Convolution of a convolution

  • Thread starter muzialis
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In summary: I tryG*ε*εand end up with(G*ε)*ε =∫ (∫G(2t-tau1 - tau2) dε dε2, the first integral being between 0 and t, and the second between 0 and t - tau2 ,the new mute variable).This appears to be the result you were looking for. Unfortunately, the original reference you were using does not have the correct notation, and the result you get is incorrect. I'm sorry I couldn't be of more help.
  • #1
muzialis
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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
 
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  • #2
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.
 
  • #3
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?
 

1. What is the definition of convolution?

The convolution of two functions is a mathematical operation that combines the information from both functions to create a new function. It represents the amount of overlap between the two functions as one function is shifted across the other.

2. How is convolution used in science?

Convolution is used in a variety of scientific fields, including signal processing, image processing, and physics. It can be used to extract useful information from data, filter out noise, and model complex systems.

3. What is the difference between convolution and correlation?

Convolution and correlation are similar operations, but they have different mathematical properties. Convolution involves flipping one function and sliding it across the other, while correlation does not involve flipping. Additionally, convolution is used to combine two functions, while correlation is used to measure the similarity between two functions.

4. How does convolution relate to the Fourier transform?

The convolution theorem states that the convolution of two functions in the time domain is equivalent to the multiplication of their Fourier transforms in the frequency domain. This relationship is often used in signal processing and image processing to simplify calculations.

5. Can convolution be applied to non-continuous functions?

Yes, convolution can be applied to both continuous and discrete functions. In fact, convolution is commonly used in discrete systems, such as digital signal processing, where the input and output signals are discrete values.

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