In article , "Dan Finn"
writes:

"Len Over 21" wrote in message
...
In article , "Dan Finn"
writes:

"Len Over 21" wrote in message
...

On most modern HF transceivers, the 3rd harmonic has the strongest
content of RF. The 3rd harmonic of 3.5 to 4.0 MHz is 10.5 to 12 MHz
and there aren't many "ham listening frequencies" there, are there?

For most modulated sine waves, the 3rd harmonic is usually the strongest
*harmonic* although it depends upon several factors. Generally, the odd
harmonic components add and the negative components subtract with
modulated sine waves.

Oh my, such interesting math (which wasn't shown)...:-)

If you do not know this to be true without seeing the math worked out for
you, then you should not be discussing harmonics in a technical context.

Your saying so does not make your statement true. We are talking technical
issues here which are a matter of common knowledge to the RF community so
your flames are irrelevant.

Oh God...Earth to Len....harmonics *are* phase shifted. If you reduce the
phase shift to zero, then you have eliminated the harmonics, stupid.

I'm not sure where you get this "phase shift reduction to zero" but it
isn't quite like that. Sigh. Read following -

OK, let's get down to basics. From "Electronic Designers' Handbook,"
by Robert W. Landee, Donovan C. Davis, Albert P. Albrecht, 1957,
McGraw-Hill Book Co., Chapter 5, here's the instantaneous voltage
at any point in time of an AM RF waveform:

e(t) = E [1+M Cos(omega_m t)] Sin (omega_c t)

Whe

e(t) = instantaneous RF voltage e at time t
t = Time position, seconds
E = Peak voltage of unmodulated RF carrier
omega_m = radian frequency of modulation
= (2 pi f_m)
f_m = Modulation frequency, Hz
omega_c = radian frequency of RF carrier
= (2 pi f_c)
f_c = Carrier frequency, Hz
M = Modulation factor, range of 0 (none) to 1 (full)

That's not intuitive. By some identities and reforming the AM equation,
the following can be found true for a single modulation frequency:

e(t) = E Sin (omega_c t) + [E M / 2]Sin[(omega_c + omega_m) t ]
+ [E M / 2]Sin[(omega_c - omega_m) t ]

The first term on the right side is the Carrier. The second term group
on the right is the Upper Sideband and the third term group (next
line) is the Lower Sideband. If, through other means, the Carrier is
suppressed and Upper Sideband is suppressed, then a "SSB"
Lower Sideband signal is formed and it can be expressed as:

e(t) = [E M / 2] Sin [(omega_c - omega_m) t ]

If there were two modulation tones, f_a and f_b, the Lower Sideband
SSB equation is then:

e(t) = [E M_a / 2] Sin [(omega_c - omega_a) t ] +
[E M_b / 2] Sin [(omega_c - omega_b) t ] + ....

whe
M_a = modulation factor for frequency a,
M_b = modulation factor for frequency b
omega_a and omega_b are the radian frequencies of modulation
frequencies f_a and f_b

The above can be expanded to include any number of frequencies
in the modulation process but they will have the most impact on
the distortion caused to e(t) through any amplifier. That distortion
will create the same as many modulation frequencies, all
harmonically related in frequency and would be a form of "super
modulation" of AM or AM on top of the SSB (AM) signal.

There is NO magic cancellation of phase terms as alleged in
such AM on top of SSB, just a lot of extra sidebands generated
as a result of the distortion. However the distortion sidebands
would be at much higher frequencies than the modulation
frequencies. Those can be synthesized in seconds using a
computer program such as WAVESPEC which I wrote in 1993
and released as Shareware on ZDNet for PCs.

Painter's "Signals, Modulation, and Noise" gives a better
description of such AM-on-top-of-AM (or SSB) but I don't have a
copy available right now to reference contents.

Landee-Davis-Albrecht take up a whole chapter on modulation
and the derivations to show Carrier-Upper and Lower Sideband
terms which are Real representations of modulation products for
AM and SSB. Distortion in a final power amplifer will result in
AM on top of the AM/SSB signal, creating modulation products
centered around carrier frequency multiples.

The Carrier-Sideband form of AM illustrated above also contains
a clue on how a Phasing method of SSB generation works for
either sideband selection. Do your own math on that...it isn't
part of this non-discussion. :-)


Pretty close, most of the time, dummy.

No need to vomit nasty names here. I've given you an opening
for discussion on the mathematics of modulation.


That old standby tube output matching circuit, the "Pi-network" is only
good for about 18 db per octave attenuation above cutoff frequency.

Since 3dB is half the power, you cut the power in half 6 times. 18dB is not
bad over one octave.

18 db per octave is approximately the cutoff slope of a 3-section
Butterworth-response L-C lowpass filter. Most "Pi-networks" are based
on Butterworth-response values.

Not
all modern transceivers have such "lowpass filters" since they rely on
Class B or AB linear amplifiers with rather broadband transformers
matching PA to load.

They had better filter it so that you do not have a significant signal at
12MHz when transmitting in the 75 meter band.

An L-C filter is not absolutely necessary. What is required basically is
to provide a "clean" signal prior to final power amplification. The final
amplifier can be improved (distortion reduced) by several methods:
negative feedback at RF; envelope detection and feedback to the
modulated carrier source; etc.


0.1mW? Hardly a significant signal.

"Significant" is a highly subjective word. -10 dbm level is FAR
higher than most received signal values whose carrier level
ranges from -120 dbm (barely above the internal noise level) to
-40 dbm (slammin-home signal of 2.2 mV into a 50 Ohm coax
from antenna).

-10 dbm or 0.1 mW is equal to about 71 mV in the 50 Ohm
coax...30 db higher than -40 dbm or nearly 32 times the
voltage.

Feel free to do your numbers in public, senior

That would be a rather boring and somewhar ominous excercize to do on a
newsgroup, dummy.

American spelling is "exercise." Exercise your word power first.

Is calling others "dummy" instead of showing math "more fun" for you?


Let us know, please, I've not seen any of those
concerning out-of-ham-band interference due to harmonic generation.

Neither have I. Most NAL's are for in band interference, referred to as QRM.

Then they must not exist, right? :-)

Is that because the FCC and ARRL signs are in opposition and
they cancel one another out? :-)

LHA

"Len Over 21" wrote in message
...
In article , "Dan Finn"
writes:

"Len Over 21" wrote in message
...
In article , "Dan Finn"
writes:

"Len Over 21" wrote in message
...

On most modern HF transceivers, the 3rd harmonic has the
strongest
content of RF. The 3rd harmonic of 3.5 to 4.0 MHz is 10.5 to 12
MHz
and there aren't many "ham listening frequencies" there, are
there?

For most modulated sine waves, the 3rd harmonic is usually the
strongest
*harmonic* although it depends upon several factors. Generally, the
odd
harmonic components add and the negative components subtract with
modulated sine waves.

Oh my, such interesting math (which wasn't shown)...:-)

If you do not know this to be true without seeing the math worked out for
you, then you should not be discussing harmonics in a technical context.

Your saying so does not make your statement true. We are talking
technical
issues here which are a matter of common knowledge to the RF community so
your flames are irrelevant.

Oh God...Earth to Len....harmonics *are* phase shifted. If you reduce the
phase shift to zero, then you have eliminated the harmonics, stupid.

I'm not sure where you get this "phase shift reduction to zero" but it
isn't quite like that. Sigh. Read following -

OK, let's get down to basics. From "Electronic Designers' Handbook,"
by Robert W. Landee, Donovan C. Davis, Albert P. Albrecht, 1957,
McGraw-Hill Book Co., Chapter 5, here's the instantaneous voltage
at any point in time of an AM RF waveform:

e(t) = E [1+M Cos(omega_m t)] Sin (omega_c t)

Whe

e(t) = instantaneous RF voltage e at time t
t = Time position, seconds
E = Peak voltage of unmodulated RF carrier
omega_m = radian frequency of modulation
= (2 pi f_m)
f_m = Modulation frequency, Hz
omega_c = radian frequency of RF carrier
= (2 pi f_c)
f_c = Carrier frequency, Hz
M = Modulation factor, range of 0 (none) to 1 (full)

That's not intuitive. By some identities and reforming the AM
equation,
the following can be found true for a single modulation frequency:

e(t) = E Sin (omega_c t) + [E M / 2]Sin[(omega_c + omega_m) t ]
+ [E M / 2]Sin[(omega_c - omega_m) t ]

The first term on the right side is the Carrier. The second term group
on the right is the Upper Sideband and the third term group (next
line) is the Lower Sideband. If, through other means, the Carrier is
suppressed and Upper Sideband is suppressed, then a "SSB"
Lower Sideband signal is formed and it can be expressed as:

e(t) = [E M / 2] Sin [(omega_c - omega_m) t ]

If there were two modulation tones, f_a and f_b, the Lower Sideband
SSB equation is then:

e(t) = [E M_a / 2] Sin [(omega_c - omega_a) t ] +
[E M_b / 2] Sin [(omega_c - omega_b) t ] + ....

whe
M_a = modulation factor for frequency a,
M_b = modulation factor for frequency b
omega_a and omega_b are the radian frequencies of modulation
frequencies f_a and f_b

The above can be expanded to include any number of frequencies
in the modulation process but they will have the most impact on
the distortion caused to e(t) through any amplifier. That distortion
will create the same as many modulation frequencies, all
harmonically related in frequency and would be a form of "super
modulation" of AM or AM on top of the SSB (AM) signal.

There is NO magic cancellation of phase terms as alleged in
such AM on top of SSB, just a lot of extra sidebands generated
as a result of the distortion. However the distortion sidebands
would be at much higher frequencies than the modulation
frequencies. Those can be synthesized in seconds using a
computer program such as WAVESPEC which I wrote in 1993
and released as Shareware on ZDNet for PCs.

Painter's "Signals, Modulation, and Noise" gives a better
description of such AM-on-top-of-AM (or SSB) but I don't have a
copy available right now to reference contents.

Landee-Davis-Albrecht take up a whole chapter on modulation
and the derivations to show Carrier-Upper and Lower Sideband
terms which are Real representations of modulation products for
AM and SSB. Distortion in a final power amplifer will result in
AM on top of the AM/SSB signal, creating modulation products
centered around carrier frequency multiples.

The Carrier-Sideband form of AM illustrated above also contains
a clue on how a Phasing method of SSB generation works for
either sideband selection. Do your own math on that...it isn't
part of this non-discussion. :-)


Pretty close, most of the time, dummy.

No need to vomit nasty names here. I've given you an opening
for discussion on the mathematics of modulation.


That old standby tube output matching circuit, the "Pi-network" is
only
good for about 18 db per octave attenuation above cutoff frequency.

Since 3dB is half the power, you cut the power in half 6 times. 18dB is
not
bad over one octave.

18 db per octave is approximately the cutoff slope of a 3-section
Butterworth-response L-C lowpass filter. Most "Pi-networks" are based
on Butterworth-response values.

Not
all modern transceivers have such "lowpass filters" since they rely
on
Class B or AB linear amplifiers with rather broadband transformers
matching PA to load.

They had better filter it so that you do not have a significant signal at
12MHz when transmitting in the 75 meter band.

An L-C filter is not absolutely necessary. What is required basically
is
to provide a "clean" signal prior to final power amplification. The
final
amplifier can be improved (distortion reduced) by several methods:
negative feedback at RF; envelope detection and feedback to the
modulated carrier source; etc.


0.1mW? Hardly a significant signal.

"Significant" is a highly subjective word. -10 dbm level is FAR
higher than most received signal values whose carrier level
ranges from -120 dbm (barely above the internal noise level) to
-40 dbm (slammin-home signal of 2.2 mV into a 50 Ohm coax
from antenna).

-10 dbm or 0.1 mW is equal to about 71 mV in the 50 Ohm
coax...30 db higher than -40 dbm or nearly 32 times the
voltage.

Feel free to do your numbers in public, senior

That would be a rather boring and somewhar ominous excercize to do on a
newsgroup, dummy.

American spelling is "exercise." Exercise your word power first.

Is calling others "dummy" instead of showing math "more fun" for you?


Let us know, please, I've not seen any of those
concerning out-of-ham-band interference due to harmonic generation.

Neither have I. Most NAL's are for in band interference, referred to as
QRM.

Then they must not exist, right? :-)

Is that because the FCC and ARRL signs are in opposition and
they cancel one another out? :-)

LHA

Len, this is bullcrap, and, it is boring. You have no idea to what the math
really means. Sorry, but I will not continue to give you a forum just so
that you can 'baffle 'em with bull****'. That's what you are doing. I doubt
that you even have a BSEE, otherwise you would not have to cite the author
when explaining what is common electrical engineering text book material
(that you have misinterpreted).

de KR4AJ

In article , "Dan Finn"
writes:


Len, this is bullcrap, and, it is boring.

Heh, I thought you'd say something like that. :-)

You have no idea to what the math really means.

Yes I do and so did my instructor over 30 years ago. Did a verbatim
oral on the whiteboard to prove the transformation from the simple AM
signal equation to the one that had the three terms for Carrier, Lower
and Upper Sideband components.

Not only that, but the next task on that oral exam was to show how
mixing the AM signal in two parallel paths with quadrature-phased
RF sources would allow selective demodulation of the sidebands.

Sorry, but I will not continue to give you a forum just so
that you can 'baffle 'em with bull****'.

It's not "bull****." Just a bit of algebra which can translate to radio
hardware values.

That's what you are doing. I doubt
that you even have a BSEE, otherwise you would not have to cite the author
when explaining what is common electrical engineering text book material
(that you have misinterpreted).

It is normal practice among ALL technical people to reference sources.
That includes PhDs, Nobel Laureates, Masters, and Bachelors.

Nothing was "misinterpreted," Finn. Nothing at all.

The only difference between what I wrote and the original work is the
inability of the Internet medium to allow Greek symbols and, especially,
subscripts and superscripts.

I can carry this a LOT farther just by spending some more typing time.

Harmonic content due to distortion in a not-quite-linear amplifier is one
huge area. That can be combined with the basic AM equation to show
how and how much harmonic content is generated. Did that in the early
1970s for RCA Corporation as part of the special FCC license
documentation required for 1.6 GHz SECANT avionics transmission
(a corporate R&D program).

I COULD do all that typing, but it would be to little avail. You would
simply say "it's a lot of bull****" and dismiss it without mentioning a
single bit of the mathematical explanation. You would probably continue
to do the gratuitous personal denigration...again without considering the
subject matter or anything under real discussion.

Meanwhile, I'm still waiting to hear your erudite explantaion that was
something about "phase changing will cancel all distortion." :-)

LHA

(Len Over 21) wrote in message ...
In article , "Dan Finn"
writes:

Oh God...Earth to Len....harmonics *are* phase shifted. If you reduce the
phase shift to zero, then you have eliminated the harmonics, stupid.

I'm not sure where you get this "phase shift reduction to zero" but it
isn't quite like that. Sigh. Read following -

OK, let's get down to basics. From "Electronic Designers' Handbook,"
by Robert W. Landee, Donovan C. Davis, Albert P. Albrecht, 1957,
McGraw-Hill Book Co., Chapter 5, here's the instantaneous voltage
at any point in time of an AM RF waveform:

e(t) = E [1+M Cos(omega_m t)] Sin (omega_c t)

Rest deleted. Interesting refresher, but not related to Amater
Radio POLICY, a subject which you, Sir Putzmaster, have been pushing
down others throats lately.

Not that I expected anything different from you. You can't help
yourself.

Steve, K4YZ

(Steve Robeson, K4CAP) wrote in message . com...
(Len Over 21) wrote in message ...
In article , "Dan Finn"
writes:

Oh God...Earth to Len....harmonics *are* phase shifted. If you reduce the
phase shift to zero, then you have eliminated the harmonics, stupid.

I'm not sure where you get this "phase shift reduction to zero" but it
isn't quite like that. Sigh. Read following -

OK, let's get down to basics. From "Electronic Designers' Handbook,"
by Robert W. Landee, Donovan C. Davis, Albert P. Albrecht, 1957,
McGraw-Hill Book Co., Chapter 5, here's the instantaneous voltage
at any point in time of an AM RF waveform:

e(t) = E [1+M Cos(omega_m t)] Sin (omega_c t)

Rest deleted. Interesting refresher, but not related to Amater
Radio POLICY, a subject which you, Sir Putzmaster, have been pushing
down others throats lately.

Len's just been politely tapping you'se guys with the back of the blade.