1. ## 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?

2. Originally Posted by mathgeekk
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

3. what is the definition of x there? a simple multiplication?

4. yep, x is just multiplication

5. can it be expanded, or re-expressed as separate terms at all?

6. no, it is not distributive over multiplication.. you can show it by definition.

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

$[f(x)*h(x)][g(x)*h(x)] = ?$

7. is it associative? ie [f(t)g(t)]*h(t) = f(t)[g(t)*h(t)]?

8. 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)