windowing question

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

"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

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...

<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