Question

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

Short answer

Transform of periodic function

$\mathcal{L}\left\{f\right(t\left)\right\}=\frac{1}{1-e^{-2\pi s}}\left[\frac{e^{-s\pi }}{s^2+1}+\frac{1}{s^2+1}\right]$

$\mathcal{L}\left\{f\right(t\left)\right\}=\frac{1}{1-e^{-2\pi s}}\left[\frac{e^{-s\pi }}{s^2+1}+\frac{1}{s^2+1}\right]$

Step-by-step solution

Define Transform of periodic function:

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

Find function on intervals:

From the graph we can conclude the period of$t=2\pi$

So the function on interval $0⩽t<2\pi$as$f\left(t\right)=\left\{\begin{array}{l}\sin t0⩽t<\pi \\ 0\pi ⩽t<2\pi \end{array}\right.$

Find ft:

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

$=\frac{1}{1-e^{-2\pi s}}\left[{\int }_0^{\pi }\sin t{e}^{-st}dt+{\int }_{\pi }^{2\pi }\left(0\right)e^{-st}dt\right]$

$$\begin{gathered} ={\left.\frac{1}{1-e^{-2\pi s}}\left[\frac{e^{-st}}{s^2+1}(-s\sin t-1\cos t)\right]\right|}_0^{\pi } \\ =\frac{1}{1-e^{-2\pi s}}\left[\frac{e^{-s\pi }}{s^2+1}(0+1)-\frac{1}{s^2+1}(0-1)\right] \end{gathered}$$

Therefore, Laplace transform is

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