1
$\begingroup$

See https://en.wikipedia.org/wiki/Convolution#Definition where the equation is mentioned:

$(f * g )(t) = \int_{0}^{t} f(\tau) g(t - \tau)\, d\tau \text{ for } f, g : [0, \infty) \to \mathbb{R}$

But the basic equation is:

$(f * g )(t) \, \stackrel{\mathrm{def}}{=}\ \int_{-\infty}^\infty f(\tau) g(t - \tau) \, d\tau \\ = \int_{-\infty}^\infty f(t-\tau) g(\tau)\, d\tau. $

So what happened to the integral from $t$ to $\infty$?

Is it just a hack to make the value more local and symmetrica? Is there an actual equivalence? Is it just a typo?

$\endgroup$

1 Answer 1

1
$\begingroup$

When I hit the submit button I figured it out. Because $f$ and $g$ are zero for all negative $t$ values, any non-local value will be zero as well because $g(t - \tau)$ will be zero for all $\tau > t$ which causes $f(\tau) g(t - \tau)$ to be zero.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .