I have a question for W5DXP, KB7QHC, W7ITM and K0TAR relating to
transmission lines.This is not a trick question. This is a straight-
forward question for which I 'm dead serious. The answer will involve
a resulting reflection coefficient.

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.

Placing a short-circuited stub having an inductive reactance of
+j57.735 ohms on the line cancels the line reactance, establishing a
new line impedance at this matching point of 50 + j0. However, a lot
more is involved than just canceling the reactances to establish the
impedance match. To begin, the stub generates a new, secondary,
canceling reflection that cancels the primary reflected wave generated
by the mismatched line termination.

Recall the voltage reflection coefficient Vñ of the line impedance
prior to placing the stub--Vñ = 0.5 at -60°. The corresponding current
reflection coefficient Iñ = 0.5 at +120°

We now look at the voltage reflection coefficient of the stub, which
is Vñ = 0.5 at +60°, and the corresponding current reflection
coefficient of the stub which is Iñ = 0.5 at -120°.

Observe that resultant angle of the voltage coefficients is 0° and
resultant angle of the current coefficients is 180°, all with the same
magnitude--0.5. It is well known that when these respective
coefficients exist at this point, the matching point, this condition
establishes a virtual open circuit to rearward traveling waves at this
point, prohibiting any further rearward travel of the reflected waves,
thus reversing their direction toward the load.

Consequently, the result is total re-reflection of the two sets of
reflected waves at the match point.

Now the question--what is the resultant voltage coefficient of
reflection at the match point? My position is that ñ = 1.0, which
indicates total reflection. However, I've been criticized on this
value, my critics saying that because this virtual open circuit was
established by wave interference, ñ cannot equal 1.0. They assert
that ñ = 1.0 can be obtained only through a physical open circuit.

So I now ask you, am I correct in saying that the reflection
coefficient in this situation is 1.0? I'm holding my breath for your
answers.

Walt

On Jun 12, 3:05*pm, walt wrote:
I have a question for W5DXP, KB7QHC, W7ITM and K0TAR relating to
transmission lines.This is not a trick question. *This is a straight-
forward question for which I 'm dead serious. *The answer will involve
a resulting reflection coefficient.

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.

Placing a short-circuited stub having an inductive reactance of
+j57.735 ohms on the line cancels the line reactance, establishing a
new line impedance at this matching point of 50 + j0. However, a lot
more is involved than just canceling the reactances to establish the
impedance match. To begin, the stub generates a new, secondary,
canceling reflection that cancels the primary reflected wave generated
by the mismatched line termination.

Recall the voltage reflection coefficient Vñ of the line impedance
prior to placing the stub--Vñ = 0.5 at -60°. The corresponding current
reflection coefficient Iñ = 0.5 at +120°

We now look at the voltage reflection coefficient of the stub, which
is Vñ = 0.5 at +60°, and the corresponding current reflection
coefficient of the stub which is Iñ = 0.5 at -120°.

Observe that resultant angle of the voltage coefficients is 0° and
resultant angle of the current coefficients is 180°, all with the same
magnitude--0.5. It is well known that when these respective
coefficients exist at this point, the matching point, this condition
establishes a virtual open circuit to rearward traveling waves at this
point, prohibiting any further rearward travel of the reflected waves,
thus reversing their direction toward the load.

Consequently, the result is total re-reflection of the two sets of
reflected waves at the match point.

Now the question--what is the resultant voltage coefficient of
reflection at the match point? My position is that ñ = 1.0, which
indicates total reflection. However, I've been criticized on this
value, my critics saying that because this virtual open circuit was
established by wave interference, ñ cannot equal 1.0. They assert
that *ñ = 1.0 can be obtained only through a physical open circuit.

So I now ask you, am I correct in saying that the reflection
coefficient in this situation is 1.0? *I'm holding my breath for your
answers.

Walt

Unfortunately, the symbol for rho came thru as 'n' with a diacritical
mark over it. That incorrect character should be replace with the
Greek symbol rho, which this venue cannot print.

Walt

On Jun 12, 3:05*pm, walt wrote:
I have a question for W5DXP, KB7QHC, W7ITM and K0TAR relating to
transmission lines.This is not a trick question. *This is a straight-
forward question for which I 'm dead serious. *The answer will involve
a resulting reflection coefficient.

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.

Placing a short-circuited stub having an inductive reactance of
+j57.735 ohms on the line cancels the line reactance, establishing a
new line impedance at this matching point of 50 + j0. However, a lot
more is involved than just canceling the reactances to establish the
impedance match. To begin, the stub generates a new, secondary,
canceling reflection that cancels the primary reflected wave generated
by the mismatched line termination.

Recall the voltage reflection coefficient Vñ of the line impedance
prior to placing the stub--Vñ = 0.5 at -60°. The corresponding current
reflection coefficient Iñ = 0.5 at +120°

We now look at the voltage reflection coefficient of the stub, which
is Vñ = 0.5 at +60°, and the corresponding current reflection
coefficient of the stub which is Iñ = 0.5 at -120°.

Observe that resultant angle of the voltage coefficients is 0° and
resultant angle of the current coefficients is 180°, all with the same
magnitude--0.5. It is well known that when these respective
coefficients exist at this point, the matching point, this condition
establishes a virtual open circuit to rearward traveling waves at this
point, prohibiting any further rearward travel of the reflected waves,
thus reversing their direction toward the load.

Consequently, the result is total re-reflection of the two sets of
reflected waves at the match point.

Now the question--what is the resultant voltage coefficient of
reflection at the match point? My position is that ñ = 1.0, which
indicates total reflection. However, I've been criticized on this
value, my critics saying that because this virtual open circuit was
established by wave interference, ñ cannot equal 1.0. They assert
that *ñ = 1.0 can be obtained only through a physical open circuit.

So I now ask you, am I correct in saying that the reflection
coefficient in this situation is 1.0? *I'm holding my breath for your
answers.

Walt

Forgot to say, the stub is placed in series with the line, not in
shunt.

Walt

On Jun 12, 3:12*pm, walt wrote:
On Jun 12, 3:05*pm, walt wrote:









I have a question for W5DXP, KB7QHC, W7ITM and K0TAR relating to
transmission lines.This is not a trick question. *This is a straight-
forward question for which I 'm dead serious. *The answer will involve
a resulting reflection coefficient.

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.

Placing a short-circuited stub having an inductive reactance of
+j57.735 ohms on the line cancels the line reactance, establishing a
new line impedance at this matching point of 50 + j0. However, a lot
more is involved than just canceling the reactances to establish the
impedance match. To begin, the stub generates a new, secondary,
canceling reflection that cancels the primary reflected wave generated
by the mismatched line termination.

Recall the voltage reflection coefficient Vñ of the line impedance
prior to placing the stub--Vñ = 0.5 at -60°. The corresponding current
reflection coefficient Iñ = 0.5 at +120°

We now look at the voltage reflection coefficient of the stub, which
is Vñ = 0.5 at +60°, and the corresponding current reflection
coefficient of the stub which is Iñ = 0.5 at -120°.

Observe that resultant angle of the voltage coefficients is 0° and
resultant angle of the current coefficients is 180°, all with the same
magnitude--0.5. It is well known that when these respective
coefficients exist at this point, the matching point, this condition
establishes a virtual open circuit to rearward traveling waves at this
point, prohibiting any further rearward travel of the reflected waves,
thus reversing their direction toward the load.

Consequently, the result is total re-reflection of the two sets of
reflected waves at the match point.

Now the question--what is the resultant voltage coefficient of
reflection at the match point? My position is that ñ = 1.0, which
indicates total reflection. However, I've been criticized on this
value, my critics saying that because this virtual open circuit was
established by wave interference, ñ cannot equal 1.0. They assert
that *ñ = 1.0 can be obtained only through a physical open circuit.

On Sun, 12 Jun 2011 12:12:46 -0700 (PDT), walt wrote:

On Jun 12, 3:05*pm, walt wrote:
I have a question for W5DXP, KB7QHC, W7ITM and K0TAR relating to
transmission lines.

Hi Walt,

What I need to see is a labeled schematic.

73's
Richard Clark, KB7QHC

On Jun 12, 2:05*pm, walt wrote:
So I now ask you, am I correct in saying that the reflection
coefficient in this situation is 1.0? *I'm holding my breath for your
answers.

Walt, some of us have previously discussed the difference between a
one-port s11 (reflection coefficient) and a two-port s11 and I think
that difference is what you are seeing. The one-port s11 (rho) is a
virtual reflection coefficient based on the available system
information. In that case, rho is 1.0. However, if we use the two-port
parameters, I believe s11 (rho) will be 0.5. The matching stub
complicates the analysis so maybe we can reduce the complexity using a
slightly different example about which you can ask the same question
and the answer will be a little easier.

--50 ohm--+--1/2WL 150 ohm--50 ohm load

It's essentially the same problem without the complexity of the stub.
There is a 50 ohm Z0-match at point '+'. The one-port s11 (rho)
looking into the Z0-match point from the source side is 0.0 because
there are no reflections. The one-port analysis is blind to any
interference that might be recognized if we were using the two-port
analysis.

The one-port s11 (rho) looking back into the Z0-match from the load is
1.0, same as your virtual open/short. The one-port analysis used for
your virtual open/short concept is unaware of the interference that is
visible in a two-port analysis. A one-port analysis cannot tell the
difference between an actual reflection and a redistribution of energy
associated with superposition interference.

However, the two-port s11 (rho) looking into the Z0-match from the
source side is 0.5, i.e. (150-50)/(150+50). Interference is not
invisible in the two-port analysis. In fact, when the voltage
reflected toward the source is zero, there is total destructive
interference in the direction of the source.

b1 = s11*a1 + s12*a2 = 0

where all math is phasor math. The two terms cancel destructively to
zero. The ExH energy components in those two terms are redistributed
back toward the load.

In a two-port analysis, the reflection coefficient looking back into
the Z0-match from the load is called s22. The matching equation is:

b2 = s21*a1 + s22*a2

All of the destructive interference energy involved toward the source
is redistributed back toward the load as constructive interference.
But such is invisible in a one-port analysis such as your virtual open/
short circuit.

There is nothing wrong with your one-port virtual open/short circuit
analysis. There is also nothing wrong with the more in-depth two-port
analysis. It is simply that more information is available during the
two-port analysis.

Over on another newsgroup, some reported using a one-port analysis
almost all of the time. Tell your critics that you are using a one-
port analysis where the interference information is simply not
available during the analysis.

What is the one-port reflection coefficient looking into a Z0-match
from the source side? 0.0
What is the one-port reflection coefficient looking into a Z0-match
from the load side? plus or minus 1.0

That is what you are doing with your virtual open/short and there's
nothing wrong with a one-port analysis.
--
73, Cecil, w5dxp.com

On Jun 12, 5:46*pm, Cecil Moore wrote:
On Jun 12, 2:05*pm, walt wrote:

So I now ask you, am I correct in saying that the reflection
coefficient in this situation is 1.0? *I'm holding my breath for your
answers.

Walt, some of us have previously discussed the difference between a
one-port s11 (reflection coefficient) and a two-port s11 and I think
that difference is what you are seeing. The one-port s11 (rho) is a
virtual reflection coefficient based on the available system
information. In that case, rho is 1.0. However, if we use the two-port
parameters, I believe s11 (rho) will be 0.5. The matching stub
complicates the analysis so maybe we can reduce the complexity using a
slightly different example about which you can ask the same question
and the answer will be a little easier.

--50 ohm--+--1/2WL 150 ohm--50 ohm load

It's essentially the same problem without the complexity of the stub.
There is a 50 ohm Z0-match at point '+'. The one-port s11 (rho)
looking into the Z0-match point from the source side is 0.0 because
there are no reflections. The one-port analysis is blind to any
interference that might be recognized if we were using the two-port
analysis.

The one-port s11 (rho) looking back into the Z0-match from the load is
1.0, same as your virtual open/short. The one-port analysis used for
your virtual open/short concept is unaware of the interference that is
visible in a two-port analysis. A one-port analysis cannot tell the
difference between an actual reflection and a redistribution of energy
associated with superposition interference.

However, the two-port s11 (rho) looking into the Z0-match from the
source side is 0.5, i.e. (150-50)/(150+50). Interference is not
invisible in the two-port analysis. In fact, when the voltage
reflected toward the source is zero, there is total destructive
interference in the direction of the source.

b1 = s11*a1 + s12*a2 = 0

where all math is phasor math. The two terms cancel destructively to
zero. The ExH energy components in those two terms are redistributed
back toward the load.

In a two-port analysis, the reflection coefficient looking back into
the Z0-match from the load is called s22. The matching equation is:

b2 = s21*a1 + s22*a2

All of the destructive interference energy involved toward the source
is redistributed back toward the load as constructive interference.
But such is invisible in a one-port analysis such as your virtual open/
short circuit.

There is nothing wrong with your one-port virtual open/short circuit
analysis. There is also nothing wrong with the more in-depth two-port
analysis. It is simply that more information is available during the
two-port analysis.

Over on another newsgroup, some reported using a one-port analysis
almost all of the time. Tell your critics that you are using a one-
port analysis where the interference information is simply not
available during the analysis.

What is the one-port reflection coefficient looking into a Z0-match
from the source side? 0.0
What is the one-port reflection coefficient looking into a Z0-match
from the load side? plus or minus 1.0

That is what you are doing with your virtual open/short and there's
nothing wrong with a one-port analysis.
--
73, Cecil, w5dxp.com


Hi Cecil and Richard,

Thank you for the insightful response, Cecil. However, when you go to
the 2-port version I’m unable to correlate that configuration with my
stubbing problem.

Here’s what my problem really involves. Referring to Eq 8 in the first
part of Steve Best’s three-part article published in QEX, the equation
is as follows: The section within the parentheses is the voltage-
increase factor when all reflections have been re-reflected. That
section is 1/1 - (rho_’a’ x rho_’s’), where ‘a’ is the mismatch at the
antenna (the load) and ‘s’ is the mismatch at the source.

We’re considering the source to be an RF power amp, where we know the
output source resistance is non-dissipative, thus re-reflects all
reflected power incident on it. I maintain that the reflection
coefficient at the source is 1.0 because of the total re-reflection
there.

Now consider working with the output of the pi-network of the RF anp,
where I believe the reflection coefficient is 1.0, and the 3:1 antenna
mismatch that yields a reflection coefficient of 0.5. Plugging these
coefficients into the equation section yields 1/1- (1 x 0.5), which
equals 2.0, which is incorrect, making that equation invalid, IMHO.
The correct value with these coefficients should be 1.1547, not 2.0.

However, mathematical experts say that the equation is correct, saying
that rho_¬’s’ cannot be equal to 1.0, because the virtual open circuit
was established by wave interference, not a physical open circuit.

On Jun 12, 8:16*pm, walt wrote:
Thank you for the insightful response, Cecil. However, when you go to
the 2-port version I’m unable to correlate that configuration with my
stubbing problem.

The reason that I didn't say anything about the stub example is
because I cannot comprehend it without a schematic. That's why I
changed examples. Do you agree with what I said about my example?
Could you post a schematic of your first example? It is the "series
stub" part that I don't understand. Such is usually called a "series
section" because a stub is usually a parallel dead end open or short
circuit. It is also difficult to comprehend how a two-port analysis
could be done at the stub connection point. Wouldn't that require a
three-port analysis?

We’re considering the source to be an RF power amp, where we know the
output source resistance is non-dissipative, thus re-reflects all
reflected power incident on it. I maintain that the reflection
coefficient at the source is 1.0 because of the total re-reflection
there.

How's about we limit the *initial* discussion and examples to a source
with zero incident reflected power so the source impedance doesn't
matter? IMO, a two-port analysis of a Z0-match point will reveal the
main ingredients of the energy flow.

However, mathematical experts say that the equation is correct, saying
that rho_¬’s’ cannot be equal to 1.0, because the virtual open circuit
was established by wave interference, not a physical open circuit.

A one-port analysis cannot tell the difference between wave
interference and reflections. You are correct that the reflection
coefficients are not necessarily the same between a one-port analysis
and a two-port analysis. Your "mathematical experts" don't seem to
understand the limitations of a one-port analysis. It's akin to not
knowing what is inside a black box, i.e. one cannot tell the
difference between a resistor and a virtual resistance. However, with
a two-port analysis, one can tell the difference. It appears that your
"mathematical experts" are insisting on a two-port analysis such as
provided by the s-parameter equations:

b1 = s11*a1 + s12*a2

b2 = s21*a1 + s22*a2

http://www.sss-mag.com/pdf/an-95-1.pdf
--
73, Cecil, w5dxp.com

On Jun 12, 9:45*pm, Cecil Moore wrote:
On Jun 12, 8:16*pm, walt wrote:

Thank you for the insightful response, Cecil. However, when you go to
the 2-port version I’m unable to correlate that configuration with my
stubbing problem.

The reason that I didn't say anything about the stub example is
because I cannot comprehend it without a schematic. That's why I
changed examples. Do you agree with what I said about my example?
Could you post a schematic of your first example? It is the "series
stub" part that I don't understand. Such is usually called a "series
section" because a stub is usually a parallel dead end open or short
circuit. It is also difficult to comprehend how a two-port analysis
could be done at the stub connection point. Wouldn't that require a
three-port analysis?

We’re considering the source to be an RF power amp, where we know the
output source resistance is non-dissipative, thus re-reflects all
reflected power incident on it. I maintain that the reflection
coefficient at the source is 1.0 because of the total re-reflection
there.

How's about we limit the *initial* discussion and examples to a source
with zero incident reflected power so the source impedance doesn't
matter? IMO, a two-port analysis of a Z0-match point will reveal the
main ingredients of the energy flow.

However, mathematical experts say that the equation is correct, saying
that rho_¬’s’ cannot be equal to 1.0, because the virtual open circuit
was established by wave interference, not a physical open circuit.

A one-port analysis cannot tell the difference between wave
interference and reflections. You are correct that the reflection
coefficients are not necessarily the same between a one-port analysis
and a two-port analysis. Your "mathematical experts" don't seem to
understand the limitations of a one-port analysis. It's akin to not
knowing what is inside a black box, i.e. one cannot tell the
difference between a resistor and a virtual resistance. However, with
a two-port analysis, one can tell the difference. It appears that your
"mathematical experts" are insisting on a two-port analysis such as
provided by the s-parameter equations:

b1 = s11*a1 + s12*a2

b2 = s21*a1 + s22*a2

http://www.sss-mag.com/pdf/an-95-1.pdf
--
73, Cecil, w5dxp.com


Cecil, the series stubbing appears in Reflections, Chapter 23, with
the same values as I presented above, with detailed diagrams shown in
each step in the progression of the explanation. I hope these diagrams
can help.

As I said in the previous post, the experts were referring to the
output of the RF amp as not establishing a reflection coefficient rho
= 1.0, which has put me in a corner.

Sorry Cecil, I can't correlate a two-port configuration using S
parameters with the problem I have. Thanks a million for the
discourse.

Walt

On Sun, 12 Jun 2011 19:09:29 -0700 (PDT), walt wrote:

the series stubbing appears in Reflections, Chapter 23, with
the same values as I presented above, with detailed diagrams shown in
each step in the progression of the explanation. I hope these diagrams
can help.

In other words, consult:
http://www.w2du.com/Chapter%2023.pdf
Figures 1 through 5

As I said in the previous post, the experts were referring to the
output of the RF amp as not establishing a reflection coefficient rho
= 1.0, which has put me in a corner.

Hi Walt,

How so? (What is the corner?)

73's
Richard Clark, KB7QHC