Fourier transform, convolution, and duality

Given that x(t)*h(t) <-> X(w)H(w),
can we obtain  x(t)h(t) <-> (1/2PI)X(w)*H(w) 
by using only the duality property of Fourier transform?
THanks!

b83503104 wrote:

> Given that x(t)*h(t) <-> X(w)H(w),
> can we obtain  x(t)h(t) <-> (1/2PI)X(w)*H(w)
> by using only the duality property of Fourier transform?

First of all, you convolve with a constant:
(1/2PI)X(w)*H(w) <-> x(t)h(t)(1/2PI)

Given your question was:
x(t)h(t) <-> DELTA(w-1/2PI)X(w)*H(w)

Then maybe.
If your signal satisfyes X(w) == X(w-PI/2) then yes, but then it not only a
matter of duality, so the answer must be no.

If we look at the signal in frequence, its periodic with 2PI. By Convolving
with DELTA(w-PI/2) you only move it PI/2.

But:
x(t)h(t) <-> DELTA(w-2PI)*X(w)*H(w)

as X(w) == X(w-n2PI) (as long as we are talking discrete signals)

/Jesper

"b83503104" <b83503104@yahoo.com> wrote in message
news:73830d2d.0312240103.215027f6@posting.google.com...
> Given that x(t)*h(t) <-> X(w)H(w),
> can we obtain  x(t)h(t) <-> (1/2PI)X(w)*H(w)
> by using only the duality property of Fourier transform?
> THanks!

It's a "given" above that x(t)h(t) <-> (1/2PI)X(w)*H(w) - which seems
perfectly OK.

Can you arrive at
x(t)h(t) <-> (1/2PI)X(w)*H(w)

from
x(t)*h(t) <-> X(w)H(w)

by using only the duality property of the Fourier Transform?

From: http://www.engineering.usu.edu/classes/ece/3640/lecture6/node6.html
I get Time/Frequency Duality Property of the Fourier Transform as

For
f(t) <-> F(w)
Then
F(t) <-> 2*pi*f(-w)

So:
f(-w) <-> (1/2*pi)*F(t)  [which uses the scaling property]

substituting t for -w and -w for t:

f(t) <-> (1/2*pi)*F(-w)

Now, I'm not sure where to go from here but note that if f(t) is real then
F(-w)=F(w) (it is an even real function of w).

Did you want someone to do the proof for you?  I don't know how you avoid
the scaling property.

Fred