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