what happens to reflected energy ?
 

On Sun, 6 Jun 2010 14:21:16 -0700 (PDT), Keith Dysart
wrote:

The 100W forward and 50W reflected have no relation to actual powers

From MECA, makers of Isolators - their application:

The isolator is placed in the measurement path of a test bench between
a signal source and the device under test (DUT) so that any
reflections caused by any mismatches will end up at the termination of
the isolator and not back into the signal source. This example also
clearly illustrates the need to be certain that the termination at the
isolated port is sufficient to handle 100% of the reflected power
should the DUT be disconnected while the signal source is at full
power. If the termination is damaged due to excessive power levels,
the reflected signals will be directed back to the receiver because of
the circular signal flow.
....
MECA offers twenty-four models of isolators and circulators in both N
and SMA-female connectors with average power ratings from 2 - 250
watts.

73's
Richard Clark, KB7QHC

what happens to reflected energy ?
 

On 7 jun, 03:04, Richard Clark wrote:
On Sun, 6 Jun 2010 15:21:59 -0700 (PDT), Wimpie
wrote:

Try the following experiment:

-sigh-

Did you do the experiment (forward power versus mismatch)?

Hi Wimpie,

This is not what I asked for. *You said what the source was NOT, but
you cannot say what the source IS.

The source is not a bipedal reptile. *The source is not an elongated
diphthong. *The source is not the resurrection of a deity (some may
argue that more than I would care to follow). *The source is not....

There are many ways to say what the source is NOT, and that will never
inform us about the source. *I see many draw deuces to this question
and try to convince everyone that the card is a pair of winning aces.

Probably you consumed something wrong here.

Now make some mismatch (for example VSWR=2 at different phase) and
read the forward power. Did it change? If so, the output impedance is
no (longer) 50 Ohms.

What IS it now? *If you could measure it once, you should be able to
tell us what it is this time too. *I did this for years to methods set
by the National Bureau of Standards. *You have drawn a deuce, not two
aces.

Some values:
9 +/-1 Ohm (real impedance), reference impedance 16 Ohms, 8 MHz
amplifier (ISM). Exact value requried because of additional filtering,
based on several IRF110, push pull.

2 Ohms (small reactance present), 10V/1A driver (sinusoidal), 700
kHz, slightly saturating class C output stage based on single mosfet.
RF feedback present to guarantee value below 2 Ohms. also here,
certain maximum value was required for the application

|RC| 0.80 (actual value depending on type of additional filtering in
between), 5kW pulsed power amplfiier, non 50 Ohm application.
Actual value not important, but is a recent project, so I knew from
memory.

See also EL34 example in posting to Walt.



Measuring method used: change in resistive load, from voltage change
you can calculate the current change, hence the output impedance. One
note, except two, all where solid state.

Yes, that is called a load pull, but you offer no data - another
deuce. *I have done a lot of load pulls. *

All high efficiency designs (class E, D) that I did have output
impedance far from the expected load impedance. With "far" I mean
factor 2 or factor 0.5. I did not measure that (as it is not
important in virtually all cases), but know it from the overload
simulation/measurement and I did the design myself.

You didn't measure it, but you can state the value - interesting state
of guessing. *So, what did the simulation/non-measurement give you as
a value? *What IS the value? *Another deuce.

Please read again last part of sentence above your text.


The reason for not being 50 Ohms

Reasons abound. *BP is giving us reasons why the Gulf Coast shouldn't
worry. *Data has proven that reasons don't work and neither does their
well. *That is a Joker draw.

High efficiency CW amplifiers

The TS830S is not such an animal. *Another Deuce.

Not al hams use TS830S, See more recent reply of Walt to Tom's
posting.

If you can drop me a link to the discussion of the TS830s, I would
appreciate that.

Google "Plate Resistance" in this group.


73's
Richard Clark, KB7QHC

I hope, further replies from you will be constructive,


Wim
PA3DJS

what happens to reflected energy ?
 

On Jun 8, 3:47*am, Richard Clark wrote:
On Sun, 6 Jun 2010 14:21:16 -0700 (PDT), Keith Dysart

wrote:
The 100W forward and 50W reflected have no relation to actual powers

From MECA, makers of Isolators - their application:

The isolator is placed in the measurement path of a test bench between
a signal source and the device under test (DUT) so that any
reflections caused by any mismatches will end up at the termination of
the isolator and not back into the signal source. This example also
clearly illustrates the need to be certain that the termination at the
isolated port is sufficient to handle 100% of the reflected power
should the DUT be disconnected while the signal source is at full
power. If the termination is damaged due to excessive power levels,
the reflected signals will be directed back to the receiver because of
the circular signal flow.
...
MECA offers twenty-four models of isolators and circulators in both N
and SMA-female connectors with average power ratings from 2 - 250
watts.

Good day Richard,

You have located several examples from reputable vendors where the
behaviour of directional couplers is described in terms of power
in a forward and reflected wave. This model of behaviour works within
its limits and allows for convenient computation and prediction of
the behaviour. But all of these papers have the appropriate
discipline and do not ask the question "where does the reflected
energy go?" which is good, for this exceeds the limits of the
model.

As soon as one assigns tangible energy to the reflected wave, it
becomes reasonable to ask for an accounting of this energy and
the model is incapable of properly accounting for the energy.

Following this weakness back through the model, the root cause
is the attempt to assign tangible energy to the reflected wave.
Think of it as a reflected voltage or current wave and all will
be well, but assign power to it and eventually incorrect
conclusions will be drawn.

For those who understand this, and know that "where did the
reflected energy go?" is an invalid question, using the power
model within its limits will not cause difficulties. But for
those who are not careful, great difficulties arise and a lot
of fancy dancing is offered to work around the difficulties,
unsuccessfully.

Just for fun, here is a simple example.

100V DC source, connected to a 50 ohm source resistor,
connected to 50 ohm transmission line, connected to a 50 ohm
load resistor. Turn on the source. A voltage step propagates
down the line to the load. The impedances are matched, so
there are no reflections. The source provides 100W. 100W is
dissipated in the source resistor. 100W is dissipated in the
load resistor. Energy moves along the transmission line
from the source to the load at the rate of 50W.
All is well.

Disconnect the load. A voltage step propagates back along
the line from the load to the source. In front of this
step current continues to flow. Behind the step, the
current is 0. When the step reaches the source there is
no longer any current flowing. The source is no longer
providing energy, the source resistor is dissipating
nothing, and neither is the load resistor.

Proponensts of the power model claim that energy is
still flowing down the line, being reflected from the
open end and flowing back to the source. Since the
source is clearly no longer providing energy, great
machinations are required to explain why the reflected
'energy' is re-reflected to provide the forward 'energy'.

But really, does anyone believe that a length of
transmission line, charged to 100V voltage with zero
current flowing, is actually simultaneously transporting
energy in both directions?

For even more fun, replace the ideal conductors in the
transmission line with some lossy conductors. How much
of the reflected and re-reflected energy flowing up and
down the line will be dissipated in the conductors?
Remember that the current is zero, everywhere along
the line.

....Keith

what happens to reflected energy ?
 

On Jun 8, 5:57*am, Keith Dysart wrote:
As soon as one assigns tangible energy to the reflected wave, it
becomes reasonable to ask for an accounting of this energy and
the model is incapable of properly accounting for the energy.

Then one is using the wrong model which is typical of many RF
engineers. Optical physicists have been accounting for reflected
energy for decades. That one is ignorant of where reflected energy
goes is not a good reason to abandon the law of physics that says that
*ALL* EM waves contain ExH energy, including reflected EM waves. EM
waves cannot exist without ExH energy so you might as well say that
reflected waves do not exist at all because that is the logical
conclusion based on your false premises.

Optical physicists have been aware of the power-density/interference
equation for a long, long time. It is covered in any good reference
book on optics, e.g. by Hecht and by Born and Wolf. I have posted the
details - why do choose to remain ignorant? At least take time to
comprehend the technical information available from the field of
optics and report back to us why many decades of that EM wave
knowledge is wrong. Here is the equation that explains where the
reflected energy goes. All you have to to is track the energy back in
time.

Ptot = P1 + P2 + 2*SQRT(P1*P2)*cos(A)

where A is the angle between the two voltages. The last term is known
as the "interference term".

When two coherent, collimated, signals interfere while traveling in
the same direction in a transmission line: If they are 90 degrees
apart, there is zero interference and the I and Q components are
easily recovered.

If they are less than 90 degrees apart, the interference is
constructive and the phasor superposition of the two waves results in
*more power in the total superposed wave* than in the arithmetic sum
of the two powers. The extra energy has to come from somewhere. In the
absence of a local source, the extra energy has to come from
destructive interference in the opposite direction, e.g. at a Z0-
match.

If they are between 90 degrees and 180 degrees apart, the interference
is destructive and the phasor superposition of the two waves results
in *less power in the total superposed wave* than in the arithmetic
sum of the two waves. The "left-over" energy has to go somewhere so it
has to be delivered in the opposite direction to the area that
supports constructive interference.

Here is what Hecht said: "Briefly then, optical (EM wave) interference
corresponds to the interaction of two or more (EM) lightwaves yielding
a resultant irradiance (power density) that deviates from the sum of
the component irradiances."

Hecht's statement hints to where the reflected energy goes. In the
case of wave cancellation at a Z0-match that eliminates ExH reflected
energy flowing toward the load, all of the ExH reflected energy is
recovered and redistributed back toward the load.

Following this weakness back through the model, the root cause
is the attempt to assign tangible energy to the reflected wave.
Think of it as a reflected voltage or current wave and all will
be well, but assign power to it and eventually incorrect
conclusions will be drawn.

Only if the model is inadequate. Optical physicists assign tangible
energy to reflected waves all the time. Many are probably reading
these postings and laughing at the collective ignorance about EM
waves.

Take a look at this web page:

http://www.teachspin.com/instruments...eriments.shtml

Scroll down to, "Using Dielectric Beamsplitters to find the "missing
energy" in destructive interference". There it is, all spelled out for
you - where the reflected energy goes. Reflected energy is never
actually missing - what are missing are a few of the brain cells that
need to be used to think about reflected energy. :-)

For those who understand this, and know that "where did the
reflected energy go?" is an invalid question, using the power
model within its limits will not cause difficulties. But for
those who are not careful, great difficulties arise and a lot
of fancy dancing is offered to work around the difficulties,
unsuccessfully.

Yes, throughout history, ignorant people have brought great
difficulties upon themselves because they refuse to alleviate their
ignorance. Those who refuse to learn from their mistakes are destined
to repeat those same mistakes. And you were making these exact same
mistakes years ago. You can lead a horse to water ...

In his Nov/Dec 2001 QEX article, Dr. Best attempted to explain where
the reflected energy goes. Shackled by ignorance, he got many things
wrong e.g. phantom EM waves that exist without energy. But the article
alludes to the conceptual path that needs to be taken to alleviate
that ignorance. In "Wave Mechanics of Transmission Lines, Part 3", Dr.
Best published the above power-density/interference equation although
he ignorantly asserted on this newsgroup that interference did not
exist.

Interference resulting from superposition at an impedance
discontinuity is the key missing link in explaining everything that
happens to the energy in a transmission line including Roy's food-for-
thought article. Why is there so much reluctance to adopt a proven law
of EM wave physics from the field of optics?
--
73, Cecil, w5dxp.com

what happens to reflected energy ?
 

On Tue, 8 Jun 2010 03:57:44 -0700 (PDT), Keith Dysart
wrote:

But all of these papers have the appropriate
discipline

Hi Keith,

But? A pejorative tied to the phrase "appropriate discipline?" What
an interesting segue into denial. Quite original.

and do not ask the question "where does the reflected
energy go?"

OK, their having stated explicitly that, and having it in black and
white from competent authority is just as insufficient as Walt's
method and data. Bench experience to this issue makes a lot of
armchair theorists uncomfortable.

To your credit, you responded to the witness of evidence. It's like
the rest threw you under the bus.

73's
Richard Clark, KB7QHC

what happens to reflected energy ?
 

On Tue, 8 Jun 2010 03:57:44 -0700 (PDT), Keith Dysart
wrote:

Just for fun, here is a simple example.

-sigh-

No wonder traffic dies here. This stuff is a wheeze:

The source is no longer
providing energy,

If it had been a battery instead, then the battery would have as much
energy as it had the moment before - (minus) the few coulombs taken
during the short step interval.

Stick with the topic, draw from authorities (that are agreeable to all
parties) that demonstrate the concepts at the bench using the ordinary
tools of the trade (well, ordinary for a Metrologist, perhaps, but is
increasingly available at Ham fests) on actual equipment (what we
actually use in the Shack).

73's
Richard Clark, KB7QHC

what happens to reflected energy ?
 

On Tue, 8 Jun 2010 02:50:17 -0700 (PDT), Wimpie
wrote:

There are many ways to say what the source is NOT, and that will never
inform us about the source. *I see many draw deuces to this question
and try to convince everyone that the card is a pair of winning aces.

Probably you consumed something wrong here.

Hi Wimpie,

Hmmm, an ad hominem usually reserved for Art, but I will let it pass
as your perceiving it as a response in kind. Fair enough.

Now make some mismatch (for example VSWR=2 at different phase) and
read the forward power. Did it change? If so, the output impedance is
no (longer) 50 Ohms.

What IS it now? *If you could measure it once, you should be able to
tell us what it is this time too. *I did this for years to methods set
by the National Bureau of Standards. *You have drawn a deuce, not two
aces.

Some values:
9 +/-1 Ohm (real impedance), reference impedance 16 Ohms, 8 MHz
amplifier (ISM).

Taking my statement and your response at face value, you are
describing how a source that formerly exhibited 50 Ohms was drawn down
to 9 +/-1 Ohm (real impedance). What "real impedance" is? is not
something that I will dwell on. Instead, I will ask how this shift
was induced. Or perhaps you are using a 16 Ohm system. If so, this
lacks too many details for consideration. When I had Walt's 300+ line
hypothesis, I also had access to his service manual for his equipment,
the final tube specification, and his measurement data. In short, I
am impressed only by data, not results.

Given the "some values" that you offer continues in disjointed
discussion as a sort of short-hand description for a project known
only to you, then these distractions draw this further away from a
conclusion. I also see your directions to wander the threads for
vague side bars of discussion. Supposedly, this is instruction for me
to draw together and connect the important details of your point for
you. No, thank you. Too many in this group contribute teasing
details in their coy writing and none of them have the power to
intrigue me with that.

Your "some values" looks vaguely interesting, but I would suggest you
put effort into describing your complete scenario with as much care as
Walt's 300+ line posting that laid bare his complete hypothesis. I
offered you a specifically new thread that I originated solely for one
issue where I see no discussion from you - quid pro quo?

And returning to the matter at hand - Up or Down:

"Does Walt's data support the evidence of a Conjugate Match?"

I cannot see pursuing your inquiry with your own example if you cannot
offer a fixed answer here. That would put me in a position of
embarking on a ponderous journey where, simultaneously, you and Walt
both DID AND DID NOT demonstrate your claims. Paradox is suitable
only for Operetta by Gilbert and Sullivan.

73's
Richard Clark, KB7QHC

what happens to reflected energy ?
 

Hello Richard,

You asked for what it IS, several times, so you virtually left me no
choice.

Especially the first one (Zout = 9 ohms where the load is 16 Ohms,
RCout = -0.28) is very well documented as it was designed to have this
value. Without the extra measures, the value was below 9 Ohms. The
amplifier is time limited fully short circuit proof, so it was not
difficult to determine Zout at various loads. I put myself into
problems when publishing the documentation over here.

When we go back to my first posting in this thread (the first reply)
where I stated that amplifiers do not show 50 Ohms in general, is not
that strange and doesn't violate Walt's conclusions.

When I understand it well, Walt assumes tuning for maximum output at a
given drive (not necessarily the maximum drive). As mentioned earlier,
I support his findings.

When you don't touch the plate and load capacitor and change from 50
Ohms to a load with VSWR = 2, you violate Walt's conditions and the
amplifier will no long behave as a 50 Ohms source, hence what happens
with the 50W reflected power from JC's posting becomes more
complicated then just a transmission line problem.

Walt made a statement in this thread that an amplifier that obeys ZL =
Zout can have higher then 50% efficiency. I fully support this
statement also (Example Class C amplifier with narrow conduction angle
at the edge of current/voltage saturation, with efficiency over
75%). However, change the load and you get a complete other
situation. You have to retune to get back to Walt's condition.

From my experience, many devices with conventional amplifiers (like
the valve with pi-filter, or single transistor stages) are now being
replaced by other topologies where ZL Zout under specified load (as
these types of amplifiers are not optimized for maximum gain, but for
maximum power added efficiency). I think Tom, KT7ITM also have this
experience.

The example given by JC (huge reflection) and my own experience with
conventional and high efficiency topologies resulted in my statement
that a PA is not a 50 Ohms source. Maybe I had to be more precise to
mention the exceptions: forward power control loop as mentioned by
Roy, -tuning to maximum output after each change of load, -specially
designed negative feedback, -adding an attenuator. The reader can now
determine the category for his PA.

In case of doubt use the forward power measurement technique, when it
changes under varying load, your amplifier doesn’t behave as a 50 Ohms
source.

Best Regards,


Wim
PA3DJS
www.tetech.nl
without abc, PM will reach me

what happens to reflected energy ?
 

On Jun 8, 2:14*pm, Cecil Moore wrote:
On Jun 8, 5:57*am, Keith Dysart wrote:

As soon as one assigns tangible energy to the reflected wave, it
becomes reasonable to ask for an accounting of this energy and
the model is incapable of properly accounting for the energy.

Then one is using the wrong model which is typical of many RF
engineers. Optical physicists have been accounting for reflected
energy for decades. That one is ignorant of where reflected energy
goes is not a good reason to abandon the law of physics that says that
*ALL* EM waves contain ExH energy, including reflected EM waves. EM
waves cannot exist without ExH energy so you might as well say that
reflected waves do not exist at all because that is the logical
conclusion based on your false premises.

Optical physicists have been aware of the power-density/interference
equation for a long, long time. It is covered in any good reference
book on optics, e.g. by Hecht and by Born and Wolf. I have posted the
details - why do choose to remain ignorant? At least take time to
comprehend the technical information available from the field of
optics and report back to us why many decades of that EM wave
knowledge is wrong. Here is the equation that explains where the
reflected energy goes. All you have to to is track the energy back in
time.

Ptot = P1 + P2 + 2*SQRT(P1*P2)*cos(A)

where A is the angle between the two voltages. The last term is known
as the "interference term".

When two coherent, collimated, signals interfere while traveling in
the same direction in a transmission line: If they are 90 degrees
apart, there is zero interference and the I and Q components are
easily recovered.

If they are less than 90 degrees apart, the interference is
constructive and the phasor superposition of the two waves results in
*more power in the total superposed wave* than in the arithmetic sum
of the two powers. The extra energy has to come from somewhere. In the
absence of a local source, the extra energy has to come from
destructive interference in the opposite direction, e.g. at a Z0-
match.

If they are between 90 degrees and 180 degrees apart, the interference
is destructive and the phasor superposition of the two waves results
in *less power in the total superposed wave* than in the arithmetic
sum of the two waves. The "left-over" energy has to go somewhere so it
has to be delivered in the opposite direction to the area that
supports constructive interference.

Here is what Hecht said: "Briefly then, optical (EM wave) interference
corresponds to the interaction of two or more (EM) lightwaves yielding
a resultant irradiance (power density) that deviates from the sum of
the component irradiances."

Hecht's statement hints to where the reflected energy goes. In the
case of wave cancellation at a Z0-match that eliminates ExH reflected
energy flowing toward the load, all of the ExH reflected energy is
recovered and redistributed back toward the load.

Following this weakness back through the model, the root cause
is the attempt to assign tangible energy to the reflected wave.
Think of it as a reflected voltage or current wave and all will
be well, but assign power to it and eventually incorrect
conclusions will be drawn.

Only if the model is inadequate. Optical physicists assign tangible
energy to reflected waves all the time. Many are probably reading
these postings and laughing at the collective ignorance about EM
waves.

Take a look at this web page:

http://www.teachspin.com/instruments...eriments.shtml

Scroll down to, "Using Dielectric Beamsplitters to find the "missing
energy" in destructive interference". There it is, all spelled out for
you - where the reflected energy goes. Reflected energy is never
actually missing - what are missing are a few of the brain cells that
need to be used to think about reflected energy. :-)

For those who understand this, and know that "where did the
reflected energy go?" is an invalid question, using the power
model within its limits will not cause difficulties. But for
those who are not careful, great difficulties arise and a lot
of fancy dancing is offered to work around the difficulties,
unsuccessfully.

Yes, throughout history, ignorant people have brought great
difficulties upon themselves because they refuse to alleviate their
ignorance. Those who refuse to learn from their mistakes are destined
to repeat those same mistakes. And you were making these exact same
mistakes years ago. You can lead a horse to water ...

In his Nov/Dec 2001 QEX article, Dr. Best attempted to explain where
the reflected energy goes. Shackled by ignorance, he got many things
wrong e.g. phantom EM waves that exist without energy. But the article
alludes to the conceptual path that needs to be taken to alleviate
that ignorance. In "Wave Mechanics of Transmission Lines, Part 3", Dr.
Best published the above power-density/interference equation although
he ignorantly asserted on this newsgroup that interference did not
exist.

Interference resulting from superposition at an impedance
discontinuity is the key missing link in explaining everything that
happens to the energy in a transmission line including Roy's food-for-
thought article. Why is there so much reluctance to adopt a proven law
of EM wave physics from the field of optics?
--
73, Cecil, w5dxp.com

interference is nothing but what you observe after you superimpose two
or more waves. the general principle at work is superposition, giving
it more specific names like destructive or constructive interference
is just describing the result that you observe.

what happens to reflected energy ?
 

On Jun 8, 10:57*am, Keith Dysart wrote:
On Jun 8, 3:47*am, Richard Clark wrote:



On Sun, 6 Jun 2010 14:21:16 -0700 (PDT), Keith Dysart

wrote:
The 100W forward and 50W reflected have no relation to actual powers

From MECA, makers of Isolators - their application:

The isolator is placed in the measurement path of a test bench between
a signal source and the device under test (DUT) so that any
reflections caused by any mismatches will end up at the termination of
the isolator and not back into the signal source. This example also
clearly illustrates the need to be certain that the termination at the
isolated port is sufficient to handle 100% of the reflected power
should the DUT be disconnected while the signal source is at full
power. If the termination is damaged due to excessive power levels,
the reflected signals will be directed back to the receiver because of
the circular signal flow.
...
MECA offers twenty-four models of isolators and circulators in both N
and SMA-female connectors with average power ratings from 2 - 250
watts.

Good day Richard,

You have located several examples from reputable vendors where the
behaviour of directional couplers is described in terms of power
in a forward and reflected wave. This model of behaviour works within
its limits and allows for convenient computation and prediction of
the behaviour. But all of these papers have the appropriate
discipline and do not ask the question "where does the reflected
energy go?" which is good, for this exceeds the limits of the
model.

As soon as one assigns tangible energy to the reflected wave, it
becomes reasonable to ask for an accounting of this energy and
the model is incapable of properly accounting for the energy.

Following this weakness back through the model, the root cause
is the attempt to assign tangible energy to the reflected wave.
Think of it as a reflected voltage or current wave and all will
be well, but assign power to it and eventually incorrect
conclusions will be drawn.

For those who understand this, and know that "where did the
reflected energy go?" is an invalid question, using the power
model within its limits will not cause difficulties. But for
those who are not careful, great difficulties arise and a lot
of fancy dancing is offered to work around the difficulties,
unsuccessfully.

Just for fun, here is a simple example.

100V DC source, connected to a 50 ohm source resistor,
connected to 50 ohm transmission line, connected to a 50 ohm
load resistor. Turn on the source. A voltage step propagates
down the line to the load. The impedances are matched, so
there are no reflections. The source provides 100W. 100W is
dissipated in the source resistor. 100W is dissipated in the
load resistor. Energy moves along the transmission line
from the source to the load at the rate of 50W.
All is well.

lets see, when steady state is reached you have 100v through 2 50ohm
resistors in series i assume since you say the source provides 100w...
that would mean the current is 1a and each resistor is dissipating
50w.


Disconnect the load. A voltage step propagates back along
the line from the load to the source. In front of this
step current continues to flow. Behind the step, the
current is 0. When the step reaches the source there is
no longer any current flowing. The source is no longer
providing energy, the source resistor is dissipating
nothing, and neither is the load resistor.

Proponensts of the power model claim that energy is
still flowing down the line, being reflected from the
open end and flowing back to the source. Since the
source is clearly no longer providing energy, great
machinations are required to explain why the reflected
'energy' is re-reflected to provide the forward 'energy'.

why would there be power flowing down the line, in this simple dc case
once the voltage in the line equals the source voltage, which in this
perfect case would be after one transient down the line, all is in
equilibrium... and art is happy.

HOWEVER!!! replace that dc source with an AC one and everything
changes, now there are forward and reflected waves continuously going
down the line and back. your DC analysis is no longer valid as it is
a very special case. now you have to go back to the general equations
and use the length of the line to transform the open circuit back to
the source based on the length of the line in wavelengths... for the
specific case of the line being 1/4 wave long the open now looks like
a short and the current in the source resistor doubles, now making it
dissipate 100w!


But really, does anyone believe that a length of
transmission line, charged to 100V voltage with zero
current flowing, is actually simultaneously transporting
energy in both directions?

For even more fun, replace the ideal conductors in the
transmission line with some lossy conductors. How much
of the reflected and re-reflected energy flowing up and
down the line will be dissipated in the conductors?
Remember that the current is zero, everywhere along
the line.

...Keith

what happens to reflected energy ?
 

On Tue, 8 Jun 2010 12:17:37 -0700 (PDT), Wimpie
wrote:

When we go back to my first posting in this thread (the first reply)
where I stated that amplifiers do not show 50 Ohms in general, is not
that strange and doesn't violate Walt's conclusions.

Hi Wimpie,

Time to rewind:

On Thu, 27 May 2010 19:32:40 -0700 (PDT), walt wrote:

The source resistance data reported in Secs 19.8 and 19.9 were
obtained using the load variation method with resistive loads. Note
that of the six measurements of output source resistance reported in
Table 19.1, the average value of the resistance is 50.3 ohms obtained
with the reference load resistance of 51.2 ohms, exhibiting an error
of only 1.8 percent.

Considering these two explicit initial conditions and findings:

1. Using a Kenwood TS-830S transceiver as the RF source, the tuning
and loading of the pi-network are adjusted to deliver all the
available power into a 50 + j0-ohm load with the grid drive adjusted
to deliver the maximum of 100 watts at 4 MHz, thus establishing the
area of the RF power window at the input of the pi-network, resistance
RLP at the plate, and the slope of the load line. The output source
resistance of the amplifier in this condition will later be shown to
be 50 ohms. In this condition the DC plate voltage is 800 v and plate
current is 260 ma. DC input power is therefore 800 v ? 0.26 a = 208 w.
Readings on the Bird 43 wattmeter indicate 100 watts forward and zero
watts reflected. (100 watts is the maximum RF output power available
at this drive level.) From here on the grid drive is left undisturbed,
and the pi-network controls are left undisturbed until Step 10.

2. The amplifier is now powered down and the load resistance RL is
measured across the input terminals of the resonant pi-network tank
circuit (from plate to ground) with an HP-4815 Vector Impedance Meter.
The resistance is found to be approximately 1400 ohms. Because the
amplifier was adjusted to deliver the maximum available power of 100
watts prior to the resistance measurement, the averaged resistance RLP
looking into the plate (upstream from the network terminals) is also
approximately 1400 ohms. Accordingly, a non-reactive 1400-ohm resistor
is now connected across the input terminals of the pi-network tank
circuit and source resistance ROS is measured looking rearward into
the output terminals of the network. Resistance ROS was found to be 50
ohms.

Quite explicit. Rated power into the rated Z load revealed a
conjugate basis Z match (afterall 50=50) or an image basis Z match
(50=50). The pi-network simply transformed the source/load. Anyone
skilled with the concept of the Smith Chart can immediately follow
that logical construct.

We even get the plate resistance which conforms to within published
specifications for the pair of tubes.

We even get the near classic result of a conjugate basis Z match
efficiency (however, only in the first pass approximation).

How asking the same question:

"Does Walt's data support the evidence of a Conjugate Match?"

migrates into responses to other issues is a strange dance. I ask "up
or down," and everyone wants to vote sideways.

When you don't touch the plate and load capacitor and change from 50
Ohms to a load with VSWR = 2, you violate Walt's conditions

I have not the vaguest idea where you got the idea that I changed
anything. The totality of my discussion never budged from steps 1 and
2 and I deliberately and clearly confined all matters there and
repeated them more than occasionally.

In case of doubt use the forward power measurement technique,

I have, and the data I've asked from you was not intended for some
one-time design for a unique application. This is an Ham radio group,
and the object under consideration is an Ham transmitter. Feel free
to substitute freely among a similar class (frequency range, power
output) such that others might compare their own.

when it
changes under varying load, your amplifier doesn’t behave as a 50 Ohms
source.

This is the most curious statement that eventually comes down the
pike. It is like I am talking to a novice to explain:
"That is what the tuner is for!"

Maybe I had to be more precise to
mention the exceptions: forward power control loop as mentioned by
Roy,

What Roy is describing is nothing close to Walt's hypothesis that I
have confined to a single point observation. No one has advanced
claims made that demonstrate:
1. Constant Z across all loads;
2. Constant Z across all frequencies;
3. Constant Power across all loads;
4. Constant Power across all frequencies.
No competent Ham expects that of any Ham transmitter.

Walt reported a 50 Ohm source Z from a tube rig. I have measured a 50
Ohm Z from two of my own transistor rigs. The difference in
technology is not an issue. I have also been responsible for
measuring RF power and source/load specifications within the chain of
national standards laboratories. Stretching the term RF, this spanned
from DC to 12GHz for power levels up to 100W. The sources across the
board exhibited 50 Ohms Source Z.

Your own national lab is represented at:
http://www.vsl.nl/
They undoubtedly use the same reference citations as our National
Institutes for Science and Technology.

***********

Putting the entirety of this discussion aside, and returning to your
very special project with an 9 Ohm source of RF at 8 Mhz. How
efficient was its operation feeding a 50 Ohm line to a 50 Ohm load?
Can you document this with schematics, parts description, and measured
data at key points from drain through to the load? This was the level
of available resources that came with Walt's discussion.

73's
Richard Clark, KB7QHC

what happens to reflected energy ?
 

On Jun 8, 7:39*pm, K1TTT wrote:
On Jun 8, 10:57*am, Keith Dysart wrote:





On Jun 8, 3:47*am, Richard Clark wrote:

On Sun, 6 Jun 2010 14:21:16 -0700 (PDT), Keith Dysart

wrote:
The 100W forward and 50W reflected have no relation to actual powers

From MECA, makers of Isolators - their application:

The isolator is placed in the measurement path of a test bench between
a signal source and the device under test (DUT) so that any
reflections caused by any mismatches will end up at the termination of
the isolator and not back into the signal source. This example also
clearly illustrates the need to be certain that the termination at the
isolated port is sufficient to handle 100% of the reflected power
should the DUT be disconnected while the signal source is at full
power. If the termination is damaged due to excessive power levels,
the reflected signals will be directed back to the receiver because of
the circular signal flow.
...
MECA offers twenty-four models of isolators and circulators in both N
and SMA-female connectors with average power ratings from 2 - 250
watts.

Good day Richard,

You have located several examples from reputable vendors where the
behaviour of directional couplers is described in terms of power
in a forward and reflected wave. This model of behaviour works within
its limits and allows for convenient computation and prediction of
the behaviour. But all of these papers have the appropriate
discipline and do not ask the question "where does the reflected
energy go?" which is good, for this exceeds the limits of the
model.

As soon as one assigns tangible energy to the reflected wave, it
becomes reasonable to ask for an accounting of this energy and
the model is incapable of properly accounting for the energy.

Following this weakness back through the model, the root cause
is the attempt to assign tangible energy to the reflected wave.
Think of it as a reflected voltage or current wave and all will
be well, but assign power to it and eventually incorrect
conclusions will be drawn.

For those who understand this, and know that "where did the
reflected energy go?" is an invalid question, using the power
model within its limits will not cause difficulties. But for
those who are not careful, great difficulties arise and a lot
of fancy dancing is offered to work around the difficulties,
unsuccessfully.

Just for fun, here is a simple example.

100V DC source, connected to a 50 ohm source resistor,
connected to 50 ohm transmission line, connected to a 50 ohm
load resistor. Turn on the source. A voltage step propagates
down the line to the load. The impedances are matched, so
there are no reflections. The source provides 100W. 100W is
dissipated in the source resistor. 100W is dissipated in the
load resistor. Energy moves along the transmission line
from the source to the load at the rate of 50W.
All is well.

lets see, when steady state is reached you have 100v through 2 50ohm
resistors in series i assume since you say the source provides 100w...
that would mean the current is 1a and each resistor is dissipating
50w.

Ouch. I can't even get the arithmetic right. You are, of course,
correct. Fortunately, the rest of the examples do not attempt
to compute actual values.

Disconnect the load. A voltage step propagates back along
the line from the load to the source. In front of this
step current continues to flow. Behind the step, the
current is 0. When the step reaches the source there is
no longer any current flowing. The source is no longer
providing energy, the source resistor is dissipating
nothing, and neither is the load resistor.

Proponensts of the power model claim that energy is
still flowing down the line, being reflected from the
open end and flowing back to the source. Since the
source is clearly no longer providing energy, great
machinations are required to explain why the reflected
'energy' is re-reflected to provide the forward 'energy'.

why would there be power flowing down the line,

Exactly. And yet, were you to attach a DC-couple directional
wattmeter to the line, it would indicate 50W forward and 50W
reflected (using your corrected numbers).

in this simple dc case
once the voltage in the line equals the source voltage, which in this
perfect case would be after one transient down the line, all is in
equilibrium... and art is happy.

HOWEVER!!! *replace that dc source with an AC one and everything
changes,

Not at all. The measurements and computations for forward and
reflected
power that work for AC, work perfectly well for DC, pulses, steps,
or pick your waveform. There is no magic related to AC.

now there are forward and reflected waves continuously going
down the line and back. *your DC analysis is no longer valid as it is
a very special case. *

Not so. And if it were, at exactly what very low frequency does
the AC analysis begin to fail?

now you have to go back to the general equations
and use the length of the line to transform the open circuit back to
the source based on the length of the line in wavelengths... for the
specific case of the line being 1/4 wave long the open

The transformation properties apply only to sinusoidal AC (note, for
example, that the line length is expressed in wavelengths), but
reflection coefficients and the the computation of forward and
reflected voltages, currents and powers are waveshape independant.

When analyzing in the time domain, use the delay of the line to decide
the values of the signals to superposed. Doing this for sinusoids
will lead to exactly the same results that are expected from, for
example, 1/4 wave lines.

....Keith

what happens to reflected energy ?
 

On Jun 8, 6:28*pm, K1TTT wrote:
interference is nothing but what you observe after you superimpose two
or more waves. *the general principle at work is superposition, giving
it more specific names like destructive or constructive interference
is just describing the result that you observe.

Yes, interference is the result of superposition. Sometimes
superposition results in wave cancellation which is destructive
interference that redistributes the energy in the canceled waves in
the opposite direction in a transmission line - the only direction
that supports constructive interference. The point is that at an
impedance discontinuity in a transmission line, in the absence of a
local source, destructive interference in one direction *requires*
constructive interference in the other direction. That's how a Z0-
match redistributes all the reflected energy back toward the load
while none reaches the source. It is exactly how a 1/4WL thin-film
coating cancels reflections on a piece of non-reflective glass.

The important thing is that a power density equation exists that
predicts the energy flow as a result of superposition.

Maybe if you reviewed "Sec 4.3 Reflection Mechanics of Stub Matching"
in "Reflections", it will be more clear?
--
73, Cecil, w5dxp.com

what happens to reflected energy ?
 

Richard Clark Inscribed thus:

How asking the same question:

"Does Walt's data support the evidence of a Conjugate Match?"

73's
Richard Clark, KB7QHC

Having struggled to keep up and understand the very different points of
view, indeed very confusing at times, I'm going to come down on the
side of "Yes, Walt's data does support evidence of a Conjugate Match."

--
Best Regards:
Baron.

what happens to reflected energy ?
 

On Wed, 09 Jun 2010 19:18:42 +0100, Baron
wrote:

Having struggled to keep up and understand the very different points of
view, indeed very confusing at times, I'm going to come down on the
side of "Yes, Walt's data does support evidence of a Conjugate Match."

Hi OM,

Thanks. Clear answers to clear questions are rare.

So:
2½ "Walt's data does support evidence of a Conjugate Match"
and
½ "Walt's data does not support evidence of a Conjugate Match"

As for confusion - the split vote must be proxy for the "silent
majority." Or, that majority cannot risk exposure. Shame? The only
issue is either with the data or the expression "Conjugate Match."
None seem anxious to avoid discussion of the expression - in fact that
conversation runs like a party line.

On the other hand, challenge the data? The numbers were cribbed? The
test gear was in the "off" position during a test? No. The data
seems to make sense. Competing data for the same initial conditions?
Hark! The scientific method rises from slumber with interest.

Nope, nada, negatory, no way, not going there.

Scientific method returns to its narcoleptic state.

73's
Richard Clark, KB7QHC

what happens to reflected energy ?
 

On Jun 9, 1:18*pm, Baron wrote:
Richard Clark Inscribed thus:
* "Does Walt's data support the evidence of a Conjugate Match?"

"Yes, Walt's data does support evidence of a Conjugate Match."

http://www.w2du.com/Appendix09.pdf

A purist might argue that, like a lossless transmission line, an ideal
conjugate match cannot exist in reality because of resistive losses
and non-linearities.

What might be a more accurate statement is: By definition, Walt's data
supports the evidence of a real-world Conjugate Match. Walt has
certainly defined in detail what he means by a "(real-world) Conjugate
Match".
--
73, Cecil, w5dxp.com

what happens to reflected energy ?
 

On Jun 9, 2:04*am, Keith Dysart wrote:
On Jun 8, 7:39*pm, K1TTT wrote:



On Jun 8, 10:57*am, Keith Dysart wrote:

On Jun 8, 3:47*am, Richard Clark wrote:

On Sun, 6 Jun 2010 14:21:16 -0700 (PDT), Keith Dysart

wrote:
The 100W forward and 50W reflected have no relation to actual powers

From MECA, makers of Isolators - their application:

The isolator is placed in the measurement path of a test bench between
a signal source and the device under test (DUT) so that any
reflections caused by any mismatches will end up at the termination of
the isolator and not back into the signal source. This example also
clearly illustrates the need to be certain that the termination at the
isolated port is sufficient to handle 100% of the reflected power
should the DUT be disconnected while the signal source is at full
power. If the termination is damaged due to excessive power levels,
the reflected signals will be directed back to the receiver because of
the circular signal flow.
...
MECA offers twenty-four models of isolators and circulators in both N
and SMA-female connectors with average power ratings from 2 - 250
watts.

Good day Richard,

You have located several examples from reputable vendors where the
behaviour of directional couplers is described in terms of power
in a forward and reflected wave. This model of behaviour works within
its limits and allows for convenient computation and prediction of
the behaviour. But all of these papers have the appropriate
discipline and do not ask the question "where does the reflected
energy go?" which is good, for this exceeds the limits of the
model.

As soon as one assigns tangible energy to the reflected wave, it
becomes reasonable to ask for an accounting of this energy and
the model is incapable of properly accounting for the energy.

Following this weakness back through the model, the root cause
is the attempt to assign tangible energy to the reflected wave.
Think of it as a reflected voltage or current wave and all will
be well, but assign power to it and eventually incorrect
conclusions will be drawn.

For those who understand this, and know that "where did the
reflected energy go?" is an invalid question, using the power
model within its limits will not cause difficulties. But for
those who are not careful, great difficulties arise and a lot
of fancy dancing is offered to work around the difficulties,
unsuccessfully.

Just for fun, here is a simple example.

100V DC source, connected to a 50 ohm source resistor,
connected to 50 ohm transmission line, connected to a 50 ohm
load resistor. Turn on the source. A voltage step propagates
down the line to the load. The impedances are matched, so
there are no reflections. The source provides 100W. 100W is
dissipated in the source resistor. 100W is dissipated in the
load resistor. Energy moves along the transmission line
from the source to the load at the rate of 50W.
All is well.

lets see, when steady state is reached you have 100v through 2 50ohm
resistors in series i assume since you say the source provides 100w...
that would mean the current is 1a and each resistor is dissipating
50w.

Ouch. I can't even get the arithmetic right. You are, of course,
correct. Fortunately, the rest of the examples do not attempt
to compute actual values.



Disconnect the load. A voltage step propagates back along
the line from the load to the source. In front of this
step current continues to flow. Behind the step, the
current is 0. When the step reaches the source there is
no longer any current flowing. The source is no longer
providing energy, the source resistor is dissipating
nothing, and neither is the load resistor.

Proponensts of the power model claim that energy is
still flowing down the line, being reflected from the
open end and flowing back to the source. Since the
source is clearly no longer providing energy, great
machinations are required to explain why the reflected
'energy' is re-reflected to provide the forward 'energy'.

why would there be power flowing down the line,

Exactly. And yet, were you to attach a DC-couple directional
wattmeter to the line, it would indicate 50W forward and 50W
reflected (using your corrected numbers).

only as you monitored the initial transient. once the transient has
passed there is no power flowing in the line. to drive power requires
a potential difference, without potential difference there is no
current and no power.


in this simple dc case
once the voltage in the line equals the source voltage, which in this
perfect case would be after one transient down the line, all is in
equilibrium... and art is happy.

HOWEVER!!! *replace that dc source with an AC one and everything
changes,

Not at all. The measurements and computations for forward and
reflected
power that work for AC, work perfectly well for DC, pulses, steps,
or pick your waveform. There is no magic related to AC.

ah, but there is... in a sine wave the voltage never stops changing so
there is always a potential difference driving the wave along the
line. with DC you can reach equilibrium in one transit of the cable,
assuming it is lossless and the source and/or load is matched to the
line. otherwise you have to do the infinite sum to approximate the
steady state as for any other transient. complex waveforms get ugly
since they are infinite sums of sine waves, though for lossless and
dispersionless cables they aren't too hard to handle. The magic of a
single frequency AC source is it lets us use simplified equations that
make use of the sinusoidal steady state approximations... this is what
lets us do things like s parameters, phasors, E=IZ, and cecil's simple
power addition equation, without the need for messy summations and
other more detailed calculations.


now there are forward and reflected waves continuously going
down the line and back. *your DC analysis is no longer valid as it is
a very special case. *

Not so. And if it were, at exactly what very low frequency does
the AC analysis begin to fail?

none, there is no magic cutoff, its just how close you want to look at
it.


now you have to go back to the general equations
and use the length of the line to transform the open circuit back to
the source based on the length of the line in wavelengths... for the
specific case of the line being 1/4 wave long the open

The transformation properties apply only to sinusoidal AC (note, for
example, that the line length is expressed in wavelengths),

true, another simplification made possible by the sinusoidal steady
state approximation.

but
reflection coefficients and the the computation of forward and
reflected voltages, currents and powers are waveshape independant.

only as long as the loads are perfectly resistive and linear. real
loads often change impedance with frequency and so distort the
reflection of complex waveforms. use a scope with a good risetime and
a decent fast rise time pulse and you'll see it.


When analyzing in the time domain, use the delay of the line to decide
the values of the signals to superposed. Doing this for sinusoids
will lead to exactly the same results that are expected from, for
example, 1/4 wave lines.

...Keith

what happens to reflected energy ?
 

On Thu, 10 Jun 2010 14:52:21 -0700 (PDT), K1TTT
wrote:

real
loads often change impedance with frequency

Real loads also change Z with power - especially dummy loads. As a
real load has some trait of dummy load in it (loss) the degree of Z
change is in the degree of temperature change.

73's
Richard Clark, KB7QHC

what happens to reflected energy ?
 

On Jun 11, 12:53*am, Richard Clark wrote:
On Thu, 10 Jun 2010 14:52:21 -0700 (PDT), K1TTT
wrote:

real
loads often change impedance with frequency

Real loads also change Z with power - especially dummy loads. *As a
real load has some trait of dummy load in it (loss) the degree of Z
change is in the degree of temperature change.

73's
Richard Clark, KB7QHC

yeah, but thats usually a much slower process... unless you get to a
point where it flashes over the insulation!

what happens to reflected energy ?
 

On Jun 10, 4:52*pm, K1TTT wrote:

but reflection coefficients and the the computation of forward
and reflected voltages, currents and powers are waveshape
independant.

only as long as the loads are perfectly resistive and linear. *real
loads often change impedance with frequency and so distort the
reflection of complex waveforms. *use a scope with a good risetime
and a decent fast rise time pulse and you'll see it.

It can be very close, though, if the system is designed/optimized
properly.

Below is a link leading to the r-f pulse measurement of a UHF TV
broadcast antenna system that I made as an RCA field engineer some 35
years ago. The incident and reflected waveforms are very similar.

The Z-match of this antenna was optimized to the transmission line
using a variable transformer at the input to the antenna, which
antenna was installed atop a ~ 1,500 foot tower (note the ~3 µs round
trip for the reflection, at ~492 feet/µs).

The H.A.D. of this sin^2 pulse denotes an r-f bandwidth approximately
as great as can be carried by a 6 MHz analog US TV channel.

http://i62.photobucket.com/albums/h8...easurement.gif

RF

what happens to reflected energy ?
 

On 9 jun, 02:07, Richard Clark wrote:
On Tue, 8 Jun 2010 12:17:37 -0700 (PDT), Wimpie
wrote:

When we go back to my first posting in this thread (the first reply)
where I stated that amplifiers do not show 50 Ohms in general, is not
that strange and doesn't violate Walt's conclusions.

Hi Wimpie,

Time to rewind:

On Thu, 27 May 2010 19:32:40 -0700 (PDT), walt wrote:

The source resistance data reported in Secs 19.8 and 19.9 were
obtained using the load variation method with resistive loads. Note
that of the six measurements of output source resistance reported in
Table 19.1, the average value of the resistance is 50.3 ohms obtained
with the reference load resistance of 51.2 ohms, exhibiting an error
of only 1.8 percent.

Considering these two explicit initial conditions and findings:



1. Using a Kenwood TS-830S transceiver as the RF source, the tuning
and loading of the pi-network are adjusted to deliver all the
available power into a 50 + j0-ohm load with the grid drive adjusted
to deliver the maximum of 100 watts at 4 MHz, thus establishing the
area of the RF power window at the input of the pi-network, resistance
RLP at the plate, and the slope of the load line. The output source
resistance of the amplifier in this condition will later be shown to
be 50 ohms. In this condition the DC plate voltage is 800 v and plate
current is 260 ma. DC input power is therefore 800 v ? 0.26 a = 208 w.
Readings on the Bird 43 wattmeter indicate 100 watts forward and zero
watts reflected. (100 watts is the maximum RF output power available
at this drive level.) From here on the grid drive is left undisturbed,
and the pi-network controls are left undisturbed until Step 10.

2. The amplifier is now powered down and the load resistance RL is
measured across the input terminals of the resonant pi-network tank
circuit (from plate to ground) with an HP-4815 Vector Impedance Meter.
The resistance is found to be approximately 1400 ohms. Because the
amplifier was adjusted to deliver the maximum available power of 100
watts prior to the resistance measurement, the averaged resistance RLP
looking into the plate (upstream from the network terminals) is also
approximately 1400 ohms. Accordingly, a non-reactive 1400-ohm resistor
is now connected across the input terminals of the pi-network tank
circuit and source resistance ROS is measured looking rearward into
the output terminals of the network. Resistance ROS was found to be 50
ohms.

Quite explicit. *Rated power into the rated Z load revealed a
conjugate basis Z match (afterall 50=50) or an image basis Z match
(50=50). *The pi-network simply transformed the source/load. *Anyone
skilled with the concept of the Smith Chart can immediately follow
that logical construct.

We even get the plate resistance which conforms to within published
specifications for the pair of tubes.

We even get the near classic result of a conjugate basis Z match
efficiency (however, only in the first pass approximation).

How asking the same question:

* "Does Walt's data support the evidence of a Conjugate Match?"

migrates into responses to other issues is a strange dance. *I ask "up
or down," and everyone wants to vote sideways.

When you don't touch the plate and load capacitor and change from 50
Ohms to a load with VSWR = 2, you violate Walt's conditions

I have not the vaguest idea where you got the idea that I changed
anything. *The totality of my discussion never budged from steps 1 and
2 and I deliberately and clearly confined all matters there and
repeated them more than occasionally.

In case of doubt use the forward power measurement technique,

I have, and the data I've asked from you was not intended for some
one-time design for a unique application. *This is an Ham radio group,
and the object under consideration is an Ham transmitter. *Feel free
to substitute freely among a similar class (frequency range, power
output) such that others might compare their own. *

when it
changes under varying load, your amplifier doesn’t behave as a 50 Ohms
source.

This is the most curious statement that eventually comes down the
pike. *It is like I am talking to a novice to explain:
* * * * "That is what the tuner is for!"

Maybe I had to be more precise to
mention the exceptions: forward power control loop as mentioned by
Roy,

What Roy is describing is nothing close to Walt's hypothesis that I
have confined to a single point observation. *No one has advanced
claims made that demonstrate:
1. *Constant Z across all loads;
2. *Constant Z across all frequencies;
3. *Constant Power across all loads;
4. *Constant Power across all frequencies.
No competent Ham expects that of any Ham transmitter.

Walt reported a 50 Ohm source Z from a tube rig. *I have measured a 50
Ohm Z from two of my own transistor rigs. *The difference in
technology is not an issue. *I have also been responsible for
measuring RF power and source/load specifications within the chain of
national standards laboratories. *Stretching the term RF, this spanned
from DC to 12GHz for power levels up to 100W. *The sources across the
board exhibited 50 Ohms Source Z.

Your own national lab is represented at:http://www.vsl.nl/
They undoubtedly use the same reference citations as our National
Institutes for Science and Technology.

***********

Putting the entirety of this discussion aside, and returning to your
very special project with an 9 Ohm source of RF at 8 Mhz. *How
efficient was its operation feeding a 50 Ohm line to a 50 Ohm load?
Can you document this with schematics, parts description, and measured
data at key points from drain through to the load? *This was the level
of available resources that came with Walt's discussion.

73's
Richard Clark, KB7QHC

Hello Richard,

We can continue mentioning measurement results, do other claims, but
that may not converge. Your results show 50 Ohms output impedance,
mine show other. Such results cannot be verified easily, so this
remains food for endless discussions.

I decided to use another approach that can be verified by people
having a (spice) simulator. I did simulations for some circuits that
can be expected in the amateur world and aren't exotic. I put them
over he http://www.tetech.nl/divers/PA_impedance.pdf . Now people
can figure out what it IS and get their own educated opinion. I used
low frequencies to avoid discussion about the validity of the
simulations (parasitic inductances, etc).

From these simulations it is clear that small changes in load or drive
level result in large change of output impedance. This knowledge
resulted in my statement: "an RF PA isn't a 50 Ohms source", what was/
is heavily disputed by you.

Of course you can design to have better output VSWR under varying load
or drive, but this is generally not done for systems that just have to
deliver the power (transmitters).

If you want to reproduce some of the results yourself, we can compare
our spice netlists. I recently added a real class C example as this
was heavily discussed in another thread.

Best regards,


Wim
PA3DJS
www.tetech.nl
When you remove abc, PM will reach me.

what happens to reflected energy ?
 

On Jun 13, 5:08*pm, Wimpie wrote:
On 9 jun, 02:07, Richard Clark wrote:



On Tue, 8 Jun 2010 12:17:37 -0700 (PDT), Wimpie
wrote:

When we go back to my first posting in this thread (the first reply)
where I stated that amplifiers do not show 50 Ohms in general, is not
that strange and doesn't violate Walt's conclusions.

Hi Wimpie,

Time to rewind:

On Thu, 27 May 2010 19:32:40 -0700 (PDT), walt wrote:

The source resistance data reported in Secs 19.8 and 19.9 were
obtained using the load variation method with resistive loads. Note
that of the six measurements of output source resistance reported in
Table 19.1, the average value of the resistance is 50.3 ohms obtained
with the reference load resistance of 51.2 ohms, exhibiting an error
of only 1.8 percent.

Considering these two explicit initial conditions and findings:

1. Using a Kenwood TS-830S transceiver as the RF source, the tuning
and loading of the pi-network are adjusted to deliver all the
available power into a 50 + j0-ohm load with the grid drive adjusted
to deliver the maximum of 100 watts at 4 MHz, thus establishing the
area of the RF power window at the input of the pi-network, resistance
RLP at the plate, and the slope of the load line. The output source
resistance of the amplifier in this condition will later be shown to
be 50 ohms. In this condition the DC plate voltage is 800 v and plate
current is 260 ma. DC input power is therefore 800 v ? 0.26 a = 208 w.
Readings on the Bird 43 wattmeter indicate 100 watts forward and zero
watts reflected. (100 watts is the maximum RF output power available
at this drive level.) From here on the grid drive is left undisturbed,
and the pi-network controls are left undisturbed until Step 10.

2. The amplifier is now powered down and the load resistance RL is
measured across the input terminals of the resonant pi-network tank
circuit (from plate to ground) with an HP-4815 Vector Impedance Meter.
The resistance is found to be approximately 1400 ohms. Because the
amplifier was adjusted to deliver the maximum available power of 100
watts prior to the resistance measurement, the averaged resistance RLP
looking into the plate (upstream from the network terminals) is also
approximately 1400 ohms. Accordingly, a non-reactive 1400-ohm resistor
is now connected across the input terminals of the pi-network tank
circuit and source resistance ROS is measured looking rearward into
the output terminals of the network. Resistance ROS was found to be 50
ohms.

Quite explicit. *Rated power into the rated Z load revealed a
conjugate basis Z match (afterall 50=50) or an image basis Z match
(50=50). *The pi-network simply transformed the source/load. *Anyone
skilled with the concept of the Smith Chart can immediately follow
that logical construct.

We even get the plate resistance which conforms to within published
specifications for the pair of tubes.

We even get the near classic result of a conjugate basis Z match
efficiency (however, only in the first pass approximation).

How asking the same question:

* "Does Walt's data support the evidence of a Conjugate Match?"

migrates into responses to other issues is a strange dance. *I ask "up
or down," and everyone wants to vote sideways.

When you don't touch the plate and load capacitor and change from 50
Ohms to a load with VSWR = 2, you violate Walt's conditions

I have not the vaguest idea where you got the idea that I changed
anything. *The totality of my discussion never budged from steps 1 and
2 and I deliberately and clearly confined all matters there and
repeated them more than occasionally.

In case of doubt use the forward power measurement technique,

I have, and the data I've asked from you was not intended for some
one-time design for a unique application. *This is an Ham radio group,
and the object under consideration is an Ham transmitter. *Feel free
to substitute freely among a similar class (frequency range, power
output) such that others might compare their own. *

when it
changes under varying load, your amplifier doesn’t behave as a 50 Ohms
source.

This is the most curious statement that eventually comes down the
pike. *It is like I am talking to a novice to explain:
* * * * "That is what the tuner is for!"

Maybe I had to be more precise to
mention the exceptions: forward power control loop as mentioned by
Roy,

What Roy is describing is nothing close to Walt's hypothesis that I
have confined to a single point observation. *No one has advanced
claims made that demonstrate:
1. *Constant Z across all loads;
2. *Constant Z across all frequencies;
3. *Constant Power across all loads;
4. *Constant Power across all frequencies.
No competent Ham expects that of any Ham transmitter.

Walt reported a 50 Ohm source Z from a tube rig. *I have measured a 50
Ohm Z from two of my own transistor rigs. *The difference in
technology is not an issue. *I have also been responsible for
measuring RF power and source/load specifications within the chain of
national standards laboratories. *Stretching the term RF, this spanned
from DC to 12GHz for power levels up to 100W. *The sources across the
board exhibited 50 Ohms Source Z.

Your own national lab is represented at:http://www.vsl.nl/
They undoubtedly use the same reference citations as our National
Institutes for Science and Technology.

***********

Putting the entirety of this discussion aside, and returning to your
very special project with an 9 Ohm source of RF at 8 Mhz. *How
efficient was its operation feeding a 50 Ohm line to a 50 Ohm load?
Can you document this with schematics, parts description, and measured
data at key points from drain through to the load? *This was the level
of available resources that came with Walt's discussion.

73's
Richard Clark, KB7QHC

Hello Richard,

We can continue mentioning measurement results, do other claims, but
that may not converge. Your results show 50 Ohms output impedance,
mine show other. Such results cannot be verified easily, so this
remains food for endless discussions.

I decided to use another approach that can be verified by people
having a (spice) simulator. *I did simulations for some circuits that
can be expected in the amateur world and aren't exotic. *I put them
over he *http://www.tetech.nl/divers/PA_impedance.pdf. *Now people
can figure out what it IS and get their own educated opinion. I used
low frequencies to avoid discussion about the validity of the
simulations (parasitic inductances, etc).

From these simulations it is clear that small changes in load or drive
level result in large change of output impedance. *This knowledge
resulted in my statement: "an RF PA isn't a 50 Ohms source", what was/
is heavily disputed by you.

Of course you can design to have better output VSWR under varying load
or drive, but this is generally not done for systems that just have to
deliver the power (transmitters).

If you want to reproduce some of the results yourself, we can compare
our spice netlists. I recently added a real class C example as this
was heavily discussed in another thread.

Best regards,

Wim
PA3DJSwww.tetech.nl
When you remove abc, PM will reach me.

Hi Wim,

I quote a statement you made in your above post:

"From these simulations it is clear that small changes in load or
drive
level result in large change of output impedance. "

Now I agree that changing the drive level will result in a change in
output impedance. However, let's say you have just adjusted the amp to
deliver all the available power into a given load at a given drive
level, and now you change the load, but leave the drive level and all
other adjustments of the amp untouched. Are you saying this change in
load changes the output impedance? If so, please explain why.

In addition, which situation has the greatest priority, data obtained
from precise measurements or data obtained solely from simulations?

Walt, W2DU

what happens to reflected energy ?
 

On Sun, 13 Jun 2010 14:08:59 -0700 (PDT), Wimpie
wrote:

This knowledge
resulted in my statement: "an RF PA isn't a 50 Ohms source", what was/
is heavily disputed by you.

Hi Wim,

Your conclusion:
"Under maximum power output matching given certain drive (not
necessarily being the maximum drive), the output impedance equals the
load resistor (the so-called "conjugated match condition")."

It was not the only conclusion, but it certainly disputes what you say
above.

Given the easy access to real amplifiers available to Hams at their
own bench, and further given that they do not present any more
complexity than your own simulation; then I have to wonder why we have
wandered into strange topologies.

I am not interested in trying to follow 23 pages of dense presentation
where you could have simulated Walt's finals, validated or rejected
his data, and THEN formed your own conclusions. If this is a problem
of simulation (you don't have that tube handy) what about topology?

It looks more like a link coupled tank circuit from pre-WWII days. The
Tank Q looks to be non-existent. When I rummage through the paper I
find 4.4! This is certainly beyond the standard for PA design - and
we have swapped out of the tube into a mosfet (would you care to stick
to one objective?). Elsewhere (wandering back in tubes) I find a
value of 10.3 (elsewhere it is 11) which inhabits the lowest margin of
good design. Then an itinerant report of a cathode tank (????) whose
Q is 0.44. Q appears to be of little concern.

Again, typical Ham equipment shows Q as low as 10 - maybe, and that
would be for a dog; but those that I have seen typically fall around
15, sometimes 20. As the Z transformation in plate/drain circuits
varies by the square of Q, this is not inconsequential and it once
again has me wondering why the strange topology?

No, your paper lacks focus by trying to be all things (tubes, mosfets,
Class AB, Class C). You examine the most inconsequential details as
if they were equally important as those details that bear on your
concern. The paper lacks structure. Headings are nothing more than
feature descriptions, not argument development. There is a lot of
discussion of the idiosyncrasies of simulation - I am not interested
and that discussion creates the impression of hidden errors. Really,
this stuff goes into an appendix not as the object of the narrative.
The graphs are pretty diversions, but they don't add much. In the
software industry, this is spaghetti design.

I would prefer a conventional topology. I suppose that has come
through clear. Pick one significant point, re-edit this into 4 pages
and then you might have something interesting. This advice comes from
one of our American writers, Mark Twain:
"If I had more time, I would have written you a shorter letter."

73's
Richard Clark, KB7QHC

what happens to reflected energy ?
 

On Jun 10, 5:52 pm, K1TTT wrote:
On Jun 9, 2:04 am, Keith Dysart wrote:

Exactly. And yet, were you to attach a DC-couple directional
wattmeter to the line, it would indicate 50W forward and 50W
reflected (using your corrected numbers).

only as you monitored the initial transient. once the transient has
passed there is no power flowing in the line.

I agree completely with the second sentence. But the first is in
error. A directional wattmeter will indeed show 50W forward and 50W
reflected on a line with constant DC voltage and 0 current. I’ll
start by providing support for your second sentence and then get
back to reflected power.

Choose an arbitrary point on the line and observe the voltage. The
observed voltage will be a function of time. Let us call it V(t).
At the same point on the line, observe the current and call it I(t).

The energy flowing (i.e. power) at this point on the line will
be P(t)=V(t)I(t). If the voltage is measured in volts and the current
in amps, then the power will be in joules/s or watts.
We can also compute the average energy flow (power) over some
interval by integrating P(t) over the interval and dividing by
the length of the interval. Call this Pavg. For repetitive
signals it is convenient to use one period as the interval.

Note that V(t) and I(t) are arbitrary functions. The analysis
applies regardless of whether V(t) is DC, a sinusoid, a step,
a pulse, etc.; any arbitrary function.

To provide some examples, consider the following simple setup:
1) 50 ohm, Thevenin equivalent circuit provides 100 V in to an
open circuit.
2) The generator is connected to a length of 50 ohm ideal
transmission line. We will stick with ideal lines for the
moment, because, until ideal lines are understood, there
is no hope of understanding more complex models.
3) The ideal line is terminated in a 50 ohm resistor.

Set the generator to DC:
V(t)=50V, wherever it is measured on the line
I(t)=1A, wherever it is measured on the line
P(t)=50W, wherever it is measured on the line
Pavg=50W
No disagreement, I hope.

Set the generator to generate a sinusoid:
V(t)=50sin(wt)
I(t)=1sin(wt)
P(t)=25+25sin(2wt)
Pavg=25W
describes the observations at any point on the line.
Measurements at different points will have a different
phase relationship to each due to the delay in the line.
Note carefully the power function. The power at any point
on the line varies from 0 to 50W with a sinusoidal
pattern at twice the frequency of the voltage sinusoid.
The power is always positive; that is, energy is always
flowing towards the load.

Now let us remove the terminating resistor. The line is
now open circuit.

Again, set the generator to DC. After one round trip
time, the observations at any point on the line will
be:
V(t)=100V
I(t)=0A
P(t)=0W
Pavg=0W

The story for a sinusoid is more interesting...
The observations now depend on the location on the
line where the measurements are performed.
At a current 0:
V(t)=100sin(Wt)
I(t)=0A
P(t)=0W
Pavg=0W
At a voltage 0:
V(t)=0V
I(t)=2sin(wt)
P(t)=0W
Pavg=0W
Halfway between the current 0 at the load and the
preceding voltage 0 (that is, 45 degrees back
from the load:
V(t)=70.7sin(wt)
I(t)=1.414cos(wt)
P(t)=50sin(2wt)
Pavg=0W
This tells us that for the first quarter cycle of the
voltage (V(t)), energy is flowing towards the load,
with a peak of 50W, for the next quarter cycle, energy
flows towards the source (peak –50W), and so on, for
an average of 0, as expected.

For a sinusoid, when the load is other than an open
or a short, energy flows forward for a greater
period than it flows backwards which results in a
net transfer of energy towards the load.

At the location of current or voltage minima on
the line, energy flow is either 0 or flows towards the
load.

All of the above was described without reference to forward
or reflected voltages, currents or power. It is the basic
circuit theory description of what happens between two
networks. It works for DC, 60 HZ, RF, sinusoids, steps,
pulses, you name it. If there is any disagreement with the
above, please do not read further until it is resolved.

Now it is time to introduce forward and reflected waves to
the discussion. Forward an reflected waves are simply a
mathematical transformation of V(t) and I(t) to another
representation which can make solving problems simpler.

The transformation begins with assuming that there is
a forward wave defined by Vf(t)=If(t)*Z0, a reflected
wave defined by Vr(t)=Ir(t)*Z0 and that at any point
on the line V(t)=Vf(t)+Vr(t) and I(t)=If(t)-Ir(t).

Since we are considering an ideal line, Z0 simplifies
to R0.

Simple algebra yields:
Vf(t)=(V(t)+R0*I(t))/2
Vr(t)=(V(t)-R0*I(t))/2
If(t)=Vf(t)/R0
Ir(t)=Vr(t)/R0

Now these algebraically transformed expressions have all
the capabilities of the original so they work for DC,
60 HZ, RF, sinusoids, steps, pulses, you name it.

Some people look at the expression If(t)=Vf(t)/R0 and
think it looks a lot like a travelling wave so they decide
to compute the energy being moved by this wave:
Pf(t)=Vf(t)*Vf(t)/R0
Pr(t)=Vr(t)*Vr(t)/R0

So let us explore the results when applied to the DC examples from
above.

Terminated with 50 ohms:
Vf(t)=50V
If(t)=1A
Vr(t)=0V
Ir(t)=0A

Pf(t)=50W
Pr(t)=0W

This all looks good, no reflected wave and the appropriate
amount of energy is being transported in the forward wave.

Let us try the open ended line:
Vf(t)=50V
If(t)=1A
Vr(t)=50V
Ir(t)=1A

which yield
V(t)=Vf(t)+Vr(t)=100V
I(t)=If(t)-Ir(t)=0A

which is in agreement with the observations. But there is
definitely a forward voltage wave and a reflected voltage
wave. Carrying on to compute powers:
Pf(t)=50W
Pr(t)=50W
and
P(t)=Pf(t)-Pr(t)=0W

So there you have it. The interpretation that the forward
and reflected wave are real and transport energy leads to
the conclusion that on a line charged to a constant DC
voltage, energy is flowing in the forward direction and
energy is flowing in the reflected direction.

Because it is obvious that there is no energy flowing on
the line, this interpretation makes me quite uncomfortable.

Now there are several ways out of this dilemma.

Some say they are only interested in RF (and optics), and
that nothing can be learned from DC, steps, pulses or other
functions. They simply ignore the data and continue with
their beliefs.

Some claim that DC is special, but looking at the equations,
there is no reason to expect the analysis to collapse with
DC.

The best choice is to give up on the interpretation that the
forward and reflected voltage waves necessarily transport
energy. Consider them to be a figment of the analysis technique;
very convenient, but not necessarily representing anything real.

As an added bonus, giving up on this interpretation will remove
the need to search for “where the reflected power goes” and
free the mind for investment in activities that might yield
meaningful results.

....Keith

what happens to reflected energy ?
 

On 13 jun, 23:43, walt wrote:
On Jun 13, 5:08*pm, Wimpie wrote:



On 9 jun, 02:07, Richard Clark wrote:

On Tue, 8 Jun 2010 12:17:37 -0700 (PDT), Wimpie
wrote:

When we go back to my first posting in this thread (the first reply)
where I stated that amplifiers do not show 50 Ohms in general, is not
that strange and doesn't violate Walt's conclusions.

Hi Wimpie,

Time to rewind:

On Thu, 27 May 2010 19:32:40 -0700 (PDT), walt wrote:

The source resistance data reported in Secs 19.8 and 19.9 were
obtained using the load variation method with resistive loads. Note
that of the six measurements of output source resistance reported in
Table 19.1, the average value of the resistance is 50.3 ohms obtained
with the reference load resistance of 51.2 ohms, exhibiting an error
of only 1.8 percent.

Considering these two explicit initial conditions and findings:

1. Using a Kenwood TS-830S transceiver as the RF source, the tuning
and loading of the pi-network are adjusted to deliver all the
available power into a 50 + j0-ohm load with the grid drive adjusted
to deliver the maximum of 100 watts at 4 MHz, thus establishing the
area of the RF power window at the input of the pi-network, resistance
RLP at the plate, and the slope of the load line. The output source
resistance of the amplifier in this condition will later be shown to
be 50 ohms. In this condition the DC plate voltage is 800 v and plate
current is 260 ma. DC input power is therefore 800 v ? 0.26 a = 208 w.
Readings on the Bird 43 wattmeter indicate 100 watts forward and zero
watts reflected. (100 watts is the maximum RF output power available
at this drive level.) From here on the grid drive is left undisturbed,
and the pi-network controls are left undisturbed until Step 10.

2. The amplifier is now powered down and the load resistance RL is
measured across the input terminals of the resonant pi-network tank
circuit (from plate to ground) with an HP-4815 Vector Impedance Meter.

what happens to reflected energy ?
 

On 14 jun, 02:13, Richard Clark wrote:
On Sun, 13 Jun 2010 14:08:59 -0700 (PDT), Wimpie
wrote:

This knowledge
resulted in my statement: "an RF PA isn't a 50 Ohms source", what was/
is heavily disputed by you.

Hi Wim,

Your conclusion:
"Under maximum power output matching given certain drive (not
necessarily being the maximum drive), the output impedance equals the
load resistor (the so-called "conjugated match condition")."

It was not the only conclusion, but it certainly disputes what you say
above.

Given the easy access to real amplifiers available to Hams at their
own bench, and further given that they do not present any more
complexity than your own simulation; then I have to wonder why we have
wandered into strange topologies.

My findings are from practice before I had the opportunity to do PA
simulations.

I am not interested in trying to follow 23 pages of dense presentation
where you could have simulated Walt's finals, validated or rejected
his data, and THEN formed your own conclusions. *If this is a problem
of simulation (you don't have that tube handy) what about topology? *

23 pages is not that much as graphs and circuit diagrams dominate. If
you can provide me a circuit diagram I will figure out whether I can
simulate it or not.


It looks more like a link coupled tank circuit from pre-WWII days. The
Tank Q looks to be non-existent. *When I rummage through the paper I
find 4.4! *This is certainly beyond the standard for PA design - and
we have swapped out of the tube into a mosfet (would you care to stick
to one objective?). *Elsewhere (wandering back in tubes) I find a
value of 10.3 (elsewhere it is 11) which inhabits the lowest margin of
good design. *Then an itinerant report of a cathode tank (????) whose
Q is 0.44. *Q appears to be of little concern.

Fully agree with the low cathode Q, therefore it doesn't affect the
simulations results significantly.


Again, typical Ham equipment shows Q as low as 10 - maybe, and that
would be for a dog; but those that I have seen typically fall around
15, sometimes 20. *As the Z transformation in plate/drain circuits
varies by the square of Q, this is not inconsequential and it once
again has me wondering why the strange topology?

Tell me what Q I should use (evt, give me component values) in the
simulation, and I will change it for you.


No, your paper lacks focus by trying to be all things (tubes, mosfets,
Class AB, Class C). *

This is to support my statement that under real world amateur
conditions many PAs do not obey ZL = Zout*. If I have some time I will
add a push-pull transfomer coupled amplifier also.

You examine the most inconsequential details as
if they were equally important as those details that bear on your
concern. *The paper lacks structure. *Headings are nothing more than
feature descriptions, not argument development. *There is a lot of
discussion of the idiosyncrasies of simulation - I am not interested
and that discussion creates the impression of hidden errors. *Really,
this stuff goes into an appendix not as the object of the narrative.
The graphs are pretty diversions, but they don't add much. *In the
software industry, this is spaghetti design.

I would prefer a conventional topology. *I suppose that has come
through clear. *Pick one significant point, re-edit this into 4 pages
and then you might have something interesting. *This advice comes from
one of our American writers, Mark Twain:
* * * * "If I had more time, I would have written you a shorter letter."

73's
Richard Clark, KB7QHC

simulation is general practice in the design flow, I know it has
limitations. Of course I could use ADS, but most people don't have
that at home, so I decided to use spice as this is used by many.

If you believe I am so wrong, why don't you present some simulations
to show that my original statement is completely wrong?

Best regards,


Wim
PA3DJS
www.tetech.nl
without abc, PM will reach me.

what happens to reflected energy ?
 

On Mon, 14 Jun 2010 06:06:06 -0700 (PDT), Wimpie
wrote:

23 pages is not that much as graphs and circuit diagrams dominate. If
you can provide me a circuit diagram I will figure out whether I can
simulate it or not.

Hi Wim,

I'm glad you wish to examine this more. I will send you Mendenhall's
400W FM design done 42 years ago to this month and nearly day. I will
also send a partial schematic for Walt's TS-830s.

Mendenhall's is not very different from your own model except with a
far more powerful VHF Tube, however, it is a tetrode design too.
Walt's is a pair of pentodes. See what you can do.

Fully agree with the low cathode Q, therefore it doesn't affect the
simulations results significantly.

Why a Common Grid design for a tetrode tube?

Tell me what Q I should use (evt, give me component values) in the
simulation, and I will change it for you.

15 is one that Terman cites, and Mendenhall cites Terman, and Walt's
(if I recall correctly) also find a similar value. Given your
especially low frequency simulation, component values are way out from
those commonly encountered in HF/VHF work.

No, your paper lacks focus by trying to be all things (tubes, mosfets,
Class AB, Class C). *

This is to support my statement that under real world amateur
conditions many PAs do not obey ZL = Zout*. If I have some time I will
add a push-pull transfomer coupled amplifier also.

Then you should do the finals deck for a transistor rig for that
topology. It would bring you into the 21st century (even if the
details still date from the 1970s). You will also find more
simulation options.

simulation is general practice in the design flow, I know it has
limitations. Of course I could use ADS, but most people don't have
that at home, so I decided to use spice as this is used by many.

One of the titans of linear design, Bob Pease of National
Semiconductor had little sympathy for those who discovered the
problems of simulation. However, those who read his work were sure to
avoid pain and trauma. Consult:
http://www.national.com/rap/

Having said that, it is somewhat ironic that we all use simulation for
antennas - so we are all aware of limitation and possibility. I have
used many simulators from a spectrum of disciplines.

If you believe I am so wrong, why don't you present some simulations
to show that my original statement is completely wrong?

You have more time and passion for it. I have already had a career in
testing these issues specifically and proof is at the bench. I was
trained to demanding methods for this and I worked with the most
accurate instruments in the world. However, I don't expect equal
results. If you wish to work further with the two schematics I send
you, then that would interest me far more than the banalities of
photonic explanations and spreadsheet solutions.

I am not saying anything you have done is wrong, I am saying that your
style is getting in the way of communication. It reads like a
detective novel, not a like presentation. You discover conclusions,
you don't seem to test for expectations. Surprise and mystery arrive
haphazardly through the narrative. Inconsequential details are mulled
over like Holmes describing the characteristics of cigar ash to
Watson.

You are trying to do too many things in one place and the topic begins
to blurs and the goal is distracted. For instance, this thread's
subject line of "what happens to reflected energy?" is never
summarized in your final conclusion - that is not good reportage. I
went to that last section and saw that discussion missing. Closer
examination revealed it by parts as seemingly trivial observations.

For style, start with a good grounding. Establish a base line. Do a
Monte Carlo analysis of a known design with known characteristics.
Then introduce a hypothesis tied to one variable. Push the variable
and note how it either conforms or strays from your hypothesis.
Present your work. Have a discussion to shake out problems of
understanding (from anyone). Then move on to either shift the
topology, or further stress the existing design. Build on a solid
foundation.

Schematics will follow as soon as I can make their file sizes suitable
for email servers (about 1-2 MB).

73's
Richard Clark, KB7QHC

what happens to reflected energy ?
 

On 14 jun, 20:00, Richard Clark wrote:
On Mon, 14 Jun 2010 06:06:06 -0700 (PDT), Wimpie
wrote:

23 pages is not that much as graphs and circuit diagrams dominate. *If
you can provide me a circuit diagram I will figure out whether I can
simulate it or not.

Hi Wim,

I'm glad you wish to examine this more. * I will send you Mendenhall's
400W FM design done 42 years ago to this month and nearly day. *I will
also send a partial schematic for Walt's TS-830s.

Mendenhall's is not very different from your own model except with a
far more powerful VHF Tube, however, it is a tetrode design too.
Walt's is a pair of pentodes. *See what you can do.

Fully agree with the low cathode Q, therefore it doesn't affect the
simulations results significantly.

Why a Common Grid design for a tetrode tube?

Because I built such circuit myself based on two PL519 television
tubes and to reduce the number of runs to tune the amplifier in
simulation (I still need some time to work also).
*

Tell me what Q I should use (evt, give me component values) in the
simulation, and I will change it for you.

15 is one that Terman cites, and Mendenhall cites Terman, and Walt's
(if I recall correctly) also find a similar value. *Given your
especially low frequency simulation, component values are way out from
those commonly encountered in HF/VHF work.

Over here 3.6 MHz is an amateur band, we may even go lower than that.
I will change the real class C ciruit to Q = 15. Q = Rload/XL?


No, your paper lacks focus by trying to be all things (tubes, mosfets,
Class AB, Class C).

This is to support my statement that under real world amateur
conditions many PAs do not obey ZL = Zout*. If I have some time I will
add a push-pull transfomer coupled amplifier also.

Then you should do the finals deck for a transistor rig for that
topology. *It would bring you into the 21st century (even if the
details still date from the 1970s). *You will also find more
simulation options.

simulation is general practice in the design flow, I know it has
limitations. Of course I could use ADS, but most people don't have
that at home, so I decided to use spice as this is used by many.

One of the titans of linear design, Bob Pease of National
Semiconductor had little sympathy for those who discovered the
problems of simulation. *However, those who read his work were sure to
avoid pain and trauma. *Consult:http://www.national.com/rap/

Having said that, it is somewhat ironic that we all use simulation for
antennas - so we are all aware of limitation and possibility. *I have
used many simulators from a spectrum of disciplines.

If you believe I am so wrong, why don't you present some simulations
to show that my original statement is completely wrong?

You have more time and passion for it. *

I have very limited time for this, so we have to keep it very
efficient.

I have already had a career in
testing these issues specifically and proof is at the bench. * I was
trained to demanding methods for this and I worked with the most
accurate instruments in the world. *However, I don't expect equal
results. *If you wish to work further with the two schematics I send
you, then that would interest me far more than the banalities of
photonic explanations and spreadsheet solutions.

I am not saying anything you have done is wrong, I am saying that your
style is getting in the way of communication. *It reads like a
detective novel, not a like presentation. *You discover conclusions,
you don't seem to test for expectations. *Surprise and mystery arrive
haphazardly through the narrative. *Inconsequential details are mulled
over like Holmes describing the characteristics of cigar ash to
Watson. *

Except the high Z value for the Common Cathode amplifier, all results
are as expected. Otherwise I did not make my statement on the output
impedance of PAs.


You are trying to do too many things in one place and the topic begins
to blurs and the goal is distracted. *For instance, this thread's
subject line of "what happens to reflected energy?" is never
summarized in your final conclusion - that is not good reportage. *I
went to that last section and saw that discussion missing. *Closer
examination revealed it by parts as seemingly trivial observations.

You disputed my statement on Zout and came up with the "what IS it"
attitude, I only provided information so that other people can get
their own opinion. I prefer to keep the focus on that.


For style, start with a good grounding. *Establish a base line. *Do a
Monte Carlo analysis of a known design with known characteristics.
Then introduce a hypothesis tied to one variable. *Push the variable
and note how it either conforms or strays from your hypothesis.
Present your work. *Have a discussion to shake out problems of
understanding (from anyone). *Then move on to either shift the
topology, or further stress the existing design. *Build on a solid
foundation.

This is not a thesis, I just did some simulation to show that things
can be different.

Schematics will follow as soon as I can make their file sizes suitable
for email servers (about 1-2 MB).

73's
Richard Clark, KB7QHC

Before I am going to spend lots into time this, some points:

1. Do you accept the "measuring" method with injecting a signal with
slight frequency difference as mentioned in the document?

2. Do you accept the way of presentation (envelope change versus time
and converted to output VSWR as used in the document)?

3. Do you accept B^2 spice A/D professional, version for as suitable
platform?

If we cannot agree on this, it is of no use to continue.

I limit the simulation to low HF frequencies as based on a circuit
diagram alone we cannot guess the parasitics for high HF/VHF use and
this may be food for more discussion. Please also mention what
components you want into the simulation as simulating a full TS830
will take to long.

All relevant info will be made available to the group, inclusive
original circuit diagrams and the diagrams I used for simulation. I
hope that I can get a suitable simulation model for the tubes.

Regarding circuit diagrams, can you reduce the color depth (gray scale
or B/W) and convert them to PNG as this is non-lossy coding.

I am looking forward to the results.

Best regards,


Wim
PA3DJS
www.tetech.nl

what happens to reflected energy ?
 

On Jun 14, 12:09*pm, Keith Dysart wrote:
On Jun 10, 5:52 pm, K1TTT wrote:

On Jun 9, 2:04 am, Keith Dysart wrote:

Exactly. And yet, were you to attach a DC-couple directional
wattmeter to the line, it would indicate 50W forward and 50W
reflected (using your corrected numbers).

now remember, this is AFTER you disconnected the load resistor.


only as you monitored the initial transient. *once the transient has
passed there is no power flowing in the line.

I agree completely with the second sentence. But the first is in
error. A directional wattmeter will indeed show 50W forward and 50W
reflected on a line with constant DC voltage and 0 current. I’ll
start by providing support for your second sentence and then get
back to reflected power.


no, it won't... as i will explain below.

Choose an arbitrary point on the line and observe the voltage. The
observed voltage will be a function of time. Let us call it V(t).
At the same point on the line, observe the current and call it I(t).


if you can choose an arbitrary point, i can choose an arbitrary
time... call it a VERY long time after the load is disconnected.

The energy flowing (i.e. power) at this point on the line will
be P(t)=V(t)I(t). If the voltage is measured in volts and the current
in amps, then the power will be in joules/s or watts.
We can also compute the average energy flow (power) over some
interval by integrating P(t) over the interval and dividing by
the length of the interval. Call this Pavg. For repetitive
signals it is convenient to use one period as the interval.

Note that V(t) and I(t) are arbitrary functions. The analysis
applies regardless of whether V(t) is DC, a sinusoid, a step,
a pulse, etc.; any arbitrary function.

To provide some examples, consider the following simple setup:
1) 50 ohm, Thevenin equivalent circuit provides 100 V in to an
* *open circuit.

this is the case we were discussing... the others are irrelevant at
this point


2) The generator is connected to a length of 50 ohm ideal
* *transmission line. We will stick with ideal lines for the
* *moment, because, until ideal lines are understood, there
* *is no hope of understanding more complex models.
3) The ideal line is terminated in a 50 ohm resistor.

Set the generator to DC:
* V(t)=50V, wherever it is measured on the line
* I(t)=1A, wherever it is measured on the line
* P(t)=50W, wherever it is measured on the line
* Pavg=50W
No disagreement, I hope.

only if this is a case WITH a load... this is not what we were
discussing above.



Set the generator to generate a sinusoid:
* V(t)=50sin(wt)
* I(t)=1sin(wt)
* P(t)=25+25sin(2wt)
* Pavg=25W
describes the observations at any point on the line.
Measurements at different points will have a different
phase relationship to each due to the delay in the line.
Note carefully the power function. The power at any point
on the line varies from 0 to 50W with a sinusoidal
pattern at twice the frequency of the voltage sinusoid.
The power is always positive; that is, energy is always
flowing towards the load.

Now let us remove the terminating resistor. The line is
now open circuit.

OK, now we are back to where i had objections above.


Again, set the generator to DC. After one round trip
time, the observations at any point on the line will
be:
* V(t)=100V
* I(t)=0A
* P(t)=0W
* Pavg=0W


ok, p(t)=0w, i=0a, i connect my directional wattmeter at some very
long time after the load is disconnected so the transients can be
ignored and i measure 0w forever. there is no power flowing in the
line.


The story for a sinusoid is more interesting...
The observations now depend on the location on the
line where the measurements are performed.
At a current 0:
* V(t)=100sin(Wt)
* I(t)=0A
* P(t)=0W
* Pavg=0W
At a voltage 0:
* V(t)=0V
* I(t)=2sin(wt)
* P(t)=0W
* Pavg=0W
Halfway between the current 0 at the load and the
preceding voltage 0 (that is, 45 degrees back
from the load:
* V(t)=70.7sin(wt)
* I(t)=1.414cos(wt)
* P(t)=50sin(2wt)
* Pavg=0W
This tells us that for the first quarter cycle of the
voltage (V(t)), energy is flowing towards the load,
with a peak of 50W, for the next quarter cycle, energy
flows towards the source (peak –50W), and so on, for
an average of 0, as expected.

For a sinusoid, when the load is other than an open
or a short, energy flows forward for a greater
period than it flows backwards which results in a
net transfer of energy towards the load.

At the location of current or voltage minima on
the line, energy flow is either 0 or flows towards the
load.

All of the above was described without reference to forward
or reflected voltages, currents or power. It is the basic
circuit theory description of what happens between two
networks. It works for DC, 60 HZ, RF, sinusoids, steps,
pulses, you name it. If there is any disagreement with the
above, please do not read further until it is resolved.

so you have derived voltages and currents at a couple of special
locations where superposition causes maximum or minimum voltages or
currents... this of course ONLY works after sinusoidal steady state
has been achieved, so your cases of connecting and disconnecting the
load must be thrown out. and how do you find a maximum or minimum
when the load is matched to the line impedance? standing waves are
always a trap... step back from the light and come back to the truth
and learn how to properly account for forward and reflected waves with
voltages or currents and all will be revealed.



Now it is time to introduce forward and reflected waves to
the discussion. Forward an reflected waves are simply a
mathematical transformation of V(t) and I(t) to another
representation which can make solving problems simpler.

The transformation begins with assuming that there is
a forward wave defined by Vf(t)=If(t)*Z0, a reflected
wave defined by Vr(t)=Ir(t)*Z0 and that at any point
on the line V(t)=Vf(t)+Vr(t) and I(t)=If(t)-Ir(t).

Since we are considering an ideal line, Z0 simplifies
to R0.

Simple algebra yields:
* Vf(t)=(V(t)+R0*I(t))/2
* Vr(t)=(V(t)-R0*I(t))/2
* If(t)=Vf(t)/R0
* Ir(t)=Vr(t)/R0

Now these algebraically transformed expressions have all
the capabilities of the original so they work for DC,
60 HZ, RF, sinusoids, steps, pulses, you name it.

hmmm, they work for all waveforms you say... even dc?? ok, open
circuit load, z0=50, dc source of 100v with 50ohm thevenin
resistance...

If(t) = 100/50 = 2a
Ir(t) = 100/50 = 2a

so there is a reflected 2a current you say? where does that go when
it gets back to the source? we know that when the 100v source is open
circuited as it must be in this case that it develops 100v across its
terminals... sure doesn't seem like any way that 2a can go back into
the source. it can't be reflected since the source impedance matches
the line impedance. so where does it go? maybe those simple
algebraic expressions have to be reconsidered a bit.

well, maybe i missed something... lets look at the voltages:

Vf(t)=(100+50*2)/2 = 100
Vr(t)=(100-50*2)/2 = 0

hmm, you do seem to get the right constant voltage... but it kind of
messes up your power flow since Vr=0 there can be no power in that
reflected wave, so it really doesn't make it much of a wave.

but wait, the equations must be consistent so

Ir(t)=0/50

So there really is no reflected current... but there is forward
current? so where does that forward current go if there is no load?
it doesn't reflect back if Ir=0... does it just build up at the open
end? going to be lots of trons sitting there after a while.





Some people look at the expression If(t)=Vf(t)/R0 and
think it looks a lot like a travelling wave so they decide
to compute the energy being moved by this wave:
* Pf(t)=Vf(t)*Vf(t)/R0
* Pr(t)=Vr(t)*Vr(t)/R0

So let us explore the results when applied to the DC examples from
above.

Terminated with 50 ohms:
* Vf(t)=50V
* If(t)=1A
* Vr(t)=0V
* Ir(t)=0A

* Pf(t)=50W
* Pr(t)=0W

This all looks good, no reflected wave and the appropriate
amount of energy is being transported in the forward wave.

Let us try the open ended line:
* Vf(t)=50V
* If(t)=1A
* Vr(t)=50V
* Ir(t)=1A

which yield
* V(t)=Vf(t)+Vr(t)=100V
* I(t)=If(t)-Ir(t)=0A

which is in agreement with the observations. But there is
definitely a forward voltage wave and a reflected voltage
wave. Carrying on to compute powers:

sorry, you can't have a 'voltage wave' without a 'current wave'...
just doesn't work i'm afraid. to have one you must have the other.

* Pf(t)=50W
* Pr(t)=50W
and
* P(t)=Pf(t)-Pr(t)=0W

So there you have it. The interpretation that the forward
and reflected wave are real and transport energy leads to
the conclusion that on a line charged to a constant DC
voltage, energy is flowing in the forward direction and
energy is flowing in the reflected direction.

obviously incorrect, by reductio ad absurdum if i remember the latin
correctly.


Because it is obvious that there is no energy flowing on
the line, this interpretation makes me quite uncomfortable.

Now there are several ways out of this dilemma.

Some say they are only interested in RF (and optics), and
that nothing can be learned from DC, steps, pulses or other
functions. They simply ignore the data and continue with
their beliefs.

Some claim that DC is special, but looking at the equations,
there is no reason to expect the analysis to collapse with
DC.

The best choice is to give up on the interpretation that the
forward and reflected voltage waves necessarily transport
energy. Consider them to be a figment of the analysis technique;
very convenient, but not necessarily representing anything real.

As an added bonus, giving up on this interpretation will remove
the need to search for “where the reflected power goes” and
free the mind for investment in activities that might yield
meaningful results.

...Keith

what happens to reflected energy ?
 

On Mon, 14 Jun 2010 12:32:02 -0700 (PDT), Wimpie
wrote:

Before I am going to spend lots into time this, some points:

1. Do you accept the "measuring" method with injecting a signal with
slight frequency difference as mentioned in the document?

Hi Wim,

I got only a hint of what you were trying to do because it was buried
in so much conflicting agendas.

However, my first impression was that it looked worth discussing. My
second impression was that it was not needed. Still, that is worth
discussing too.

2. Do you accept the way of presentation (envelope change versus time
and converted to output VSWR as used in the document)?

I hadn't the slightest idea what that was all about. Like I said, too
many things going on. From your email, it sounds like too much
complexity to accomplish what was done with meter indications.

For others, this might be found in the section called "Output Tuning"
in:
"Fine Tuning FM Final Stages," by Mendenhall (google for a link or
email me for a copy).

3. Do you accept B^2 spice A/D professional, version for as suitable
platform?

I have no problem with what you are comfortable with. I am sure a
suitable simulation will evolve.

If we cannot agree on this, it is of no use to continue.

I limit the simulation to low HF frequencies as based on a circuit
diagram alone we cannot guess the parasitics for high HF/VHF use and
this may be food for more discussion. Please also mention what
components you want into the simulation as simulating a full TS830
will take to long.

All relevant info will be made available to the group, inclusive
original circuit diagrams and the diagrams I used for simulation. I
hope that I can get a suitable simulation model for the tubes.

Regarding circuit diagrams, can you reduce the color depth (gray scale
or B/W) and convert them to PNG as this is non-lossy coding.

I am looking forward to the results.

We will resolve the details offline. I would encourage you to save
effort (we both have things to do) by doing a "quick and dirty" first
pass just to see how little we may have left to do (it could be that
simple). There is more to read than there is to simulate.

For others, the material I have linked Wim to probably tips the
download scale at 50MB. 44 MB is one document alone that others here
might find useful:
"EIMAC Care and Feeding of Power Grid Tubes" 44MB at:
http://www.g8wrb.org/data/Eimac/care...grid_tubes.pdf

It will brush away the cobwebs.

73's
Richard Clark, KB7QHC

what happens to reflected energy ?
 

On Jun 14, 11:04*pm, K1TTT wrote:
read more »

no, i've read enough thank you.

I think there are 2 basic things you are missing... V and I are
functions not only of time, but of distance along the line, both for
the forward and reflected waves. looking at them at specific points
can lead to incorrect assumptions and over simplified conclusions.
Also, there are very good reasons why the sinusoidal steady state
condition is used to simplify the equations they way they are normally
shown, and why your over simplifications won't hack it in the long
run. to do the full analysis of a distributed system (except in very
special cases) for an arbitrary waveform requires summations of
reflections both ways on the line over many reflection periods.. the
results of this for step functions can be seen with simple tdr's on
lines with multiple discontinuities where you can see the reflections
ringing down.

what happens to reflected energy ?
 

Keith Dysart wrote:

. . .

So there you have it. The interpretation that the forward
and reflected wave are real and transport energy leads to
the conclusion that on a line charged to a constant DC
voltage, energy is flowing in the forward direction and
energy is flowing in the reflected direction.

Because it is obvious that there is no energy flowing on
the line, this interpretation makes me quite uncomfortable.

Now there are several ways out of this dilemma.

Some say they are only interested in RF (and optics), and
that nothing can be learned from DC, steps, pulses or other
functions. They simply ignore the data and continue with
their beliefs.

Some claim that DC is special, but looking at the equations,
there is no reason to expect the analysis to collapse with
DC.

The best choice is to give up on the interpretation that the
forward and reflected voltage waves necessarily transport
energy. Consider them to be a figment of the analysis technique;
very convenient, but not necessarily representing anything real.

As an added bonus, giving up on this interpretation will remove
the need to search for “where the reflected power goes” and
free the mind for investment in activities that might yield
meaningful results.

...Keith

Well done, Keith. Thanks.

Roy Lewallen, W7EL

what happens to reflected energy ?
 

On Jun 16, 4:11*pm, Roy Lewallen wrote:
Well done, Keith. Thanks.

Two ignorant people patting each other on the back. If DC methods
worked on distributed networks, there would never have been any need
to invent distributed network analysis - yet it was invented because
DC methods failed miserably for distributed network analysis. How does
a 1/4WL impedance transformer perform at DC? How does a DC analysis
work for light waves? Good grief!
--
73, Cecil, w5dxp.com

what happens to reflected energy ?
 

On Jun 16, 5:26*pm, Cecil Moore wrote:
On Jun 16, 4:11*pm, Roy Lewallen wrote:

Well done, Keith. Thanks.

Two ignorant people patting each other on the back. If DC methods
worked on distributed networks, there would never have been any need
to invent distributed network analysis - yet it was invented because
DC methods failed miserably for distributed network analysis. How does
a 1/4WL impedance transformer perform at DC? How does a DC analysis
work for light waves? Good grief!

Ahhh. Your standard answer. You are an RF dude and there is nothing
that can possibly be learned from simple DC circuits.
Especially when the results challenge your beliefs.

Still, I observe that you have discovered no errors in the exposition.
Which is good.

....Keith

what happens to reflected energy ?
 

On Jun 17, 5:38*am, Keith Dysart wrote:
You are an RF dude and there is nothing
that can possibly be learned from simple DC circuits.

We studied DC circuits in EE101. Later we were forced to expand our
knowledge into distributed networks because the DC circuit model
failed at RF frequencies.

Still, I observe that you have discovered no errors in the exposition.

I have not wasted my time time trying to discover any errors. In a
nutshell, what new knowledge have you presented?
--
73, Cecil, w5dxp.com

what happens to reflected energy ?
 

Keith Dysart wrote:
. . .
By substitution, it is easy to show that the rules used to
derive Vf and Vr results in
P(t)=Pf(t)-Pr(t)

So, while superposing powers, does not work in general, it does
for this case. Directional wattmeters take advantage of this
ideosyncracy to let the user compute the power being delivered
to the load.
. . .

And it works only if Z0 is purely real. When it's not, P(t) also
includes Vf*Ir and Vr*If terms which are neither "forward" nor "reverse"
power. It's unfortunate in a way that Z0 is close enough to being real
for decent lines at Amateur frequencies that these terms are small.
Otherwise they'd have to be confronted and some mistaken assumptions
abandoned.

Roy Lewallen, W7EL

what happens to reflected energy ?
 

Roy Lewallen wrote in
:

Keith Dysart wrote:
. . .
By substitution, it is easy to show that the rules used to
derive Vf and Vr results in
P(t)=Pf(t)-Pr(t)

So, while superposing powers, does not work in general, it does
for this case. Directional wattmeters take advantage of this
ideosyncracy to let the user compute the power being delivered
to the load.
. . .

And it works only if Z0 is purely real. When it's not, P(t) also
includes Vf*Ir and Vr*If terms which are neither "forward" nor
"reverse" power. It's unfortunate in a way that Z0 is close enough to
being real for decent lines at Amateur frequencies that these terms
are small. Otherwise they'd have to be confronted and some mistaken
assumptions abandoned.

While Zo of transmission lines might not be purely real, if the sampler
element is calibrated for V/I being real, then the power is given by
'forward power' - 'reflected power'. This is true even if the
calibration impedance is different to the transmission line in which the
measurements are made, though significant departure will impact
measurement uncertainty.

For example, if I insert a 50 ohm Bird 43 in a 75 ohm line at 1.8MHz,
and measure Pf=150W and Pr=50W, then the power is 150-50=100W. The
insertion VSWR due to the Bird 43 is trivial in this case, so it hardly
disturbs the thing being measured. Each of the power measurements is of
little value alone, no inference can be made (in this case) of the
actual line VSWR, but the difference of the Pf and Pr readings does give
the power at that point.

I discuss this in my article entitled "http://vk1od.net/blog/?p=1004" at
http://vk1od.net/blog/?p=1004 .

Owen

what happens to reflected energy ?
 

Correction:

I discuss this in my article entitled "Power in a mismatched transmission
line" at http://vk1od.net/blog/?p=1004 .

Owen

what happens to reflected energy ?
 

Owen Duffy wrote:

While Zo of transmission lines might not be purely real, if the sampler
element is calibrated for V/I being real, then the power is given by
'forward power' - 'reflected power'. This is true even if the
calibration impedance is different to the transmission line in which the
measurements are made, though significant departure will impact
measurement uncertainty.

For example, if I insert a 50 ohm Bird 43 in a 75 ohm line at 1.8MHz,
and measure Pf=150W and Pr=50W, then the power is 150-50=100W. The
insertion VSWR due to the Bird 43 is trivial in this case, so it hardly
disturbs the thing being measured. Each of the power measurements is of
little value alone, no inference can be made (in this case) of the
actual line VSWR, but the difference of the Pf and Pr readings does give
the power at that point.

I discuss this in my article entitled "http://vk1od.net/blog/?p=1004" at
http://vk1od.net/blog/?p=1004 .

Owen

Adding a directional wattmeter to the mix raises an interesting issue.

Recalling my earlier example of a 100 watt transmitter connected to a
half wavelength 200 ohm transmission line and then to a 50 ohm resistive
load, the "forward power" on the line was 156.25 watts and the "reverse
power" or "reflected power" 56.25 watts. The subject of this topic asks
where the "reflected energy" goes. Does it go back into the transmitter?
Well, let's put a 50 ohm directional wattmeter between the transmitter
and the line. It measures 100 watts forward and 0 watts reverse. So what
happened to the "reverse power" on the transmission line? If it was
going back into the transmitter, we've now fixed the problem just by
adding the wattmeter. In fact, we could replace the wattmeter with a
piece of 50 ohm transmission line of any length, as short or long as we
want, and the "forward power" on that line will be 100 watts and
"reverse power" zero. So we've protected the transmitter from this
horrible, damaging "reverse power" just by adding a couple of inches of
50 ohm line -- or a directional wattmeter. Is this cool or what?

Or we can put the wattmeter at the far end of the line and read 100
watts forward and zero watts reverse, eliminating "reverse power" at the
load -- although we've lost 56.25 watts of "forward power". Or put it at
the center of the line, where we'd read a whopping 451.6 watts forward
and 351.6 reverse(*).

These are the "forward power" and "reverse power" on the short 50 ohm
transmission line (or lumped equivalent) inside the wattmeter. So we can
create "forward power" and "reverse power" just by moving the wattmeter
around. And most importantly, we can use it to isolate the transmitter
from that bad "reverse power". What a powerful tool!

I should point out that the example doesn't even mention, or need to
mention, the transmitter output impedance. The entire analysis holds for
any transmitter impedance, whether "dissipative", "non-dissipative",
linear, or nonlinear. Wherever the "forward power" and "reverse power"
come from and do to, it doesn't depend on any particular value or kind
of transmitter impedance.

(*) This brings up a great idea. Drop down to Radio Shack
or HRO and pick up one of those circulator things that separates forward
and reverse power. Put it into the middle of the line where you have
451.6 watts of "forward power" and 351.6 watts of "reverse power". Take
the 351.6 watts of "reverse power", rectify it, send it to an inverter,
and use it to run the transmitter -- you'll have plenty, even with poor
efficiency, and you can unplug the transmitter from the mains. There
should even be enough power left over to run your cooler and keep a
six-pack cold. Then work DX with your 451.6 watts of forward power while
you enjoy a cool one. Goodbye to electric bills! Hello to DX, free
power, and a tall cool one!

Roy Lewallen, W7EL

what happens to reflected energy ?
 

On Jun 17, 10:31*am, Keith Dysart wrote:
On Jun 14, 7:04 pm, K1TTT wrote:

On Jun 14, 12:09 pm, Keith Dysart wrote:
Choose an arbitrary point on the line and observe the voltage. The
observed voltage will be a function of time. Let us call it V(t).
At the same point on the line, observe the current and call it I(t).

if you can choose an arbitrary point, i can choose an arbitrary
time... call it a VERY long time after the load is disconnected.

That works for me, as long as the time is at least twice the
propagation
time down the line.



The energy flowing (i.e. power) at this point on the line will
be P(t)=V(t)I(t). If the voltage is measured in volts and the current
in amps, then the power will be in joules/s or watts.
We can also compute the average energy flow (power) over some
interval by integrating P(t) over the interval and dividing by
the length of the interval. Call this Pavg. For repetitive
signals it is convenient to use one period as the interval.

Note that V(t) and I(t) are arbitrary functions. The analysis
applies regardless of whether V(t) is DC, a sinusoid, a step,
a pulse, etc.; any arbitrary function.

To provide some examples, consider the following simple setup:
1) 50 ohm, Thevenin equivalent circuit provides 100 V in to an
* *open circuit.

this is the case we were discussing... the others are irrelevant at
this point

2) The generator is connected to a length of 50 ohm ideal
* *transmission line. We will stick with ideal lines for the
* *moment, because, until ideal lines are understood, there
* *is no hope of understanding more complex models.
3) The ideal line is terminated in a 50 ohm resistor.

snip the experiments you do not consider relevent



Now let us remove the terminating resistor. The line is
now open circuit.

OK, now we are back to where i had objections above.

Again, set the generator to DC. After one round trip
time, the observations at any point on the line will
be:
* V(t)=100V
* I(t)=0A
* P(t)=0W
* Pavg=0W

ok, p(t)=0w, i=0a, i connect my directional wattmeter at some very
long time after the load is disconnected so the transients can be
ignored and i measure 0w forever.

Not correct. See the equations for a directional wattmeter below.

there is no power flowing in the line.

I agree with this, though a directional wattmeter indicates
otherwise. See below.



Now it is time to introduce forward and reflected waves to
the discussion. Forward an reflected waves are simply a
mathematical transformation of V(t) and I(t) to another
representation which can make solving problems simpler.

The transformation begins with assuming that there is
a forward wave defined by Vf(t)=If(t)*Z0, a reflected
wave defined by Vr(t)=Ir(t)*Z0 and that at any point
on the line V(t)=Vf(t)+Vr(t) and I(t)=If(t)-Ir(t).

Since we are considering an ideal line, Z0 simplifies
to R0.

Simple algebra yields:
* Vf(t)=(V(t)+R0*I(t))/2
* Vr(t)=(V(t)-R0*I(t))/2
* If(t)=Vf(t)/R0
* Ir(t)=Vr(t)/R0

The above equations are used in a directional wattmeter.

Now these algebraically transformed expressions have all
the capabilities of the original so they work for DC,
60 HZ, RF, sinusoids, steps, pulses, you name it.

hmmm, they work for all waveforms you say... even dc?? *ok, open
circuit load, z0=50, dc source of 100v with 50ohm thevenin
resistance...

You have made some arithmetic errors below. I have re-arranged
your prose a bit to allow correction, then comment.

well, maybe i missed something... lets look at the voltages:

Vf(t)=(100+50*2)/2 = 100
Vr(t)=(100-50*2)/2 = 0

You made a substitution error here. On the transmission line
with a constant DC voltage:
* V(t)=100V
* I(t)=0A

substituting in to
* Vf(t)=(V(t)+R0*I(t))/2
* Vr(t)=(V(t)-R0*I(t))/2

yields
* Vf(t)=50V
* Vr(t)=50V

and
* If(t) = 50/50 = 1a
* Ir(t) = 50/50 = 1a

Recalling that the definition of forward and reflected is
* V(t)=Vf(t)+Vr(t)
* I(t)=If(t)-Ir(t)
we can check our arithmetic
* V(t)= 50+50 = 100V
* I(t)= 1-1 * = 0A
Exactly as measured.

If you are disagree with the results, please point out
my error in the definitions, derivations or arithmetic.

except i can measure If and Ir separately and can see they are both
zero. this is where your argument about putting a resistor between
two batteries falls apart... there is a length of cable with a finite
time delay in the picture here. so it is possible to actually watch
the waves reflect back and forth and see what really happens. in the
case where the line is open at the far end we can measure the voltage
at both ends of the line and see the constant 100v, so where is the
voltage difference needed to support the forward and reverse current
waves? you can't have current without a voltage difference. plus i
can substitute in a lossy line and measure the line heating and
measure no i^2r loss that would exist if there were currents flowing
in the line, so there are none.

try this case, which is something we do at work... take a 14ga piece
of copper wire maybe 50' above ground, its characteristic impedance is
probably a few hundred ohms, and put 100kv on it... how much power is
it dissipating from your If and Ir heating losses?? lets say 200 ohms
and 100kv gives 100,000/200a=500a, shouldn't take long to burn up that
wire should it? and yet it never does... so why doesn't it have your
circulating currents?