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AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency

J

Jim Kelley

craigm said:
Jim Kelley wrote:




Much of the discussion in the thread was on two closely spaced frequencies.
Try your evaluation with closely spaced frequencies.

You seem to be laboring under the impression that, like you, this
subject matter is new to me. It is difficult to resolve the changes
in period when the frequencies are close. Could even lead one to
believe that it doesn't change and post claims accordingly on a
newsgroup.

jk
 
H

Hein ten Horn

craigm said:
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)
 
S

stan

John said:
---
That's not true. The original allegation was mine, and was that
since the ear is a device with an "output" which doesn't change
linearly with linearly changing input amplitudes, it's a non-linear
device, is incapable of _not_ producing harmonics and heterodynes
and, as such, is where the mixing occurs.

My contention was that zero-beat was the difference frequency
between two input tones close to unison, and I still maintain that's
true and that that difference frequency is in there. However, your
contention that zero-beat is the result of the vector summation of
two tones close to unison is also valid, since a non-linear detector
is capable of doing that summation well enough to allow that be the
dominant phenomenon as evidenced by the fact that the ear is
incapable of directly detecting (say) a 1Hz tone but is fully
capable of hearing the 1Hz amplitude warble which would result from
the vector addition of the two tones.

Beats are actually a well known phenomenon in physics and actually they
require no mixing at all. In fact all that is required is a very linear
operation of addition. A quick check of google turns up many
explanations of how two aubible frequencies can interfere in a very
linear fashion and result in beats. In fact beats don't require a human
head involvement in any way. Wikipedia has a pretty good description
with a trig identity for the math hungry.

As the identity shows, when two sin waves are added (linearly) the
result is equivalent to the average of the two tones with an amplitude
that varies at the frequency of the difference between the two tones.

About the amplitude warble, this can easily be demonstrated by anyone
with an actual volume knob ( surprisingly rare these days ). It's
possible to vary the volume at frequencies well under 20hz and hearing
the changes even with a human ear that can't hear a tone at frequencies
under 20hz.
 
R

Ron Baker, Pluralitas!

If you are talking about the beat between two close
audio frequencies then one can easily hear a beat way
below 1 Hz.

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.
 
N

NotMe

|
| | > On Wed, 11 Jul 2007 22:52:17 -0700, Jeff Liebermann wrote:
| >
| >> "NotMe" <[email protected]> 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.
 
R

Ron Baker, Pluralitas!

Hein ten Horn said:
In this particular example nothing is heard
because 25003 Hz is an ultrasonic frequency.



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.
 
B

Brenda Ann

|
| | > On Wed, 11 Jul 2007 22:52:17 -0700, Jeff Liebermann wrote:
| >
| >> "NotMe" <[email protected]> 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.
 
H

Hein ten Horn

Ron said:
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) said:
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.

gr, Hein
 
H

Hein ten Horn

Jim said:
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. ;-)

gr, Hein
 
J

John Fields

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.
 
J

John Fields

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.
---
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.
---
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.
---
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:

+-------+ +--------+
| OSC 1 |----| SPKR 1 |---/AIR/---> TO EAR
+-------+ +--------+

+-------+ +--------+
| OSC 2 |----| SPKR 2 |---/AIR/---> TO EAR
+-------+ +--------+

+-------+ +--------+
| OSC 3 |----| SPKR 3 |---/AIR/---> TO EAR
+-------+ +--------+

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.
 
J

John Fields

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.
 
R

Ron Baker, Pluralitas!

Hein ten Horn said:
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().
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.
 
C

craigm

Jim Kelley said:
You seem to be laboring under the impression that, like you, this subject
matter is new to me. It is difficult to resolve the changes in period
when the frequencies are close. Could even lead one to believe that it
doesn't change and post claims accordingly on a newsgroup.

jk


It is difficult to resolve changes that don't exist. Your generalization on
changing frequency/period isn't correct.

Here is a perl script to look at

#!/usr/bin/perl
use strict;
###############################################################################
&main;
###############################################################################
sub main
{
my $f1 = 46;
my $f2 = 55;

my $nSteps = 10000000;
my $stepSize = 0.000001;

my $i;
my $ta;
my $tb;
my $prevZero = 0;
my $diffSteps;

my $sumValue;
my $prevSum;
my $zeroFreq;

for ($i=0; $i < $nSteps; $i++)
{
$ta = $i * $stepSize * $f1 * 2 * 3.1415926535;
$tb = $i * $stepSize * $f2 * 2 * 3.1415926535;

$sumValue = sin($ta) + sin($tb);
if (($sumValue*$prevSum)<0)
{
$diffSteps = $i - $prevZero;
$zeroFreq = 0.5/($diffSteps*$stepSize);
printf "zero found at step %8d, steps from prev zero %6d, inferred freq
%8.1f\n",$i,$diffSteps,$zeroFreq;
$prevZero = $i;
}
$prevSum = $sumValue;
# print "$i $sumValue \n";
}
}


Some of the output:

zero found at step 9901, steps from prev zero 9901, inferred freq
50.5
zero found at step 19802, steps from prev zero 9901, inferred freq
50.5
zero found at step 29703, steps from prev zero 9901, inferred freq
50.5
zero found at step 39604, steps from prev zero 9901, inferred freq
50.5
zero found at step 49505, steps from prev zero 9901, inferred freq
50.5
zero found at step 55556, steps from prev zero 6051, inferred freq
82.6
zero found at step 59406, steps from prev zero 3850, inferred freq
129.9
zero found at step 69307, steps from prev zero 9901, inferred freq
50.5
zero found at step 79208, steps from prev zero 9901, inferred freq
50.5
zero found at step 89109, steps from prev zero 9901, inferred freq
50.5
zero found at step 99010, steps from prev zero 9901, inferred freq
50.5
zero found at step 108911, steps from prev zero 9901, inferred freq
50.5
zero found at step 118812, steps from prev zero 9901, inferred freq
50.5
zero found at step 128713, steps from prev zero 9901, inferred freq
50.5
zero found at step 138614, steps from prev zero 9901, inferred freq
50.5
zero found at step 148515, steps from prev zero 9901, inferred freq
50.5
zero found at step 158416, steps from prev zero 9901, inferred freq
50.5
zero found at step 166667, steps from prev zero 8251, inferred freq
60.6
zero found at step 168317, steps from prev zero 1650, inferred freq
303.0
zero found at step 178218, steps from prev zero 9901, inferred freq
50.5
zero found at step 188119, steps from prev zero 9901, inferred freq
50.5
zero found at step 198020, steps from prev zero 9901, inferred freq
50.5
zero found at step 207921, steps from prev zero 9901, inferred freq
50.5
zero found at step 217822, steps from prev zero 9901, inferred freq
50.5
zero found at step 227723, steps from prev zero 9901, inferred freq
50.5
zero found at step 237624, steps from prev zero 9901, inferred freq
50.5
zero found at step 247525, steps from prev zero 9901, inferred freq
50.5
zero found at step 257426, steps from prev zero 9901, inferred freq
50.5
zero found at step 267327, steps from prev zero 9901, inferred freq
50.5
zero found at step 277228, steps from prev zero 9901, inferred freq
50.5
zero found at step 277778, steps from prev zero 550, inferred freq
909.1
zero found at step 287129, steps from prev zero 9351, inferred freq
53.5
zero found at step 297030, steps from prev zero 9901, inferred freq
50.5
zero found at step 306931, steps from prev zero 9901, inferred freq
50.5
zero found at step 316832, steps from prev zero 9901, inferred freq
50.5
zero found at step 326733, steps from prev zero 9901, inferred freq
50.5
zero found at step 336634, steps from prev zero 9901, inferred freq
50.5
zero found at step 346535, steps from prev zero 9901, inferred freq
50.5
zero found at step 356436, steps from prev zero 9901, inferred freq
50.5
zero found at step 366337, steps from prev zero 9901, inferred freq
50.5
zero found at step 376238, steps from prev zero 9901, inferred freq
50.5
zero found at step 386139, steps from prev zero 9901, inferred freq
50.5
The input frequencies were 46 and 55 Hz, Note that the inferred frequency
from the zero crossings is the average of the two and is very consistent.
The indications other than 50.5 are form the zeo crossings added by the
difference frequency components.

You can try other frequencies, but the zero crossings are defined by the sin
and cos terms in the trig identity that has been previously posted.
 
J

Jeff Liebermann

I can barely stomach it the _first_ time around! ;)

My rants have been accused of being all manner of things. However,
this is the first time they've been accused of being indigestible. If
your stomach can't take it, and you feel the need to regurgitate an
apparently involuntary one line response, I can offer a suitable
therapeutic regime. If you read my writings, rants, stories, humor,
and poetry in much smaller portions, you will eventually find my stuff
more agreeable. Given sufficient time, you will develop a tolerance
for my stuff. Incidentally, my writings are rather dry and might
require a few grains of salt to be considered tasteful.

If only they would pay my time,
to write this stuff in verse and rhyme.
 
R

Ron Baker, Pluralitas!

Hein ten Horn said:
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.

It does. Not identically but it does follow all of them.
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.

Your idea of frequency is informal and leaves out
essential aspects of how physical systems work.

You have looked at a segment of the waveform
and judged "frequency" based on a few peaks.
Your method is incomplete and cannot be applied
generally.

<snip>
 
R

Ron Baker, Pluralitas!

Brenda Ann said:
But what you hear below ~20 Hz is not the beat note, but changes in sound

Semantics.
The "beat" heard in tuning a guitar is commonly
referred to as a "beat".
Yes, it is not the same thing as the "beat" from
a BFO in a radio receiver.
 
H

Hein ten Horn

Ron said:
It does. Not identically but it does follow all of them.

Impossible. Remember, we're talking about sound. Mechanical
forces only. Suppose you're driving, just going round the corner.
From the outside a fistful of forces is working on your body,
downwards, upwards, sidewards. It is absolutely impossible
that your body's centre of gravity is following different forces in
different directions at the same time. Only the resulting force is
changing your movement (according to Newton's second law).
Your idea of frequency is informal and leaves out
essential aspects of how physical systems work.

Nonsense. Mechanical oscillations are fully determined by
forces acting on the vibrating mass. Both mass and resulting force
determine the frequency. It's just a matter of applying the laws of physics.


Question

Is our auditory system in some way acting like a spectrum analyser?
(Is it able to distinguish the composing frequencies from a vibration?)

Ron?
Somebody else?

Thanks

gr, Hein
 
J

Jim Kelley

Impossible. Remember, we're talking about sound. Mechanical
forces only. Suppose you're driving, just going round the corner.
From the outside a fistful of forces is working on your body,
downwards, upwards, sidewards. It is absolutely impossible
that your body's centre of gravity is following different forces in
different directions at the same time. Only the resulting force is
changing your movement (according to Newton's second law).



Nonsense. Mechanical oscillations are fully determined by
forces acting on the vibrating mass. Both mass and resulting force
determine the frequency. It's just a matter of applying the laws of physics.

Question

Is our auditory system in some way acting like a spectrum analyser?
(Is it able to distinguish the composing frequencies from a vibration?)

Ron?
Somebody else?

Thanks

gr, Hein
 
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