"John S" wrote in message ...

On 7/4/2015 12:32 PM, Wayne wrote:


"John S" wrote in message ...
On 7/4/2015 10:55 AM, Wayne wrote:


By the way, Wayne...

Are you aware of a companion Excel application for EZNEC called
AutoEZ? You can run many test cases in a few seconds using it. You can
find it on the EZNEC site.

It is how I generated the data I posted.

Thanks, I'll look for that. I run the old wood burning version 3.0.


BTW, Wayne, what are the dimensions of your metal roof? And how high?

It's about 20 by 35 feet, 9 feet above ground. it's a big patio cover.

The underside is relatively smooth, but the top has all the ribbed sections.
Ribs are about 20 feet long and separated by about 2 feet.

Also on top are various runs of conduit and three large rectangular covers
for the underside lights.

I know this is TMI, but I'm just illustrating why I haven't tried to model
the entire metal roof....it's complicated

And I don't really need accurate patterns.

On 29/06/15 16:48, Wayne wrote:
As a lead in, I use a 16 ft vertical on 20-10 meters, mounted on a flat
metal roof. The antenna is fed with about 25 feet of RG-8, and there is
a tuner at the transmit end.

While I'm pretty happy with the antenna, I'd like to simplify the matching.

Thus, the question: what is the purpose of a 1:4 unun on a 43 foot
vertical? ( I assume the "4" side is on the antenna side.)

I'd expect a better coax to antenna match when the antenna feedpoint is
a high Z (example, at 30 meters), but I'd also expect a worse coax to
antenna match when the feedpoint is a low Z (example, at 10 meters).

Is that the way it works, or is there other magic involved?


I'm not going to disagree with Dave Platt's post re how the matching on
a 43' vertical works.

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.

Remember, you really want a low SWR for two reasons, one because modern
radios demand it but also to reduce coax (or feeder) loss. With an
matcher at the antenna feed point, coax losses are minimised. An ATU at
the TX end does nothing to reduce coax losses in real terms.

wrote:

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.

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.



The impedance of a transmitter output will be nothing like 50 ohms
resistive, as this would result in an efficiency well below 50%, with
all the normal amplfier losses plus the actual RF power produced being
50% dissipated in the PA. This is why matching in the forward direction
coexists with a mjaor mismatch in the reverse direction. This is good
because if there is any reflected wave we don't want it to add yet more
to the PA dissipation. But it does explain what is happening, and why
there are increased losses in the feeder as well as the matching
networks.



--
Roger Hayter

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.

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.

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!)

--
Roger Hayter

On 5.7.15 18:56, rickman wrote:
On 7/4/2015 9:43 PM, Jeff Liebermann wrote:
On Sat, 04 Jul 2015 19:33:30 -0400, Jerry Stuckle
wrote:

On 7/4/2015 7:22 PM, Jeff Liebermann wrote:
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."

And you believe everything Wikipedia says? ROFLMAO.
But that also explains your ignorance.

Let's see if I understand you correctly. You claim that with a power
amplifier (source) output impedance that is perfectly matched to the
coax cable, but not necessarily the load (antenna), any reflected
power from the load (antenna) is bounced back to the load (antenna) by
the perfectly matched source (power amp). Is that what you're saying?

Yet, when I have a perfectly matched load (antenna), all the power it
is fed is radiated and nothing is reflected. You can't have it both
ways because the reflected power from the load (antenna), becomes the
incident power going towards the source (power amp). Matched and
mismatched loads do NOT act differently depending on the direction of
travel. If you claim were true, then transmitting into a matched
antenna or dummy load would reflect all the power back towards the
transmitter.

I think this is one of those situations where a casual explanation won't
work. You can use a "casual" explanation when the various
qualifications for a simplification apply. But to do that, the
qualifiers have to be fully understood and no one here is showing what
the qualifiers are much less that they are met. So until we get a real
explanation I will stick with what I recall. In the end, to settle this
we may have to use the math.

I'm sure someone in s.e.d could explain this properly. Some of them may
be purely argumentative, but some really know their stuff. I believe
the description of a conjugate match is the mathematical inverse of the
complex impedance of the antenna "viewed" through the feed line, but I
have to admit I don't really know what that implies or if it is even an
accurate description.


You're right. To get a 1:1 match for the piece of feedline between
the transmitter and the antenna tuning unit, the tuning unit has
to present a conjugate match to the feedline from the tuning unit
to the antenna. If there is a reflected wave from the antenna, it
will be re-reflected back toward the antenna from the tuning unit.
The ping-ponging signal will die out by antenna radiation or feedline
losses. The situation is quite OK with slow modulations (like voice),
but the ping-ponging is unacceptable for fast signals (like analog TV).

--

-TV

Roger Hayter wrote:
wrote:

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.

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.



The impedance of a transmitter output will be nothing like 50 ohms
resistive, as this would result in an efficiency well below 50%, with
all the normal amplfier losses plus the actual RF power produced being
50% dissipated in the PA. This is why matching in the forward direction
coexists with a mjaor mismatch in the reverse direction. This is good
because if there is any reflected wave we don't want it to add yet more
to the PA dissipation. But it does explain what is happening, and why
there are increased losses in the feeder as well as the matching
networks.

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.


--
Jim Pennino

On 7/5/2015 12:21 PM, Jeff Liebermann wrote:
On Sat, 04 Jul 2015 22:37:41 -0400, Jerry Stuckle
wrote:

On 7/4/2015 9:43 PM, Jeff Liebermann wrote:
Let's see if I understand you correctly. You claim that with a power
amplifier (source) output impedance that is perfectly matched to the
coax cable, but not necessarily the load (antenna), any reflected
power from the load (antenna) is bounced back to the load (antenna) by
the perfectly matched source (power amp). Is that what you're saying?

With a perfect matching network and a perfect feedline (which is what we
are discussing), that is true.

Ok, so we agree on that part.

Yet, when I have a perfectly matched load (antenna), all the power it
is fed is radiated and nothing is reflected. You can't have it both
ways because the reflected power from the load (antenna), becomes the
incident power going towards the source (power amp). Matched and
mismatched loads do NOT act differently depending on the direction of
travel. If you claim were true, then transmitting into a matched
antenna or dummy load would reflect all the power back towards the
transmitter.

Nothing wrong with it at all

Ah, but there's plenty wrong with your view. At the load (antenna)
end of the coax, we both agree that with a perfect match, perfect
coax, and perfect load, there's is no reflection. Yet when you look
at the other end of the same coax, the same perfectly matched coax
(S21 back into the PA) suddenly decides to reflect any RF that might
be returned from a load (antenna) mismatch. It seems rather odd that
RF would act differently at opposing ends of the coax cable. In the
forward direction, a matched load either absorbs or radiate. In the
reverse direction, a matched load changes its mind and decides to
reflect? I don't think so.


That's because you don't understand a perfect matching network or
feedline - which is what we are discussing. Your knowledge of theory is
sadly lacking.

There's another problem with your view of how VSWR works. If I
transmitted into an open or short circuit load (antenna), all the
forward RF would be reflected back to the source (PA). That would
mean that the PA will need to protect itself from over voltage or over
current using the traditional VSWR protection circuit. Yet, if the PA
were perfectly matched into a perfect coax cable, all that reflected
RF would bounce off the PA and back to the load (antenna). The PA
would not see any of that RF, and there would not be any need for a
VSWR protection circuit. I don't think so.


Once again, you're not discussing the same subject.

Personal experience with blowing up finals has demonstrated to my
satisfaction that a perfectly matched PA is quite capable of being
blown up by transmitting into an open or short with no VSWR
protection. Presumably, the damage was done by the reflected RF
(causing over voltage or over current in the PA) which would not be
present in your scheme of things, with a perfectly matched PA and
coax.


See above.

Another problem is IMD (intermodulation) products produced in the
power amplifier (PA). This is not a major problem with HF radios, but
is a serious problem with mountain top repeater sites. The antennas
on the towers tend to be rather close together. RF from a nearby
transmitter can couple into adjacent antennas, travel down the coax to
the PA, mix with the transmit signal in the PA, get amplified by the
PA, and get re-radiated by the antenna. The effect is typically
blocked by cavity filters and one-way isolators or circulators. The
problem here is that if the perfectly matched PA really did reflect
anything coming down from the antenna back to the antenna, there would
be no need for such IMD protection. The RF from the adjacent antenna
would simply bounce back towards the antenna and be re-radiated
without any mixing taking place. It would be a wonderful world if it
worked that way, but it obviously does not.


Once again, see above.

There are other problems, but they require math to explain, which
requires more time than I have available right now.



Because it's obvious you don't understand the theoretical aspects of the
system. Sure, they don't exist in the real world. But they provide a
simplified system for a start to the math; real world deficiencies can
then be added to the math to define the real-world aspects.

But you don't have the background to understand the theoretical aspects
we are discussing. I suggest you quit showing your ignorance.

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

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

On 7/5/2015 12:10 PM, rickman wrote:
On 7/5/2015 10:48 AM, Ralph Mowery wrote:
"Ian Jackson" wrote in message
...
class-C PAs supposed to be around 66.%?). Also, as much power would be
dissipated in the PA stage as in the load.

I think this is easy to disprove in practice. I have an amp that is
probably class B, but it does not mater about the class. If I adjust
it to
an input of 2000 watts from the DC power supply, I get out 1200 watts
to a
resistive dummy load. If the above is true, I should have to input 2400
watts to the final stage. Now can someone tell me where the extra 400
watts
are comming from ? This 400 extra watts is not even counting on any
loss in
the circuits.

The idea of matched impedance transferring maximum power is one of those
"simplified" descriptions that has preconditions that some people forget
about. It is not a universal truth.

If you have a transmitter output with a fixed impedance you can get
maximum power transferred to the feed line by matching the feed line
impedance to the transmitter output impedance. But if your feed line
impedance is the constant, you get maximum power transfer by minimizing
the transmitter output impedance, meaning zero ohms.

So you could in theory get 1200 watts into your feed line while drawing
only 1200 watts from the power supply.


So why don't manufacturers design transmitters with 1 ohm output
impedance, Rick? Hint: Sophomore-level AC Circuits course in virtually
any EE degree, and anyone claiming an EE degree should be able to tell why.

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

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

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.

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

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

On 05/07/2015 21:17, wrote:
Roger Hayter wrote:
wrote:

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.

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.



The impedance of a transmitter output will be nothing like 50 ohms
resistive, as this would result in an efficiency well below 50%, with
all the normal amplfier losses plus the actual RF power produced being
50% dissipated in the PA. This is why matching in the forward direction
coexists with a mjaor mismatch in the reverse direction. This is good
because if there is any reflected wave we don't want it to add yet more
to the PA dissipation. But it does explain what is happening, and why
there are increased losses in the feeder as well as the matching
networks.

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.

They are designed to drive into a 50 ohm load, that doesn't mean they
have a 50 ohm source impedance. Otherwise efficiency would be rather
'disappointing'.

The PA stages are designed to operate safely with a load equivalent to a
SWR of (typically) 1.5:1 . Any higher, and it means the load is out of
spec, and the PA leaves its safe area of operation (assuming there is no
mechanism to reduce the power). This is were the myth of RF 'entering'
the PA came from - people thinking that a high SWR meant the reflected
RF was getting into the PA and causing damage. In fact, it 'sees' a
mismatch and therefore can't enter the PA.