# Laplace transform of |sin(t)| ?

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• Dec 30th 2010, 01:47 PM
Liverpool
Laplace transform of |sin(t)| ?
Hello

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

that absolute value makes my life hard.
how can i deal with it?
• Dec 30th 2010, 01:58 PM
pickslides
How about $\displaystyle \mathcal{L} ( |sin(t) | ) =\int_0^{\infty} e^{-st}\sqrt{\sinh^2t}~dt$
• Dec 30th 2010, 02:02 PM
Liverpool
$|sin(t)| = \sqrt{sinh^2(t)}$ ?
• Dec 30th 2010, 02:24 PM
pickslides

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

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

$\chi$ $\sigma$
• Dec 31st 2010, 04:22 AM
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.
• Dec 31st 2010, 12:29 PM
Liverpool
Quote:

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?
• Dec 31st 2010, 01:03 PM
Drexel28
Quote:

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:

Problem: Compute $\displaystyle \int_0^{\infty}e^{-st}|\sin(t)|\text{ }dt$.

Solution: We note that

\displaystyle \begin{aligned}\int_0^{\infty}e^{-st}|\sin(t)|\text{ }dt &= \sum_{k=0}^{\infty}\int_{\pi k}^{\pi(k+1)}e^{-st}|\sin(t)|\text{ }dt\\ &=\sum_{k=0}^{\infty}\frac{e^{-s\pi k}\left(e^{\pi s}+1\right)}{s^2+1}\\ &= \frac{e^{\pi s}+1}{(s^2+1)(1-e^{-\pi s})}\\ &=\frac{\text{coth}\left(\frac{\pi s}{2}\right)}{s^2+1}\end{aligned}

• Dec 31st 2010, 01:05 PM
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.
• Dec 31st 2010, 01:10 PM
Drexel28
Quote:

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 $\displaystyle \sum_{k=0}^{\infty}s(k)\int_{k\omega}^{(k+1)\omega }f(x)g(x)\text{ }dx$.
• Dec 31st 2010, 01:34 PM
Random Variable
Does the first line need to be justified, or is it always valid?

Quote:

Originally Posted by Drexel28
This is the correct method:

Spoiler:

Problem: Compute $\displaystyle \int_0^{\infty}e^{-st}|\sin(t)|\text{ }dt$.

Solution: We note that

\displaystyle \begin{aligned}\int_0^{\infty}e^{-st}|\sin(t)|\text{ }dt &= \sum_{k=0}^{\infty}\int_{\pi k}^{\pi(k+1)}e^{-st}|\sin(t)|\text{ }dt\\ &=\sum_{k=0}^{\infty}\frac{e^{-s\pi k}\left(e^{\pi s}+1\right)}{s^2+1}\\ &= \frac{e^{\pi s}+1}{(s^2+1)(1-e^{-\pi s})}\\ &=\frac{\text{coth}\left(\frac{\pi s}{2}\right)}{s^2+1}\end{aligned}

• Dec 31st 2010, 01:38 PM
Liverpool
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 PM
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.
• Dec 31st 2010, 01:45 PM
Random Variable
Quote:

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.
• Dec 31st 2010, 01:48 PM
Liverpool
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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