Hello
Evaluate $\displaystyle \mathcal{L} ( |sin(t) | )$
that absolute value makes my life hard.
how can i deal with it?
The L-transform of the 'full wave rectified' function $\displaystyle |\sin t|$ is...
$\displaystyle \displaystyle \mathcal{L} \{|\sin t| \} = \frac{\coth \frac{\pi s}{2}}{1+s^{2}}$ (1)
The (1) can be obtained applying the 'periodically repetead function's rule' of the L-tranform...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
i had in mind transforming the integral into $\displaystyle \displaystyle\int_{0}^{\infty }{{{e}^{-st}}\left| \sin t \right|\,dt}=\sum\limits_{k=0}^{\infty }{\int_{\pi k}^{(k+1)\pi }{{{e}^{-st}}\left| \sin t \right|\,dt}},$ i don't know if it works though.
Yeah. The general rule of thumb is that if $\displaystyle f:[0,\infty)\to\mathbb{R}$ is $\displaystyle \omega$-periodic such that $\displaystyle \text{sgn}\left(f(x)\right),\text{ }x\in[k\omega,(k+1)\omega]=s(k)$ is some nice function then one always writes $\displaystyle \displaystyle \int_0^{\infty}|f(x)|g(x)\text{ }dx$ as $\displaystyle \displaystyle \sum_{k=0}^{\infty}s(k)\int_{k\omega}^{(k+1)\omega }f(x)g(x)\text{ }dx$.
OK Guys
those solutions are too advanced to me
what I know that is |sin(t)| is periodic and I use a program that told me its period is pi
So Am able now to solve that laplace using the formula for finding laplace transform for periodic functions
but the problem here
is How did we know that the function is periodic?
it's an easy mix of telescoping sums and use of FTC.
we claim that $\displaystyle \displaystyle\sum\limits_{k=1}^{n-1}{\int_{k}^{k+1}{f(t)\,dt}}=\displaystyle\int_{1} ^{n}{f(t)\,dt}.$
all we need to have is that $\displaystyle f$ is given continuous, so it has an antiderivative, then
$\displaystyle \begin{aligned}
\sum\limits_{k=1}^{n-1}{\int_{k}^{k+1}{f(t)\,dt}}&=\sum\limits_{k=1}^{n-1}{\left( F(k+1)-F(k) \right)} \\
& =F(n)-F(1) \\
& =\int_{1}^{n}{f(t)\,dt}.
\end{aligned}$
it's a very useful trick.