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

[craigm]

And requires, for proper reception, that a carrier be recreated at the
receiver which has not only the amplitude of the original, but also its
exact phase. Absent some sort of "pilot" to get things synchronized,
this makes reception very difficult.

Isaac

Try a Costas loop.

 

[Don Bowey]

Costas loop

What has that to do with demodulating a signal that has no transmitted
reference signal?

 

[Tommy Tootles]

This is your last question of 1001 questions:

Since the information (modulation/voice) is repeated in both sides of
the modulation envelope in am, ssb "chops off" one half of the envelope.
The receiver is responsible for "mirroring" the other and reproducing
the information on that end--this allows for almost doubling the
effective range of am.

Get off the drugs, get a job and become a productive citizen!

JS

Uh, John...respectfully, I have to wonder just who is on drugs.

The original poster *ASKED* about *DSB* vs. AM

YOU *ANSWERED* about *SSB*. Here is the correct answer...

There are two broad types of DSB (double sideband) transmission:

DSB-RC and DSB-SC, meaning BOTH sidebands are transmitted, but with
either a (R)educed (C)arrier or a (S)urpressed (C)arrier. AM sends both
sidebands and full carrier.

Hope that answers the OP's question>

Tommy T.

 

[John Smith I]

Tommy Tootles wrote:

Yeah, you just discovered that for all intents and purposes double
sideband is am, and suppressed carrier is just like suppressed carrier am?

Oh well, better late than never ...

JS

 

[craigm]

This is your last question of 1001 questions:

Since the information (modulation/voice) is repeated in both sides of
the modulation envelope in am, ssb "chops off" one half of the envelope.
The receiver is responsible for "mirroring" the other and reproducing
the information on that end

Huh? this makes no sense. There is no need to regenerate the missing
sideband to recover the signal. What circuit does this? What's the math?

--this allows for almost doubling the
effective range of am.

Increased range comes from transmit bandwidth, power distribution vs
frequency, and receiver bandwidth. Again, what's the math that explains
this?

 

[Don Bowey]

Uh, John...respectfully, I have to wonder just who is on drugs.

The original poster *ASKED* about *DSB* vs. AM

Which means he asked about a process (AM), and one of the output
permutations of AM (DSB), which as you point out can have several options.

YOU *ANSWERED* about *SSB*. Here is the correct answer...

There are two broad types of DSB (double sideband) transmission:

DSB-RC and DSB-SC, meaning BOTH sidebands are transmitted, but with
either a (R)educed (C)arrier or a (S)urpressed (C)arrier. AM sends both
sidebands and full carrier.

All of them are AM in different forms using different added circuitry.
There are many ways to achieve them.

 

[Jim Kelley]

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. (Note that if the beat
frequency was a separate, difference signal as you suggest, at this
frequency it would certainly be audible.)

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.

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.

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.

jk

 

[isw]

This is your last question of 1001 questions:

Since the information (modulation/voice) is repeated in both sides of
the modulation envelope in am, ssb "chops off" one half of the envelope.
The receiver is responsible for "mirroring" the other and reproducing
the information on that end--this allows for almost doubling the
effective range of am.

The range is increased, but not for the reason you stated (and BTW, he
asked about DSB; you responded about SSB).

There are two primary reasons why suppressed carrier SSB has more
"range":

1) The receiver can have half the bandwidth as compared to AM; that
reduces the amount of noise the receiver will pick up.

2) For AM at 100% modulation, 50% of the total transmitted power is in
the carrier (carrying no information), and 25% is in each of the (upper
and lower) sidebands, where they carry redundant information.

So the total power that's actually useful to the receiver is a pretty
small fraction of the total that is transmitted. The advantage of AM, of
course, is that it requires only a very simple receiver.

For suppressed-carrier SSB, all the radiated power is in the one set of
sidebands being transmitted. Assuming that in both cases the
transmitters draw equal power from the "wall outlet", then there's a
clear win for SSB. That advantage comes at the expense of a more complex
receiver, and the need for a more skillful operator to tune it.

The advantage of SSB is considerably more than the "almost double the
range" you mention -- more like 10 to 12 dB, where doubling would imply
3 dB.

Isaac

 

[isw]

On 7/5/07 10:27 PM, in article [email protected],


On 7/5/07 12:00 AM, in article [email protected],


On 7/4/07 8:42 PM, in article [email protected],
"Ron


On 7/4/07 10:16 AM, in article
[email protected],


On 7/4/07 7:52 AM, in article
[email protected],
"Ron

<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.

What was the subject of your question?

Copying from my original post:

Suppose you have a 1 MHz sine wave whose amplitude
is multiplied by a 0.1 MHz sine wave.
What would it look like on an oscilloscope?
What would it look like on a spectrum analyzer?

Then suppose you have a 1.1 MHz sine wave added
to a 0.9 MHz sine wave.
What would that look like on an oscilloscope?
What would that look like on a spectrum analyzer?




So the first (1) is an AM question and the second (2) is a non-AM
question......

What is the difference between AM and DSB?




AM is a process. DSB (double sideband), with carrier, is it's most simple
result. DSB without carrier (suppressed carrier dsb) requires using, at
least, a balanced mixer as the AM multiplier.

And requires, for proper reception, that a carrier be recreated at the
receiver which has not only the amplitude of the original,

There is no need at all to match the carrier amplitude of the original
signal. You can use an excessively high carrier injection amplitude with no
detrimental affect, but if the injected carrier is too little, the
demodulated signal will be over modulated and sound distorted.

but also its exact phase.

Exact, not required. The closer the better, however.

Well, OK, the phase must at least bear a constant relationship to the
one that created the signal. If you inject a carrier that has a
quadrature relationship to the one that created the DSB signal, the
output will be PM (phase modulation). In between zero and 90 degrees,
the output is a combination of the two. If the injected carrier is not
at precisely the proper frequency, the phase will roll around and the
output will be unintelligible.

Isaac

 

[isw]

Which of them is linear?

A well-designed filter running into a bolometer would be. You can make
the filter narrow enough to respond to only one frequency component at
the time, and a bolometer just turns the signal power into heat; nothing
nonlinear there...

Isaac

 

[Mika Lindblad]

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.

Unless.. The amplifier wasn't so good after all, and the higher tone was
created by intermodulation distortion? That'd be the sum of the frequencies..

Or maybe it was some harmonic of the original signal, created by either the
amplifier or the signal source/generator?

 

[isw]

That doesn't explain why the effect would come and go.

I don't understand what effect you're referring to here.

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
maintained that if the only way one can understand physical phenomena is
by solving the differential equations that describe them, then one does
not understand the phenomena at all. If you can express a thing in
words, such that a person with little mathematical ability can
understand what's going on, *then* you have a good grasp of it.

It looks to me that the tripple frequency sidebands
are there but the basic sidebands dominate.
Especially at lower modulation indexes.

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"?

Isaac

 

[Rich Grise]

Suppose you have a 1 MHz sine wave whose amplitude
is multiplied by a 0.1 MHz sine wave.
What would it look like on an oscilloscope?

This is close, but not to scale:
http://en.wikipedia.org/wiki/Amplitude_modulation
The animation shows the "envelope".

What would it look like on a spectrum analyzer?

One vertical "spike" at 1 MHz with smaller spikes at .9 and 1.1 MHz. The
height of the two side spikes, depends on the depth of modulation.
In this case, the carrier is in the middle, and the sidebands are on the
sides.

Then suppose you have a 1.1 MHz sine wave added
to a 0.9 MHz sine wave.
What would that look like on an oscilloscope?

whatever 0.9 MHz superimposed on 1.1 MHz looks like. ;-)

What would that look like on a spectrum analyzer?

One spike at each input frequency, 0.9 and 1.1 MHz. If they're mixed
nonlinearly, then you get modulation, as above.

Hope This Helps!
Rich

 

[Rich Grise]

Huh? this makes no sense. There is no need to regenerate the missing
sideband to recover the signal. What circuit does this? What's the math?

You really don't need it, because with your BFO the other sideband
would give you negative frequencies anyway. ;-)

Cheers!
Rich

 

[Don Bowey]

On 7/5/07 10:27 PM, in article [email protected],


On 7/5/07 12:00 AM, in article [email protected],


On 7/4/07 8:42 PM, in article [email protected],
"Ron


On 7/4/07 10:16 AM, in article
[email protected],


On 7/4/07 7:52 AM, in article
[email protected],
"Ron

<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.

What was the subject of your question?

Copying from my original post:

Suppose you have a 1 MHz sine wave whose amplitude
is multiplied by a 0.1 MHz sine wave.
What would it look like on an oscilloscope?
What would it look like on a spectrum analyzer?

Then suppose you have a 1.1 MHz sine wave added
to a 0.9 MHz sine wave.
What would that look like on an oscilloscope?
What would that look like on a spectrum analyzer?




So the first (1) is an AM question and the second (2) is a non-AM
question......

What is the difference between AM and DSB?




AM is a process. DSB (double sideband), with carrier, is it's most simple
result. DSB without carrier (suppressed carrier dsb) requires using, at
least, a balanced mixer as the AM multiplier.

And requires, for proper reception, that a carrier be recreated at the
receiver which has not only the amplitude of the original,

There is no need at all to match the carrier amplitude of the original
signal. You can use an excessively high carrier injection amplitude with no
detrimental affect, but if the injected carrier is too little, the
demodulated signal will be over modulated and sound distorted.

but also its exact phase.

Exact, not required. The closer the better, however.

Well, OK, the phase must at least bear a constant relationship to the
one that created the signal. If you inject a carrier that has a
quadrature relationship to the one that created the DSB signal, the
output will be PM (phase modulation). In between zero and 90 degrees,
the output is a combination of the two. If the injected carrier is not
at precisely the proper frequency, the phase will roll around and the
output will be unintelligible.

Not unintelligible.... Donald Duckish.

On a more practical side, however, most receiver filters for ssb will
essentially remove one sideband if there are two, and can attenuate a
carrier so the local product detector can do it's job resulting in improved
receiving conditions. But this is more advanced than the Ops questions.

Don

 

[Rich Grise]

Increased range comes from transmit bandwidth, power distribution vs
frequency, and receiver bandwidth. Again, what's the math that explains
this?

Say you've got a transmitter that outputs 1000 watts. So, apply a 1 MHz
sine wave - that's CW (Continuous Wave). Now to modulate that, you have
to add in the audio power through a modulator. This adds energy, at
frequencies called "sidebands". These ride along with the carrier and for
100% modulation you also need 1000W of audio, which shows up in two
sidebands with a total power content of 500W each. That's a total of
2000W; the modulator provides the additional 1000W to the xmtr's.

The scope waveform would show a 1MHz sine wave that goes from 0 to 2X
the original amplitude in step with the modulation.

Now, if you could filter out the carrier, you could send the same
amount of information the same distance with only 1000W output, which
would all be in the sidebands. Boost this to 2000W, and you'll hear
a 3 db increase in the audio's signal strength.

Now, if you take away one of those sidebands, you only need 1000W to
send the same amount of information, since the sidebands are simply
mirror images of each other, you're back to 1000W again, and if you
up that to 2KW, your audio is 6dB stronger than it would be with an AM
signal of the same wattage, since no power is wasted on the carrier or
extraneous sideband.

NTSC TV video uses "vestigial sideband", but I don't know why - maybe
they need more power down at the V. sync freq. or something.

Cheers!
Rich

 

[Rich Grise]

Sorry, John - while the ear's amplitude response IS nonlinear, it
does not act as a mixer. "Mixing" (multiplication) occurs when
a given nonlinear element (in electronics, a diode or transistor, for
example) is presented with two signals of different frequencies.
But the human ear doesn't work in that manner - there is no single
nonlinear element which is receiving more than one signal.

Sure there is - the cochlea. (well, the whole middle ear/inner ear
system.)

What would the output look like if you summed a 300Hz tone and a 400Hz
tone and sent the sum to a log amp and spectrum analyzer/fft?

Thanks,
Rich

 

[Don Bowey]

Say you've got a transmitter that outputs 1000 watts. So, apply a 1 MHz
sine wave - that's CW (Continuous Wave). Now to modulate that, you have
to add in the audio power through a modulator. This adds energy, at
frequencies called "sidebands". These ride along with the carrier and for
100% modulation you also need 1000W of audio,

Small woops here....... 1000 Watts Tx carrier requires 500 watts of
modulator power for 100 % modulation, which puts 250 Watts in each sideband.

 

[Don Bowey]

On 7/4/07 7:52 AM, in article [email protected],
"Ron

<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])
)

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?

Post your intention; it might help.

 

[Bob Myers]

Sure there is - the cochlea. (well, the whole middle ear/inner ear
system.)

Nope - the point had to do with the inner workings of the cochlea.
You can't consider it as a single element, as the inner workings
consists of what are essentially thousands of very narrowband
individual sensors. There is no *single* nonlinear element in which
mixing of, say, the hypothetical 300 Hz and 400 Hz tones would
take place. John responded that the eardrum (typmanic membrane)
would act as such an element, but I would suggest that any mixing
which might in theory go on here is not a signifcant factor in how we
perceive such tones. The evidence for this is obvious - if presented
with, say, a pure 440 Hz "A" from a tuning fork, and the note from the
slightly flat instrument we're trying to tune (let's say 438 Hz), we DO
hear the 2 Hz "beat" that results from the interference (in the air)
between these two sounds. What we do NOT hear to any significant
degree is the 878 Hz sum that would be expected if there were much
contribution from a multiplicative ("mixing") process.

Bob M.