The Rest of the Story
 

Keith Dysart wrote:
The simulator at that web site does seem to have
its issues. Ask it to simulate 700 nm + 680 nm
at the same amplitude and see if the result
represents reality.

The duration of each calculation appears to be
about one second and then a reboot.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Mar 27, 2:06 am, Roger Sparks wrote:
Cecil Moore wrote:
Roger Sparks wrote:
You need to take a look at the spreadsheets.

Roger, in a nutshell, what is the bottom line?

The bottom line in a nutshell? I'll try.

First, I added a note to both spreadsheets indicating that zero degrees
is CURRENT zero degrees. This because the source turns out to be
reactive, with current peak 45 degrees from voltage peak.

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

I was not able to discern the derivation of the various equations
though the data in the columns looked somewhat reasonable. Were
the equation really based on the opening paragraph statement of
100 Vrms, or is it scaled to a different source voltage. The sin
functions have an amplituted of 100, which suggests a source of
200 volts or Vrms of 141.4 volts.

The spreadsheet addresses the following issues:

Does the traveling wave carry power? Yes. The spreadsheet was built
assuming that power is carried by traveling waves. Because the
resulting wave form and powers seem correct, the underlaying assumption
seems correct.

It was not obvious which columns were used to draw this correlation.

However, even if this experiment is consistent with the hypothesis
it only takes one experiment which is not to disprove the hypothesis.

Is power conserved on the transmission line, meaning, can the energy
contained in power be conserved and located over time on the
transmission line? Yes, the spreadsheet was built assuming that power
could be conserved and traced over time so the underlaying assumption
seems correct.

Does interference occur in this example? The spreadsheet was built
assuming that voltage and currents from superpose in a manner consistent
with constructive and destructive interference, so the underlaying
assumption seems correct.

Is power stored in the reactive component for release in later in the
cycle or during the next half cycle? Yes, power is stored on the
transmission line during the time it takes for power to enter the line,
travel to the end and return. The time of wave travel on the
transmission line is related to the value of the reactive component.

Does the direction of wave travel affect the measurement of voltage and
the application of power to a device? Yes. A wave loses energy (and
therefore voltage) as it travels through a resistance. As a result,
power from the prime source is ALWAYS applied across the sum of the
resistance from the resistor AND transmission line.

I am not sure that I would describe this as the wave losing energy,
but rather as the voltage dividing between the two impedances.
If the source resistance was replaced by another transmission,
which could easily be set to provide a 50 ohm impedance, would
you still describe it as the wave losing energy?

The spreadsheet was
built using this assumption and seems correct. (At times during the
cycle, the forward and reflected waves oppose, resulting in very little
current through the resistor. During those times, the power applied to
the transmission line is much HIGHER because the reflected wave reflects
from the load and source, and merges/adds to the forward wave from the
source.)

I am not convinced. When there is very little current through the
resistor,
there is also very little current into the transmission line. This
suggests to me that the power applied to the transmission line is low.

....Keith

The Rest of the Story
 

On Mar 27, 11:10 am, Cecil Moore wrote:
Keith Dysart wrote:
So you are having difficulty doing the math to justify
your hypothesis.

Actually no, the math is not difficult. I'm pre-
occupied with something else and think it's just
time to agree with Hecht that instantaneous power
is "of limited utility".

Have you taken a look at Roger's spreadsheets?

Yes.

Conservation of energy requires that the total
quantity of energy in the system not change.

:-) Isn't the whole purpose of a transmitting
antenna to radiate energy away from the antenna
system? And that radiation continues to be "lost"
from the system space for some time after the
source power is removed?

Wouldn't you have to define the "system" as the
entire universe for your statement to be true?

Sort of, but there is an easy work-around. Since we
are not particularly interested in what happens to
the energy going to the source resistor, for
example, we define the boundary of the system to
pass through the resistor and simply account for
the energy as a flow out of the system. And we
do the same for the source since we are not
particularly interested in where the energy for
the source comes from; we just account for it as
a flow into the system. And lastly, we don't
really care where the energy on the line goes to
or comes from, so we just account for it as a flow
out of the system. You may have noticed that we
have removed all the components from the system,
it is a null system, and all we are doing is
accounting for the flows into and out of the
system. Hence
Ps(t) = Prs(t) + Pg(t)

The powers must sum to 0 to satisfy conservation
of energy.

That may be true, but there's still no conservation
of instantaneous power principle.

To conserve energy however, i.e. to have no energy
accumulate in the null system described above, the
flows must balance at all times, that is,
instantaneously.

A hot resistor
continues to radiate heat long after any power
source is removed.

If you want to be that complicated you can; energy
is delivered instantaneously to the resistor according
to Vr*Ir (or equivalently, V**2/R), stored in the
resistor as heat and dissipated to the environment
later.

But for our purposes, we can stop that analysis at
the point where the energy enters the resistor and
use Vr(t) * Ir(t) to compute the instantaneous flow
into the resistor.

You could simply do the derivation for an example that
demonstrates your hypothesis.

Already done on my web page. My only actual hypothesis
concerns average power. I've wasted too much time
bantering about something that Hecht says is "of
limited utility".

Unfortunately for your hypothesis, average power is
insufficient to account for energy which might be in
the reflected wave. An average power analysis agrees
with your hypothesis, while a more detailed instaneous
analysis disproves it.

the idea that Pfor and Pref describe actual energy
flows is very dubious.

Again, look yourself in the mirror and tell yourself
that what you are seeing contains no energy. The
theory that some EM waves contain energy and some
do not is not new to you. Dr. Best was the first to
theorize that canceled waves continue to propagate
forever devoid of energy. Someone else asserted that
canceled waves never contained any energy to start
with. I strongly suspect that what you are seeing
in the mirror are the waves that didn't cancel and
that do contain energy. :-)

If it was just a 'suspicion' you could probably let
go of the idea long enough to learn what is really
happening and why it is not inconsistent with the
idea that reflected wave energy is a dubious concept.

So the energy is not being stored
in the reactive component of the line input
impedance.

Assuming you have not made an error,

You *have* found it hard to the do the math; otherwise,
you could detect an error, if there was one.

so what? Energy
stored in the reactance is only one of the possibilities
that I listed earlier. As I said in an earlier posting
which you declared a non-sequitor (sic), one or more of
the following is true:

1. The source adjusts to the energy requirements.

2. The reactance stores and delivers energy.

3. Wave energy is redistributed during superposition.

4. Something I haven't thought of.

Your explanation is not complete until you can identify
the element that stores and returns the energy and its
energy transfer function.

The ExH reflected wave energy exists and cannot
be destroyed. It goes somewhere and its average
value is dissipated in the source resistor in
my special case example.

Or not, as has been shown with the detailed analysis.

You are attempting to
destroy the reflected wave energy using words
and math presumably knowing all along that
reflected wave energy cannot be destroyed.

Energy can not be destroyed. This leads inexorably
to the conclusion that the reflected wave does not
necessarily contain energy.

....Keith

The Rest of the Story
 

On Mar 27, 11:37 am, Cecil Moore wrote:
Keith Dysart wrote:
I had not realized that you had these alternate sources for
the interference energies, not having seen that in your papers.

I only posted it three times here and you chose to
ignore all of those postings. I have published only
one paper with three more to go. The special case
Part 1 contains zero average interference so there
is no alternate source for average interference and
indeed, none is needed for Part 1.

But it is one way to sidestep the issue; different rules for
the expectations of superposition and interference in different
scenarios.

That's why I have four parts only one of which has
been published. The rules are not different but the
conditions within the examples are different. Part 2
will be an example with the condition of average
destructive interference existing at the source
resistor. Although there are no ordinary reflections
because the reflection coefficient is 0.0, there will
exist something that looks a lot like a reflection
caused by superposition/interference. The FSU web page
calls it a "redistribution", not a "reflection".
I am satisfied with FSU's word "redistribution" for
the results of coherent wave interaction.

I am surprised then, for the example of Fig 1-1 with 12.5 ohms,
that you don't just say "There is a source nearby, that *must*
be where the unaccounted energy comes from", and leave it at
that.

Since my special case example contains zero average
interference, the average power output of the source
is constant and unaffected by zero interference.
There is zero average energy unaccounted for.

Part 2 will illustrate the source adjusting its power
output to compensate for destructive interference.

Perhaps you should complete part one so that it fully
accounts for the energy flows before progressing to
writing part two. Average is not a full accounting.

....Keith

The Rest of the Story
 

On Sat, 29 Mar 2008 12:45:48 -0700 (PDT)
Keith Dysart wrote:

On Mar 27, 2:06 am, Roger Sparks wrote:
Cecil Moore wrote:
Roger Sparks wrote:
You need to take a look at the spreadsheets.

Roger, in a nutshell, what is the bottom line?

The bottom line in a nutshell? I'll try.

First, I added a note to both spreadsheets indicating that zero degrees
is CURRENT zero degrees. This because the source turns out to be
reactive, with current peak 45 degrees from voltage peak.

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

I was not able to discern the derivation of the various equations
though the data in the columns looked somewhat reasonable. Were
the equation really based on the opening paragraph statement of
100 Vrms, or is it scaled to a different source voltage. The sin
functions have an amplituted of 100, which suggests a source of
200 volts or Vrms of 141.4 volts.

I simplified and changed the phase for the equation in column C from '100sin(wt-90) + 100sin(wt-180)' to '100sin(wt+90) + 100sin(-wt)' so that the total voltage in column D better matches with the voltages from my formula in column G.

The equation for column B is just the sine wave for the base wave, showing the voltage resulting to the source resistor from the peak current. The equation in column C is the same (from current) but recognizes that current from the reflected wave will cancel the current from the forward wave coming through Rs. The two waves are traveling in opposite directions so one angle must be labeled with a negative sign so that it will rotate in the opposite direction. In column C, the first term is the reflected wave and the second term is the base wave.

My formula, displayed in column G, uses the applied voltage of 141.4 volts. I thought it would be easier to see how the spreadsheet was built using current to find the voltages. You can see how the results of column D match well with those from column G.


The spreadsheet addresses the following issues:

Does the traveling wave carry power? Yes. The spreadsheet was built
assuming that power is carried by traveling waves. Because the
resulting wave form and powers seem correct, the underlaying assumption
seems correct.

It was not obvious which columns were used to draw this correlation.


Column E and column F display power. Column F recognizes that if 100 watts is applied continueously (on the average) over an entire 360 degree cycle, the final power applied over time would be 360 * 100 = 36000 watt-degrees. Part of the power comes from the reflection, part comes directly. Obviously, interference is very much at work in this example.

However, even if this experiment is consistent with the hypothesis
it only takes one experiment which is not to disprove the hypothesis.


True!

Is power conserved on the transmission line, meaning, can the energy
contained in power be conserved and located over time on the
transmission line? Yes, the spreadsheet was built assuming that power
could be conserved and traced over time so the underlaying assumption
seems correct.

Does interference occur in this example? The spreadsheet was built
assuming that voltage and currents from superpose in a manner consistent
with constructive and destructive interference, so the underlaying
assumption seems correct.

Is power stored in the reactive component for release in later in the
cycle or during the next half cycle? Yes, power is stored on the
transmission line during the time it takes for power to enter the line,
travel to the end and return. The time of wave travel on the
transmission line is related to the value of the reactive component.

Does the direction of wave travel affect the measurement of voltage and
the application of power to a device? Yes. A wave loses energy (and
therefore voltage) as it travels through a resistance. As a result,
power from the prime source is ALWAYS applied across the sum of the
resistance from the resistor AND transmission line.

I am not sure that I would describe this as the wave losing energy,
but rather as the voltage dividing between the two impedances.
If the source resistance was replaced by another transmission,
which could easily be set to provide a 50 ohm impedance, would
you still describe it as the wave losing energy?


It should be OK to think of the voltage dividing between two impedances. The important thing is to consider how the reflected voltage sums with the forward voltage where it is measured. Because the two waves are traveling in opposite directions, the measured voltage is not the voltage applied to either Rs or the transmission line. This is why columns B and C must be added to find the total voltage across Rs.

The spreadsheet was
built using this assumption and seems correct. (At times during the
cycle, the forward and reflected waves oppose, resulting in very little
current through the resistor. During those times, the power applied to
the transmission line is much HIGHER because the reflected wave reflects
from the load and source, and merges/adds to the forward wave from the
source.)

I am not convinced. When there is very little current through the
resistor,
there is also very little current into the transmission line. This
suggests to me that the power applied to the transmission line is low.

I don't follow you here. Right, power into the transmission line is low when the current in is low, and it is high when the current is high. It is clear that peak current and peak voltage do not occur at the same time except when measured across the resistor in column D.


...Keith


This is just one example, but it seems like the power is accounted for here. We need another example where power can NOT be accounted for.

--
73, Roger, W7WKB

The Rest of the Story
 

On Mar 29, 7:18 pm, Roger Sparks wrote:
On Sat, 29 Mar 2008 12:45:48 -0700 (PDT)

Keith Dysart wrote:
On Mar 27, 2:06 am, Roger Sparks wrote:
Cecil Moore wrote:
Roger Sparks wrote:
You need to take a look at the spreadsheets.

Roger, in a nutshell, what is the bottom line?

The bottom line in a nutshell? I'll try.

First, I added a note to both spreadsheets indicating that zero degrees
is CURRENT zero degrees. This because the source turns out to be
reactive, with current peak 45 degrees from voltage peak.

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

I was not able to discern the derivation of the various equations
though the data in the columns looked somewhat reasonable. Were
the equation really based on the opening paragraph statement of
100 Vrms, or is it scaled to a different source voltage. The sin
functions have an amplituted of 100, which suggests a source of
200 volts or Vrms of 141.4 volts.

I simplified and changed the phase for the equation in column C from '100sin(wt-90) + 100sin(wt-180)' to '100sin(wt+90) + 100sin(-wt)' so that the total voltage in column D better matches with the voltages from my formula in column G.

The equation for column B is just the sine wave for the base wave, showing the voltage resulting to the source resistor from the peak current. The equation in column C is the same (from current) but recognizes that current from the reflected wave will cancel the current from the forward wave coming through Rs. The two waves are traveling in opposite directions so one angle must be labeled with a negative sign so that it will rotate in the opposite direction. In column C, the first term is the reflected wave and the second term is the base wave.

My formula, displayed in column G, uses the applied voltage of 141.4 volts. I thought it would be easier to see how the spreadsheet was built using current to find the voltages. You can see how the results of column D match well with those from column G.

I am still having difficulty matching these equations with mine. Just
to make sure we are discussing the same problem....

The circuit is a voltage source
Vs(t) = 141.4 sin(wt-45)
driving a source resistor of 50 ohms and 45 degress of 50 ohm
transmission line that is shorted at the end.

My calculations suggest that
Vrs.total(t) = 100 sin(wt-90)
Irs.total(t) = 2 sin(wt-90)

which agrees with you column D but not the introduction which
says that the zero current is at 0 degrees. With -100 volts
across the source resistor at 0 degrees, the current should
be -2 amps through the source resistor at 0 degrees; in other
words, a current maximum.

Using superposition, I compute the the contribution of the
source to be
Vrs.source(t) = 70.7 sin(wt-45)
and the contribution from the reflected wave to be
Vrs.reflected(t) = -70.7 sin(wt+45)
which sum to
Vrs.total(t) = 100 sin(wt-90)
as expected.

So while I can construct an expression for Column D, the
contributing values do not agree with those you have
provided in Columns B and C.

Column E follows from Column D and my calculations agree.
And also with Column F.

The spreadsheet addresses the following issues:

Does the traveling wave carry power? Yes. The spreadsheet was built
assuming that power is carried by traveling waves. Because the
resulting wave form and powers seem correct, the underlaying assumption
seems correct.

It was not obvious which columns were used to draw this correlation.

Column E and column F display power. Column F recognizes that if 100 watts is applied continueously (on the average) over an entire 360 degree cycle, the final power applied over time would be 360 * 100 = 36000 watt-degrees. Part of the power comes from the reflection, part comes directly. Obviously, interference is very much at work in this example.

Column E is the power dissipated in the resistor, and Column F
is the integral of Column and represents the total energy which
has flowed in to the resistor over the cycle. It is also the
average energy per cycle.

If you were to extend your analysis to compute the energy
in each degree of the reflected wave and add it to the energy
in each degree of Vrs.source(t) and sum these, you would
find that the instantaneous energy from Vrs.source and
Vrs.reflected does not agree with the instantaneous energy
dissipated in the source resistor. It is this disagreement
that is the root of my argument that the power in the
reflected wave is a dubious concept.

Using averages, the computed powers support the hypothesis,
but when examined with finer granularity, they do not.

However, even if this experiment is consistent with the hypothesis
it only takes one experiment which is not to disprove the hypothesis.

True!

Is power conserved on the transmission line, meaning, can the energy
contained in power be conserved and located over time on the
transmission line? Yes, the spreadsheet was built assuming that power
could be conserved and traced over time so the underlaying assumption
seems correct.

Does interference occur in this example? The spreadsheet was built
assuming that voltage and currents from superpose in a manner consistent
with constructive and destructive interference, so the underlaying
assumption seems correct.

Is power stored in the reactive component for release in later in the
cycle or during the next half cycle? Yes, power is stored on the
transmission line during the time it takes for power to enter the line,
travel to the end and return. The time of wave travel on the
transmission line is related to the value of the reactive component.

Does the direction of wave travel affect the measurement of voltage and
the application of power to a device? Yes. A wave loses energy (and
therefore voltage) as it travels through a resistance. As a result,
power from the prime source is ALWAYS applied across the sum of the
resistance from the resistor AND transmission line.

I am not sure that I would describe this as the wave losing energy,
but rather as the voltage dividing between the two impedances.
If the source resistance was replaced by another transmission,
which could easily be set to provide a 50 ohm impedance, would
you still describe it as the wave losing energy?

It should be OK to think of the voltage dividing between two impedances. The important thing is to consider how the reflected voltage sums with the forward voltage where it is measured. Because the two waves are traveling in opposite directions, the measured voltage is not the voltage applied to either Rs or the transmission line. This is why columns B and C must be added to find the total voltage across Rs.

Whether one needs to add or subtract is more a matter of the
convention
being used for the signs of the values. When Vf and Vr are derived
using
Vtot = Vf + Vr; Itot = If + Ir
one would expect to have to add the negative of Vr to the contribution
from Vs to arrive at the total voltage across the source resistor.

The spreadsheet was
built using this assumption and seems correct. (At times during the
cycle, the forward and reflected waves oppose, resulting in very little
current through the resistor. During those times, the power applied to
the transmission line is much HIGHER because the reflected wave reflects
from the load and source, and merges/adds to the forward wave from the
source.)

I am not convinced. When there is very little current through the
resistor,
there is also very little current into the transmission line. This
suggests to me that the power applied to the transmission line is low.

I don't follow you here. Right, power into the transmission line is low when the current in is low, and it is high when the current is high. It is clear that peak current and peak voltage do not occur at the same time except when measured across the resistor in column D.

Power into the transmission line is low when either the voltage or
the current is low; when either is zero, the power is zero.

Since the highest voltage occurs with zero current and the highest
current occurs with zero voltage, maximum power into the transmission
line occurs when the voltage and current are both medium; more
precisely, when they are both at .707 of their maximum values.

This is just one example, but it seems like the power is accounted for here. We need another example where power can NOT be accounted for.

I suggest it is the same example, but the granularity of the
analysis needs to be increased.

....Keith

The Rest of the Story
 

Keith Dysart wrote:
Unfortunately for your hypothesis, average power is
insufficient to account for energy which might be in
the reflected wave. An average power analysis agrees
with your hypothesis, while a more detailed instaneous
analysis disproves it.

One more time: My hypothesis doesn't apply to instantaneous
powers at all so there is nothing to disprove. Please
leave me out of any discussion of instantaneous powers.

If it was just a 'suspicion' you could probably let
go of the idea long enough to learn what is really
happening and why it is not inconsistent with the
idea that reflected wave energy is a dubious concept.

One look in a mirror should convince you otherwise.

You *have* found it hard to the do the math;

Nope, I just think that instantaneous power math is a
waste of my time. Seems to also be a waste of your
time.

Your explanation is not complete until you can identify
the element that stores and returns the energy and its
energy transfer function.

I simply don't have anything at all to say about
instantaneous powers. In my opinion, such is
a waste of my time. You might as well be demanding
that I produce the math for how many angels can
dance on the head of a pin.

Energy can not be destroyed.

Yet, you are trying your best to destroy the energy
in the reflected wave. Since you cannot destroy
reflected energy at the average power level, you
are trying to destroy it at the instantaneous
level.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
Perhaps you should complete part one so that it fully
accounts for the energy flows before progressing to
writing part two. Average is not a full accounting.

Average is the only accounting that I consider to
be important and the only accounting that I am going
to do. I have added a disclaimer about instantaneous
power to my Part 1 article. I personally don't think
that anyone else cares about instantaneous powers.

If you need an instantaneous power article written,
please feel free to write it yourself. I wish you luck
but I personally consider it to be a waste of time.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
Using averages, the computed powers support the hypothesis,
but when examined with finer granularity, they do not.

Well Keith, yours also falls apart at finer granularity
where you are required to determine the position and
momentum of the individual charge carriers.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Mar 30, 8:28*pm, Cecil Moore wrote:
Keith Dysart wrote:
Unfortunately for your hypothesis, average power is
insufficient to account for energy which might be in
the reflected wave. An average power analysis agrees
with your hypothesis, while a more detailed instaneous
analysis disproves it.

One more time: My hypothesis doesn't apply to instantaneous
powers at all so there is nothing to disprove. Please
leave me out of any discussion of instantaneous powers.

You state that your hypothesis is that for this specific
circuit, "the energy in the reflected wave is dissipated in
the source resistor". This claim is amenable to analysis
using instantaneous energy flows. When so analyzed, the
hypothesis fails.

I can see why you want to limit the kinds of analysis
applied to your hypothesis, but that is not how science
works.

If you wish to narrow your hypothesis to "the average energy
in the reflected wave is simply numerically equal to the
increase in the average dissipation in the source
resistor" I will not object since that hypothesis would
be completely accurate and not misleading.

As long as you wish to claim that there is actual energy
in the reflected wave, and that this energy is dissipated
in the source resistor, you must be prepared to offer a
full accounting.

If it was just a 'suspicion' you could probably let
go of the idea long enough to learn what is really
happening and why it is not inconsistent with the
idea that reflected wave energy is a dubious concept.

One look in a mirror should convince you otherwise.

You *have* found it hard to the do the math;

Nope, I just think that instantaneous power math is a
waste of my time. Seems to also be a waste of your
time.

Not at all. I've learned a new (at least to me) technique
for analyzing where the energy goes. I've increased my skills
with complex numbers in Excel. I have a firmer grasp about
power computations. I've explored the tautology of
Ptotal = Pforward - Preflected. I've learned a bit about
macros in Excel. I've tried Calc and found it wanting.
I've looked at energy flows in circulators. I've explored
active and passive circulators. And I've proved your
hypothesis to be false.

Not a waste at all.

Your explanation is not complete until you can identify
the element that stores and returns the energy and its
energy transfer function.

I simply don't have anything at all to say about
instantaneous powers. In my opinion, such is
a waste of my time.

A higly convenient opinion since it allows you to continue
to believe your hypothesis.

You might as well be demanding
that I produce the math for how many angels can
dance on the head of a pin.

No. Instantaneous flows *can* be computed. Angels on a pin
are a little more problematic.

Energy can not be destroyed.

Yet, you are trying your best to destroy the energy
in the reflected wave.

Since you start with an unshakeable belief in the
existance of energy in the reflected wave, this would be
your natural conclusion.

Those of us whose beliefs are not so fixed, evaluate
the proofs and come to the conclusion that energy in
the reflected wave is a dubious concept.

Since you cannot destroy
reflected energy at the average power level, you
are trying to destroy it at the instantaneous
level.

"Average" is just a data reduction from "instantaneous".
An agreement of averages is not a proof of anything
except that the "averages are numerically equal".

...Keith