The Rest of the Story
 

Keith Dysart wrote:
To be convincing, the various functions of time need to
align appropriately.

And one can tell that they indeed do "align appropriately"
just by looking at the graphs of the two voltages. We know
that the average power in the reflected wave is dissipated
in the source resistor. All that is left to understand is
how long the destructive interference energy is stored in
the transmission line before being dissipated in the source
resistor as constructive interference.

Graphing in ASCII is pretty difficult. Let's see if we
can do it with words. Take a piece of transparent film
and draw the forward voltage to scale. Take another piece
of transparent film and draw the reflected voltage to
scale. Now we have graphs of two voltages that can be
varied by phase. In ASCII, the best I can do is:

------ ----
/ \ /
/ \ forward voltage /
/________________\____________________/_________
\ /
\ /
\ /
------

----------
/ \ reflected voltage
__________/________________\_____________________
/ \ /
/ \ /
----- ----------

For the first 90 degrees of the above graph, the forward voltage
is positive and the reflected voltage is negative. That is
destructive interference so there's an excess of energy that
is stored in the transmission line. For the second 90 degrees,
the forward voltage is positive and the reflected voltage is
positive. That is constructive interference so the excess energy
from the first 90 degrees is sucked back out of the transmission
line and dissipated in the source resistor. Given that the average
reflected energy is dissipated in the source resistor and assuming
we honor the conservation of energy principle, nothing else is
possible.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Mar 11, 11:07*pm, Cecil Moore wrote:
Keith Dysart wrote:
To be convincing, the various functions of time need to
align appropriately.

And one can tell that they indeed do "align appropriately"
just by looking at the graphs of the two voltages. We know
that the average power in the reflected wave is dissipated
in the source resistor.

Actually, we have shown that that is not the case. If the
instantaneous power is not dissipated in the source resistor,
then neither is the average of the instantaneous power. And
we have shown that the instantaneous power is not dissipated
in the source resistor.

You might like to visit
http://keith.dysart.googlepages.com/radio6
for a graph that shows the actual power dissipated in the
source resistor along with the power in the reflected
wave. It can be visually seen that the power dissipated in
the source resistor has no relationship to the sum of the
reflected power and the power that would be dissipated in
the source resistor if there was no reflection.

All that is left to understand is
how long the destructive interference energy is stored in
the transmission line before being dissipated in the source
resistor as constructive interference.

Graphing in ASCII is pretty difficult. Let's see if we
can do it with words. Take a piece of transparent film
and draw the forward voltage to scale. Take another piece
of transparent film and draw the reflected voltage to
scale. Now we have graphs of two voltages that can be
varied by phase. In ASCII, the best I can do is:

* * * *------ * * * * * * * * * * * * * * * *----
* * */ * * * *\ * * * * * * * * * * * * * */
* */ * * * * * *\ * *forward voltage * * /
/________________\____________________/_________
* * * * * * * * * * \ * * * * * * * */
* * * * * * * * * * * \ * * * * * */
* * * * * * * * * * * * \ * * * */
* * * * * * * * * * * * * ------

* * * * * * * *----------
* * * * * * */ * * * * * *\ * reflected voltage
__________/________________\_____________________
* * * * */ * * * * * * * * * *\ * * * * * * * */
* * * */ * * * * * * * * * * * *\ * * * * * */
----- * * * * * * * * * * * * * *----------

For the first 90 degrees of the above graph, the forward voltage
is positive and the reflected voltage is negative. That is
destructive interference so there's an excess of energy that
is stored in the transmission line. For the second 90 degrees,
the forward voltage is positive and the reflected voltage is
positive. That is constructive interference so the excess energy
from the first 90 degrees is sucked back out of the transmission
line and dissipated in the source resistor.

You have shown that for some of the time energy is delivered to
the line and sometimes it returns energy. This is
Pg(t) = 32 + 68cos(2wt)
which has nothing to do with reflected wave energy dissipated in
the source resistor.

Pg(t) is equal, however, to Pf.g(t) + Pr.g(t).

Given that the average
reflected energy is dissipated in the source resistor and assuming
we honor the conservation of energy principle, nothing else is
possible.

But the premise in the above sentence (average reflected energy is
dissipated in the source resistor) is wrong, so the conclusion
(nothing else is possible) is as well.

...Keith

The Rest of the Story
 

Keith Dysart wrote:
Actually, we have shown that that is not the case.

If the reflected energy is not being dissipated in
the source resistor, where does it go? It is not
dissipated in the load, by definition, and there is
no other source of dissipation in the network besides
the source resistor. There are no reflections and
no average interference to redistribute the reflected
energy back toward the load. The reflected wave
possesses energy and momentum which must be conserved.
Do your reflected waves obey your every whim and just
disappear and reappear as willed by you in violation
of the conservation of energy principle?

You might like to visit
http://keith.dysart.googlepages.com/radio6
for a graph that shows the actual power dissipated in the
source resistor along with the power in the reflected
wave. It can be visually seen that the power dissipated in
the source resistor has no relationship to the sum of the
reflected power and the power that would be dissipated in
the source resistor if there was no reflection.

The difference in your energy levels is in the reactance.
The reactance stores energy and delivers it later. Why
are you having so much trouble with that age-old concept?
The reflected energy that is not dissipated in the source
resistor at time 't' is stored in the reactance and dissipated
90 degrees later. Until you choose to account for that time
delay, your energy equations are not going to balance.

Your equations would be valid only if there were no reactance
and no delays in the network. The lumped circuit model strikes
again.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
Actually, we have shown that that is not the case. If the
instantaneous power is not dissipated in the source resistor,
then neither is the average of the instantaneous power. And
we have shown that the instantaneous power is not dissipated
in the source resistor.

Actually, we have shown exactly the opposite. Maybe a
different example will help. We are going to replace
the 23.5+j44.1 ohm load with a *phase-locked* signal
generator equipped with a circulator and 50 ohm load.
The signal generator supplies 18 watts back to the
original source and dissipates all of the incident
power of 50 watts, i.e. all of the forward power from
the original source. The original source sees
*exactly the same 23.5+j44.1 ohms as a load*.

Rs Vg
+----/\/\/-----+----------------+--2---1-----+
| 50 ohm \ / |
| 3 18 watt
Vs 1 wavelength | Signal
100v RMS 50 ohm line 50 Generator
| ohms |
| | |
+--------------+----------------+----+-------+
gnd gnd


The conditions at the original source are identical.
The circulator load resistor is dissipating the 50 watts
of forward power supplied by the original source. The
signal generator is sourcing 18 watts.

Your instantaneous power equations have not changed.
Are you going to try to tell us that the 18 watts from
the signal generator are not dissipated in Rs???? If
not in Rs, where????
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
If the
instantaneous power is not dissipated in the source resistor,
then neither is the average of the instantaneous power.

Let's see what happens when we use that same logic with my
AC wall sockets:

If the instantaneous AC voltage at my QTH is ever
non-zero, then the average voltage cannot be zero.

Agree? Disagree?
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Mar 12, 9:49 am, Cecil Moore wrote:
Keith Dysart wrote:
Actually, we have shown that that is not the case.

If the reflected energy is not being dissipated in
the source resistor, where does it go?

Good question.

It is not
dissipated in the load, by definition, and there is
no other source of dissipation in the network besides
the source resistor.

Not true. The voltage source also absorbs energy whenever
its voltage is positive but its current is negative.

There are no reflections and
no average interference to redistribute the reflected
energy back toward the load. The reflected wave
possesses energy and momentum which must be conserved.
Do your reflected waves obey your every whim and just
disappear and reappear as willed by you in violation
of the conservation of energy principle?

It would be nice, but, unfortunately, no.

You might like to visit
http://keith.dysart.googlepages.com/radio6
for a graph that shows the actual power dissipated in the
source resistor along with the power in the reflected
wave. It can be visually seen that the power dissipated in
the source resistor has no relationship to the sum of the
reflected power and the power that would be dissipated in
the source resistor if there was no reflection.

The difference in your energy levels is in the reactance.
The reactance stores energy and delivers it later. Why
are you having so much trouble with that age-old concept?

I have no trouble with the concept, but the exposition you
have offered is weak.

Show me the reactance and its power function of time such that
it stores and releases the energy at the correct time.

Otherwise... Just handwaving.

The reflected energy that is not dissipated in the source
resistor at time 't' is stored in the reactance and dissipated
90 degrees later. Until you choose to account for that time
delay, your energy equations are not going to balance.

I see no reactance that performs this function.

But the actual answer is that it is the voltage source which is
absorbing the energy for part of the cycle and delivering the extra
energy to the source resistor for the other part of the cycle.

So again, it is not the energy in the reflected wave that accounts
for the change of the dissipation in the source resistor.

....Keith

The Rest of the Story
 

On Mar 12, 10:17 am, Cecil Moore wrote:
Keith Dysart wrote:
Actually, we have shown that that is not the case. If the
instantaneous power is not dissipated in the source resistor,
then neither is the average of the instantaneous power. And
we have shown that the instantaneous power is not dissipated
in the source resistor.

Actually, we have shown exactly the opposite. Maybe a
different example will help. We are going to replace
the 23.5+j44.1 ohm load with a *phase-locked* signal
generator equipped with a circulator and 50 ohm load.
The signal generator supplies 18 watts back to the
original source and dissipates all of the incident
power of 50 watts, i.e. all of the forward power from
the original source. The original source sees
*exactly the same 23.5+j44.1 ohms as a load*.

Rs Vg
+----/\/\/-----+----------------+--2---1-----+
| 50 ohm \ / |
| 3 18 watt
Vs 1 wavelength | Signal
100v RMS 50 ohm line 50 Generator
| ohms |
| | |
+--------------+----------------+----+-------+
gnd gnd

The conditions at the original source are identical.
The circulator load resistor is dissipating the 50 watts
of forward power supplied by the original source. The
signal generator is sourcing 18 watts.

Your instantaneous power equations have not changed.
Are you going to try to tell us that the 18 watts from
the signal generator are not dissipated in Rs???? If
not in Rs, where????

As you note, the conditions at the original source have
not changed, so, since the energy in the reflected wave
was not dissipated in the source resistor in the original
circuit, it is not dissipated there in the revised circuit.

So, no changes at the original source.

Let us look at the circulator. Assuming a circulator
which does not accumulate energy, the net energy into
circulator must be zero.

0 = Pcp1(t) + Pcp2(t) + Pcp3(t)

Let us start with the 18 W producing 0 output.

Let us turn on the original source
Pg(t) = 50 + 50cos(2wt)

After one cycle, the wave reaches the circulator,
so at the circulator
Pcp1(t) = 0
Pcp2(t) = 50 + 50cos(2wt)
Pcp3(t) = -50 - 50cos(2wt)
which satisifies
0 = Pcp1(t) + Pcp2(t) + Pcp3(t)

Now turn on the right signal generator set to produce
18 W. Using the phase relationship needed to reproduce
the conditions from the original experiment
Pcp1(t) = 18 - 18cos(2wt)

This alters the voltage and current conditions at Port 2
of the circulator so that power delivered to this port is
now
Pcp2(t) = 32 + 68cos(wt)

and the power into Port 3 remains
Pcp3(t) = -50 - 50cos(2wt)

which still satisfies
0 = Pcp1(t) + Pcp2(t) + Pcp3(t)

Since turning on the 18 W generator altered the conditions
at Port 2, this change in line state propagates back towards
the original source and this state change reaches point Vg
Pg.before(t) = 50 + 50cos(2wt)
changes to
Pg(t) = 32 + 68cos(2wt)
which is good since this is the same power extracted from
the line at the circulator.

The changed load conditions at Vg alter the dissipation in
the source resistor as well as changing the power delivered
by the voltage source.

So where is the 18 Watts from the generator dissipated?
This is obvious. When the 18 W generator was turned on,
the power delivered to the circulator from the line (i.e.
the power delivered into Port 2, Pcp2(t) above) dropped,
but the power dissipated in the 50 ohm resistor stayed
the same. So it must be that the 18 W from Port 1 is now
being used to heat the resistor attached to Port 3.

....Keith

The Rest of the Story
 

On Mar 13, 7:02*pm, Cecil Moore wrote:
Keith Dysart wrote:
If the
instantaneous power is not dissipated in the source resistor,
then neither is the average of the instantaneous power.

Let's see what happens when we use that same logic with my
AC wall sockets:

If the instantaneous AC voltage at my QTH is ever
non-zero, then the average voltage cannot be zero.

Agree? Disagree?

Agree with what? That it is the same logic? You will have to
expand on your question.

Regardless, the average voltage is zero.

...Keith

The Rest of the Story
 

Keith Dysart wrote:
Show me the reactance and its power function of time such that
it stores and releases the energy at the correct time.

You've got to be kidding - EE201.

I see no reactance that performs this function.

Have you no ideal what the +j44.1 ohms means?
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
So where is the 18 Watts from the generator dissipated?
This is obvious. When the 18 W generator was turned on,
the power delivered to the circulator from the line (i.e.
the power delivered into Port 2, Pcp2(t) above) dropped,
but the power dissipated in the 50 ohm resistor stayed
the same. So it must be that the 18 W from Port 1 is now
being used to heat the resistor attached to Port 3.

Keith, I'm going to now leave you alone with your new
religion. Good grief!
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
Regardless, the average voltage is zero.

No, that contradicts what you said before which
was, if it ever wasn't zero instantaneously, it
cannot possibly average out to zero.

I'm sorry, Keith. Your assertions have gotten
just as irrational as my 96 year old aunt's.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
Show me the reactance and its power function of time such that
it stores and releases the energy at the correct time.
I see no reactance that performs this function.

Huh??? When the instantaneous source voltage is zero,
the instantaneous source power is zero. Yet, there
is instantaneous power dissipation in every resistive
element in a circuit with a reactive component. Where
do you reckon that power is coming from? It is apparent
that you are now just trying to pull my leg with your
ridiculous assertions. That's too bad. For awhile, I
thought you were serious.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
On Mar 13, 7:02 pm, Cecil Moore wrote:
Keith Dysart wrote:
If the
instantaneous power is not dissipated in the source resistor,
then neither is the average of the instantaneous power.
Let's see what happens when we use that same logic with my
AC wall sockets:

If the instantaneous AC voltage at my QTH is ever
non-zero, then the average voltage cannot be zero.

Agree? Disagree?

Agree with what? That it is the same logic? You will have to
expand on your question.

Regardless, the average voltage is zero.

...Keith

Keith;

Put your fingers across the bare wires and then tell me the results, if
you are still around that is. ;^)

Dave N

The Rest of the Story
 

Keith Dysart wrote:
So where is the 18 Watts from the generator dissipated?
This is obvious. When the 18 W generator was turned on,
the power delivered to the circulator from the line (i.e.
the power delivered into Port 2, Pcp2(t) above) dropped,

Sorry Keith, that's not the way circulators work. The
power incident upon port 2 does NOT change when the
18w generator is turned on.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
Cecil Moore wrote:
The difference in your energy levels is in the reactance.
The reactance stores energy and delivers it later. Why
are you having so much trouble with that age-old concept?

Show me the reactance and its power function of time such that
it stores and releases the energy at the correct time.

I don't have time to teach you AC circuit theory which
you really do need to understand. The power function
for the reactance is of the same form as it is for a
resistance, i.e. V(t)*I(t). I haven't seen you include
that reactance power function anywhere. When you do,
you will find your "missing" power and balance your
energy equations.

The reflected energy that is not dissipated in the source
resistor at time 't' is stored in the reactance and dissipated
90 degrees later. Until you choose to account for that time
delay, your energy equations are not going to balance.

I see no reactance that performs this function.

Please go take an AC circuits course. That's exactly
what a reactance does. The source terminals see +j44.1
ohms of reactance looking into the transmission line.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Mar 13, 10:33*pm, Cecil Moore wrote:
Keith Dysart wrote:
Regardless, the average voltage is zero.

No, that contradicts what you said before which
was, if it ever wasn't zero instantaneously, it
cannot possibly average out to zero.

If you could kindly point out which of my previous
writings you misinterpreted to be say this, I will
gladly correct your misunderstanding.

...Keith

The Rest of the Story
 

On Mar 13, 11:48*pm, "David G. Nagel"
wrote:
Keith Dysart wrote:
On Mar 13, 7:02 pm, Cecil Moore wrote:
Keith Dysart wrote:
If the
instantaneous power is not dissipated in the source resistor,
then neither is the average of the instantaneous power.
Let's see what happens when we use that same logic with my
AC wall sockets:

If the instantaneous AC voltage at my QTH is ever
non-zero, then the average voltage cannot be zero.

Agree? Disagree?

Agree with what? That it is the same logic? You will have to
expand on your question.

Regardless, the average voltage is zero.

...Keith

Keith;

Put your fingers across the bare wires and then tell me the results, if
you are still around that is. ;^)

Dave N- Hide quoted text -

- Show quoted text -

The average of a sine wave is 0. Positive half the time,
negative half the time, sums to 0.

Perhaps you are confusing average with root-mean-square?

...Keith

The Rest of the Story
 

On Mar 13, 10:28*pm, Cecil Moore wrote:
Keith Dysart wrote:
Show me the reactance and its power function of time such that
it stores and releases the energy at the correct time.

You've got to be kidding - EE201.

I see no reactance that performs this function.

Have you no ideal what the +j44.1 ohms means?

This is the third opportunity you have had to clearly
identify the component and show that its energy flow
as a function of time is exactly that needed to store
and release the energy in the reflected wave.

Since you have not done so, I conclude that you can't
find it either.

...Keith

The Rest of the Story
 

On Mar 13, 11:22*pm, Cecil Moore wrote:
Keith Dysart wrote:
Show me the reactance and its power function of time such that
it stores and releases the energy at the correct time.
I see no reactance that performs this function.

Huh??? When the instantaneous source voltage is zero,
the instantaneous source power is zero.

So true.
Yet, there
is instantaneous power dissipation in every resistive
element in a circuit with a reactive component. Where
do you reckon that power is coming from? *

Fairly obviously, the equations I have presented show
that it comes from energy stored in the line.

Recall that
Pg(t) = 32 + 68cos(2wt)

For some of the cycle, energy flow is from the line
towards the resistor and the voltage source.

But this is not the energy in the reflected wave
which has the function
Pr.g(t) = -18 + cos(2wt)
and only flows in one direction, towards the source.
And it is this supposed energy that can not be
accounted for in the dissipation of the source
resistor.

It is apparent
that you are now just trying to pull my leg with your
ridiculous assertions. That's too bad. For awhile, I
thought you were serious.

Completely serious, I am.

...Keith

The Rest of the Story
 

On Mar 14, 12:49*am, Cecil Moore wrote:
Keith Dysart wrote:
So where is the 18 Watts from the generator dissipated?
This is obvious. When the 18 W generator was turned on,
the power delivered to the circulator from the line (i.e.
the power delivered into Port 2, Pcp2(t) above) dropped,

Sorry Keith, that's not the way circulators work. The
power incident upon port 2 does NOT change when the
18w generator is turned on.

You need to read more carefully. I made no statement about
the energy incident upon port 2, only about the energy
flowing into port 2, which, after the 18 W generator is
turned on, is
Pcp2(t) = 32 + 68cos(wt)

...Keith

The Rest of the Story
 

On Mar 14, 12:59*am, Cecil Moore wrote:
Keith Dysart wrote:
Cecil Moore wrote:
The difference in your energy levels is in the reactance.
The reactance stores energy and delivers it later. Why
are you having so much trouble with that age-old concept?

Show me the reactance and its power function of time such that
it stores and releases the energy at the correct time.

I don't have time to teach you AC circuit theory which
you really do need to understand. The power function
for the reactance is of the same form as it is for a
resistance, i.e. V(t)*I(t). I haven't seen you include
that reactance power function anywhere. When you do,
you will find your "missing" power and balance your
energy equations.

A fourth opportunity missed.

The reflected energy that is not dissipated in the source
resistor at time 't' is stored in the reactance and dissipated
90 degrees later. Until you choose to account for that time
delay, your energy equations are not going to balance.

I see no reactance that performs this function.

Please go take an AC circuits course. That's exactly
what a reactance does. The source terminals see +j44.1
ohms of reactance looking into the transmission line.

Yes indeed. The reactance looking into the line. But
the reflected wave is not going into the line so this
is not a reactance that it sees.

Recall, the whole premise of forward and reverse waves
is that they see the line characteristic impedance, in
these examples, a real 50 ohms with no reactance.

...Keith

The Rest of the Story
 

Keith Dysart wrote:
Yes indeed. The reactance looking into the line. But
the reflected wave is not going into the line so this
is not a reactance that it sees.

Part of the reflected wave energy is going into the
reactance along with part of the forward wave energy
when the instantaneous interference between those two
waves is destructive at the source resistor. Anything
else would violate the conservation of energy principle.
That same energy is returned to the source resistor 90
degrees later as constructive interference. Your missing
energy is in the reactance. Now you know why Hecht said
instantaneous power is "of limited utility". Where the
instantaneous energy is at any point in time is a
complicated mess that you haven't solved.

It is the average power that really matters and all the
average reflected power is dissipated in the source
resistor when the reflected wave is 90 degrees out of
phase with the forward wave but under no other conditions.

Instantaneous power is completely irrelevant to the
average power data posted on my web page. You are
saying that because my pickup has black tires, it
is not a white pickup. My article stands as written
with a disclaimer about any importance being attached
to instantaneous power.

Walter Maxwell didn't deal with instantaneous powers.
Steven Best didn't deal with instantaneous powers.
To the best of my knowledge, the instantaneous power
straw man was invented by you for the purpose of
muddying the waters.

Your instantaneous power analysis is also incorrect
because you completely ignored the instantaneous
power in the system reactance. You have completely
ignored the role of destructive and constructive
interference. You cannot possibly understand where
the energy goes until you understand interference.

We have reached the end of the discussion road and
hashed it to death. If we are still in disagreement,
we are just going to have to agree to disagree.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
This is the third opportunity you have had to clearly
identify the component and show that its energy flow
as a function of time is exactly that needed to store
and release the energy in the reflected wave.

P(t).reactance = [V(t).reactance][I(t).reactance]

Where is that term in your equations?
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
You need to read more carefully. I made no statement about
the energy incident upon port 2, only about the energy
flowing into port 2, which, after the 18 W generator is
turned on, is
Pcp2(t) = 32 + 68cos(wt)

If we have two pipes each carrying one gallon of water
per minute in opposite directions, we can agree that
the net flow of water is zero. But you are taking it
one step farther and arguing there is no water flowing
at all which is a ridiculous assertion. I'm going to
ignore this latest obvious diversion.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Mar 14, 8:08*am, Cecil Moore wrote:
Keith Dysart wrote:
You need to read more carefully. I made no statement about
the energy incident upon port 2, only about the energy
flowing into port 2, which, after the 18 W generator is
turned on, is
Pcp2(t) = 32 + 68cos(wt)

If we have two pipes each carrying one gallon of water
per minute in opposite directions, we can agree that
the net flow of water is zero. But you are taking it
one step farther and arguing there is no water flowing
at all which is a ridiculous assertion. I'm going to
ignore this latest obvious diversion.
--
73, Cecil *http://www.w5dxp.com

If there were two transmission lines, then I could see
why you might want two pipes in an analogy.

But since there is only one transmission line, an
analogy with one pipe makes more sense.

So you want to argue that when there is one pipe
with no water flowing, what is really happening is
that one gallon per minute is simultaneously flowing
in each direction. In the same pipe. At the same
time. I don't buy it.

You should think a bit more about
Pcp2(t) = 32 + 68cos(wt)

It is the time rate of energy flow into the port.
It can trivially be computed from the voltage and
current functions at that port.

It sums with the energy flows into the other ports
appropriately to satisfy the principle of conservation
of energy.

All is well.

And there is only one pipe for each port.

...Keith

The Rest of the Story
 

On Mar 14, 7:59*am, Cecil Moore wrote:
Keith Dysart wrote:
This is the third opportunity you have had to clearly
identify the component and show that its energy flow
as a function of time is exactly that needed to store
and release the energy in the reflected wave.

P(t).reactance = [V(t).reactance][I(t).reactance]

Where is that term in your equations?

It is unnecessary. But if you believe me wrong, show me
where it goes, compute the values, and show how it
accounts for the energy that is not dissipated in the
source resistor.

...Keith

The Rest of the Story
 

On Mar 14, 7:53*am, Cecil Moore wrote:
Keith Dysart wrote:
Yes indeed. The reactance looking into the line. But
the reflected wave is not going into the line so this
is not a reactance that it sees.

Part of the reflected wave energy is going into the
reactance along with part of the forward wave energy
when the instantaneous interference between those two
waves is destructive at the source resistor. Anything
else would violate the conservation of energy principle.
That same energy is returned to the source resistor 90
degrees later as constructive interference. Your missing
energy is in the reactance.

Still handwaving. Show the expressions and the numbers
that make it balance. Otherwise, just handwaving.

Now you know why Hecht said
instantaneous power is "of limited utility".

I've known that for quite a while. It is because it is so
difficult to measure with optical signals.

Where the
instantaneous energy is at any point in time is a
complicated mess that you haven't solved.

Not at all. Follow the spreadsheet for a full accounting.

It is the average power that really matters and all the
average reflected power is dissipated in the source
resistor when the reflected wave is 90 degrees out of
phase with the forward wave but under no other conditions.

Averaging is a mathetical operation applied to a function.
In this case a function of time.

The underlying function of time conveys more information than
does just the average which is why just dealing with averages
can lead one astray.

Instantaneous power is completely irrelevant to the
average power data posted on my web page.

Well, it does disprove some of your claims so I can see
why you like to belittle it.

You are
saying that because my pickup has black tires, it
is not a white pickup. My article stands as written
with a disclaimer about any importance being attached
to instantaneous power.

See previous comment.

Walter Maxwell didn't deal with instantaneous powers.
Steven Best didn't deal with instantaneous powers.
To the best of my knowledge, the instantaneous power
straw man was invented by you for the purpose of
muddying the waters.

Your instantaneous power analysis is also incorrect
because you completely ignored the instantaneous
power in the system reactance. You have completely
ignored the role of destructive and constructive
interference. You cannot possibly understand where
the energy goes until you understand interference.

I observe that you have not provided any expansion
based on either reactance or interference that accounts
for the differences.
Most probably because it is not possible.

We have reached the end of the discussion road and
hashed it to death. If we are still in disagreement,
we are just going to have to agree to disagree.

Are you really going to let me be the last man standing
this time?

We shall see.

...Keith

The Rest of the Story
 

Keith Dysart wrote:
If there were two transmission lines, then I could see
why you might want two pipes in an analogy.

But since there is only one transmission line, an
analogy with one pipe makes more sense.

Unfortunately for your argument, molecular water and
EM waves are considerably different animals. Water
energy traveling in opposite directions in a pipe
interact. EM waves traveling in opposite directions
in a transmission line interact only at an impedance
discontinuity or at an impedor. As long as only a
constant Z0 environment exists, the forward wave and
reflected wave pass like ships in the night. For you
to prove otherwise, you are going to have to define
the position and momentum of a single photon. Good
luck on that one.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
On Mar 14, 7:59 am, Cecil Moore wrote:
P(t).reactance = [V(t).reactance][I(t).reactance]
Where is that term in your equations?

It is unnecessary. But if you believe me wrong, show me
where it goes, compute the values, and show how it
accounts for the energy that is not dissipated in the
source resistor.

It is unnecessary to account for all of the instantaneous
power???? Your problem is greater than just a simple
misunderstanding of the laws of physics by which we must
all abide.

The DC energy is stored in your vehicle's battery until
it is needed to start your vehicle. That delay between
stored energy and needed energy is related to the
(undefined) wavelength. Think about it.

In an AC circuit, the reactance has no say as to when
to store the energy and when it is delivered back to the
system. It is also related to wavelength which is defined.

When the source voltage is zero at its zero crossing
point/time, the instantaneous power dissipation in
the source resistor is NOT zero! Doesn't that give
you pause to wonder where the instantaneous power
is coming from when the instantaneous power delivered
by the source is zero???? Why do you ignore that power
and try to sweep it under the rug?

You are making the mistakes that your EE 201 professor
warned you not to make. You are superposing powers,
something that all the gurus on this newsgroup will
condemn. Until you learn not to superpose powers, you
will remain forever ignorant. Richard C. made the same
mistake when he declared that the reflections from non-
reflective thin-film coatings on glass are "brighter
than the surface of the sun". If one ignores the laws
of physics, anything is possible.

There is a condition where it is valid to superpose
powers. That is when (V1^2 + V2^2) = (V1 + V2)^2
You have obviously not satisfied that condition and
are paying for your violations of those laws of physics.
Until you give up on your superposition of powers, I
cannot help you.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

Keith Dysart wrote:
Still handwaving. Show the expressions and the numbers
that make it balance. Otherwise, just handwaving.

Show the expressions and the numbers for how many angles
can dance on the head of a pin???? Shirley, you jest.
This is not the proper venue to try to establish your
new religion.

Are you really going to let me be the last man standing
this time?

No, you died in action a few weeks ago and don't realize
it. It happened the first time you superposed powers when
(V1^2 + V2^2) is not equal to (V1 + V2)^2. Too bad you
are incapable of comprehending exactly what that means.
You can easily work it out for yourself but you haven't
yet attempted to do so. Hopefully, one of the resident
gurus whom you trust will explain it to you.

Until you correct your ignorance on that subject, *you* are
just handwaving. I suggest you contact someone who is more
knowledgeable than you on the subject. Ask Richard C. to
prove his assertion that the reflections from a non-reflective
thin-film surface are brighter than the sun. If you detect
what is wrong with his argument, you will know what is wrong
with yours.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Mar 14, 6:00*pm, Cecil Moore wrote:
Keith Dysart wrote:
Still handwaving. Show the expressions and the numbers
that make it balance. Otherwise, just handwaving.

Show the expressions and the numbers for how many angles
can dance on the head of a pin???? Shirley, you jest.
This is not the proper venue to try to establish your
new religion.

Are you really going to let me be the last man standing
this time?

No, you died in action a few weeks ago and don't realize
it. It happened the first time you superposed powers when
(V1^2 + V2^2) is not equal to (V1 + V2)^2. Too bad you
are incapable of comprehending exactly what that means.
You can easily work it out for yourself but you haven't
yet attempted to do so. Hopefully, one of the resident
gurus whom you trust will explain it to you.

Until you correct your ignorance on that subject, *you* are
just handwaving. I suggest you contact someone who is more
knowledgeable than you on the subject. Ask Richard C. to
prove his assertion that the reflections from a non-reflective
thin-film surface are brighter than the sun. If you detect
what is wrong with his argument, you will know what is wrong
with yours.
--
73, Cecil *http://www.w5dxp.com

The Rest of the Story
 

On Mar 14, 5:29*pm, Cecil Moore wrote:
Keith Dysart wrote:
If there were two transmission lines, then I could see
why you might want two pipes in an analogy.

But since there is only one transmission line, an
analogy with one pipe makes more sense.

Unfortunately for your argument, molecular water and
EM waves are considerably different animals. Water
energy traveling in opposite directions in a pipe
interact. EM waves traveling in opposite directions
in a transmission line interact only at an impedance
discontinuity or at an impedor. As long as only a
constant Z0 environment exists, the forward wave and
reflected wave pass like ships in the night. For you
to prove otherwise, you are going to have to define
the position and momentum of a single photon. Good
luck on that one.

I accept your retraction of your analogy.

...Keith

The Rest of the Story
 

On Mar 14, 6:00*pm, Cecil Moore wrote:
Keith Dysart wrote:
Still handwaving. Show the expressions and the numbers
that make it balance. Otherwise, just handwaving.

Show the expressions and the numbers for how many angles
can dance on the head of a pin???? Shirley, you jest.
This is not the proper venue to try to establish your
new religion.

First you proffer reactance or interference or both as the
solution to the missing energy, and then you characterize
it as 'angels dancing'. Which is it?

Are you really going to let me be the last man standing
this time?

No,

I didn't really expect so.

...Keith

The Rest of the Story
 

On Mar 14, 5:51*pm, Cecil Moore wrote:
Keith Dysart wrote:
On Mar 14, 7:59 am, Cecil Moore wrote:
P(t).reactance = [V(t).reactance][I(t).reactance]
Where is that term in your equations?

It is unnecessary. But if you believe me wrong, show me
where it goes, compute the values, and show how it
accounts for the energy that is not dissipated in the
source resistor.

It is unnecessary to account for all of the instantaneous
power???? Your problem is greater than just a simple
misunderstanding of the laws of physics by which we must
all abide.

The DC energy is stored in your vehicle's battery until
it is needed to start your vehicle. That delay between
stored energy and needed energy is related to the
(undefined) wavelength. Think about it.

In an AC circuit, the reactance has no say as to when
to store the energy and when it is delivered back to the
system. It is also related to wavelength which is defined.

When the source voltage is zero at its zero crossing
point/time, the instantaneous power dissipation in
the source resistor is NOT zero! Doesn't that give
you pause to wonder where the instantaneous power
is coming from when the instantaneous power delivered
by the source is zero????

Previously explained, as you may have forgotten, with
the energy cyclically returned from the line.

For your convenience, I copy from a previous post:
----
Recall that
Pg(t) = 32 + 68cos(2wt)

For some of the cycle, energy flow is from the line
towards the resistor and the voltage source.

But this is not the energy in the reflected wave
which has the function
Pr.g(t) = -18 + cos(2wt)
and only flows in one direction, towards the source.
And it is this supposed energy that can not be
accounted for in the dissipation of the source
resistor.
----

And, as expected
Ps(t) = Prs(t) + Pg(t)

Energy is conserved.

...Keith

The Rest of the Story
 

"Keith Dysart" wrote in message
...
On Mar 14, 5:51 pm, Cecil Moore wrote:
Energy is conserved.

qed.

now, can we get back to antennas?? its much more fun mocking art than the
endless discussions of electrons and holes sloshing back and forth.

The Rest of the Story
 

Keith Dysart wrote:
And, as expected
Ps(t) = Prs(t) + Pg(t)

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?
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Mar 16, 10:21*am, Cecil Moore wrote:
Keith Dysart wrote:
And, as expected
Ps(t) = Prs(t) + Pg(t)

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?

Vs(t) = 1.414cos(wt)

After settling, from some circuit analysis...

Vg(t) = 100 cos(wt+45degrees)
Ig(t) = 2 cos(wt-45degrees)

Vrs(t) = Vs(t) - Vg(t)
= 100 cos(wt-45degrees)
Irs(t) = Ig(t)
= 2 cos(wt-45degrees)

Is(t) = Ig(t)
= 2 cos(wt-45degrees)

Ps(t) = Vs(t) * Is(t)
= 100 + 141.4213562 cos(2wt-45degrees)

Prs(t) = Vrs(t) * Irs(t)
= 100 + 100 cos(2wt-90degrees)

Pg(t) = Vg(t) * Ig(t)
= 0 + 100 cos(2wt)

For confirmation of conservation of energy, the above
is in agreement with
Ps(t) = Prs(t) + Pg(t)

Ps(t) = 100 + 141.4213562 cos(2wt-45degrees)
so, Ps(t) = 0, occurs whenever
141.4213562 cos(2wt-45degrees)
is equal to
-100
which, for example, would happen when 2wt = 180 degrees
or wt = 90 degrees.

At this time, again for example, the energy being
dissipated in the source resistor
Prs(t) = 100 + 100 cos(2wt-90degrees)
= 100 + 0
= 100

Since no energy is being delivered from the source,
(Ps(t) is 0), then given
Ps(t) = Prs(t) + Pg(t)
the energy must be coming from the line. Let us
check the energy flow at the point Vg at this
time
Pg(t) = 0 + 100 cos(2wt)
= -100
as required.

So the energy being dissipated in the source
resistor at this time is being returned from
the line.

Just for completeness we can compute the line
state at the point Vg in terms of forward and
reverse waves...

Vf.g(t) = 70.71067812 cos(wt)
Vr.g(t) = 70.71067812 cos(wt+90degrees)
Vg(t) = Vf.g(t) + Vr.g(t)
= 100 cos(wt+45degrees)

If.g(t) = 1.414213562 cos(wt)
Ir.g(t) = 1.414213562 cos(wt-90degrees)
Ig(t) = If.g(t) + Ir.g(t)
= 2 cos(wt-45degrees)

Pf.g(t) = Vf.g(t) * If.g(t)
= 50 + 50 cos(2wt)

Pr.g(t) = Vr.g(t) * Ir.g(t)
= -50 + 50 cos(2wt)

And since
Pg(t) = Pf.g(t) + Pr.g(t)
= 0 + 100 cos(2wt)
confirming the previously computed values.

It is valuable to examine Pr.g(t) at the time when
Ps(t) is zero. Substituting wt = 90degrees
into Pr.g(t)...

Pr.g(t) = -50 + 50 cos(2wt)
= -50 -50
= -100
which would appear to be the 100 watts needed to
heat the source resistor. This is misleading.
When all the values of t are examined it will be
seen that only the sum of Pf.g(t) and Pr.g(t),
that is, Pg(t), provides the energy not provided
from the source that heats the source resistor.
wt=90 is a special case in that Pf.g(t) is 0 at
this particular time.

The line input impedance does have a reactive
component and it is this reactive component that
can store and return energy. The energy flow
into and out of this impedance (pure reactance
for the example under consideration) is described
by Pg(t).

In summary,
Ps(t) = Prs(t) + Pg(t)

and by substitution, if the solution is preferred
in terms of Pf and Pr,
Ps(t) = Prs(t) + Pf.g(t) + Pr.g(t)

Ps and Pr alone are insufficient to explain the
heating of the source resistor.

...Keith

The Rest of the Story
 

Keith Dysart wrote:
Pr.g(t) = -50 + 50 cos(2wt)
= -50 -50
= -100
which would appear to be the 100 watts needed to
heat the source resistor.

Which, contrary to your previous assertions, agrees
with what I have said previously. This is the
destructive interference energy stored in the
feedline 90 degrees earlier and being returned
as constructive interference to the source
resistor. It's impossible to sweep it under the
rug when the source power is zero, huh?

All power comes from the source. Since power is
being delivered to the source resistor at a time
when the source is delivering zero power, it must
have been previously been stored in the reactance
of the transmission line.
--
73, Cecil http://www.w5dxp.com

The Rest of the Story
 

On Mar 17, 1:06*am, Cecil Moore wrote:
Keith Dysart wrote:
Pr.g(t) = -50 + 50 *cos(2wt)
* * * * = -50 -50
* * * * = -100
which would appear to be the 100 watts needed to
heat the source resistor.

Which, contrary to your previous assertions, agrees
with what I have said previously. This is the
destructive interference energy stored in the
feedline 90 degrees earlier and being returned
as constructive interference to the source
resistor. It's impossible to sweep it under the
rug when the source power is zero, huh?

The intriguing question is:
1. Do you just stop reading as soon as you find a snippet
that aligns with your claim?
2. Do you keep reading, but do not understand because you
are so gleeful at finding a snippet that aligns with
your claim?
3. Or do you read, understand, but choose to disingenuously
ignore that which follows the snippet that aligns with
your claim?

Regardless, if you use the snippet above to support
your claim, you have effectively modified your claim.
You are now saying that the reflected energy is
dissipated in the source resistor only at particular
times, such as when the source voltage is 0, and that
you are not interested in what happens during the rest
of the cycle.

For example, if you were to do the same analysis,
except do it for t such that wt equals 100 degrees,
instead of 90, you would find that you need more terms
than just Pr.g to make the energy flows balance.

When wt equals 100 degrees
Ps = -28.170
Prs = 65.798
Pg = -93.96
so, as expected
Ps = Prs + Pg

And
Pr.g = -96.985
Pf.g = 3.015
Ooooppppss, no way to make those add to 65.798.

But
Ps = Prs + Pf.g + Pr.g
since
Pg = Pf.g + Pr.g

So all is well with world.

All power comes from the source. Since power is
being delivered to the source resistor at a time
when the source is delivering zero power, it must
have been previously been stored in the reactance
of the transmission line.

This is true. But it is, of course, Pg(t) that
describes the energy flow in and out of the line,
not Pr.g(t).

So are you now agreeing that it is not the energy
in the reflected wave that accounts for the
difference in the heating of the source resistor
but, rather, the energy stored and returned from
the line, i.e. Pg(t)?

Note also, that when wt is 100 degrees, not only is
the source resistor being heated by energy being
returned from the line, the source is also absorbing
some of the energy being returned from the line
(Ps = -28.170).

...Keith

The Rest of the Story
 

Keith Dysart wrote:
Regardless, if you use the snippet above to support
your claim, you have effectively modified your claim.

False. My claim is what it has always been which is:
An amateur radio antenna system obeys the conservation
of energy principle and abides by the principles of
superposition (including interference) and the wave
reflection model. Everything I have claimed falls out
from those principles.

Your claims, however, are in direct violation of the
principles of superposition and of the wave reflection
model, e.g. waves smart enough to decide to be reflected
when the physical reflection coefficient is 0.0. Your
claims even violate the principles of AC circuit theory,
e.g. a reactance doesn't store energy and deliver it back
to the system at a later time in the same cycle.

You are now saying that the reflected energy is
dissipated in the source resistor only at particular
times, such as when the source voltage is 0, and that
you are not interested in what happens during the rest
of the cycle.

You haven't read my article yet, have you? Here's a quote:
"For this *special case*, it is obvious that the reflected energy
from the load is flowing through the source resistor, RS, and
is being dissipated there. But remember, we chose a special
case (resistive RL and 1/8 wavelength feedline) in order to
make that statement true and it is *usually not true* in the
general case."

If there is one case where your assertion is wrong, then
your assertion is false. I found that special case when
the source voltage is zero that makes your assertions
false.

For example, if you were to do the same analysis,
except do it for t such that wt equals 100 degrees,
instead of 90, you would find that you need more terms
than just Pr.g to make the energy flows balance.

Yes, you have realized that destructive and constructive
interference energy must be accounted for to balance the
energy equations. I have been telling you that for weeks.

I repeat: The *only time* that reflected energy is 100%
dissipated in the source resistor is when the two component
voltages satisfy the condition: (V1^2 + V2^2) = (V1 + V2)^2.
None of your examples have satisfied that necessary condition.
All it takes is one case to prove the following assertion false:

"Reflected energy is *always* re-reflected from the
source and redistributed back toward the load."

You appear to think that if you can find many cases where
an assertion is true, then you can simply ignore the cases
where it is not true. I have presented some special cases
where it is not true. It may be true for 99.9% of cases,
but that nagging 0.1% makes the statement false overall.

Equally false is the assertion: "Reflected energy is
always dissipated in the source resistor." The amount
of reflected energy dissipated in the source resistor
can vary from 0% to 100% depending upon network
conditions. That statement has been in my article from
the beginning.

So are you now agreeing that it is not the energy
in the reflected wave that accounts for the
difference in the heating of the source resistor
but, rather, the energy stored and returned from
the line, i.e. Pg(t)?

I have already presented a case where there is *zero*
power dissipated in the source resistor in the presence
of reflected energy so your statement is obviously just
false innuendo, something I have come to expect from
you when you lose an argument.
--
73, Cecil http://www.w5dxp.com