Not replying to anyone in particular - Because I can post graphics
directly to QRZ, I have started a thread on the antenna forum at:

http://forums.qrz.com/showthread.php...er-Brainteaser

It is my contention that reflections are eliminated by a Z0-match
which, in a tuner system, occurs at the *input* to the tuner.
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

On May 11, 2:40*pm, Cecil Moore wrote:
Not replying to anyone in particular - Because I can post graphics
directly to QRZ, I have started a thread on the antenna forum at:

http://forums.qrz.com/showthread.php...er-Brainteaser

It is my contention that reflections are eliminated by a Z0-match
which, in a tuner system, occurs at the *input* to the tuner.
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

Cecil, you are correct!!!

Walt

On May 11, 3:09*pm, walt wrote:
Cecil, you are correct!!!

Thanks Walt, it would be interesting to see what else about which I am
correct. W7EL has a food-for-thought publication with one section
about reflections from the source. His source has a 50 ohm source
impedance, by design, so any dynamic load pulling experiment should
result in zero reflections from the source. Question is, does it?

http://eznec.com/misc/Food_for_thought.pdf

The "forward and reverse power section" is at the bottom of the page.
Using the dynamic load pulling method, do the results always indicate
50 ohms for the source impedance no matter what is the load and/or the
transmission line length? If not, the method yields invalid results.

IMO, it is a gross error to presume that all redistribution of energy
is a result of reflections. In W7EL's example, there are no
reflections from the source yet it is obvious that energy is being
redistributed from the source back toward the load in some cases but
not in other cases. IMO, destructive/constructive interference must be
taken into account in order to explain the results. Yet, no one except
yours truly has even mentioned interference effects as a method of
redistributing energy.

Anyone interested in understanding the role of interference at
impedance discontinuities in transmission lines is welcome to read my
article at:

http://www.w5dxp.com/energy.htm
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

On May 13, 4:09*pm, Cecil Moore wrote:
On May 11, 3:09*pm, walt wrote:

Cecil, you are correct!!!

Thanks Walt, it would be interesting to see what else about which I am
correct. W7EL has a food-for-thought publication with one section
about reflections from the source. His source has a 50 ohm source
impedance, by design, so any dynamic load pulling experiment should
result in zero reflections from the source. Question is, does it?

http://eznec.com/misc/Food_for_thought.pdf

The "forward and reverse power section" is at the bottom of the page.
Using the dynamic load pulling method, do the results always indicate
50 ohms for the source impedance no matter what is the load and/or the
transmission line length? If not, the method yields invalid results.

IMO, it is a gross error to presume that all redistribution of energy
is a result of reflections. In W7EL's example, there are no
reflections from the source yet it is obvious that energy is being
redistributed from the source back toward the load in some cases but
not in other cases. IMO, destructive/constructive interference must be
taken into account in order to explain the results. Yet, no one except
yours truly has even mentioned interference effects as a method of
redistributing energy.

Anyone interested in understanding the role of interference at
impedance discontinuities in transmission lines is welcome to read my
article at:

http://www.w5dxp.com/energy.htm
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK


Right on, Cecil. The article you URL'ed is the most clear and accurate
paraphrasing of my presentation of the subject in 'Reflections'.
Nowhere else is the subject clearly defined concerning destructive and
constructive interference in relation to impedance matching and 100%
re-reflection. I hope RF engineers everywhere can realize that optics
and RF are indeed using the same principles of electromagnetics, and
therefore follow exactly the same rules.

However, I would like for you to add one more reference to your
article, an article I published in QEX that proves Steve Best's QEX
three articles wrong. It appeared in the Jul/Aug 2004 issue, entitled,
"A tutorial Dispelling Certain Misconceptions Concerning Wave
Interference In Impedance Matching."

Let me know here whether you have a copy of my QEX article. If not,
I'll email you a copy.

Walt, W2DU

If anyone else on this thread would like a copy of my QEX article
please let me know by email. My email address is shown above.

On May 13, 5:39*pm, walt wrote:
On May 13, 4:09*pm, Cecil Moore wrote:









On May 11, 3:09*pm, walt wrote:

Cecil, you are correct!!!

Thanks Walt, it would be interesting to see what else about which I am
correct. W7EL has a food-for-thought publication with one section
about reflections from the source. His source has a 50 ohm source
impedance, by design, so any dynamic load pulling experiment should
result in zero reflections from the source. Question is, does it?

http://eznec.com/misc/Food_for_thought.pdf

The "forward and reverse power section" is at the bottom of the page.
Using the dynamic load pulling method, do the results always indicate
50 ohms for the source impedance no matter what is the load and/or the
transmission line length? If not, the method yields invalid results.

IMO, it is a gross error to presume that all redistribution of energy
is a result of reflections. In W7EL's example, there are no
reflections from the source yet it is obvious that energy is being
redistributed from the source back toward the load in some cases but
not in other cases. IMO, destructive/constructive interference must be
taken into account in order to explain the results. Yet, no one except
yours truly has even mentioned interference effects as a method of
redistributing energy.

Anyone interested in understanding the role of interference at
impedance discontinuities in transmission lines is welcome to read my
article at:

http://www.w5dxp.com/energy.htm
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

Right on, Cecil. The article you URL'ed is the most clear and accurate
paraphrasing of my presentation of the subject in 'Reflections'.
Nowhere else is the subject clearly defined concerning destructive and
constructive interference in relation to impedance matching and 100%
re-reflection. I hope RF engineers everywhere can realize that optics
and RF are indeed using the same principles of electromagnetics, and
therefore follow exactly the same rules.

However, I would like for you to add one more reference to your
article, an article I published in QEX that proves Steve Best's QEX
three articles wrong. It appeared in the Jul/Aug 2004 issue, entitled,
"A tutorial Dispelling Certain Misconceptions Concerning Wave
Interference In Impedance Matching."

Let me know here whether you have a copy of my QEX article. If not,
I'll email you a copy.

Walt, W2DU

If anyone else on this thread would like a copy of my QEX article
please let me know by email. My email address is shown above.


Sorry guys, my error, my email address is .

Walt

On 13 mayo, 22:09, Cecil Moore wrote:
On May 11, 3:09*pm, walt wrote:

Cecil, you are correct!!!

Thanks Walt, it would be interesting to see what else about which I am
correct. W7EL has a food-for-thought publication with one section
about reflections from the source. His source has a 50 ohm source
impedance, by design, so any dynamic load pulling experiment should
result in zero reflections from the source. Question is, does it?

http://eznec.com/misc/Food_for_thought.pdf

The "forward and reverse power section" is at the bottom of the page.
Using the dynamic load pulling method, do the results always indicate
50 ohms for the source impedance no matter what is the load and/or the
transmission line length? If not, the method yields invalid results.

IMO, it is a gross error to presume that all redistribution of energy
is a result of reflections. In W7EL's example, there are no
reflections from the source yet it is obvious that energy is being
redistributed from the source back toward the load in some cases but
not in other cases. IMO, destructive/constructive interference must be
taken into account in order to explain the results. Yet, no one except
yours truly has even mentioned interference effects as a method of
redistributing energy.

Anyone interested in understanding the role of interference at
impedance discontinuities in transmission lines is welcome to read my
article at:

http://www.w5dxp.com/energy.htm
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

Hello Cecil,

When a source behaves like a 50 Ohms source, it will not rereflect the
reflected power back to the load. In other words the forward power
generated by the source will not change when changing the load. The
reflected power that is absorbed by the source may result in increase
or reduction of power dissipation of the active device. It all depends
on the change of the integral of V*I (for the active device).

With regards to load pulling, output impedance of an amplifier may
change depending on the rate of change of the load. As I mentioned
earlier, manual load pulling may result in change of supply and bias
voltage and current. Average Voltage and current have time to settle
to the new load condition.

In case of load pulling where the load changes relatively fast, but
well within the matching section's bandwidth (as is the case for the
off-carrier signal injection method) supply and bias voltage change
may be less due to decoupling capacitance.

In simulation this can easily be seen by applying a load step change
and observe the envelope and phase response versus time. Of course
this requires you to model your power supply and bias circuit
correctly. You may see a slow step response that cannot be explained
by the bandwidth of the matching network.

To walt:
I didn’t see your announced comment to my last post (regarding the
forward/reverse power mathematics), did I missed it?

With kind regards,


Wim
PA3DJS
www.tetech.nl

On May 13, 6:51*pm, Wimpie wrote:
On 13 mayo, 22:09, Cecil Moore wrote:









On May 11, 3:09*pm, walt wrote:

Cecil, you are correct!!!

Thanks Walt, it would be interesting to see what else about which I am
correct. W7EL has a food-for-thought publication with one section
about reflections from the source. His source has a 50 ohm source
impedance, by design, so any dynamic load pulling experiment should
result in zero reflections from the source. Question is, does it?

http://eznec.com/misc/Food_for_thought.pdf

The "forward and reverse power section" is at the bottom of the page.
Using the dynamic load pulling method, do the results always indicate
50 ohms for the source impedance no matter what is the load and/or the
transmission line length? If not, the method yields invalid results.

IMO, it is a gross error to presume that all redistribution of energy
is a result of reflections. In W7EL's example, there are no
reflections from the source yet it is obvious that energy is being
redistributed from the source back toward the load in some cases but
not in other cases. IMO, destructive/constructive interference must be
taken into account in order to explain the results. Yet, no one except
yours truly has even mentioned interference effects as a method of
redistributing energy.

Anyone interested in understanding the role of interference at
impedance discontinuities in transmission lines is welcome to read my
article at:

http://www.w5dxp.com/energy.htm
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

Hello Cecil,

When a source behaves like a 50 Ohms source, it will not rereflect the
reflected power back to the load. In other words the forward power
generated by the source will not change when changing the load. The
reflected power that is absorbed by the source may result in increase
or reduction of power dissipation of the active device. It all depends
on the change of the integral of V*I (for the active device).

With regards to load pulling, output impedance of an amplifier may
change depending on the rate of change of the load. As I mentioned
earlier, manual load pulling may result in change of supply and bias
voltage and current. *Average Voltage and current have time to settle
to the new load condition.

In case of load pulling where the load changes relatively fast, but
well within the matching section's bandwidth (as is the case for the
off-carrier signal injection method) supply and bias voltage change
may be less due to decoupling capacitance.

In simulation this can easily be seen by applying a load step change
and observe the envelope and phase response versus time. Of course
this requires you to model your power supply and bias circuit
correctly. You may see a slow step response that cannot be explained
by the bandwidth of the matching network.

To walt:
I didn’t see your announced comment to my last post (regarding the
forward/reverse power mathematics), did I missed it?

With kind regards,

Wim
PA3DJSwww.tetech.nl


Hello Wim,

I hadn't yet given you my final understanding of it. I understand the
voltage divider action, but I don't yet understand the following quote
from you:

"When terminated according to the numbers above the socket ("100W
into
50 Ohms"), this results in 100W ( Vout = 212.1*50/(50+100) = 70.7V,
just a voltage divider consisting of 100 Ohms and 50 Ohms )."

The portion I don't understand is the mathematical basis for:

"( Vout = 212.1*50/(50+100) = 70.7V,"

And Wim, on your last post above I agree that if the source is 50 ohms
with a real resistor, as in the classical generator used in text
books. However, are you not aware that the output resistance of the RF
power amp is R = E/R that appears at the output. Since this R is non-
dissipative it re-reflects all the reflected power. I have proved this
to be true in the experiment I presented with the Kenwood TS-830S
presentation that I'm sure you have a copy of.

Consequently, the RF power amp is not going to absorb the reflected
power.

At this point my email server is down, but I'll forward a copy of my
QEX article ASAP.

Walt

On May 11, 4:21*am, Richard Fry wrote:
On May 10, 9:35 pm, walt wrote:

If you are so confident that the number is 88.8889%,
please derive the conditions that yield that number.

à = R(load) - R(source) / R(load) + R(source)

SWR = (1 + |Ã|) / (1 - |Ã|)

Power Accepted by the Load = Incident Power * (1 - Ã^2)

For a 50 ohm source connected to a 100 ohm load:

à = (100 - 50) / (100 + 50) = 50 / 150 = 0.333333...

SWR = (1 + 0.333333) / (1 - 0.333333) = 1.333333 / 0.666666 = 2:1

Load Power = Incident Power * (1 - 0.333333^2) = Incident Power *
0.888889 (or 88.8889%)

The answers are the same for a 50 ohm source with a 25 ohm load.

RF

Richard F is, of course, correct. You can just as well do it for DC:
2V O.C., 1 ohm source will deliver 1A, 1V, 1 watt to a 1 ohm load.
For a 2 ohm load, the current is 2/3 amp; the power is 8/9 watt. For
an 0.5 ohm load, the current is 4/3 amp; the power is 8/9 watt. It's
not terribly difficult to show, for the AC case, that the same is true
for any 2:1 load, regardless of phase angle--trivial, in fact, if you
accept that the SWR along a lossless TEM line is constant.

Cheers,
Tom

On May 13, 7:07*pm, K7ITM wrote:
On May 11, 4:21*am, Richard Fry wrote:









On May 10, 9:35 pm, walt wrote:

If you are so confident that the number is 88.8889%,
please derive the conditions that yield that number.

à = R(load) - R(source) / R(load) + R(source)

SWR = (1 + |Ã|) / (1 - |Ã|)

Power Accepted by the Load = Incident Power * (1 - Ã^2)

For a 50 ohm source connected to a 100 ohm load:

à = (100 - 50) / (100 + 50) = 50 / 150 = 0.333333...

SWR = (1 + 0.333333) / (1 - 0.333333) = 1.333333 / 0.666666 = 2:1

Load Power = Incident Power * (1 - 0.333333^2) = Incident Power *
0.888889 (or 88.8889%)

The answers are the same for a 50 ohm source with a 25 ohm load.

RF

Richard F is, of course, correct. *You can just as well do it for DC:
2V O.C., 1 ohm source will deliver 1A, 1V, 1 watt to a 1 ohm load.
For a 2 ohm load, the current is 2/3 amp; the power is 8/9 watt. *For
an 0.5 ohm load, the current is 4/3 amp; the power is 8/9 watt. *It's
not terribly difficult to show, for the AC case, that the same is true
for any 2:1 load, regardless of phase angle--trivial, in fact, if you
accept that the SWR along a lossless TEM line is constant.

Cheers,
Tom


My apology, Wim, the equation I quoted in my last post is incorrect--I
mean't to say R = E/I, not E/R. Sorry.

Walt

On May 11, 7:15*am, Wimpie wrote:
On 10 mayo, 23:37, walt wrote:

[deleted]



Now Wim, your math is very challenging: You state 100w delivered by
the source, but at one point you also state 111w is delivered to the
100-load--at another point you state that 100w is delivered to the 100-
ohm load. Which is it?

You also state that voltage out of the source is 212.1v--sum ting wong
here. 212.1v across 50 ohms yields 899.73w, and 212.1v across 100 ohm
yields 449.86w. These power values are nowhere near the values
appearing in you statements.

In my calculations, with 100w the voltage across a 50-ohm line is
70.71v, and across a 100-ohm line the voltage is 100v.

With a 50-ohm line terminated in 100 ohms, the vrc is 0.3333 as you
stated, thus the power-reflection coefficient is 0.1111. This means,
as I've stated continually, that with 100w delivered by the source,
the power reflected at the 2:1 mismatch is 11.111w, which when added
to the 100w supplied by the source, makes the forward power 111.11w.
Now with 11.111w of power reflected at the mismatch, this leaves 100w
delivered to the 100-ohm load, not 111w. Please tell me where the 111w
came from. You also haven't told us why you calculate 130w forward and
15w reflected. Someone else may have made these calculations, but it
was you I asked for an explanation, because you repeated those
calculations.

When using the correct physics and math in this example, how can the
results *be so different?

To close, let me present the procedure I use to calculate the total
forward power--I guess it is related to Ohm's Law:

With the power-reflection coefficient as prc, the forward power PF = 1/
(1 - prc).

So Wim, can you clarify the confusion you appear to have made?

Walt

Hello Walt,

The source (that is the PA in this case) produces an EMF (as mentioned
before) of 212.1Vrms, Source impedance is 100 Ohms (so forget the
"conjugated match" thing for this case).

When terminated according to the numbers above the socket ("100W into
50 Ohms"), this results in 100W *( Vout = 212.1*50/(50+100) = 70.7V,
just a voltage divider consisting of 100 Ohms and 50 Ohms ).

70.7V into 50 Ohms makes 100W.

Now we remove the 50 Ohms load and create a mismatch (referenced to 50
Ohms), by connecting a 100 Ohms load (VSWR = 100/50 = 2)

Now the output voltage will be:

Vout = 212.1*100/(100 +100) = 106V

106V into 100 Ohms makes 112W (forgive me the one Watt difference).

This all without any transmission line theory.

Now the forward and reflected power balance (this is exactly the same
as in my previous posting, except for using 112W instead of 111W):

100 Ohms equals |rc| = 0.3333 (50 Ohms reference).

Hence "reflected power" = 0.333^2*"incident power" * = 0.111*"incident
power".

"net delivered power" = "incident power" – "reflected power" = 112W..

"net delivered power" = (1-0.11111)*"incident power" = 112W.

"incident power"= 126W, "reflected power" = 14 W

As far as I can see, there is nothing new with respect to the previous
calculation, except for the truncation/rounding.

Similar reasoning can be applied to my class-E PA. It is designed for
500W into 4.5 Ohm, but the output impedance is below 1 Ohms. If you
try to achieve conjugate match, you fry the active devices. *Same is
valid for most audio amplifiers, they may mention: *"80W into 8 Ohm",
but that does not mean that Zout = 8 Ohms.

Awaiting your comment,

Wim
PA3DJSwww.tetech.nl

Of course, with pretty much any decently designed power amplifier, no
matter the frequency, included protection circuits will limit the
dissipation and voltage in the critical areas (e.g., the output active
devices). Loads become disconnected or shorted sometimes; winds blow
stray wires or tree limbs across antennas. Through a piece of
transmission line, such a change can reflect back to become any
possible phase angle, and any of a very wide range of magnitudes, at
the amplifier's output.

Though it's a red herring typical of audio-speak, most modern high
fidelity audio amplifiers have a very low output impedance, a small
fraction of an ohm, so they can claim a high damping factor. Off
topic: the reason it's a red herring is that the impedance of the
speaker connected to the amplifier must be included to figure the
damping, and that impedance (even just the DC resistance of voice
coils) changes by considerably more than the amplifier's output
impedance (resistance) just because of heating on audio peaks.

Especially when reactive components (inductors and capacitors) couple
power between a source and a load, you can get stresses--voltages and/
or currents--well beyond what's safe when you try to operate the
source into a load it's not intended to handle. That's true even when
the net power delivered to the load is considerably LESS than the
rated output power of the source. Wim's example of the class-E
amplifier is true enough, but it's not necessary to ask the source to
deliver more net load power than it's rated to deliver, to establish
conditions that cause trouble. Thus, even sources that have an output
impedance at or very close to the rated load impedance will have
circuits to protect against loads that could destroy things inside the
amplifier.

Cheers,
Tom