The Rest of the Story
 

Keith Dysart wrote:
Ps(t) = 100 + 116.6190379cos(2wt-30.96375653)
Prs(t) = 68 + 68cos(2wt-61.92751306)
Pg(t) = 32 + 68cos(2wt)
Pf.g(t) = 50 + 50cos(2wt)
Pr.g(t) = -18 + 18cos(2wt)

Now it is time for you to explain exactly why you
believe in a conservation of power principle. Do
you demand that the instantaneous power delivered
by a battery charger be instantaneously dissipated
in the battery being charged? If not, why do you
require such for the example under discussion?

The correct equation for adding the powers above is?

Prs(t) = Pf.g(t) + Pr.g(t) +/- 2*SQRT[Pf.g(t)*Pr.g(t)]

The last term is the interference term. The sign of
the interference term is negative if Vf.g(t) and
Vr.g(t) are out of phase. The sign of the interference
term is positive if Vf.g(t) and Vr.g(t) are in phase.
Vf.g(t) and Vr.g(t) are in phase for half of the cycle.
Vf.g(t) and Vr.g(t) are out of phase for the other half
of the cycle. The "excess" energy from the destructive
interference is dissipated in the source resistor as
constructive interference after being delayed by 90
degrees.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
To me it was a complete
surprise that summing the voltages produces the correct
total voltages and, at the same time, summing the powers
(which are a squared function of the voltage) also
produce the correct result.

Why is that a surprise to you? I have been telling you
for many days about that special case whe

(V1^2 + V2^2) = (V1 + V2)^2 for zero interference

This applies equally well to phasors or instantaneous
values of voltage. The above special case is what my
Part 1 article is all about.

Since a 45 degree long transmission line forces the
above RMS voltage equation to be true for all values of
resistive loads, the *average* reflected power based on
RMS voltage values is *always* dissipated in the source
resistor. Since a conservation of power principle does
not exist, it is perfectly acceptable for destructive
interference energy to be stored for part of the cycle
and be dissipated later in the cycle.

So
Pg(t) = Pf.g(t) + Pr.g(t)
is always true.

Of course it is true but it says absolutely nothing about
the dissipation of the reflected power in the source resistor
which is the subject of the discussion. The above equation
remains always true while the dissipation of the reflected
power in the source resistor varies from 0% to 100%.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Fri, 21 Mar 2008 17:08:39 -0700 (PDT)
Keith Dysart wrote:

On Mar 21, 9:12*am, Roger Sparks wrote:
On Fri, 21 Mar 2008 05:10:01 -0700 (PDT)

Keith Dysart wrote:
On Mar 20, 1:07*pm, Roger Sparks wrote:
Keith Dysart wrote:
On Mar 16, 10:21 am, Cecil Moore wrote:
snip
Would you please explain how energy is conserved in the
following example at the zero-crossing point for Vs?

* * * * * * * *Rs * * * Vg * * * * * * * * * * Vl
* * * * *+----/\/\/-----+----------------------+
* * * * *| * *50 ohm * * * * * * * * * * * * * |
* * * * *| * * * * * * * * * * * * * * * * * * |
* * * * Vs * * * * * * * * 45 degrees * * * * *| Shorted
* * *100v RMS * * * * * * *50 ohm line * * * * | Stub
* * * * *| * * * * * * * * * * * * * * * * * * |
* * * * *| * * * * * * * * * * * * * * * * * * |
* * * * *+--------------+----------------------+
* * * * gnd

At the zero-crossing of Vs, Ps(t) = 0, i.e. the source
is supplying zero watts at that time but Prs(t) = 100w.
Where is the 100 watts coming from?

For the first 90 degrees of time, the circuit can be represented as

* * * * * * * * *Rs * * * Vg * * * * * * * * * * Vl
* * * * * *+----/\/\/-----+----------------------+
* * * * * *| * *50 ohm * * * * * * * * * * * * * |
* * * * * *| * * * * * * * * * * * * * * * * * * /
* * * * * Vs * * * * * * * * * * * * * * * * * * \ 50 ohm resistor
* * * *100v RMS * * * * * * * * * * * * * * * * */
* * * * * *| * * * * * * * * * * * * * * * * * * \
* * * * * *| * * * * * * * * * * * * * * * * * * |
* * * * * *+--------------+----------------------+
* * * * * gnd

After 90 degrees of time has passed, the circuit can be represented as

* * * * * * * * *Rs * * * Vg * * * * * * * * * * Vl
* * * * * *+----/\/\/-----+----------------------+
* * * * * *| * *50 ohm * * * * * * * * * * * * * |
* * * * * *| * * * * * * * * * * * * * * * * * * |
* * * * * Vs * * * * * * * * * * * * * * * * * *---
* * * *100v RMS * * * * * * * * * * * * * * * * --- 50 ohm inductive
* * * * * *| * * * * * * * * * * * * * * * * * * |
* * * * * *| * * * * * * * * * * * * * * * * * * |
* * * * * *+--------------+----------------------+
* * * * * gnd
HEAVILY CLIPPED
We differ on how to utilize the term Pg(t). *I think it describes the standing wave, not the voltage across Rs. *

I describe Vg(t) as the actual voltage that would be measured by an
oscilloscope at the point g; i.e. the beginning of the transmission
line. It is certainly not the voltage across Rs. Perhaps it IS what
you mean by standing wave. In any case the voltage across the resistor
is
Vs(t) = Vrs(t) + Vg(t)
Vrs(t) = Vs(t) - Vg(t)

Pg(t) is the power being delivered to the transmission line at any
instance t.
Pg(t) = Vg(t) * Ig(t)

...Keith

You gave a very thoughtful reply Keith. It looks to me like two perspectives, both right so far as the analysis has proceeded.
--
73, Roger, W7WKB

The Rest of the Story
 

Keith Dysart wrote:
Cecil Moore wrote:
The net total voltage at point g is the standing wave
voltage that is present at that point.

I suppose that is a name you could use, but do you still
call it "standing wave voltage" when there is no
reflected wave, i.e. Vr is 0 for all time. There is
no standing wave when Vr is always 0.

We are not discussing matched systems. If the total
RMS voltage differs from the RMS forward voltage,
then Vr is NOT zero.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Sat, 22 Mar 2008 03:48:51 -0700 (PDT)
Keith Dysart wrote:

On Mar 21, 9:32*pm, Roy Lewallen wrote:
Keith Dysart wrote:
. . .
The equation
* Pg(t) = Pf.g(t) + Pr.g(t)
is more interesting. The basis for this is superposition.
The forward and reverse voltage and current are superposed
to derive the actual voltage and current. It would seem
invalid to also sum the powers. To me it was a complete
surprise that summing the voltages produces the correct
total voltages and, at the same time, summing the powers
(which are a squared function of the voltage) also
produce the correct result.

But by starting with the equations used to derive forward
and reverse voltage and current, it can be easily shown
with appropriate substitution that Ptot is always equal
to Pforward + Preverse (Or Pf - Pr if you use the other
convention for the direction of the energy flow)s. It
simply falls out from the way that Vf and Vr are derived
from Vactual and Iactual.

It only holds true when Z0 is purely real. Of course, when it isn't,
time domain analysis becomes very much more cumbersome. But it's not
hard to show the problem using steady state sinusoidal analysis, and
that's where the cos term appears and is appropriate.

*
* So
* * Pg(t) = Pf.g(t) + Pr.g(t)
* is always true. For any arbitrary waveforms. Inclusion
* of cos(theta) terms would be incorrect.

Thanks for providing the limitation.

But I am having difficulty articulating where the math in the
following
derivation fails.

Starting by measuring the actual voltage and current at a single
point on the line, and wishing to derive Vf and Vr we have the
following four equations:

V = Vf + Vr
I = If - Ir
Zo = Vf / If
Zo = Vr / Ir

rearranging and substituting
Vf = V - Vr
= V - Zo * Ir
= V - Zo * (If - I)
= V - Zo * (Vf/Zo - I)
= V - Vf + Zo * I
= (V + Zo * I)/2
similarly
Vr = (V - Zo * I)/2

Pf = Vf * If
= Vf**2 / Zo
= ((V + Zo * I)(V + Zo * I)/4)/Zo
= (V**2 + 2(V * Zo * I) + Zo**2 * I**2) / (4 * Zo)

Pr = Vr * Ir
= (V**2 - 2(V * Zo * I) + Zo**2 * I**2) / (4 * Zo)

So, comtemplating that
P = Pf - Pr
and substituting
P = (V**2 + 2(V * Zo * I) + Zo**2 * I**2) / (4 * Zo)
- (V**2 - 2(V * Zo * I) + Zo**2 * I**2) / (4 * Zo)
= 4(V * Zo * I) / (4 * Zo)
= V * I
as required.

So when Zo is real, i.e. can be represented by R, it is
clear that P always equals Pf - Pr. And it does not even
matter which value of R is used for R. It does not have
to be the characteristic impedance of the transmission
line, the subtraction of powers still produces the correct
answer.

But when Zo has a reactive component, it still cancels
out of the equations. So why is this not a proof that
also holds for complex Zo.

I suspect it has to do with complex Zo being a concept
that only works for single frequency sinusoids, but am
having difficulty discovering exactly where it fails.

And if it is related to Zo and single frequency sinusoids,
does that mean that P = Pf - Pr also always works for
single frequency sinusoids?

...Keith

I am very impressed with this series of equations/relationships. These equations clarify your previous postings and provide a basis for future enrichment.

I think that a complex Zo would not be a transmission line, but would be an end point. Any complex end point could be represented by a length of transmission line with a resistive termination. Once that substitution was made, the problem should come back to the basic equations you presented here.
--
73, Roger, W7WKB

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On Sat, 22 Mar 2008 14:14:35 GMT
Cecil Moore wrote:

Keith Dysart wrote:
Ps(t) = 100 + 116.6190379cos(2wt-30.96375653)
Prs(t) = 68 + 68cos(2wt-61.92751306)
Pg(t) = 32 + 68cos(2wt)
Pf.g(t) = 50 + 50cos(2wt)
Pr.g(t) = -18 + 18cos(2wt)

Now it is time for you to explain exactly why you
believe in a conservation of power principle. Do
you demand that the instantaneous power delivered
by a battery charger be instantaneously dissipated
in the battery being charged? If not, why do you
require such for the example under discussion?

The correct equation for adding the powers above is?

Prs(t) = Pf.g(t) + Pr.g(t) +/- 2*SQRT[Pf.g(t)*Pr.g(t)]

Prs(t) = Pf.g(t) + Pr.g(t) +/- 2*SQRT[Pf.g(t)*Pr.g(t)]

= [SQRT(pf.g(t) + SQRT(pr.g(t)]^2

and/or= [SQRT(pf.g(t) - SQRT(pr.g(t)]^2

How can we justify calling the +/- 2*SQRT[Pf.g(t)*Pr.g(t) term the "interference term"?

Am I correct in assuming that this equation describes the instantaneous power delivered to Rs?

The last term is the interference term. The sign of
the interference term is negative if Vf.g(t) and
Vr.g(t) are out of phase. The sign of the interference
term is positive if Vf.g(t) and Vr.g(t) are in phase.
Vf.g(t) and Vr.g(t) are in phase for half of the cycle.
Vf.g(t) and Vr.g(t) are out of phase for the other half
of the cycle. The "excess" energy from the destructive
interference is dissipated in the source resistor as
constructive interference after being delayed by 90
degrees.
--
73, Cecil http://www.w5dxp.com


--
73, Roger, W7WKB

The Rest of the Story
 

On Mar 22, 10:14*am, Cecil Moore wrote:
Keith Dysart wrote:
Ps(t) = 100 + 116.6190379cos(2wt-30.96375653)
Prs(t) = 68 + 68cos(2wt-61.92751306)
Pg(t) = 32 + 68cos(2wt)
Pf.g(t) = 50 + 50cos(2wt)
Pr.g(t) = -18 + 18cos(2wt)

Now it is time for you to explain exactly why you
believe in a conservation of power principle. Do
you demand that the instantaneous power delivered
by a battery charger be instantaneously dissipated
in the battery being charged?

Are you making this a trick question by using the
word "dissipated"?

If not, then yes. Consider my laptop which has an
external power supply connected by a cord to the
laptop.

Conservation of energy means that the instantaneous
energy flow (i.e. power) along this cord into the
laptop is always exactly equal to the sum of
- the sum of energy being dissipated as heat
in each individual component
- the increase in energy being stored in
components such as capacitors, inductors
and batteries, minus any stored energy
being returned from components such as
capacitors, inductors and batteries
- energy being emitted such as
- light (e.g. display)
- sound (e.g. fans, speakers)
- RF (e.g. Wifi antennas, RFI)
- etc.

So yes, the phrase "conservation of power" is
appropriately descriptive and follows from
conservation of energy.

If not, why do you
require such for the example under discussion?

I require it for both.

The correct equation for adding the powers above is?

Prs(t) = Pf.g(t) + Pr.g(t) +/- 2*SQRT[Pf.g(t)*Pr.g(t)]

The last term is the interference term. The sign of
the interference term is negative if Vf.g(t) and
Vr.g(t) are out of phase. The sign of the interference
term is positive if Vf.g(t) and Vr.g(t) are in phase.
Vf.g(t) and Vr.g(t) are in phase for half of the cycle.
Vf.g(t) and Vr.g(t) are out of phase for the other half
of the cycle. The "excess" energy from the destructive
interference is dissipated in the source resistor as
constructive interference after being delayed by 90
degrees.

Where is "cos(theta)" in this?
And what "theta" is to be used?
Is it the same theta that was used to conclude that
this was a "no interference" example?
See "A Simple Voltage Source - No Interference" at
http://www.w5dxp.com/intfr.htm

So the plus/minus in the equation aboves means that
for part of the cycle we should add and part of the
cycle we should subtract.
This equation is ambiguous and is therefore incomplete
since it does not tell us when we should add and when
we should subtract.
Could you provide a mathematically precise description
of when we should do each?

Where is this destructive interference energy stored
while waiting to be dissipated constructively later?
If it is in a capacitance, which one? Does the
voltage on the capacitance increase appropriately
to account for the energy being stored?
Similarly if it is stored in an inductance.

...Keith

The Rest of the Story
 

On Mar 22, 10:55*am, Cecil Moore wrote:
Keith Dysart wrote:
Cecil Moore wrote:
The net total voltage at point g is the standing wave
voltage that is present at that point.

I suppose that is a name you could use, but do you still
call it "standing wave voltage" when there is no
reflected wave, i.e. Vr is 0 for all time. There is
no standing wave when Vr is always 0.

We are not discussing matched systems. If the total
RMS voltage differs from the RMS forward voltage,
then Vr is NOT zero.

True for this example. But it is generally unwise to pick
names or descriptions that will have to change when
the component values are changed.

...Keith

The Rest of the Story
 

On Mar 22, 11:17*am, Roger Sparks wrote:
I think that a complex Zo would not be a transmission line, but would be an end point. *Any complex end point could be represented by a length of transmission line with a resistive termination. *Once that substitution was made, the problem should come back to the basic equations you presented here..

The characteristic impedance for a transmission line is
Zo = sqrt( (R + jwL) / (G + jwC) )

For a lossline (no resistance in the conductors, and no
conductance between the conductors), this simplifies to
Zo = sqrt( L / C )

So real lines actually have complex impedances. But the
math is simpler for ideal (lossless) lines and there is
much to be learned from studying the simplified examples.

But caution is needed when taking these results to
the real world of lines with loss.

...Keith

The Rest of the Story
 

On Sun, 23 Mar 2008 03:42:36 -0700 (PDT)
Keith Dysart wrote:

On Mar 22, 11:17*am, Roger Sparks wrote:
I think that a complex Zo would not be a transmission line, but would be an end point. *Any complex end point could be represented by a length of transmission line with a resistive termination. *Once that substitution was made, the problem should come back to the basic equations you presented here.

The characteristic impedance for a transmission line is
Zo = sqrt( (R + jwL) / (G + jwC) )

For a lossline (no resistance in the conductors, and no
conductance between the conductors), this simplifies to
Zo = sqrt( L / C )

So real lines actually have complex impedances. But the
math is simpler for ideal (lossless) lines and there is
much to be learned from studying the simplified examples.

But caution is needed when taking these results to
the real world of lines with loss.

...Keith

Yes, I concur with these comments.

The characteristic impedance can also be found from

Zo = 1/(C*Vel)

where C is the capacitance of the line per unit distance and Vel is the velocity of the wave.

This second solution for Zo demonstrates the power storage capabilities of the transmission line over time.

But as you say, real lines also have resistance losses and other losses so use great care when taking these results into the real world

--
73, Roger, W7WKB

The Rest of the Story
 

Roger Sparks wrote:
How can we justify calling the +/- 2*SQRT[Pf.g(t)*Pr.g(t) term the "interference term"?

Page 388, "Optics", by Eugene Hecht, 4th edition:
"The interference term becomes I12 = 2*SQRT(I1*I2)cos(A)"
where 'I' is the Irradiance (power density)[NOT Current]
Later Hecht says +2*SQRT(I1*I2) is the total constructive
interference term and -2*SQRT(I1*I2) is the total
destructive interference term. Chapter 9 is titled
"Interference" - recommended reading.

Am I correct in assuming that this equation describes the instantaneous power delivered to Rs?

Yes, if Tom, K7ITM, is correct about the equation working
for instantaneous power densities, not just for average
power densities as I had first assumed.

Let's say the instantaneous forward voltage dropped across
the source resistor is +50 volts and the instantaneous
reflected voltage across the source resistor is -30 volts.
The source resistor is 50 ohms.
Pf.rs(t) = (+50v)^2/50 = 50w
Pr.rs(t) = (-30v)^2/50 = 18w
Prs(t) = Pf.rs(t) + Pr.rs(t) - interference
Prs(t) = 50w + 18w - 2*SQRT(50*18) = 8 watts

If Tom is correct, that should be the actual dissipation
in the source resistor at that time which includes 60 watts
of destructive interference that will be dissipated 90 degrees
later when 2*SQRT(50*18) = +60 watts.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
So yes, the phrase "conservation of power" is
appropriately descriptive and follows from
conservation of energy.

You have a contradiction built into your concepts.
You have argued that the instantaneous power dissipated
in the source resistor is not equal to the instantaneous
forward power component plus the instantaneous reflected
power because power must be conserved at each instant of
time. That is simply not true.

I'm telling that energy must be conserved at each instant
of time but power does not have to be conserved at each
instant of time. Energy can obviously be stored in a
battery or network reactance for dissipation later in
time.

Do you require that the power used to charge a battery
be instantaneously dissipated in the battery? Of course
not! That's true for a dummy load but NOT for a battery.
There is no reason to require dissipation of power at
each instant of time to balance. Since energy can be
stored, there is no such thing as conservation of
instantaneous power, only of instantaneous energy.

Where is "cos(theta)" in this?
And what "theta" is to be used?

How many times do I have to explain this? For instantaneous
values of voltage, if the sign of the two interfering voltages
are the same, theta is zero degrees and the cosine of theta is
+1.0. If the sign of the two interfering voltages are opposite,
theta is 180 degrees and the cosine of theta is -1.0.

Is it the same theta that was used to conclude that
this was a "no interference" example?
See "A Simple Voltage Source - No Interference" at
http://www.w5dxp.com/intfr.htm

That theta is the phase angle of the phasor. If you
would switch over to phasor notation for your values
of instantaneous voltages, it might make things clearer
for you. You have been dealing only with real part of
the phasors which limits theta to either zero or 180
degrees.

So the plus/minus in the equation aboves means that
for part of the cycle we should add and part of the
cycle we should subtract.
This equation is ambiguous and is therefore incomplete
since it does not tell us when we should add and when
we should subtract.

I have told you about five times now. If the sign of the
two interfering voltages are the same, the interference
term has a positive sign (constructive). If the sign of
the two interfering voltages are opposite, the interference
term has a negative sign (destructive). Instead of me
having to keep posting this over and over how about you
keep reading it over and over until you understand it?

Could you provide a mathematically precise description
of when we should do each?

See above. If we deal with phasors, theta is the phase
angle between the two interfering voltages. If we deal
only with real values, theta is limited to either zero
(cosine = +1.0) or 180 degrees (cosine = -1.0).

Where is this destructive interference energy stored
while waiting to be dissipated constructively later?

In our 45 degree shorted stub example, it is stored in
the equivalent inductive reactance of the stub. For each
instant in which destructive interference energy is stored,
there is an instant in time 90 degrees later when that same
energy is dissipated as constructive interference power
in the source resistor. That's why 100% of the average
reflected power is dissipated in the source resistor for
these special cases of zero average interference.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
Cecil Moore wrote:
We are not discussing matched systems. If the total
RMS voltage differs from the RMS forward voltage,
then Vr is NOT zero.

True for this example. But it is generally unwise to pick
names or descriptions that will have to change when
the component values are changed.

Picking nits won't change the laws of physics.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Sun, 23 Mar 2008 09:03:31 -0500
Cecil Moore wrote:

Roger Sparks wrote:
How can we justify calling the +/- 2*SQRT[Pf.g(t)*Pr.g(t) term the "interference term"?

Page 388, "Optics", by Eugene Hecht, 4th edition:
"The interference term becomes I12 = 2*SQRT(I1*I2)cos(A)"
where 'I' is the Irradiance (power density)[NOT Current]
Later Hecht says +2*SQRT(I1*I2) is the total constructive
interference term and -2*SQRT(I1*I2) is the total
destructive interference term. Chapter 9 is titled
"Interference" - recommended reading.

Am I correct in assuming that this equation describes the instantaneous power delivered to Rs?

Yes, if Tom, K7ITM, is correct about the equation working
for instantaneous power densities, not just for average
power densities as I had first assumed.

Let's say the instantaneous forward voltage dropped across
the source resistor is +50 volts and the instantaneous
reflected voltage across the source resistor is -30 volts.
The source resistor is 50 ohms.
Pf.rs(t) = (+50v)^2/50 = 50w
Pr.rs(t) = (-30v)^2/50 = 18w
Prs(t) = Pf.rs(t) + Pr.rs(t) - interference
Prs(t) = 50w + 18w - 2*SQRT(50*18) = 8 watts

If Tom is correct, that should be the actual dissipation
in the source resistor at that time which includes 60 watts
of destructive interference that will be dissipated 90 degrees
later when 2*SQRT(50*18) = +60 watts.
--
73, Cecil http://www.w5dxp.com

Thanks for your thoughtful reply.

TanH(30/50) = 30.96 degrees. This takes us back to the 12.5 ohm load example.

Is it possible that in your example here, the reflected voltage acts in series with Rs but arrives 90 degrees out of phase with the forward voltage? If so, then

Vrs = sqrt(50^2 + 30^2) (the reflected voltage should ADD to the source voltage)

= sqrt(3400) = 58.31v

The power to Rs would be

Prs = (V^2)/50 = 3400/50 = 68w

We previously found that 32w was used at the 12.5 ohm load, so 32 + 68 = 100w. The entire output from the source is accounted for.

If this is the case, we have here an example of constructive interference, and complete accounting for the power.

You might wonder why I would consider this alternative. If the destructive interference included a 90 degree delay, how would I know whether the 30v was the delayed voltage or exactly in phase with the source? It must be delayed by 90 degrees because the forward voltage is always 90 degrees ahead of the reflected wave (in 45 degree line length example).

Your example certainly works as written, but it also introduces a dilemma. Where is the power stored for 90 degrees?

To answer that question, I see two possiblities: The source voltage causes a reflection so the 60w is stored as an additional reflected wave on the transmission line.

Or second, the 60w is stored in the source.

--
73, Roger, W7WKB

The Rest of the Story
 

On Sun, 23 Mar 2008 05:10:26 -0700, Roger Sparks
wrote:

Zo =3D 1/(C*Vel)

where C is the capacitance of the line per unit distance and Vel is the vel=
ocity of the wave.

This second solution for Zo demonstrates the power storage capabilities of =
the transmission line over time.

What is old is new again.

Hasn't this "formula" been put in the ground once before? For one,
what is velocity but something that has to be computed first only to
find us refilling all the terms back into this shortcut? For two,
"power storage capabilities... over time?" Apparently the stake
missed the heart of this one.

= A0 =A 0 = A0 = A 0 Vs = A 0 =A 0 = A0 = A0 =A 0 = A0 = A 0

73's
Richard Clark, KB7QHC

The Rest of the Story
 

Roger Sparks wrote:
(the reflected voltage should ADD to the source voltage)

If you graph the two voltages you will find that half the time
the reflected voltage adds to the source voltage and half the
time the reflected voltage subtracts from the source voltage.
Both are true half the time. You can point out either case on
the graph. That's why the average interference term is zero
for this special case and therefore why 100% of the average
reflected power is dissipated in the source resistor for
this special case.

You might wonder why I would consider this alternative.
If the destructive interference included a 90 degree delay,
how would I know whether the 30v was the delayed voltage or
exactly in phase with the source?

By looking at the graphs?

Where is the power stored for 90 degrees?

In the equivalent reactance of the transmission line.

That's what reactances do in AC circuits. They store
energy and deliver it back to the system some time
later.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Mar 23, 10:31 am, Cecil Moore wrote:
Keith Dysart wrote:
So yes, the phrase "conservation of power" is
appropriately descriptive and follows from
conservation of energy.

You have a contradiction built into your concepts.
You have argued that the instantaneous power dissipated
in the source resistor is not equal to the instantaneous
forward power component plus the instantaneous reflected
power because power must be conserved at each instant of
time.
That is simply not true.

Bzzzzzzzzzzztt.

I'm telling that energy must be conserved at each instant
of time but power does not have to be conserved at each
instant of time.

Bzzzzzzzzzzztt.

Energy can obviously be stored in a
battery or network reactance for dissipation later in
time.

Indeed. And forgetting to include such flows in the summation
would be a serious error.

For example, from the example,
Ps(t) = Prs(t) + Pg(t)
includes Pg(t) which accounts for the energy stored in the
line and later returned. There are no missing flows in
this equation. And Ps(t) also accounts for energy absorbed
in the voltage source. Only the source resistor has a
unidirectional energy flow.

Do you require that the power used to charge a battery
be instantaneously dissipated in the battery?

The energy flow into the battery is exactly and always
accounted for by the energy flow that heats the battery
and the energy flow consumed in the reversable chemical
reaction.

The instantaneous flows always sum appropriately to
satisfy conservation of energy. Of course, if one forgets
a flow, then the sum will not balance.

Of course
not! That's true for a dummy load but NOT for a battery.
There is no reason to require dissipation of power at
each instant of time to balance. Since energy can be
stored, there is no such thing as conservation of
instantaneous power,

Of course there is, but you must include the flows into
the elements that store energy as I have done.

only of instantaneous energy.

Where is "cos(theta)" in this?
And what "theta" is to be used?

How many times do I have to explain this? For instantaneous
values of voltage, if the sign of the two interfering voltages
are the same, theta is zero degrees and the cosine of theta is
+1.0. If the sign of the two interfering voltages are opposite,
theta is 180 degrees and the cosine of theta is -1.0.

A strange of way of looking at it. It seems easier just
to say that there is no theta. And add the voltages.

But no matter, I have figured out where your extra term
comes from.

Let us a consider a simple circuit with two voltage sources
(V.s1 and V.s2) in series, connected to a resistor R.

Using superposition we have
V.s1 = R * Ir.s1
and
V.s2 = R * Ir.s2

So
Vr.tot = V.s1 + V.s2
and
Ir.tot = I.s1 + I.s2
This is superposition, and all is well.

The power dissipated in the resistor is
Pr = (Vr.tot)**2 / R
but we could also derive Pr in terms of V.s1 and V.s2
Pr = (V.s1 + V.s2) (V.s1 + V.s2) / R
= ((V.s1)**2 + (V.s2)**2 + (2 * V.s1 * V.s2) ) / R
= (V.s1)**2 / R + (V.s2)**2 / R + (2 * V.s1 * V.s2) / R

Now some people attempt to compute a power for each of
the contributing voltages across the resistor and obtain
Pr.s1 = (V.s1)**2 / R
Pr.s2 = (V.s1)**2 / R
When these are added one obtains
Pr.false = (V.s1)**2 / R + (V.s2)**2 / R
which, by comparison with Pr above can easily be seen
not to be the power dissipated in the resistor. Pr.false
is missing the term ((2 * V.s1 * V.s2) / R) from Pr. It
is for this reason that it is said that one can not
superpose powers. Simply stated, when powers are dervived
from the constituent voltages that are superposed, it is
not valid to add the powers together to derive the total
power.

Of course for the most part, powers being added are not
powers derived from the constituent voltages of a total
voltage, so in most cases it is quite valid to add powers
and expect them to sum to the total power.

But what do you do if a circuit is superposing two
voltages and you are presented with information about
the circuit in terms of powers. Well then you can add
the powers and include a correction term.

Assume
Pr = Pr.false + Pr.correction
= Pr.s1 + Pr.s2 + Pr.correction
But can we find a Pr.correction? It has to correct for
the term missing from Pr.false but present in Pr,
i.e. ((2 * V.s1 * V.s2) / R).

Restated in terms of power, ((2 * V.s1 * V.s2) / R)
becomes
2 * sqrt(P.s1 * P.s2)
But sqrt has two solutions so we have to write
Pr = Pr.s1 + Pr.s2 +/- 2 * sqrt(P.s1 * P.s2)
which should look very familiar.

As you have correctly pointed out, the sign of
Pr.correction is negative when the signs of the
constituent voltages are different and positive when
they are the same. The reason for this can easily
be seen from the derivation of Pr.correction.

This Pr.correction term has nothing to do with
interference, it is the correction required when it is
desired to add two powers computed from the superposing
constituent voltages of an actual total voltage across
an element and derive the energy flow into the
element.

Note that there is no hint that Pr.correction needs to be
stored when it is negative nor come from somewhere when
it is positive. It is, after all, just a correction that
needs to be applied when one wants to compute the total
power given two powers derived from the constituent
voltages of superposition.

Since one needs to know the constituent voltages to
determine the sign of Pr.correction, why not just use
superposition to compute the total voltage and then
derive the power? It would be much simpler. With no
need for Pr.correction, interference, storage and
release of interference 'energy', ...
(Of course, over in optics land it is difficult to
measure the voltage, so suffering the pain of using
powers is probably appropriate).

This analysis also makes clear the nature of powers
computed from the constituent voltages of superposition.
These powers do not represent real energy flows. As
discussed far above, real energy flows can be summed
to test for conservation of energy. When energy flows
do not sum appropriately, then either an energy flow
is missing, or one is attempting to sum powers which
are not real, for example, having been computed from
the constituent voltages of superposition (e.g. Pfor
and Pref in a transmission line).

....Keith

The Rest of the Story
 

On Sun, 23 Mar 2008 19:57:24 GMT
Cecil Moore wrote:

Roger Sparks wrote:
(the reflected voltage should ADD to the source voltage)

If you graph the two voltages you will find that half the time
the reflected voltage adds to the source voltage and half the
time the reflected voltage subtracts from the source voltage.
Both are true half the time. You can point out either case on
the graph. That's why the average interference term is zero
for this special case and therefore why 100% of the average
reflected power is dissipated in the source resistor for
this special case.

You might wonder why I would consider this alternative.
If the destructive interference included a 90 degree delay,
how would I know whether the 30v was the delayed voltage or
exactly in phase with the source?

By looking at the graphs?

Where is the power stored for 90 degrees?

In the equivalent reactance of the transmission line.

That's what reactances do in AC circuits. They store
energy and deliver it back to the system some time
later.
--
73, Cecil http://www.w5dxp.com

I did not use a graph, but created a spreadsheet that calculated Vrs for the short circuit, 45 degree long line. It shows the 90 degree transfer of power that you described. I posted the spreadsheet in PDF format at http://www.fairpoint.net/~rsparks/Reflect_short.pdf.

To me, this shows that my traveling wave analysis on an instant basis is not correct because the energy can not be located precisely on a degree-by-degree scale. Yes, it is correct on the average over 360 degrees, but not instantaneously. We are missing something.

Central to traveling waves is the assumption that the wave is not compressable. The energy is assumed to flow in a consistantly predictable mannor that is linear and described by a sine wave. That assumption is violated when energy is delayed for reasons other than distance of travel, which is demonstrated in this example.

I am not ready to suggest a cure for my traveling wave analysis. I only see that it does not work to my expectations.

Thanks for providing the examples and comments.
--
73, Roger, W7WKB

The Rest of the Story
 

Keith Dysart wrote:
The energy flow into the battery is exactly and always
accounted for by the energy flow that heats the battery
and the energy flow consumed in the reversable chemical
reaction.

Point is, energy can be stored and released at a
later time. You earlier said that reactances do not
store energy for release at a later time yet that
is exactly what reactances do.

A strange of way of looking at it. It seems easier just
to say that there is no theta. And add the voltages.

Saying there is no theta is a shortcut that can get
one into trouble as it did with you. Since there is
no such thing as negative energy, there is also no
such thing as negative power. Note there are no
negative power signs in the power density equation
where 'theta' is the phase angle between the two
interfering voltages:

Ptot = P1 + P2 + 2*SQRT(P1*P2)cos(theta)

The last term is known as the "interference term",
page 388 of "Optics" by Hecht, 4th edition. When
90 theta 180, the sign of the last term is negative
indicating destructive interference. When
0 = theta 90, the sign of the last term is positive
indicating constructive interference. When theta = 90,
there is zero interference which is what Part 1 of my
web articles is based upon.

This Pr.correction term has nothing to do with
interference, ...

Your argument is not with me but rather is with
Eugene Hecht who defined that term as the "interference
term" in "Optics". Have you even read his chapter on
interference? If not, I would suggest that you do so.
Two other enlightening chapters are on "Superposition"
and "Coherency".

Note that there is no hint that Pr.correction needs to be
stored when it is negative nor come from somewhere when
it is positive.

You're correct, there's no hint. It is spelled out in
detail in "Optics". The possibilities are listed below.

Your above statement is a conceptual violation of the
conservation of energy principle. In the absence of
any other energy source or energy sink, localized
destructive interference must exactly match the
localized constructive interference magnitude in
order to avoid a violation of the conservation of
energy principle. This is why a Z0-match works.

Since one needs to know the constituent voltages to
determine the sign of Pr.correction, why not just use
superposition to compute the total voltage and then
derive the power?

That is what has extended this discussion to arguments
over the past quarter century. That 30,000 foot method
says nothing about where the ExH energy in the reflected
wave goes. The irradiance (power density) equation with
its defined "interference term" tells us exactly where
all the energy goes and answers the question: What happens
to the ExH energy in the reflected wave?

Here are the basic principles:

When destructive interference occurs, there is "extra"
energy left over from that isolated event. That energy
must go somewhere. Here are the possibilities in a
typical lossless RF transmitting system.

1. The source can throttle back on its energy output
to compensate for the destructive interference energy.

2. Reactive components can store the destructive
interference energy and return it to the network at
a later time.

3. In the absence of (1) and (2) above, an RF energy
wave is launched in a direction that allows the
"extra" energy to leave the destructive event area.

The last possibility is why we can observe reflected
energy being redistributed back toward the load in
the complete absence of single-wave reflections.

When constructive interference occurs, there is "missing"
energy needed to be supplied into that isolated event. That
energy must come from somewhere. Here are the possibilities
in a typical lossless RF transmitting system.

1. The source can simply supply the energy needed by
the constructive interference event.

2. Reactive components can return stored energy to
the network.

3. In the absence of (1) and (2) above, constructive
interference energy *must* be supplied in real time
by destructive interference between two other waves.

************************************************** *******
* The last possibility is how a Z0-match redistributes *
* all of the reflected energy back toward the load when *
* the physical reflection coefficient is not 1.0. *
************************************************** *******

The two-step process of redistributing 100% of the ExH reflected
wave energy back toward the load is covered in my other energy
analysis article on my web page at:

http://www.w5dxp.com/energy.htm

This analysis also makes clear the nature of powers
computed from the constituent voltages of superposition.
These powers do not represent real energy flows. As
discussed far above, real energy flows can be summed
to test for conservation of energy.

Translation: Don't bother trying to ascertain where the
ExH component wave energy goes. Since the conservation
of energy principle cannot be violated in reality, it
is OK to violate it conceptually. Now where have I
heard that argument before? :-)

"I personally don't have a compulsion to understand where
this power 'goes'."

Do you really think that the ExH energy in a reflection
from a mirror does not represent real energy flow?
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Roger Sparks wrote:
To me, this shows that my traveling wave analysis on an instant basis
is not correct because the energy can not be located precisely on a
degree-by-degree scale. Yes, it is correct on the average over 360
degrees, but not instantaneously. We are missing something.

What you are missing is the localized interference patterns
within the individual cycles. The interference changes from
destructive to constructive every 90 degrees. For every
negative (destructive) interference term, there is an equal
magnitude positive (constructive) interference term 90 degrees
later. These, of course, average out to zero. Exactly the
same thing happens when a coil or capacitor is present in
a circuit. When the instantaneous voltage of a source is
zero and thus delivering zero instantaneous power, a
circuit capacitor is delivering energy back into the
circuit that can be dissipated by a resistor.

Central to traveling waves is the assumption that the wave is not
compressable. The energy is assumed to flow in a consistantly
predictable mannor that is linear and described by a sine wave.
That assumption is violated when energy is delayed for reasons
other than distance of travel, which is demonstrated in this example.

Power is certainly compressible. One can stuff 100 amphere-
hours into a battery in 2 hours and take 20 hours to remove it.
Why can't 60 watts of instantaneous power be stuffed into
a reactance and be removed 90 degrees later?

I am not ready to suggest a cure for my traveling wave analysis. I
only see that it does not work to my expectations.

Your expectations seem to be based on a conservation of
power principle which doesn't exist. There is no violation
of linearity if the energy dissipation is delayed by 90
degrees or by ten billion years.

I don't recall any published material where anyone tried
to explain where the instantaneous energy goes while at
the same time denying the possibility of interference.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Mar 24, 10:46 am, Cecil Moore wrote:
Keith Dysart wrote:
The energy flow into the battery is exactly and always
accounted for by the energy flow that heats the battery
and the energy flow consumed in the reversable chemical
reaction.

Point is, energy can be stored and released at a
later time. You earlier said that reactances do not
store energy for release at a later time yet that
is exactly what reactances do.

Yes indeed. And what I have said, is that when this is
happening it is always possible to identify the element
which is storing the energy and provide the function
that describes the energy flow in and out of the
element. It is this identification and function that
I keep asking for to back up the handwaving claim that
you have been making.

A strange of way of looking at it. It seems easier just
to say that there is no theta. And add the voltages.

Saying there is no theta is a shortcut that can get
one into trouble as it did with you. Since there is
no such thing as negative energy, there is also no
such thing as negative power.

Bzzt. Power is the rate of change of energy. The
quantity of energy can be dropping (i.e. negative power),
without the quantity of energy ever going below zero.

Note there are no
negative power signs in the power density equation
where 'theta' is the phase angle between the two
interfering voltages:

Ptot = P1 + P2 + 2*SQRT(P1*P2)cos(theta)

Unfortunately, I took a small shortcut in my last
post and left out the "(t)" from all the functions.
You immediately jumped to an RMS interpretation.

Please re-read all the equations with "(t)". There
is no "cos(theta)" factor when "(t)" is present.

The last term is known as the "interference term",
page 388 of "Optics" by Hecht, 4th edition. When
90 theta 180, the sign of the last term is negative
indicating destructive interference. When
0 = theta 90, the sign of the last term is positive
indicating constructive interference. When theta = 90,
there is zero interference which is what Part 1 of my
web articles is based upon.

But this applies to RMS voltages and average powers.
You have extended this to instantaneous, for which a
"cos(theta)" factor is inappropriate.

This Pr.correction term has nothing to do with
interference, ...

Your argument is not with me but rather is with
Eugene Hecht who defined that term as the "interference
term" in "Optics". Have you even read his chapter on
interference? If not, I would suggest that you do so.
Two other enlightening chapters are on "Superposition"
and "Coherency".

Read it as Pr.correction(t) to emphasize that it is not
average power of which I am writing. Then it is not
interference.

Note that there is no hint that Pr.correction needs to be
stored when it is negative nor come from somewhere when
it is positive.

You're correct, there's no hint. It is spelled out in
detail in "Optics". The possibilities are listed below.
Your above statement is a conceptual violation of the
conservation of energy principle.

Of course not. Because the powers imputed to the
constituent voltages of superposition do not represent
actual energy flows. Conservation of energy only
applies to powers that represent actual energy flows.

In the absence of
any other energy source or energy sink, localized
destructive interference must exactly match the
localized constructive interference magnitude in
order to avoid a violation of the conservation of
energy principle. This is why a Z0-match works.

But you have to be cautious that you are applying
conservation to powers that represent actual energy
flows.

Since one needs to know the constituent voltages to
determine the sign of Pr.correction, why not just use
superposition to compute the total voltage and then
derive the power?

That is what has extended this discussion to arguments
over the past quarter century. That 30,000 foot method
says nothing about where the ExH energy in the reflected
wave goes. The irradiance (power density) equation with
its defined "interference term" tells us exactly where
all the energy goes and answers the question: What happens
to the ExH energy in the reflected wave?

It would be more valuable were you to thoroughly study
and understand what is happening in a transmission line
and then apply those learnings to ExH. The transmission
line is easier to understand. The voltages, currents and
time relationships can easily be precisely computed and
measured. Once you have gained a full understanding of
what power means in this easier to follow environment,
extend that understanding to the meaning of power in
an ExH, or optics environment where calculation and
measurement is much more difficult.

Here are the basic principles:

When destructive interference occurs, there is "extra"
energy left over from that isolated event. That energy
must go somewhere. Here are the possibilities in a
typical lossless RF transmitting system.

1. The source can throttle back on its energy output
to compensate for the destructive interference energy.

2. Reactive components can store the destructive
interference energy and return it to the network at
a later time.

3. In the absence of (1) and (2) above, an RF energy
wave is launched in a direction that allows the
"extra" energy to leave the destructive event area.

Or perhaps, these powers of which you speak do not
represent actual energy flows and therefore your
requirement that they need accounting is incorrect
and all of your attempts to explain them, unnecessary.

The difficulty of accounting for these powers is entirely
consistent with them not representing the actual flow
of energy.

The last possibility is why we can observe reflected
energy being redistributed back toward the load in
the complete absence of single-wave reflections.

When constructive interference occurs, there is "missing"
energy needed to be supplied into that isolated event. That
energy must come from somewhere. Here are the possibilities
in a typical lossless RF transmitting system.

1. The source can simply supply the energy needed by
the constructive interference event.

2. Reactive components can return stored energy to
the network.

3. In the absence of (1) and (2) above, constructive
interference energy *must* be supplied in real time
by destructive interference between two other waves.

Or possibly, the premise that these powers represent
actual energy flows is flawed.

************************************************** *******
* The last possibility is how a Z0-match redistributes *
* all of the reflected energy back toward the load when *
* the physical reflection coefficient is not 1.0. *
************************************************** *******

The two-step process of redistributing 100% of the ExH reflected
wave energy back toward the load is covered in my other energy
analysis article on my web page at:

http://www.w5dxp.com/energy.htm

This turns out, however, just to be an ideosyncracy of the math,
much like the way Pf-Pr is the actual energy flow in the transmission
line because of the way that Vf and Vr are derived from Vactual
and Iactual.

This analysis also makes clear the nature of powers
computed from the constituent voltages of superposition.
These powers do not represent real energy flows. As
discussed far above, real energy flows can be summed
to test for conservation of energy.

Translation: Don't bother trying to ascertain where the
ExH component wave energy goes. Since the conservation
of energy principle cannot be violated in reality, it
is OK to violate it conceptually. Now where have I
heard that argument before? :-)

"I personally don't have a compulsion to understand where
this power 'goes'."

Do you really think that the ExH energy in a reflection
from a mirror does not represent real energy flow?

What can I say? That is what the math proves. The
reflected power is a power computed from partial E and
H fields that are being superposed, and we know that when
you superpose, you need to compute the total voltage and
current (or E and H) and then use that to compute the
actual energy flow.

It would be good, if just for a day, you let go of the
idea that Preflected represents an actual energy flow.
Explore the actual measureable behaviour of transmission
lines without using the idea that Preflected represents
an actual energy flow.

Everything works. There is no violation of conservation
of energy or any other fundamental physical law.

And the explanations are much simpler. You will no longer
find the question "where does the reflected power go?"
relevant. You can terminate your quest.

And as for "2*SQRT(P1*P2)cos(theta)", this will just be
an idiosyncracy of the math that allows you to compute
the total power, if you are presented with P1 and P2 (not
being actual powers) that were computed from the
constituent voltages of superposition; a useful tool
when you can not measure the voltages (e.g. in optics),
but not to be confused with reality.

....Keith

The Rest of the Story
 

On Mon, 24 Mar 2008 16:15:21 GMT
Cecil Moore wrote:

Roger Sparks wrote:
To me, this shows that my traveling wave analysis on an instant basis
is not correct because the energy can not be located precisely on a
degree-by-degree scale. Yes, it is correct on the average over 360
degrees, but not instantaneously. We are missing something.

What you are missing is the localized interference patterns
within the individual cycles. The interference changes from
destructive to constructive every 90 degrees. For every
negative (destructive) interference term, there is an equal
magnitude positive (constructive) interference term 90 degrees
later. These, of course, average out to zero. Exactly the
same thing happens when a coil or capacitor is present in
a circuit. When the instantaneous voltage of a source is
zero and thus delivering zero instantaneous power, a
circuit capacitor is delivering energy back into the
circuit that can be dissipated by a resistor.

Central to traveling waves is the assumption that the wave is not
compressable. The energy is assumed to flow in a consistantly
predictable mannor that is linear and described by a sine wave.
That assumption is violated when energy is delayed for reasons
other than distance of travel, which is demonstrated in this example.

Power is certainly compressible. One can stuff 100 amphere-
hours into a battery in 2 hours and take 20 hours to remove it.
Why can't 60 watts of instantaneous power be stuffed into
a reactance and be removed 90 degrees later?

I am not ready to suggest a cure for my traveling wave analysis. I
only see that it does not work to my expectations.

Your expectations seem to be based on a conservation of
power principle which doesn't exist. There is no violation
of linearity if the energy dissipation is delayed by 90
degrees or by ten billion years.

I don't recall any published material where anyone tried
to explain where the instantaneous energy goes while at
the same time denying the possibility of interference.
--
73, Cecil http://www.w5dxp.com

Hi Cecil,

I feel better today. I think I have connected the dots and now have the spreadsheet showing that we really can use the traveling waves to solve the shorted transmission line problem on a instantaneous basis without the delay of energy into the next half cycle.

Here is a link to the new spreadsheet.

http://www.fairpoint.net/~rsparks/Sm...Reflection.pdf

I used the logic and formula presented in my post "Subject: The Rest of the Story
Date: Thu, 20 Mar 2008 10:07:44 -0700"

You called it interference. Keith used your power equation and called the interference term a mathematical correction. It looks to me like the correction can be avoided by choosing the correct sin wave offset.

Ultimately, the waves can be resolved into one more powerful wave carrying the power described by Keith's "false power" equation. This is demonstrated in a spreadsheet found at

http://www.fairpoint.net/~rsparks/Re...em%20Power.pdf

You need to take a look at the spreadsheets. I think they support the theory that we can track the power on an instant basis using traveling waves.
--
73, Roger, W7WKB

The Rest of the Story
 

Keith Dysart wrote:
Cecil Moore wrote:
Point is, energy can be stored and released at a
later time. You earlier said that reactances do not
store energy for release at a later time yet that
is exactly what reactances do.

Yes indeed. And what I have said, is that when this is
happening it is always possible to identify the element
which is storing the energy and provide the function
that describes the energy flow in and out of the
element. It is this identification and function that
I keep asking for to back up the handwaving claim that
you have been making.

Good grief, Keith, do you not know how to track the
energy flow into and out of a reactance during an RF
cycle? Isn't that covered in EE201, "Alternating
Current Circuits", by Kerchner and Corcoran, 3rd
edition (c)1951? Quoting page 19:

"The implication is that the inductive element
receives energy from the source during one-quarter
of a cycle of the applied voltage and returns exactly
the same amount of energy to the driving source during
the next one-quarter of a cycle."

The equations are provided if you really need them.
Hint: A shorted 1/8WL stub is inductive.

Bzzt. Power is the rate of change of energy.

Sorry, you are wrong about that.
From the IEEE Dictionary: "power - the rate of
generating, transferring, or using energy".

Power is a rate, not a rate of change. The
energy flow can be constant, i.e. rate of change
equal zero.

Please re-read all the equations with "(t)". There
is no "cos(theta)" factor when "(t)" is present.

I assume that exponential (phasor) notation for the
instantaneous values of the interfering voltages could
be used in which case there would indeed be a cos(theta)
present.

But this applies to RMS voltages and average powers.
You have extended this to instantaneous, for which a
"cos(theta)" factor is inappropriate.

Nope, it's not. See above. Your same argument could
extend to the real part of phasors in which case you
could argue that the irradiance equation is bogus.
Good luck on that one. If the math didn't work, it
would have been discarded long ago and Hecht wouldn't
have an entire chapter devoted to "Interference".

Read it as Pr.correction(t) to emphasize that it is not
average power of which I am writing. Then it is not
interference.

That statement makes it obvious that you don't understand
interference. When instantaneous values are being used,
if [V1(t)^2 + V2(t)^2] NOT= [V1(t) + V2(t)]^2, then
interference is present. Did you miss Physics 201?

Because the powers imputed to the
constituent voltages of superposition do not represent
actual energy flows.

That statement is a violation of the wave reflection
model. Do you really believe that when you look
yourself in the mirror that those reflections are
devoid of energy? If so, please feel free to prove
your assertion.

But you have to be cautious that you are applying
conservation to powers that represent actual energy
flows.

Reflected waves contain energy whether from your mirror
or from a mismatched load at the end of a transmission
line. You are arguing that the wave reflection model
is wrong. Please prove it.

Or perhaps, these powers of which you speak do not
represent actual energy flows and therefore your
requirement that they need accounting is incorrect
and all of your attempts to explain them, unnecessary.

Yes, perhaps the wave reflection model is wrong but
that makes your argument not with me, but with Ramo,
Whinnery, Johnson, Chipman, Slater, Hecht, and Walter
Maxwell. Good luck on winning that one.

The difficulty of accounting for these powers is entirely
consistent with them not representing the actual flow
of energy.

No, it is perfectly consistent with a large degree
of ignorance which few people desire to alleviate.

Ignoring the role of interference and lumping all the
energy components into a mashed potato salad is one
method of sweeping everything under the rug so you
can ignore the problem instead of solving it.

Or possibly, the premise that these powers represent
actual energy flows is flawed.

Feel free to prove the wave reflection model wrong.

This turns out, however, just to be an ideosyncracy of the math,
much like the way Pf-Pr is the actual energy flow in the transmission
line because of the way that Vf and Vr are derived from Vactual

Feel free to prove the math wrong.

It would be good, if just for a day, you let go of the
idea that Preflected represents an actual energy flow.

I will do that the day that you prove those reflections
from your mirror, that allow you to see yourself each
morning, contain zero energy. Your concepts seem more
like a religion than anything associated with reality.
Your mantra seems to be: "If I don't understand it,
it doesn't exist."
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Roger Sparks wrote:
You need to take a look at the spreadsheets.

Roger, in a nutshell, what is the bottom line?
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Cecil Moore wrote:

[snip]


That statement is a violation of the wave reflection
model. Do you really believe that when you look
yourself in the mirror that those reflections are
devoid of energy? If so, please feel free to prove
your assertion.


[snip]


Yes, perhaps the wave reflection model is wrong but
that makes your argument not with me, but with Ramo,
Whinnery, Johnson, Chipman, Slater, Hecht, and Walter
Maxwell. Good luck on winning that one.

Cecil,

I have completely stayed away from any comments in this thread, but one
thing has caught my attention. You repeatedly refer to the "wave
reflection model" almost as if it was the equivalent of the Grand
Unified Theory.

I did a little bit of the cheap modern day replacement for research by
Googling "wave reflection model". Some 455 references came back. Of
course many of those references are irrelevant to the topic at hand, but
of the ones that seemed to be related to transmission lines and/or RF,
it appears that all of the references come back to you.

I tried adding Ramo, Whinnery, Johnson, Chipman, Slater, Hecht, and
Walter Maxwell to the search (one at a time) and came up with no hits,
other than your references.

Perhaps your specific interpretation of the writings of the masters to
form your "wave reflection model" is part of the source of the disagreement.

Do you have a concise reference to exactly what you mean by "wave
reflection model"?

There is no need to explain reflection, interference, conservation of
energy, or any other standard wave phenomenon. I am merely seeking the
definition of "wave reflection model" or a lead to any other reference
that uses that terminology.

73,
Gene
W4SZ

The Rest of the Story
 

Gene Fuller wrote:
I am merely seeking the
definition of "wave reflection model" or a lead to any other reference
that uses that terminology.

The wave reflection model is, of course, the body of
mathematics encompassing the reflection mechanics for
EM waves. Everywhere except on r.r.a.a, it is a subset
of the distributed network model.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Gene Fuller wrote:
I did a little bit of the cheap modern day replacement for research by
Googling "wave reflection model". Some 455 references came back.

That's strange. When I did the identical thing, 1,970,000
references came back. Wonder what is wrong with your computer?
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Cecil Moore wrote:
Gene Fuller wrote:
I did a little bit of the cheap modern day replacement for research by
Googling "wave reflection model". Some 455 references came back.

That's strange. When I did the identical thing, 1,970,000
references came back. Wonder what is wrong with your computer?

When quotes are used, I get 455. Without quotes, I get about 297,000.
Since I am looking for the exact expression, not just the three words
somewhere in a web page, I use the quotes.

I am using Firefox with plain 'ol Google. No special setups.

In any case, you answered my question. "Wave reflection model" means
nothing beyond ordinary electromagnetic theory. Big deal.

73,
Gene
W4SZ

The Rest of the Story
 

On Mar 9, 6:33 pm, Chuck wrote:

On Sun, 9 Mar 2008 15:07:26 -0700 (PDT), K7ITM wrote:
Note that, as far as I've been able to determine, Michelson did not
have a coherent light source to shine into his interferometer, but
still he saw interference patterns. Perhaps he had invented lasers

It is said he used sodium vapor gas light (~589 nm). Coherent enough.
Chuck

Just a slight addition here. Before lasers, the way to get a
coherent light source was to bottle-up a high-intensity,
monochromatic source, such as the aforementioned sodium-
vapor light, in a reflective cavity with a very small pinhole in
its side. As the photons dribble out through the pinhole, they
are forced into a somewhat phase-coherent wave train. This
source was used in optical processors for synthetic-aperture
imagery back in the 50's.... Jim, K7JEB

The Rest of the Story
 

On Mar 25, 10:28*am, Cecil Moore wrote:
Keith Dysart wrote:
Cecil Moore wrote:
Point is, energy can be stored and released at a
later time. You earlier said that reactances do not
store energy for release at a later time yet that
is exactly what reactances do.

Yes indeed. And what I have said, is that when this is
happening it is always possible to identify the element
which is storing the energy and provide the function
that describes the energy flow in and out of the
element. It is this identification and function that
I keep asking for to back up the handwaving claim that
you have been making.

Good grief, Keith, do you not know how to track the
energy flow into and out of a reactance during an RF
cycle? Isn't that covered in EE201, "Alternating
Current Circuits", by Kerchner and Corcoran, 3rd
edition (c)1951? Quoting page 19:

"The implication is that the inductive element
receives energy from the source during one-quarter
of a cycle of the applied voltage and returns exactly
the same amount of energy to the driving source during
the next one-quarter of a cycle."

The equations are provided if you really need them.
Hint: A shorted 1/8WL stub is inductive.

Instead of writing two lines identifying the element
and providing the function describing its energy flow,
you write 14 lines tell me I should do it.

But my explanations do not require this element to
store and return the "interference energy".

You should consider that perhaps your inability to
identify the element and its energy function really
calls into question your concept of "interference
energy" being stored and returned later.

Bzzt. Power is the rate of change of energy.

Sorry, you are wrong about that.
*From the IEEE Dictionary: "power - the rate of
generating, transferring, or using energy".

Power is a rate, not a rate of change. The
energy flow can be constant, i.e. rate of change
equal zero.

When energy is transferred, the quantity is decreasing
in the supplier and increasing in the receiver. From
the supplier's perspective, this is a negative flow and
from the receiver's perspective, a positive flow.

In calculus terms, energy flow is the derivative of the
quantity of energy, i.e. the rate of change of the
amount of energy. The slope of the curve recording the
amount of energy can be negative, even though the amount
of energy is always positive.

Please re-read all the equations with "(t)". There
is no "cos(theta)" factor when "(t)" is present.

I assume that exponential (phasor) notation for the
instantaneous values of the interfering voltages could
be used in which case there would indeed be a cos(theta)
present.

No. "cos(theta)" only appears in the equations describing
the average, and not in those equations that describe the
actual function of time.

[snip]
Read it as Pr.correction(t) to emphasize that it is not
average power of which I am writing. Then it is not
interference.

That statement makes it obvious that you don't understand
interference. When instantaneous values are being used,
if [V1(t)^2 + V2(t)^2] NOT= [V1(t) + V2(t)]^2, then
interference is present. Did you miss Physics 201?

I suppose, if you want to rename superposition as interference.
But none of my basic circuit theory books use the word
interference when discussing superposition.

Because the powers imputed to the
constituent voltages of superposition do not represent
actual energy flows.

That statement is a violation of the wave reflection
model. Do you really believe that when you look
yourself in the mirror that those reflections are
devoid of energy? If so, please feel free to prove
your assertion.

If the powers imputed to the constituent voltages of
superposition did represent actual energy flows, then
you would be able to simply add them to get the total
flow, since energy can not be created or destroyed.

The fact that a correction needs to be applied when
adding them is proof that they can not be actual energy
flows.

But you know that, and that is why you have to search
for where this correction, that which you call the
"interference energy", goes. Because only if you can
account for it, can you claim that it is an actual
energy flow, which is needed to make you explanations
agree with conservation of energy. But in this example
you can not account for this "interference energy".
You have not identified the element that stores it nor
being able to obtain a function which describes the flow
into that element. You should take this as a reason to
call into question the whole idea that this "interference
energy" is an actual energy flow.

But you have to be cautious that you are applying
conservation to powers that represent actual energy
flows.

Reflected waves contain energy whether from your mirror
or from a mismatched load at the end of a transmission
line. You are arguing that the wave reflection model
is wrong. Please prove it.

If *your* "wave reflection model" includes the idea that
Pref always represents an actual energy flow, then *your*
"wave reflection model" is wrong.

Or perhaps, these powers of which you speak do not
represent actual energy flows and therefore your
requirement that they need accounting is incorrect
and all of your attempts to explain them, unnecessary.

Yes, perhaps the wave reflection model is wrong but
that makes your argument not with me, but with Ramo,
Whinnery, Johnson, Chipman, Slater, Hecht, and Walter
Maxwell. Good luck on winning that one.

I am not convinced. It is clear that *your* "wave reflection
model" is wrong, but I have not seen these other authors
invest any effort in trying to explain where the reflected
power goes. Perhaps they realized it was a meaningless
question and their "wave reflection models" do not require
that the Pref represent an actual energy flow.

The difficulty of accounting for these powers is entirely
consistent with them not representing the actual flow
of energy.

No, it is perfectly consistent with a large degree
of ignorance which few people desire to alleviate.

Ignoring the role of interference and lumping all the
energy components into a mashed potato salad is one
method of sweeping everything under the rug so you
can ignore the problem instead of solving it.

I am still waiting for the simple answer as to which
element stores and returns this "interference energy" and
the function that describes the flow into this element.

Or possibly, the premise that these powers represent
actual energy flows is flawed.

Feel free to prove the wave reflection model wrong.

That your "wave reflection model" is inconsistent with
conservation of energy (until you identify the storage
element and its energy transfer function) should be proof
enough.

This turns out, however, just to be an ideosyncracy of the math,
much like the way Pf-Pr is the actual energy flow in the transmission
line because of the way that Vf and Vr are derived from Vactual

Feel free to prove the math wrong.

The math is correct. It is the interpretation that is in error.

Pf-Pr is always equal to Pnet simply because of the way that Vfor
and Vref are computed. Even though Pf-Pr adds to the actual measured
energy flow, it does not mean that Pf and Pr are actual energy flows.
They MUST add simply becase of the way they are computed.

The same is true for some of the "proofs" in your other papers. The
successful equalities are simply a consequence of the way the numbers
being added are computed. A successful equality does not necessarily
prove an interpretation.

It would be good, if just for a day, you let go of the
idea that Preflected represents an actual energy flow.

I will do that the day that you prove those reflections
from your mirror, that allow you to see yourself each
morning, contain zero energy.

This is indeed the root of the problem. You need to let go
of the mirror just long enough to get over the hump. As it
stands, whenever you approach the hump, you think about the
mirror and refuse to see what might be on the other side of
the hump. That is why I suggest letting go of the mirror
just for a short while. Explore to see if there exists a
completely self consistent set of explanations on the other
side of the hump. You will find it to be so. But this can
only happen if you let go of the mirror long enough to get
over the hump.

If it doesn't work out, you can always pick up the mirror
again. There is nothing to lose by temporarily doing some
exploration without the mirror.

[snip]

...Keith

The Rest of the Story
 

On Mar 25, 10:28 am, Cecil Moore wrote:
Keith Dysart wrote:
Read it as Pr.correction(t) to emphasize that it is not
average power of which I am writing. Then it is not
interference.

That statement makes it obvious that you don't understand
interference. When instantaneous values are being used,
if [V1(t)^2 + V2(t)^2] NOT= [V1(t) + V2(t)]^2, then
interference is present. Did you miss Physics 201?

Let us build a slightly better example that complies with
your "NOT =" expression above.

+-----------------------------------+
| |
Vs1(t) = 141.4cos(wt) \
| = 100 Vrms Rload /
| 50 ohms \
Vs2(t) = 70.7cos(wt) /
| = 50 Vrms |
+-----------------------------------+

Using superposition the contribution from source 1 is
Vload.s1 = 100 Vrms
Iload.s1 = 2 Arms
and from source 2 is
Vload.s2 = 50 Vrms
Iload.s2 = 1 Arms
combining
Vload = 150 Vrms
Iload = 3 Arms

From
Pload = Vload * Iload
= 150 * 3
= 450 Waverage

As can be seen, this example satisfies your requirement
for interference:
[V1(t)^2 + V2(t)^2] NOT= [V1(t) + V2(t)]^2

Computing the imputed powers for the waves from each source
we have
Pload.s1 = 100 * 2
= 200 Waverage
Pload.s2 = 50 * 1
= 50 Waverage
To obtain the power in the load from these imputed powers
we need to use
Pload = Pload.s1 + Pload.s2 + Pload.correction
450 = 200 + 50 + Pload.correction
200 = Pload.correction

From previous analysis
Pcorrection = 2 * sqrt(P1 * P2)cos(theta)
(the cos(theta) term is appropriate here because these
are average powers being used)
Pcorrection = 2 * sqrt(10000) * 1
= 200
as required from above.

So according to your energy analsysis, the power in
the load comes from
the wave from source 1 = 200 W
the wave from source 2 = 50 W
"interference energy" = 200 W
for a total of
450 W as required.

Now if the 200 W from the wave from source 1 and the
50 W from the wave from source 2 represent actual
energy flows, then the "interference energy" must
also be an actual energy flow to satisfy conservation
of energy.

What element provides the energy for this "interference
energy" flow? Note that in this analysis, this energy
is an average energy flow, so it can not be saved during
part of the cycle and returned during another part.

In other posts, you have suggested that this would be
a constructive interference energy and that there would
be an equal destructive interference energy to provide
it. If you still claim this, where is this destructive
interference happening?

....Keith

The Rest of the Story
 

Gene Fuller wrote:
In any case, you answered my question. "Wave reflection model" means
nothing beyond ordinary electromagnetic theory. Big deal.

It is a big deal when someone asserts that there is
no energy in a reflected EM wave or that the energy
in a reflected EM wave doesn't have to be conserved
or that a reflected wave doesn't obey the rules
of the wave reflection model.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

"Jim, K7JEB" wrote in message
...
....This
source was used in optical processors for synthetic-aperture
imagery back in the 50's....

That should have read "synthetic-aperture radar imagery"...
Just trying to not let my mind outrun my typing speed.
K7JEB

The Rest of the Story
 

Cecil Moore wrote:
Gene Fuller wrote:
I did a little bit of the cheap modern day replacement for research by
Googling "wave reflection model". Some 455 references came back.

That's strange. When I did the identical thing, 1,970,000
references came back. Wonder what is wrong with your computer?

I used the quote marks around the words and got 266 references.
Few of them, other than yours, relate to electronics at all, much less
radio.

Dave K8MN

The Rest of the Story
 

Keith Dysart wrote:
You should consider that perhaps your inability to
identify the element and its energy function really
calls into question your concept of "interference
energy" being stored and returned later.

I have previously multiple times identified the element
as the network reactance and pointed you to a reference.

If you were asking me to teach you the English
language on r.r.a.a, you would get the same response.

I suppose, if you want to rename superposition as interference.
But none of my basic circuit theory books use the word
interference when discussing superposition.

I'm not renaming superposition. I'm using the
definition of "interference" provided by Hecht
in "Optics". Superposition can occur with or
without interference. The present discussion is
about superposition with interference present.
Interference is just a word which identifies
the special case of superposition that is
under discussion.

If the powers imputed to the constituent voltages of
superposition did represent actual energy flows, then
you would be able to simply add them to get the total
flow, since energy can not be created or destroyed.

There you go again, confusing power and energy.
There is *NO* conservation of power principle.
Until you give up on superposing powers, you
are doomed to failure. There is absolutely nothing
wrong with storing energy and turning it into
power later in time. Do you think that backup
storage batteries are a violation of the
conservation of energy principle?

The fact that a correction needs to be applied when
adding them is proof that they can not be actual energy
flows.

There is *NO CORRECTION TO THE ENERGY COMPONENTS*. There
is only a correction to the power components to account
for the time the energy is being stored before it is
dissipated. You really need to learn the difference
between energy and power.

You should take this as a reason to
call into question the whole idea that this "interference
energy" is an actual energy flow.

Your argument is not with me - it is with experts
like Eugene Hecht. Please read his *57 page* Chapter
9 on "Interference" and then get back to us. With 57
pages devoted to the subject, Hecht doesn't seem to
share your problems with it.

If *your* "wave reflection model" includes the idea that
Pref always represents an actual energy flow, then *your*
"wave reflection model" is wrong.

When you can prove that reflected traveling waves contain
zero energy, i.e. that ExH=0, I will accept your assertion
but not before. Exactly how does a TDR detect zero energy?

In fact, the thing you need to do is forget the transmission
line and deal with light waves encountering boundaries with
different indexes of refraction. The problem is identical,
but dealing with light out in the open prohibits you from
pushing your mashed-potatoes energy religion.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
What element provides the energy for this "interference
energy" flow? Note that in this analysis, this energy
is an average energy flow, so it can not be saved during
part of the cycle and returned during another part.

Already answered a few postings ago. Sorry you
missed it. Here it is again:

Here are the basic principles:

When destructive interference occurs, there is "extra"
energy left over from that isolated event. That energy
must go somewhere. Here are the possibilities in a
typical lossless RF transmitting system.

1. The source can throttle back on its energy output
to compensate for the destructive interference energy.

2. Reactive components can store the destructive
interference energy and return it to the network at
a later time.

3. In the absence of (1) and (2) above, an RF energy
wave is launched in a direction that allows the
"extra" energy to leave the destructive event area.

When constructive interference occurs, there is "missing"
energy needed to be supplied into that isolated event. That
energy must come from somewhere. Here are the possibilities
in a typical lossless RF transmitting system.

1. The source can simply supply the energy needed by
the constructive interference event.

2. Reactive components can return stored energy to
the network.

3. In the absence of (1) and (2) above, constructive
interference energy *must* be supplied in real time
by destructive interference between two other waves.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
Now if the 200 W from the wave from source 1 and the
50 W from the wave from source 2 represent actual
energy flows, then the "interference energy" must
also be an actual energy flow to satisfy conservation
of energy.

One other observation: Although the interference model
will work for a lumped circuit example, there is no
reason to use it as it complicates the computations
and adds nothing to the solution. The wave reflection
model also works for circuits but there is simply no
good reason to use it for lumped circuit analysis.

Where interference becomes a useful tool is when it
happens away from any compensating source. An analysis
of the interference of two EM light waves in free space
far removed from any source leaves us with two constant
sources of energy, the total energy of which has to go
somewhere. The following two web pages tell us exactly
where the energy goes.

http://www.mellesgriot.com/products/optics/oc_2_1.htm

http://micro.magnet.fsu.edu/primer/j...ons/index.html

Your theory seems to require that the EM waves must
know beforehand whether to carry energy or not from
a star light years away. You apparently have invented
a rather curious "smart wave theory".

The question is: How did those two interfering waves
from Alpha Centauri know whether to arrive at the
planet Earth ten years later carrying ExH energy
or not carrying ExH energy?
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
Let us build a slightly better example that complies with
your "NOT =" expression above.

Your constant voltage sources are NOT a better example
and not even a good example. To confront the subject
being discussed, you should use constant power sources.

The reason is obvious. A constant steady-state EM
wave is a constant average power source, not a constant
voltage source. Constant voltage source examples
just won't do.

Try your example with constant power sources and
see what happens.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Mar 26, 11:43*am, Cecil Moore wrote:
Keith Dysart wrote:
You should consider that perhaps your inability to
identify the element and its energy function really
calls into question your concept of "interference
energy" being stored and returned later.

I have previously multiple times identified the element
as the network reactance and pointed you to a reference.

Indeed you have, but you have not answered the second clause
in the question. Until you can provide the energy flow
function of the element you claim is storing the energy
there is no reason to believe that it is the element.

Can it be that hard to provide the function?
Or perhaps the element you have identified does not have
the appropriate energy flow function? (It doesn't.)

I suppose, if you want to rename superposition as interference.
But none of my basic circuit theory books use the word
interference when discussing superposition.

I'm not renaming superposition. I'm using the
definition of "interference" provided by Hecht
in "Optics". Superposition can occur with or
without interference. The present discussion is
about superposition with interference present.
Interference is just a word which identifies
the special case of superposition that is
under discussion.

Sure. OK.

If the powers imputed to the constituent voltages of
superposition did represent actual energy flows, then
you would be able to simply add them to get the total
flow, since energy can not be created or destroyed.

There you go again, confusing power and energy.
There is *NO* conservation of power principle.

Conservation of energy arises from the inability
to create or destroy energy. Energy can only flow from
one element to another.

This requires that the sum of the flows out of the
elements providing energy equals the sum of the flows
into the elements receiving the energy. Choose
whatever name you want for it, but this is the
reason that
Ps(t) must equal Prs(t) + Pg(t)

And it is just as powerful a concept as conservation
of energy since it follows directly from that principle.

The fact that a correction needs to be applied when
adding them is proof that they can not be actual energy
flows.

There is *NO CORRECTION TO THE ENERGY COMPONENTS*. There
is only a correction to the power components to account
for the time the energy is being stored before it is
dissipated. You really need to learn the difference
between energy and power.

And we are still waiting for the energy flow function
for the element that you claim is doing the storing
of the energy.

You should take this as a reason to
call into question the whole idea that this "interference
energy" is an actual energy flow.

Your argument is not with me - it is with experts
like Eugene Hecht. Please read his *57 page* Chapter
9 on "Interference" and then get back to us. With 57
pages devoted to the subject, Hecht doesn't seem to
share your problems with it.

Actually, we are debating *your* interpretation.

If *your* "wave reflection model" includes the idea that
Pref always represents an actual energy flow, then *your*
"wave reflection model" is wrong.

When you can prove that reflected traveling waves contain
zero energy, i.e. that ExH=0, I will accept your assertion
but not before.

You see. There you go again. Refusing to set aside an
assumption, even temporarily. It does make it difficult
to explore alternate explanations if you stop before you
start.

Exactly how does a TDR detect zero energy?

Does it detect energy? Are you sure?
Or is it voltage that it detects? Or current?

In fact, the thing you need to do is forget the transmission
line and deal with light waves encountering boundaries with
different indexes of refraction. The problem is identical,
but dealing with light out in the open prohibits you from
pushing your mashed-potatoes energy religion.

No. Light, in a 3 dimensional space and at such high frequency
makes the math and measurements so complicated that it is
extremely difficult to follow the energy. Much better to learn
from a one-dimensional transmission line and then see if the
solutions also apply to light. Which they do. But the analysis
is tractable in a transmission line.

...Keith

The Rest of the Story
 

On Mar 26, 11:49*am, Cecil Moore wrote:
Keith Dysart wrote:
What element provides the energy for this "interference
energy" flow? Note that in this analysis, this energy
is an average energy flow, so it can not be saved during
part of the cycle and returned during another part.

Already answered a few postings ago. Sorry you
missed it. Here it is again:

Since there was no destructive interference in the example,
the rest of your post is snipped as being a non-sequitor.

...Keith