Question

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

Short answer

The Laplace transform of the given periodic function by using the theorem of transform of a periodic function is $\mathcal{L}\left\{f\right(t\left)\right\}=\frac{1}{1-e^{-sT}}{\int }_0^T{e}^{-st}f\left(t\right)dt$.

Step-by-step solution

Define the theorem transform of a periodic function:

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

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

Consider the given graph and determine the given function by using the interval:

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

Thus, the function can be defined on the interval as,

$f\left(t\right)=\left\{\begin{array}{l}t0⩽t<1\\ 2-t1⩽t<2\end{array}\right.$

Consider the theorem transform of a periodic function,

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

Apply the intervals in the equation,

$=\frac{1}{1-e^{-2s}}\left[{\int }_0^{1}t{e}^{-st}dt-{\int }_1^2(2-t)e^{-st}dt\right]$

$=\frac{1}{1-e^{-2s}}\left[{\left.\left(-\frac{t{e}^{-st}}{s}+\frac{e^{-st}}{s^2}\right)\right|}_0^1+{\left.\left(-\frac{(2-t)e^{-st}}{s}-\frac{e^{-st}}{s^2}\right)\right|}_1^2\right]$

Substitute the interval values in $t$,

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

Evaluate the equation,

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

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