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Math Help - convolution help

  1. #1
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    Exclamation convolution help

    does anybody know...

    can (f(t) x g(t))*h(t) be expressed as f(t)*h(t) x g(t)*h(t)? where * is convolution?
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  2. #2
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    Quote Originally Posted by mathgeekk View Post
    does anybody know...

    can (f(t) x g(t))*h(t) be expressed as f(t)*h(t) x g(t)*h(t)? where * is convolution?
    No.

    Counter example let h(t) be a Gaussian kernel with unit spread parameter (normalise it as you wish). Let f(t) and g(t) both be 1 on disjoint intervals and zero elsewhere.

    Then f(t) \times g(t) \equiv 0, and so:

    [(f(.) \times g(.))*h](t) \equiv 0,

    but (f*h)(t) (g*h)(t) >0

    RonL
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  3. #3
    MHF Contributor kalagota's Avatar
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    what is the definition of x there? a simple multiplication?
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  4. #4
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    yep, x is just multiplication
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  5. #5
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    can it be expanded, or re-expressed as separate terms at all?
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  6. #6
    MHF Contributor kalagota's Avatar
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    no, it is not distributive over multiplication.. you can show it by definition.

    [f(x)g(x)]*h(x)=(fg)(x)*h(x) = ? and

    [f(x)*h(x)][g(x)*h(x)] = ?
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  7. #7
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    is it associative? ie [f(t)g(t)]*h(t) = f(t)[g(t)*h(t)]?
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  8. #8
    MHF Contributor kalagota's Avatar
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    no.. in LHS: f and g is inside the integral sign. in RHS, f will be outside of the integral sign.

    these are the some properties (as far as i know) that the convolution function have:

    symmetric: f*g=g*f
    associative (in convolution): (f*g)*h=f*(g*h)
    distributive over addition: f*(g+h) = f*g + f*h
    associative and symmetric in real multiplication: a(f*g)=(af)*g=f*(ag)
    also: (f*g)=(Df)*g = f* (Dg)
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