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!





--
Ian

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.

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).

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.

No, honestly, you're not getting more power output when you match the
load to the source. *If* you have a *given* voltage generator with a
*given* source impedance, then yes: that situation arises, for
instance, if you have a very low noise amplifier with given output
characteristics and you want to extract the maximum signal power in
order to maintain the best noise factor through stages of amplification.
But when you are designing a PA you start with a pile of components (or
a catalogue of same) and you choose your voltage swing and current
capacity to put as much power in the load as you want to (limited
largely by the heat dissipation of the output devices in a practical
circuit) and design the circuit to dissipate as little power in the
amplifier as you can. You are *not* interested in transferring as much
power as you can from a given circuit. It may only be tenth of the
power output that you could get (ignoring practical dissipation limits)
from a certain voltage with a different load, or a higher voltage with
the same load (which you could achieve with a transformer), but that is
irrelevant. Among other things, you are likely to end up with a low
source impedance compared with the load and that makes no difference to
the operation of the transmission line or aerial.



--
Roger Hayter

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?

--

Rick

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?

--

Rick

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.

rickman wrote:

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?

You are, of course, right. I suspect that VSWR was defined to give
technicians a nice easy number to aim for, rather than infinite return
loss, to indicate no reflections.


--
Roger Hayter

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.

--

Rick

On 06/07/15 01:21, 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.

That is true but is doesn't address the point. There should still be
somewhere to represent the source impedance, albeit normalised.

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

Likewise, that is a sweeping statement which evades the point.

On 06/07/15 17:48, wrote:
John S wrote:
On 7/6/2015 12:19 AM, wrote:
John S wrote:
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.


So, it can't be used in a 50 ohm environment? What does that have to do
with anything? The chart has a SWR graph and nowhere does it need source
impedance. If you disagree, please link to one.


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

Please show the EZNEC statement that "assumes the transmission line
matches the transmitter". Look in the help section if you have EZNEC and
can cut and paste or just refer me to the chapter and verse. Also, if
you have EZNEC, you can insert a transmission line with arbitrary
characteristic impedance, put a load on the far end matching the line,
and look at the SWR. It will still be 1:1 because the LOAD matches the
LINE. Not because EZNEC assumes a source impedance. Try it with and
report back here.

There is no way that a source initiates reflections. That is a property
of the line and load only. It may re-reflect a wave reflected from the
load, but that is all.

You can also verify this in LTSPICE if you wish.

What happens if you take any off the shelf commercial amateur radio
transmitter that does not have a built in tuner and:

Attach a 10 Ohm load.

Attach a 200 Ohm load.

Attach a 1,000 Ohm load.

Attach a 1 Ohm load.

Attach a 50 Ohm load.



Please address my questions first before setting up another strawman.

Start with Electromagnetics by Kraus and Carver, Chapter 13.



Do the experiment.

On 07/07/15 00:36, Dave Platt wrote:
The only case I am aware of that will give total reflection is when
the terminal is open circuit with infinite impedance absorbing *no*
signal.

Also when it is a zero impedance (short circuit).

I'm trying to picture this.

In the case of an open circuit a matched driver drives the transmission
line to 50% of the driving voltage. The wave reaches the open
termination and is reflected with the same polarity resulting in a
return wave that reaches 100% of the driving voltage.

In the same vein, if the wave hits the short circuit the reflected wave
will be the opposite polarity making the reflected wave 0% of the
driving voltage resulting in the short circuit eventually showing to the
drive circuit.

Yup. You can see this happen on a time-domain reflectometer (an
o'scope and a pulse generator will do).

(although, to pick nits, I'd clarify your latter paragraph to read
"will be of the opposite polarity, making the sum of the forward and
reflected wave 0% of the driving voltage...")




There is a good demo on YouTube of this. The presenter built simple
little, rather neat, pulse generator to demonstrate such things. I built
one for demos. It works very well.

It is quite useful for measurements etc.

https://www.youtube.com/watch?v=9cP6w2odGUc