This is a really dumb question but it dawned on me that I did not know
the correct answer.

In terms of old transmitters from the 20s/30s...In a crystal oscillator
I understand the concept of setting the oscillator output tank to
favor the harmonic from the crystal. (Stop me if I'm wrong already...)

But in a doubling amplifier stage am I counting on having enough
harmonic content at the input or am I creating the harmonic with the
non-linearity of the amplifier?

TIA
-Bill WX4A

"exray" wrote in message
...
In terms of old transmitters from the 20s/30s...In a crystal oscillator I
understand the concept of setting the oscillator output tank to favor the
harmonic from the crystal. (Stop me if I'm wrong already...)

But in a doubling amplifier stage am I counting on having enough harmonic
content at the input or am I creating the harmonic with the non-linearity
of the amplifier?

TIA
-Bill WX4A

Hi Bill.

Remember that single ended frequency multiplier stages are usually operated
in Class-C where the nonlinear operation of the stage produces the
harmonics. In a Class-C stage, the grid (of the tube since we are talking
about vintage transmitters) is biased such that the plate current only flows
in short pulses. The narrower the pulse width, the greater the harmonic
generation of the stage. If you look in the old RCA Transmitting Tube
Manual, there is a design procedure where the "conduction angle" of the tube
is chosen for proper harmonic generation.

Frequency doubling is unique in that two Class-B stages may be used in a
push-push arrangement. Here the grids are driven in push-pull while the
plates are connected together in parallel. The resulting waveform will
essentially be the equivalent of full-wave rectification of the input
signal. Without going into Fourier series, the resultant waveform only
contains even harmonics of the input signal while the fundamental driving
frequency is cancelled out.

Fortunately I was already a ham operator when my high school math class
taught Fourier series *. I immediately saw the practical value of this
mathematical concept and it made good sense to me. Your question is a good
one and reading some of the tutorials on Fourier series (do a Google search)
will be very useful to your understanding of harmonic generation and
intermodulation distortion. I hope that my simple explanation will start
you in your own exploration.

73, Barry L. Ornitz WA4VZQ

* More years ago that I care to admit! :-)

NoSPAM wrote:


Remember that single ended frequency multiplier stages are usually operated
in Class-C where the nonlinear operation of the stage produces the
harmonics. In a Class-C stage, the grid (of the tube since we are talking
about vintage transmitters) is biased such that the plate current only flows
in short pulses.

Thanks Barry, I get it. I was becoming distracted by some of the old
1930s articles touting the tritet osc for its ability to create more
harmonic output (true) for directly driving following stages.

I suppose thats just another way of reaching the same goal.

(btw, haven't heard from you in years - I recall your great input over
on r.a.r. +p. No, it hasn't changed )

-Bill

On Nov 21, 8:24*pm, "NoSPAM" wrote:
"exray" wrote in message

...

*In terms of old transmitters from the 20s/30s...In a crystal oscillator I
understand the concept of setting the oscillator output tank to favor the
harmonic from the crystal. *(Stop me if I'm wrong already...)

*But in a doubling amplifier stage am I counting on having enough harmonic
content at the input or am I creating the harmonic with the non-linearity
of the amplifier?

*TIA
*-Bill *WX4A

Hi Bill.

Remember that single ended frequency multiplier stages are usually operated
in Class-C where the nonlinear operation of the stage produces the
harmonics. *In a Class-C stage, the grid (of the tube since we are talking
about vintage transmitters) is biased such that the plate current only flows
in short pulses. *The narrower the pulse width, the greater the harmonic
generation of the stage. *If you look in the old RCA Transmitting Tube
Manual, there is a design procedure where the "conduction angle" of the tube
is chosen for proper harmonic generation.

Frequency doubling is unique in that two Class-B stages may be used in a
push-push arrangement. *Here the grids are driven in push-pull while the
plates are connected together in parallel. *The resulting waveform will
essentially be the equivalent of full-wave rectification of the input
signal. *Without going into Fourier series, the resultant waveform only
contains even harmonics of the input signal while the fundamental driving
frequency is cancelled out.

Fortunately I was already a ham operator when my high school math class
taught Fourier series *. *I immediately saw the practical value of this
mathematical concept and it made good sense to me. *Your question is a good
one and reading some of the tutorials on Fourier series (do a Google search)
will be very useful to your understanding of harmonic generation and
intermodulation distortion. *I hope that my simple explanation will start
you in your own exploration.

* * 73, *Barry L. Ornitz * WA4VZQ

* More years ago that I care to admit! *:-)

Wow, there's a name I haven't seen for a while. I must be frequenting
the wrong groups. I was just thinking about you a couple days ago.
Hi Barry!

More about what Barry wrote: in the limit as the conduction angle
goes to zero and you generate a very narrow pulse of current, the
harmonics end up all the same amplitude. That's for an impulse of
zero width. Unfortunately, given limited amplitude of the current in
that very narrow pulse, the total energy becomes small. As you widen
the pulse, you'll see that the "comb" of harmonics no longer has
constant amplitude, but the amplitudes as you go up the "comb" (higher
in frequency)drop, and there will be a frequency at which they go to
zero, and then increase again (and go to zero again, and increase
again). The magnitudes follow a sin(x)/x shape, for perfectly
rectangular pulses. This becomes interesting for a the design of
frequency multiplier stages: if for example you want to get x4 out of
a stage you better NOT run it at a conduction angle that results in
nulling of the fourth harmonic! I think I was bit by this a time or
two in my youth when I didn't understand this. (I'm working on
something right now where I want that comb of harmonics to be all very
nearly equal amplitude up to about 100MHz, and that tells me how
narrow the pulse must be.)

Cheers,
Tom

On Fri, 21 Nov 2008 22:30:24 -0400, exray
wrote:

This is a really dumb question but it dawned on me that I did not know
the correct answer.

In terms of old transmitters from the 20s/30s...In a crystal oscillator
I understand the concept of setting the oscillator output tank to
favor the harmonic from the crystal. (Stop me if I'm wrong already...)

In an overtone oscillator, the resonator actually forces the crystal
to mechanically resonate at 1/3, 1/5, 1/7, 1/9 etc. of the crystal
width. Due to the end effects, the frequency is *not* _exactly_ 3, 5,
7, 9 etc. times the fundamental frequency, but quite close. In
principle, the oscillator is producing a single frequency, the
(harmonic) mechanical resonance frequency of the crystal.

But in a doubling amplifier stage am I counting on having enough
harmonic content at the input or am I creating the harmonic with the
non-linearity of the amplifier?

The non-linearity of the stage will produce the harmonics, which are
_exact_ integer multipliers of the input frequency. Symmetrically
clipping stages generate strong odd harmonics, while asymmetric
stages create strong even harmonics. The following stages need to
filter out the desired harmonics.

So if you need exactly 30.000... MHz, you either have to use a
10.000... MHz fundamental crystal oscillator followed by a tripler
(and filtering stage) or order an _overtone_ crystal for exactly
30.000... MHz.

Running a nominally 10.000... MHz fundamental mode crystal in an
overtone oscillator tuned at 30 MHz will not produce exactly 30.000...
MHz but something quite closely, due to the end effect.

Paul OH3LWR

On Fri, 21 Nov 2008, exray wrote:

Date: Fri, 21 Nov 2008 22:30:24 -0400
From: exray
Newsgroups: rec.radio.amateur.homebrew
Subject: Doubling

This is a really dumb question but it dawned on me that I did not know the
correct answer.

In terms of old transmitters from the 20s/30s...In a crystal oscillator I
understand the concept of setting the oscillator output tank to favor the
harmonic from the crystal. (Stop me if I'm wrong already...)

I think this is correct, but the books say that tuning the output of the
oscillator can "pull" the frequency of the oscillating crystal. I have
sometimes seen this.

But in a doubling amplifier stage am I counting on having enough harmonic
content at the input or am I creating the harmonic with the non-linearity of
the amplifier?

Despite what at least one other person responding to this said, I can rest
assure you that if you run a doubler/multiplier stage even in a linear
mode, AND if you tune the output of that stage to the multiple harmonic,
you will definitely get output at that harmonic frequency which is
stronger than the input drive voltage. In the last few years I have built
many tube stages and observed the harmonic voltage output on a wideband
oscilloscope. As a matter of fact if you ever get a wideband scope and
look at the locked output waveform as you tune through the both the
fundamental and the harmonic frequency you will be very surprised at what
you will see. All of the descriptions in all of the handbooks I have read
(a few) explain this from a theoretical perspective but don't bother to
actually show, with photographs of actual scope traces, how this works.
It would just take an extra page or two and would make people think about
what they are doing.

All amplifiers have some non-linearity, the question is what effect this
has on you meeting "purity" of emissions requirements. The more important
question is whether you are getting the gain/drive that you want from a
given stage of amplification. Reducing unwanted spurious emissions might
require more tuned circuits or measurement using a receive with an S-meter
and operated many wavelengths from your antenna. Most "appliance
operators" just buy a commercial rig and don't worry about anything;
homebrewers might not worry either if their signals go through a tuned
circuit, an antenna tuner, and an antenna for a narrow frequency range.

If you really want to blow your mind, then hook an oscilloscope to the
output of a mixer with two low harmonic content input sine waves to be
mixed. The raw output will look like hell on a scope. The only way to see
the mixed (say, difference) frequency will be to go through at least a
couple of tuned circuits that are tuned for the wanted sine wave
frequency.

I've done this stuff. There are a couple of other minor matters that are
not quite correct in our ham handbooks, too.


TIA
-Bill WX4A

On Nov 22, 8:43*pm, Stray Dog wrote:
Despite what at least one other person responding to this said, I can rest
assure you that if you run a doubler/multiplier stage even in a linear
mode, AND if you tune the output of that stage to the multiple harmonic,
you will definitely get output at that harmonic frequency which is
stronger than the input drive voltage.

Huh? No way... you MUST have non-linearities to make a doubler.

On Sun, 14 Dec 2008, Telstar Electronics wrote:

Date: Sun, 14 Dec 2008 08:20:56 -0800 (PST)
From: Telstar Electronics
Newsgroups: rec.radio.amateur.homebrew
Subject: Doubling

On Nov 22, 8:43*pm, Stray Dog wrote:
Despite what at least one other person responding to this said, I can rest
assure you that if you run a doubler/multiplier stage even in a linear
mode, AND if you tune the output of that stage to the multiple harmonic,
you will definitely get output at that harmonic frequency which is
stronger than the input drive voltage.

Huh? No way... you MUST have non-linearities to make a doubler.

All tubes (and transistors, etc) have non-linearities (if the transfer
characteristics are non-straight lines) if that is what you are talking
about.

However, I have observed output on a scope of second harmonics (and, yes,
the time base was set right and auto-self triggering) and the
amplifier was running no higher than Class B. You should actually try this
yourself and see for yourself. Tune the output to the second harmonic and
you will see grow out of the vally new "peaks" corresponding to that
second harmonic.

I don't know what the solid state gear is doing, but from many schematics
of the vintage tube gear I'm familiar with show, and measure, biasing for
linear operation, even in stages meant to multiply frequency.

"Telstar Electronics" wrote in message ...
On Nov 22, 8:43 pm, Stray Dog wrote:
? Despite what at least one other person responding to this said, I can rest
assure you that if you run a doubler/multiplier stage even in a linear
mode, AND if you tune the output of that stage to the multiple harmonic,
you will definitely get output at that harmonic frequency which is
stronger than the input drive voltage.

Huh? No way... you MUST have non-linearities to make a doubler.

Actually you do not need any nonlinearity to make a doubler (quadrupler, etc.).

Assume you have two Class B (or AB) stages that are driven in push-pull. The outputs are connected in parallel. And to make things even more linear, let each stage have a resistive load. Each stage will produce a linearly amplified (but inverted) version of the input signal FOR THE POSITIVE HALF of the driving waveform only. Being driven 180 degrees out of phase with the input signal, the second stage will produce a linearly amplified but (again inverted) version of the input signal FOR THE NEGATIVE HALF of the driving waveform. Both outputs will have a DC offset of the plate (collector, drain) voltage.

The resultant waveform with the outputs in parallel will look like much like a full wave rectified version of the input signal subtracted from the plate voltage. To express this mathematically, let the input signal be expressed as:

Vin = A sin(wt)

Now let the voltage gain of each stage be "-k" and the plate voltage be "B". The resultant waveform of the two stages connected in parallel will be:

Vout = B - abs[A*k sin(wt)] where "abs" is the absolute value

Vout = B - A*k sin(wt) for 0 wt Pi and
= B + A*k sin(wt) for Pi wt 2Pi or alternately for -Pi wt 0

We can then calculate the Fourier series of this waveform to determine its spectrum. I will not present the calculations here as it is too difficult to show the integration over defined integrals using only plain text (and I doubt many readers will have math fonts anyway). If you wish to see the math for the Fourier series for a number of functions, read:
http://www.maths.qmul.ac.uk/~agp/calc3/notes2.pdf or
http://www.physics.hku.hk/~phys2325/notes/chap7.doc.

Vout = B - 2*A*K/Pi * [1 - SUMMATION {2*cos(nwt)/(n*n - 1)] for n=2, 4, 6, 8...

Note that the original frequency has been eliminated and that only even order harmonics are present, and that the amplitudes drop off quite rapidly. For example, the fourth harmonic will be one fifth of the second harmonic.

For those that need a simplified explanation of Fourier series, Don Lancaster wrote a good article that can be found at:
http://www.tinaja.com/glib/muse90.pdf. I always thought Don had a ham license but I could not find one.

In a real implementation of this multiplier, a tuned circuit would be used as the plate load. The Q of this tuned circuit will assure that only the second harmonic is present in the output. The two stages would need to be well balanced if cancellation of odd harmonics and the fundamental is required.

73, Dr. Barry L. Ornitz WA4VZQ


POSTSCRIPT:

Now let me describe how it is possible to produce ONLY the second harmonic. Instead of using two Class B or AB stages, it is possible to use triodes operating where their plate current is proportional to the square of the grid voltage. Driving the two such stages in push-pull with the outputs in parallel with a resistive load, the output waveform will be:

Vout = B - A*A*k sin(wt)*sin(wt)

Using a trigonometric identity {see:
http://en.wikipedia.org/wiki/List_of_trigonometric_identities},

sin(x)*sin(x) = sin(x)^2 = 0.5[1-cos(2x)]

thus Vout = B - A*A*K/2 + A*A*k/2 cos(2wt)

This shows that only the second harmonic is found at the output.

NoSPAM wrote:
"Telstar Electronics"
wrote in message
...
On Nov 22, 8:43 pm, Stray Dog
wrote:
? Despite what at least one other person responding to this said, I
can rest
assure you that if you run a doubler/multiplier stage even in a linear
mode, AND if you tune the output of that stage to the multiple
harmonic,
you will definitely get output at that harmonic frequency which is
stronger than the input drive voltage.

Huh? No way... you MUST have non-linearities to make a doubler.

Actually you do not need any nonlinearity to make a doubler
(quadrupler, etc.).

Assume you have two Class B (or AB) stages that are driven in
push-pull. The outputs are connected in parallel. And to make things
even more linear, let each stage have a resistive load. Each stage
will produce a linearly amplified (but inverted) version of the input
signal FOR THE POSITIVE HALF of the driving waveform only. Being
driven 180 degrees out of phase with the input signal, the second
stage will produce a linearly amplified but (again inverted) version
of the input signal FOR THE NEGATIVE HALF of the driving waveform.
Both outputs will have a DC offset of the plate (collector, drain)
voltage.
Class B or even Class AB in the circuit you described are non-linear.
Try that circuit
with Class A biasing.

Bill K7NOM


The resultant waveform with the outputs in parallel will look like
much like a full wave rectified version of the input signal subtracted
from the plate voltage. To express this mathematically, let the input
signal be expressed as:

Vin = A sin(wt)

Now let the voltage gain of each stage be "-k" and the plate voltage
be "B". The resultant waveform of the two stages connected in
parallel will be:

Vout = B - abs[A*k sin(wt)] where "abs" is the absolute value

Vout = B - A*k sin(wt) for 0 wt Pi and
= B + A*k sin(wt) for Pi wt 2Pi or
alternately for -Pi wt 0

We can then calculate the Fourier series of this waveform to determine
its spectrum. I will not present the calculations here as it is too
difficult to show the integration over defined integrals using only
plain text (and I doubt many readers will have math fonts anyway). If
you wish to see the math for the Fourier series for a number of
functions, read:
http://www.maths.qmul.ac.uk/~agp/calc3/notes2.pdf
http://www.maths.qmul.ac.uk/%7Eagp/calc3/notes2.pdfhttp://www.physics.hku.hk/%7Ephys2325/notes/chap7.doc or
_http://www.physics.hku.hk/~phys2325/notes/chap7.doc
http://www.physics.hku.hk/%7Ephys2325/notes/chap7.doc._
Vout = B - 2*A*K/Pi * [1 - SUMMATION {2*cos(nwt)/(n*n - 1)] for n=2,
4, 6, 8...

Note that the original frequency has been eliminated and that only
even order harmonics are present, and that the amplitudes drop off
quite rapidly. For example, the fourth harmonic will be one fifth of
the second harmonic.

For those that need a simplified explanation of Fourier series, Don
Lancaster wrote a good article that can be found at:
http://www.tinaja.com/glib/muse90.pdf. I always thought Don had a ham
license but I could not find one.

In a real implementation of this multiplier, a tuned circuit would be
used as the plate load. The Q of this tuned circuit will assure that
only the second harmonic is present in the output. The two stages
would need to be well balanced if cancellation of odd harmonics and
the fundamental is required.

73, Dr. Barry L. Ornitz WA4VZQ


POSTSCRIPT:

Now let me describe how it is possible to produce ONLY the second
harmonic. Instead of using two Class B or AB stages, it is possible
to use triodes operating where their plate current is proportional to
the square of the grid voltage. Driving the two such stages in
push-pull with the outputs in parallel with a resistive load, the
output waveform will be:

Vout = B - A*A*k sin(wt)*sin(wt)

Using a trigonometric identity {see:
http://en.wikipedia.org/wiki/List_of_trigonometric_identities},

sin(x)*sin(x) = sin(x)^2 = 0.5[1-cos(2x)]

thus Vout = B - A*A*K/2 + A*A*k/2 cos(2wt)

This shows that only the second harmonic is found at the output.