# convolution help

#### 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?

#### CaptainBlack

MHF Hall of Fame
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 $$\displaystyle h(t)$$ be a Gaussian kernel with unit spread parameter (normalise it as you wish). Let $$\displaystyle f(t)$$ and $$\displaystyle g(t)$$ both be $$\displaystyle 1$$ on disjoint intervals and zero elsewhere.

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

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

but $$\displaystyle (f*h)(t) (g*h)(t) >0$$

RonL

#### kalagota

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

#### mathgeekk

yep, x is just multiplication

#### mathgeekk

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

#### kalagota

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

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

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

#### mathgeekk

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

#### kalagota

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)