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Associative Property of Convolution?

  1. Mar 8, 2012 #1

    I have a quick question about certain algebraic properties of convolution. If I have 3 functions x(f), y(f) and z(f), is the following true?

    [x(f) . g(f)] * z(f) = [x(f) * z(f)].g(f)

    I looked on Wikipedia but there's only a property like this if one of the terms is a scalar, so most likely I can't do relation described above?

  2. jcsd
  3. Mar 9, 2012 #2


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    Your notation is confusing. (What is *, what is .?) However, in general, convolution is associative and communicative. The easiest way to see it is by looking at the relationship of the transforms, where convolution becomes multiplication, which is both associative and communicative.
  4. Mar 9, 2012 #3
    Hey DWill.

    On top of what mathman said, you can prove it has these properties by resorting to the definition of convolution.

    Also if you aren't convinced, take a look at probability theory for finding the cumulative distribution for X,Y,Z where they are all independent (but not necessarily identically distributed) which is given by the convolution of all three pdf's.

    Because X + Y + Z = (X + Y) + Z = X + (Y + Z) = Y + X + Z = Z + X + Y and so on, you intuitively get the idea once you accept the theorem in probability that convolution must be associative and commutative.
  5. Mar 9, 2012 #4
    Ok thanks mathman and chiro! That makes sense. I just wondered if the associative property still applied with three arbitrary functions, because in all the places I've looked so far there is only two functions and a scalar used for the associative property. I'll think about it a bit further but I think this clears it up.

    And sorry if my notation was confusing, I was just going by the convention of the "." being multiplication and "*" being convolution.
  6. Mar 10, 2012 #5


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    Now I am very confused. I assumed you were interested in a three function convolution. However your notation, as you just defined it, seems to involved a convolution and a product.
  7. Mar 10, 2012 #6
    That's correct, the operation I was asking involves a convolution and a product. I looked at it more myself and tried it out on a few functions, and I think this might not be possible?

    To further clarify, I was wondering if the product of the convolution of x(f) and g(f) with z(f) is equal to the product of the convolution of x(f) and z(f) with g(f)?
  8. Mar 10, 2012 #7
    Ohh! That might not work. I was under the impression that both the . and the * were convolutions.
  9. Mar 10, 2012 #8


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    Indeed! Sometimes it tells us lot!

    :smile: Sorry, I couldn't resist.
  10. Mar 11, 2012 #9


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    Tooshay - I need to proofread better.
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