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Old March 25th 08, 12:11 PM posted to rec.radio.amateur.antenna
Keith Dysart[_2_] Keith Dysart[_2_] is offline
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First recorded activity by RadioBanter: May 2007
Posts: 492
Default 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