On Mar 7, 5:31*pm, Cecil Moore wrote:
Keith Dysart wrote:
My understanding of your claim was that for the special case of
a 45 degree line supplied from a matched source, the energy
in the reflected wave is dissipated in the source resistor.

I have told you time and again that your understanding is
wrong. My claim is that for the special case of a 45 degree
phase difference between the forward wave and reflected wave,
the *average* power in the reflected wave is dissipated in the
source resistor. Irradiance is an *average* power density as
defined by Hecht, in "Optics". I have told you previously (many
times) that when I use the word "power", I am talking about
*average* power. I agree with Hecht that instantaneous power
is "of limited utility" and is therefore mostly irrelevant.

For the record - for the umteenth time: When I say "power", I
am talking about "*average* power". If I ever talk about
instantaneous power, I will say "instantaneous power". If
you still don't understand, you need professional help.

This sentence fragment from your document suggests this:
"reflected energy from the load is flowing through the source
resistor, RS, and is being dissipated there".

I left no doubt as to what I meant in my document. Here is
a quote from the second paragraph in my document:
"Please note that any power referred to in this paper
is an *average power*. Nothing is being asserted or implied about
instantaneous powers. In fact, instantaneous powers are completely
irrelevant to the following discussion."

I simply don't know how to say it any plainer than that.

I really resent your lack of ethics in this matter.
If you are forced to create a Big Lie about what I have
said in order to try to win, is it really worth it?

OK. I think I've got it now.

You are *not* claiming that the *energy* from the reflected wave
is dissipated in the source resistor, because for the *energy*
in the reflected wave to be dissipated in the source resistor,
the *energy* would have to dissipate at the same time that the
reflected wave delivered the *energy*, and the analysis of
instantaneous *energy* flows shows that this is not the case.

Rather, you are saying that the average reflected power is
numerically equal to the increase in the average dissipation
in the source resistor.

I can accept that as correct.

You might consider rewriting the sentence "reflected energy
from the load is flowing through the source resistor, RS, and
is being dissipated there" since it refers to the energy in
the reflected wave and may mislead others in the same way it
mislead me.

...Keith

Keith Dysart wrote:
You are *not* claiming that the *energy* from the reflected wave
is dissipated in the source resistor, because for the *energy*
in the reflected wave to be dissipated in the source resistor,
the *energy* would have to dissipate at the same time that the
reflected wave delivered the *energy*, and the analysis of
instantaneous *energy* flows shows that this is not the case.

I clearly stated that my claim is based on a special case
zero interference condition - it's even in the title of the
article. The instantaneous energy that you (not I) introduced,
does not meet the zero interference precondition. Therefore,
anything that does not meet the zero interference precondition
that I enumerated is an irrelevant diversion. You introduced
that irrelevant straw man and tried to make hay out it. :-)

you are saying that the average reflected power is
numerically equal to the increase in the average dissipation
in the source resistor.

I can accept that as correct.

Finally, after a million words. :-)

You might consider rewriting the sentence "reflected energy
from the load is flowing through the source resistor, RS, and
is being dissipated there" since it refers to the energy in
the reflected wave and may mislead others in the same way it
mislead me.

Since "average reflected power" is dependent upon the
"reflected energy", I don't see any problem. Would
"average reflected energy" work for you? It should be
obvious that, associated with interference during each
cycle, destructive interference energy is stored during
part of the cycle and delivered back as constructive
interference energy during another part of the cycle.
The intra-cycle interference averages out to zero.

Many have objected to the term "reflected power" saying
it is not power that is reflected but instead is "reflected
energy". So I stopped talking about "reflected power" and
started talking about "reflected energy". Now you object
to the use of the term "reflected energy". Would you and
the rest of the guru attack gang please get together on
what term you would like for me to use?
--
73, Cecil http://www.w5dxp.com

On Mar 8, 2:30 pm, Cecil Moore wrote:
Keith Dysart wrote:
You are *not* claiming that the *energy* from the reflected wave
is dissipated in the source resistor, because for the *energy*
in the reflected wave to be dissipated in the source resistor,
the *energy* would have to dissipate at the same time that the
reflected wave delivered the *energy*, and the analysis of
instantaneous *energy* flows shows that this is not the case.

I clearly stated that my claim is based on a special case
zero interference condition - it's even in the title of the
article. The instantaneous energy that you (not I) introduced,
does not meet the zero interference precondition. Therefore,
anything that does not meet the zero interference precondition
that I enumerated is an irrelevant diversion. You introduced
that irrelevant straw man and tried to make hay out it. :-)

you are saying that the average reflected power is
numerically equal to the increase in the average dissipation
in the source resistor.

I can accept that as correct.

Finally, after a million words. :-)

You might consider rewriting the sentence "reflected energy
from the load is flowing through the source resistor, RS, and
is being dissipated there" since it refers to the energy in
the reflected wave and may mislead others in the same way it
mislead me.

Since "average reflected power" is dependent upon the
"reflected energy", I don't see any problem. Would
"average reflected energy" work for you? It should be
obvious that, associated with interference during each
cycle, destructive interference energy is stored during
part of the cycle and delivered back as constructive
interference energy during another part of the cycle.
The intra-cycle interference averages out to zero.

Now you have me quite confused. One moment you agree that
your claim is mere numerical eqivalency and the next you
seem to again be claiming that the energy in the reflected
wave is dissipated in the source resistor.

Can you clarify which is really your claim?

Many have objected to the term "reflected power" saying
it is not power that is reflected but instead is "reflected
energy". So I stopped talking about "reflected power" and
started talking about "reflected energy". Now you object
to the use of the term "reflected energy". Would you and
the rest of the guru attack gang please get together on
what term you would like for me to use?

I have never been particularly fussy about the terminology.
"Energy flow", "power flow": The latter is often used when,
strictly, the former is meant, but there is seldom confusion,
except for those excessive pedantics who choose to be
confused.

My issue was that you seemed, in that sentence, to be saying
that the reflected energy was dissipated in the source
resistor. But earlier you had stated that was not your claim.
See my request for clarification above.

....Keith

Keith Dysart wrote:
Now you have me quite confused. One moment you agree that
your claim is mere numerical eqivalency and the next you
seem to again be claiming that the energy in the reflected
wave is dissipated in the source resistor.

I have never used the words "mere numerical equivalency"
so I have never claimed any such thing. The energy in
the reflected wave is dissipated in the source resistor
but you are confused about the timing of that dissipation.
It doesn't happen when you are saying it happens. Some
magnitude of energy is stored and dissipated later in
the same cycle.

Your analysis so far has completely ignored
*one additional source of instantaneous energy*, namely
the reactive component in the network, i.e. the reactance
of the transmission line. When you account for the temporary
storing of the intra-cycle destructive interference energy
in the transmission line followed by its later release as
constructive interference, you will find that all of the
energy in the reflected wave is dissipated in the source
resistor.

During part of a cycle, energy is lost from the source
resistor into the transmission line. During the following
part of the same cycle, that same energy is recovered
from the transmission line back to the source resistor.

instantaneous power dissipated in the source resistor =
1. instantaneous forward power plus
2. instantaneous reflected power plus
3. instantaneous interference (can be plus or minus)

Your calculations so far have completely ignored that
third term which is a source (or sink) of energy depending
upon what part of the cycle exists.
--
73, Cecil http://www.w5dxp.com

"Keith Dysart" wrote in message
...
On Mar 8, 2:30 pm, Cecil Moore wrote:
Keith Dysart wrote:
You are *not* claiming that the *energy* from the reflected wave
is dissipated in the source resistor, because for the *energy*
in the reflected wave to be dissipated in the source resistor,
the *energy* would have to dissipate at the same time that the
reflected wave delivered the *energy*, and the analysis of
instantaneous *energy* flows shows that this is not the case.

I clearly stated that my claim is based on a special case
zero interference condition - it's even in the title of the
article. The instantaneous energy that you (not I) introduced,
does not meet the zero interference precondition. Therefore,
anything that does not meet the zero interference precondition
that I enumerated is an irrelevant diversion. You introduced
that irrelevant straw man and tried to make hay out it. :-)

you are saying that the average reflected power is
numerically equal to the increase in the average dissipation
in the source resistor.

I can accept that as correct.

Finally, after a million words. :-)

You might consider rewriting the sentence "reflected energy
from the load is flowing through the source resistor, RS, and
is being dissipated there" since it refers to the energy in
the reflected wave and may mislead others in the same way it
mislead me.

Since "average reflected power" is dependent upon the
"reflected energy", I don't see any problem. Would
"average reflected energy" work for you? It should be
obvious that, associated with interference during each
cycle, destructive interference energy is stored during
part of the cycle and delivered back as constructive
interference energy during another part of the cycle.
The intra-cycle interference averages out to zero.

Now you have me quite confused. One moment you agree that
your claim is mere numerical eqivalency and the next you
seem to again be claiming that the energy in the reflected
wave is dissipated in the source resistor.

Can you clarify which is really your claim?

Many have objected to the term "reflected power" saying
it is not power that is reflected but instead is "reflected
energy". So I stopped talking about "reflected power" and
started talking about "reflected energy". Now you object
to the use of the term "reflected energy". Would you and
the rest of the guru attack gang please get together on
what term you would like for me to use?

NEITHER! they are both confusing. use the most fundamental things that you
can measure, either voltage or current. either one is completely defined in
the basic maxwell equations, and either one is completely sufficient to
describe ALL effect on a cable or in any circuit.

Dave wrote:
"Keith Dysart" wrote in message
Cecil Moore wrote:
So I stopped talking about "reflected power" and
started talking about "reflected energy". Now you object
to the use of the term "reflected energy". Would you and
the rest of the guru attack gang please get together on
what term you would like for me to use?

NEITHER! they are both confusing. use the most fundamental things that you
can measure, either voltage or current. either one is completely defined in
the basic maxwell equations, and either one is completely sufficient to
describe ALL effect on a cable or in any circuit.

If Maxwell's equations could be used to answer the
questions that we are asking, why haven't they been
answered a long time ago?

How can Maxwell's equations be used to track the path
and fate of the energy in a reflected wave?
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:
Dave wrote:
"Keith Dysart" wrote in message
Cecil Moore wrote:
So I stopped talking about "reflected power" and
started talking about "reflected energy". Now you object
to the use of the term "reflected energy". Would you and
the rest of the guru attack gang please get together on
what term you would like for me to use?

NEITHER! they are both confusing. use the most fundamental things
that you can measure, either voltage or current. either one is
completely defined in the basic maxwell equations, and either one is
completely sufficient to describe ALL effect on a cable or in any
circuit.

If Maxwell's equations could be used to answer the
questions that we are asking, why haven't they been
answered a long time ago?

How can Maxwell's equations be used to track the path
and fate of the energy in a reflected wave?

Cecil,

There may be some terminology confusion here.

Maxwell's equations are really the only relevant physical equations
there are to work with, at least in the classical regime. The discussion
about constructive, destructive, superposition, linearity, etc.
represents merely mathematical manipulation of the basic physical
entities embodied in Maxwell's equations.

All of this *math* is of course very important and very useful. However,
it is not a replacement for the *physical* laws known as the Maxwell
equations.

(And we all know that these endless threads are closely parallel to the
blind men describing an elephant puzzle.)

73,
Gene
W4SZ

Keith Dysart wrote:
My issue was that you seemed, in that sentence, to be saying
that the reflected energy was dissipated in the source
resistor. But earlier you had stated that was not your claim.

My earlier claim was that the average power in a reflected
wave is dissipated in the source resistor when the forward
wave is 90 degrees out of phase with the reflected wave at
the source resistor. In that earlier claim, I didn't care
to discuss instantaneous power and thus excluded instantaneous
power from that claim.

For instantaneous values, it will be helpful to change the
example while leaving the conditions at the source resistor
unchanged. Here's the earlier example:

Rs Vg Vl
+----/\/\/-----+----------------------+
| 50 ohm |
| 1/8 WL |
Vs 45 degrees 12.5 ohm
100v RMS 50 ohm line Load
| |
| |
+--------------+----------------------+
gnd

Here's the present example:

Rs Vg Vl
+----/\/\/-----+----------------------+
| 50 ohm |
| 1 WL |
Vs 360 degrees 23.5+j44.1
100v RMS 50 ohm line ohm Load
| |
| |
+--------------+----------------------+
gnd

If I haven't made some stupid mistake, the conditions at
the source resistor are identical in both examples. But
in the second example, it is obvious that energy can
be stored in the transmission line during part of a cycle
(thus avoiding dissipation at that instant in time) and be
delivered back to the source resistor during another part
of the cycle (to be dissipated at a later instant in time).
That is the nature of interference energy and is exactly
equal to the difference between the two powers that you
calculated. You neglected to take into account the ability
of the network reactance to temporarily store energy and
dissipate it later in time.

All of the reflected energy is dissipated in the source
resistor, just not at the time you thought it should be.

While performing this analysis, you will discover the ability
of superposing waves to redistribute energy (even in the
complete absence of ordinary reflections). Not possible in
the above special case examples, but under other conditions,
the redistribution of energy can attain perpetual steady-state
status (even in the complete absence of ordinary reflections).

There are two mechanisms involved in the re-routing of
steady-state reflected energy back toward the load.

1. Ordinary reflection of single EM waves governed by the
rules of the wave reflection model.

2. A redistribution of reflected energy back toward the
load as a result of superposition of waves accompanied
by an average level of steady-state interference.
--
73, Cecil http://www.w5dxp.com

On Sun, 09 Mar 2008 14:35:25 GMT
Cecil Moore wrote:

Keith Dysart wrote:
My issue was that you seemed, in that sentence, to be saying
that the reflected energy was dissipated in the source
resistor. But earlier you had stated that was not your claim.

My earlier claim was that the average power in a reflected
wave is dissipated in the source resistor when the forward
wave is 90 degrees out of phase with the reflected wave at
the source resistor. In that earlier claim, I didn't care
to discuss instantaneous power and thus excluded instantaneous
power from that claim.

For instantaneous values, it will be helpful to change the
example while leaving the conditions at the source resistor
unchanged. Here's the earlier example:

Rs Vg Vl
+----/\/\/-----+----------------------+
| 50 ohm |
| 1/8 WL |
Vs 45 degrees 12.5 ohm
100v RMS 50 ohm line Load
| |
| |
+--------------+----------------------+
gnd

Here's the present example:

Rs Vg Vl
+----/\/\/-----+----------------------+
| 50 ohm |
| 1 WL |
Vs 360 degrees 23.5+j44.1
100v RMS 50 ohm line ohm Load
| |
| |
+--------------+----------------------+
gnd

If I haven't made some stupid mistake, the conditions at
the source resistor are identical in both examples. But
in the second example, it is obvious that energy can
be stored in the transmission line during part of a cycle
(thus avoiding dissipation at that instant in time) and be
delivered back to the source resistor during another part
of the cycle (to be dissipated at a later instant in time).
That is the nature of interference energy and is exactly
equal to the difference between the two powers that you
calculated. You neglected to take into account the ability
of the network reactance to temporarily store energy and
dissipate it later in time.

All of the reflected energy is dissipated in the source
resistor, just not at the time you thought it should be.

I think we can all agree to what would happen in the turn off situation following a long period of stable power flow. When the source voltage steps from 100v to 0v, the power on the transmission line reflects to the 50 ohm source resistor and is all absorbed.

On the other hand, in the case of steady power flow, the source is presented with a reactive load of 73.5 + J44.1 ohms. The reactive part of this load results from returning power from the transmission line.

For the resistor Rs, it will have power applied from two sources, the source and the reflected power from the transmission line, i.e., the source power and reflected power are in series when considered in relationship with the resistor Rs. The problem is that 'what looks like two sources to resistor Rs, is really only one source, Vs'.

I think you are looking for solutions that show how power to Rs peaks at a different time from when power into the transmission line peaks, which is yet a different time from when power from Vs peaks. You are looking for the instantaneous timing of 3 peaks.


While performing this analysis, you will discover the ability
of superposing waves to redistribute energy (even in the
complete absence of ordinary reflections). Not possible in
the above special case examples, but under other conditions,
the redistribution of energy can attain perpetual steady-state
status (even in the complete absence of ordinary reflections).

There are two mechanisms involved in the re-routing of
steady-state reflected energy back toward the load.

1. Ordinary reflection of single EM waves governed by the
rules of the wave reflection model.

2. A redistribution of reflected energy back toward the
load as a result of superposition of waves accompanied
by an average level of steady-state interference.
--
73, Cecil http://www.w5dxp.com


--
73, Roger, W7WKB

Roger Sparks wrote:
For the resistor Rs, it will have power applied from two sources, the source and the reflected power from the transmission line, i.e., the source power and reflected power are in series when considered in relationship with the resistor Rs. The problem is that 'what looks like two sources to resistor Rs, is really only one source, Vs'.

Actually, it will have energy applied from three sources, the
source, the reflected energy, and the reactive energy stored
in the transmission line. The energy that Keith is missing
comes from the reactance in the transmission line.
--
73, Cecil http://www.w5dxp.com