On 6/24/2011 1:24 PM, dave wrote:
On Jun 24, 1:52 pm, Cecil wrote:
On Jun 23, 4:41 pm, wrote:

but what is your second source? you can always represent the second
source in that case in terms of the transmitter output so the second
input can be eliminated giving you a single port model.

a1 is the normalized forward voltage on the 50 ohm feedline from the
source. a2 is the normalized reflected voltage on the 291.4 ohm
feedline from the load. Those are the two sources associated with the
impedance discontinuity inside the black box. a2 could just as easily
be from a second generator instead of a reflection.

When the single-port model is used, if the impedance is not an
impedor, i.e. if the impedance is virtual, the reflection coefficients
are virtual reflection coefficients that do not reflect anything and
do not absorb power. I will repeat an earlier assertion:

Since a virtual impedance is result of the superposition of a forward
wave and a reflected wave, a virtual impedance cannot re-reflect the
reflected wave, i.e. one cannot re-reflect the reflected wave while at
the same time the reflected wave is being used to generate an
impedance. It has to be one or the other. Otherwise, there is a
violation of the conservation of energy principle. RF EM ExH energy
cannot be used simultaneously to generate a virtual impedance while at
the same time being re-reflected.

If the reflected wave is re-reflected, it must be by an impedance
other than the virtual impedance generated by the reflected wave
itself. If the reflected wave is being used to generate a virtual
impedance, it cannot at the same time be being re-reflected.

On Jun 24, 6:27 am, wrote:

p.s. if the separation between the two ports is just the discontinuity
connection 'point' then the voltages must be the same and the currents
are exact opposites only because of the direction convention defined,
there can be no difference measuring on one side of a point to the
other.

The total voltage and total current on both sides of the impedance
discontinuity must be equal. But the superposition components do not
have to be equal and, in fact, cannot be equal. In the case of the Z0-
matched example, the forward voltage on the 50 ohm side is 70.7 volts
while the forward voltage on the 291.4 ohm side is 241.4 volts. In
order for the total voltage to be the same, the reflected voltage on
the 291.4 ohm side, which is 170.7 volts, must be subtracted from the
241.4 volts of forward voltage to yield a total of 70.7 volts. For the
Z0-matched example:

Vfwd1 = Vfwd2 - Vref2

70.7v = 241.4v - 170.7v

Please note that the Z0-match point is at a voltage minimum on the
291.4 ohm feedline. 1/4WL toward the load, the total voltage is
241.4+170.7=412.1 volts (in a lossless system).
--
73, Cecil, w5dxp.com

meaningless hair splitting. if i put a meter on one side of the stub
connection point i will measure the exact same voltage as on the other
side of the connection point. why don't you guys do something
practical instead of arguing about split hairs and things that can't
be measured?

It can be measured. Why don't you go to another group or thread? Nobody
is forcing you to read this one.

On 6/24/2011 8:52 AM, Cecil Moore wrote:

When the single-port model is used, if the impedance is not an
impedor, i.e. if the impedance is virtual, the reflection coefficients
are virtual reflection coefficients that do not reflect anything and
do not absorb power. I will repeat an earlier assertion:

Since a virtual impedance is result of the superposition of a forward
wave and a reflected wave, a virtual impedance cannot re-reflect the
reflected wave, i.e. one cannot re-reflect the reflected wave while at
the same time the reflected wave is being used to generate an
impedance. It has to be one or the other. Otherwise, there is a
violation of the conservation of energy principle. RF EM ExH energy
cannot be used simultaneously to generate a virtual impedance while at
the same time being re-reflected.

If the reflected wave is re-reflected, it must be by an impedance
other than the virtual impedance generated by the reflected wave
itself. If the reflected wave is being used to generate a virtual
impedance, it cannot at the same time be being re-reflected.

I disagree. There are 100W supplied by the source and 100W consumed by
the load. There are 200W in the 291.4 ohm line. 100W of that is just
"passing through". The other 100W is circulating, that is, stored energy
which was put there by the start-up transient. If it is circulating,
then it must be reflected from each end of the 291.4 ohm line.

Cheers,
John

On Jun 24, 2:20*pm, John S wrote:
If the reflected wave is re-reflected, it must be by an impedance
other than the virtual impedance generated by the reflected wave
itself. If the reflected wave is being used to generate a virtual
impedance, it cannot at the same time be being re-reflected.

I disagree. There are 100W supplied by the source and 100W consumed by
the load. There are 200W in the 291.4 ohm line. 100W of that is just
"passing through". The other 100W is circulating, that is, stored energy
which was put there by the start-up transient. If it is circulating,
then it must be reflected from each end of the 291.4 ohm line.

Let's assume that the 100 watts is just "passing through". It would
change from 70.7 volts in the 50 ohm environment to 170.7 volts in the
291.4 ohm environment. The reflected power is 100 watts so the
reflected voltage is also 170.7 volts. Those two voltages would have
to add together to get the forward voltage. The forward voltage is
known to be 241.4 volts. Exactly how do you add two 170.7 volt in-
phase waves to get a total of 241.4 volts?

Here is actually what happens. Since the physical power reflection
coefficient at the Z0-match point is 0.5, only 50 watts of the source
power makes it through the impedance discontinuity. That voltage is
120.7 volts. The same thing applies to the reflected power - only 50
watts is re-reflected by the 0.5 power reflection coefficient. So the
re-reflected voltage is also 120.7 volts. Adding those two voltages
together yields 241.4 volts which we know is the correct forward
voltage.

Now you are going to ask how two 50 watt waves can add up to 200 watts
forward power. That's just the nature of constructive interference
since power is proportional to voltage squared. If we add two 50 watt
waves in phase, we get a 200 watt wave. Where did the extra 100 watts
come from? Why, from the 100 watts of destructive interference toward
the source that eliminated the reflections toward the source.
--
73, Cecil, w5dxp.com

On Jun 24, 1:24*pm, dave wrote:
meaningless hair splitting.

That's my attitude toward religion so I don't frequent any religious
newsgroups.

That meaningless hair splitting is the answer to the apparent
contradiction with which Walt is wrestling.
--
73, Cecil, w5dxp.com

On Jun 24, 7:05*pm, John S wrote:
On 6/24/2011 1:24 PM, dave wrote:









On Jun 24, 1:52 pm, Cecil *wrote:
On Jun 23, 4:41 pm, *wrote:

but what is your second source? *you can always represent the second
source in that case in terms of the transmitter output so the second
input can be eliminated giving you a single port model.

a1 is the normalized forward voltage on the 50 ohm feedline from the
source. a2 is the normalized reflected voltage on the 291.4 ohm
feedline from the load. Those are the two sources associated with the
impedance discontinuity inside the black box. a2 could just as easily
be from a second generator instead of a reflection.

When the single-port model is used, if the impedance is not an
impedor, i.e. if the impedance is virtual, the reflection coefficients
are virtual reflection coefficients that do not reflect anything and
do not absorb power. I will repeat an earlier assertion:

Since a virtual impedance is result of the superposition of a forward
wave and a reflected wave, a virtual impedance cannot re-reflect the
reflected wave, i.e. one cannot re-reflect the reflected wave while at
the same time the reflected wave is being used to generate an
impedance. It has to be one or the other. Otherwise, there is a
violation of the conservation of energy principle. RF EM ExH energy
cannot be used simultaneously to generate a virtual impedance while at
the same time being re-reflected.

If the reflected wave is re-reflected, it must be by an impedance
other than the virtual impedance generated by the reflected wave
itself. If the reflected wave is being used to generate a virtual
impedance, it cannot at the same time be being re-reflected.

On Jun 24, 6:27 am, *wrote:

p.s. if the separation between the two ports is just the discontinuity
connection 'point' then the voltages must be the same and the currents
are exact opposites only because of the direction convention defined,
there can be no difference measuring on one side of a point to the
other.

The total voltage and total current on both sides of the impedance
discontinuity must be equal. But the superposition components do not
have to be equal and, in fact, cannot be equal. In the case of the Z0-
matched example, the forward voltage on the 50 ohm side is 70.7 volts
while the forward voltage on the 291.4 ohm side is 241.4 volts. In
order for the total voltage to be the same, the reflected voltage on
the 291.4 ohm side, which is 170.7 volts, must be subtracted from the
241.4 volts of forward voltage to yield a total of 70.7 volts. For the
Z0-matched example:

Vfwd1 = Vfwd2 - Vref2

70.7v = 241.4v - 170.7v

Please note that the Z0-match point is at a voltage minimum on the
291.4 ohm feedline. 1/4WL toward the load, the total voltage is
241.4+170.7=412.1 volts (in a lossless system).
--
73, Cecil, w5dxp.com

meaningless hair splitting. *if i put a meter on one side of the stub
connection point i will measure the exact same voltage as on the other
side of the connection point. *why don't you guys do something
practical instead of arguing about split hairs and things that can't
be measured?

It can be measured. Why don't you go to another group or thread? Nobody
is forcing you to read this one.

try it! you will read the exact same voltage on either side of that
connection point!

On Jun 25, 5:53*am, dave wrote:
try it! *you will read the exact same voltage on either side of that
connection point!

I already told you that only applies to the total voltage and total
current. You will NOT read the same forward voltage on either side,
you will NOT read the same forward current on either side, you will
NOT read the same reflected voltage on either side, and you will NOT
read the same reflected current on either side. The total voltage and
total current are the results of the superposition of the four
component voltages and currents that obey the rules of wave reflection
mechanics. Recognizing the interference patterns when two phasor
voltages are superposed is the key to understanding exactly what is
happening to the energy in the waves. At an impedance discontinuity in
a transmission line some distance from any active source, the average
destructive interference power in one direction MUST equal the average
constructive interference power in the opposite direction in order to
avoid a violation of the conservation of energy principle.

So why isn't the forward current flowing into the impedance
discontinuity equal to the forward current flowing out of the
impedance discontinuity? The answer to that question will solve Walt's
apparent contradiction between voltages and powers. Look at the Z0-
match again.

source--50 ohm--+--1/2WL Z050 ohm--50 ohm load

The total current on the 50 ohm side of point '+' is equal to the
total current on the Z050 ohm side but the current on the 50 ohm side
is a flat traveling wave *constant* current while the current on the
Z050 ohm side is a standing-wave current maximum, i.e. the total
current on the Z050 ohm side is a *variable* that changes with a
change in the measurement point. A variable current is NOT the same as
a constant current.

The total voltage on the 50 ohm side is a flat traveling wave
*constant* voltage while the voltage on the Z050 ohm side is a
standing wave voltage minimum, i.e. the total voltage on the Z050 ohm
side is a *variable* that changes with a change in the measurement
point.

The power on the 50 ohm side is V*I where V and I are constant values.
The power on the Z050 ohm side is V*I*cos(A) where A is the angle
between the current phasor and the voltage phasor and, because of the
standing waves, all three parameters vary with location on the
feedline.
--
73, Cecil, w5dxp.com

On 6/24/2011 7:41 PM, Cecil Moore wrote:
On Jun 24, 2:20 pm, John wrote:
If the reflected wave is re-reflected, it must be by an impedance
other than the virtual impedance generated by the reflected wave
itself. If the reflected wave is being used to generate a virtual
impedance, it cannot at the same time be being re-reflected.

I disagree. There are 100W supplied by the source and 100W consumed by
the load. There are 200W in the 291.4 ohm line. 100W of that is just
"passing through". The other 100W is circulating, that is, stored energy
which was put there by the start-up transient. If it is circulating,
then it must be reflected from each end of the 291.4 ohm line.

Let's assume that the 100 watts is just "passing through". It would
change from 70.7 volts in the 50 ohm environment to 170.7 volts in the
291.4 ohm environment. The reflected power is 100 watts so the
reflected voltage is also 170.7 volts. Those two voltages would have
to add together to get the forward voltage. The forward voltage is
known to be 241.4 volts. Exactly how do you add two 170.7 volt in-
phase waves to get a total of 241.4 volts?

Here is actually what happens. Since the physical power reflection
coefficient at the Z0-match point is 0.5, only 50 watts of the source
power makes it through the impedance discontinuity. That voltage is
120.7 volts. The same thing applies to the reflected power - only 50
watts is re-reflected by the 0.5 power reflection coefficient. So the
re-reflected voltage is also 120.7 volts. Adding those two voltages
together yields 241.4 volts which we know is the correct forward
voltage.

Now you are going to ask how two 50 watt waves can add up to 200 watts
forward power. That's just the nature of constructive interference
since power is proportional to voltage squared. If we add two 50 watt
waves in phase, we get a 200 watt wave. Where did the extra 100 watts
come from? Why, from the 100 watts of destructive interference toward
the source that eliminated the reflections toward the source.
--
73, Cecil, w5dxp.com

If the reflections toward the source is eliminated, how is it that it
appears to be 50 ohms at that point rather than 291.4 ohms?

John

On Jun 25, 12:02*pm, John S wrote:
If the reflections toward the source is eliminated, how is it that it
appears to be 50 ohms at that point rather than 291.4 ohms?

You answered your own question - if reflections toward the source are
eliminated in a Z0=50 ohm environment, the apparent (virtual)
impedance cannot be anything except 50 ohms. When you look at yourself
in the mirror, your reflected apparent (virtual) distance behind the
mirror is the same as your actual distance from the mirror. The
reflection is NOT where it appears to be, i.e. it is virtual.
--
73, Cecil, w5dxp.com

On 6/25/2011 12:41 PM, Cecil Moore wrote:
On Jun 25, 12:02 pm, John wrote:
If the reflections toward the source is eliminated, how is it that it
appears to be 50 ohms at that point rather than 291.4 ohms?

You answered your own question - if reflections toward the source are
eliminated in a Z0=50 ohm environment, the apparent (virtual)
impedance cannot be anything except 50 ohms

You said "Since a virtual impedance is result of the superposition of a
forward
wave and a reflected wave, a virtual impedance cannot re-reflect the
reflected wave, i.e. one cannot re-reflect the reflected wave while at
the same time the reflected wave is being used to generate an
impedance."

But, it does. First, it causes the 50 ohms line (looking into the 291.4
ohms line to see a match due to the reflection. Second, the
re-reflection from that discontinuity is half of what maintains the
circulating energy on the line. The other half is the discontinuity of
the non-virtual load.

On Jun 23, 12:50*am, John Smith wrote:
On 6/22/2011 3:24 PM, dave wrote:









On Jun 22, 1:49 pm, Cecil *wrote:
Let's return to an earlier example and compare a single-port analysis
with a dual-port analysis.

100w
source--50 ohm--+--1/2WL 291.4 ohm--50 ohm load

The 50 ohm Z0-match point is at '+'. The forward power on the 50 ohm
line is 100 watts and the reflected power on the 50 ohm line is zero
watts. The forward power on the 291.4 ohm line is 200 watts and the
reflected power on the 291.4 ohm line is 100 watts. 100 watts is being
sourced and delivered to the 50 ohm load.

The voltage reflection coefficient, rho, at the load is (50-291.4)/
(50+291.4)=0.7071. The power reflection coefficient, rho^2, at the
load is 0.5, i.e. half of the power incident upon the load (200w) is
reflected (100w). Since the load is a single-port, these parameters
are consistent with a single-port analysis. In a single-port analysis,
we cannot tell the difference between a virtual reflection coefficient
and a physical reflection coefficient.

The problem comes when we use a single-port analysis on the Z0-match
point. Since the reflected power on the 50 ohm line is zero, a single-
port analysis would yield rho=0.0 and rho^2=0.0 when viewing the Z0-
match from the source side. When we perform a dual-port analysis, we
get different values for rho and rho^2, i.e. we get the complement of
the reflection coefficients at the load which is a characteristic of
any simple Z0-match similar to the above example.

For a dual-port analysis, rho looking into the Z0-match from the
source side is (291.4-50)/(291.4+50)=0.7071 and rho^2 looking into the
Z0-match from the source side is 0.5, the same as at the load. Looking
back into the Z0-match from the load side, the sign of rho is negative
just as it is at the load with rho^2=0.5, the same as at the load.

Since the two analyses yield different values for the reflection
coefficients, which analysis is correct? The answer gives the clue to
the resolution of this discussion.
--
73, Cecil, w5dxp.com

ok, i'm afraid i'm going to have to ask the simple question... if you
blackbox the load and stub and look at just the one connection to it
and that gives you no reflected power... where do you define the
second port, and why?

Logic, immediately, suggests to me, that varying the frequency and
measuring voltage, amperage, and SWR would begin to immediately point
the answer(s.)

Regards,
JS

but you can't do that! they don't like real measurements, only
arguing about their virtual reflections and the separate currents and
voltages.