Laplace transform $f(t) = f(t + \frac{2\pi }{a}) = \frac{\sin at}{|\sin at|}$

Question

I had done the laplace transform with the period fucntion (T): $F(s) = \frac{1}{1 - e^{ - sT}}\int_o^T {e^{ - st}}f(t) \, dt $ Howerver, in this fuction:

$$f(t) = f \left(t + \frac{2\pi}{a}\right) = \frac{\sin at}{|\sin at|}$$

What is the period? And You can give me detail of solution

Answer 1

Note that $T = 2 \pi/a$, and

$$\frac{\sin{a t}}{|\sin{a t}|} = \begin{cases} \\ +1 & a\, t \in (0,\pi) \\ -1 & a \,t \in (\pi,2 \pi)\end{cases}$$

Then

$$\begin{align}\int_0^{2 \pi/a} dt\, \frac{\sin{a t}}{|\sin{a t}|} \, e^{-s t} &= \int_0^{\pi/a} dt \, e^{-s t} - \int_{\pi/a}^{2 \pi/a} dt \, e^{-s t}\\ &= \frac{1-e^{-\pi s/a}}{s} -\frac{e^{-\pi s/a}-e^{-2 \pi s/a}}{s} \\ &= \frac{(1-e^{-\pi s/a})^2}{s} \end{align}$$

The full Laplace transform is then

$$\int_0^{\infty} dt\, \frac{\sin{a t}}{|\sin{a t}|} \, e^{-s t} = \frac{(1-e^{-\pi s/a})^2}{s(1-e^{-2\pi s/a})} = \frac{1}{s}\tanh{\frac{\pi s}{2 a}}$$

Answer 2

I did by let u = at, so the period T = 2pi

$\begin{array}{l} f(t) = f(t + \frac{{2\pi }}{a}) = \frac{{\sin at}}{{|\sin at|}} \\ Let:f(u) = (u + 2\pi ) = \frac{{\sin u}}{{|\sin u|}} \\ F(s) = L{\rm{[f(u)]}} = \frac{1}{{1 - {e^{ - 2\pi s}}}}\int_o^{2\pi } {{e^{ - st}}\frac{{\sin {a^2}u}}{{|\sin {a^2}u|}}du} \\ = \frac{1}{{1 - {e^{ - 2\pi s}}}}{\rm{\{ }}\int_o^\pi {{e^{ - st}}du} - \int_\pi ^{2\pi } {{e^{ - st}}du} {\rm{\} }} \\ = \frac{{{e^{\pi s}} - 1}}{{s({e^{\pi s}} + 1)}} \\ L{\rm{[f(u)]}} = \frac{{{e^{\pi s}} - 1}}{{s({e^{\pi s}} + 1)}} \\ = > L{\rm{[f(at)] = }}\frac{1}{a}F(\frac{s}{a}) = \frac{{{e^{\frac{{\pi s}}{a}}} - 1}}{{a\frac{s}{a}({e^{\frac{{\pi s}}{a}}} + 1)}} = \frac{{{e^{\frac{{\pi s}}{a}}} - 1}}{{s({e^{\frac{{\pi s}}{a}}} + 1)}} = Yours\\ \end{array}$

So did i work corecttly??