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.