Richard Clark wrote in
:

On Tue, 27 Feb 2007 01:07:12 GMT, Owen Duffy wrote:

Richard Clark wrote in
Let's treat this like the Chinese Box problem.

If you didn't know what the load was, could you explain it any
differently? No. Apriori knowledge is not a proof.

Richard, I content that:

Contend or offer in contention.

Richard

Yes, my spelling mistake.


- the power output of the PA; and
- the efficiency of the PA may be (and usually are) sensitive to the
load impedance.

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". 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!)

suggest that a PA's sensitivity is somehow exhalted in the face of
identical, ordinary behavior of a passive component is hardly
seperable. Consider the simple substitution to your quote:
- the power output at the terminus of the line; and
- the efficiency at the terminus of the line may be (and usually are)
sensitive to the load impedance.

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


....

Though it is often asserted that the PA will get hotter as a result of
"reflected power" being dissipated in the dynamic output impedance of
the PA, and that this may / will damage the PA, the power explanation
doesn't work numerically in the general case.

Heat is the outward proof of power and is always demonstrable in both
specific and general cases. Occurrences of other, significant
radiation from the source (as long as that source physically occupies
a substantially minor region of wavelength) is exceedingly difficult
to achieve.

You don't offer a numerical proof of a general case, and given that
the general case must allow for the specific cases already allowed in
your discussion above - that may be an untenable assertion for you.
Those specific cases are demonstrably caloric and must follow the same
math you suggest.

I suspect you are trying to argue differences by degree (no pun
intended as to heat); but I seriously doubt you can produce the math
to do that. The arguments that flow from that involve what is called
source resistance, and those arguments are legion in this forum (where
naysayers embrace a refusal to accept or name ANY value - a curious
paradox and an engineering nihilism I enjoy to watch).

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.

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

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.

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.

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

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.

Owen

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

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:
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 ...

Yep, the great majority of amateur radio antenna systems
are matched by a tuner. That act of matching prohibits
reflected load energy from reaching the PA. Except for
overall efficiency, when an antenna system is matched,
the PA impedance doesn't matter. A 5 ohm PA, a 50 ohm PA,
and a 500 ohm PA all output the same power if the output
voltage is the same into the same load.
--
73, Cecil http://www.w5dxp.com

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.

On Wed, 28 Feb 2007 03:10:28 GMT, Owen Duffy wrote:

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

Hi Owen,

I already anticipated that, didn't I? Certainly parsing my last
discussion is hardly necessary.

I note you have no solution that answers for the loss in a fairly
typical instance for a fairly typical line condition. It couldn't
have been any more difficult than your former computations, could it?
(In fact it is, but not conceptually.) And yet the absence of that
effort is notable (OK, so you've been ambushed). I may have stumbled
on a novelty application but I didn't trip over a boulder of a common
usage.

Given this sub-thread flowed from my response that a source does
dissipate a reverse power flow (both of which, the direction and
dissipation, are held in contention); and further given my "perverse"
challenge fully specifies such a condition and has a real solution, it
stands to reason that if your general computation is in fact general,
then it can resolve the contention to one or both of our satisfaction.
All it requires is that your math treatment accepts both directions of
power flow, and loss in the source. This is not unreasonable,
especially when any number of references encompass just such concepts.

73's
Richard Clark, KB7QHC

Richard Clark wrote in
:

....
I note you have no solution that answers for the loss in a fairly
typical instance for a fairly typical line condition. It couldn't
have been any more difficult than your former computations, could it?
(In fact it is, but not conceptually.) And yet the absence of that
effort is notable (OK, so you've been ambushed). I may have stumbled
on a novelty application but I didn't trip over a boulder of a common
usage.
....

Richard,

No, it is just that I have not posted a solution as yet.

Your problem was described as "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?"

Rather than sit down an write a bunch of stuff to calculate it, I took
the lazy way out and played with a scenario using Belden 8262 (RG58C/U)
in my line loss calculator that was pretty close to your scenario (Zo is
a touch different at 50-j0.24, most lines with loss will have a non-zero
jX component in Zo). My result for 15.8m of 8262 at 67MHz with a 200+j0
load is a transmission line loss (power into the load divided by power
into the line at the source end) is 3.3dB. This value of line loss is
independent of the source equivalent impedance.

If I reproduced the algorithms with the exact propagation constant and Zo
for your problem scenario, the answer would be more accurate, but I think
similar, and still obtained independently of the source impedance, and
not worth the time.

The figures from the calculator are below if someone wants to play with
it.

Now, are you prepared to post your solution?

Owen

Parameters
Transmission Line Belden 8262 (RG-58C/U)
Code B8262
Data source Belden
Frequency 67.000 MHz
Length 15.800 metres
Zload 200.00+j0.00 ?
Yload 0.005000+j0.000000 ?
Results
Zo 50.00-j0.24 ?
Velocity Factor 0.660
Length 1924.75 °, 5.347 ?
Line Loss (matched) 1.997 dB
Line Loss 3.279 dB
Efficiency 47.00%
Zin 30.48+j24.99 ?
Yin 0.019622-j0.016087 ?
VSWR(source end) 2.22
VSWR(load end) 4.00
? 1.46e-2+j2.13e+0
k1, k2 1.30e-5, 2.95e-10
Correlation coefficient (r) 0.999884

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).



VSWR 4.75
Loss 3.55dB

73
Jeff