FPGA Central

I am stuck on the following Fourier Transform problem.


Suppose that I have two time-domain signals a(t) and b(t) that are
trains of impulses, one with spacing ln(2) seconds and one with spacing
ln(3) seconds. The Nth impulse of a(t) therefore occurs at time N*ln(2)
= ln(2^N), and likewise for b(t) at t=ln(3^N).

Since 2 and 3 are relatively prime, these two signals NEVER overlap in
time except for the N=0 case (because 2^K is never equal to 3^N, K and
N both integers).

Let the Fourier transform of a(t) = A(w) and likewise the FT of b(t) =
B(w). A(w) is clearly a train of impulses with spacing 2*PI/ln(2) and
B(w) is a train of impulses with spacing 2*PI/ln(3).

Now suppose I multiply there two signals together. The result is
trivially a single impulse at t=0, since the time sequences only
coincide at t=0. This implies that the convolution of A(w) and B(w) is
also unity (independant of w).

I would like to show that A convolved with B is unity without resorting
to the time-domain non-overlap argument. Is this possible? The
convolution seems difficult because the amount that you shift one
spectrum by before an alignment occurs between an impulse in A(w) and
an impulse in B(w) is non-trivial.


Thanks for any pointers.

Bob Adams

"Robert Adams" <[email protected]> writes:

> I am stuck on the following Fourier Transform problem.
>
>
> Suppose that I have two time-domain signals a(t) and b(t) that are
> trains of impulses, one with spacing ln(2) seconds and one with spacing
> ln(3) seconds. The Nth impulse of a(t) therefore occurs at time N*ln(2)
> = ln(2^N), and likewise for b(t) at t=ln(3^N).
>
> Since 2 and 3 are relatively prime, these two signals NEVER overlap in
> time except for the N=0 case (because 2^K is never equal to 3^N, K and
> N both integers).
>
> Let the Fourier transform of a(t) = A(w) and likewise the FT of b(t) =
> B(w). A(w) is clearly a train of impulses with spacing 2*PI/ln(2) and
> B(w) is a train of impulses with spacing 2*PI/ln(3).
>
> Now suppose I multiply there two signals together. The result is
> trivially a single impulse at t=0, since the time sequences only
> coincide at t=0. This implies that the convolution of A(w) and B(w) is
> also unity (independant of w).
>
> I would like to show that A convolved with B is unity without resorting
> to the time-domain non-overlap argument. Is this possible? The
> convolution seems difficult because the amount that you shift one
> spectrum by before an alignment occurs between an impulse in A(w) and
> an impulse in B(w) is non-trivial.
>
>
> Thanks for any pointers.
>
> Bob Adams

Gosh, Bob, what an intriguing thought problem! The only thing I'd like
to point out is that a(t) * b(t) is delta^2(t), and I'm not sure that
the FT of delta^2(t) is the same as the FT of delta(t). I.e., you're
multiplying a distribution by another distribution and who knows what
the result should be?!
--
Randy Yates
Sony Ericsson Mobile Communications
Research Triangle Park, NC, USA
[email protected], 919-472-1124

"Robert Adams" <[email protected]> wrote in message
news:[email protected] ups.com...
>I am stuck on the following Fourier Transform problem.
>
>
> Suppose that I have two time-domain signals a(t) and b(t) that are
> trains of impulses, one with spacing ln(2) seconds and one with spacing
> ln(3) seconds. The Nth impulse of a(t) therefore occurs at time N*ln(2)
> = ln(2^N), and likewise for b(t) at t=ln(3^N).
>
> Since 2 and 3 are relatively prime, these two signals NEVER overlap in
> time except for the N=0 case (because 2^K is never equal to 3^N, K and
> N both integers).
>
> Let the Fourier transform of a(t) = A(w) and likewise the FT of b(t) =
> B(w). A(w) is clearly a train of impulses with spacing 2*PI/ln(2) and
> B(w) is a train of impulses with spacing 2*PI/ln(3).
>
> Now suppose I multiply there two signals together. The result is
> trivially a single impulse at t=0, since the time sequences only
> coincide at t=0. This implies that the convolution of A(w) and B(w) is
> also unity (independant of w).
>
> I would like to show that A convolved with B is unity without resorting
> to the time-domain non-overlap argument. Is this possible? The
> convolution seems difficult because the amount that you shift one
> spectrum by before an alignment occurs between an impulse in A(w) and
> an impulse in B(w) is non-trivial.
>
>
> Thanks for any pointers.
>
> Bob Adams

Bob,

I am going to come at this in a way that may seem circuitous but I believe
helps in the end:

a and b are discrete as you have defined them - and, apparently of infinite
extent?
Therefore, A and B are each discrete and are periodic - albeit with
different periods.
The way they are defined, the periods are relatively prime.

I'm not sure whether you intended to multiply a X b or A X B. I'm not sure
that it matters either :-)

Here's what I get from drawing "cartoons" of the transform pairs,
multiplying and convolving:
It's an arm-waving description that can no doubt be more rigorously
justified:

Above, you said:
>The
> convolution seems difficult because the amount that you shift one
> spectrum by before an alignment occurs between an impulse in A(w) and
> an impulse in B(w) is non-trivial.

I think it's easy to assert the opposite:
The amount that you shift one spectrum before an alignment occurs is
infinitesimally short.

Here's how I justify this hypothesis:

From a practical point of view either of the impulse trains *nearly*
coincides with the other after some period of time. You define "nearly".
This means that there is an *effective* period. But, it's not so much to
the point here - only to go to the next step.
As you define "nearly" to be a smaller and smaller difference in position,
the time span over which near coincidence occurs must get longer and longer.
In infinite time there is a pair of impulses that are infinitely close.
So, if we convolve these two waveforms (by definition over infinite time)
then there will be one pair and only one pair of impulses that coincide. At
the next time shift there will be another and so on ad infinitum....

QED sorta

Fred

Interesting observation; it seems like this reasoning might lead one to
conclude that the convolution is approaching being continuous!


Bob Adams

"Robert Adams" <[email protected]> wrote in message
news:[email protected] ups.com...
> Interesting observation; it seems like this reasoning might lead one to
> conclude that the convolution is approaching being continuous!
>
>
> Bob Adams
>

Exactly

Fred