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I'm not totally sure I understand the concept, maybe an easy example will help me understand it.

Let f be $ f(x) = 1 $ if $ 0 \le x \le 1 $ and $f(x) =0$ elsewhere.

So the convolution is defined to be

$\int_{-\infty}^{\infty}f(x-t)f(t)dt$

So, basically, if x is not defined, what is $(f*f)$? I get that $ 0 \le t \le 1 $ is a necessary condition for the integral not to be 0 on a point, but what about x-t, if x is not specified?

aga7689
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1 Answers1

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Ok, $f\ast f$ is mean to be a function of $x$ which is why $x$ is not a fixed value.

$f\ast f=\int_{-\infty}^\infty f(x-t)f(t)dt=\int_{0}^1 f(x-t)dt$ because of the definition of $f$ being $0$ outside of $[0,1]$.

Jürgen Sukumaran
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