convolution help

Jul 2008
4
0
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

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Nov 2005
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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
 
Oct 2007
1,026
278
Taguig City, Philippines
what is the definition of x there? a simple multiplication?
 
Jul 2008
4
0
can it be expanded, or re-expressed as separate terms at all?
 
Oct 2007
1,026
278
Taguig City, Philippines
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

\(\displaystyle [f(x)*h(x)][g(x)*h(x)] = ?\)
 
Jul 2008
4
0
is it associative? ie [f(t)g(t)]*h(t) = f(t)[g(t)*h(t)]?
 
Oct 2007
1,026
278
Taguig City, Philippines
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)