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Math Help - Convolution: sin(t) * u(t)

  1. #1
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    Convolution: sin(t) * u(t)

    A computer gives the answer as -\cos t, but by hand, I seem to be getting a non-converging integral. What am I doing wrong?

    \sin t * u(t) = \int_{-\infty}^\infty \sin \tau \cdot u(t - \tau) d\tau

    Since t > \tau for result to be non-zero, we have:

    = \int_{-\infty}^t \sin \tau d\tau

    = (- \cos \tau)|_{-\infty}^t

    Which does not converge.
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  2. #2
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    Re: Convolution: sin(t) * u(t)

    Quote Originally Posted by VinceW View Post
    A computer gives the answer as -\cos t, but by hand, I seem to be getting a non-converging integral. What am I doing wrong?

    \sin t * u(t) = \int_{-\infty}^\infty \sin \tau \cdot u(t - \tau) d\tau

    Since t > \tau for result to be non-zero, we have:

    = \int_{-\infty}^t \sin \tau d\tau

    = (- \cos \tau)|_{-\infty}^t

    Which does not converge.
    You're meant to assume sin(t) = 0 for t < 0. In which case the answer is 1 - cos(t).
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  3. #3
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    Re: Convolution: sin(t) * u(t)

    Why? I don't see any validity behind that assumption at all.
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    Re: Convolution: sin(t) * u(t)

    Quote Originally Posted by VinceW View Post
    Why? I don't see any validity behind that assumption at all.
    Well, I guess it will depend on whether t < 0 makes sense in the context from which the question has come, won't it.
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  5. #5
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    Re: Convolution: sin(t) * u(t)

    Quote Originally Posted by mr fantastic View Post
    Well, I guess it will depend on whether t < 0 makes sense in the context from which the question has come, won't it.
    You are not helping with this problem. This is a pure math problem so of course t < 0 makes sense.

    I'm wondering whether that convolution genuinely doesn't exist for \sin t * u(t) and a converging convolution only exists for [(\sin t) \cdot u(t)] * u(t)
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