Question

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

Short answer

The resultant solution for the Laplace transform of a periodic function by using a theorem is found to be $\mathcal{L}\left\{f\right(t\left)\right\}=\frac{1}{1-e^{-2as}}\left[\frac{1}{s}-\frac{e^{-as}}{s}\right]$.

Step-by-step solution

Define the transform of a periodic function theorem:

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

.$\mathbf{L}\mathbf{\left\{}\mathbf{f}\mathbf{\right(}\mathbf{t}\mathbf{\left)}\mathbf{\right\}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{1}\mathbf{-}{\mathbf{e}}^{\mathbf{-}\mathbf{sT}}}{\int }_0^T{\mathbf{e}}^{\mathbf{-}\mathbf{st}}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{dt}$

Evaluate the periodic function using the theorem to find a solution:

Consider the given periodic function graph,

Hence, from the graph, the given function is periodic with $t=2a$.

Thus, the function can be defined on the interval $0⩽t<2a$,

$f\left(t\right)=\left\{\begin{array}{l}1,0⩽t<a\\ 0,a⩽t<2a\end{array}\right.$

Apply the periodic function theorem to find the Laplace transform of a periodic function,

$\mathcal{L}\left\{f\right(t\left)\right\}=\frac{1}{1-e^{-sT}}{\int }_0^T{e}^{-st}f\left(t\right)dt$

Therefore, the equation can be written as,

$$" width="9" height="19" role="math">=11-e-2as∫0ae-stdt-∫a2a(0)e-stdt

$=\frac{1}{1-e^{-2as}}\left[-{\left.\frac{e^{-st}}{s}\right|}_0^a\right]$

$=\frac{1}{1-e^{-2as}}\left[\frac{1}{s}-\frac{e^{-sa}}{s}\right]$

$=\frac{1}{1-e^{-2as}}\left[\frac{1}{s}-\frac{e^{-as}}{s}\right]$

Hence, the Laplace transform of a periodic function is $\mathcal{L}\left\{f\right(t\left)\right\}=\frac{1}{1-e^{-2as}}\left[\frac{1}{s}-\frac{e^{-as}}{s}\right]$.