How Does Modifying Parameters Affect Convolution and Fourier Transforms?

[AxiomOfChoice]

I know that the convolution of two functions f(x) and g(x) is given by

<br /> (f * g)(y) = \int_{\mathbb R} f(x)g(y-x) dx.<br />

But what if I'm trying to convolve a function f(x) with a function g(x + az), where a is some constant? Is it just

<br /> (f*g)(y) = \int_{\mathbb R} f(x)g(y - x + az) dx.<br />

If so, why? I can't seem to find a definition of the convolution that makes this obvious.

 

[Mute]

Convolutions are symmetric in f and g:

$$(f \star g)(y) = (g \star f)(y);$$

you could thus write the convolution of f(x) and g(x + az) as

$$\int_{\mathbb{R}} dx~f(y-x)g(x+az),$$

which is perhaps easier to see.

 

[mathman]

Your question is a little puzzling. How does z relate to x in g(x+az)?

 

[AxiomOfChoice]

Okay, let me see if I can be more specific. What I'm really trying to do is to take the Fourier transform...in the x variable...of the product f(x) g(ax + y), where x,y\in \mathbb R are independent (hence unrelated) variables. So I'm interested in

<br /> \int_{-\infty}^{\infty} e^{-i p x} f(x) \cdot g(ax + y) \ dx.<br />

I know that Fourier transforms turn products into convolutions...but is it that straightforward in this case?

 

[mathman]

Okay, let me see if I can be more specific. What I'm really trying to do is to take the Fourier transform...in the x variable...of the product f(x) g(ax + y), where x,y\in \mathbb R are independent (hence unrelated) variables. So I'm interested in

<br /> \int_{-\infty}^{\infty} e^{-i p x} f(x) \cdot g(ax + y) \ dx.<br />

I know that Fourier transforms turn products into convolutions...but is it that straightforward in this case?

What you need is the opposite. Fourier transform of a convolution is product of Fourier transforms of the individual items. Once you've done that, take the back transform to get what you are looking for.