Cecil Moore wrote:
. . .
Yes, you are right about that. But one can visualize the interference
by inserting 1WL of lossless 200 ohm feedline between the source and
the 50 ohm line which changes virtually nothing outside of the 200
ohm line.

100v/50ohm source with 80v at the output terminals.

80v--1WL lossless 200 ohm line--+--1/2WL 50 ohm line--200 ohm load

The source sees the same impedance as before. The same impedance as
before is seen looking back toward the source. Voltages, currents,
and powers remain the same. But now the interference patterns are
external to the source and can be easily analyzed.

Hm, I wasn't having any trouble analyzing the system without the 200 ohm
line. Why do you have to make the system more complex in order to apply
your theory? Looking back at previous postings, it appears that any time
anyone presents a model that gives you difficulty, you simply modify it
to suit yourself, and deflect the discussion. I'm not interested in
whether you can explain your theories in models of your choice. What
remains to be shown is whether you can do so for the extremely simple
model I proposed. You might recall that the first example in my posting
in response to H's claim did indeed have source resistor dissipation
equal to the "reverse power" -- it's much easier to apply a defective
theory if you're free to choose special cases that support it. A valid
theory should be able to work on all models, unless you clearly give the
boundaries of its validity and why it has those limitations.

It baffles me how you think you can calculate the line's stored energy
without knowing its time delay.


The time delay was given at one second, Roy. I really wish you would
read my postings. Here is the quote from the earlier posting:

************************************************** *****************
* If we make Roy's lossless 50 ohm feedline one second long (an *
* integer number of wavelengths), during steady-state, the source *
* will have supplied 68 joules of energy that has not reached the *
* load. That will continue throughout steady state. The 68 joules *
* of energy will be dissipated by the system during the power-off *
* transient state. *
************************************************** *****************

I DIDN'T EVER TRY TO CALCULATE THE STORED ENERGY IN YOUR LINE. But a
VF could be assumed for your lossless line and a delay calculated
from the length. Or you can just scale my one second line down to a
one microsecond line. The results will conceptually be the same. Of
course, the one microsecond line would have to be defined as an integer
number of half wavelengths but the frequency could be chosen for that
result.

So 68 microjoules would be be stored in that one microsecond feedline,
50 microjoules in the forward wave and 18 microjoules in the reflected
wave. The source is still supplying 32 microjoules per microsecond and
there is exactly enough energy stored in the feedline to support the
energy in the forward wave and reflected wave as predicted by the wave
reflection model or S-parameter analysis.

The calculation of stored energy is simple enough, but it requires
knowledge of the line's time delay.


The time delay was given, Roy, at one second. See the above quote.
That technique changes watts to joules. Buckets of joules are not
as easy to hide as watts.

I apologize for not having read your posting more carefully. When I get
snowed or when the discussion deviates from the point in question, I do
tend to not read the rest. Because the stored energy has nothing to do
with what happens to the waves of bouncing average power in steady
state, I did ignore your details about it. Double my line length and the
waves of bouncing average power have the same values as before, although
the stored energy has doubled.


If the forward power and reflected power remain the same, doubling the
length of the feedline must necessarily double the amount of energy
stored in the forward and reflected waves. That fact supports my side
of the argument, not yours.

I'm not sure what you think my side of the argument is. That a
transmission line doesn't contain stored energy? Of course it does. (See
below, where I calculate it using conventional wave mechanics.) I'm
simply asking where your imaginary waves of bouncing average power go in
a painfully simple steady state system.

Your argument is that there are waves of bouncing average power. I asked
where they went in the simple circuit I described. Calculation of the
energy stored in the line does nothing to explain it. All it does is
create the necessary diversion to deflect the discussion from the fact
that you don't know or at best have only a vague and general idea.

Here's a derivation of the energy stored in the line using conventional
wave mechanics:

T is the time it takes a wave to traverse the line in one direction. The
analysis begins at t = 0, with the line completely discharged, at which
time the source is first turned on.

1. From time t = 0 to T, the impedance seen looking into the line is 50
ohms, so the 100 volt source sees 100 ohms. Source current = 1 amp,
source is delivering 100 watts, source resistor is dissipating 50 watts,
and load resistor is dissipating zero. Therefore the energy being put
into the line is 100 - 50 = 50 watts * t.
2. From time t = T to 2T, all the conditions external to the line are
the same as above, except that the load resistor is now dissipating 32
watts, so the net energy being put into the line from the source is 100
- 50 - 32 = 18 watts * (t - T).
3. After time = 2T, steady state occurs, with the conditions I gave
originally. From that time onward, the power into the line equals the
power out, so no additional energy is being stored.

It's obvious from the above that the energy stored in the line from 0 to
T = 50 * T joules, and from T to 2T = 18 * T joules, for a total energy
storage of 68 * T joules.

No bouncing waves of average power are required; this can be completely
solved, as can all other transmission line problems, by looking at
voltage and current waves, along with simple arithmetic. I didn't bother
doing this earlier simply because it's irrelevant; it has nothing to do
with steady state conditions any more than the DC charge on a capacitor
has to do with an AC circuit analysis. The only reason it's important is
that it provided you with yet another way to divert the discussion from
the question of what happens to those imaginary waves of bouncing
average power in the steady state.

In the steady state we've got energy stored in the line (of course), and
18 watts of "reverse power". 8 watts is being dissipated in the source
resistor. We can store just as much or little energy in the line as we
choose by changing its length; as long as the line remains an integral
number of half wavelengths long, there's no change to the line's
"forward power", "reverse power" or any external voltage, current, or
power. In fact, if we choose a line length that's not an integral number
of half wavelengths long, we change the dissipation in the source and
load resistors without any change in the "forward power", "reverse
power", or reflection coefficients at either end of the line.

Your postings indicate that in your mind you've completely explained
where the bouncing waves of average power are going, what they do, and
why. If I'm correct and this is indeed all you have to offer, I'll once
again bow out. I look forward to a careful reading of the QEX article.

Roy Lewallen, W7EL