Calculate Coaxial Cable Loss & Input Impedance at 1 MHz

[Averagesupernova]

the_emi_guy, you are causing the output of the transmitter to see a 10 ohm load correct? Looking into the LC network? If this is your aim, here is what happens which gets up back to what sophie and I are saying is something to avoid:
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Take a transmitter with Zout of 10 ohms as you described. Feed it into a 50 ohm load. We will say there is a voltage across the load of 10 VAC. Thevenize and you will find the thevenin voltage is 12 VAC. The load is dissipating 2 watts, the Zout of the transmitter dissipates .4 watts. Use an impedance matching network and now the Zout of the transmitter dissipates 3.6 watts. From the thevenin voltage you find this to be the case, and assuming a lossless matching network the power dissipated in the load will also be 3.6 watts. This kind of power dissipated in the output stage is considerably higher than the .4 watts when driving the load 'mismatched' which is exactly what sophie was saying is to be avoided in the first place.

 

[the_emi_guy]

Sorry, my input was not well thought out.

The output stage of an RF transmitter is significantly more complex than what I proposed. It includes a tank circuit with a carefully selected Q (too low and we get distortion from class B or C amplifiers since it provides the freewheeling when active devices are off), too high and peak currents are excessive.

Then there is the matching network. This is required since the RF amp itself wants to see a specific load, usually lower than 50 ohms for transistor amplifiers (often higher than 50 ohms for tube amps).

I *believe* that this same matching network (and it is more than two components) is designed to make the RF source look like 50 ohms for a reverse flowing signal.

I will have to try to find (or work out) a better example...

 

[Baluncore]

The transmitter output stage has an impedance transformer called a tank circuit. To cancel the inductive component of the wound transformer it has a parallel capacitor, adjustable for different frequencies.

The load line or V:I ratio of the amplifier's active element is transformed to the impedance of the transmission line by the tank.

A reflected signal returning from the output t'line is reverse transformed by the tank from a V:I ratio of 50 ohm in the t'line to the V:I ratio of the amplifier's active element, and so is not reflected.

 

[sophiecentaur]

A few basic points would not go amiss here.
Reactive components do not dissipate any power. Their effect may be to change the maximum volts of current for a given power, which may, by implication, be problematic for the equipment and cause further internal dissipation (hence the Power Factor correction which is done in AC power systems). This is not relevant to the basic issue. Specific details of tank circuits and the like, just cloud things.

Anyone who cares to look at this link can see what my point has been. It's only one of many sources which back me up in this. A transmitter (any electrical / mechanical power source) that is designed with efficiency in mind will not be matched to its load. Yes, you will get maximum power into your load for a given emf but that will demand twice as much input power (DC power supply, in the end) and it will require you to dissipate half of this power in the amplifying device.

The confusion that other contributors are showing is that they are making assumption about what their practical results are telling them. What actual evidence do they have that the output impedance of their transmitter is '50 - or whatever'? How have they measured it? Efficient transmitters tend to be non-linear so it is hard to determine internal resistance. You can't just measure Open Circuit and then put 50Ω across the output to see how may volts are lost - as you can with the mains or a car battery. I guess you could see the differential effect of 50Ω and 51Ω loads on the output volts. Have any of you done this - or equivalent?

The idea of a conjugate match is not of much significance because the feeder is assumed to be providing the transmitter with a 'good enough '50Ω' at its output.

 

[sophiecentaur]

Here are a few links on the subject of matching. Read what they are actually saying about the significance of transmitter / load matching and efficiency. They all agree with what I have been saying.
http://www.jaycar.co.uk/images_uploaded/impmatch.pdf
http://users.tpg.com.au/users/ldbutler/OutputLoadZ.htm
http://urgentcomm.com/test-amp-measurement-mag/maximum-power-transfer

 

[Averagesupernova]

Thank you sophie. I will take a little more time later too look at the links you have provided. I skimmed through them very quickly.

 

[the_emi_guy]

Here are a few links on the subject of matching. Read what they are actually saying about the significance of transmitter / load matching and efficiency. They all agree with what I have been saying.
http://www.jaycar.co.uk/images_uploaded/impmatch.pdf
http://users.tpg.com.au/users/ldbutler/OutputLoadZ.htm
http://urgentcomm.com/test-amp-measurement-mag/maximum-power-transfer

Thank you sophie. I will take a little more time later too look at the links you have provided. I skimmed through them very quickly.

Agreed, I would like to digest these but it will take me some time. There does seem to be a divergence of opinion on this interesting topic (or maybe I am just not up to speed).

In the meanwhile here is what I have done:

I used Motorola AN267 which was created to help impedance match RF power amplifiers to transmission line. I designed two cases:

Case1: 1 ohm source impedance to 50 ohm line.
Case2: 5+5j source impedance (10 ohms parallel with 10j capacitive reactance) to 50 ohm line.

The results are attached (clipped from Mathcad).

Images 1,2,3 are case1.
Images 4,5,6 are case2.

Note that in both cases the network presents the desired impedance to the output of the amp while simultaneously presenting 50 ohms to any signal traveling in the reverse direction from far end reflection.

I would say that this is a "first order" method of matching. More elaborate schemes are used that take into account non-linearities and load pull characterization, but it seems that the basic scheme provides backmatch.

 

[sophiecentaur]

@the_emi_guy
The calculations are probably right; you can clearly drive the software correctly and your network will present the 50 Ohm load to the transmitter as a 1 Ohm load and vice versa. That would be a match for maximum power. It would result in an equal amount of power being dissipated in the 1 Ohm source resistance. No problem for a low power amplifier and it would be a well behaved piece of equipment - like a good Test Signal Generator, for instance but, with 50% efficiency, a high power transmitter would produce loads of embarrassing heat in its output stage and cost a lot of money to run. That was the point I have been making. I have never claimed that you can't match power amplifiers to deliver maximum power but I know that high power transmitters are built to be efficient (90% or so) and so they cannot be matched in that way - else, where do they get the 40% lost power back from and why do they not melt into the ground? Would you comment on that issue please and not on the nuts and bolts of matching networks?

 

[the_emi_guy]

@the_emi_guy
Would you comment on that issue please and not on the nuts and bolts of matching networks?

I hear you sophie, I understand your point and I need to think about this some more...

 

[sophiecentaur]

OK
No problem.
I think this is just to do with the directions we are coming from. I have no problem with your last post at all.

 

[the_emi_guy]

Here is what I am trying to get my arms around:

Assuming that the impedance that the amp want to see hanging on its output for optimum, high efficiency, performance is higher than its impedance as seen from outside looking in (these were assumed to be the same in my previous analysis).

For example an RF amp with 1ohm output impedance specified to drive 1Kohm load (not maximum power transfer but really efficient).

We need a matching network to make 50 ohm cable look like 1K load. Is it not possible to design this network to simultaneously provide 50 ohm backmatch?

Maybe not, that is what I am hung up on.

 

[Averagesupernova]

Here is what I am trying to get my arms around:

Assuming that the impedance that the amp want to see hanging on its output for optimum, high efficiency, performance is higher than its impedance as seen from outside looking in (these were assumed to be the same in my previous analysis).

For example an RF amp with 1ohm output impedance specified to drive 1Kohm load (not maximum power transfer but really efficient).

We need a matching network to make 50 ohm cable look like 1K load. Is it not possible to design this network to simultaneously provide 50 ohm backmatch?

Maybe not, that is what I am hung up on.

Any time we match with a transformer, Q-section, pi network, T-network, L-network, etc. we will match in both directions. If something is spec'd to drive a 1000 ohm load and you have a 50 ohm antenna system you will most certainly need to transform the impedance.

 

[sophiecentaur]

Here is what I am trying to get my arms around:

Assuming that the impedance that the amp want to see hanging on its output for optimum, high efficiency, performance is higher than its impedance as seen from outside looking in (these were assumed to be the same in my previous analysis).

For example an RF amp with 1ohm output impedance specified to drive 1Kohm load (not maximum power transfer but really efficient).

We need a matching network to make 50 ohm cable look like 1K load. Is it not possible to design this network to simultaneously provide 50 ohm backmatch?

Maybe not, that is what I am hung up on.

If you think of any matching network as a transformer and if you ignore reactances then the ratio looking one way will be the inverse of the ratio looking the other way. If you transform the 50Ω feeder to look like 1kΩ from the transmitter's point of view (that's 1:√20 turns ratio) then the 1Ω output resistance of the transmitter will look like 1/20Ω (20:1 times 1), from the feeder side of the transformer. Afaics, you are stuck with that and it means that designers have to take that sort of thing into account and make allowance for reflected power. But, as long as the antenna match is good, there shouldn't be any signals arriving back at the Tx.
For UHF frequencies and above, it is possible to include an isolator or circulator to avoid the problems associated with mismatch at the transmitter. But then you need to accept a 1dB (iirc) or more extra loss.

 

[Baluncore]

My analysis shows that there are two quite distinct situations involved here.

Firstly; For the RF signal path through multiple stages to beyond the output transmission line, I believe that my assertion of “impedance matching is required for optimum efficiency” is correct for any RF signal path.

Secondly; for regional AC power distribution grids and local DC power supply rails, including 12V car batteries, I believe that sophicentaur's assertion of “zero impedance source is required for optimum efficiency” is correct.


RF signal paths.

Firstly, we should not dismiss the reactive coupling systems that transform the complex impedance between RF stages as being irrelevant. They are vitally important to the RF energy economy. It is the reactive components in switching power supplies and class C or D amplifiers that make those devices so highly efficient.

The second point I want to make regards a fundamental principle of RF design, one that has now been in use successfully for almost a century. When considering the design of a two stage RF amplifier, the load line of the first stage output is known and the input impedance of the second stage is known. A network that will efficiently match those quite different complex impedances is required in order to fully utilise the capabilities of the active elements in both stages. Likewise, we know the load line of the final output stage and we must match that to the transmission line impedance. I agree that without careful matching it will still work, but it will be wasteful of RF energy and equipment resources. The argument that the output impedance of signal path modules must be as low as possible for optimum efficiency is clearly false.

I therefore make the substantial claim that the RF impedances throughout the system need to be matched. Failure to efficiently match at RF represents an underutilisation of the components available and a reduction in the maximum RF energy that can be passed to the transmission line.

The characteristic impedance of a linear transmission line does not effect the energy it will dissipate as heat. It simply sets the relative phase and magnitudes of the voltage and current propagating in its two independent directions. Likewise, in free space, the ratio of the E to M field strengths represents an impedance of approximately 120*Pi ohms. That does not in any way represent a dissipation of power. An amplifier with a load line having a negative Gm represents power gain, not dissipation.


Power Distribution Systems.

A power supply can be derived from the regional AC distribution grid which has an extremely low impedance. The impedance of a power supply should not be matched to the grid, for obvious reasons. So let's buy a 90% efficient switching power supply that generates a DC supply voltage with an effectively zero resistance or impedance at the maximum current we specify, and pay for. We all should recognise that a switching power supply is a class D amplifier and as such it can have very high efficiency. But all this purchase has done is extended the “zero impedance source needed for efficiency” situation from the AC distribution grid to an internal DC supply distribution rail, and yes, as expected, it has done it efficiently without any need to waste 50% of the energy in the process.

Wherever modules share a common energy source, or distribute power to many users or stages with variable requirements, efficiency is gained by having a low supply resistance. Apart from thick wires and plenty of parallel capacitance, the low output impedance is often achieved by providing the source with some form of voltage control feedback. Two examples are the control of field current of an alternator, or the error voltage comparator that adjusts the duty cycle of a switching voltage regulator.


Summary.

The boundary between these two distinct regions occurs near to or within the RF amplifiers or modules. Just where the regions approach is actually determined by the operating class of the amplifiers employed. Close examination of the transition within an amplifier module reveals that for optimal efficiency the two regions are separated by a reactive network. That reactive isolation prevents the low impedance power distribution supply from short circuiting the signal path. It also prevents the RF signal from influencing other modules through the power supply rails.

Class A amplifiers will always be inefficient, Class B will be better while classes C and D can be very efficient. By mismatching the power distribution supplies, while matching the RF signal path, it is possible, (by using class C or D amplifiers), to make equipment that operates at significantly better than 50% efficiency.

To ascribe to the RF signal path the “zero impedance = high efficiency” concepts applicable to power distribution systems is a mistake.

 

[sophiecentaur]

You clearly have lots of experience and knowledge about transmitter practice but the notion that the Maximum Power theorem somehow only works under certain circumstances is a step too far. It involves Conjugate matching, of course, Introducing reactance is a red herring. Resistive components are the only ones that dissipate Power.
"Underuse of" components is not as heinous a crime as 'overusing' Power. Good Engineering aims at minimising the appropriate losses. Electrical supply costs soon outweigh the cost of a powerful broadcast transmitter - the electricity bills would make your eyes water. No one would ever chuck away 500kW of RF power into the cooling water.

Likewise, in free space, the ratio of the E to M field strengths represents an impedance of approximately 120*Pi ohms. That does not in any way represent a dissipation of power.

If you look at the Impedance, looking into a transmitting antenna, you will come across a resistive component (if not, then it is not radiating any power). That resistive component is referred to as Radiation Resistance. The Characteristic impedance of a 50Ω line is the ratio of Reactances, as is the impedance of free space, effectively. But, in both cases, the Energy transfer is due to where the energy goes and not to the characteristic impedance.

 

[Averagesupernova]

I have a question for you baluncore. If zero output impedance is not applying to RF transmitters, then why are we not dissipating the same amount of power in the class C amplifier that drives the antenna system that we are radiating in the antenna? You said it yourself in a previous post.

A 1 kW RF transmitter driving a matched line does not generate >= 1 kW internally.
Class C and D amplifiers can operate at efficiencies better than 80% into matched lines.

I call you on this by saying: The reason it does not generate >= 1kW internall is because it has a Zout below 50 ohms which is exactly what sophie has been saying. The load at the end of the line may be 50 ohms and the line may be 50 ohms and we can call that matched but that does not mean the Zout of the transmitter is 50 ohms. It may well be spec'd to drive a 50 ohm load, but that is irrelevant.
...and then you go on to say:

The transmitter output stage has an impedance transformer called a tank circuit. To cancel the inductive component of the wound transformer it has a parallel capacitor, adjustable for different frequencies.

The load line or V:I ratio of the amplifier's active element is transformed to the impedance of the transmission line by the tank.

A reflected signal returning from the output t'line is reverse transformed by the tank from a V:I ratio of 50 ohm in the t'line to the V:I ratio of the amplifier's active element, and so is not reflected.

The reflected signal is irrelevant since the assumption is that there is no reflected signal in this discussion. In the real world there will always be some reflected power but it is low enough to be disregarded with respect to the subject being discussed.
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As for the tank circuit behaving as an impedance transformer. I suppose one can look at it in this manner. However, you cannot have it both ways. If the tank is acting as a transformer and matching the low impedance line to the transistor/tube, then you have to treat it as an actual transformer in that it is only changing the V:I ratio and not the power. So, by your logic the active element would once again be dissipating the same amount of power that the antenna is radiating and you clearly said this is not the case:

Class C and D amplifiers can operate at efficiencies better than 80% into matched lines.

 

[Baluncore]

As this subject is fraught by the many devils in the details, any general statement in a non-mathematical language must be expected to fail under some interpretation. So, full speed ahead and damn the torpedoes.

Unlike a linear supply regulator, a switching power supply can drop the voltage without the need to dissipate energy. It does it by employing a series inductor that acts as a reactive current limiter. The duty cycle of the switch, in conjunction with another switch or diode, effectively pulse-width-modulates the average value of the fixed inductive reactance at the switching frequency. That regulates the average current and so can be used to regulate the voltage efficiently.

The load line of an active device can also be modulated by oscillating between two points on the line rather than operating on the linear middle part of the line. That is a difference between the class A,B and class C,D amplifiers. The tank circuit will be different for different class amplifiers. That is because the effective output impedance of class A and B will appear resistive while the output impedance of classes C and D will be a very low resistance in series with a reactive network that functions as a flywheel.

Class A is inefficient because it operates in the linear mode. It forms a simple model of an externally matched amplifier in which a significant proportion of the energy is wasted in the output of the active device. Class D is efficient because the switching is between a point with high voltage with very low current and a point of high current with very low voltage, neither of which dissipates high power in the output of the active device. The output impedance is effectively the ratio of the pulsed current average to the pulsed average voltage. The lack of in-phase voltage and current is consistent with the use of a reactive element to limit the output power.

There is another way of viewing the design of a linear RF power amplifier. We can consider it to be two amplifiers in one. One is a voltage amplifier, the other is a current amplifier. The one active element performs both those tasks. The operating conditions of the active element can be chosen to maximise the output power, W_out, which is the product of V_out and I_out. The selection of active element operating conditions decides the input impedance. Matching between stages then becomes a case of juggling the voltage and current gain to maximise power output for hardware investment and/or energy expenditure. It should be clear that the ratio of V to I is the terminal impedance, and that it does not matter what V to I ratio is used so long as output power is optimised.

It should also be clear that when considering power output, neither voltage nor current is more important than the other. As such, arguing for a minimum ratio of voltage to current is unrealistic. What we actually need for economy is a minimum real loss resistance in series with the output. The output impedance Zo is then the ratio of output voltage to output current, which is quite different to the lossy real Rs component.

This demonstrates that a fixed voltage power distribution system should have a minimum series loss resistance. The output impedance as an output voltage to output current ratio is then unimportant.

On the other hand, the transfer of power between amplifier stages needs to transfer all available signal V and I for maximum power. That requires the V to I ratio, the impedance, be matched between sources and their loads for efficient power transfer. I agree that Rs << Zo to minimise the waste of the hard-won RF signal energy.

 

[AlephZero]

As this subject is fraught by the many devils in the details, any general statement in a non-mathematical language must be expected to fail under some interpretation.

Indeed - so why don't you back up your assertions with some math? (Give some references, if you don't want to duplicate standard theory)

Otherwise, you are imposing on the rest of the PF community to pick through your non-mathematical assertions and figure out which of them are correct and/or relevant.

For example, IMO much of your last post is confused through mixing different definitions of "efficiency", but I'm not interested in spending an hour of my time (literally) unpicking it in detail.

 

[sophiecentaur]

@Baluncore
You are totally missing the wood for the trees.
1. You can never ignore the maximum power theorem. It applies everywhere - with or without the detail.
2. Reactive components dissipate no power.
3. A low efficiency transmitter can be probably be well matched from the feeder as well as into the feeder
4. High efficiency transmitters (amplifiers) must have very low (or very high, aamof) source resistances. The way they are tuned to the feeder will (usually) assume the antenna is matched to the feeder and will be done to achieve efficiency. If the 1Ohm source is matched to the 50 Ohm feeder then they cannot be working at 90% efficiency because they will be seeing a 1 Ohm Load.
5. Introducing switched mode PSUs into your argument can only go against your ideas because they are designed to have a low internal resistance as possible (i.e. they don't get hot).


You are introducing so many irrelevant factors to back up the ideas you got from somewhere about matching. The only way you can prove this for yourself, and get it right, is to do the actual calculations with some real values. The analysis will be very hard. At any particular frequency, you can reduce all your networks to a simple emf and one source Z so you don't need to over-complicate things to find the ultimate truth in this.

I have very little argument with you except when you claim that matching both ways is possible to achieve with high efficiency. That just has to be wrong on basic, almost philosophical, grounds.

 

[Averagesupernova]

I just wish my questions would be answered.

 

[Baluncore]

I just wish my questions would be answered.

In post #46 you asked only one question.

I have a question for you baluncore. If zero output impedance is not applying to RF transmitters, then why are we not dissipating the same amount of power in the class C amplifier that drives the antenna system that we are radiating in the antenna?

My reply was;

The load line of an active device can also be modulated by oscillating between two points on the line rather than operating on the linear middle part of the line. That is a difference between the class A,B and class C,D amplifiers. The tank circuit will be different for different class amplifiers. That is because the effective output impedance of class A and B will appear resistive while the output impedance of classes C and D will be a very low resistance in series with a reactive network that functions as a flywheel.

Class A is inefficient because it operates in the linear mode. It forms a simple model of an externally matched amplifier in which a significant proportion of the energy is wasted in the output of the active device. Class D is efficient because the switching is between a point with high voltage with very low current and a point of high current with very low voltage, neither of which dissipates high power in the output of the active device. The output impedance is effectively the ratio of the pulsed current average to the pulsed average voltage. The lack of in-phase voltage and current is consistent with the use of a reactive element to limit the output power.

 

[Baluncore]

You can never ignore the maximum power theorem. It applies everywhere - with or without the detail.

I do not ignore it. I simply do not misapply it to concepts such as matching characteristic impedance.
The MPTT applies to the real dissipative internal resistance of the output of power distribution systems, NOT to the complex ratio of the voltage and current waves propagating toward the end of a lossless transmission line.

The problem arises because Impedance has two meanings that share the same dimensions and unit, the ohm.
Firstly, the complex vector sum; Impedance = Resistance + Reactance.
And secondly, the complex ratio;
Impedance = Voltage / Current, for example, the characteristic impedance of a transmission line, or;
Impedance = Electric / Magnetic field strengths, for example the intrinsic impedance of free space.

The MPTT indicates, that to be efficient, the internal series resistance, Rs, of a line must be much less than the characteristic impedance, Zo. I have absolutely no argument with that.

I have very little argument with you except when you claim that matching both ways is possible to achieve with high efficiency. That just has to be wrong on basic, almost philosophical, grounds.

It depends on the situation you are referring to. A fixed voltage power source, with unlimited current capability should never be matched to any load in any direction. I am talking about “matching both ways” along a signal path that has a limited RF energy available. The example I used was between two stages of an RF amplifier.

I see no reason why “efficient matching both ways” must be philosophically impossible. Where two ports are coupled by a lossless transformer that has an appropriate ratio, and any reactive mismatch is neutralised with a conjugate reactance, then there can be no real loss, hence there must be high efficiency.

Perfectly matched interfaces are invisible from both sides. We live in a world composed of mismatched interfaces.
It is all the more beautiful because we can see that many of the mismatches are EM wavelength dependent.

 

[sophiecentaur]

I see no reason why “efficient matching both ways” must be philosophically impossible. Where two ports are coupled by a lossless transformer that has an appropriate ratio, and any reactive mismatch is neutralised with a conjugate reactance, then there can be no real loss, hence there must be high efficiency.

That statement includes nothing about the source impedance relative to the load impedance, which is what the whole of the MPT is about - so we can ignore it. Obviously, a lossless transformer is 100% efficient but it can't impose efficiency onto anything else.

If you see no reason why not then you must be able to give an example where it actually happens, involving some numbers.

It strikes me that you were either mis-taught or mis understood this particular thing, a long time ago and the misconception has stuck in your mind. You are treating it as an article of faith and you keep trying to justify it by digging further and further into irrelevant complexities of non-linearity and Impedance transformation rather than aiming at the sort of simplification which is the essence of good theory.

This nonsensical distinction between the two 'sorts of' Impedance doesn't help in any way. If you were really across this stuff you would see there is no such distinction. 'Purple passages' do not cut any ice in Science and Engineering proofs.

Give us a credible reference to all this. That's the acid test, as usual.

 

[Baluncore]

This nonsensical distinction between the two 'sorts of' Impedance doesn't help in any way. If you were really across this stuff you would see there is no such distinction.

I'm sorry that you are blind to the distinction.

 

[sophiecentaur]

I'm sorry that you are blind to the distinction.

Haha
I am sorry you are blind to the equivalence.

Still waiting for a reference, btw.

 

[Baluncore]

This thread is quite confused. A reference to precisely what. Life, the universe and everything ?

 

[sophiecentaur]

A reference to the meaning / relevance / possibility of matching for high efficiency and perfect termination, that you claim. A reference that provides an exception to the MPT, even.

 

[Baluncore]

A reference to the meaning / relevance / possibility of matching for high efficiency and perfect termination , that you claim.

Matching what ? for high efficiency. Matching at one frequency or broadband ? One way or reciprocal matching ?

 

[sophiecentaur]

I think you are being disingenuous here. I thought we were well aware of the difference of opinion.

The root of the disagreement is that you seem to claim that a transmitter can be operated 1.) at high efficiency 2.) see the load as the same as the transmitter internal resistance 3.) present a perfect termination when looking back from the load.
The three are not mutually compatible. A reciprocal match can only be achieved with 50% efficiency (as the MPT tells us)
The MPT can be derived in three lines andI don't think is in doubt.

If the above is not what you are claiming and I have misunderstood your posts then we have no argument.

 

[Baluncore]

I think you are being disingenuous here. I thought we were well aware of the difference of opinion.

It has been quite clear to me that there has been a misunderstanding for some time. That is why I have been trying to isolate a simple situation where we can either clarify the misunderstanding or resolve the difference. You call it “disingenuous”, I call it reductionist.

The root of the disagreement is that you seem to claim that a transmitter can be operated
1.) at high efficiency
2.) see the load as the same as the transmitter internal resistance
3.) present a perfect termination when looking back from the load.

I do claim points 1 and 3. I don't understand precisely what you mean by your point 2.
By “transmitter internal resistance”, do you mean the “load line” of the active device ?

To “match” something actually requires the definition of two things, the output port of a source and the input port to a load.
The hypothetical matching network is placed between those two defined ports.
What are those ports in your example ?