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

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

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

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

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

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

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

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