Reflection coefficient for total re-reflection
 

On Tue, 14 Jun 2011 11:26:41 -0700 (PDT), walt wrote:

Hi Walt,

The focus seems to have shifted away from the matching hardware
previously described to:
What I'm really concerned about is the reflection coefficient at the
output of the tank circuit of the RF power amp.

It includes a point argued by many, and which you make a condition of
success for your argument:
Being non-dissipative
it cannot absorb the reflected power, so it must re-reflect it.

Your argument now applied at two points, one not illustrated in your
references (the source):
Therefore, my position is that its reflection coefficient rho = 1.0,

The counter argument:
My critics say that a rho = 1.0 cannot be established when the virtual
short is caused by wave interference.
which they express NOT as a condition of the tank circuit, but of the
matching hardware.

However, if Best's Eq 8 is
valid,

You have acknowledge that experts in math whom you trust say it is
valid. This yields a paradox. By all the shifts in focus, it is
apparent that:
1. You are both right;
2. You are both wrong.
3. One of you is right, and one of you is wrong;
more likely given the drift of focus:
4. You are at cross purposes and arguing two different things.

Now that I've narrowed the problem down, what do you believe is the
answer?

You have not narrowed the problem down at all. It is composed of a
succession of premises that all have arguments for-and-against them.
This compounds the uncertainty. It also brings the hazard of any
single premise being rejected bringing down the entire goal you are
pursuing.

* * * * * * * * * * * * * * * * * * *

To put this bluntly, let's test the assertion repeated he
output of the tank circuit of the RF power amp.
Being non-dissipative
it cannot absorb the reflected power, so it must re-reflect it.

Odds are, you will lose support from at least half of the readership
for this statement alone - and it isn't even in your written work that
I've provided links for, and it certainly doesn't appear in Johnson.
It is not discussed by the experts in math either. It is a flyer, a
leap of faith whose validity is now emeshed with the outcome of
another problematic discussion.

Steve's opinion:
viz.
"A total re-reflection of power at the match
point is not necessary for the impedance
match to occur."
offers NOTHING to the matter of non-dissipation.

Odds are, Steve could easily lose support from at least half of the
readership for this statement alone. Unfortunately, that doesn't
leverage your position one iota.

There are too many irons in the fire.

73's
Richard Clark, KB7QHC

Reflection coefficient for total re-reflection
 

On Jun 14, 2:58*pm, Richard Clark wrote:
On Tue, 14 Jun 2011 11:26:41 -0700 (PDT), walt wrote:

Hi Walt,

The focus seems to have shifted away from the matching hardware
previously described to:

What I'm really concerned about is the reflection coefficient at the
output of the tank circuit of the RF power amp.

It includes a point argued by many, and which you make a condition of
success for your argument:

Being non-dissipative
it cannot absorb the reflected power, so it must re-reflect it.

Your argument now applied at two points, one not illustrated in your
references (the source):

Therefore, my position is that its reflection coefficient rho = 1.0,

The counter argument:My critics say that a rho = 1.0 cannot be established when the virtual
short is caused by wave interference.

which they express NOT as a condition of the tank circuit, but of the
matching hardware.

However, if *Best's Eq 8 is
valid,

You have acknowledge that experts in math whom you trust say it is
valid. *This yields a paradox. *By all the shifts in focus, it is
apparent that:
1. *You are both right;
2. *You are both wrong.
3. *One of you is right, and one of you is wrong;
more likely given the drift of focus:
4. *You are at cross purposes and arguing two different things.

Now that I've narrowed the problem down, what do you believe is the
answer?

You have not narrowed the problem down at all. *It is composed of a
succession of premises that all have arguments for-and-against them.
This compounds the uncertainty. *It also brings the hazard of any
single premise being rejected bringing down the entire goal you are
pursuing.

* * * * * * * * * * * * * * * * * * *

To put this bluntly, let's test the assertion repeated he

output of the tank circuit of the RF power amp.
Being non-dissipative
it cannot absorb the reflected power, so it must re-reflect it.

Odds are, you will lose support from at least half of the readership
for this statement alone - and it isn't even in your written work that
I've provided links for, and it certainly doesn't appear in Johnson.
It is not discussed by the experts in math either. *It is a flyer, a
leap of faith whose validity is now emeshed with the outcome of
another problematic discussion.

Steve's opinion:viz.
* *"A total re-reflection of power at the match
* *point is not necessary for the impedance
* *match to occur."

offers NOTHING to the matter of non-dissipation.

Odds are, Steve could easily lose support from at least half of the
readership for this statement alone. *Unfortunately, that doesn't
leverage your position one iota.

There are too many irons in the fire.

73's
Richard Clark, KB7QHC


Thank you for your insightful response, Richard. However, I must plod
on, and introduce the results of my own measurements, the data of
which appear in Chapter 19, Sec 19.14 of Reflections 3. Using a
Kenwood TS-830S I measured the parameters required to prove the output
source resistance of the tank circuit of its RF power amp is non-
dissipative. The measurements also prove that the reflected power
incident on its output is totally re-reflected. From reading an
earlier post of yours, Richard, I know you have read this portion of
Chapter 19, and you agreed with it. My measured data conflicts with
Best's Eq 8, so there's got to be a valid answer to this dilemma, but
where?

Walt

Reflection coefficient for total re-reflection
 

On Tue, 14 Jun 2011 12:25:33 -0700 (PDT), walt wrote:

Thank you for your insightful response, Richard. However, I must plod
on, and introduce the results of my own measurements, the data of
which appear in Chapter 19, Sec 19.14 of Reflections 3. Using a
Kenwood TS-830S I measured the parameters required to prove the output
source resistance of the tank circuit of its RF power amp is non-
dissipative. The measurements also prove that the reflected power
incident on its output is totally re-reflected. From reading an
earlier post of yours, Richard, I know you have read this portion of
Chapter 19, and you agreed with it. My measured data conflicts with
Best's Eq 8, so there's got to be a valid answer to this dilemma, but
where?

Walt

Hi Walt,

Yes, I am very familiar with your bench work. I am very annoyed by
those who trivialize it. I am further annoyed by those who patronize
you to then shift the parameters to explain something else -
completely abandoning you.

However, I also acknowledge that I can hurt your feelings about where
I stand on these matters. Yes, to a limited degree, I do agree with
much of your assessment, but I do not follow it into the arena of
where you express it as being "non-dissipative."

However, my denying this proposition does NOT mean I dispute your
other proposition of complete reflection. This cross connection of
several topics is where things get overly complex and gives the entire
discussion the appearance of a house of cards.

I cannot tell what your agenda is, but the material you quote as
source, and for which I have provided links to in this thread - all
seem a subtext to non-dissipative sources. Attention directed towards
this pursuit of matching points rendered in stubs seems like tea leaf
reading to resolve a larger issue.

As a student, tech, and engineer employing a LOT of precision
waveguide technology, I am quite comfortable with the components and
topologies you describe. Arguments for wave interferences being
rendered into useful circuit constructs (that is, constructed on the
basis of wave interference) has been with us easily since the early
40s. One component, the ATR tube, is one I have handled and replaced
to insure the proper operation of RF paths being steered in
waveguides. One proviso, however = for any of these
"virtualizations" to work, they require a physical (and dissipative)
element as the initiator. Waves do not interfere without the presence
of a physical element (often some form of detector, or minimally a
load).

73's
Richard Clark, KB7QHC

Reflection coefficient for total re-reflection
 

On Jun 14, 3:53*pm, Richard Clark wrote:
On Tue, 14 Jun 2011 12:25:33 -0700 (PDT), walt wrote:
Thank you for your insightful response, Richard. However, I must plod
on, and introduce the results of my own measurements, the data of
which appear in Chapter 19, Sec 19.14 of Reflections 3. Using a
Kenwood TS-830S I measured the parameters required to prove the output
source resistance of the tank circuit of its RF power amp is non-
dissipative. The measurements also prove that the reflected power
incident on its output is totally re-reflected. From reading an
earlier post of yours, Richard, I know you have read this portion of
Chapter 19, and you agreed with it. My measured data conflicts with
Best's Eq 8, so there's got to be a valid answer to this dilemma, but
where?

Walt

Hi Walt,

Yes, I am very familiar with your bench work. *I am very annoyed by
those who trivialize it. *I am further annoyed by those who patronize
you to then shift the parameters to explain something else -
completely abandoning you.

However, I also acknowledge that I can hurt your feelings about where
I stand on these matters. *Yes, to a limited degree, I do agree with
much of your assessment, but I do not follow it into the arena of
where you express it as being "non-dissipative." *

However, my denying this proposition does NOT mean I dispute your
other proposition of complete reflection. *This cross connection of
several topics is where things get overly complex and gives the entire
discussion the appearance of a house of cards.

I cannot tell what your agenda is, but the material you quote as
source, and for which I have provided links to in this thread - all
seem a subtext to non-dissipative sources. *Attention directed towards
this pursuit of matching points rendered in stubs seems like tea leaf
reading to resolve a larger issue.

As a student, tech, and engineer employing a LOT of precision
waveguide technology, I am quite comfortable with the components and
topologies you describe. *Arguments for wave interferences being
rendered into useful circuit constructs (that is, constructed on the
basis of wave interference) has been with us easily since the early
40s. *One component, the ATR tube, is one I have handled and replaced
to insure the proper operation of RF paths being steered in
waveguides. *One proviso, however = for any of these
"virtualizations" to work, they require a physical (and dissipative)
element as the initiator. *Waves do not interfere without the presence
of a physical element (often some form of detector, or minimally a
load).

73's
Richard Clark, KB7QHC

Thanks again Richard, for your insightful response. And goodness no,
Richard, you won't hurt my feelings. I like being correct, but if I'm
not I surely want to be told about it. This situation is no different.

However, if you're not following me on the 'non-dissipative' path, I'd
like you to review the last portion of the TS-830S experiment, and
follow the numbers. If you don't have that material in front of you,
you can find it again on my web page at www.w2du.com. Click on
'Preview Chapters from Reflections 3', and then click on Chapter 19A,
which is a part of Chapter 19 in the 3rd edition.

Observe that when the load impedance is changed from 50+j0 to the
complex impedance 17.98 + j8.77 ohms, the plate current rose
expectedly from 260ma to 290ma. This change occurred because the amp
is now mismatched, and the pi-network is also detuned from resonance.
The unwashed would conclude that the reflected power caused the
increase in plate current. However, one will also observe that after
the pi-network has been retuned and adjusted to again deliver all the
available power into the otherwise 'mismatched' load (which is now
matched to the source), the plate current went back to it's original
value, 260ma. In addition, the amp returned to deliver 100w into the
complex impedance at the line input, with the 30.6w of reflected power
in the line, and adding to the 100w of source power, making 130.6w
incident on the mismatched load, absorbing 100w and reflecting 30.6w.

IMHO, these data prove beyond doubt that the pi-network not only
didn't absorb any of the reflected power, but totally re-reflected it.
In my book this says the pi-network reflection coefficient rho = 1.0.
How can anyone disagree with this?

Walt

Reflection coefficient for total re-reflection
 

On Tue, 14 Jun 2011 14:18:53 -0700 (PDT), walt wrote:

Thanks again Richard, for your insightful response. And goodness no,
Richard, you won't hurt my feelings. I like being correct, but if I'm
not I surely want to be told about it. This situation is no different.

However, if you're not following me on the 'non-dissipative' path, I'd
like you to review the last portion of the TS-830S experiment, and
follow the numbers. If you don't have that material in front of you,
you can find it again on my web page at www.w2du.com. Click on
'Preview Chapters from Reflections 3', and then click on Chapter 19A,
which is a part of Chapter 19 in the 3rd edition.

Observe that when the load impedance is changed from 50+j0 to the
complex impedance 17.98 + j8.77 ohms, the plate current rose
expectedly from 260ma to 290ma. This change occurred because the amp
is now mismatched, and the pi-network is also detuned from resonance.
The unwashed would conclude that the reflected power caused the
increase in plate current. However, one will also observe that after
the pi-network has been retuned and adjusted to again deliver all the
available power into the otherwise 'mismatched' load (which is now
matched to the source), the plate current went back to it's original
value, 260ma. In addition, the amp returned to deliver 100w into the
complex impedance at the line input, with the 30.6w of reflected power
in the line, and adding to the 100w of source power, making 130.6w
incident on the mismatched load, absorbing 100w and reflecting 30.6w.

Hi Walt,

There is absolutely NOTHING that I can dispute in your numbers or
method. However, as to "dissipation" this says nothing.

As for what it says about the reflection coefficient, I would agree
with you. However, the math and the math experts you speak of - they
are not answering the model we are examining above. You distracted
them with stubs and reflecting waves.

If you cast this agreement in the reflection coefficient back into
effects purported to exist in the matching with stubs model, we may
yet argue.

IMHO, these data prove beyond doubt that the pi-network not only
didn't absorb any of the reflected power, but totally re-reflected it.
In my book this says the pi-network reflection coefficient rho = 1.0.
How can anyone disagree with this?

*** Nothing further useful about the topic is to be found below ***

There is a curious side bar to this found at:
http://www.w5big.com/purchase4170c.htm
Observe the second data screen that purports to examine a mismatch
through a length of both RG58 and RG59 terminated with 100 Ohms.

Of particular note is the distinct transition from the Zc of 50 Ohms
to the Zc of 75 Ohms and the similarly distinct transition from the Zc
of 75 Ohms to the termination R of 100 Ohms. Note that the
transitions both span a distance of 2 feet. 2 feet is not
insubstantial compared to connection technology that spans, probably,
no more than one tenth that distance. Whence the extra 11 inches on
both sides of the connection plane of either connection?

In the physics I've been studying for the past 10 years, near fields,
this would not be unusual. In fact it would be expected. There is a
transition zone (in waveguide design, there would be a taper or a
sweep section to anticipate this and they would be physically large in
terms of wavelength) not a transition cliff.

Similarly, at the connector to your TS-830S, there is a zone that
surrounds it that only approximates the 50 Ohm which occurs (in the
terms of this link's demonstration) some distance away, deep inside
your TS-830S.

To me, dissipation inhabits this zone (certainly large enough for the
tube and tank to occupy) and embraces the match with a complex
addition of phases that could result in loss or even gain. This is
also to say that I do not subscribe to dissipation being all about
loss - especially when the hand of man is on the tuning knob instead
of letting the chips fall where they may.

73's
Richard Clark, KB7QHC

Reflection coefficient for total re-reflection
 

On 06/12/2011 03:05 PM, walt wrote:


Assume a 50-ohm line terminated in a purely resistive 150-ohm load,
yielding a 3:1 mismatch with a voltage reflection coefficient Vñ equal
to 0.5 at 0°. We want to place a stub on the line at the position
relating to the unit-resistance circle on the Smith Chart. For a 3:1
mismatch this position is exactly 30° rearward from the load, and has
a voltage reflection coefficient Vñ equal to 0.5 at -60°. The
normalized line resistance at this point is 1.1547 chart ohms, or
57.735 ohms on the 50-ohm line, for a line impedance of 50 - j57.735
ohms.


Walt, I can only get your 1.0 - j1.1547 normalized impedance value
(looking into the line towards the normalized load of 3.0) if I move 60
degrees, not 30, from the load towards the generator. I'm also assuming
a lossless line. Sincerely, and 73s from N4GGO,


--
J. B. Wood e-mail:

Reflection coefficient for total re-reflection
 

On Jun 15, 8:22*am, "J.B. Wood" wrote:
On 06/12/2011 03:05 PM, walt wrote:



Assume a 50-ohm line terminated in a purely resistive 150-ohm load,
yielding a 3:1 mismatch with a voltage reflection coefficient Vñ equal
to 0.5 at 0°. We want to place a stub on the line at the position
relating to the unit-resistance circle on the Smith Chart. For a 3:1
mismatch this position is exactly 30° rearward from the load, and has
a voltage reflection coefficient Vñ equal to 0.5 at -60°. The
normalized line resistance at this point is 1.1547 chart ohms, or
57.735 ohms on the 50-ohm line, for a line impedance of 50 - j57.735
ohms.

Walt, *I can only get your 1.0 - j1.1547 normalized impedance value
(looking into the line towards the normalized load of 3.0) if I move 60
degrees, not 30, from the load towards the generator. *I'm also assuming
a lossless line. *Sincerely, and 73s from N4GGO,

--
J. B. Wood * * * * * * * * *e-mail:

Hello JB, thank you for the response. I'm sure I understand what's
going on. When you view the Smith Chart at the unity resistance circle
with a 3:1 mismatch the normalized impedance there is 1.0 - j1.1547,
as you stated. However, a radius through that point yields a voltage
reflection coefficient rho = 0.5 @ -60°. In other words, the radial
line intersects the periphery of the Smith Chart at -60°.

As I'm sure you know, reflection degrees equals two electrical
degrees. Therefore, the normalized impedance 1.0 -j1.1547 ohms occurs
at 30° rearward of the load. That OK?

In the absence of a Smith Chart here's an easy way to determine the
reactance appearing at the unity resistance circle for any given
degree of mismatch:

1/(sqrt SWR/
SWR-1)

Walt, W2DU

Reflection coefficient for total re-reflection
 

On Jun 15, 10:57*am, walt wrote:
On Jun 15, 8:22*am, "J.B. Wood" wrote:









On 06/12/2011 03:05 PM, walt wrote:

Assume a 50-ohm line terminated in a purely resistive 150-ohm load,
yielding a 3:1 mismatch with a voltage reflection coefficient Vñ equal
to 0.5 at 0°. We want to place a stub on the line at the position
relating to the unit-resistance circle on the Smith Chart. For a 3:1
mismatch this position is exactly 30° rearward from the load, and has
a voltage reflection coefficient Vñ equal to 0.5 at -60°. The
normalized line resistance at this point is 1.1547 chart ohms, or
57.735 ohms on the 50-ohm line, for a line impedance of 50 - j57.735
ohms.

Walt, *I can only get your 1.0 - j1.1547 normalized impedance value
(looking into the line towards the normalized load of 3.0) if I move 60
degrees, not 30, from the load towards the generator. *I'm also assuming
a lossless line. *Sincerely, and 73s from N4GGO,

--
J. B. Wood * * * * * * * * *e-mail:

Hello JB, thank you for the response. I'm sure I understand what's
going on. When you view the Smith Chart at the unity resistance circle
with a 3:1 mismatch the normalized impedance there is 1.0 - j1.1547,
as you stated. However, a radius through that point yields a voltage
reflection coefficient rho = 0.5 @ -60°. In other words, the radial
line intersects the periphery of the Smith Chart at -60°.

As I'm sure you know, reflection degrees equals two electrical
degrees. Therefore, the normalized impedance 1.0 -j1.1547 ohms occurs
at 30° rearward of the load. That *OK?

In the absence of a Smith Chart here's an easy way to determine the
reactance appearing at the unity resistance circle for any given
degree of mismatch:

* * * * * * * * * * * * * * * * * * * * * * * * * * * * *1/(sqrt SWR/
SWR-1)

Walt, W2DU


Fixing the equation: 1/(sqrt SWR/SWR-1)

Walt

Reflection coefficient for total re-reflection
 

On 06/15/2011 10:57 AM, walt wrote:

As I'm sure you know, reflection degrees equals two electrical
degrees. Therefore, the normalized impedance 1.0 -j1.1547 ohms occurs
at 30° rearward of the load. That OK?


Well, I'm not clear what you mean by "rearward of the load". At 60
degrees toward the generator the normalized impedance looking into the
line is 1.0 - j1.1547 ohms and that corresponds to a reflection
coefficient of -j1.1547/(2-j1.1547) or alternatively, 0.5 @ -60 degrees
(sorry my char set doesn't do Steinmetz notation). So I think we're in
agreement here. What more needs to be said? Now put in a shorted stub
(of appropriate length at the desired frequency) in series to tune out
(cancel) the reactive part of the line impedance. The combination of
the series stub and the line will appear as 1.0 + j0 to the
generator/source and the reflection coefficient is now 0. How could it
be anything else (at the desired frequency) in this scenario? You don't
need a discussion of travelling waves to arrive at the correct answer.
The reflection coefficient corresponding to a normalized (say to 50
ohms) impedance zx = Zx/50 is rho = (zx - 1)/(zx +1). Clearly a values
of rho = 1 + j0 or -1 + j0 correspond to a zx values of infinity (open
circuit) or zero (short circuit), respectively.


You might want to think of the shorted line stub as inside one box and
the transmission line + 150 ohm load in another black box. Let's assume
someone hands you these two boxes and you have no idea what's inside but
you do have access to set of terminals (input port) on each box from
which you can measure steady-state impedance of each. At some frequency
you find the normalized input impedance on one box is +j1.1547 ohms and
on the other box at that same frequency it is 1.0 - j1.1547 ohms. Now
you connect the boxes is series and measure the impedance of the series
combination again at that frequency. The point I'm making here is that
at the terminals of either of the boxes you measured the impedance at
some frequency - it doesn't matter what's inside the box. (Of course
the impedance behavior over a range of frequencies certainly is
dependent on the circuit topology within the box.) Sincerely,

--
J. B. Wood e-mail:

Reflection coefficient for total re-reflection
 

On Jun 15, 2:55*pm, "J.B. Wood" wrote:
On 06/15/2011 10:57 AM, walt wrote:

As I'm sure you know, reflection degrees equals two electrical
degrees. Therefore, the normalized impedance 1.0 -j1.1547 ohms occurs
at 30 rearward of the load. That *OK?

Well, I'm not clear what you mean by "rearward of the load". *At 60
degrees toward the generator the normalized impedance looking into the
line is 1.0 - j1.1547 ohms and that corresponds to a reflection
coefficient of -j1.1547/(2-j1.1547) or alternatively, 0.5 @ -60 degrees
(sorry my char set doesn't do Steinmetz notation). *So I think we're in
agreement here. *What more needs to be said? *Now put in a shorted stub
(of appropriate length at the desired frequency) in series to tune out
(cancel) the reactive part of the line impedance. *The combination of
the series stub and the line will appear as 1.0 + j0 to the
generator/source and the reflection coefficient is now 0. *How could it
be anything else (at the desired frequency) in this scenario? *You don't
need a discussion of travelling waves to arrive at the correct answer.
The reflection coefficient corresponding to a normalized (say to 50
ohms) impedance zx = Zx/50 is rho = (zx - 1)/(zx +1). *Clearly a values
of rho = 1 + j0 or -1 + j0 correspond to a zx values of infinity (open
circuit) or zero (short circuit), respectively.

You might want to think of the shorted line stub as inside one box and
the transmission line + 150 ohm load in another black box. *Let's assume
someone hands you these two boxes and you have no idea what's inside but
you do have access to set of terminals (input port) on each box from
which you can measure steady-state impedance of each. *At some frequency
you find the normalized input impedance on one box is +j1.1547 ohms and
on the other box at that same frequency it is 1.0 - j1.1547 ohms. *Now
you connect the boxes is series and measure the impedance of the series
combination again at that frequency. *The point I'm making here is that
at the terminals of either of the boxes you measured the impedance at
some frequency - it doesn't matter what's inside the box. *(Of course
the impedance behavior over a range of frequencies certainly is
dependent on the circuit topology within the box.) *Sincerely,

--
J. B. Wood * * * * * * * * *e-mail:

Thanks again for the response, JB. What I mean by rearward of the load
is simply going toward the source from the load. But are we in
agreement? We probably are as long as we agree on terminology
regarding the Smith Chart. The unity resistance circle for a 3:1
mismatch yields a normalized impedance of 1 - j1.1547 as you know. As
I said in my previous post, a radial line passing through this point
on the unity-resistance circle joins the periphery of the Chart at
-60°. It important to note that this -60° is reflection degrees, which
is two-times greater than degrees along the transmission line.
Therefore, the normalized line impedance of 1 -j1.1547 appears at 30°
from the load toward the source on the xmsn line, not at -60°. Hope
we're in agreement on this.

Walt