On Thu, 05 Jul 2007 07:32:20 -0700, Keith Dysart
wrote:

On Jul 5, 10:01 am, John Fields 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.


--
JF

On Fri, 06 Jul 2007 19:04:00 -0000, Jim Kelley
wrote:

On Jul 5, 9:38 pm, John Fields wrote:
On Thu, 05 Jul 2007 18:37:21 -0700, Jim Kelley
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:

+-------+ +--------+
| 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.



--
JF

On Sat, 14 Jul 2007 15:14:28 +0900, "Brenda Ann"
wrote:


"Ron Baker, Pluralitas!" wrote in message
...
|
| "Rich Grise" wrote in message
| news | On Wed, 11 Jul 2007 22:52:17 -0700, Jeff Liebermann wrote:
|
| "NotMe" 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.

Other than that, Bingo!!!


--
JF

"Hein ten Horn" wrote in message
...
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

"Hein ten Horn" wrote in message
...
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.

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

"Brenda Ann" wrote in message
...

"Ron Baker, Pluralitas!" wrote in message
...
|
| "Rich Grise" wrote in message
| news | On Wed, 11 Jul 2007 22:52:17 -0700, Jeff Liebermann wrote:
|
| "NotMe" 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

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.

pressure (volume) as the mixing product goes in and out of phase. This
actually becomes easier to hear as you near zero beat.

Ron Baker, Pluralitas! wrote:
Hein ten Horn wrote:
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.

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).

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.

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

On Jul 14, 12:42 pm, "Hein ten Horn"
wrote:
Ron Baker, Pluralitas! wrote:
Hein ten Horn wrote:
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.

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).

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.

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

On Sat, 14 Jul 2007 19:51:40 -0000, Jim Kelley
wrote:

On Jul 14, 12:42 pm, "Hein ten Horn"
wrote:
Ron Baker, Pluralitas! wrote:
Hein ten Horn wrote:
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.

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).

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.

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?)

---
Yes, of course.

The cilia in the cochlea are different lengths and, consequently,
"tuned" to different frequencies to which they respond by undulating
and sending electrical signals to the brain when the nerves to which
they're connected fire. See:

http://en.wikipedia.org/wiki/Organ_of_Corti



--
JF

Ron Baker, Pluralitas! wrote:
Hein ten Horn wrote:
Ron Baker, Pluralitas! wrote:

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?

Oh, nice question. Well, usually (in my case) the functions
are quite simple (like the ones we're here discussing) so that
I see the beat in a picture of a rough plot in my mind.

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.

Mathematical terms like linear, logarithmic, etc. are familiar
to me, but the guys here use linear and nonlinear in another
sense. Something to do with harmonics or so? Anyway,
that's why the hint isn't working here.

(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().

Got the (co)sin-stuff. But the unequallities are still there.
It's easy to understand: the left-hand term is sooner or later
greater then one, the right-hand term not (in both unequalities).
As a consequence we've two different x's.

So which is it really? Hint: If all you have is x then
you can't tell how it was generated.

Yep.

What you do with it afterwards can make a
difference.

Sure (but nature doesn't mind).

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.

I would say we need the math to work with it, to get our
things done. Understanding nature is not self-evident the
same thing. I would really appreciate it if you would take
the time to read my UTC 9:57 reply to JK once again,
but then with a more open mind. Thanks.

gr, Hein