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Math Help - Convolution

  1. #1
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    Convolution

    Hi,

    I have the following problem:

    h(t)=(sin t+cost)\theta(t)
    w(t)=\theta(t)-\theta(t-2\pi)

    How do I compute the following convolution::
    h(t)\ast w(t)

    I know that the definition for convolution is as follow:

    (1)

    But how do I apply (1) to the given problem?

    Do I distribute the parentheses

    w(t)=\theta(t)-\theta(t-2\pi)=\theta(t)-\theta(t)-\theta2\pi

    Any help would be greatly appreciate it.
    Thank you
    Last edited by 4Math; March 21st 2011 at 05:02 AM.
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  2. #2
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    If you apply the definition of convolution then you get \displaystyle(h*w)(t) = \int_{-\infty}^\infty\!\!\!(\sin\tau + \cos\tau)\theta(\tau)\bigl(\theta(t-\tau) - \theta(t-\tau-2\pi)\bigr)\,d\tau. You cannot simplify that any further unless you know something about the nature of the function \theta.
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  3. #3
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    Quote Originally Posted by Opalg View Post
    If you apply the definition of convolution then you get \displaystyle(h*w)(t) = \int_{-\infty}^\infty\!\!\!(\sin\tau + \cos\tau)\theta(\tau)\bigl(\theta(t-\tau) - \theta(t-\tau-2\pi)\bigr)\,d\tau. You cannot simplify that any further unless you know something about the nature of the function \theta.
    How do I integrate this integral?

    Kind regards
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  4. #4
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    I guess that integration is solved by partialintegration.

     \int udv=uv-\int vdu

    My question is this:

    what is u and dv in this integral: \displaystyle \int_{-\infty}^\infty\!\!\!(\sin\tau + \cos\tau)\theta(\tau)\bigl(\theta(t-\tau) - \theta(t-\tau-2\pi)\bigr)\,d\tau.

    u=...
    dv=....
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  5. #5
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    May be \theta(t) is the Heaviside step function ?
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