"Ron Baker, Pluralitas!" wrote in message
...

"Bob Myers" wrote in message
...

"Ron Baker, Pluralitas!" wrote in message
...

First of all, do you think you could possibly learn to trim your posts?

Apparently, no, you can't. Too lazy to take the trouble to
perform this common courtesy, or what?

An audible beat tone is produced by the constructive and destructive
interference between two sound waves in air. Look at a pictorial
representation (in the time domain) of the sum of sine waves,of similar
amplitudes, one at, say, 1000 Hz and the other at 1005, and you'll
see it.

Bob M.


How come you don't hear a 200 Hz beat
with a 1000 Hz tone and a 1200 Hz tone?

For the simple reason that there isn't actually a "tone" involved -
in other words, there is no actual signal at the difference frequency.
There can't be, since there is no "mixing" (multiplication) of the
two original tones. The "beat" is really just the perception of
the amplitude variation caused by the interference previously
mentioned. You cannot sense such variations if they occur
rapidly enough, any more than you can detect the flicker of a
light source which is varying rapidly enough.

Bob M.

That's easy. Radium has never actually had an original
question. All of his questions are either rehashing very
well-understood situations, or are utterly nonsensical and
therefore not answerable AS legitimate questions.

Bob M.

RHF wrote:

WHAT WAS "RADIUM'S" ORIGINAL QUESTION ?
-and- HAS IT BEEN ANSWERED ?

Does it matter ?

It's only attention seeking.

Graham

On Jul 7, 4:08 pm, Eeyore
wrote:
RHF wrote:
WHAT WAS "RADIUM'S" ORIGINAL QUESTION ?
-and- HAS IT BEEN ANSWERED ?


- Does it matter ?
-
- It's only attention seeking.
-
- Graham

Graham - Thank You for the Attention ) ~ RHF
- - - You are Reply # 228 . . . and still Counting.

"Bob Myers" wrote in message
...

"Ron Baker, Pluralitas!" wrote in message
...

"Bob Myers" wrote in message
...

"Ron Baker, Pluralitas!" wrote in message
...

First of all, do you think you could possibly learn to trim your posts?

Apparently, no, you can't. Too lazy to take the trouble to
perform this common courtesy, or what?

You could always plonk me.


An audible beat tone is produced by the constructive and destructive
interference between two sound waves in air. Look at a pictorial
representation (in the time domain) of the sum of sine waves,of similar
amplitudes, one at, say, 1000 Hz and the other at 1005, and you'll
see it.

Bob M.


How come you don't hear a 200 Hz beat
with a 1000 Hz tone and a 1200 Hz tone?

For the simple reason that there isn't actually a "tone" involved -
in other words, there is no actual signal at the difference frequency.
There can't be, since there is no "mixing" (multiplication) of the
two original tones.

There is no multiplication of 1000 Hz and 1005 Hz
either, is there? Why don't you hear 1000 Hz and
1005 Hz rather than a single tone varying in amplitude?

The "beat" is really just the perception of
the amplitude variation caused by the interference previously
mentioned. You cannot sense such variations if they occur
rapidly enough, any more than you can detect the flicker of a
light source which is varying rapidly enough.

Bob M.

Could it be that the human auditory system is not
linear?

"Don Bowey" wrote in message
...
On 7/4/07 8:42 PM, in article ,
"Ron
Baker, Pluralitas!" wrote:


"Don Bowey" wrote in message
...
On 7/4/07 10:16 AM, in article ,
"Ron Baker, Pluralitas!" wrote:


"Don Bowey" wrote in message
...
On 7/4/07 7:52 AM, in article
,
"Ron
Baker, Pluralitas!" wrote:

snip


cos(a) * cos(b) = 0.5 * (cos[a+b] + cos[a-b])

Basically: multiplying two sine waves is
the same as adding the (half amplitude)
sum and difference frequencies.

No, they aren't the same at all, they only appear to be the same
before
they are examined. The two sidebands will not have the correct phase
relationship.

What do you mean? What is the "correct"
relationship?


One could, temporarily, mistake the added combination for a full
carrier
with independent sidebands, however.




(For sines it is
sin(a) * sin(b) = 0.5 * (cos[a-b]-cos[a+b])
= 0.5 * (sin[a-b+90degrees] - sin[a+b+90degrees])
= 0.5 * (sin[a-b+90degrees] + sin[a+b-90degrees])
)

--
rb





When AM is correctly accomplished (a single voiceband signal is
modulated

The questions I posed were not about AM. The
subject could have been viewed as DSB but that
wasn't the specific intent either.

You should take some time to more carefully frame your questions.

Do you understand that a DSB signal *is* AM?

So all the AM broadcasters are wasting money by
generating a carrier?


Post your intention; it might help.


onto a carrier via a non-linear process), at an envelope detector the
two
sidebands will be additive. But if you independe ntly place a carrier
at
frequency ( c ), another carrier at ( c-1 khz) and another carrier at
(c+
1
kHz), the composite can look like an AM signal, but it is not, and only
by
the most extreme luck will the sidebands be additive at the detector.
They
would probably cycle between additive and subtractive since they have no
real relationship and were not the result of amplitude modulation.

"Rich Grise" wrote in message
news
On Tue, 03 Jul 2007 22:42:20 -0700, isw wrote:

After you get done talking about modulation and sidebands, somebody
might want to take a stab at explaining why, if you tune a receiver to
the second harmonic (or any other harmonic) of a modulated carrier (AM
or FM; makes no difference), the audio comes out sounding exactly as it
does if you tune to the fundamental? That is, while the second harmonic
of the carrier is twice the frequency of the fundamental, the sidebands
of the second harmonic are *not* located at twice the frequencies of the
sidebands of the fundamental, but rather precisely as far from the
second harmonic of the carrier as they are from the fundamental.

Have you ever actually observed this effect?

Thanks,
Rich


I have.
I tuned to the third harmonic of a strong local
AM broadcast station. There it was. Quite
a surprise. It is a bit distorted but intelligible.
Another odd thing is that it comes and goes
somewhat abruptly.

In article ,
"Ron Baker, Pluralitas!" wrote:


--snippety-snip--

You said you are a physicist/engineer.
What does "linear" mean?

Let's not get too far off the subject here. We were discussing whether
the "tuning beat" that you use to tune a musical instrument involved a
nonlinear process (ie. "modulation").

Then linearity is at the core of the matter.
What does "linear" (or "nonlinear") mean to you?

OK, if you insist -- *in this case* it means "linear enough to not
produce IM products of significant amplitude".

I said that it does not, and that
it could be detected by instrumentation which was proveably linear (i.e.
not "perfectly" linear, because that's not required, but certainly
linear enough to discount the requirement for "modulation").

No nonlinearity is necessary in order to hear
a beat?
Where does the beat come from?

As the phase of the two nearly equal waves move past each other, there
is simple vector summation which varies the amplitude.

Consider two sine waves of precisely the same frequency, where one of
them is adjustable in phase -- use a goniometer, for instance. Use a set
of resistors to sum the two signals, and observe the summing point with
a 'scope or a loudspeaker. By altering the phase of one source, you can
get any amplitude you want from zero up to twice the amplitude of either
one.

Now just twiddle that phase knob around and around as fast as you can.

You've just slightly altered the instantaneous frequency of one of the
generators (but only while you twiddle), and accomplished pretty much
the same effect as listening to the beat between two guitar strings at
nearly zero frequency offset. With no nonlinear processes in sight.

Isaac

On 7/7/07 9:17 PM, in article , "Ron
Baker, Pluralitas!" wrote:


"Don Bowey" wrote in message
...
On 7/4/07 8:42 PM, in article ,
"Ron
Baker, Pluralitas!" wrote:


"Don Bowey" wrote in message
...
On 7/4/07 10:16 AM, in article ,
"Ron Baker, Pluralitas!" wrote:


"Don Bowey" wrote in message
...
On 7/4/07 7:52 AM, in article
,
"Ron
Baker, Pluralitas!" wrote:

snip


cos(a) * cos(b) = 0.5 * (cos[a+b] + cos[a-b])

Basically: multiplying two sine waves is
the same as adding the (half amplitude)
sum and difference frequencies.

No, they aren't the same at all, they only appear to be the same
before
they are examined. The two sidebands will not have the correct phase
relationship.

What do you mean? What is the "correct"
relationship?


One could, temporarily, mistake the added combination for a full
carrier
with independent sidebands, however.




(For sines it is
sin(a) * sin(b) = 0.5 * (cos[a-b]-cos[a+b])
= 0.5 * (sin[a-b+90degrees] - sin[a+b+90degrees])
= 0.5 * (sin[a-b+90degrees] + sin[a+b-90degrees])
)

--
rb





When AM is correctly accomplished (a single voiceband signal is
modulated

The questions I posed were not about AM. The
subject could have been viewed as DSB but that
wasn't the specific intent either.

You should take some time to more carefully frame your questions.

Do you understand that a DSB signal *is* AM?

So all the AM broadcasters are wasting money by
generating a carrier?


You are an ignorant, useless troll, and not worth my time




Post your intention; it might help.


onto a carrier via a non-linear process), at an envelope detector the
two
sidebands will be additive. But if you independe ntly place a carrier
at
frequency ( c ), another carrier at ( c-1 khz) and another carrier at
(c+
1
kHz), the composite can look like an AM signal, but it is not, and only
by
the most extreme luck will the sidebands be additive at the detector.
They
would probably cycle between additive and subtractive since they have no
real relationship and were not the result of amplitude modulation.

"Ron Baker, Pluralitas!" wrote in message
...
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.