Chris Bore wrote:
(snip)

> By the way, if you evaluate the Fourier integral over a finite period
> of time, you find that the signal whose transform you obtained is the
> signal during that period of time, repeated ad infinitum - it is not
> zero outside the integration time.

The Fourier series represents periodic functions.

A finite length integral has to be mathematically the
same as an infinite one that is zero outside a finite range.

The difference is that the Fourier series uses a different
basis set.

-- glen

On Aug 14, 9:21 am, Chris Bore <chris.b...@gmail.com> wrote:

> > Since we are integrating with respect to time, we have to decide what
> > period of time we are interested in evaluating.  Evaluate your Fourier
> > transform for that amount of time, and evaluate your receiver's AM
> > audio output for the same amount of time, by integrating it's output.
>
> By the way, if you evaluate the Fourier integral over a finite period
> of time, you find that the signal whose transform you obtained is the
> signal during that period of time, repeated ad infinitum - it is not
> zero outside the integration time.

In that case, the output of your receiver other than in the instant at
which you sample it is not zero, but is what you sampled repeated ad
infinitum.  Or if you've finally realized the need to integrate the
receiver output, the output outside of the inteveral of the
integration is not zero, but a repeate of whatever you integrated.

You can't make a fair comparison unless you apply the same treatment
in both cases.

You can however change the methodology of the transform and the
receiver case by specifying that you will use a window function to
fade in and fade out your data.  In that case you know that the signal
outside the period of the integration is effectively zero, because
your window function made it so.

On Aug 12, 8:08�pm, cs_post...@hotmail.com wrote:
> On Aug 12, 9:17 am, Chris Bore <chris.b...@gmail.com> wrote:
>
> > Surely the correct answer is, as has been said, that the Fourier
> > Transform is a lousy model for short-time signals. So the prof is
> > wrong to use it to explain the effect of a short signal. That's the
> > only point I am trying to make. I understand all the 'short time =
> > broadband' stuff, I don't dispute it. What I do dispute, is using a
> > Fourier Transform frequency domain, where there is no time, to explain
> > interference with narrow band receivers that is localized in time. It
> > is mixing two orthogonal domains.
>
> Actually, I don't think there's a problem at all, if we remember that
> the fourier transform is formally an integral. &#4294967295;If we are going from
> time domain to frequency domain, it's an _integral with respect to
> time_.
>
> Since we are integrating with respect to time, we have to decide what
> period of time we are interested in evaluating. &#4294967295;Evaluate your Fourier
> transform for that amount of time, and evaluate your receiver's AM
> audio output for the same amount of time, by integrating it's output.

By the way, if you evaluate the Fourier integral over a finite period
of time, you find that the signal whose transform you obtained is the
signal during that period of time, repeated ad infinitum - it is not
zero outside the integration time.

> With a lot of handwaving of details, the AM output's integrated power
> should equal the power in the fourier transform bin corresponding to
> the receiver bandwidth.

On Aug 13, 4:01 pm, glen herrmannsfeldt <g...@ugcs.caltech.edu> wrote:
> cs_post...@hotmail.com wrote:

> > You cannot for example integrate the fourier integral over inifinite
> > time, and then sample the receiver output at a single instant in time,
> > which is what you seemed to want to do.
>
> The effect of the lightning strike will bounce off the moon.
> With the appropriate detector, you should be able to detect it.

Only if you are looking for it at a multiple of the earth-moon travel
time after the strike occurs.  This isn't really part of the
discussion at all.

The first issue is that while the frequency components of a perfect
impulse last for all time (negative and positive), lighting is not a
perfect impulse so it does not last for all time.  Especially not very
far into negative time.

The second issue is that people are forgetting that the fourier
transform is an integral evaluated over a (possible infinite) period
of time, and so cannot be compared to a snapshot at a single instant
in time.

This is why a fourier integral evaluated over a period of time
bracketing a lightning strike will show its frequency components as if
they are present for all time, but a receiver tuned to one of the
evaluated frequencies will only show a brief pulse of energy.  The
transform is an integral, while the receiver output is not - to make a
fair comparison, you must integrate the receiver output over the same
period of time.

People like to think of the fourier transform (especially the DFT) as
being a filter bank.  But we forget that it's not an instantaneous
one.  Instead, it's a bunch of filters with integrators on their
outputs - which is to say, a bunch of narrow band receivers with
integrators on their outputs.  (For the DFT, while the continuous one
is an infinite number of infinitely narrow receivers integrating their
outputs infinitely long).

Perhaps the clearest demonstration is this:  If you evaluate a DFT
over a period of say 10 seconds, and inject into it a 1 KHz sine wave,
you will get reading in the 1 KHz bin.  If you pulse that sinewave on
for only 1 of the seconds (and ramp it on and off via a nice window
function), you will get a smaller reading.  If you turn it on for two
seconds, you get a larger reading proportional to the amount of time
it was one.  The transform is fundamentally an integral over time.
And while it treats the components as if they were present for all
time, the output of the transform is in fact proportional to the
amount of time that they were present, provided of course that goot
windowing has been performed to step functions at the turn on and turn
off points.

cs_posting@hotmail.com wrote:
(snip)

> What you cannot do is integrate the fourier integral and the receiver
> AM audio over different periods of time and necessarily expect their
> results to match.

> You cannot for example integrate the fourier integral over inifinite
> time, and then sample the receiver output at a single instant in time,
> which is what you seemed to want to do.

The effect of the lightning strike will bounce off the moon.
With the appropriate detector, you should be able to detect it.

That is the result of some terms not canceling at the
appropriate time, the interference pattern from reflections
off different parts of the moon (and everything else).

-- glen

On Aug 13, 4:32 am, Chris Bore <chris.b...@gmail.com> wrote:

> The Fourier Transform is from -infinity to +infinity.
> You can't choose a time interval over which to evaluate it, infinity
> is in its definition.

If you wish to evaluate it over a period of infinity, then you will
also have to integrate the audio output from your AM receiver over
that same infinite interval.

When you do that (theoretically of course), you will in fact find your
interfering signal exists in the receiver output, just as it exists in
the fourier transform.

What you cannot do is integrate the fourier integral and the receiver
AM audio over different periods of time and necessarily expect their
results to match.

You cannot for example integrate the fourier integral over inifinite
time, and then sample the receiver output at a single instant in time,
which is what you seemed to want to do.

On Aug 12, 8:08�pm, cs_post...@hotmail.com wrote:
> On Aug 12, 9:17 am, Chris Bore <chris.b...@gmail.com> wrote:
>
> > Surely the correct answer is, as has been said, that the Fourier
> > Transform is a lousy model for short-time signals. So the prof is
> > wrong to use it to explain the effect of a short signal. That's the
> > only point I am trying to make. I understand all the 'short time =
> > broadband' stuff, I don't dispute it. What I do dispute, is using a
> > Fourier Transform frequency domain, where there is no time, to explain
> > interference with narrow band receivers that is localized in time. It
> > is mixing two orthogonal domains.
>
> Actually, I don't think there's a problem at all, if we remember that
> the fourier transform is formally an integral. &#4294967295;If we are going from
> time domain to frequency domain, it's an _integral with respect to
> time_.
>
> Since we are integrating with respect to time, we have to decide what
> period of time we are interested in evaluating. &#4294967295;Evaluate your Fourier
> transform for that amount of time, and evaluate your receiver's AM
> audio output for the same amount of time, by integrating it's output.

The Fourier Transform is from -infinity to +infinity.
You can't choose a time interval over which to evaluate it, infinity
is in its definition.

> With a lot of handwaving of details, the AM output's integrated power
> should equal the power in the fourier transform bin corresponding to
> the receiver bandwidth.

On Aug 12, 9:17 am, Chris Bore <chris.b...@gmail.com> wrote:
> Surely the correct answer is, as has been said, that the Fourier
> Transform is a lousy model for short-time signals. So the prof is
> wrong to use it to explain the effect of a short signal. That's the
> only point I am trying to make. I understand all the 'short time =
> broadband' stuff, I don't dispute it. What I do dispute, is using a
> Fourier Transform frequency domain, where there is no time, to explain
> interference with narrow band receivers that is localized in time. It
> is mixing two orthogonal domains.

Actually, I don't think there's a problem at all, if we remember that
the fourier transform is formally an integral.  If we are going from
time domain to frequency domain, it's an _integral with respect to
time_.

Since we are integrating with respect to time, we have to decide what
period of time we are interested in evaluating.  Evaluate your Fourier
transform for that amount of time, and evaluate your receiver's AM
audio output for the same amount of time, by integrating it's output.
With a lot of handwaving of details, the AM output's integrated power
should equal the power in the fourier transform bin corresponding to
the receiver bandwidth.

Chris Bore wrote:
>>>> Since lightning has a short duration in time we know it has a broad
>>>> spectrum.  It therefore interferes with many receivers.  Q.E.D.?
>>> Accepted, but...
>>> The specrum is broad for ever. So why, then, is the interference not
>>> also eternal?
>>         The story is the frequencies cancel out for most of eternity.
> 
> I'm not making myself clear. Any given narrow band does not cancel
> out. All of them together do, to make the short tie signal - that is
> what the Fourier Transform is about. But the Prof's question was why
> does the lightning interfere with a range of bands such as AM and FM
> radio, etc. These are all narrow-band receivers. So they don't cancel
> within their band.

The tricky point is that even that part of the signal in a narrow band 
cancels out if the original spectrum is continuous. The narrower the 
band, the longer the remnant.

> OK. Let's take an example. I listen to BBC Radio 4 on 92.5 MHz FM.
> 
> The flat spectrum of the lightning surely has a frequency component at
> 92.5 MHz, so it will interfere with the radio receiver.
> 
> The radio is a narrow band receiver, so it picks up a narrow range of
> frequency components. Let's idealize and say it selects just one.

Aye, there's the rub! If it selected just one, it would ring forever. 
These idealizations obscure the true nature of the case. To take that 
further, the receiver wouldn't respond at all if it were capable of 
isolating a single frequency. Just as the output would ring forever, it 
would take forever for the response to begin.

>                                                   The
> radio is in effect acting like one bin of a Fourier Transform,
> modelled as a set of narrow band filters. So it is detecting this
> frequency component. And we agree that the frequency component is
> present for all time, with fixed amplitude. Because it is a narrow
> band receiver, you can't argue that this frequency somehow cancels
> itself out. It is always there, yet it interferes with this receiver
> for only a short time.
> 
> Surely the correct answer is, as has been said, that the Fourier
> Transform is a lousy model for short-time signals. So the prof is
> wrong to use it to explain the effect of a short signal. That's the
> only point I am trying to make. I understand all the 'short time =
> broadband' stuff, I don't dispute it. What I do dispute, is using a
> Fourier Transform frequency domain, where there is no time, to explain
> interference with narrow band receivers that is localized in time. It
> is mixing two orthogonal domains.

The Fourier transform is a lousy model for short-term signals because 
the subtleties that we can usually ignore become highly relevant. The 
math still holds, though, if we really do it instead of the arm waving 
we've become accustomed to.

   ...

Jerry
-- 
Engineering is the art of making what you want from things you can get.
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On Aug 11, 3:12�pm, cs_post...@hotmail.com wrote:
> On Aug 8, 1:08 pm, "dvsarw...@yahoo.com" <dvsarw...@gmail.com> wrote:
>
> > On Aug 7, 11:44 am, cs_post...@hotmail.com wrote
> > (among other things) that
>
> > > Or to put it bluntly, the frequency domain is practically speaking a
> > > bad way to represent one-shot or random (non-periodic) events.

Yes!

> You might also note that I was saying the frequency domain is a bad
> way to localize one-shot events.

Yes!

You must see what I am banging on about in this thread.

Please help me. I am not getting through about this.. :-)

I think frequency in the Fourier Transform is not what frequency is in
common usage.

Chris
=================
Chris Bore
BORES Signal Processing
www.bores.com