Question

Use Theorem 7.4.3 to find the Laplace transform of the given periodic function.

Short answer

Laplace transform of periodic function$f\left(t\right)=\frac{a}{b{s}^2}-\frac{a}{s\left(e^{sb}-1\right)}$

Step-by-step solution

Define Transform of a periodic function:

If $\mathit{f}\left(t\right)\mathbf{}$is piecewise continuous on$\left(0,x\right)$of exponential order and periodic with period T then,

$\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{1}\mathbf{-}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}}}{\int }_0^T{\mathbf{e}}^{\mathbf{-}\mathbf{8}}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{dt}$

Find ft :

$f\left(t\right)=\frac{a}{b}t$

$\frac{1}{1-e^{-sb}}{\int }_0^b{e}^{-st}\frac{a}{b}tdt$Integrate by parts,

$=\frac{1}{1-e^{-sb}}\left(\frac{-abs{e}^{-bs}-a{e}^{-bs}+a}{b{s}^2}\right)$

=$\frac{a}{b{s}^2\left(1-e^{-sb}\right)}\left(1-bs{e}^{-sb}-e^{-sb}\right)$

Therefore, Laplace transform is

$f\left(t\right)=\frac{a}{b{s}^2}-\frac{a}{s\left(e^{sb}-1\right)}$