Reply by R.G. Stockwell April 28, 20062006-04-28
<robert.w.adams@verizon.net> wrote in message 
news:1145980027.476608.310590@j33g2000cwa.googlegroups.com...
> Suppose that I have a continous-time signal that consists of a finite > number of Dirac impulses at integer time points. For example, lets say > the signal starts at t=0 and lasts for 100 seconds, with 100 dirac > impulses, one per second.
Interesting question. It seems to me we can simplify this and still get to the same issues. Suppose your continous time signal is only a single impulse. delta(t - t0) Your spectrum is: exp(-i2 pi f t0) Now multiply your signal by any boxcar function which has a '1' at time t0. (it can be any pretty much length). You still have the same signal, but as you say you have the original spectrum convolved with the boxcar function's spectrum. Another related interesting question is to 1) take any time signal. 2) Multiply by a boxcar. So in the fourier domain, you convolve the original spectrum with your sinc function and get a "new spectrum". 3) Now, multiply by the boxcar again. The time signal does not change. However, your "new spectrum" is now convolved by a sinc function a second time, and yet it should not change. Hmmmm. So what does the FT of a time signal multiplied by two boxcars (the larger completely overlapping the smaller) look like? Lots of fun, the answer to these problems is of course,.... oh, lunch time is over, back to work. Cheers, bob PS these look like unitary operators
Reply by Dmitry Utyansky April 27, 20062006-04-27
Hi Robert,

robert.w.adams@verizon.net wrote:

> So this means that whe you do the convolution, the decreasing lobes of > the sinc function are picking up energy in the original spectrum at all > multiples of FS. So if you slightly change the window length, the > sinc(x) nulls move, and the energy picked up in every one of the N*FS > regions of the original signal would change slightly. But somehow, whn > youu add all these infinite contributions together, you get an answer > that does not change until the window function passes over the next > integer time period, and then suddenly everything changes all at once
I guess this is easier to visualize when one considers a single pulse in time domain at 0 (not a train), and its obviuos Fourier transfer image (without loss of generality). Multiplication of the pulse in time domain with arbitrary rectangular window (as long as the pulse is within) == convolution of thie corresponding spectrum with sinc. Which will not change the spectrum, since it is flat. Like, vice versa, in time domain, when one filters DC function with low-pass filter of any width, the function does not change. I.e., from what I understand, both methods give the same result, not surpisingly... Regards, Dmitry. PS. Hope the Google interface works...
Reply by Clay S. Turner April 27, 20062006-04-27
"robert bristow-johnson" <rbj@audioimagination.com> wrote in message 
news:1146103097.715117.79580@y43g2000cwc.googlegroups.com...
> Andor wrote: >> robert bristow-johnson wrote: >> >> > method 1 requires convolution in the frequency domain with the >> > FT of the window function (which is a sinc() function). doesn't >> > appear easy at first blush to me. >> >> Convolution with the sinc function is equivalent to ideal bandlimiting. >> If the convolution is in the frequency domain, then the bandlimiting >> takes place in the time domain, ie. windowing. Is that hard to accept? >> If you don't believe it, you can easily work out the integrals and >> prove the Fourier pair pulse <-> sinc. > > whoa, Andor... > > "... _easily_ work out the integrals..." > > so please easily work out this one: > > +inf > rect(t) ?= integral{ sinc(f*t) * exp(j*2*pi*f*t) df} > -inf > > > where sinc(u) = sin(pi*u)/(pi*u) > > { 1 |u| < 1/2 > and rect(u) = { > { 0 |u| > 1/2 > > i'm not too anal about what rect(+/- 1/2) is (but if i where, it would > be 1/2), just as long as it's not a dirac impulse function. > > i know it's true only because of the duality theorem of the F.T. and > knowing what the F.T. of rect(t) is. but i can't work out that > integral directly for some arbitrary t. > > r b-j >
Hello Robert, I worked it out using simple integration. You will even see I dispensed with complex numbers in the 1st step. As they say it is easy once you know the trick. The trick I used in this is quite powerful and since it can be used in more than one problem it is actually a technique. http://personal.atl.bellsouth.net/p/h/physics/FTofSync.pdf Clay
Reply by Andor April 27, 20062006-04-27
robert bristow-johnson wrote:

> Andor wrote: > > robert bristow-johnson wrote: > > > > > method 1 requires convolution in the frequency domain with the > > > FT of the window function (which is a sinc() function). doesn't > > > appear easy at first blush to me. > > > > Convolution with the sinc function is equivalent to ideal bandlimiting. > > If the convolution is in the frequency domain, then the bandlimiting > > takes place in the time domain, ie. windowing. Is that hard to accept? > > If you don't believe it, you can easily work out the integrals and > > prove the Fourier pair pulse <-> sinc. > > whoa, Andor... > > "... _easily_ work out the integrals..."
By "easily" I meant to compute the FT of the rect(t) and use duality to determine the inverse FT of the sinc(t). It occured to me later on, that a proof of the Convolution Theorem would also be required (although this doesn't look too hard, once you know it). Yet later, it also occured to me that for this specific instance (Fourier Transform of Dirac pulses), you also need a working knwoledge of funcional analysis and measure theory, if you don't believe the theorems and want to work out everything yourself. In retrospect, "easily" might have been a misnomer :-).
> so please easily work out this one: > > +inf > rect(t) ?= integral{ sinc(f*t) * exp(j*2*pi*f*t) df} > -inf
I agree, this one is a bummer (BTW, the sinc function in the integrand should only have "f" as an argument, and not "f * t"). If I really had to work out this integral (homework :-), I would probably try to using complex contour integration (sinc = re exp(z)/z) using two half circles around 0, and letting their radius go to zero and infinity, respectively. Regards, Andor
Reply by robert bristow-johnson April 27, 20062006-04-27
robert.w.adams@verizon.net wrote:
> No Internet connection in the Hotel, but, hey, I am in Copenhagen so > who is gonna complain!
geez, i wouldn't. but what ungodly hour is it there, now? and i can't imagine who you're seeing in Denmark! r b-j
Reply by April 27, 20062006-04-27
No Internet connection in the Hotel, but, hey, I am in Copenhagen so
who is gonna complain!


Bob

Reply by robert bristow-johnson April 27, 20062006-04-27
robert.w.adams@verizon.net wrote:
>... I am making no > assumptions about whether or not there is aliasign going on; remember, > the impulse train IS my sgnal and does not represent samples of > anything. > Now lets consider the window function. It is NOT a series of impulses; > it is ecxactly 1.0 all the way up to some time interval (NOT > necessarily an integer value), and therefore its Fourier Transform is > NOT periodic with Fs. In fact it is a true sinc function whose lobes > extend to infinity, with decreasing energy. > > So this means that whe you do the convolution, the decreasing lobes of > the sinc function are picking up energy in the original spectrum at all > multiples of Fs. So if you slightly change the window length, the > sinc(x) nulls move, and the energy picked up in every one of the N*Fs > regions of the original signal would change slightly. But somehow, when > you add all these infinite contributions together, you get an answer > that does not change until the window function passes over the next > integer time period, and then suddenly everything changes all at once
Bob, i think this is mathematically the same issue as the one regarding reconstruction of the oversampled signal. there are multiple ideal brick-wall LPFs that can be used to reconstruct the very same bandlimited (to something well below Fs/2) input, x(t), from the very same samples, x[n] = x(n*T). same exact samples, different sinc() functions attached to them, but somehow they add up to the same x(t). the only reason i brought it up is that is seems to me to be a simpler setting for this same phenom.
> I would like to understand this more deeply,
me too.
> ... for reasons that are too involved to explain here.
well, i don't have any of those. my reasons are not involved: i'm just anal.
> I apologize for any typos; I am using one of those horrible Hotel-room > "internet-on-your-tv-monitor" things!
gee, Bob, i had thunked that i was the only luddite that didn't have a wireless laptop that traveled with him to everywhere and every hotel he goes. :-) r b-j
Reply by robert bristow-johnson April 26, 20062006-04-26
Andor wrote:
> robert bristow-johnson wrote: > > > method 1 requires convolution in the frequency domain with the > > FT of the window function (which is a sinc() function). doesn't > > appear easy at first blush to me. > > Convolution with the sinc function is equivalent to ideal bandlimiting. > If the convolution is in the frequency domain, then the bandlimiting > takes place in the time domain, ie. windowing. Is that hard to accept? > If you don't believe it, you can easily work out the integrals and > prove the Fourier pair pulse <-> sinc.
whoa, Andor... "... _easily_ work out the integrals..." so please easily work out this one: +inf rect(t) ?= integral{ sinc(f*t) * exp(j*2*pi*f*t) df} -inf where sinc(u) = sin(pi*u)/(pi*u) { 1 |u| < 1/2 and rect(u) = { { 0 |u| > 1/2 i'm not too anal about what rect(+/- 1/2) is (but if i where, it would be 1/2), just as long as it's not a dirac impulse function. i know it's true only because of the duality theorem of the F.T. and knowing what the F.T. of rect(t) is. but i can't work out that integral directly for some arbitrary t. r b-j
Reply by Jerry Avins April 26, 20062006-04-26
robert.w.adams@verizon.net wrote:
> Thanks for all the input. > > > Here is how I am thinking of this. > > Yes, this is truly a continuous time input, and for the moment I dont > even want to consider that the Dirac functions are samples of some > other analog signal; the dirac functions actually ARE the signal of > interest. > Now because the pulse-train is on a fixed time grid, the spectrum of > the pulse train is exactly periodic with period FS. I am making no > assumptions about whether or not there is aliasign going on; remember, > the impulse train IS my sgnal and does not represent samples of > anything. > Now lets consider the window function. It is NOT a series of impulses; > it is ecxactly 1.0 all the way up to some time interval (NOT > necessarily an integer value), and therefore its Fourier Transform is > NOT periodic with FS. In fact it is a true sinc function whose lobes > extend to infinity, with decreasing energy. > > So this means that whe you do the convolution, the decreasing lobes of > the sinc function are picking up energy in the original spectrum at all > multiples of FS.
What FS? You just wrote that there's no sampling going on.
> So if you slightly change the window length, the > sinc(x) nulls move, and the energy picked up in every one of the N*FS > regions of the original signal would change slightly. But somehow, whn > youu add all these infinite contributions together, you get an answer > that does not change until the window function passes over the next > integer time period, and then suddenly everything changes all at once > > I would like to understand this more deeply, for reasons that are too > involved to explain here. > > I apologize for any typos; I am using one of those horrible Hotel-room > "internet-on-your-tv-monitor" things! > > Bob Adams >
-- Engineering is the art of making what you want from things you can get. &#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;
Reply by April 26, 20062006-04-26
Thanks for all the input.


Here is how I am thinking of this.

Yes, this is truly a continuous time input, and for the moment I dont
even want to consider that the Dirac functions are samples of some
other analog signal; the dirac functions actually ARE the signal of
interest.
Now because the pulse-train is on a fixed time grid, the spectrum of
the pulse train is exactly periodic with period FS. I am making no
assumptions about whether or not there is aliasign going on; remember,
the impulse train IS my sgnal and does not represent samples of
anything.
Now lets consider the window function. It is NOT a series of impulses;
it is ecxactly 1.0 all the way up to some time interval (NOT
necessarily an integer value), and therefore its Fourier Transform is
NOT periodic with FS. In fact it is a true sinc function whose lobes
extend to infinity, with decreasing energy.

So this means that whe you do the convolution, the decreasing lobes of
the sinc function are picking up energy in the original spectrum at all
multiples of FS. So if you slightly change the window length, the
sinc(x) nulls move, and the energy picked up in every one of the N*FS
regions of the original signal would change slightly. But somehow, whn
youu add all these infinite contributions together, you get an answer
that does not change until the window function passes over the next
integer time period, and then suddenly everything changes all at once

I would like to understand this more deeply, for reasons that are too
involved to explain here.

I apologize for any typos; I am using one of those horrible Hotel-room
"internet-on-your-tv-monitor" things!

Bob Adams