Cecil Moore wrote:
Roy Lewallen wrote:

[No, I didn't. Cecil wrote the following paragraph.]

We are not going to get anywhere until you admit there is 68
joules/sec in the feedline that haven't yet made it to the load. Once
you admit that fact, everything else will be moot.

I never said a word about how much energy is stored in the feedline;
it's irrelevant. It's stored there during the initial charging to the
steady state condition, and the same amount remains there until the
steady state condition no longer exists. It's exactly like the DC charge
on a capacitor or a DC current through an inductor (which, in fact, is
exactly what the feedline stored energy consists of) -- it doesn't have
any effect on an AC analysis.

[I did write this one.]

Let's try again. The source is providing 40 watts, 32 watts of
which is delivered to the transmission line. The transmission line
is transferring this 32 watts of power to the load. In the
transmission line, we can calculate that there's 50 watts of
"forward power", and 18 watts of "reverse power".


And it is easy to prove that the source has generated 50+18=68 watts
that have not been delivered to the load.

Surely even you can do the basic circuit analysis which shows that the
source is continuously generating 40 watts, not 60. 32 of those are
delivered to the load and 8 to the source resistor. I guess you mean 68
joules -- but as I said, it's irrelevant.

So I ask you: Where are
those 68 joules/sec located during steady-state if not in the forward
and reflected power waves? Why will 68 joules/sec be dissipated in
the system *after* the source power is turned off?

The line's stored energy will be dissipated in either the source or load
resistor or both when the source power is turned off. But we're doing a
steady state analysis here.


If those 68 joules/sec that have been generated by the source but not
delivered to the load are not in the forward and reflected power
waves, exactly where are they located? There's really no sense in
continuing this discussion until you answer that question. Everything
else is just a side argument.

You tell me -- they can be anywhere you'd like. Just answer the simple
questions about the power "waves".


The answer to that question will expose the errors in your premises.

What exactly is my premise, please? All I've done is to give the
currents and powers at significant points in the circuit. Are any of the
values incorrect? It's you who has the premise, not me.

You are apparently assuming there is not enough energy in the system
during steady-state to support the forward and reflected power waves.

I'm making no such assumption. I'm questioning the existence of
traveling waves of average power, and so far you've failed to give any
evidence to convince me otherwise.

But that exact amount of energy was supplied during the power-on
transient state and will be dissipated during the power-off transient
state. If it's not in the forward and reflected power waves, you are
going to have to store it somewhere else. Where is that somewhere
else?

You tell me. My analysis doesn't need to consider the stored energy at
all. Apparently yours does, so have at it.

The source has supplied 68 joules/sec that has not reached the load.

I think you mean 68 joules.

The forward and reflected power waves require 68 joules/sec.

The "forward power" is 50 watts. The "reverse power" is 18 watts. It
requires 32 watts to sustain this. That's the amount of power flowing
through the transmission line, from source to load. That's 32
joules/second, not 68. If the line were open circuited, the forward and
reverse powers would be equal, and it would take no power to sustain them.

That you
don't see the logical connection between those two equal energy
values is amazing.

But I will get you started on an understanding of the component
powers using an S-parameter analysis.


Where will you find to put those all-important 68 joules in an
s-parameter analysis? That's a steady state analysis.

How much of that 18 watts of reverse power is going through the
source resistor to reach the source to "engage in destructive
interference"?


reference the S-parameter equation: b1 = s11*a1 + s12*a2

I calculate 11.52 watts. (s12*a2)^2 = 11.52 watts


Finally, an actual answer. So of the 18 watts of "reverse power", 11.52
watts is making it to the source to "engage in constructive
interference". Does any of it get dissipated in the source resistor, or
does it just slide through unscathed?

The other 6.48 watts are in (s22*a2)^2 where |a2|^2 = 18 watts

What's (s22*a2)^2? The forward power wave? The reverse power wave? If
it's something else, does it have a name? Where does it go? Is it
getting dissipated in the source resistor, reflected at the transmission
line/resistor interface, reflected at the source/resistor interface, get
radiated, or what?

What does it interfere with?


From the S-parameter equation above, it obviously interferes with
s11*a1 .

What's s11*a1? It must be something inside the source. The source is
just that, a source. It has (AC) voltage and current, 100 volts of
voltage and 0.4 amps of current. Does s11*a1 reside inside every source,
or only some special ones? Apparently 11.52 watts of this s11*a1 gets
cancelled by the reverse power wave. How much of it is left over?


Please reference HP App Note 95-1, available on the web. It should
answer most of your questions, in particular pages 16 & 17. |a1|^2 =
Power incident on the input of the network |a2|^2 = Power incident on
the output of the network |b1|^2 = Power reflected from the input
port of the network |b2|^2 = Power reflected from the output port of
the network

Nope.

Enough hand waving and evasion, we've been here before. [Of all the
questions, the sole quantitative answer was "(s12*a2)^2 = 11.52 watts".]
I'll leave you to the folks who regard this kind of gobbledegook as
convincing evidence. Have fun -- I've got actual work to do while you
take care of the visionary leadership part.

Roy Lewallen, W7EL