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

  1. #16
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    yup yup ignore the poster and destroy his thread.
    The problem is that 2 are expert and one is a moderator.
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  2. #17
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    Krizalid's Avatar
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    we're respecting you, and i gave you an alternate way to solve your problem which Drexel has completed giving you a full solution, absolutely, understandable, so if you don't get it, the only easy thing you need to do, is to ask, and that's all.

    i actually complemented why i "broke" the integral into a sum and why using that bounds, so if still having any questions, ask, since at least my method it's pretty easy to get if you fully understand every content, which is not hard to digest!!!
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  3. #18
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    I told you those solutions are too advanced to me , and I ask a question
    this is my reply:

    Quote Originally Posted by Liverpool View Post
    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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  4. #19
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Liverpool View Post
    I told you those solutions are too advanced to me , and I ask a question
    this is my reply:
    How did we know that \left|\sin(t)\right| was periodic? Because \sin\left(t+\pi\right)=-\sin(t) and so \left|\sin(t+\pi)\right|=|\sin(t)|. Does that help?
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  5. #20
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    This is what I want
    Thanks
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  6. #21
    Super Member Random Variable's Avatar
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    for  \displaystyle \matcal{L} \{|\cos t|\} would you write it as  \displaystyle \int_{0}^{\frac{\pi}{2}} e^{-st} |\cos t| \ dt + \sum_{k=1}^{\infty}\int_{\frac{\pi (2k-1)}{2}}^{\frac{\pi (2k+1)}{2}} e^{-st} |\cos t | \ dt ?
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  7. #22
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Random Variable View Post
    for  \displaystyle \matcal{L} \{|\cos t|\} would you write it as  \displaystyle \int_{0}^{\frac{\pi}{2}} e^{-st} |\cos t| \ dt + \sum_{k=1}^{\infty}\int_{\frac{\pi (2k-1)}{2}}^{\frac{\pi (2k+1)}{2}} e^{-st} |\cos t | \ dt ?
    Right, to get \displaystyle \frac{s+e^{\frac{-\pi s}{2}}}{s^2+1}+\sum_{k=1}^{\infty}\frac{e^{\frac{-\pi(2k+1)}{2}}\left(e^{\pi s}+1\right)}{s^2+1} and just do the geometric series.
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  8. #23
    Math Engineering Student
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    well the plus there is actually an equal sign.
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  9. #24
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    Thanks all.
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