 |
Search Sci.Electronics.Basics |
|
|
|
|
|
|
 |
|
|
Sci.Electronics.Basics -> AM electromagnetic waves: 20 KHz modulation frequency on an astronomically-low carrier frequency
There are 269 messages in this thread.
You are currently looking at messages 220 to 240.
|
Author: Bob MyersDate: 15:43 07-07-07
|
|
"Ron Baker, Pluralitas!" <this@aint.me> wrote in message
news:468fe7df$0$16560$4c368faf@roadrunner.com...
First of all, do you think you could possibly learn to trim your posts?
> No nonlinearity is necessary in order to hear
> a beat?
> Where does the beat come from?
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.
|
|
|
|
Author: John Smith IDate: 15:44 07-07-07
|
|
Tommy Tootles wrote:
> [stuff]
Your expansion of the original and simple question into a convoluted and
obfuscated mess shows an outstanding knack for skills related to the
psychotic ... however, it also shows you to be an idiot.
Hey, are you attempting to fake a mental disorder so you can get off
welfare and onto SSI?
Sharpen your razor blade, return to the mental hospital--begin splitting
hairs ...
JS
|
|
|
|
Author: iswDate: 15:56 07-07-07
|
|
In article <468f0515$0$30690$4c368faf@roadrunner.com>,
"Ron Baker, Pluralitas!" <this@aint.me> wrote:
--snippage--
> >> That doesn't explain why the effect would come and go.
> >
> > I don't understand what effect you're referring to here.
>
> When I was tuned to the 3rd harmonic sometimes
> I would hear it and sometimes not.
> It would come and go rather abruptly. It didn't seem
> to be gradual fading.
Especially if the RF field is strong, there are a lot of mechanisms
which can create harmonics after the signal leaves the transmitter --
rusty fencing, or tooth fillings, for example. I can see how one of
those could be intermittent.
> >> But once again you have surprised me.
> >> Your explanation of the non-multiplied sidebands,
> >> while qualitative and incomplete, is sound.
> >
> > I'm a physicist/engineer, and have been for a long time. I have always
>
> The you understand Fourier transforms and convolution.
I suppose so; I've spent over fifteen years poking around in the
entrails of MPEG...
> > I don't understand what you are saying here either. And in my
> > experience, the term "modulation index" is more likely to show up in
a
> > discussion of FM or PM than AM; are you using it interchangeably with
> > "modulation percentage"?
As I suspected -- just different words for the same thing.
So:
>> It looks to me that the tripple frequency sidebands
>> are there but the basic sidebands dominate.
>> Especially at lower modulation indexes.
With well-designed gear (or theoretically), for AM there will be no
other frequencies present except for the carrier and the ones
represented by the Fourier spectrum of the modulation -- one set either
side of the carrier. That is only true, of course, as long as there is
no overmodulation; that creates a *lot* of other junk, because there are
periods where the carrier is entirely cut off.
So I still don't understand what you mean by "triple frequency
sidebands" or "basic sidebands".
As I said in another post, modulation is a "rate effect", so there never
should be any frequencies generated at multiples of the sidebands
surrounding the fundamental; instead they are always identically as far
from the harmonics as they are from the fundamental. Is that what you
are calling "triple frequency sidebands"?
Isaac
|
|
|
|
Author: RHFDate: 15:58 07-07-07
|
|
On Jun 29, 7:41 pm, Radium <gluceg...@gmail.com> wrote:
> Hi:
>
> Please don't be annoyed/offended by my question as I decreased the
> modulation frequency to where it would actually be realistic.
>
> I have a very weird question about electromagnetic radiation,
> carriers, and modulators.
>
> Is it mathematically-possible to carry a modulator signal [in this
> case, a pure-sine-wave-tone] with a frequency of 20 KHz and an
> amplitude of 1-watt-per-meter-squared on a AM carrier signal whose
> frequency is 10^-(1,000,000,000-to-the-power-10^1,000,000,000)
> nanocycle* every 10^1,000,000,000-to-the-power-10^1,000,000,000 giga-
> eons and whose amplitude is a minimum of 10^1,000,000,000-to-the-
> power-10^1,000,000,000 gigaphotons per 10^-(1,000,000,000-to-the-
> power-10^1,000,000,000) nanosecond?
>
> If it is not mathematically-possible, then please explain why.
>
> 10^-(1,000,000,000-to-the-power-10^1,000,000,000) second is an
> extremely short amount of time. 10^-(1,000,000,000-to-the-
> power-10^1,000,000,000) nanosecond is even shorter because a
> nanosecond is shorter than a second.
>
> Giga-eon = a billion eons
>
> Eon = a billion years
>
> *nanocycle = billionth of a cycle
>
> Gigaphoton = a billion photons
>
> 10^1,000,000,000-to-the-power-10^1,000,000,000 -- now that is one
> large large number.
>
> 10^1,000,000,000 = 10-to-the-power-1,000,000,000
>
> So you get:
>
> (10-to-the-power-1,000,000,000) to the power (10-to-the-
> power-1,000,000,000)
>
> 10^-(1,000,000,000-to-the-power-10^1,000,000,000) = 10^-(10-to-the-
> power-1,000,000,000)-to-the-power-(10-to-the-power-1,000,000,000)
>
> 10^-(10-to-the-power-1,000,000,000) to the power (10-to-the-
> power-1,000,000,000) is an extremely small number at it equals 10-to-
> the-power-NEGATIVE-[(10-to-the-power-1,000,000,000) to the power (10-
> to-the-power-1,000,000,000)]
>
> No offense but please respond with reasonable answers & keep out the
> jokes, off-topic nonsense, taunts, insults, and trivializations. I am
> really interested in this.
>
> Thanks,
>
> Radium
WHAT WAS "RADIUM'S" ORIGINAL QUESTION ?
-and- HAS IT BEEN ANSWERED ?
Hi:
Please don't be annoyed/offended by my question as I decreased the
modulation frequency to where it would actually be realistic.
I have a very weird question about electromagnetic radiation,
carriers, and modulators.
Is it mathematically-possible to carry a modulator signal [in this
case, a pure-sine-wave-tone] with a frequency of 20 KHz and an
amplitude of 1-watt-per-meter-squared on a AM carrier signal whose
frequency is 10^-(1,000,000,000-to-the-power-10^1,000,000,000)
nanocycle* every 10^1,000,000,000-to-the-power-10^1,000,000,000 giga-
eons and whose amplitude is a minimum of 10^1,000,000,000-to-the-
power-10^1,000,000,000 gigaphotons per 10^-(1,000,000,000-to-the-
power-10^1,000,000,000) nanosecond?
If it is not mathematically-possible, then please explain why.
10^-(1,000,000,000-to-the-power-10^1,000,000,000) second is an
extremely short amount of time. 10^-(1,000,000,000-to-the-
power-10^1,000,000,000) nanosecond is even shorter because a
nanosecond is shorter than a second.
Giga-eon = a billion eons
Eon = a billion years
*nanocycle = billionth of a cycle
Gigaphoton = a billion photons
10^1,000,000,000-to-the-power-10^1,000,000,000 -- now that is one
large large number.
10^1,000,000,000 = 10-to-the-power-1,000,000,000
So you get:
(10-to-the-power-1,000,000,000) to the power (10-to-the-
power-1,000,000,000)
10^-(1,000,000,000-to-the-power-10^1,000,000,000) = 10^-(10-to-the-
power-1,000,000,000)-to-the-power-(10-to-the-power-1,000,000,000)
10^-(10-to-the-power-1,000,000,000) to the power (10-to-the-
power-1,000,000,000) is an extremely small number at it equals 10-to-
the-power-NEGATIVE-[(10-to-the-power-1,000,000,000) to the power (10-
to-the-power-1,000,000,000)]
No offense but please respond with reasonable answers & keep out the
jokes, off-topic nonsense, taunts, insults, and trivializations. I am
really interested in this.
Thanks,
Radium
|
|
|
|
Author: Ron Baker, Pluralitas!Date: 16:11 07-07-07
|
|
"Bob Myers" <nospamplease@address.invalid> wrote in message
news:f6oqdv$npt$1@usenet01.boi.hp.com...
>
> "Ron Baker, Pluralitas!" <this@aint.me> wrote in message
> news:468fe7df$0$16560$4c368faf@roadrunner.com...
>
> First of all, do you think you could possibly learn to trim your posts?
>
>> No nonlinearity is necessary in order to hear
>> a beat?
>> Where does the beat come from?
>
> 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?
|
|
|
|
Author: DTCDate: 16:37 07-07-07
|
|
isw wrote:
> Especially if the RF field is strong, there are a lot of mechanisms
> which can create harmonics after the signal leaves the transmitter --
> rusty fencing, or tooth fillings, for example.
What we used to call miscellaneous metallic junction intermod.
|
|
|
|
Author: Bob MyersDate: 16:41 07-07-07
|
|
"Ron Baker, Pluralitas!" <this@aint.me> wrote in message
news:468ff303$0$24708$4c368faf@roadrunner.com...
>
> "Bob Myers" <nospamplease@address.invalid> wrote in message
> news:f6oqdv$npt$1@usenet01.boi.hp.com...
>>
>> "Ron Baker, Pluralitas!" <this@aint.me> wrote in message
>> news:468fe7df$0$16560$4c368faf@roadrunner.com...
>>
>> First of all, do you think you could possibly learn to trim your posts?
Apparently, no, you can't. Too lazy to take the trouble to
perform this common courtesy, or what?
>> An audible beat tone is produced by the constructive and destructive
>> interference between two sound waves in air. Look at a pictorial
>> representation (in the time domain) of the sum of sine waves,of similar
>> amplitudes, one at, say, 1000 Hz and the other at 1005, and you'll
>> see it.
>>
>> Bob M.
>>
>
> How come you don't hear a 200 Hz beat
> with a 1000 Hz tone and a 1200 Hz tone?
For the simple reason that there isn't actually a "tone" involved -
in other words, there is no actual signal at the difference frequency.
There can't be, since there is no "mixing" (multiplication) of the
two original tones. The "beat" is really just the perception of
the amplitude variation caused by the interference previously
mentioned. You cannot sense such variations if they occur
rapidly enough, any more than you can detect the flicker of a
light source which is varying rapidly enough.
Bob M.
|
|
|
|
Author: Bob MyersDate: 16:46 07-07-07
|
|
That's easy. Radium has never actually had an original
question. All of his questions are either rehashing very
well-understood situations, or are utterly nonsensical and
therefore not answerable AS legitimate questions.
Bob M.
|
|
|
|
Author: Elmo P. ShagnastyDate: 18:02 07-07-07
|
|
In article <f6ou2s$r1s$1@usenet01.boi.hp.com>,
"Bob Myers" <nospamplease@address.invalid> wrote:
> That's easy. Radium has never actually had an original
> question. All of his questions are either rehashing very
> well-understood situations, or are utterly nonsensical and
> therefore not answerable AS legitimate questions.
>
> Bob M.
Are you saying that John Navas *also* posts under the name "Radium"?
|
|
|
|
Author: EeyoreDate: 19:08 07-07-07
|
|
RHF wrote:
> WHAT WAS "RADIUM'S" ORIGINAL QUESTION ?
> -and- HAS IT BEEN ANSWERED ?
Does it matter ?
It's only attention seeking.
Graham
|
|
|
|
Author: RHFDate: 19:31 07-07-07
|
|
On Jul 7, 4:08 pm, Eeyore <rabbitsfriendsandrelati...@hotmail.com>
wrote:
> RHF wrote:
> > WHAT WAS "RADIUM'S" ORIGINAL QUESTION ?
> > -and- HAS IT BEEN ANSWERED ?
>
- Does it matter ?
-
- It's only attention seeking.
-
- Graham
Graham - Thank You for the Attention :o) ~ RHF
- - - You are Reply # 228 . . . and still Counting.
|
|
|
|
Author: Ron Baker, Pluralitas!Date: 00:15 08-07-07
|
|
"Bob Myers" <nospamplease@address.invalid> wrote in message
news:f6otp9$qtt$1@usenet01.boi.hp.com...
>
> "Ron Baker, Pluralitas!" <this@aint.me> wrote in message
> news:468ff303$0$24708$4c368faf@roadrunner.com...
>>
>> "Bob Myers" <nospamplease@address.invalid> wrote in message
>> news:f6oqdv$npt$1@usenet01.boi.hp.com...
>>>
>>> "Ron Baker, Pluralitas!" <this@aint.me> wrote in message
>>> news:468fe7df$0$16560$4c368faf@roadrunner.com...
>>>
>>> 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?
|
|
|
|
Author: Ron Baker, Pluralitas!Date: 00:17 08-07-07
|
|
"Don Bowey" <dbowey@comcast.net> wrote in message
news:C2B403C2.6DC94%dbowey@comcast.net...
> On 7/4/07 8:42 PM, in article 468c6838$0$4664$4c368faf@roadrunner.com,
> "Ron
> Baker, Pluralitas!" <this@aint.me> wrote:
>
>>
>> "Don Bowey" <dbowey@comcast.net> wrote in message
>> news:C2B16AE5.6D5BC%dbowey@comcast.net...
>>> On 7/4/07 10:16 AM, in article 468bd5ad$0$16531$4c368faf@roadrunner.com,
>>> "Ron Baker, Pluralitas!" <this@aint.me> wrote:
>>>
>>>>
>>>> "Don Bowey" <dbowey@comcast.net> wrote in message
>>>> news:C2B1129D.6D573%dbowey@comcast.net...
>>>>> On 7/4/07 7:52 AM, in article
>>>>> 468bb3c0$0$24780$4c368faf@roadrunner.com,
>>>>> "Ron
>>>>> Baker, Pluralitas!" <this@aint.me> 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.
>>>
>>
>>
>
|
|
|
|
Author: Ron Baker, Pluralitas!Date: 00:24 08-07-07
|
|
"Rich Grise" <rich@example.net> wrote in message
news:pan.2007.07.05.23.42.45.964093@example.net...
> 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.
|
|
|
|
Author: iswDate: 00:49 08-07-07
|
|
In article <468fe7df$0$16560$4c368faf@roadrunner.com>,
"Ron Baker, Pluralitas!" <this@aint.me> 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
|
|
|
|
Author: Don BoweyDate: 00:52 08-07-07
|
|
On 7/7/07 9:17 PM, in article 469064fd$0$16540$4c368faf@roadrunner.com, "Ron
Baker, Pluralitas!" <this@aint.me> wrote:
>
> "Don Bowey" <dbowey@comcast.net> wrote in message
> news:C2B403C2.6DC94%dbowey@comcast.net...
>> On 7/4/07 8:42 PM, in article 468c6838$0$4664$4c368faf@roadrunner.com,
>> "Ron
>> Baker, Pluralitas!" <this@aint.me> wrote:
>>
>>>
>>> "Don Bowey" <dbowey@comcast.net> wrote in message
>>> news:C2B16AE5.6D5BC%dbowey@comcast.net...
>>>> On 7/4/07 10:16 AM, in article
468bd5ad$0$16531$4c368faf@roadrunner.com,
>>>> "Ron Baker, Pluralitas!" <this@aint.me> wrote:
>>>>
>>>>>
>>>>> "Don Bowey" <dbowey@comcast.net> wrote in message
>>>>> news:C2B1129D.6D573%dbowey@comcast.net...
>>>>>> On 7/4/07 7:52 AM, in article
>>>>>> 468bb3c0$0$24780$4c368faf@roadrunner.com,
>>>>>> "Ron
>>>>>> Baker, Pluralitas!" <this@aint.me> 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.
>>>>
>>>
>>>
>>
>
>
|
|
|
|
Author: DanaDate: 00:56 08-07-07
|
|
"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.
|
|
|
|
Author: m IIDate: 01:08 08-07-07
|
|
Ron Baker, Pluralitas! wrote:
> Could it be that the human auditory system is not
> linear?
No. Humans had to evolve to incorporate a non linear response to sound
when the electronics manufacturers started supplying ONLY non linear
potentiometers for audio equipment use.
This mutation, which is now the norm, was completely unknown before the
start of the twentieth century.
We, here at Densa Labs, call it Darwinian Decibelism
mike
|
|
|
|
Author: RHFDate: 01:16 08-07-07
|
|
On Jul 7, 9:56 pm, "Dana" <raff...@yahoo.com> wrote:
> "Ron Baker, Pluralitas!" <t...@aint.me> wrote in
messagenews: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.
Somewhere between the Original Post #1
and the 236 Replies to date. ~ RHF
|
|
|
|
Author: Fred AbseDate: 10:32 08-07-07
|
|
On Sat, 07 Jul 2007 20:37:14 +0000, DTC wrote:
> isw wrote:
>> Especially if the RF field is strong, there are a lot of mechanisms
>> which can create harmonics after the signal leaves the transmitter --
>> rusty fencing, or tooth fillings, for example.
>
> What we used to call miscellaneous metallic junction intermod.
AKA "Rusty Bolt Effect"
--
"Electricity is of two kinds, positive and negative. The difference
is, I presume, that one comes a little more expensive, but is more
durable; the other is a cheaper thing, but the moths get into it."
(Stephen Leacock)
|
|
|
|
|
|
|
|
|
Contact | Electronic Portal
|
|
|
