AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
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Re: AM electromagnetic waves: 20 KHz modulation frequencyonanastronomically-low carrier frequency
In article <46935ee4$0$30624$4c368faf@roadrunner.com>,
"Jimmie D" <GFENDER@carolina.rr.com> wrote:
> "Ron Baker, Pluralitas!" <this@aint.me> wrote in message
> news:4693037c$0$31221$4c368faf@roadrunner.com...
> >
> > "Dana" <raff242@yahoo.com> wrote in message
> > news:80b7c$46906e83$944e306e$20475@STARBAND.NET...
> >>
> >> "Ron Baker, Pluralitas!" <this@aint.me> wrote in message
> >> news:469064fd$0$16540$4c368faf@roadrunner.com...
> >>>> Do you understand that a DSB signal *is* AM?
> >>>
> >>> So all the AM broadcasters are wasting money by
> >>> generating a carrier?
> >>
> >> How did you jump to that conclusion.
> >
> > Is "DSBSC" DSB?
> >
>
> There have been attempts to remove the carrier but receivers could not be
> manufatured at a reasonable price that would demodulate the signal with the
> fidelity of an AM BCB signal. Probably could be done today but what would
> you l do with all those AM rx that suddenly dont work when the transition is
> made.
There's no advantage to DSB-SC that SSB-SC doesn't have and several that
SSB-SC alone has. Getting rid of one of the redundant sets of sidebands
halves the required bandwidth, for one. Also, if the two sideband sets
of DSB-SC experience differing phase alteration due to propagation
effects (not too uncommon), the signal can become unintelligible; that
effect is minimized with SSB-SC.
If all broadcasters used SSB-SC and precision frequency control (easy
and inexpensive these days) then SSB-SC receivers are pretty easy. But
that doesn't solve the problem of all those AM receivers...
Things seem to be moving in the direction of digital modulation and even
more complex receivers; whether that's a Good Thing or not, I'm not sure.
Re: AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
Jim Kelley wrote:
> Hein ten Horn wrote:
>>
>> We hear the average of two frequencies if both frequencies
>> are indistinguishably close, say with a difference of some few
>> hertz. For example, the combination of a 220 Hz signal and
>> a 224 Hz signal with the same amplitude will be perceived as
>> a 4 Hz beat of a 222 Hz tone.
>
> I have also read this accounting, but from what I've been able to determine
> it lacks mathematical and phenomenological support. Here's why. As two
> audio frequencies are moved closer and closer together, there is no point
> where an average of the two frequencies can be perceived. There is however
> a point where no difference in the two frequencies is perceived. Obviously
> if we cannot discern the difference between 220Hz and 224Hz (as an example),
> we are not going to be able to discern half their difference either. I
> suspect the notion may have originated from a trigonometric identity which
> has what could be interpreted as an average term in it.
>
> sin(a) + sin(b) = 2sin(.5(a+b))cos(.5(a-b))
>
> A plot of the function reveals that cos(.5(a-b)) describes the envelope.
> The period of the 'enveloped' waveform (or the arcane, beat modulated
> waveform) then can be seen to vary continuously and repetitiously over
> time - from 1/a at one limit to 1/b at the other. At a particular instant in
> time the period does in fact equal the average of the two. But this is true
> only for an instant every 1/(a-b) seconds.
The math is perfectly describing what is happening in the
course of time at an arbitrary location in the air or in the
medium inside the cochlea. Concerning the varying
amplitude it does a good job.
But does someone (here) actually know how our hearing
system interprets both indistinguishable(!) frequencies (or
even a within a small range rapidly varying frequency) and
how the resulting 'signal' is translated into what we call the
perception? Evidently the math given above doesn't
reckon with any hearing mechanism at all. Hence it cannot
rule out perceiving an average frequency.
For the rest I don't get your point on a varying period.
From a mathematical point of view the function
sin( pi * (f_2 + f_1) * t )
has a constant frequency of (f_2 + f_1)/2
and a constant period of 2/(f_2 + f_1).
This frequency is indeed the arithmetical average and
it is not affected by a multiplication of the function by
a relatively slow varying amplitude.
> An interesting related experiment can be performed by setting a sweep
> generator to sweep over a narrow range of frequencies. The range can be
> adjusted as well as the sweep time. One can then study what sorts of
> effects are discernible.
>
> I have found that it is very difficult to fool the ear in some of the ways
> that have been suggested. It does not appear, for example, that the claim
> for 'perceiving the average' is valid for two arbitrarily close frequencies
> any more than it is for any two other frequencies. But I would appreciate
> learning of any contradictory research that you might be able to cite.
Apart from the mathematical support, I saw the average
frequency mentioned in several books on physics, unfortunately
without further enclosed proof (as far as I remember).
However, getting some empirical evidence should be a
rather easy piece of work.
Re: AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
"Hein ten Horn" <tenhornRemovE@ThiSraketnet.nl> wrote in message
news:4695b657$0$79139$e4fe514c@news.xs4all.nl...
| Jim Kelley wrote:
| > Hein ten Horn wrote:
| >>
| >> We hear the average of two frequencies if both frequencies
| >> are indistinguishably close, say with a difference of some few
| >> hertz. For example, the combination of a 220 Hz signal and
| >> a 224 Hz signal with the same amplitude will be perceived as
| >> a 4 Hz beat of a 222 Hz tone.
| >
| > I have also read this accounting, but from what I've been able to
determine
| > it lacks mathematical and phenomenological support. Here's why. As two
| > audio frequencies are moved closer and closer together, there is no
point
| > where an average of the two frequencies can be perceived. There is
however
| > a point where no difference in the two frequencies is perceived.
Obviously
| > if we cannot discern the difference between 220Hz and 224Hz (as an
example),
| > we are not going to be able to discern half their difference either. I
| > suspect the notion may have originated from a trigonometric identity
which
| > has what could be interpreted as an average term in it.
| >
| > sin(a) + sin(b) = 2sin(.5(a+b))cos(.5(a-b))
| >
| > A plot of the function reveals that cos(.5(a-b)) describes the envelope.
| > The period of the 'enveloped' waveform (or the arcane, beat modulated
| > waveform) then can be seen to vary continuously and repetitiously over
| > time - from 1/a at one limit to 1/b at the other. At a particular
instant in
| > time the period does in fact equal the average of the two. But this is
true
| > only for an instant every 1/(a-b) seconds.
|
| The math is perfectly describing what is happening in the
| course of time at an arbitrary location in the air or in the
| medium inside the cochlea. Concerning the varying
| amplitude it does a good job.
| But does someone (here) actually know how our hearing
| system interprets both indistinguishable(!) frequencies (or
| even a within a small range rapidly varying frequency) and
| how the resulting 'signal' is translated into what we call the
| perception? Evidently the math given above doesn't
| reckon with any hearing mechanism at all. Hence it cannot
| rule out perceiving an average frequency.
|
| For the rest I don't get your point on a varying period.
| From a mathematical point of view the function
|
| sin( pi * (f_2 + f_1) * t )
|
| has a constant frequency of (f_2 + f_1)/2
| and a constant period of 2/(f_2 + f_1).
| This frequency is indeed the arithmetical average and
| it is not affected by a multiplication of the function by
| a relatively slow varying amplitude.
|
| > An interesting related experiment can be performed by setting a sweep
| > generator to sweep over a narrow range of frequencies. The range can be
| > adjusted as well as the sweep time. One can then study what sorts of
| > effects are discernible.
| >
| > I have found that it is very difficult to fool the ear in some of the
ways
| > that have been suggested. It does not appear, for example, that the
claim
| > for 'perceiving the average' is valid for two arbitrarily close
frequencies
| > any more than it is for any two other frequencies. But I would
appreciate
| > learning of any contradictory research that you might be able to cite.
|
| Apart from the mathematical support, I saw the average
| frequency mentioned in several books on physics, unfortunately
| without further enclosed proof (as far as I remember).
| However, getting some empirical evidence should be a
| rather easy piece of work.
|
| gr, Hein
Actually the human ear can detect a beat note down to a few cycles.
Re: AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
"NotMe" <me@privacy.net> hath wroth:
(Please learn to trim quotations)
>Actually the human ear can detect a beat note down to a few cycles.
No, you cannot. Figure on 20Hz to 20KHz for human hearing:
<http://hypertextbook.com/facts/2003/ChrisDAmbrose.shtml>
What happens when you zero beat something is that your brain is
filling in the missing frequencies. As you tune across the frequency,
and the beat note goes down in frequency, most people overshoot to the
other side, and then compensate by splitting the different. As you
approach zero beat, your perception of the sound drops. If the lack
of hearing below 20Hz doesn't make it disappear, the frequency rolloff
in the audio amplifier stages will probably also drop off at about
20-300Hz depending on whether it's a hi-fi or communications radio. I
have a home made DC coupled hi-fi and can see the speaker moving in
and out slowly at very low frequencies. I don't hear a thing.
However, you don't have to hear it to detect infrasonic sounds.
<http://en.wikipedia.org/wiki/Infrasound>
Your inner ear, which is responsible for your sense of balance, can do
that for you. You don't actually hear the tone, but your body
certainly responds to it. Depending on frequency and level, tones
below about 20Hz will bring on confusion, nausia, disorientation, and
all manner of sensory anomalies. It's been used for effects in music,
sound tracks, and military weapon systems. I've experienced the
effects personally and can assure you that it was not pleasant.
Re: AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
On Wed, 11 Jul 2007 22:52:17 -0700, Jeff Liebermann wrote:
> "NotMe" <me@privacy.net> hath wroth:
>
> (Please learn to trim quotations)
>
>>Actually the human ear can detect a beat note down to a few cycles.
>
> No, you cannot. Figure on 20Hz to 20KHz for human hearing:
> <http://hypertextbook.com/facts/2003/ChrisDAmbrose.shtml>
>
> What happens when you zero beat something is that your brain is filling in
> the missing frequencies. As you tune across the frequency, and the beat
> note goes down in frequency, most people overshoot to the other side, and
> then compensate by splitting the different.
No, you've got it all wrong. The beat note happens because, when the
signals are close to 180 degrees out of phase, they cancel out such that
there is, in fact, no sound. This is what your ear detects. Now, if
you're zero-beating, say, 400 Hz against 401 Hz, I don't know if the
801 Hz component is audible or if it's even really there, but
mathematically, it kinda has to, doesn't it?
Re: AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
Rich Grise <rich@example.net> hath wroth:
>On Wed, 11 Jul 2007 22:52:17 -0700, Jeff Liebermann wrote:
>
>> "NotMe" <me@privacy.net> hath wroth:
>>
>> (Please learn to trim quotations)
>>
>>>Actually the human ear can detect a beat note down to a few cycles.
>>
>> No, you cannot. Figure on 20Hz to 20KHz for human hearing:
>> <http://hypertextbook.com/facts/2003/ChrisDAmbrose.shtml>
>>
>> What happens when you zero beat something is that your brain is filling in
>> the missing frequencies. As you tune across the frequency, and the beat
>> note goes down in frequency, most people overshoot to the other side, and
>> then compensate by splitting the different.
>No, you've got it all wrong.
Sorry, I'm perfect and never make misteaks.
>The beat note happens because, when the
>signals are close to 180 degrees out of phase, they cancel out such that
>there is, in fact, no sound. This is what your ear detects. Now, if
>you're zero-beating, say, 400 Hz against 401 Hz, I don't know if the
>801 Hz component is audible or if it's even really there, but
>mathematically, it kinda has to, doesn't it?
Ok, I'll bite. I think you'll find that if you actually do that with
a non-distorting audio mixer[1], and look at an oscilloscope, you'll
see the 1Hz envelope, but the 400 and 401 Hz tones will still be
there. Same on a spectrum analyzer, where the two carriers (400/401)
are still there. If they're there, you'll hear them. The tones may
be going up and down once per second (1Hz), but you'll still hear
tones in between. No way are they going to disappear with a 1Hz
separation. However, if they're exactly on the same frequency, and
exactly 180 degrees otto phase, they will cancel.
The zero beat example I offered is more a psychology problem than
acoustics or hearing. Our ears and brain expect the sweep through
zero beat to be continuous, that we fill in the missing frequencies.
It's really apparent in ham radio, where tuning across carriers is a
common event. I've watched how people do it, and noticed that they
always overshoot and come back to the perceived center. If you ask
them to nail the frequency to within 10Hz without overshooting, they
usually have a difficult time.
[1] no compression, limiting, fuzz box, reverb, equalizer, etc.
Re: AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
Hein ten Horn wrote:
>
> < quote >
> We hear the average of two frequencies if both frequencies
> are indistinguishably close, say with a difference of some few
> hertz. For example, the combination of a 220 Hz signal and
> a 224 Hz signal with the same amplitude will be perceived as
> a 4 Hz beat of a 222 Hz tone.
> < unquote >
> (..)
> From the example: there's no 222 Hz tone in the air.
That one I'd like to take back.
Obviously the superposition didn't cross my mind.
The matter is actually vibrating at the frequency
of 222 Hz. Not at 220 Hz or 224 Hz.
Re: AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
Ron Baker, Pluralitas! wrote:
> David L. Wilson wrote:
>> Hein ten Horn wrote:
>>> ...
>>> So take another example: 25000 Hz and 25006 Hz.
>>> Again, constructive and destructive interference produce 6 Hz
>>> amplitude variations in the air.
>>> But, as we can't hear ultrasonic frequencies, we will not produce
>>> a 25003 Hz perception in our brain. So there's nothing to hear,
>>> no tone and consequently, no beat.
>>
>> If one looks at an oscilloscope of the audio converted to voltage, one
>> still can see the 6Hz variations on the 25003 Hz and still refers to those
>> as tone and beat. These exist in mathematically formulation of the
>> resulting waveforms
Right.
>> not just as something in the brain.
In this particular example nothing is heard
because 25003 Hz is an ultrasonic frequency.
> What is the mathematical formulation?
sin(2 * pi * f_1 * t) + sin(2 * pi * f_2 * t)
or
2 * cos( pi * (f_1 - f_2) * t ) * sin( pi * (f_1 + f_2) * t )
So every cubic micrometre of the air (or another medium)
is vibrating in accordance with
2 * cos( 2 * pi * 3 * t ) * sin(2 * pi * 25003 * t ),
thus having a beat frequency of 2*3 = 6 Hz
and a vibration frequency of 25003 Hz
(let alone phase differences of neighbouring
vibrating elements).
Re: AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
On Jul 1, 11:11 am, Jeff Liebermann <je...@cruzio.com> wrote:
> John Smith I <assemblywiz...@gmail.com> hath wroth:
>
> >RHF wrote:
> > > ...
> >> Because "Radium" Touched Them With A Thirst
> >> For Knowledge And A Quest For Answers.
> >> ...
> >I don't know, according to any instructor I have ever had respect for:
> >"There are NO stupid questions, only stupid people who are afraid to ask
> >questions."
>
> I beg to differ. My favorite mentor/instructor/employer had a
> different philosophy regarding questions and answers. His line was
> something like "If you don't understand the problem, no solution is
> possible". His method was to concentrate on understanding the
> problem, refining the corresponding questions, and only then
> concentrating on finding the answer. I would spend much more time
> thinking about "what problem am I trying to solve" instead of
> blundering prematurely toward some potentially irrelevant solution.
>
> My problem with the original question is that it fails to associate
> itself with anything recognizable as a real problem to solve or a
> theory to expound. In my never humble opinion, if there was a
> question under all that rubbish, it was quite well hidden and severely
> muddled. He also introduced a substantial number of "facts" that
> varied from irrelevant to incoherent to just plain wrong. The problem
> for us in not in finding the answer, but in decoding the question.
>
> There may not be any stupid questions, but there seem to be a
> substantial number of marginal people asking questions. I answer some
> techy questions in alt.internet.wireless. What I see, all too often,
> are people that seem to think that no effort on their part is
> necessary to obtain an answer. They exert no effort to read the FAQ,
> no effort to supply what problem they are trying to solve, and no
> effort to supply what they have to work with. In this case, Mr Radium
> has either exerted no effort to compose his question in a form that
> can be answered, or if there was such an effort, it has failed
> miserably. He couldn't even find a suitable collection of newsgroups
> for his question.
>
> There may not be any stupid questions, but there certainly are
> questions not worth the time attempting to answer. If Mr Radium had
> left the question at the subject line:
> "AM electromagnetic waves: 20 KHz modulation frequency on
> an astronomically-low carrier frequency"
> the question would have been easy to answer, as several people have
> done. However, those that answered and I all did the same thing. We
> extracted from the word salad question what we thought was something
> resembling a coherent question, and ignored the rest of the rubbish.
> In other words, we did the necessary simplification and problem
> reduction, and discarded the bulk of the incoherent residue. There
> may not be any stupid questions, but if you bury it under a sufficient
> number of words, it may closely resemble a stupid question.
>
> >Depends ... I guess.
> >JS
>
> Well, let's see:
> <http://groups.google.com/groups?as_q=%22guess%28tm%29%22&as_uauthors=...>
> 533 guesses, out of about 16,000 postings, which I guess(tm) isn't all
> that bad.
>
> --
> Jeff Liebermann je...@cruzio.com
> 150 Felker St #D http://www.LearnByDestroying.com
> Santa Cruz CA 95060http://802.11junk.com
> Skype: JeffLiebermann AE6KS 831-336-2558
thankyou, for the wonderfully varied responses.
here's my question?
as one who simply asks the questions!
can fm waves ( any kind) PIGGY BACK ON AM WAVES?
THE IMPLICATIONS ARE FAR REACHING!!!!!!!!!
REMEBER AS OUR GREAT ANCESTORS SO ELOCENTLY PUT IT (PARAPHRASED)
WHEN CONSIDERING THE LIGHT BULB
" I FOUND 2000 NEW WAYS OF THINKING"
BUT THIS GREAT MAN DIDNT FINNISH UNTIL THE GOAL WAS REACHED
ADMIRABLE QUALITIES.
personally speaking, I have no formal or imformal education that can
match the depths of this scientific quorum.
I do have a vision
I hope to find new ways every day of connecting so i can get to the
beaches where the waves and the people think outside the box of normal
surfing and envision a whole new world metaphoricaly speaking.
Please learn to operate a text editor and kindly trim the surplus
quotes from your ranting. I can't stand to read my own stuff twice.
>can fm waves ( any kind) PIGGY BACK ON AM WAVES?
Sure. It's called QAM (quadrature amplitude muddlation). The
quadrature part is actually PM (phase modulation), which is a form of
FM (freak modulation). The amplitude also varies at the same time.
Look for the constellation diagrams. Which one is a pig on which back
is an open question.
>personally speaking, I have no formal or imformal education that can
>match the depths of this scientific quorum.
Yeah, it shows.
>I do have a vision
Your vision is not 20-20. I suggest corrective glasses.
>I hope to find new ways every day of connecting so i can get to the
>beaches where the waves and the people think outside the box of normal
>surfing and envision a whole new world metaphoricaly speaking.
Never mind thinking outside the box. Work on thinking in the first
place. Once you master that, you can worry whether it works better
inside or outside a box.
Re: AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-lowcarrier frequency
Hein ten Horn wrote:
> Hein ten Horn wrote:
>
>>< quote >
>>We hear the average of two frequencies if both frequencies
>>are indistinguishably close, say with a difference of some few
>>hertz. For example, the combination of a 220 Hz signal and
>>a 224 Hz signal with the same amplitude will be perceived as
>>a 4 Hz beat of a 222 Hz tone.
>>< unquote >
>>(..)
>
>
>>From the example: there's no 222 Hz tone in the air.
>
>
> That one I'd like to take back.
> Obviously the superposition didn't cross my mind.
> The matter is actually vibrating at the frequency
> of 222 Hz. Not at 220 Hz or 224 Hz.
>
> gr, Hein
You were correct before. It might be correct to say that matter is
vibrating at an average, or effective frequency of 222 Hz. But the
only sine waves present in the air are vibrating at 220 Hz and 224 Hz.
Obviously. It's a very simple matter to verify this by experiment.
You really ought to perform it (as I just did) before posting
further on the subject.
Re: AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
craigm wrote:
> Jim Kelley wrote:
>> David L. Wilson wrote:
>>> Jim Kelley wrote:
>>>>
>>>> At a particular instant in time the period does in fact equal the average
>>>> of the two. But this is true only for an instant every 1/(a-b) seconds.
>>>
>>> How do you come up with anything but a period of of the average of the
>>> two for the enveloped waveform?
>>
>> The error here is in assuming that the sin and cos terms in the
>> equivalent expression are representative of individual waves. They
>> are not. The resultant wave can only be accurately described as the
>> sum of the constituent waves sin(a) and sin(b), or as the function
>> 2sin(.5(a+b))cos(.5(a-b)). That function, plotted against time
>> appears exactly as I have described. I have simply reported what is
>> readily observable.
>
> I would submit you plotted it wrong and/or misinterpreted the results.
Jim, if you'd like me to send you an Excel sheet about this,
please let me know.
gr, Hein
I've sent this post already once. For some strange reason it didn't
come up in rec.radio.shortwave (craigm?).
I only read rec.radio.shortwave these days.
(repost to: sci.electronics.basics, rec.radio.shortwave,
rec.radio.amateur.antenna, alt.cellular.cingular,
alt.internet.wireless)
Re: AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
"Rich Grise" <rich@example.net> wrote in message
newsan.2007.07.12.17.28.11.148983@example.net...
> On Wed, 11 Jul 2007 22:52:17 -0700, Jeff Liebermann wrote:
>
>> "NotMe" <me@privacy.net> hath wroth:
>>
>> (Please learn to trim quotations)
>>
>>>Actually the human ear can detect a beat note down to a few cycles.
If you are talking about the beat between two close
audio frequencies then one can easily hear a beat way
below 1 Hz.
>>
>> No, you cannot. Figure on 20Hz to 20KHz for human hearing:
>> <http://hypertextbook.com/facts/2003/ChrisDAmbrose.shtml>
>>
>> What happens when you zero beat something is that your brain is filling
>> in
>> the missing frequencies. As you tune across the frequency, and the beat
>> note goes down in frequency, most people overshoot to the other side, and
>> then compensate by splitting the different.
If you are talking about beat frequency heard when
tuning to a carrier with a radio with a BFO or in SSB mode
then one can't hear any beat below 50 Hz or so.
The audio section of the receiver blocks anything
below about 50 Hz.
>
> No, you've got it all wrong. The beat note happens because, when the
> signals are close to 180 degrees out of phase, they cancel out such that
> there is, in fact, no sound. This is what your ear detects. Now, if
> you're zero-beating, say, 400 Hz against 401 Hz, I don't know if the
> 801 Hz component is audible or if it's even really there, but
> mathematically, it kinda has to, doesn't it?
Are you talking radios or guitars?
With a guitar you might beat 400 Hz against 401 Hz.
With a radio you'd more likely beat 455 kHz against
455.001 kHz.
Re: AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
"Ron Baker, Pluralitas!" <this@aint.me> wrote in message
news:46984f9d$0$8004$4c368faf@roadrunner.com...
|
| "Rich Grise" <rich@example.net> wrote in message
| newsan.2007.07.12.17.28.11.148983@example.net...
| > On Wed, 11 Jul 2007 22:52:17 -0700, Jeff Liebermann wrote:
| >
| >> "NotMe" <me@privacy.net> hath wroth:
| >>
| >> (Please learn to trim quotations)
| >>
| >>>Actually the human ear can detect a beat note down to a few cycles.
|
| If you are talking about the beat between two close
| audio frequencies then one can easily hear a beat way
| below 1 Hz.
Based on studies done at Tulane Department of Neurology (mid 60's) the
detection is not in the ear but in the brain. The process can be taught and
refined though bio-feedback.
Re: AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
"Ron Baker, Pluralitas!" <this@aint.me> wrote in message
news:46984f9d$0$8004$4c368faf@roadrunner.com...
|
| "Rich Grise" <rich@example.net> wrote in message
| newsan.2007.07.12.17.28.11.148983@example.net...
| > On Wed, 11 Jul 2007 22:52:17 -0700, Jeff Liebermann wrote:
| >
| >> "NotMe" <me@privacy.net> hath wroth:
| >>
| >> (Please learn to trim quotations)
| >>
| >>>Actually the human ear can detect a beat note down to a few cycles.
|
| If you are talking about the beat between two close
| audio frequencies then one can easily hear a beat way
| below 1 Hz.
Based on studies done at Tulane Department of Neurology (mid 60's) the
detection is not in the ear but in the brain. The process can be taught and
refined though bio-feedback.
Re: AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
"Hein ten Horn" <tenhornRemovE@ThiSraketnet.nl> wrote in message
news:46976788$1$331$e4fe514c@news.xs4all.nl...
> Ron Baker, Pluralitas! wrote:
>> David L. Wilson wrote:
>>> Hein ten Horn wrote:
>>>> ...
>>>> So take another example: 25000 Hz and 25006 Hz.
>>>> Again, constructive and destructive interference produce 6 Hz
>>>> amplitude variations in the air.
>>>> But, as we can't hear ultrasonic frequencies, we will not produce
>>>> a 25003 Hz perception in our brain. So there's nothing to hear,
>>>> no tone and consequently, no beat.
>>>
>>> If one looks at an oscilloscope of the audio converted to voltage, one
>>> still can see the 6Hz variations on the 25003 Hz and still refers to
>>> those
>>> as tone and beat. These exist in mathematically formulation of the
>>> resulting waveforms
>
> Right.
>
>>> not just as something in the brain.
>
> In this particular example nothing is heard
> because 25003 Hz is an ultrasonic frequency.
>
>
>> What is the mathematical formulation?
>
> sin(2 * pi * f_1 * t) + sin(2 * pi * f_2 * t)
> or
> 2 * cos( pi * (f_1 - f_2) * t ) * sin( pi * (f_1 + f_2) * t )
>
> So every cubic micrometre of the air (or another medium)
> is vibrating in accordance with
> 2 * cos( 2 * pi * 3 * t ) * sin(2 * pi * 25003 * t ),
> thus having a beat frequency of 2*3 = 6 Hz
How do you arrive at a "beat"?
Hint: Any such assessment is nonlinear.
(And kudos to you that you can do the math.)
Simplifying the math:
x = cos(a) * cos(b) = 0.5 * (cos[a+b] + cos[a-b])
(Where a = 2 * pi * f_1 * t and b = same but f_2.)
All three of the above are equivalent. There is no difference.
You get x if you add two sine waves or if you
multiply two (different) sine waves.
So which is it really? Hint: If all you have is x then
you can't tell how it was generated.
What you do with it afterwards can make a
difference.
> and a vibration frequency of 25003 Hz
> (let alone phase differences of neighbouring
> vibrating elements).
>
> gr, Hein
>
>
Re: AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
> "Ron Baker, Pluralitas!" <this@aint.me> wrote in message
> news:46984f9d$0$8004$4c368faf@roadrunner.com...
> |
> | "Rich Grise" <rich@example.net> wrote in message
> | newsan.2007.07.12.17.28.11.148983@example.net...
> | > On Wed, 11 Jul 2007 22:52:17 -0700, Jeff Liebermann wrote:
> | >
> | >> "NotMe" <me@privacy.net> hath wroth:
> | >>
> | >> (Please learn to trim quotations)
> | >>
> | >>>Actually the human ear can detect a beat note down to a few cycles.
> |
> | If you are talking about the beat between two close
> | audio frequencies then one can easily hear a beat way
> | below 1 Hz.
But what you hear below ~20 Hz is not the beat note, but changes in sound
pressure (volume) as the mixing product goes in and out of phase. This
actually becomes easier to hear as you near zero beat.
Re: AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
Ron Baker, Pluralitas! wrote:
> Hein ten Horn wrote:
>> Ron Baker, Pluralitas! wrote:
>>> David L. Wilson wrote:
>>>> Hein ten Horn wrote:
>>>>>
>>>>> So take another example: 25000 Hz and 25006 Hz.
>>>>> Again, constructive and destructive interference produce 6 Hz
>>>>> amplitude variations in the air.
>>>>> But, as we can't hear ultrasonic frequencies, we will not produce
>>>>> a 25003 Hz perception in our brain. So there's nothing to hear,
>>>>> no tone and consequently, no beat.
>>> What is the mathematical formulation?
>>
>> sin(2 * pi * f_1 * t) + sin(2 * pi * f_2 * t)
>> or
>> 2 * cos( pi * (f_1 - f_2) * t ) * sin( pi * (f_1 + f_2) * t )
>>
>> So every cubic micrometre of the air (or another medium)
>> is vibrating in accordance with
>> 2 * cos( 2 * pi * 3 * t ) * sin(2 * pi * 25003 * t ),
>> thus having a beat frequency of 2*3 = 6 Hz
>
> How do you arrive at a "beat"?
Not by train, neither by UFO.
Sorry. English, German and French are only 'second'
languages to me.
Are you after the occurrence of a beat?
Then: a beat appears at constructive interference, thus
when the cosine function becomes maximal (+1 or -1).
Or are you after the beat frequency (6 Hz)?
Then: the cosine function has two maxima per period
(one being positive, one negative) and with three
periodes a second it makes six beats/second.
> Hint: Any such assessment is nonlinear.
> (And kudos to you that you can do the math.)
>
> Simplifying the math:
> x = cos(a) * cos(b) = 0.5 * (cos[a+b] + cos[a-b])
> (Where a = 2 * pi * f_1 * t and b = same but f_2.)
> All three of the above are equivalent. There is no difference.
> You get x if you add two sine waves or if you
> multiply two (different) sine waves.
??
sin(a) + sin(b) <> sin(a) * sin(b)
In case you mistyped cosine:
sin(a) + sin(b) <> cos(a) * cos(b)
> So which is it really? Hint: If all you have is x then
> you can't tell how it was generated.
>
> What you do with it afterwards can make a
> difference.
Referring to the physical system, what's now your point?
That system contains elements vibrating at 25003 Hz.
There's no math in it. More on that in my posting to
JK at nearly the same sending time.
Re: AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
Jim Kelley wrote:
> Hein ten Horn wrote:
>> Hein ten Horn wrote:
>>>
>>> < quote >
>>> We hear the average of two frequencies if both frequencies
>>> are indistinguishably close, say with a difference of some few
>>> hertz. For example, the combination of a 220 Hz signal and
>>> a 224 Hz signal with the same amplitude will be perceived as
>>> a 4 Hz beat of a 222 Hz tone.
>>> < unquote >
>>
>>>From the example: there's no 222 Hz tone in the air.
>>
>> That one I'd like to take back.
>> Obviously the superposition didn't cross my mind.
>> The matter is actually vibrating at the frequency
>> of 222 Hz. Not at 220 Hz or 224 Hz.
>
> You were correct before.
That's a misunderstanding.
A vibrating element here (such as a cubic micrometre
of matter) experiences different changing forces. Yet
the element cannot follow all of them at the same time.
As a matter of fact the resulting force (the resultant) is
fully determining the change of the velocity (vector) of
the element.
The resulting force on our element is changing at the
frequency of 222 Hz, so the matter is vibrating at the
one and only 222 Hz.
> It might be correct to say that matter is vibrating at an
> average, or effective frequency of 222 Hz.
No, it is correct. A particle cannot follow two different
harmonic oscillations (220 Hz and 224 Hz) at the same
time.
> But the only sine waves present in the air are vibrating
> at 220 Hz and 224 Hz.
If so, we have a very interesting question...
What is waving here? A vacuum?
But don't take the trouble to answer.
You'd better distinguish the behaviour of nature and the
way we try to understand and describe all things.
As long as both sound sources are vibrating there are
no sine waves (220 Hz, 224 Hz) present, yet we do
use them to find the frequency of 222 Hz (and the
displacement of a vibrating element at a particular
location in space on a particular point in time).
> Obviously. It's a very simple matter to verify this by experiment.
Indeed, it is. But watch out for misinterpretations of
the measuring results! For example, if a spectrum
analyzer, being fed with the 222 Hz signal, shows
that the signal can be composed from a 220 Hz and
a 224 Hz signal, then that won't mean the matter is
actually vibrating at those frequencies.
> You really ought to perform it (as I just did) before
> posting further on the subject.
I did happen to see interference of waterwaves
including some beautiful (changing) hyperbolic structures,
but no sign of any sine wave at all. So, with your kind
permission, here's my posting. ;-)
Re: AM electromagnetic waves: 20 KHz modulation frequencyonanastronomically-low carrier frequency
On Thu, 05 Jul 2007 07:32:20 -0700, Keith Dysart
<Keith.Dysart@gmail.com> wrote:
>On Jul 5, 10:01 am, John Fields <jfie...@austininstruments.com> wrote:
>> ---
>> The first example is amplitude modulation precisely _because_ of the
>> multiplication, while the second is merely the algebraic summation
>> of the instantaneous amplitudes of two waveforms.
>>
>> The circuit lists I posted earlier will, when run using LTSPICE,
>> show exactly what the signals will look like using an oscilloscope
>> and, using the "FFT" option on the "VIEW" menu, give you a pretty
>> good approximation of what they'll look like using a spectrum
>> analyzer.
>>
>> If you don't have LTSPICE it's available free at:
>>
>> http://www.linear.com/designtools/software/
>>
>> --
>> JF
>
>Since your modulator version has a DC offset applied to
>the 1e5 signal, some of the 1e6 signal is present in the
>output, so your spectrum has components at .9e6, 1e6 and
>1.1e6.
---
Yes, of course, and 1e5 as well. That offset will make sure that
the output of the modulator contains both of the original signals as
well as their sums and differences. That is, it'll be a classic
mixer.
---
>To generate the same signal with the summing version you
>need to add in some 1e6 along with the .9e6 and 1.1e6.
---
That wouldn't be the same signal since .9e6 and 1.1e6 wouldn't have
been created by heterodyning and wouldn't be sidebands at all.
---
>The results will be identical and the results of summing
>will be quite detectable using an envelope detector just
>as they would be from the modulator version.
---
The results would certainly _not_ be identical, since the 0.9e6 and
1.1e6 signals would bear no cause-and-effect relationship to the 1e6
and 1e5 signals, not having been spawned by them in a mixer.
Moreover, using an envelope detector would be pointless since there
would be no information in the .9e6 and 1.1e6 signals which would
relate to either the 1e6 or the 0.1e6 signals. Again, because no
mixing would have occurred in your scheme, only a vector addition.
---
>Alternatively, remove the bias from the .1e6 signal on
>the modulator version. The spectrum will have only
>components at .9e6 and 1.1e6. Of course, an envelope
>detector will not be able to recover this signal,
>whether generated by the modulator or summing.
---
Hogwash.
If the envelope detector you're talking about is a rectifier
followed by a low-pass filter and neither f1 nor f2 were DC offset,
then if the sidebands were created in a modulator they'll largely
cancel, (except for the interesting fact that the diode rectifier
looks like a small capacitor when it's reverse biased) so you're
almost correct on that count.
However, If f1 and f2 were created by independent oscillators and
algebraically added in a linear system, the output of the envelope
detector would be the vector sum of f1 and f2 either above or below
zero volts, depending on how the diode was wired.
Re: AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
On Fri, 06 Jul 2007 19:04:00 -0000, Jim Kelley <jwkelley@uci.edu>
wrote:
>On Jul 5, 9:38 pm, John Fields <jfie...@austininstruments.com> wrote:
>> On Thu, 05 Jul 2007 18:37:21 -0700, Jim Kelley <jwkel...@uci.edu>
>> wrote:
>>
>> >John Fields wrote:
>>
>> >> You missed my point, which was that in a mixer (which the ear is,
>> >> since its amplitude response is nonlinear) as the two carriers
>> >> approach each other the difference frequency will go to zero and the
>> >> sum frequency will go to the second harmonic of either carrier,
>> >> making it largely appear to vanish into the fundamental.
>>
>> >Hi John -
>>
>> >Given two sources of pure sinusoidal tones whose individual amplitudes
>> >are constant, is it your claim that you have heard the sum of the two
>> >frequencies?
>>
>> ---
>> I think so.
>
>So if you have for example, a 300 Hz signal and a 400 Hz signal, your
>claim is that you also hear a 700 Hz signal? You'd better check
>again. All you should hear is a 300 Hz signal and a 400 Hz signal.
>The beat frequency is too high to be audible.
---
Well, I'm just back from the Panama Canal Society's 75th reunion and
I haven't read through the rest of the thread, but it case someone
else hasn't already pointed it out to you, it seems you've missed
the point that a non-linear detector, (the human ear, for example)
when presented with two sinusoidal carriers, will pass the two
carrier frequencies through, as outputs, as well as two frequencies
(sidebands) which are the sum and difference of the carriers.
In your example, with 300Hz and 400Hz as the carriers, the sidebands
would be located at:
f3 = f1 + f2 = 300Hz + 400Hz = 700Hz
and
f4 = f2 - f1 = 400Hz - 300Hz = 100Hz
both of which are clearly within the range of frequencies to which
the human ear responds.
---
>(Note that if the beat
>frequency was a separate, difference signal as you suggest, at this
>frequency it would certainly be audible.)
---
Your use of the term "beat frequency" is confusing since it's
usually used to describe the products of heterodyning, not the
audible warble caused by the vector addition of signals close to
unison.
---
>> A year or so ago I did some casual experiments with pure tones being
>> fed simultaneously into individual loudspeakers to which I listened,
>> and I recall that I heard tones which were higher pitched than
>> either of the lower-frequency signals. Subjective, I know, but
>> still...
>
>Excessive cone excursion can produce significant 2nd harmonic
>distortion. But at normal volume levels your ear does not create
>sidebands, mixing products, or anything of the sort. It hears the
>same thing that is shown on both the oscilloscope and on the spectrum
>analyzer.
---
No, it doesn't.
Since the response of the ear is non-linear in amplitude it has no
choice _but_ to be a mixer and create sidebands.
What you see on an oscilloscope are the time-varying amplitude
variations caused by the linear vector summation of two signals
walking through each other in time, and what you see on a spectrum
analyzer is the two spectral lines caused by two signals adding, not
mixing. If you want to see what happens when the two signals hit
the ear, run them through a non-linear amp before they get to the
spectrum analyzer and you'll see at least the two original signals
plus their two sidebands.
---
>> Interestingly, this afternoon I did the zero-beat thing with 1kHz
>> being fed to one loudspeaker and a variable frequency oscillator
>> being fed to a separate loudspeaker, with me as the detector.
>
>My comments were based on my results in that experiment, common
>knowledge, and professional musical and audio experience.
---
Your "common knowledge" seems to not include the fact that a
non-linear detector _is_ a mixer.
---
>> I also connected each oscillator to one channel of a Tektronix
>> 2215A, inverted channel B, set the vertical amps to "ADD", and
>> adjusted the frequency of the VFO for near zero beat as shown on the
>> scope.
>>
>> Sure enough, I heard the beat even though it came from different
>> sources, but I couldn't quite get it down to DC even with the
>> scope's trace at 0V.
>
>Of course you heard beats. What you didn't hear is the sum of the
>frequencies. I've had the same setup on my bench for several months.
>It's also one of the experiments the students do in the first year
>physics labs. Someone had made the claim a while back that what we
>hear is the 'average' of the two frequencies. Didn't make any sense
>so I did the experiment. The results are as I have explained.
---
The "beat" heard wasn't an actual beat frequency, it was the warble
caused by the change in amplitude of the summed signals and isn't a
real, spectrally definable signal.
The reason you didn't hear the real difference frequency is because
it was below the range of audible frequencies and the reason you
didn't hear the sum frequency is because it was close enough to the
second harmonic of the output of either oscillator (with the
oscillators close to unison) that you couldn't discern it from the
fundamental(s).
There also seems to be a reticence, on your part, to believe that
the ear is, in fact, a mixer and, consequently, you hear what you
want to.
But...
In order to bring this fol-de-rol to an end,I propose an experiment
to determine whether the ear does or does not create sidebands:
1. Set OSC 1 and OSC 2 to two harmonically unrelated frequencies
such that their frequencies and the sum and difference of their
frequencies lie within the ear's audible range of frequencies.
2. Slowly tune OSC 3 so that its output crosses the sum and
difference frequencies of OSC 1 and OSC 2.
If a warble is heard in the vicinity of either frequency, the ear is
creating sidebands.
I'll do the experiment sometime today, if I get a chance, and post
my results here. Since you're all set up you may want to do the
same thing.
Re: AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
On Fri, 13 Jul 2007 09:53:21 -0700, Jeff Liebermann
<jeffl@cruzio.com> wrote:
>shawn.cormican@gmail.com hath wroth:
>
>>thankyou, for the wonderfully varied responses.
>
>Please learn to operate a text editor and kindly trim the surplus
>quotes from your ranting. I can't stand to read my own stuff twice.
---
I can barely stomach it the _first_ time around!
>
>> "Ron Baker, Pluralitas!" <this@aint.me> wrote in message
>> news:46984f9d$0$8004$4c368faf@roadrunner.com...
>> |
>> | "Rich Grise" <rich@example.net> wrote in message
>> | newsan.2007.07.12.17.28.11.148983@example.net...
>> | > On Wed, 11 Jul 2007 22:52:17 -0700, Jeff Liebermann wrote:
>> | >
>> | >> "NotMe" <me@privacy.net> hath wroth:
>> | >>
>> | >> (Please learn to trim quotations)
>> | >>
>> | >>>Actually the human ear can detect a beat note down to a few cycles.
>> |
>> | If you are talking about the beat between two close
>> | audio frequencies then one can easily hear a beat way
>> | below 1 Hz.
>
>But what you hear below ~20 Hz is not the beat note, but changes in sound
>pressure (volume) as the mixing product goes in and out of phase. This
>actually becomes easier to hear as you near zero beat.
---
It's not a mixing product, it's a sum. Actually, the vector
_addition_ of two signals varying in phase.
Re: AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
"Hein ten Horn" <tenhornRemovE@ThiSraketnet.nl> wrote in message
news:4698a055$0$320$e4fe514c@news.xs4all.nl...
> Ron Baker, Pluralitas! wrote:
>> Hein ten Horn wrote:
>>> Ron Baker, Pluralitas! wrote:
>>>> David L. Wilson wrote:
>>>>> Hein ten Horn wrote:
>>>>>>
>>>>>> So take another example: 25000 Hz and 25006 Hz.
>>>>>> Again, constructive and destructive interference produce 6 Hz
>>>>>> amplitude variations in the air.
>>>>>> But, as we can't hear ultrasonic frequencies, we will not produce
>>>>>> a 25003 Hz perception in our brain. So there's nothing to hear,
>>>>>> no tone and consequently, no beat.
>
>>>> What is the mathematical formulation?
>>>
>>> sin(2 * pi * f_1 * t) + sin(2 * pi * f_2 * t)
>>> or
>>> 2 * cos( pi * (f_1 - f_2) * t ) * sin( pi * (f_1 + f_2) * t )
>>>
>>> So every cubic micrometre of the air (or another medium)
>>> is vibrating in accordance with
>>> 2 * cos( 2 * pi * 3 * t ) * sin(2 * pi * 25003 * t ),
>>> thus having a beat frequency of 2*3 = 6 Hz
>>
>> How do you arrive at a "beat"?
>
> Not by train, neither by UFO.
> Sorry. English, German and French are only 'second'
> languages to me.
> Are you after the occurrence of a beat?
Another way to phrase the question would have been:
Given a waveform x(t) representing the sound wave
in the air how do you decide whether there is a
beat in it?
> Then: a beat appears at constructive interference, thus
> when the cosine function becomes maximal (+1 or -1).
> Or are you after the beat frequency (6 Hz)?
> Then: the cosine function has two maxima per period
> (one being positive, one negative) and with three
> periodes a second it makes six beats/second.
>
>> Hint: Any such assessment is nonlinear.
>> (And kudos to you that you can do the math.)
>>
>> Simplifying the math:
>> x = cos(a) * cos(b) = 0.5 * (cos[a+b] + cos[a-b])
>> (Where a = 2 * pi * f_1 * t and b = same but f_2.)
>> All three of the above are equivalent. There is no difference.
>> You get x if you add two sine waves or if you
>> multiply two (different) sine waves.
>
> ??
> sin(a) + sin(b) <> sin(a) * sin(b)
> In case you mistyped cosine:
> sin(a) + sin(b) <> cos(a) * cos(b)
It would have been more proper of me to say
"sinusoid" rather than "sine wave". I called
cos() a "sine wave". If you look at cos(2pi f1 t)
on an oscilloscope it looks the same as sin(2pi f t).
In that case there is essentially no difference.
Yes, there are cases where it makes a difference.
But at the beginning of an analysis it is rather
arbitrary and the math is less cluttered with
cos().
>
>> So which is it really? Hint: If all you have is x then
>> you can't tell how it was generated.
>>
>> What you do with it afterwards can make a
>> difference.
>
> Referring to the physical system, what's now your point?
> That system contains elements vibrating at 25003 Hz.
> There's no math in it.
Whoops. You'll need math to understand it.
> More on that in my posting to
> JK at nearly the same sending time.
>
> gr, Hein
>
>
Re: AM electromagnetic waves: 20 KHz modulation frequencyonanastronomically-low carrier frequency 
an.2007.07.12.17.28.11.148983@example.net...