On 7/7/2015 2:32 PM, Ian Jackson wrote:
In message , Jerry Stuckle
writes
On 7/7/2015 6:25 AM, Ian Jackson wrote:
In message , Jerry Stuckle
writes


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

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




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

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

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

The point I'm trying to make is not technical. It's simply one of verbal
logic. Without the presence of a standing wave, you can't possibly have
something called a "standing wave ratio". But, like all RF engineers, an
SWR of 1-to-1 is something I too strive to achieve!
"Yesterday, upon the stair,
I met a man who wasn't there.
He wasn't there again today,
I wish, I wish he'd go away..."

It doesn't matter what your verbal logic is, Ian. The correct term is
"Standing Wave Ratio", and an SWR of 1:1 means there is no reflected
power and you have a perfect match.

That is the technical definition of a technical term.

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

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

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.

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


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

Why don't you email the author and get his take on your assumptions?





--
Jim Pennino

On 7/7/2015 11:09 AM, Ian Jackson wrote:
In message , rickman writes
On 7/7/2015 6:25 AM, Ian Jackson wrote:
In message , Jerry Stuckle
writes


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

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

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

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

Sounds great, but that is not how the VSWR is defined.

--

Rick

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

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

writes
Wayne wrote:


"Jeff Liebermann" wrote in message
...

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

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

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

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

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

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

Both the antenna and the source are terminations.

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

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

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


The wikipedia entry is correct as written.

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

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

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

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

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

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

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

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

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


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

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


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

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


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

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

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

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

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

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

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


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

Can you provide a reference to any of this?


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

Anything more accessible?


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

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


Yes, there is.

I did NOT say that a wire was dissipationless.

More communication failure. I never said *that*.

Please reread the above.

Ditto.


I said "There can be a lossless resistive part of source impedance".

What you said is clearly quoted above.


Source impedance may or may not contain resistance which dissipates. A
50 ohm piece of coax has an impedance of approximately 50+j0 ohms. So,
does the 50 ohms dissipate? Look at the specifications of the coax. How
much loss?

I've never seen a coax specified with a complex impedance. But I'll
take your word that this is accurate.

However, if you drive a 50 ohm cable, how much current flows? If it is
unterminated, I believe the result is none because it all flows back
from the reflection and is returned to the source. If you terminate it
with a 50 ohm load it is all absorbed and dissipates the appropriate
wattage.

So I don't see how the 50 ohm impedance of the cable is in any way a
useful analogy to say the real part of the impedance does not dissipate.
A transmission line is not a load.


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

If you don't want to discuss this, why do you type so many words?

Maybe it would be useful to explain how an impedance matching network
would provide a conjugate match. That is, show an example with real
values for components. If this is too much work and you don't wish to
do that, fine. I'm ok with it. But no need to insult me. Better to
just not respond.

--

Rick

On 7/7/2015 1:44 PM, wrote:
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!

Despite the name, VSWR is defined in terms of complex impedances
and wavelengths, not "waves" of any kind.



Actually, VSWR is defined as the ratio of Vmax/Vmin.

On 7/7/2015 12:47 PM, Dave Platt wrote:
In article , rickman wrote:

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.

Yah.

It's a question of terminology. Unfortunately, one term has come to
be used for two (related but different) concepts.

There is "resistance", as in the E=I^2*R sort. If I recall correctly,
Maxwell refers to this as "dissipative" impedance. If you put current
through a dissipative resistance, a voltage drop develops across the
resistance, and power is dissipated.

There are plenty of examples of this, with which I'm sure you're
familiar.

There is also "resistance", as in "the 'real', non-reactive component
of a complex impedance, in which current is in phase with voltage."
This type of "resistance" is fundamentally non-dissipative - that is,
you can run power through it without dissipating the power as heat.

There are also good examples of this. One "textbook" example would be
a perfectly-lossless transmission line... say, one made out of a wire
and tube of a superconductor, cooled to below the superconducting
temperature.

You can (in principle) build such a superconducting coax to have
almost any convenient impedance... 50 or 75 ohms, for example. Since
we're theorizing, let's assume we can built one a few trillion miles
long... so long that the far end is light-years away.

If you hook a transmitter to one end of this and start transmitting,
it will "look" to the transmitter like a 50-ohm dummy load. The
transmitter itself won't be able to tell the difference. The
transmitter puts out an RF voltage, and the line "takes current"
exactly in phase with the voltage, in a ratio of one RF ampere per 50
RF volts.

But, there's a fundamental difference between this "resistance" and
that of a dummy load. A 50-ohm dummy load's resistance is
dissipative... all of the power going into it turns into heat, and is
dissipated in accordance with the fundamental laws of thermodynamics.

*None* of the power being fed into the superconducting coax, is
dissipated as heat in the coax. All of the power still exists, in its
original RF form. It's being stored/propagated down the coax without
loss.

When it hits a load at the other end, it may be dissipated as heat
there.

Or, perhaps not. What if what's at the other end of the
superconducting coax is a superconducting antenna, tweaked to present
an impedance of exactly 50 ohms? The RF will be radiated into space.

And, "free space" is another great example of a medium that has a
well-defined "resistance" (in the non-dissipated sense).

https://en.wikipedia.org/wiki/Impedance_of_free_space

One of the fundamental jobs of an antenna, is to match the impedance
of its feedline to the impedance of free space.

Now, any coax you can buy at the store has *both* types of
"resistance", of course. It has a dissipative component, and a
non-dissipative component. Typically, the more you spend and the more
you have to strain your back carrying it around, the lower the amount
of dissipative resistance (which is only good for keeping the pigeons'
feet warm) and the more predictable and precisely-defined the
non-dissipative part.

Excellent explanation, Dave. I don't have the ability to educate like
you do. Thanks.

John

On 7/7/2015 1:58 PM, 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.

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


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

Why don't you email the author and get his take on your assumptions?



Why don't YOU? You are the one in need of knowledge. If I do it and
report back here you will just doubt it or find something else to argue
about. Better you should do it first-hand.

On 7/7/2015 1:37 PM, wrote:
John S wrote:
On 7/6/2015 1:03 PM, wrote:
John S wrote:
On 7/6/2015 11:01 AM, Jerry Stuckle wrote:
On 7/6/2015 4:20 AM, Ian Jackson wrote:
In message , rickman writes



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

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

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

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

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

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

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

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

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


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

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

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

What part of "you get a whole new set of laws" was it you failed
to understand?

Here's a clue for you; at DC the reactive components of a length
of wire are irrelevant but at AC they are not.

So, at 1Hz the law has changed, eh? What new law do I need to use?

On 7/7/2015 10:05 AM, Jerry Stuckle wrote:
On 7/7/2015 3:05 AM, John S wrote:
On 7/6/2015 1:03 PM, wrote:
John S wrote:
On 7/6/2015 11:01 AM, Jerry Stuckle wrote:
On 7/6/2015 4:20 AM, Ian Jackson wrote:
In message , rickman
writes



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

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

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

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

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

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

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

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

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


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

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

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


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

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

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

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

There is a huge difference between ac and dc!


Yes, but the LAWS have not changed. The components have changed. So,
changing the components changes the laws of physics?

Suppose you apply .01Hz AC RMS to the components you specified. What then?

On 7/7/2015 1:52 PM, wrote:
Brian Reay wrote:

Do the experiment.

Did it decades ago in electromagnetics lab with calibrated test equipmemnt,
not with amateur radio equipment.

Post the original lab notes, please. That way we cannot challenge the
accuracy of your memory.