Question

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

Short answer

The solution of the Laplace transform of the given periodic function by using the theorem is found to be $-\frac{\left(e^{-as}-1\right)}{s\left(e^{-as}+1\right)}$.

Step-by-step solution

Define the transform of a periodic function theorem:

If$\mathit{f}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{}$is a piecewise continuous on $\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{x}\mathbf{)}$of exponential order and periodic with period$\mathit{T}$then, $\mathbf{L}\mathbf{\left\{}\mathbf{f}\mathbf{\right(}\mathbf{t}\mathbf{\left)}\mathbf{\right\}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{1}\mathbf{-}{\mathbf{e}}^{\mathbf{-}\mathbf{sT}}}{\int }_0^T{\mathbf{e}}^{\mathbf{-}\mathbf{st}}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{dt}$

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Determine the solution for the given periodic function:

Consider the periodic function,

Hence, use the theorem of 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$

Evaluate with $T=2a$,

$=\frac{1}{1-e^{-2as}}\left[{\int }_0^a{e}^{-st}dt-{\int }_a^{2a}{e}^{-st}dt\right]$

$=\frac{1}{1-e^{-2as}}\left(\frac{e^{-2as}}{s}-\frac{2e^{-as}}{s}+\frac{1}{s}\right)$

$=\frac{1}{s-s{e}^{-2as}}{\left(e^{-as}-1\right)}^2$

Simplify by taking the common factor out,

$=-\frac{{\left(e^{-as}-1\right)}^2}{s\left(e^{-2as}-1\right)}$

$=-\frac{\left(e^{-as}-1\right)}{s\left(e^{-as}+1\right)}$

Therefore, the solution for the Laplace transform of a periodic function isdata-custom-editor="chemistry" $-\frac{\left(e^{-as}-1\right)}{s\left(e^{-as}+1\right)}$ .