Thread Subject:

Convolution

Hello,

Denote convolution by *.

h(x) = f(x)*g(x)

If f(x0) = 0, then h(x0) = 0?

Thanks.

"Frank " <allinone_2003@yahoo.com.hk> wrote in message <j6e5hb$qjh$1@newscl01ah.mathworks.com>...
> Hello,
>
> Denote convolution by *.
>
> h(x) = f(x)*g(x)
>
> If f(x0) = 0, then h(x0) = 0?

Generally no.

Bruno

Thanks.

"Bruno Luong" <b.luong@fogale.findmycountry> wrote in message <j6ebrs$f6e$1@newscl01ah.mathworks.com>...
> "Frank " <allinone_2003@yahoo.com.hk> wrote in message <j6e5hb$qjh$1@newscl01ah.mathworks.com>...
> > Hello,
> >
> > Denote convolution by *.
> >
> > h(x) = f(x)*g(x)
> >
> > If f(x0) = 0, then h(x0) = 0?
>
> Generally no.
>
> Bruno

"Frank " <allinone_2003@yahoo.com.hk> wrote in message <j6e5hb$qjh$1@newscl01ah.mathworks.com>...
> Hello,
>
> Denote convolution by *.
>
> h(x) = f(x)*g(x)
>
> If f(x0) = 0, then h(x0) = 0?
==================

It usually won't be true for finite signals. Recall that

duration(h) = duration(f) +duration(g)

where duration() denotes the support length.Your hypothesis, however, would imply that

duration(h)<=duration(f)