On 7/7/2015 6:17 AM, Ian Jackson wrote:
In message , Jerry Stuckle
writes

So why don't manufacturers design transmitters with 1 ohm output
impedance, Rick?

They probably would - if they could (at least for some applications).
That would then enable you to step up the TX output voltage (using a
transformer), so that you could drive more power into a higher (eg 50
ohm) load.


Oh, it's completely possible. It's just a matching network, anyway -
one which has to be in place anyway, because the output of a tube amp is
relatively high impedance, and the output of a transistor amp is
relatively low impedance. In fact, a 144W transistor amp running on 12V
wouldn't even need a matching network. It's output would have 1 ohm
impedance.

But of course, the overall output impedance would then become
correspondingly higher. You would also be drawing correspondingly more
current from the original 1 ohm source, and if you used too high a
step-up, you would risk exceeding the permitted internal power
dissipation (and other performance parameters).


And why would the output impedance change just because the load
impedance changes? They are two separate things.

But according to you, no step-up is required - you can drive any
(comparatively) high impedance source most efficiently from a low impedance.

So yes, you are getting more power output when you match* the source
impedance to the load - but it doesn't necessarily mean you always can
(or should) go the whole hog.
*Or, at least, partially match.


So, which is it? First you say the output impedance of the transmitter
should be very low for maximum power to the antenna. Now you say it
should be matched. Then you say it shouldn't be matched.

Which is it?

--
==================
Remove the "x" from my email address
Jerry, AI0K

==================

On 7/7/2015 8:33 AM, rickman wrote:
On 7/7/2015 8:52 AM, John S wrote:
On 7/7/2015 7:19 AM, rickman wrote:
On 7/7/2015 3:19 AM, John S wrote:
On 7/6/2015 12:02 PM, rickman wrote:
On 7/6/2015 12:50 PM, John S wrote:
On 7/5/2015 11:39 PM, rickman wrote:
On 7/5/2015 4:45 PM, Jerry Stuckle wrote:
On 7/5/2015 3:58 PM, Roger Hayter wrote:
Jerry Stuckle wrote:

On 7/5/2015 9:23 AM, Ian Jackson wrote:
In message ,

writes
Wayne wrote:


"Jeff Liebermann" wrote in message
...

On Sat, 04 Jul 2015 19:04:01 -0400, Jerry Stuckle
wrote:
Think of it this way, without the math. On the transmitter
side of
the
network, the match is 1:1, with nothing reflected back to
the
transmitter.

So you have a signal coming back from the antenna. You
have a
perfect
matching network, which means nothing is lost in the
network.
The
feedline is perfect, so there is no loss in it. The only
place
for
the
signal to go is back to the antenna.

Wikipedia says that if the source is matched to the line, any
reflections that come back are absorbed, not reflected
back to
the
antenna:

https://en.wikipedia.org/wiki/Impedance_matching
"If the source impedance matches the line, reflections
from the load end will be absorbed at the source end.
If the transmission line is not matched at both ends
reflections from the load will be re-reflected at the
source and re-re-reflected at the load end ad infinitum,
losing energy on each transit of the transmission line."

Well, I looked at that section of the writeup.
And, I have no idea what the hell they are talking about.
Looks like a good section for a knowledgeable person to edit.

If the termination matches the line impedance, there is no
reflection.

Both the antenna and the source are terminations.

This is a bit difficult to visualze with an RF transmitter,
but is
more easily seen with pulses.

Being essentially a simple soul, that's how I sometimes try to
work
out
what happening.

You'll be better off if you killfile the troll. You'll get a lot
less
bad information and your life will become much easier.


The wikipedia entry is correct as written.

In the real world, the output of an amateur transmitter will
seldom
be exactly 50 Ohms unless there is an adjustable network of
some
sort.

I've always understood that the resistive part of a TX output
impedance
was usually less than 50 ohms.

If a transmitter output impedance WAS 50 ohms, I would have
thought
that
the efficiency of the output stage could never exceed 50% (and
aren't
class-C PAs supposed to be around 66.%?). Also, as much power
would be
dissipated in the PA stage as in the load.

Fixed output amateur transmitters are a nominal 50 ohms. It can
vary,
but that is due to normal variances in components, and the
difference
can be ignored in real life.

But output impedance has little to do with efficiency. A Class C
amplifier can run 90%+ efficiency. It's output may be anything,
i.e.
high with tubes but low with solid state. But the output
impedance
can
be converted to 50 ohms or any other reasonable impedance
through a
matching network.

A perfect matching network will have no loss, so everything the
transmitter puts out will go through the matching network. Of
course,
nothing is perfect, so there will be some loss. But the
amount of
loss
in a 1:1 match will not be significant.

Also, the amplifier generates the power; in a perfect world,
100% of
that power is transferred to the load. The transmitter doesn't
dissipate 1/2 of the power and the load the other 1/2. It's not
like
having two resistors in a circuit where each will dissipate
1/2 of
the
power.

BTW - the resistive part of the impedance is not the same as
resistance.
For a simple case - take a series circuit of a capacitor, an
inductor
and a 50 ohm resistor. At the resonant frequency, the impedance
will
be 50 +j0 (50 ohms from the resistor, capacitive and inductive
reactances cancel). But the DC resistance is infinity. Again, a
simple
example, but it shows a point.

The resistive part of the impedance is exactly the same as a
resistance
as far as the frequency you are using is concerned. And if the
amplifier output impedance *and* the feeder input resistance were
*both*
matched to 50 ohms resistive then 50% of the power generated
(after
circuit losses due of inefficiency of generation) would be
dissipated in
the transmitter. It would, at the working frequency, be
*exactly*
like
having two equal resistors in the circuit each taking half the
generated
power. So the amplifier has a much lower output impedance than
50ohms
and no attempt is made to match it to 50 ohms.


OK, so then please explain how I can have a Class C amplifier with
1KW
DC input and a 50 ohm output, 50 ohm coax and a matching network at
the
antenna can show 900 watt (actually about 870 watts due to feedline
loss)? According to your statement, that is impossible. I should
not
be able get more than 450W or so at the antenna.

It is true that if the transmitter is thought of as a fixed
voltage
generator in series with, say, a 5 ohm resistor then the maximum
power
transfer in theory would occur with a 5 ohm load. But to achieve
this
the output power would have to be 100 times higher (ten times the
voltage) and half of it would be dissipated in the PA. Not the
best
way
to run things, it is better to have a voltage generator chosen to
give
the right power with the load being much bigger than its generator
resistance.


So why do all fixed-tuning amateur transmitters have a nominal 50
ohm
output instead of 1 or two ohms? And why do commercial radio
stations
spend tens of thousands of dollars ensuring impedance is matched
throughout the system?

The maximum power transfer at equal impedance theorem only
applies if
you started with a *fixed* output voltage generator. We
don't; we
start with a load impedance (50 ohm resistive), then we decide
what
power output we want, and we choose the voltage to be generated
accordingly. (Thank you for giving me the opportunity to think
about
this!)


Actually, it doesn't matter if it's a fixed or a variable output
voltage
- maximum power transfer always occurs when there is an impedance
match.

How about we quit with the speculation and come up with some
numbers?

Here is a simulation of a 50 ohm load with a 50 ohm matched series
output impedance and a voltage source of 200 VAC peak. Power into
the
load is 100 W.

http://arius.com/sims/Matched%20Load%20Power.png

Same exact circuit with the series impedance of just 1 ohm, power
into
the load is 385 W.

http://arius.com/sims/UnMatched%20Load%20Power.png

I'd say that is pretty clear evidence that matched loads are not the
way
to maximize power transfer when the load impedance is fixed and the
output impedance is controllable.


There can be a lossless resistive part of source impedance
according to
the IEEE (and most every other well educated EEs). After all, a
transmission line has a resistance but it's loss resistance is much
lower.

Can you provide a reference to any of this?


IEEE Standard Dictionary of Electrical and Electron Terms, IEEE Std
100-1972. For a complete treatment on the topic, see Reflections III by
Walter Maxwell, W2DU, Appendix 10.

Anything more accessible?


I will not lead you by the hand. I have supplied the requested
references. If you can't find the referenced material, then you need to
learn how to research, buy books, go to the library, etc. It is your
loss. Education is more than online chatter.

Lol. You are a trip. I'm not going to spend $100 on a book just to see
if you are right. I was intrigued by the idea that a wire could carry a
signal without the resistance dissipating power according to P = I^2 R.
I guess there is some communication failure.


Yes, there is.

I did NOT say that a wire was dissipationless. Please reread the above.
I said "There can be a lossless resistive part of source impedance".
Source impedance may or may not contain resistance which dissipates. A
50 ohm piece of coax has an impedance of approximately 50+j0 ohms. So,
does the 50 ohms dissipate? Look at the specifications of the coax. How
much loss?

If you are unwilling to go to the library or do research or buy a book
on you own, then you are beyond help. What is your ignorance worth to you?

On 7/7/2015 3:05 AM, John S wrote:
On 7/6/2015 1:03 PM, wrote:
John S wrote:
On 7/6/2015 11:01 AM, Jerry Stuckle wrote:
On 7/6/2015 4:20 AM, Ian Jackson wrote:
In message , rickman
writes



How about we quit with the speculation and come up with some numbers?

Here is a simulation of a 50 ohm load with a 50 ohm matched series
output impedance and a voltage source of 200 VAC peak. Power into
the
load is 100 W.

http://arius.com/sims/Matched%20Load%20Power.png

Same exact circuit with the series impedance of just 1 ohm, power
into
the load is 385 W.

http://arius.com/sims/UnMatched%20Load%20Power.png

I'd say that is pretty clear evidence that matched loads are not the
way to maximize power transfer when the load impedance is fixed and
the output impedance is controllable.

Quite simply, if your prime objective is to get maximum power out of a
power (energy?) source, the source having an internal resistance is a
BAD THING. You don't design the source to have an internal resistance
equal to its intended load resistance. No one designs lead-acid
batteries that way (do they?), so why RF transmitters?

While theoretically you can extract the maximum power available
from the
source when the load resistance equals the source resistance, you can
only do so provided that the heat you generate in the source does not
cause the source to malfunction (in the worst case, blow up).

Because DC power transfer is not the same as AC power transfer.


Why not? Does something happen to the laws of physics with AC?

Yes, quite a lot, you get a whole new set of laws.

If you apply 1vDC to a 1 ohm resistor, you get 1A of current. If you
apply 1vAC RMS (at any frequency) to a 1 ohm resistor, you get 1A of
current. How does the AC change the law?


You apply 1vdc to a 0.159 microfarad capacitor and you get 0 amps
flowing (open circuit).
You apply 1vac at 1MHz to that same capacitor and you get 1 amp flowing,
with the current leading the voltage by 90 degrees.

You apply 1vdc to a 0.159 microhenry inductor and you get infinite amps
flowing (short circuit).
You apply 1vdc at 1MHz to that same inductor, and you get 1 amp flowing
with the voltage leading the current by 90 degrees.

You place the capacitor and inductor in series.
Fed with DC, you get 0 amps flowing (open circuit).
Fed with 1MHz AC, you get infinite current flowing (short circuit).

You place the capacitor and inductor in parallel.
Fed with DC, you get infinite current flowing (short circuit).
Fed with 1MHz AC you get 0 amps flowing (open circuit).

There is a huge difference between ac and dc!

--
==================
Remove the "x" from my email address
Jerry, AI0K

==================

On 7/7/2015 3:37 AM, Roger Hayter wrote:
Jerry Stuckle wrote:

On 7/6/2015 7:59 PM, Roger Hayter wrote:
Jerry Stuckle wrote:

On 7/6/2015 12:41 PM, John S wrote:
On 7/6/2015 11:01 AM, Jerry Stuckle wrote:
On 7/6/2015 4:20 AM, Ian Jackson wrote:
In message , rickman
writes



How about we quit with the speculation and come up with some numbers?

Here is a simulation of a 50 ohm load with a 50 ohm matched series
output impedance and a voltage source of 200 VAC peak. Power into the
load is 100 W.

http://arius.com/sims/Matched%20Load%20Power.png

Same exact circuit with the series impedance of just 1 ohm, power into
the load is 385 W.

http://arius.com/sims/UnMatched%20Load%20Power.png

I'd say that is pretty clear evidence that matched loads are not the
way to maximize power transfer when the load impedance is fixed and
the output impedance is controllable.

Quite simply, if your prime objective is to get maximum power out of a
power (energy?) source, the source having an internal resistance is a
BAD THING. You don't design the source to have an internal resistance
equal to its intended load resistance. No one designs lead-acid
batteries that way (do they?), so why RF transmitters?

While theoretically you can extract the maximum power available from the
source when the load resistance equals the source resistance, you can
only do so provided that the heat you generate in the source does not
cause the source to malfunction (in the worst case, blow up).

Because DC power transfer is not the same as AC power transfer.


Why not? Does something happen to the laws of physics with AC?


Yup, AC has reactance. DC does not. Big difference.


If what you say is correct, then it wouldn't matter what antenna
impedance I had, as long as it matches the transmission line. VSWR
would be immaterial.


There is no VSWR nor ISWR if the load matches the line.


Sure, there is ALWAYS VSWR. It may be 1:1, but it's always there.


That is demonstrably false.

Please demonstrate this for us as we wish to learn.



OK, take your amateur transmitter. Connect it through a 1:1 balun to
300 ohm feedline. Connect that to a 300 ohm antenna.

The VSWR would be near to 1:1. Of course, you would need a 300 ohm
meter to measure it. It would be the same whether or not you connected
your amateur transmitter


That's right. It's a 1:1 SWR on the transmission line. And it matches
your requirements.




According to you, you should get full power output at the antenna. In
reality, you will get a 6:1 SWR and about 49% of the power at the
antenna, minus transmission line loss (assuming, of course, your
transmitter hasn't cut it's power back).


You probably won't get full power, because the transmitter would have to
produce 2.4 times it's usual output voltage to achieve it. But the
system would be gratifyingly low in SWR and transmission line loss.



Not according to you. Since the transmitter has a comparatively low
impedance, it should dissipate very little power and virtually all of
the power should go to the antenna. After all, you do have a 1:1 VSWR
on the transmission line/antenna, fed by a low impedance transmitter.

True. I agree. But the power it can generate depends on its load
impedance. Consider a very high or even open circuit load. It can put
little or no power into it, while having its normal voltage output.




No, the power the transmitter can generate has nothing to do with the
load impedance. The transmitter generates what it can generate.

The amount of power transferred to the load is dependent on the load
impedance, with the maximum power transfer occurring when the impedances
match.

--
==================
Remove the "x" from my email address
Jerry, AI0K

==================

In message , rickman
writes
On 7/7/2015 6:25 AM, Ian Jackson wrote:
In message , Jerry Stuckle
writes


Sure, there is ALWAYS VSWR. It may be 1:1, but it's always there.

If there's no reflection, there can be no standing wave. So, being
pedantic, there's no such thing as an SWR of 1:1!

Why do you say that? If there is no reflection the voltage on the line
is purely due to the forward signal and so the VSWR is 1:1. What's
wrong with that?

A standing wave is caused by a reflection. If there IS no reflection,
there is NO standing wave. So while you can have an SWR of
1.00000000000001-to-1 (because a standing wave DOES exist), you can't
really have one of 1-to-1 (because there IS no standing wave). ;o))
[Just a bit of pedantic, lateral thinking on my part. Don't worry too
much about it. It has absolutely no bearing whatsoever on the current
discussions.]
--
Ian

On 7/5/2015 7:21 PM, wrote:
John S wrote:
On 7/5/2015 5:24 PM, wrote:
Roger Hayter wrote:
wrote:


The output impedance of an amateur transmitter IS approximately 50 Ohms
as is trivially shown by reading the specifications for the transmitter
which was designed and manufactured to match a 50 Ohm load.

Do you think all those manuals are lies?

You are starting with a false premise which makes everything after that
false.


A quick google demonstrates dozens of specification sheets that say the
transmitter is designed for a 50 ohm load, and none that mention its
output impedance.

If the source impedance were other than 50 Ohms, the SWR with 50 Ohm
coax and a 50 Ohm antenna would be high. It is not.

Where is the source impedance found on a Smith chart? Also, if you have
EZNEC, you will not find a place to specify source impedance but it will
show the SWR.

A Smith chart is normalized to 1.

EZNEC allows you to set the impedance to anything you want and assumes
the transmission line matches the transmitter.


The EZNEC help file is very comprehensive. Please find any reference to
your assertion that there is an assumption of source impedance there and
provide information for us to verify your assertion.

"John S" wrote in message ...

On 7/5/2015 7:08 PM, Wayne wrote:


"Jeff Liebermann" wrote in message
...

On Sun, 05 Jul 2015 20:22:19 +0100, Brian Reay wrote:

As for a simpler way, I'd recommend a remote auto-matcher like an SGC at
the antenna base. It will minimise coax losses and should give you a
good match, at least for most bands. I've used a similar set up (with
radials) and achieved a good match even on 80m.

If your radio has a built in tuner, then it can be used to 'tweak' the
match in the event the radio isn't 'seeing' 1.5:1. Turn it off
initially. Let the SGC find a match. If it isn't ideal, use the local
ATU for a final tweak. I never found this was required but YMMV.

Not everyone is a true believer in antenna tuners:
http://www.qsl.net/g3tso/Hombrew-Mobile%20Antennas.html


Interesting.
I'm off on a different approach.

I have an RF ammeter mounted in a box. The box is in the shack between
the ATU and the antenna.

I simply adjust the ATU for max current on the ammeter.

Hey, Wayne -

As a matter of curiosity on my part, can you find a way to measure the
ammeter's resistance and let me know the full-scale value?

No, I don't have enough test equipment to easily do that.

With a DVM it measures 0.4 ohms and with a VOM measures 28 ohms. And the
VOM gives no needle movement.

It is a O. D. McClintock Signal Corp typs I S-III with full scale of 2.5
amps.
Since it was salvaged from some WW II equipment back in the 1950s, it
probably isn't calibrated.

But, it gives a useable relative reading.

On 7/7/2015 6:25 AM, Ian Jackson wrote:
In message , Jerry Stuckle
writes


Sure, there is ALWAYS VSWR. It may be 1:1, but it's always there.

If there's no reflection, there can be no standing wave. So, being
pedantic, there's no such thing as an SWR of 1:1!




Wrong. An SWR of 1:1 indicates a perfect match, with no reflected
power. It is recognized by all electronics texts and experts.

My suggestion would be for you to learn some transmission line theory.
Your statement here just showed you have no knowledge of it at all.

Even when I took my novice test many years ago I had to understand SWR
better than that.

--
==================
Remove the "x" from my email address
Jerry, AI0K

==================

On 7/7/2015 6:34 AM, Ian Jackson wrote:
In message , Jerry Stuckle
writes




But if what you say is correct, then I should be able to get a lot of
power out of my 100 watt transmitter feeding a 1 ohm antenna.

But probably not for long!






Why not? The 100 W transmitter will only put out 100 watts!

As a matter of fact, due to the 50:1 SWR, I will only get about 4 watts
to the antenna, assuming, of course, the transmitter doesn't shut down
from having to dissipate and additional 96 watts.

--
==================
Remove the "x" from my email address
Jerry, AI0K

==================

"rickman" wrote in message ...

On 7/7/2015 3:19 AM, John S wrote:
On 7/6/2015 12:02 PM, rickman wrote:
On 7/6/2015 12:50 PM, John S wrote:
On 7/5/2015 11:39 PM, rickman wrote:
On 7/5/2015 4:45 PM, Jerry Stuckle wrote:
On 7/5/2015 3:58 PM, Roger Hayter wrote:
Jerry Stuckle wrote:

On 7/5/2015 9:23 AM, Ian Jackson wrote:
In message ,

writes
Wayne wrote:


"Jeff Liebermann" wrote in message
...

On Sat, 04 Jul 2015 19:04:01 -0400, Jerry Stuckle
wrote:
Think of it this way, without the math. On the transmitter
side of
the
network, the match is 1:1, with nothing reflected back to the
transmitter.

So you have a signal coming back from the antenna. You have a
perfect
matching network, which means nothing is lost in the network.
The
feedline is perfect, so there is no loss in it. The only place
for
the
signal to go is back to the antenna.

Wikipedia says that if the source is matched to the line, any
reflections that come back are absorbed, not reflected back to
the
antenna:

https://en.wikipedia.org/wiki/Impedance_matching
"If the source impedance matches the line, reflections
from the load end will be absorbed at the source end.
If the transmission line is not matched at both ends
reflections from the load will be re-reflected at the
source and re-re-reflected at the load end ad infinitum,
losing energy on each transit of the transmission line."

Well, I looked at that section of the writeup.
And, I have no idea what the hell they are talking about.
Looks like a good section for a knowledgeable person to edit.

If the termination matches the line impedance, there is no
reflection.

Both the antenna and the source are terminations.

This is a bit difficult to visualze with an RF transmitter, but is
more easily seen with pulses.

Being essentially a simple soul, that's how I sometimes try to work
out
what happening.

You'll be better off if you killfile the troll. You'll get a lot
less
bad information and your life will become much easier.


The wikipedia entry is correct as written.

In the real world, the output of an amateur transmitter will
seldom
be exactly 50 Ohms unless there is an adjustable network of some
sort.

I've always understood that the resistive part of a TX output
impedance
was usually less than 50 ohms.

If a transmitter output impedance WAS 50 ohms, I would have thought
that
the efficiency of the output stage could never exceed 50% (and
aren't
class-C PAs supposed to be around 66.%?). Also, as much power
would be
dissipated in the PA stage as in the load.

Fixed output amateur transmitters are a nominal 50 ohms. It can
vary,
but that is due to normal variances in components, and the
difference
can be ignored in real life.

But output impedance has little to do with efficiency. A Class C
amplifier can run 90%+ efficiency. It's output may be anything,
i.e.
high with tubes but low with solid state. But the output impedance
can
be converted to 50 ohms or any other reasonable impedance through a
matching network.

A perfect matching network will have no loss, so everything the
transmitter puts out will go through the matching network. Of
course,
nothing is perfect, so there will be some loss. But the amount of
loss
in a 1:1 match will not be significant.

Also, the amplifier generates the power; in a perfect world, 100% of
that power is transferred to the load. The transmitter doesn't
dissipate 1/2 of the power and the load the other 1/2. It's not
like
having two resistors in a circuit where each will dissipate 1/2 of
the
power.

BTW - the resistive part of the impedance is not the same as
resistance.
For a simple case - take a series circuit of a capacitor, an
inductor
and a 50 ohm resistor. At the resonant frequency, the impedance
will
be 50 +j0 (50 ohms from the resistor, capacitive and inductive
reactances cancel). But the DC resistance is infinity. Again, a
simple
example, but it shows a point.

The resistive part of the impedance is exactly the same as a
resistance
as far as the frequency you are using is concerned. And if the
amplifier output impedance *and* the feeder input resistance were
*both*
matched to 50 ohms resistive then 50% of the power generated (after
circuit losses due of inefficiency of generation) would be
dissipated in
the transmitter. It would, at the working frequency, be *exactly*
like
having two equal resistors in the circuit each taking half the
generated
power. So the amplifier has a much lower output impedance than
50ohms
and no attempt is made to match it to 50 ohms.


OK, so then please explain how I can have a Class C amplifier with 1KW
DC input and a 50 ohm output, 50 ohm coax and a matching network at
the
antenna can show 900 watt (actually about 870 watts due to feedline
loss)? According to your statement, that is impossible. I should not
be able get more than 450W or so at the antenna.

It is true that if the transmitter is thought of as a fixed voltage
generator in series with, say, a 5 ohm resistor then the maximum
power
transfer in theory would occur with a 5 ohm load. But to achieve
this
the output power would have to be 100 times higher (ten times the
voltage) and half of it would be dissipated in the PA. Not the best
way
to run things, it is better to have a voltage generator chosen to
give
the right power with the load being much bigger than its generator
resistance.


So why do all fixed-tuning amateur transmitters have a nominal 50 ohm
output instead of 1 or two ohms? And why do commercial radio stations
spend tens of thousands of dollars ensuring impedance is matched
throughout the system?

The maximum power transfer at equal impedance theorem only applies if
you started with a *fixed* output voltage generator. We don't; we
start with a load impedance (50 ohm resistive), then we decide what
power output we want, and we choose the voltage to be generated
accordingly. (Thank you for giving me the opportunity to think about
this!)


Actually, it doesn't matter if it's a fixed or a variable output
voltage
- maximum power transfer always occurs when there is an impedance
match.

How about we quit with the speculation and come up with some numbers?

Here is a simulation of a 50 ohm load with a 50 ohm matched series
output impedance and a voltage source of 200 VAC peak. Power into the
load is 100 W.

http://arius.com/sims/Matched%20Load%20Power.png

Same exact circuit with the series impedance of just 1 ohm, power into
the load is 385 W.

http://arius.com/sims/UnMatched%20Load%20Power.png

I'd say that is pretty clear evidence that matched loads are not the
way
to maximize power transfer when the load impedance is fixed and the
output impedance is controllable.


There can be a lossless resistive part of source impedance according to
the IEEE (and most every other well educated EEs). After all, a
transmission line has a resistance but it's loss resistance is much
lower.

Can you provide a reference to any of this?


IEEE Standard Dictionary of Electrical and Electron Terms, IEEE Std
100-1972. For a complete treatment on the topic, see Reflections III by
Walter Maxwell, W2DU, Appendix 10.

Anything more accessible?


Just a comment....
Independent of discussions on this thread, Walt's book is an excellent
reference to have.
It's available on Amazon.