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

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 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 Jun 25, 1:18*pm, Cecil Moore wrote:
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.

when someone gives me a voltmeter i can touch to that connection point
and measure the 4 components then we can talk. as far as designing
anything i need i can do it without giving those s parameters or your
4 components a second thought. it is very easy to transform and
combine the impedances to tell me what the load seen by the
transmitter is, or to figure out the needed stub for providing a
proper match without all that stuff.

On Jun 26, 6:43*am, dave wrote:
when someone gives me a voltmeter i can touch to that connection point
and measure the 4 components then we can talk.

I could design an expensive device that will do exactly that but it is
a lot easier to just calculate the values using Mathcad.
--
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