On Jun 21, 5:40 pm, K1TTT wrote:
On Jun 21, 1:03 am, Keith Dysart wrote:

You might consider the following experiment:

- a sinusoidal generator that will produce 50W in to 50 ohms
- attach one half wavelength of transmission line
- leave the far end of the transmission line open

After the line settles, a directional "wattmeter" anywhere on
the line will indicate 50W forward and 50W reflected.
The generator will not be putting any energy in to the line
since the line input appears as an open circuit at the
generator output.

Insert a circulator (with the reflected port terminated in
50 ohms) between the generator and the line. The circulator
termination will be dissipating 50W and the generator will
now be delivering 50W in to the circulator.

1. Please explain how inserting the circulator did not change
the circuit conditions. The generator went from delivering
0W to delivering 50W.

2. Where do you think the 50W being dissipated in the circulator
termination resistor is coming from? The line? Or the
generator (which is now outputting 50W)?

well, its coming from the generator via the far end of the line of
course.

That is the classic answer from the “reflected power” model, but
consider the following...

We construct two experiments similar to the above.
The first one:
- a sinusoidal generator that will produce 50W in to 50 ohms
- attach one half wavelength of transmission line
- leave the far end of the transmission line open
- the generator is constructed in the Thevenin style with a
voltage source and 50 ohm resistor

The second experiment:
- same as above except there is a circulator between the
generator and the line and the circulator is terminated
with 50 ohms

Examine experiment 1. After the line settles, a directional
wattmeter indicates 50W “forward power” and 50W “reflected
power”. The current in the voltage source and source resistor
is 0 so no energy is dissipated in the source resistor and
the voltage source is delivering no energy to the system.
The “reflected power” is not being dissipated in the source
resistor or voltage source so where is it going. Cecil will
offer some explanation where the “reflected energy”
is “redistributed” in the other direction as the “forward
energy”, which “must” be happening since it is not dissipated
in the source and it can not be reflected because there is
no impedance discontinuity.

Examine experiment 2. After the line settles, a directional
wattmeter indicates 50W “forward power” and 50W “reflected
power”. But with a circulator, the voltage source is
delivering 100W, the source resistor is dissipating 50W
and the circulator resistor is also dissipating 50W.
You offer that the 50W being dissipated in the circulator
is the 50W of “reflected power” coming from the line.
Presumably then, the 50W of “forward power” is coming from
the generator. This agrees numerically since the voltage
source is providing 100W and the source resistor is
dissipating 50W.

In both experiments, conditions on the line are identical.
There is 50W of “forward and reflected power” indicated.
There is no energy flowing past the end of the line since
it is open circuited.
Because the line is one-half wavelength long, the line
presents an open circuit to the generator (exp 1) or
circulator (exp 2).

In both experiments, the line is presented with a 50 ohm
source impedance, either from the generator or from the
circulator.

What puzzles me then, is how the “reflected power” knows
that in experiment 1, it should stay out of the generator
so that it is not dissipated in the source resistor
but in experiment 2, it should enter the circulator so
that it can be dissipated in the circulator load resistor.

Can you explain how the “reflected power” “knows”?

....Keith

On Jun 22, 5:49*am, Keith Dysart wrote:
What puzzles me then, is how the “reflected power” knows
that in experiment 1, it should stay out of the generator
so that it is not dissipated in the source resistor
but in experiment 2, it should enter the circulator so
that it can be dissipated in the circulator load resistor.

Can you explain how the “reflected power” “knows”?

Reflected "power" doesn't know anything. Reflected voltages and
currents simply respond to the known laws of physics involving wave
reflection and wave superposition, and obey the conservation of energy
principle.

In 1, it doesn't stay out of the generator. It is redistributed from
the generator back toward the load by superposition associated with
*destructive interference) which happens when two superposed coherent
waves are between 90 deg and 180 deg out of phase. The entire
experiment is set up in a perfect 50 ohm environment so power =
V^2/50. What you guys are missing is that step which carries the
energy along with the voltage, i.e. the voltage (and current) require
a certain energy level which is being ignored.

In 2, the source wave never encounters the reflected wave so there is
no superposition or interference at the source resistor.

You can add a couple of more configurations. Number 3 could be when
the interference between the source wave and the reflected wave occurs
and they are less than 90 deg out of phase. More power than the
average reflected power plus average load power will be dissipated in
the source resistor because of superposition associated with
*constructive interference*.

Number 4 could be the special case when the source wave and the
reflected wave are 90 degrees out of phase. This is the condition of
zero interference (neither destructive nor constructive) and 100% of
the average reflected power is dissipated in the source resistor. This
is the easiest case to understand because there are no reflections and
no interference.
--
73, Cecil, w5dxp.com

On 22 jun, 11:45, Cecil Moore wrote:
On Jun 22, 5:49*am, Keith Dysart wrote:

What puzzles me then, is how the “reflected power” knows
that in experiment 1, it should stay out of the generator
so that it is not dissipated in the source resistor
but in experiment 2, it should enter the circulator so
that it can be dissipated in the circulator load resistor.

Can you explain how the “reflected power” “knows”?

Reflected "power" doesn't know anything. Reflected voltages and
currents simply respond to the known laws of physics involving wave
reflection and wave superposition, and obey the conservation of energy
principle.

In 1, it doesn't stay out of the generator. It is redistributed from
the generator back toward the load by superposition associated with
*destructive interference) which happens when two superposed coherent
waves are between 90 deg and 180 deg out of phase. The entire
experiment is set up in a perfect 50 ohm environment so power =
V^2/50. What you guys are missing is that step which carries the
energy along with the voltage, i.e. the voltage (and current) require
a certain energy level which is being ignored.

In 2, the source wave never encounters the reflected wave so there is
no superposition or interference at the source resistor.

You can add a couple of more configurations. Number 3 could be when
the interference between the source wave and the reflected wave occurs
and they are less than 90 deg out of phase. More power than the
average reflected power plus average load power will be dissipated in
the source resistor because of superposition associated with
*constructive interference*.

Number 4 could be the special case when the source wave and the
reflected wave are 90 degrees out of phase. This is the condition of
zero interference (neither destructive nor constructive) and 100% of
the average reflected power is dissipated in the source resistor. This
is the easiest case to understand because there are no reflections and
no interference.
--
73, Cecil, w5dxp.com

Dear Cecil:

I understand you proposition about energy redistribution on
reflections and constructive/destructive interference phenomena. It is
easy to me understand tridimensional waves examples interfering on a
surface, rendering strips of nulls and maximuns. In a TL I also
understand interference must render -for example- a destructive
composition towar one direction and a constructive towards opposite
direction. Full destructive interference in one directions implies
full constructive on the opposite one, rendering a unidirectional
energy flow towards the constructive direction, and zero flow toward
the other. However I can not visualize a simple mechanism to generate
such system in a TL. Tridimensional examples are easily because RF or
light source can be set very close to each other as in physics
traditional ligth examples and render the interference pattern, but in
a TL they need be in the same place: I managed to explain my idea?
Give me a hand. (It is a pitty newsgroups do not have image
capabilities)

I am studying your paper published in World Radio Oct 2005. Have you a
demonstration of the equation cited fro "Optics" book = Pfor=Pi
+P2+2[sqrt(P1P2)]cos(theta)?

73 - Miguel - LU6ETJ

On Jun 23, 10:06*pm, lu6etj wrote:
On 22 jun, 11:45, Cecil Moore wrote:





On Jun 22, 5:49*am, Keith Dysart wrote:

What puzzles me then, is how the “reflected power” knows
that in experiment 1, it should stay out of the generator
so that it is not dissipated in the source resistor
but in experiment 2, it should enter the circulator so
that it can be dissipated in the circulator load resistor.

Can you explain how the “reflected power” “knows”?

Reflected "power" doesn't know anything. Reflected voltages and
currents simply respond to the known laws of physics involving wave
reflection and wave superposition, and obey the conservation of energy
principle.

In 1, it doesn't stay out of the generator. It is redistributed from
the generator back toward the load by superposition associated with
*destructive interference) which happens when two superposed coherent
waves are between 90 deg and 180 deg out of phase. The entire
experiment is set up in a perfect 50 ohm environment so power =
V^2/50. What you guys are missing is that step which carries the
energy along with the voltage, i.e. the voltage (and current) require
a certain energy level which is being ignored.

In 2, the source wave never encounters the reflected wave so there is
no superposition or interference at the source resistor.

You can add a couple of more configurations. Number 3 could be when
the interference between the source wave and the reflected wave occurs
and they are less than 90 deg out of phase. More power than the
average reflected power plus average load power will be dissipated in
the source resistor because of superposition associated with
*constructive interference*.

Number 4 could be the special case when the source wave and the
reflected wave are 90 degrees out of phase. This is the condition of
zero interference (neither destructive nor constructive) and 100% of
the average reflected power is dissipated in the source resistor. This
is the easiest case to understand because there are no reflections and
no interference.
--
73, Cecil, w5dxp.com

Dear Cecil:

I understand you proposition about energy redistribution on
reflections and constructive/destructive interference phenomena. It is
easy to me understand tridimensional waves examples interfering on a
surface, rendering strips of nulls and maximuns. In a TL I also
understand interference must render -for example- a destructive
composition towar one direction and a constructive towards opposite
direction. Full destructive interference in one directions implies
full constructive on the opposite one, rendering a unidirectional
energy flow towards the constructive direction, and zero flow toward
the other. However I can not visualize a simple mechanism to generate
such system in a TL. Tridimensional examples are easily *because RF or
light source can be set very close to each other as in physics
traditional ligth examples and render the interference pattern, but in
a TL they need be in the same place: I managed to explain my idea?
Give me a hand. (It is a pitty newsgroups do not have image
capabilities)

I am studying your paper published in World Radio Oct 2005. Have you a
demonstration of the equation cited fro "Optics" book = Pfor=Pi
+P2+2[sqrt(P1P2)]cos(theta)?

This equation is problematic. Firstly, it mixes power with voltage
since
theta is the angle between the voltage waveforms. As K1TTT said, "why
not just do the whole thing with voltages?" This mixing is bad form
and
clearly demonstrates the incompleteness of the power based analysis.

The second is that there are two solutions depending on whether the
positive or negative root is used? Why is one discarded? Is this
numerology at work?

Thirdly, it only produces the correct answer for average energy flows.
If the instantaneous energy flows are examined, the results using
this
equation do not align with observations.

....Keith

On Jun 24, 5:27*am, Keith Dysart wrote:
This equation is problematic. Firstly, it mixes power with voltage
since theta is the angle between the voltage waveforms.

You apparently don't understand what happens when one takes the dot
product of two voltage phasors (and divides by Z0). The result is
watts but the math involves the cosine of the angle between the two
voltage phasors. All competent EEs should already know that.

As K1TTT said, "why not just do the whole thing with voltages?"

Because the title of this thread is: "What happens to reflected
energy?", not what happens to reflected voltages? Doing the whole
thing with voltages allows the obfuscation of interference to be swept
under the rug.

Because some of you guys don't recognize interference when it is
staring you in face? I was taught to recognize interference between
voltage phasors at Texas A&M in the 1950s. What happened to you guys?
Here's a short lesson about dot products of voltage phasors and the
resulting interference between the two voltages.

Vtot = V1*V2

Vtot^2 = (V1*V2)^2

Vtot^2/Z0 watts = (V1*V2)^2/Z0 watts

There will be the two obvious power terms, V1^2/Z0 and V2^2/Z0,
representing the powers in the individual waves before superposition.
There will be a third, additional interference term whose dimension is
watts that *requires the dot product* between the two phasor voltages.
Therefore, your objection is apparently just based on ignorance of the
dot product of two voltage phasors.

This mixing is bad form and
clearly demonstrates the incompleteness of the power based analysis.

Good grief! This "bad form" has been honored in the field of optical
physics for at least a century. It was taught in EE courses 60 years
ago. I don't know what has happened in the meantime. Walter Maxwell
explains interference in section 4.3 in "Reflections" and obviously
understands the role of interference in the redistribution of energy.

The second is that there are two solutions depending on whether the
positive or negative root is used? Why is one discarded? Is this
numerology at work?

Negative power is just a convention for "negative" direction of energy
flow. All EEs are taught in our engineering courses to ignore the
imiginary root when calculating resistance, energy, or power.

For instance, the Z0 for the 1/4WL matching section between R1 and R2
needs to be SQRT(R1*R2). When you perform that math function, do you
really go on a world-wide search demanding a transmssion line with a
negative Z0? Please get real.

Thirdly, it only produces the correct answer for average energy flows.
If the instantaneous energy flows are examined, the results using
this equation do not align with observations.

You forgot to add that instantaneous energy is as useless as tits on a
boar hog, or as Hecht said, putting it mildly: "of limited utility".
It appears to me that instantaneous energy is just a mathematical
artifact inside a process requiring integration in order to bear any
resemblence to reality. Omit the integration and the process loses
touch with reality. Instantaneous energy has zero area under the curve
until the intergration process has been performed. A zero area
represents zero energy. Otherwise, when you integrate from zero to
infinity, the result would be infinite energy.

Do you have any kind of reference for your treatment of instantaneous
power?
--
73, Cecil, w5dxp.com

On Jun 24, 9:39*am, Cecil Moore wrote:
Vtot^2/Z0 watts = (V1*V2)^2/Z0 watts

There will be the two obvious power terms, V1^2/Z0 and V2^2/Z0,

Sorry, I lost my train-of-thought here and changed horses in mid-
stream. All I was trying to illustrate is that the dot product of two
voltage phasors divided by Z0 has the dimensions of watts and contains
the cos(A) term. Again, my apology for wandering off the logical path.
--
73, Cecil, w5dxp.com

On Jun 24, 10:39*am, Cecil Moore wrote:
On Jun 24, 5:27*am, Keith Dysart wrote:

This equation is problematic. Firstly, it mixes power with voltage
since theta is the angle between the voltage waveforms.

You apparently don't understand what happens when one takes the dot
product of two voltage phasors (and divides by Z0). The result is
watts but the math involves the cosine of the angle between the two
voltage phasors. All competent EEs should already know that.

As K1TTT said, "why not just do the whole thing with voltages?"

Because the title of this thread is: "What happens to reflected
energy?", not what happens to reflected voltages? Doing the whole
thing with voltages allows the obfuscation of interference to be swept
under the rug.

Because some of you guys don't recognize interference when it is
staring you in face? I was taught to recognize interference between
voltage phasors at Texas A&M in the 1950s. What happened to you guys?
Here's a short lesson about dot products of voltage phasors and the
resulting interference between the two voltages.

Vtot = V1*V2

Vtot^2 = (V1*V2)^2

Vtot^2/Z0 watts = (V1*V2)^2/Z0 watts

There will be the two obvious power terms, V1^2/Z0 and V2^2/Z0,
representing the powers in the individual waves before superposition.
There will be a third, additional interference term whose dimension is
watts that *requires the dot product* between the two phasor voltages.
Therefore, your objection is apparently just based on ignorance of the
dot product of two voltage phasors.

This mixing is bad form and
clearly demonstrates the incompleteness of the power based analysis.

Good grief! This "bad form" has been honored in the field of optical
physics for at least a century. It was taught in EE courses 60 years
ago. I don't know what has happened in the meantime. Walter Maxwell
explains interference in section 4.3 in "Reflections" and obviously
understands the role of interference in the redistribution of energy.

The second is that there are two solutions depending on whether the
positive or negative root is used? Why is one discarded? Is this
numerology at work?

Negative power is just a convention for "negative" direction of energy
flow.

Still does not explain why you choose only the positive root.
Especially when you let the result be negative as a consequence
of cos(theta).

All EEs are taught in our engineering courses to ignore the
imiginary root when calculating resistance, energy, or power.
For instance, the Z0 for the 1/4WL matching section between R1 and R2
needs to be SQRT(R1*R2). When you perform that math function, do you
really go on a world-wide search demanding a transmssion line with a
negative Z0? Please get real.

Thirdly, it only produces the correct answer for average energy flows.
If the instantaneous energy flows are examined, the results using
this equation do not align with observations.

You forgot to add that instantaneous energy is as useless as tits on a
boar hog, or as Hecht said, putting it mildly: "of limited utility".

I always get a chuckle when you write this. It makes me think of a
kid,
who, upon being told that he can not have his favourite dessert until
he finishes his brussel sprouts, declares that he has always hated
that dessert.

You might study why the real power folk prefer three phase to single.
It all has to do with instantaneous power.

Might not make a difference for light, but it sure helps the
understanding at RF and lower.

And your analysis still only produces the correct answers for the
average and still gets the instantaneous wrong. I do see the benefit
of repeating "of limited utility".

It appears to me that instantaneous energy is just a mathematical
artifact inside a process requiring integration in order to bear any
resemblence to reality. Omit the integration and the process loses
touch with reality. Instantaneous energy has zero area under the curve
until the intergration process has been performed. A zero area
represents zero energy. Otherwise, when you integrate from zero to
infinity, the result would be infinite energy.

Do you have any kind of reference for your treatment of instantaneous
power?

As I suggested above, have a look at the benefit of three phase over
single.

....Keith

On Jun 24, 8:58*pm, Keith Dysart wrote:
Still does not explain why you choose only the positive root.

Of course it does. In the power density equation, choosing the
negative root would lead to a violation of the conservation of energy
principle. When one of the roots is obviously impossible in reality, a
rational person chooses the other root.

You might study why the real power folk prefer three phase to single.
It all has to do with instantaneous power.

I am a "real power folk", Keith. My first EE degree was in power
generation and transmission. Three-phase puts less stress on the
system by eliminating the hills and valleys in the energy flow common
with traveling waves. But why do you believe that three-phase power
transmission is relevant to ham radio? Are you running three-phase RF?

Maxwell's equations don't even work for your mashed-potatoes version
of energy. That should tell you something.
--
73, Cecil, w5dxp.com

On Jun 23, 9:06*pm, lu6etj wrote:
However I can not visualize a simple mechanism to generate
such system in a TL.

I have given the equations for what happens at an impedance
discontinuity in a transmission line. The s-parameter equations are
the same equations in a different format. Simply visualize the voltage
phasors resulting from reflections-from and transmissions-through the
impedance discontinuity. Let's start with voltages instead of power.

source------Z01=50 ohms------+------Z02=300 ohms--------load

measured Vfor1 = 50v, Vref1 = 0v
calculated rho1 = (300-50)/(300+50) = 0.7143

Vfor1*rho1 = 35.7v at zero degrees = portion of Vfor1 reflected by the
impedance discontinuity at '+'.

Vref2*tau2 = 35.7v at 180 degrees = portion of Vref2 transmitted back
through the impedance discontinuity at '+'.

Vref1 = Vfor*rho1 + Vref2*tau2 = 0, measured Vref1. This is the same
as the s-parameter equation (1).

b1 = s11*a1 + s12*a2, all phasor math

These are the two voltage components that superpose to zero volts.
Both of those wavefronts are phasors with phase angles referenced to
the Vfor1 phase angle.

Now let's look at the power in the component phasor wavefronts.

(Vfor1)^2/Z01 = 50^2/50 = 50w, power in the Vfor1 forward wave

(Vref1)^2/Z01 = 0^2/50 = 0w, power in the Vref1 reflected wave

(Vfor1*rho1)^2/Z01 = 35.7^2/50 = 25.49w

(Vref2*tau2)^2/Z01 = 35.7^2/50 = 25.49w

Pref1 = 25.49w + 25.49w + 2*SQRT(25.49*25.49)cos(180)

The corresponding s-parameter power equation is:

b1^2 = (s11*a1 + s12*a2)^2

b1^2 = (s11*a1)^2 + (s12*a2)^2 + 2(s11*a1)(s12*a2)

The two wavefronts that cancel toward the source contain energy before
they superpose to zero. Where does that energy go? Superposing to zero
indicates total destructive interference toward the source. The
conservation of energy principle says that energy cannot be destroyed
so it must appear as an equal magnitude of constructive interference
in the only other direction possible, i.e. toward the load.

The above equations deal only with the destructive interference toward
the source. There is a second complimentary set of voltage/energy
equations that deal with constructive interference toward the load.

I am studying your paper published in World Radio Oct 2005. Have you a
demonstration of the equation cited fro "Optics" book = Pfor=Pi
+P2+2[sqrt(P1P2)]cos(theta)?

See above.
--
73, Cecil, w5dxp.com

On 24 jun, 10:32, Cecil Moore wrote:
On Jun 23, 9:06*pm, lu6etj wrote:

However I can not visualize a simple mechanism to generate
such system in a TL.

I have given the equations for what happens at an impedance
discontinuity in a transmission line. The s-parameter equations are
the same equations in a different format. Simply visualize the voltage
phasors resulting from reflections-from and transmissions-through the
impedance discontinuity. Let's start with voltages instead of power.

source------Z01=50 ohms------+------Z02=300 ohms--------load

measured Vfor1 = 50v, Vref1 = 0v
calculated rho1 = (300-50)/(300+50) = 0.7143

Vfor1*rho1 = 35.7v at zero degrees = portion of Vfor1 reflected by the
impedance discontinuity at '+'.

Vref2*tau2 = 35.7v at 180 degrees = portion of Vref2 transmitted back
through the impedance discontinuity at '+'.

Vref1 = Vfor*rho1 + Vref2*tau2 = 0, measured Vref1. This is the same
as the s-parameter equation (1).

b1 = s11*a1 + s12*a2, all phasor math

These are the two voltage components that superpose to zero volts.
Both of those wavefronts are phasors with phase angles referenced to
the Vfor1 phase angle.

Now let's look at the power in the component phasor wavefronts.

(Vfor1)^2/Z01 = 50^2/50 = 50w, power in the Vfor1 forward wave

(Vref1)^2/Z01 = 0^2/50 = 0w, power in the Vref1 reflected wave

(Vfor1*rho1)^2/Z01 = 35.7^2/50 = 25.49w

(Vref2*tau2)^2/Z01 = 35.7^2/50 = 25.49w

Pref1 = 25.49w + 25.49w + 2*SQRT(25.49*25.49)cos(180)

The corresponding s-parameter power equation is:

b1^2 = (s11*a1 + s12*a2)^2

b1^2 = (s11*a1)^2 + (s12*a2)^2 + 2(s11*a1)(s12*a2)

The two wavefronts that cancel toward the source contain energy before
they superpose to zero. Where does that energy go? Superposing to zero
indicates total destructive interference toward the source. The
conservation of energy principle says that energy cannot be destroyed
so it must appear as an equal magnitude of constructive interference
in the only other direction possible, i.e. toward the load.

The above equations deal only with the destructive interference toward
the source. There is a second complimentary set of voltage/energy
equations that deal with constructive interference toward the load.

I am studying your paper published in World Radio Oct 2005. Have you a
demonstration of the equation cited fro "Optics" book = Pfor=Pi
+P2+2[sqrt(P1P2)]cos(theta)?

See above.
--
73, Cecil, w5dxp.com

Oh, I'm so sorry Cecil, I should have written "However I can not
visualize a simple PHYSICAL mechanism/example to generate
such system in a TL". Anyway, your additional info it is very usefu to
me. Thanks.

Miguel