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Math Help - sin(t)*cos(t) convolution

  1. #1
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    sin(t)*cos(t) convolution

    Hello everyone, what is the convolution of sin(t) with cos(t). The provided answer seems to be merely t/2.sin(wt)... However, the integral im getting includes this expression along with other sin and cos terms.

    Another question, whats the integral of sin(X) FROM -inf to +inf....

    Any help is appreciated...
    Last edited by CaptainBlack; March 21st 2010 at 11:19 AM.
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  2. #2
    MHF Contributor chisigma's Avatar
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    You can compute the convolution \sin t * \cos t or from definition...

    \sin t * \cos t = \int_{0}^{t} \sin u \cdot \cos (t-u)\cdot du (1)

    ... or using the Laplace Transform. In that case is \mathcal{L} \{\sin t\} = \frac{1}{1+s^{2}} , \mathcal{L} \{\cos t\} = \frac{s}{1+s^{2}} so that...

    \mathcal {L} \{\sin t * \cos t \}= \frac{s}{(1+s^{2})^{2}} (2)

    ... and ...

    \sin t * \cos t = \mathcal{L}^{-1} \{\frac {s}{(1+s^{2})^{2}}\} = \frac{t\cdot \sin t}{2} (3)

    Kind regards

    \chi \sigma
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by chisigma View Post
    You can compute the convolution \sin t * \cos t or from definition...

    \sin t * \cos t = \int_{0}^{t} \sin u \cdot \cos (t-u)\cdot du (1)
    Convolution is defined as:

    (f*g)(t)=\int_{-\infty}^{\infty} f(\tau)g(t-\tau)\; d\tau

    So in this case the convolution cannot be defined by this integral.

    What you have defined is the convolution of u(t)\sin(t) and u(t)\cos(t), which may be what the OP wants, but is not what they asked for.

    CB
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  4. #4
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    Thanks guy for the help, well turns out that i've done a horrible calculation mistake, the result worked out fine. Well i would have loved to use laplace transform but we (as students) are obliged to resolve to "convolution" for the time being. And concerning the 0 --> t or the -inf ---> +inf dispute, the result i got was using the 1st set of limits...

    And i still dont get why it can't be defined.

    (thanks once more)
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