On Tue, 27 Feb 2007 16:51:21 +0000, Ian White GM3SEK
wrote:

For that matter, almost all measuring instruments abstract some energy
or power from whatever they are measuring - but that is usually
incidental. It certainly does not make every instrument into a power
meter.

Can you not see this?

Hi Ian,

As a trained metrologist, I've seen it far closer up than you. I've
accurately measured the physics of technology out to more decimal
places than most, and there is nothing incidental about it - it is
quite fundamental. In fact I've encountered a spectrum of conflicting
and interfering physical principles to every measurement that had to
be accounted for to obtain that accuracy.

I dare say none here can equal the accuracy of my bench work at RF
Power measurement here, even those with a shelf full of digital meters
used as bookends for their library.

To wave a
hand and say NOTHING does not make it so.

To wave another hand and say traveling waves of power exist does not
make that so, either.

Your counter is not an argument for their inexistence.

The examples of "traveling waves of power" abound, even to the trade
craft of antennas. Examples of "traveling wave of power" are classic
in whole to the explanation for the operation of the standard
Directional Couplers (to distinguish them from the Bruene design).

There are more such examples, but the Directional Coupler is quite
suitable to their class, and its operation satisfies the proof of
power flow in both directions. There are other examples outside of
this class that also provide proofs; and they are based on physical
principles as well (polarization and magnetic moment).

Traditionally, these points pass in silence until some later time when
we visit them once again.

73's
Richard Clark, KB7QHC

Richard Clark wrote in
:

....
We've been through this before. No instrument operates in the absence
of power. Simply because you and Owen are graced with instruments
that demand so little, does not negate what power they do rob from
what is available. Even the humble electrometer has to overcome the
force of gravity to open its foil leaves, and climbing that potential
energy hill is work over time - power.

Hammer down the directivity as much as you want, and it will still
resolve to some diminution of power available to the load. To wave a
hand and say NOTHING does not make it so.

The fact that an instrument may consume power from the circuit does not
imply that it measures power.

For example, a "voltmeter" that samples the voltage and draws a small
current from the cicuit under test does consume some power, but it cannot
"measure" power in the circuit under test without knowledge of the
complex load impedance.

I wrote an analysis of the Breune circuit and show that the meter
deflection is a response to the Vf or Vr component, and I dealt with how
that information can be used (including applicable conditions). If you
(or others) think there are flaws in the article, I welcome constructive
feedback.

Owen

Cecil Moore wrote:
The joules/sec are real quantities but whether joules/sec
is power depends upon the definition of "power".

In our case here on the internet, it depends on whether or not you
choose to equate 'units of power' with the definition of power.

73 ac6xg

On Tue, 27 Feb 2007 21:14:38 GMT, Owen Duffy wrote:

This is not contending nor contention and is content only for a non
sequitur. The line following a tuner exhibits considerable loss (poor
efficiency) that can only occur on the basis of power and mismatch.
You yourself offered in other correspondence that it exceeds cable
attenuation specifications found only in a matching condition. To

I am being picky, but "it *may* exceed cable attenuation specifications
found only in a matching condition, it may also be lower".

Hi Owen,

Lower? That is rather astonishing in light of responding to my
comment.

If I said it
as you stated, I made an error. The common statement (and I have no doubt
made it) that VSWR exacerbates line loss is actually wrong in the general
case. (Having Googled my own web site I see one statement along those
lines which needs further qualification!)

This is even more astonishing. Irrespective of you being the source,
why inject this confusing comment? SWR always exacerbates line loss!
Give me any normal line attenuation and SWR at the load, and I will
tell you exactly how much additional loss will occur. There's a
general solution for you.

Something tells me that your comments are based on a confusion between
the power loss of a cable, and its mismatch loss. They are not the
same thing although they are usually tightly twined in discussion. You
later exhibit a confusion between a conjugate match and an impedance
match. They are not the same thing either. The confusion on both
these points have abounded in this group in past "debates."

I meant the output at the PA terminals where an lumped constant load
would be attached for comparison.

This then removes the reflection from the argument, doesn't it? It
actually doesn't; but this unwarranted substitution is like Zen
Archery in that the line already demonstrates the validity of
reflected power as distinct from that "power" just being a
mathematical fiction.

Putting the lumped load at the PA terminal merely casts the proof back
into the box, it doesn't negate reflected power. As the proof is
already supported in the line, then removing it is not strictly a
valid counter argument. However, we will explore it further:

PAs can be designed to behave as an equivalent fixed voltage or current
source with fixed source impedance of Zo, but HF PAs are not usually
designed in that way.

OK so we are now in my sidebar of source resistance. Even so, it has
nothing to do with the concept of reflected power except insofar as
that resistance's ability to reveal that power's dissipation.

Other's should ponder how the reflected power has a caloric proof in
the line, and then question why it wouldn't prove out when it arrives
back in the box where the temperature rises on its return.

Same source, same power, same reflection, same loss. The only thing
that varies is the capacity of any point along this signal chain to
support that heat burden. Let's skip these as choices of design.

I know that there is a vein of thought that the process of adjusting a PA
for maximum output always, somewhat magically, creates a match condition
where the source impedance is the conjugate of the load at the PA
terminals, but it is contentious.

That contention arises out of mistaking Z0 Matches with Conjugate
Matches. This is a common affliction among "debaters" here. Let's
skip their prejudices.

What of broadband PA designs with no
such adjustment, are they source matched over a broad range of
frequencies?

Having had designed broadband amps, this is simply accomplished with
the proper feedback such that, yes, they are matched over a broad
range. The math is quite simple, the cost is another matter. Can you
afford one? Probably not. The lack of commercial examples available
to the Ham is not proof they do not exist. Let's pass on from issues
of economy.

On the other side of the aisle, I've worked with active loads that
will absorb as much power (up to a limit) at any frequency (up to a
limit) that you care to throw at it.

Observations are that experiments to discover the source
impedance by incrementally changing load current can produce a range of
values for the same PA on different frequencies, and at different power
levels.

This is called "Load Pulling," and is a classic technique to
demonstrate source Z. Thevenin first described it and Norton followed
suit. I cannot, for the life of me, recall any other intellectual
giants that have pulled these apart.

I have done this with my own gear. The variation from a source Z of
50 Ohms wandered the SWR range of 1.5:1 over all bands and most power
levels. Given this conformed to the manufacturer's specification, I
was not particularly surprised. Where it deviated the most, the rig
also operated the worst. What can we say about experience and
performance design converging?

Why do amplifiers with say tetrodes and triodes which exhibit
such different dynamic plate resistance but requiring the same load
impedance deliver the same equivalent source impedance?

A cable connector instead of binding posts? Let's dismiss this as
being obvious.

I am also aware that supporters of the inherent source match position
assert that you must be selective in choosing tests for source impedance.
It is all rather unconvincing when only some of the implications of a
particular source impedance are effective.

Where is the rig specified to exhibit this condition?

Is your rig a VW that stalls trying to pull a trailer from a stop in
3rd gear? Or is it a Mack truck trying to park in the handicap zone
in an underground mall parking lot? Arguing other's incapabilities is
something I like doing, but with more flair. Let's skip these
Tritonic minnows.

It is my view that modelling the PA as a fixed voltage or current source
with fixed source impedance of Zo, and where reflected waves on a
transmission line are absorbed by the matched source is not a good
general model for HF PAs.

You have already said as much. I see nothing new so far.

The application of small signal analysis to amplifiers that sweep from
near cutoff to near saturation is suspect.

If it is near cutoff or saturation, it is suspect small signal
analysis. Certainly, anyone can conspire to fail gracelessly. Would
you care to elaborate the suspicion beyond the evidence of gross
negligence? Why don't we skip this minor excursion?

I believe that it is sound (in the steady state) to resolve the forward
and reflected wave voltages and currents at the source end of the
transmission line, calculate the complex impedance, and predict the
effects of that impedance as a PA load using the same techniques that
were used to design the PA.

Sound though it may be, if I were to line up another transmitter
boresight down the antenna connector of the first, light it up to
provide power with no equivocation of it being fictional; then yes,
all things may appear to be the same. ...and yet I have just
demonstrated reverse power arriving at the antenna terminal (where did
it go?). My having experience in doing just this (aka active load
already described above) fully conforms to your sound idea, and yet,
as for myself, it is not an idea I would rely on to deny the existence
of reverse power nor its capacity to fry the innards of a transmitter
(active loads are heavily heat-sinked and fan driven).

73's
Richard Clark, KB7QHC

On Tue, 27 Feb 2007 21:37:15 GMT, Owen Duffy wrote:

The fact that an instrument may consume power from the circuit does not
imply that it measures power.

Hi Owen,

If that power is from a reflection, then we've come to the logical
conclusion of the argument.

73's
Richard Clark, KB7QHC

Richard Clark wrote in
:

On Tue, 27 Feb 2007 21:14:38 GMT, Owen Duffy wrote:

This is not contending nor contention and is content only for a non
sequitur. The line following a tuner exhibits considerable loss
(poor efficiency) that can only occur on the basis of power and
mismatch. You yourself offered in other correspondence that it
exceeds cable attenuation specifications found only in a matching
condition. To

I am being picky, but "it *may* exceed cable attenuation
specifications found only in a matching condition, it may also be
lower".

Hi Owen,

Lower? That is rather astonishing in light of responding to my
comment.

If I said it
as you stated, I made an error. The common statement (and I have no
doubt made it) that VSWR exacerbates line loss is actually wrong in
the general case. (Having Googled my own web site I see one statement
along those lines which needs further qualification!)

This is even more astonishing. Irrespective of you being the source,
why inject this confusing comment? SWR always exacerbates line loss!
Give me any normal line attenuation and SWR at the load, and I will
tell you exactly how much additional loss will occur. There's a
general solution for you.

Something tells me that your comments are based on a confusion between
the power loss of a cable, and its mismatch loss. They are not the
same thing although they are usually tightly twined in discussion. You
later exhibit a confusion between a conjugate match and an impedance
match. They are not the same thing either. The confusion on both
these points have abounded in this group in past "debates."

Just to deal with this issue, line loss under VSWR, which at first seems
a side issue, but it illustrates one of the problems of a "power
perspective" in analysing a transmission line.

By "line loss" I mean the ratio of power at the load end of the line to
power at the source end of the line, not "forward power" or "reflected
power", but the average rate of flow of energy at those points.

So, to your challenge:

The problem is 1m of Belden 8262 (RG58C/U type) at 3.5MHz with three
loads, 50+j0, 5+j0, and 500+j0.

The loss for 50+j0 load is 0.025dB (equivalent to the Matched Line Loss).

What Line Loss to you get for the other two cases?

(I make it 0.24dB for 5+j0, and 0.014 for 500+j0.)

Owen

On Wed, 28 Feb 2007 01:01:55 GMT, Owen Duffy wrote:

By "line loss" I mean the ratio of power at the load end of the line to
power at the source end of the line, not "forward power" or "reflected
power", but the average rate of flow of energy at those points.

Hi Owen,

What's wrong with conventional terms so that we BOTH know what you
mean? The convention would call this Mismatch Loss. If you dispute
this, then it serves my complaint. Further, convention has no
interest in "forward power" nor "reflected power" except as expressed
as SWR. I thought I was quite terse in this regard.

So, to your challenge:
The problem is 1m of Belden 8262 (RG58C/U type) at 3.5MHz with three
loads, 50+j0, 5+j0, and 500+j0.

Well, as I've pointed out, it is not strictly in the terms of my
challenge, is it?

The loss for 50+j0 load is 0.025dB (equivalent to the Matched Line Loss).

Sigh... parentheticals?

What Line Loss to you get for the other two cases?

(I make it 0.24dB for 5+j0, and 0.014 for 500+j0.)

An additional 0.1dB However, this example strains the utility of the
challenge.

Let's try a perverse challenge.

Presume a source of 100+j0 Ohms impedance sees a 50 Ohm line that is
5.35 wavelengths long and is terminated with a load of 200+j0 Ohms.
The normal attenuation of the line is 2.00 dB. What is the loss in
the line?

Can your general solution solve this? It uniquely describes both the
kinetics of reverse power flow AND the impact of source resistance.

No one has every answered this one correctly, by the way (and I can
anticipate you are ready to spring that observation on me with your
source feeding essentialy a voltage oriented high Z load as opposed to
the current oriented low Z load).

73's
Richard Clark, KB7QHC

On Tue, 27 Feb 2007 17:51:37 -0800, Richard Clark
wrote:

Presume a source of 100+j0 Ohms impedance sees a 50 Ohm line that is
5.35 wavelengths long and is terminated with a load of 200+j0 Ohms.
The normal attenuation of the line is 2.00 dB. What is the loss in
the line?

For the others,

Any who complain about their transmitter having:
1. No source resistance;
2. Not this much resistance:
3. Not this little resistance;
4. None of the above (the usual response).
can take heart that if you simply substitute a tuner which presents
the equivalent SWR at the plane of the input to the line, then you can
progress to the solution with equal fluidity (which is to say like
molasses in December).

This particular example has been around for at least 40 years if not
since WWII. No one has rushed to answer it here in at least a quarter
of that time, I don't expect a cascade of guesses soon either.

73's
Richard Clark, KB7QHC

Richard Clark wrote in
:

On Wed, 28 Feb 2007 01:01:55 GMT, Owen Duffy wrote:

By "line loss" I mean the ratio of power at the load end of the line to
power at the source end of the line, not "forward power" or "reflected
power", but the average rate of flow of energy at those points.

Hi Owen,

What's wrong with conventional terms so that we BOTH know what you
mean? The convention would call this Mismatch Loss. If you dispute
this, then it serves my complaint. Further, convention has no
interest in "forward power" nor "reflected power" except as expressed
as SWR. I thought I was quite terse in this regard.

So, to your challenge:
The problem is 1m of Belden 8262 (RG58C/U type) at 3.5MHz with three
loads, 50+j0, 5+j0, and 500+j0.

Well, as I've pointed out, it is not strictly in the terms of my
challenge, is it?

Richard,

Your challenge was "Give me any normal line attenuation and SWR at the
load, and I will tell you exactly how much additional loss will occur."

I didn't state the VSWR, but it is 10:1 in both cases. The "normal line
attenation" you refer to is I expect the Matched Line Loss" which I have
given you.


The loss for 50+j0 load is 0.025dB (equivalent to the Matched Line
Loss).

Sigh... parentheticals?

What Line Loss to you get for the other two cases?

(I make it 0.24dB for 5+j0, and 0.014 for 500+j0.)

An additional 0.1dB However, this example strains the utility of the
challenge.

Is that your answer, an additional 0.1db due to the 10:1 VSWR? We do not
agree on either answer.

BTW, my figures were not additional loss, but total Line Loss as I
defined it.

You will note that my calculation for the 5+j0 case is less than the
Matched Line Loss, not higher.

In practical transmission lines, most of the loss is in current flowing
in the R component of an RLGC equivalent of the line, the loss in the
copper conductors forming the line. For VSWR1, the net current varies
along the line forming the classic standing wave pattern, and the loss in
incremental lengths of the line varies approximately with the square of
current in that increment.

So in the two cases above, even though the load VSWR is the same, the
loss is quite different due to the different current distribution in both
cases, one is near a current maximum, and the other is near a current
minimum. Any adjustment of Matched Line Loss for VSWR1 using only the
VSWR cannot take the location of the standing wave pattern into account,
and is an inaccurate approximation in some situations.

Many books showing a VSWR based formula for "additional loss due to
VSWR" don't spell out the assumptions underlying the formula. Phillip
Smith does in his book "Electronic Applications of the Smith Chart", he
says "If a waveguide is one or more wavelengths long, the average loss
due to standing waves in a region extending plus or minus a half
wavelength from the point of observation may be expressed as a
coefficient or factor of the one way transmission loss per unit length."
and he gives the ratio as (1+S^2)/(2*S). Though the ARRL shows graphs and
formulas they don't always (if ever) spell out the assmptions.

So, yes I assert that the Line Loss under mismatch conditions may be less
than the Matched Line Loss.

Owen

PS: I calculated my answers using http://www.vk1od.net/tl/tllc.php ,
Dan's TLDETAILS.EXE and ARRL's TLW3.EXE give similar results.

Richard Clark wrote in
:

On Tue, 27 Feb 2007 21:37:15 GMT, Owen Duffy wrote:

The fact that an instrument may consume power from the circuit does not
imply that it measures power.

Hi Owen,

If that power is from a reflection, then we've come to the logical
conclusion of the argument.

Richard,

If you read and understood the article, you would see that the instrument
is based on sampling the V/I ratio at a point, and that being surrounded by
transmission line is not important to the principle of operation, in other
words, it does not directly measure a reflected wave.

Owen