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I kind of want to clear this up once and for all:

$g (t) * f(t) = \int g(u)f(t-u)du$

$g(at) * f(bt) = \int g(au)f(bt-u)du$ ???

Benjamin Lindqvist
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1 Answers1

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The convolution $g*f$ is a single function, by definition: $(g*f)(t) = \int g(u)f(t-u)\,du$. Therefore $g(t)*f(t)$ doesn't really make sense. In particular, you can't plug in two different values at the same time to a function of one variable; so $g(at)*f(bt)$ definitely doesn't make sense. If you want to define new functions $G(t)=g(at)$ and $F(t)=f(bt)$ and consider $(G*F)(t) = \int G(u) F(t-u)\,du = \int g(au) f(b(t-u))\,du$, that's a reasonable thing to do.

Greg Martin
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  • Yes, I know it's kind of hand-wavy, but it's kind of accepted shorthand I thought. The 'correct' method does make the solution kind of obvious though, thanks a lot. – Benjamin Lindqvist Nov 09 '15 at 08:05