Laplace transform of |sin(t)| ?

**Liverpool** · December 30th 2010, 01:47 PM

Hello

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

that absolute value makes my life hard.
how can i deal with it?

**pickslides** · December 30th 2010, 01:58 PM

How about $\displaystyle \mathcal{L} ( |sin(t) | ) =\int_0^{\infty} e^{-st}\sqrt{\sinh^2t}~dt$

**Liverpool** · December 30th 2010, 02:02 PM

$|sin(t)| = \sqrt{sinh^2(t)}$ ?

**pickslides** · December 30th 2010, 02:24 PM

No it doesn't, I miss read your first post.

I meant to say $\displaystyle \mathcal{L} ( |\sin(t) | ) =\int_0^{\infty} e^{-st}\sqrt{\sin^2t}~dt$

**chisigma** · December 31st 2010, 01:51 AM

Originally Posted by Liverpool

Hello

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

that absolute value makes my life hard.
how can i deal with it?

The L-transform of the 'full wave rectified' function $|\sin t|$ is...

$\displaystyle \mathcal{L} \{|\sin t| \} = \frac{\coth \frac{\pi s}{2}}{1+s^{2}}$ (1)

The (1) can be obtained applying the 'periodically repetead function's rule' of the L-tranform...

Kind regards

$\chi$ $\sigma$

**Krizalid** · December 31st 2010, 04:22 AM

i had in mind transforming the integral into $\displaystyle\int_{0}^{\infty }{{{e}^{-st}}\left| \sin t \right|\,dt}=\sum\limits_{k=0}^{\infty }{\int_{\pi k}^{(k+1)\pi }{{{e}^{-st}}\left| \sin t \right|\,dt}},$ i don't know if it works though.

**Liverpool** · December 31st 2010, 12:29 PM

Originally Posted by chisigma

The (1) can be obtained applying the 'periodically repetead function's rule' of the L-tranform...

Ok Good
I know that rule
but what is the period of it?

**Drexel28** · December 31st 2010, 01:03 PM

Originally Posted by Krizalid

i had in mind transforming the integral into $\displaystyle\int_{0}^{\infty }{{{e}^{-st}}\left| \sin t \right|\,dt}=\sum\limits_{k=0}^{\infty }{\int_{\pi k}^{(k+1)\pi }{{{e}^{-st}}\left| \sin t \right|\,dt}},$ i don't know if it works though.

This is the correct method:

Spoiler:

**Krizalid** · December 31st 2010, 01:05 PM

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.

**Drexel28** · December 31st 2010, 01:10 PM

Originally Posted by Krizalid

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.

Yeah. The general rule of thumb is that if $f:[0,\infty)\to\mathbb{R}$ is $\omega$ -periodic such that $\text{sgn}\left(f(x)\right),\text{ }x\in[k\omega,(k+1)\omega]=s(k)$ is some nice function then one always writes $\displaystyle \int_0^{\infty}|f(x)|g(x)\text{ }dx$ as $\displaystyle \sum_{k=0}^{\infty}s(k)\int_{k\omega}^{(k+1)\omega }f(x)g(x)\text{ }dx$ .

**Random Variable** · December 31st 2010, 01:34 PM

Does the first line need to be justified, or is it always valid?

Originally Posted by Drexel28

This is the correct method:

Spoiler:

**Liverpool** · December 31st 2010, 01:38 PM

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?

**Krizalid** · December 31st 2010, 01:39 PM

it's an easy mix of telescoping sums and use of FTC.

we claim that $\displaystyle\sum\limits_{k=1}^{n-1}{\int_{k}^{k+1}{f(t)\,dt}}=\displaystyle\int_{1} ^{n}{f(t)\,dt}.$

all we need to have is that $f$ is given continuous, so it has an antiderivative, then

$\begin{aligned} \sum\limits_{k=1}^{n-1}{\int_{k}^{k+1}{f(t)\,dt}}&=\sum\limits_{k=1}^{n-1}{\left( F(k+1)-F(k) \right)} \\ & =F(n)-F(1) \\ & =\int_{1}^{n}{f(t)\,dt}. \end{aligned}$

it's a very useful trick.

**Random Variable** · December 31st 2010, 01:45 PM

Originally Posted by Krizalid

it's an easy mix of telescoping sums and use of FTC.

we claim that $\displaystyle\sum\limits_{k=1}^{n-1}{\int_{k}^{k+1}{f(t)\,dt}}=\displaystyle\int_{1} ^{n}{f(t)\,dt}.$

all we need to have is that $f$ is given continuous, so it has an antiderivative, then

$\begin{aligned} \sum\limits_{k=1}^{n-1}{\int_{k}^{k+1}{f(t)\,dt}}&=\sum\limits_{k=1}^{n-1}{\left( F(k+1)-F(k) \right)} \\ & =F(n)-F(1) \\ & =\int_{1}^{n}{f(t)\,dt}. \end{aligned}$

it's a very useful trick.

Thanks. I've seen it done before, but was never quite sure of the justification.

**Liverpool** · December 31st 2010, 01:48 PM

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!!!!

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