what happens to reflected energy ?
 

On Jun 21, 7:25*pm, Keith Dysart wrote:
The dissipation seems to correlate much more strongly with the
circuit design than it does with the magnitude of the "reflected
power".

Given a 50 ohm source resistor resulting in zero re-reflections, *the
dissipation correlates perfectly with the level and type of
interference*. I am amazed at the level of ignorance concerning
interference and the conservation of energy principle. I studied
interference in detail in my 1950's college physics courses including
what happens to the energy during interference events. Although Dr.
Best didn't realize it, in his QEX Nov/Dec 2001 article on
transmission lines, he presented the equations governing energy
redistribution associated with interference. What didn't help was his
equation of the form:

Ptot = 75w + 8.33w = 133.33w

which is obviously false. What he left out was the constructive
interference term of:

2*SQRT(75w*8.33w) = 50w

Given two coherent waves, each associated with a Poynting Vector, when
those two waves are superposed, interference will result. I'm sure
everyone has seen interference rings and pictures of interference
rings where the energy in the interference rings has obviously been
redistributed away from a homogeneous power density. The same thing
happens in a transmission line and/or at a source, just at longer
wavelengths where we cannot "see" them with our eyes.

Constructive interference requires "extra" energy. Superpose two 50
watt coherent waves in phase and the result is a 200 watt wave.
Destructive interference has energy left over. Superpose two 50 watt
coherent waves 180 degrees out of phase and the result is a zero watt
wave. At a source, the source is capable of supplying extra power or
throttling back on its power output depending on the level and kind of
interference. If one simply takes the time to understand interference,
one knows where the energy components go even when there is zero re-
reflection.

Given a single traveling wave in a lossless transmission line with a
purely resistive characteristic impedance of Z0, the power (joules/
second) measured at a point on the line is P=V^2/Z0=I^2*Z0=V*I. This
is classic transmission line math no matter which direction the wave
is traveling. Superposing two coherent traveling waves cannot create
or destroy the energy pre-existing in the two waves, but it can (and
often does) redistribute the energy according to the conservation of
energy principle. If the total energy in the two superposed waves is
different from the sum of the two energy components before
superposition, then interference exists and some interference energy
has been redistributed in the system. That some people are ignorant of
where the constructive interference energy came from or where the
destructive interference energy went is not a good reason to deny its
existence (which violates the accepted laws of physics).

The thing that makes an interference event so easy to analyze in a
transmission line is the fact that the transmission line is
essentially one-dimensional. At an impedance discontinuity in a
transmission line (away from any active source) if interference
exists, the destructive interference in one direction must equal the
constructive interference in the other direction. Any other outcome
would violate the conservation of energy principle and that is a
common occurence on this newsgroup.

In my earlier example, the two results of *(1) reflection* are
traveling waves containing ExH power densities:

Pfor1(rho^2) toward the source and

Pref2(rho^2) toward the load

The complimentary transmissions a

Pfor1(1-rho^2) toward the load and

Pref2(1-rho^2) toward the source

Those two reflections *(2) superpose* with the two complimentary
transmissions in each same direction according to the power density
equation:

0 = Pref1 = Pfor1(rho^2) + Pref2(1-rho^2) minus total destructive
interference toward the source

Pfor2 = Pfor1(1-rho^2) + Pref2(rho^2) plus constructive interference
toward the load

When one assumes that reflection is the only phenomenon that can
redistribute wave energy, one is ignoring the superposition process
which is also known to be perfectly capable of redistributing wave
energy. In Roy's food-for-thought example, he designed it so
reflections would not exist at the source. That leaves superposition
as the phenomenon that is accomplishing the obvious redistribution of
energy. It certainly does NOT mean that the laws of physics have been
suspended due to the ignorance of the writer.

Here's a simple exercise in phasor superposition. Given two 50w
(joules/sec) coherent EM waves traveling in the same direction in
Z0=50 ohm coax:

1. What is the magnitude of the two voltages? ________

2. If one superposes those two voltages in phase in the Z0=50 ohm
transmission line, what is the resulting power (joules/sec) in the
total wave? ___________

3. If one superposes those two voltages 180 degrees out of phase in
the Z0=50 ohm transmission line, what is the resulting power (joules/
sec) in the total wave? ____________

In #2, if there is no source around, where does the "extra" wave
energy come from?

In #3, if there is no source or sink around, where does the "excess"
wave energy go?
--
73, Cecil, w5dxp.com

what happens to reflected energy ?
 

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

what happens to reflected energy ?
 

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

what happens to reflected energy ?
 

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

what happens to reflected energy ?
 

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

what happens to reflected energy ?
 

On Jun 24, 5:17*am, Keith Dysart wrote:
I assume you will claim that there is now “constructive
interference” rather than the previous “destructive interference”,
but the line conditions are the same. How does the “reflected
power” know if it should construct or destroy? The phase is the
same.

That's easy to answer. The Norton equivalent is a current source so
currents should be used in the calculations. The phase angles between
the two current components are 180 degrees different from the phase
angles between the two voltage components. If the interference between
voltages is constructive, the interference between currents will be
destructive. Hint: the reflected current phasor is 180 degrees out of
phase with the reflected voltage phasor because of the direction of
travel of the reflected wave. As a result of directional convention,
the power in the reflected wave is negative.

So destructive interference for forward/reverse voltages is
constructive interference for forward/reverse currents and vice versa.
An SWR voltage maximum (constructive voltage interference) is an SWR
current minimum (destructive current interference) and an SWR voltage
minimum (destructive current interference) is an SWR current maximum
(constructive voltage interference).
--
73, Cecil, w5dxp.com

what happens to reflected energy ?
 

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

what happens to reflected energy ?
 

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

what happens to reflected energy ?
 

On Jun 24, 9:20*am, lu6etj wrote:
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 useful to
me. Thanks.

The physical mechanism is the Z01==Z02 impedance discontinuity with
its associated reflection coefficient, rho. We can see that reflection
on a TDR so it is indeed a PHYSICAL mechanism.
--
73, Cecil, w5dxp.com

what happens to reflected energy ?
 

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