Reply by glen herrmannsfeldt●September 14, 20052005-09-14
Jerry Avins wrote:
(snip, I wrote)
>> There is, but how do you find it in a time varying envelope of
>> the wave? As in my previous example, say a gaussian wave packet.
>> Any inaccuracy in finding the DC reference is an uncertainty in the
>> frequency measured by zero crossings. That is, even with a noiseless
>> signal.
> Certainly, but not being able to use it isn't the same as its not being
> there. I don't mind loose terminology as long as everyone involved knows
> what's real. I hate it when a sidekick complains, "But you said ..." so
> I've gotten uptight in that department.
I agree. I was trying, but I think I give up now. There are no
noiseless sources and no noiseless detectors. There are no filters
with infinite edges, either. Back to ordinary problems...
thanks,
-- glen
Reply by glen herrmannsfeldt●September 14, 20052005-09-14
Rune Allnor wrote:
(snip originally regarding noiseless measurements)
>>Well, sodium already has two lines without any modifying.
> I don't understand the basis for your post.
> If you know everything about an experiment (like with
> sodium) there is no reason for making a measurement.
> If you state a model case wher you have access to the
> model parameters, there is no reason to make a measurement.
Some of the discussion had to do with the ability to measure
the frequency of a single sine wave, in the absence of
noise, with only three points. The case with two sines
was also discussed. I was using the sodium doublet as a
two sine case at least a little bit real.
> What I am discussing, are real-world contraints where one
> aims to learn something about the process being measured,
> and where one have to deal with noise and system
> imperfections.
I was trying to add a little real world into an idealized
problem. One can never have an infinite duration sine in
a real problem, so it must have finite duration. I was then
trying to understand the measurement of a finite duration
noiseless sine or small number of sine source. To understand
whether the finite duration limits the ability to measure the
frequency even without any noise.
> The first factors to consider (there can be far more)
> is the finite-duration measurement in the presence of
> noise. In that case, the resolution principle applies
> (if one aims to find if there is one or two lines
> present by mean of the DFT) or the Cramer Rao limit
> applies, if one uses a parametric approach.
First I wanted to consider a finite duration noiseless
source. To add more realism, though still without noise,
I didn't allow a perfect rectangle envelope, as that
would require excessive bandwidth. (Even putting
the edges on sine zero crossings still requires a
discontinuous, and unrealistic, derivative.)
> Whatever one tries to do, it is crucial to have a clear
> appreciation of the problem one tries to solve. What
> method will work an what will not, depend on the
> question one seks the answer for.
That is true, and it wasn't so clear here. I wanted
to add realism to a measurement of a noiseless source,
but how much realism is needed? How about detector
noise? Mixer noise for a heterodyne detector?
There are many possibilities.
-- glen
Reply by Rune Allnor●September 14, 20052005-09-14
glen herrmannsfeldt wrote:
> Rune Allnor wrote:
>
> > glen herrmannsfeldt wrote:
>
> (snip)
>
> >>Now consider sodium at higher temperature and pressure.
> >>The lines are broadened due to doppler shift and collisions,
> >>and there is also reabsorption by other atoms.
>
> > Now I am confused. Are you capable of studying each line in
> > its own right? Or do you just see a fuzzy wnergy distribution
> > with respect to frequency?
>
> Well, the question had to do with being able to measure
> frequency exactly when there is no noise. I was trying to
> find real world examples that, even without noise, one would
> not be able to measure exactly. You can study each line in
> its own right if you can build a filter to separate them.
> That filter is then part of the system. It can be considered
> noiseless, but otherwise should have properties that real
> filters have.
>
> > If you are not able to study single absorption/emission lines,
> > then I would say you are dealing with a power spectrum density
> > estimation problem.
>
> So, in the noiseless case can you do an exact power
> spectral density measurement?
>
> > If you can see single lines and want to find their frequency
> > as exact as possible, then you are dealing with a frequency
> > estimation problem.
>
> The claim was exact. I am looking for reasons why it
> won't be exact.
>
> > If you want to see if one line has split in two when you
> > modify the material, you are dealing with a line resolution
> > problem.
>
> Well, sodium already has two lines without any modifying.
I don't understand the basis for your post.
If you know everything about an experiment (like with
sodium) there is no reason for making a measurement.
If you state a model case wher you have access to the
model parameters, there is no reason to make a measurement.
What I am discussing, are real-world contraints where one
aims to learn something about the process being measured,
and where one have to deal with noise and system
imperfections.
The first factors to consider (there can be far more)
is the finite-duration measurement in the presence of
noise. In that case, the resolution principle applies
(if one aims to find if there is one or two lines
present by mean of the DFT) or the Cramer Rao limit
applies, if one uses a parametric approach.
Whatever one tries to do, it is crucial to have a clear
appreciation of the problem one tries to solve. What
method will work an what will not, depend on the
question one seks the answer for.
Rune
Reply by Jerry Avins●September 13, 20052005-09-13
glen herrmannsfeldt wrote:
> Jerry Avins wrote:
>
> (snip regarding electromagnetic fields and zero crossing)
>
>> The analogy goes further. Just as the voltage on a tank's capacitor
>> crosses zero, so does the EM wave's E field. There is no DC (the
>> wavelength would be too long :-)!) but there is a "DC" reference.
>
>
> There is, but how do you find it in a time varying envelope of
> the wave? As in my previous example, say a gaussian wave packet.
> Any inaccuracy in finding the DC reference is an uncertainty in the
> frequency measured by zero crossings. That is, even with a noiseless
> signal.
Certainly, but not being able to use it isn't the same as its not being
there. I don't mind loose terminology as long as everyone involved knows
what's real. I hate it when a sidekick complains, "But you said ..." so
I've gotten uptight in that department.
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
Reply by glen herrmannsfeldt●September 13, 20052005-09-13
Jerry Avins wrote:
(snip regarding electromagnetic fields and zero crossing)
> The analogy goes further. Just as the voltage on a tank's capacitor
> crosses zero, so does the EM wave's E field. There is no DC (the
> wavelength would be too long :-)!) but there is a "DC" reference.
There is, but how do you find it in a time varying envelope of
the wave? As in my previous example, say a gaussian wave packet.
Any inaccuracy in finding the DC reference is an uncertainty in the
frequency measured by zero crossings. That is, even with a noiseless
signal.
-- glen
Reply by glen herrmannsfeldt●September 13, 20052005-09-13
Jerry Avins wrote:
(snip)
>> If you recall the Poynting vector is simply S = E cross H. In free
>> space both E and H are in phase but both are time varying at the
>> frequency of the E-M wave. Hence S varies at twice the frequency of
>> the wave. If a 30 Hz E-M wave passes through you, you are actually hit
>> with 60 pulses of energy per second. For all but the lowest
>> frequencies, this pulsing is almost unobservable. So most deal with
>> the time averaged power, which for a plane wave is constant. And the
>> average works out to be 1/2 of the peak.
Also, the frequency that most people think of as "60Hz" is
actually 120Hz. Most vibrating systems, such as transformer
windings, move on both halves of the cycle.
-- glen
Reply by rhnl...@yahoo.com●September 13, 20052005-09-13
Rune Allnor wrote:
> kroger@princeton.edu wrote:
> > 2. If I sample less than one period, then zero-fill up to one period
> > (or more?), can I get a good result with an FFT?
>
> It depends what you want to do. You will not be able to separate
> several close sinusoidals by zero-filling.
Depends on how close. If the sinusoids are farther apart than a
few bins, then zero-filling and using a longer FFT may tell you how
far apart they are with a higher resolution, perhaps even with
more accuracy that a 3-point polynomial interpolation. Whether
the S/N ratio is high enough for that difference to be statistically
significant, and whether it's worth the computational cost
is another question.
IMHO. YMMV.
--
rhn A.T nicholson d.O.t C-o-M
Reply by Jerry Avins●September 13, 20052005-09-13
Clay S. Turner wrote:
> "Jerry Avins" <jya@ieee.org> wrote in message
> news:orydnfiM27QwSrveRVn-gg@rcn.net...
>
>>glen herrmannsfeldt wrote:
>>
>> ...
>>
>>
>>>In considering zero crossing of an optical signal, how do we
>>>know where zero is? Optical signals are always AC coupled,
>>>and so require a baseline restore. Any errors in the baseline
>>>restore shift the reference for zero crossing.
>>
>>Whatever applies to optics in this regard applies to all electromagnetic
>>radiation. There are zero crossings of the E and H waves, but it's hard to
>>measure them. What's more, the instantaneous power remains constant.
>>
>
>
> Hello Jerry,
>
> You actually have to be careful here. For plane waves in space, the
> instantaneous power as viewed by a stationary observer pulses.
>
> If you recall the Poynting vector is simply S = E cross H. In free space
> both E and H are in phase but both are time varying at the frequency of the
> E-M wave. Hence S varies at twice the frequency of the wave. If a 30 Hz E-M
> wave passes through you, you are actually hit with 60 pulses of energy per
> second. For all but the lowest frequencies, this pulsing is almost
> unobservable. So most deal with the time averaged power, which for a plane
> wave is constant. And the average works out to be 1/2 of the peak.
I wasn't thinking of the Poynting vector, but of the cyclic energy
transfer between the E and H fields. The energy relations in a lossless
tank circuit (an idealization, to be sure) in which the energy remains
constant but shuttles between the electric field in the capacitor and
the magnetic field of the inductor the analog of what I meant to convey.
The analogy goes further. Just as the voltage on a tank's capacitor
crosses zero, so does the EM wave's E field. There is no DC (the
wavelength would be too long :-)!) but there is a "DC" reference.
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
Reply by Clay S. Turner●September 13, 20052005-09-13
"Jerry Avins" <jya@ieee.org> wrote in message
news:orydnfiM27QwSrveRVn-gg@rcn.net...
> glen herrmannsfeldt wrote:
>
> ...
>
>> In considering zero crossing of an optical signal, how do we
>> know where zero is? Optical signals are always AC coupled,
>> and so require a baseline restore. Any errors in the baseline
>> restore shift the reference for zero crossing.
>
> Whatever applies to optics in this regard applies to all electromagnetic
> radiation. There are zero crossings of the E and H waves, but it's hard to
> measure them. What's more, the instantaneous power remains constant.
>
Hello Jerry,
You actually have to be careful here. For plane waves in space, the
instantaneous power as viewed by a stationary observer pulses.
If you recall the Poynting vector is simply S = E cross H. In free space
both E and H are in phase but both are time varying at the frequency of the
E-M wave. Hence S varies at twice the frequency of the wave. If a 30 Hz E-M
wave passes through you, you are actually hit with 60 pulses of energy per
second. For all but the lowest frequencies, this pulsing is almost
unobservable. So most deal with the time averaged power, which for a plane
wave is constant. And the average works out to be 1/2 of the peak.
Clay
Reply by Jerry Avins●September 13, 20052005-09-13
glen herrmannsfeldt wrote:
...
> In considering zero crossing of an optical signal, how do we
> know where zero is? Optical signals are always AC coupled,
> and so require a baseline restore. Any errors in the baseline
> restore shift the reference for zero crossing.
Whatever applies to optics in this regard applies to all electromagnetic
radiation. There are zero crossings of the E and H waves, but it's hard
to measure them. What's more, the instantaneous power remains constant.
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������