How Do You Compute the Convolution of a Convolution with Variable Limits?

  • Thread starter Thread starter muzialis
  • Start date Start date
  • Tags Tags
    Convolution
muzialis
Messages
156
Reaction score
1
Hello there,

I can not work out a computation i found, involving the convolution of a convolution.

G is a function, as well as ε, and using the notation

G*ε = ∫G(t-tau) dε (the integral being performed between 0 and t)

I want to compute

G*ε*ε

I try

(G*ε)*ε

and end up with

(G*ε)*ε =∫ (∫G(t-tau) dε ) dε2,

the first integral being between 0 and t, and the second between 0 and t - tau2 ,the new mute variable).

In a paper I find a different result, of the type

(G*ε)*ε =∫ ∫G(2t-tau1 - tau2) dε dε2,

I really cann ot get my head around it, any help so appreciated.

Thanks
 
Physics news on Phys.org
convolution definition is incorrect. It should be integral of G(t-tau)epsilon(tau) dtau.

You can't integrate over depsilon. If epsilon is a function. It means nothing. So, try with proper definition.

Your use of convolution property is correct, so it should work out. If not then try first doing epsilon with epsilon and then the result with g.
 
utkarsh1 said:
convolution definition is incorrect. It should be integral of G(t-tau)epsilon(tau) dtau.

You can't integrate over depsilon. If epsilon is a function. It means nothing. So, try with proper definition.

Your use of convolution property is correct, so it should work out. If not then try first doing epsilon with epsilon and then the result with g.

I think the OP is using a measure theory notation.
muzialis said:
Hello there,

I can not work out a computation i found, involving the convolution of a convolution.

G is a function, as well as ε, and using the notation

G*ε = ∫G(t-tau) dε (the integral being performed between 0 and t)

Why are the integral limits only between 0 and t? Typically convolution integrals should run from -infinity to +infinity. Are you sure the limits in the reference you looked at are the same as yours?