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 periodic function is $\mathcal{L}\left\{f\right(t\left)\right\}=\coth \frac{\pi s}{2}\frac{1}{s^2+1}$

Step-by-step solution

Define the theorem of transform of 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 withperiod$\mathit{T}$, 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 graph and define the function using the interval:

Hence, the function $f\left(t\right)$can be defined on the interval $0⩽t<\pi$.

Thus, from the given graph, it is identified that the curve is sinusoidal.

Therefore,$f\left(t\right)=\sin t,0⩽t<\pi$

Define the function$f\left(t\right)$ outside the interval,$f(t+\pi )=f\left(t\right)$

Consider the periodicity is,$T=\pi$.

Evaluate the function using the transform of a periodic function theorem,

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

Substitute,$f\left(t\right)=\sin t$and $T=\pi$in the function,

$\mathcal{L}\left\{f\right(t\left)\right\}=\frac{1}{1-e^{-\pi s}}{\int }_0^{\pi }{e}^{-st}\sin t\text{d}t$----(1)

Find the integration of${\int }_0^{\pi }{e}^{-st}\sin t\text{d}t$ by using the formula $uv-\int v\text{d}u$.

role="math" localid="1664171177781" $$\begin{gathered} Integrationbyparts \\ u=e^{-st}du=-s{e}^{-st} \\ \text{d}v=\sin tv=\int \sin t\text{d}t=-\cos t \end{gathered}$$

$$\begin{gathered} u=e^{-st}du=-s{e}^{-st} \\ \text{d}v=\cos tv=\int \cos t\text{d}t=\sin t \end{gathered}$$

${\int }_0^{\pi }{e}^{-st}\sin t\text{d}t=-{\left.e^{-st}\cos t\right|}_0^{\pi }-s{\int }_0^{\pi }\cos t{e}^{-st}\text{d}t$

$=-{\left.e^{-st}\cos t\right|}_0^{\pi }-{\left.s{e}^{-st}\sin t\right|}_0^{\pi }-s^2{\int }_0^{\pi }{e}^{-st}\sin t\text{d}t$

Apply the intervals in,

$=\left(-e^{-s\pi }\cos \pi +e^0\cos 0\right)+\left(-s{e}^{-s\pi }\sin \pi +s{e}^0\sin 0\right)-s^2{\int }_0^{\pi }{e}^{-st}\sin t\text{d}t$

$=e^{-s\pi }+1-s^2{\int }_0^{\pi }{e}^{-st}\sin t\text{d}t(\cos \pi =-1,\sin \pi =0,\cos 0=1,\sin 0=0)$

Evaluate the integral function to find the Laplace transform of a given periodic function:

Thus,${\int }_0^{\pi }{e}^{-st}\sin t\text{d}t=e^{-s\pi }+1-s^2{\int }_0^{\pi }{e}^{-st}\sin t\text{d}t$

$\left(1+s^2\right){\int }_0^{\pi }{e}^{-st}\sin t\text{d}t=e^{-s\pi }+1$

${\int }_0^{\pi }{e}^{-st}\sin t\text{d}t=\frac{e^{-s\pi }+1}{1+s^2}$

Substitute${\int }_0^{\pi }{e}^{-st}\sin t\text{d}t=\frac{e^{-s\pi }+1}{1+s^2}$inequation$\left(1\right)$

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

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

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

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

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

Therefore, by using the theorem, the Laplace transform of a given periodic function is found to be.$\mathcal{L}\left\{f\right(t\left)\right\}=\coth \frac{\pi s}{2}\frac{1}{s^2+1}$