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Math Help - Laplace transform of |sin(t)| ?

  1. #1
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    Laplace transform of |sin(t)| ?

    Hello

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

    that absolute value makes my life hard.
    how can i deal with it?
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  2. #2
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    How about \displaystyle \mathcal{L} ( |sin(t) | ) =\int_0^{\infty}  e^{-st}\sqrt{\sinh^2t}~dt
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  3. #3
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    |sin(t)| = \sqrt{sinh^2(t)} ?
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  4. #4
    Master Of Puppets
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    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
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  5. #5
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Liverpool View Post
    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
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  6. #6
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    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.
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  7. #7
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    Quote Originally Posted by chisigma View Post
    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?
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  8. #8
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Krizalid View Post
    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}


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  9. #9
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    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.
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  10. #10
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Krizalid View Post
    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.
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  11. #11
    Super Member Random Variable's Avatar
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    Does the first line need to be justified, or is it always valid?

    Quote Originally Posted by Drexel28 View Post
    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}


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  12. #12
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    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?
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  13. #13
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    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}<br />
   \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)} \\ <br />
 & =F(n)-F(1) \\ <br />
 & =\int_{1}^{n}{f(t)\,dt}. <br />
\end{aligned}

    it's a very useful trick.
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  14. #14
    Super Member Random Variable's Avatar
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    Quote Originally Posted by Krizalid View Post
    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}<br />
   \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)} \\ <br />
 & =F(n)-F(1) \\ <br />
 & =\int_{1}^{n}{f(t)\,dt}. <br />
\end{aligned}

    it's a very useful trick.
    Thanks. I've seen it done before, but was never quite sure of the justification.
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  15. #15
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    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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