Hello

Evaluate $\displaystyle \mathcal{L} ( |sin(t) | )$

that absolute value makes my life hard.

how can i deal with it?

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- Dec 30th 2010, 01:47 PMLiverpoolLaplace transform of |sin(t)| ?
Hello

Evaluate $\displaystyle \mathcal{L} ( |sin(t) | )$

that absolute value makes my life hard.

how can i deal with it? - Dec 30th 2010, 01:58 PMpickslides
How about $\displaystyle \displaystyle \mathcal{L} ( |sin(t) | ) =\int_0^{\infty} e^{-st}\sqrt{\sinh^2t}~dt$

- Dec 30th 2010, 02:02 PMLiverpool
$\displaystyle |sin(t)| = \sqrt{sinh^2(t)}$ ?

- Dec 30th 2010, 02:24 PMpickslides
No it doesn't, I miss read your first post.

I meant to say $\displaystyle \displaystyle \mathcal{L} ( |\sin(t) | ) =\int_0^{\infty} e^{-st}\sqrt{\sin^2t}~dt$ - Dec 31st 2010, 01:51 AMchisigma
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

$\displaystyle \chi$ $\displaystyle \sigma$ - Dec 31st 2010, 04:22 AMKrizalid
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.

- Dec 31st 2010, 12:29 PMLiverpool
- Dec 31st 2010, 01:03 PMDrexel28
- Dec 31st 2010, 01:05 PMKrizalid
haha, i knew it, it just that the web sites i often use to compute integrals are not computing them so it's too lazy for me to compute them by hand.

- Dec 31st 2010, 01:10 PMDrexel28
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 $\displaystyle \displaystyle \sum_{k=0}^{\infty}s(k)\int_{k\omega}^{(k+1)\omega }f(x)g(x)\text{ }dx$.

- Dec 31st 2010, 01:34 PMRandom Variable
- Dec 31st 2010, 01:38 PMLiverpool
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? - Dec 31st 2010, 01:39 PMKrizalid
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. - Dec 31st 2010, 01:45 PMRandom Variable
- Dec 31st 2010, 01:48 PMLiverpool
Its better to answer the original poster then answer anyone else, and the latter should post a new thread if he have any question not asking them in the same thread.

I don't like this.

Respect me!!!!