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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?

Ethan
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2 Answers2

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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?

AxelT
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  • Thanks for the quick reply. But sometimes we still need to write down the convolution equation as an integral to see what happens... – Ethan Nov 29 '22 at 12:45
  • In what way do you mean to see what happens? Maybe you could clarify?

    If you take your above functions, e.g. $f(ax+b)$, you can calculate the Fourier transform of both and then invert the expression. It gives the same result as calculating the convolution directly, but is in some cases simpler to handle. The effect of adding a scale or so to the arguments is handled simply in that case.

     – AxelT Nov 29 '22 at 12:58
  • As we know, when f(x) convolves with a delta function delta(x), the product is f(x). What I wanted to know is, what happens when f(ax+b) convolves with delta(cx+d) and what the function we end up with is? I don't want to go to the Fourier transform route. I just want to see what f(x) will become. – Ethan Nov 29 '22 at 13:13
  • I see, I thought you weren't interested in merely the Dirac delta functions but any function in general as you changed from $\delta$ to $g$. If you keep the $\delta$, as in the Dirac delta function, you only introduce a shift; $\delta(xk-a)\neq 0$ at $x=a/k$. Calculating the convolution after that should be able to use the same reasoning as you provided. – AxelT Nov 29 '22 at 13:23
  • Thanks. I've changed the question a bit to make it clearer. – Ethan Nov 29 '22 at 13:34
  • I edited my post at the end, I believe it answers your question now. – AxelT Nov 30 '22 at 08:00
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I just found a textbook. I should be something like this?

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

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

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

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

Ethan
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