{\left(p^2+b^2\right)}\right]=\frac{1}{2a}\left(\frac{2a\cos at-\cos bt}{a^2-b^2}\right) \\ =\frac{\cos at-\cos bt}{a^2-b^2} \end{gathered}$$

Thus, the inverse transform of given equation is role="math" localid="1659270365004" $\frac{\cos at-\cos bt}{a^2-b^2}$.