Convolution of functions with modified variables

Question

I come across with a convolution like this: $f(at+b)\ast\delta(t-k)$, and I am puzzled.
I am wondering in general how to convolute two functions both with scaled/complicated variables, like $f(ax+b)\ast g(cx+d)$. What is exactly the formula for this convolution?

Answer 1

At least one way would be to use the Fourier transform, as you can treat f and g separately. I.e. use $\mathcal{F}(f(t)*g(t))(s) = F(s)G(s)$ where $F(s)=\mathcal{F}(f(t))(s)$ and similarly for $G(s)$. In this case, you are still left with the inversion which hopefully isn't too difficult.

Furthermore, there are some identities which may be useful, e.g. that if you want the Fourier transform of $f(at)$, one hast that $\mathcal{F}(f(at))(s) = \frac{1}{|a|}F(\frac{s}{a})$.

I hope this is what you were asking for.

Edit: Due to the clarification, I guess you could simply write as the following. Let $u(x)=ax+b, v(x)=cx+d)$, then

$ f(u(x)) * g(v(x)) = \int_{-\infty}^{\infty} f(u(\tau))g(v(t-\tau))d\tau $

is what you are looking for?

Answer 2

I just found a textbook. I should be something like this?

Let $u(x)=f(ax+b), \\v(x) = g(cx+d)$

$f(ax+b)\ast g(cx+d)=u(x)\ast v(x)$

$\qquad \qquad \qquad \qquad \quad =\int_{-\infty}^{\infty} u(x-y)\;v(y)dy$

$\qquad \qquad \qquad \qquad \quad =\int_{-\infty}^{\infty} f(a(x-y)+b)\;g(cy+d)dy$