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