On May 24, 6:58*pm, Richard Clark wrote:
On Mon, 24 May 2010 07:06:44 -0700 (PDT), Keith Dysart

wrote:
I have often wondered if the manufacturer's
tuning
procedures had anything to do with maximizing output power transfer,
or
were they, in fact, optimizing some other aspect.

This resolves quickly in measurement - no need to wonder unless it
offers some secondary benefit of not measuring things. *

An alternative is to simply examine conventional design
considerations. *One can add to Plate current by throwing a lot of
power into the grid. *More plate current yields more output power
results, but grid lifetime plumments.

One can do innumerable things to force an artificial outcome that
strains to prove a distorted logic. *Examining a suite of sources, in
initial conditions that are average for their application quickly
reveals a common design paradigm.

******

The fundamental answer to your question is the manufacturer ultimately
designs for market domination, or maximum investment return (the two
don't necessarily converge). *Thus the marketplace gives us a spectrum
of choice and the norm of the distribution reveals cautious design
that has its eye on a value exchange expressed in money. *THAT is the
only optimization you can expect = in an honest barter, you get what
you pay for.

73's
Richard Clark, KB7QHC

You have gone to a bit higher level than I intended with my question
and
I agree with you conclusions at that level. But my question was more
basic.

When designing the filter for a PA, among other things, one uses the
desired load to be applied to the tube and the disired load impedance
to be supported and selects filter components to perform the desired
transformation.

When operating the radio, the operator has meters that measure some
values, some knobs that control some component values and a procedure
for adjusting these knobs.

It is not at all obvious what exactly the result of performing the
procedure is. Does it result in the same load being applied to the
tube that was computed by the designer? There are some hints that
the procedure will result in the load applied to the tube being
real, but beyond that, what exactly are the circuit conditions
that result?

....Keith

On Mon, 24 May 2010 16:23:26 -0700 (PDT), Keith Dysart
wrote:

It is not at all obvious what exactly the result of performing the
procedure is. Does it result in the same load being applied to the
tube that was computed by the designer?

Hi Keith,

By and large, Yes.

There are some hints that
the procedure will result in the load applied to the tube being
real, but beyond that, what exactly are the circuit conditions
that result?

I am a little lost on that. The load applied is the load applied
(sorry for the Zen). If you mean that the load is transformed by
tuning to a real R for the Plate to see, then, yes, that is operative.

However, that is not the end of it. That R is seen as the loss of a
now-poorer Q for the Plate tank. This is the distinction between
loaded and unloaded Q. The Plate tank Q expressed in terms of loaded
Q, to be effective, is quite low in comparison to its unloaded value.
This value of loaded Q is roughly between 10 and 20 where the
components in isolation (unloaded) could easily achieve 10 to 30 times
that.

The term "loaded" includes BOTH the plate and the applied load
(whatever is presented to the antenna connection). The only time the
unloaded Q of the Plate tank is at peak value is when it is sitting in
isolation from the chassis, circuitry, and even mounts - which means
it is not very useful in that configuration, except as a trophy. Many
silver plate their tanks as trophies (because this rarely results in
better operation).

Now, let's return to my statement about what Q is "effective" AND that
it measures out at roughly 10 to 20. This is straight out of Terman
if you need a citation. As for explanation (also found in Terman),
you have to consider that the Plate tank is the gate-keeper (as well
as transformer of Z) of power. If you have too high a Q, the power is
not getting THROUGH the tank as it must, and necessarily it remains in
the tank (as energy, albeit).

Consider further that ALL resonant circuits can be cast from series
circuits to parallel circuits or parallel to series (a fact lost on
some inventors of antennas). To describe the Plate tank in series
terms as I do, then the plate resistance and load resistance combine
in series through a simple circular path through ground. There are
parallel tank designs where the resistances combine in parallel. The
net result is the same insofar as Q is concerned.

Consult Terman if that is confusing. No doubt others will either more
clearly cite him, or add to the confusion.

73's
Richard Clark, KB7QHC

Hello Keith,

Thank you for your response. I’m starting my answer to your statements
by first quoting from one of your posts:

“Try as I might, I have not been able to derive a mechanism to explain
the observations in Reflections. But the explanations offered in
Reflections require large chunks of linear circuit theory to be
discarded, so this does not seem to be an appropriate path.”

That you have been unable to derive a mechanism that explains the
action in an RF power amplifier is evidence that you do not understand
it. So let’s examine the action that follows an appropriate path that
does not require any linear circuitry to be discarded. Further
evidence that you do not understand it is that you used a bench power
supply to describe the action, which you state has an infinite source
resistance when the load exceeds 50 ohm, and zero source resistance
when the load is less than 50 ohms. Unfortunately, this power supply
in no way resembles an RF power amplifier, either in components or
action.

We’ll begin by stipulating that the ‘filter’ is a pi-network tank
circuit, having a tuning capacitor at the input and a loading-
adjustment capacitor at the output. We’ll also stipulate that the
plate voltage and the grid bias are set to provide the desired
conditions at the input of the tank circuit, which means that the
desired grid voltage is that which results in the desired conduction
time for the applied plate voltage. The result provides a dynamic
resistance RL, which is determined by the average plate voltage VPavg
and the average plate current IPavg appearing at the terminals leading
to the input of the tank circuit. In other words, RL = VPavg/IPavg.
To permit delivery of all available power to be delivered by the
dynamic resistance RL, we want the input impedance appearing at the
input of the tank circuit to be equal to RL.

We’ll now go to the output of the tank circuit. We’ll assume the load
to be the input of a transmission line on which there are reflections.
The result is that the input to the line contains a real component R
and a reactance jX. The output terminals of the tank circuit are the
two terminals of the output-loading capacitor. When the line is
connected to the output terminals of the tank circuit the reactance
appearing at the line input is reflected into the tank circuit. This
reactance is then cancelled by the tuning capacitor at the input of
the tank circuit, resulting in a resonant tank circuit. We now need to
adjust the output-loading capacitor to apply the correct voltage
across the input of the transmission line so that the real component R
appearing at the line input is reflected into the tank circuit such
that the resistance RL appears at the input of the tank circuit, thus
allowing all the available power to enter the tank circuit. In other
words, adjusting the loading capacitor to deliver all the available
power into the line also makes the output resistance of the tank
circuit equal to the real component R appearing at the line input.
With any other value of output resistance of the source, all the
available power would not be delivered to the line. A corollary to
that condition follows from the Maximum Power Transfer Theorem that
for a given output resistance of the source (the tank circuit), if the
load resistance is either increased for decreased from the value of
the source resistance, the delivery of power will decrease. This
condition also accurately describes the condition for the conjugate
match.
Keep in mind that the input impedance of the line is complex, or
reactive, but the reactance of the correctly-adjusted tuning capacitor
has introduced the correct amount of the opposite reactance to cancel
the reactance appearing at the line input. Thus the line input
impedance is R + jX and the output impedance of the source is R – jX,
providing the conjugate match.

You stated in one of your posts that the phase of the reflected wave
in relation to that of the source wave results in a non-linear
condition. This is totally untrue. The tuning action of the input
capacitor in the tank circuit that cancels the line reactance caused
by the reflection on the line in no way introduces any non-linearity
in the circuit, and the condition in the vicinity of the output of the
tank circuit is totally linear. Thus, circuit theorems that require
linearity to be valid are completely valid when used with the RF power
amplifier as described above. This applies to all RF power amplifiers,
Class A, AB, B and C.

I hope my comments above assist in understanding the action that
occurs in RF power amplifiers.

Walt Maxwell, W@DU

On May 25, 4:20 pm, walt wrote:
Hello Keith,

Thank you for your response. I’m starting my answer to your statements
by first quoting from one of your posts:

“Try as I might, I have not been able to derive a mechanism to explain
the observations in Reflections. But the explanations offered in
Reflections require large chunks of linear circuit theory to be
discarded, so this does not seem to be an appropriate path.”

That you have been unable to derive a mechanism that explains the
action in an RF power amplifier is evidence that you do not understand
it. So let’s examine the action that follows an appropriate path that
does not require any linear circuitry to be discarded. Further
evidence that you do not understand it is that you used a bench power
supply to describe the action, which you state has an infinite source
resistance when the load exceeds 50 ohm, and zero source resistance
when the load is less than 50 ohms. Unfortunately, this power supply
in no way resembles an RF power amplifier, either in components or
action.

“No way” is a bit strong. The RF PA is constructed from a constant
voltage source (the power supply) and a constant current controller
(the tube), both aspects present in the bench supply example
previously
offered. A tube is often modelled as an ideal variable constant
current
source but unlike an ideal source, which can produce whatever voltage
is needed to drive the current, the current produced by the tube is
limited by the power supply voltage. Thus, assertions of linear
behaviour need to be tempered by ensuring that such voltage limits
are not exceeded.

We’ll begin by stipulating that the ‘filter’ is a pi-network tank
circuit, having a tuning capacitor at the input and a loading-
adjustment capacitor at the output. We’ll also stipulate that the
plate voltage and the grid bias are set to provide the desired
conditions at the input of the tank circuit, which means that the
desired grid voltage is that which results in the desired conduction
time for the applied plate voltage. The result provides a dynamic
resistance RL, which is determined by the average plate voltage VPavg
and the average plate current IPavg appearing at the terminals leading
to the input of the tank circuit. In other words, RL = VPavg/IPavg.
To permit delivery of all available power to be delivered by the
dynamic resistance RL, we want the input impedance appearing at the
input of the tank circuit to be equal to RL.

Most references use Vpeak and Ipeak, though they are usually related
to average values with constants of proportionality so the computed
RL will be the same. None-the-less, I prefer peak values since it
ties better to the choices made in the design.

The power that can be controlled by a control device (be it a switch,
tube or transistor) is related to device limitations. So, for example,
the maximum power that can be controlled by a 250V 1A switch is 250W.
This occurs with a supply voltage of 250V and a load of 250 ohms.
Increasing the supply voltage exceeds the switch capability as does
reducing the load resistance.

If the supply voltage is less than 250V then the maximum power occurs
with a load that causes 1A to flow and is now a limit based on
circuit choices and device capabilities.

Note that these power limits have nothing to do with maximum power
transfer in a linear circuit.

Similarly in a tube circuit, the maximum power is limited by the
supply
voltage and the tube drive level (which sets the current that will
flow in the tube). Maximum controlled power is then V*I and occurs
with a load resistance of V/I. Increasing the load resistance reduces
the power because there is insufficient voltage from the supply to
drive more current through the load and reducing the load resistance
reduces the power because less voltage is impressed across the load.

Note that neither of these effects is related to the maximum power
transfer in a linear circuit.

We’ll now go to the output of the tank circuit. We’ll assume the load
to be the input of a transmission line on which there are reflections.
The result is that the input to the line contains a real component R
and a reactance jX. The output terminals of the tank circuit are the
two terminals of the output-loading capacitor. When the line is
connected to the output terminals of the tank circuit the reactance
appearing at the line input is reflected into the tank circuit. This
reactance is then cancelled by the tuning capacitor at the input of
the tank circuit, resulting in a resonant tank circuit. We now need to
adjust the output-loading capacitor to apply the correct voltage
across the input of the transmission line so that the real component R
appearing at the line input is reflected into the tank circuit such
that the resistance RL appears at the input of the tank circuit, thus
allowing all the available power to enter the tank circuit. In other
words, adjusting the loading capacitor to deliver all the available
power into the line also makes the output resistance of the tank
circuit equal to the real component R appearing at the line input.
With any other value of output resistance of the source, all the
available power would not be delivered to the line. A corollary to
that condition follows from the Maximum Power Transfer Theorem that
for a given output resistance of the source (the tank circuit), if the
load resistance is either increased for decreased from the value of
the source resistance, the delivery of power will decrease. This
condition also accurately describes the condition for the conjugate
match.

While a conjugate match does result in a situation where altering
the load will reduce the power transfer, it is not true that any
situation where altering the load reduces the power transfer is
also a conjugate match. The two examples above (bench power supply,
tube in a circuit) amply demonstrate this.

Keep in mind that the input impedance of the line is complex, or
reactive, but the reactance of the correctly-adjusted tuning capacitor
has introduced the correct amount of the opposite reactance to cancel
the reactance appearing at the line input. Thus the line input
impedance is R + jX and the output impedance of the source is R – jX,
providing the conjugate match.

This is quite in error, unless, by happenstance, RL is equal to Rp
(plus
the other contributors to source impedance).

You stated in one of your posts that the phase of the reflected wave
in relation to that of the source wave results in a non-linear
condition. This is totally untrue. The tuning action of the input
capacitor in the tank circuit that cancels the line reactance caused
by the reflection on the line in no way introduces any non-linearity
in the circuit, and the condition in the vicinity of the output of the
tank circuit is totally linear. Thus, circuit theorems that require
linearity to be valid are completely valid when used with the RF power
amplifier as described above. This applies to all RF power amplifiers,
Class A, AB, B and C.

For any circuit with a conduction angle of less than 360 degrees, my
simulations indicate otherwise. The reflection coefficient experienced
by the reflected wave when it arrives at the amplifier output varies
with the phase of the reflected wave. Since the reflection coefficient
is a function of source impedance and line impedance, and the line
impedance is not changing, this means that the source impedance is
changing with the phase of the reflected wave. This is not a behaviour
that is consistent with a linear circuit. Given the non-linearities
in a circuit with a conduction angle of less than 360 degrees, this
should not be a surprise. More, it would be a surprise if such a
circuit did behave as a linear circuit.

I hope my comments above assist in understanding the action that
occurs in RF power amplifiers.

Thank you. They have indeed helped clarify my thinking.

We are still left with the puzzle of why the observations documented
in Reflections report a reduction in power transfer when the load is
changed in either direction. It seems unlikely that RL is, by
happenstance, equal to Rp, which would be one explanation.

It seems plausible that it is related to the behaviours associated
with the examples I provided above, but I can not articulate a
mechanism that satisfies.

…Keith

On Wed, 26 May 2010 04:50:30 -0700 (PDT), Keith Dysart
wrote:

a conjugate match does result in a situation where altering
the load will reduce the power transfer

....

We are still left with the puzzle of why the observations documented
in Reflections report a reduction in power transfer when the load is
changed in either direction.

Hi Keith,

Stripping away everything that you offer as objections to what is not
in Walt's premise (I cannot vouch for his attempts to explain the
universality of it), your statements come into conflict.

If you offer you find a puzzle about measurements, then that is simply
researched at the bench instead of in expansive wanderings in myriad
qualifications. Do you have documented measurements under initial
conditions identical to Walt's that run counter to Walt's quantitative
results?

I suspect not, or we would be talking about competing bench results
instead. This would be a more productive and genuine debate seeking
explanation for what you describe as the "puzzle."

Barring quantitative evidence, anything that continues this rag-chew
is a simple example of "modeling is doomed to succeed."

73's
Richard Clark, KB7QHC

On May 26, 11:50*am, Richard Clark wrote:
On Wed, 26 May 2010 04:50:30 -0700 (PDT), Keith Dysart

wrote:
a conjugate match does result in a situation where altering
the load will reduce the power transfer

...

We are still left with the puzzle of why the observations documented
in Reflections report a reduction in power transfer when the load is
changed in either direction.

Hi Keith,

Stripping away everything that you offer as objections to what is not
in Walt's premise (I cannot vouch for his attempts to explain the
universality of it), your statements come into conflict.

If you offer you find a puzzle about measurements, then that is simply
researched at the bench instead of in expansive wanderings in myriad
qualifications. *Do you have documented measurements under initial
conditions identical to Walt's that run counter to Walt's quantitative
results?

I suspect not, or we would be talking about competing bench results
instead. *This would be a more productive and genuine debate seeking
explanation for what you describe as the "puzzle." *

Barring quantitative evidence, anything that continues this rag-chew
is a simple example of "modeling is doomed to succeed."

73's
Richard Clark, KB7QHC

Hi Keith,

Sorry, OM, but you still misunderstand various aspects of RF power amp
operation.

First, the power supply is not the limiting factor concerning plate
current. The grid drive is what determines the plate current, and thus
the output power.

Second, the tank circuit is an energy storage device that isolates the
non-linear input from the linear output. That the output is linear is
because the voltage and current are in phase at the output of the tank
circuit. The effect of the energy storage of the tank results in the
tank becoming the source of the energy appearing at the output.

Third, the action of plate resistance Rp occurs only in the formation
of RL, and has no further effect on any action downstream of the input
of the tank circuit. Thus, it has no bearing on the development of the
conjugate match that occurs at the junction of the tank output and the
input of the transmission line.

Fourth, as I said earlier, the the action of the bench power supply
that you presented in no way models the action of the RF power
amplifier. Furthermore, you are incorrect when you say that when
varying the load in either direction causing the power deliver to
decrease there is no conjugate match. In saying what you did violates
the theorem of Maximum Transfer of Power.

Fifth, as I stated earlier, when the reactance appearing at the input
of the load (the transmission line with reflections) is canceled by
the opposite reactance introduced by the pi-network tuning capacitor,
the output impedance of the source (the tank circuit) is the conjugate
of the line-input impedance. If you cannot accept this as fact you
have a problem.

Sixth, your understanding of the effect of the reflected wave on the
source wave is flawed. The non-linearity of the plate current when the
conduction time is less than 360° has no relation to the action
downstream of the input to the tank circuit, because from that point
on the voltage current relationship is linear. If you cannot accept
this as fact you have still another problem.

Seventh, your belief that because there is a conjugate match at the
output of the tank there must be a conjugate match at the input of the
tank is also not true. The effect of the energy storage in the tank
isolates the non-linearity af the input from the linear operation at
the output, permitting a conjugate match at the output, while not
allowing it to occur at the input.

These seven comments are born out (proven) by the results of many
measurements I made using laboratory grade instruments, HP and General
Radio. If you check my record as a professional electrical engineer
regarding the measurements I've made that led to successful hardware
flying on various Earth-orbiting platforms, you must accept the
validity of the measurements I made on RF power amplifiers that prove
my position.

As I said earlier, no one but you has considered my position on this
subject incorrect. Therefore, if you cannot agree with the comments I
made above, but still consider my statements in Reflections flawed,
then there is no point in my making any further comments. I hope
someday you'll finally understand what's really happening within the
RF amplifier.

Walt Maxwell, W2DU

On May 26, 1:42 pm, walt wrote:
Hi Keith,

Sorry, OM, but you still misunderstand various aspects of RF power amp
operation.

First, the power supply is not the limiting factor concerning plate
current. The grid drive is what determines the plate current, and thus
the output power.

Of course the grid drive is one of the factors which controls the
current flowing in the load.

But the power supply is also one of the limiting factors. Reducing
the power supply voltage below that which is necessary to cause the
desired current to flow in the load will reduce the power output.
Similarly, increasing the load resistance will eventually raise it to
the point where the voltage is no longer adequate to cause the desired
current to flow.

Second, the tank circuit is an energy storage device that isolates the
non-linear input from the linear output. That the output is linear is
because the voltage and current are in phase at the output of the tank
circuit.

Can one not have a linear circuit where the current and voltage are
not
in phase? Also, if one loads any tank circuit with a resistance, the
output current and voltage will be in phase and if it is loaded with
a
reactance, they won’t be.

The effect of the energy storage of the tank results in the
tank becoming the source of the energy appearing at the output.

Yes, but that does not make the output independent of the input.

Third, the action of plate resistance Rp occurs only in the formation
of RL, and has no further effect on any action downstream of the input
of the tank circuit. Thus, it has no bearing on the development of the
conjugate match that occurs at the junction of the tank output and the
input of the transmission line.

I do not understand what is being said here.

Fourth, as I said earlier, the the action of the bench power supply
that you presented in no way models the action of the RF power
amplifier. Furthermore, you are incorrect when you say that when
varying the load in either direction causing the power deliver to
decrease there is no conjugate match. In saying what you did violates
the theorem of Maximum Transfer of Power.

The definition I use for conjugate match is one where the source
impedance is the complex conjugate of the load impedance. When this
situation occurs between linear networks, maximum power is transferred
between the networks.

None-the-less, just because maximum power is being transferred between
two networks does not mean they are complex conjugates of each other.
This is demonstrated with the non-linear behaviour of the bench power
supply example. Maximum power is transferred but the source and load
impedance are not complex conjugates.

Fifth, as I stated earlier, when the reactance appearing at the input
of the load (the transmission line with reflections) is canceled by
the opposite reactance introduced by the pi-network tuning capacitor,
the output impedance of the source (the tank circuit) is the conjugate
of the line-input impedance. If you cannot accept this as fact you
have a problem.

Perhaps I am not computing the impedances correctly. Let us see if I
have done so for the following example.

Consider a generator constructed of current source in parallel with
a resistor, driving a PI network, connected to a load.

generator filter load
6.945uH
+-------+------- ----+---/\/\/\/---+---- ---+
| | | |1.398 |
+---+ \ | | nF \
3.75 | I | /8000 ----- ----- / 50 ohm
MHz | | \ ----- ----- \
+---+ / |295.5 | /
| | | pF | |
+-------+------- ----+-------------+---- ---+

Looking into the input of the filter, then impedance is 1500 ohms.
This is the load applied to the generator and is computed by
applying the rules for series and parallel components to
the 50 ohm load, and the two capacitors and inductor in the PI
network.

It is, I hope, generally accepted that the generator will have an
output impedance of 8000 ohms.
The output impedance of the filter is computed by applying the
rules for series and parallel components to the 8000 ohm
generator impedance and the 3 components in the filter.
The result is 58.00 /_ 68.60 ohms.

Note that the component values were taken from a PA design where
the desired load for the tube was 1500 ohms. And 8000 is not an
unreasonable slope for the plate E-I curve of a tube.

This has not resulted in a conjugate match.

Sixth, your understanding of the effect of the reflected wave on the
source wave is flawed. The non-linearity of the plate current when the
conduction time is less than 360° has no relation to the action
downstream of the input to the tank circuit, because from that point
on the voltage current relationship is linear. If you cannot accept
this as fact you have still another problem.

It is, perhaps, this claim of isolation that is most strange. It seems
quite at odds with the rules for connected networks.

Seventh, your belief that because there is a conjugate match at the
output of the tank there must be a conjugate match at the input of the
tank is also not true. The effect of the energy storage in the tank
isolates the non-linearity af the input from the linear operation at
the output, permitting a conjugate match at the output, while not
allowing it to occur at the input.

It was my understanding that in a sequence of connected linear
networks, if any connection exhibited a conjugate match, then they
all were conjugately matched. Is this not correct?
Are you saying that if a conjugate match is present between the line
and the antenna, it might not be present between the transmitter and
the line?

These seven comments are born out (proven) by the results of many
measurements I made using laboratory grade instruments, HP and General
Radio. If you check my record as a professional electrical engineer
regarding the measurements I've made that led to successful hardware
flying on various Earth-orbiting platforms, you must accept the
validity of the measurements I made on RF power amplifiers that prove
my position.

I quite believe your measurements. It is the conclusion that they
prove a conjugate match that I find impossible to accept. Both
because there are other situations that can lead to power behaviours
that may appear similar to the power behaviour of a conjugate match
and the method proposed for computing source impedance is quite at
odds with linear theory.

But the quality of the measurements suggest it is worthwhile to
explore other explanations.

....Keith

On May 26, 8:53*pm, Keith Dysart wrote:
It was my understanding that in a sequence of connected linear
networks, if any connection exhibited a conjugate match, then they
all were conjugately matched. Is this not correct?

The theorem requires linear *lossless* networks which do not exist in
reality, i.e. networks containing only reactances. Therefore an
*ideal* system-wide conjugate match cannot exist in reality just as a
lossless transmission line cannot exist in reality. In low-loss
systems, we can only achieve a system-wide near-conjugate match with
an ideal conjugate match existing at one point, e.g. the Z0-match
point where reflected energy flowing toward the source is eliminated.

Are you saying that if a conjugate match is present between the line
and the antenna, it might not be present between the transmitter and
the line?

Yes, speaking for me, in the real world, it is easy to prove that the
system-wide impedance looking in one direction is not always exactly
the conjugate of the impedance looking in the other direction. Thus
the "maximum power transfer" assertion has to be modified to "maximum
*available power* transfer". In the real world, ohmic and dielectric
losses reduce the power available to be delivered to the load.

It's easy to see. Let's say we have a completely flat 50 ohm system;
50 ohm source, 50 ohm coaxial feedline, and 50 ohm antenna. Now assume
we install an antenna tuner between the source and the feedline that
exhibits some series impedance and we adjust the tuner such that the
source sees 50 ohms. At the output of the tuner looking toward the
antenna, we will see 50 ohms. Looking back through the tuner toward
the source, we will see the tuner impedance in series with 50 ohms.
That proves it is not an *ideal* (lossless) conjugate match although
it may be considered to be a near-conjugate match, as close as we can
come in the real world.

What I don't know is how close a real-world conjugate match has to be
to an ideal lossless conjugate to be called a "conjugate match". A
purist might argue that an ideal conjugate match cannot exist in
reality. A realist might argue that if we are within 10% of an ideal
conjugate match, then it is a real-world conjugate match, by
definition.

Note that I am not speaking for Walt here, just for myself.
--
73, Cecil, w5dxp.com