Roy Lewallen wrote:
Cecil Moore wrote:
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.

Do you think it is random coincidence that the amount of energy
stored in the feedline is *EXACTLY* the amount of energy required
by the forward and reflected waves???? The fact that you think
it is irrelevant is simply a flight into a wet dream fantasy.

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 ...

What a coincidence! It's *EXACTLY* the amount of energy required
by the forward and reflected waves that you say don't exist - and
it's RF photons, not DC, so it must travel at the speed of light!
You cannot store photons in a capacitor.

Surely even you can do the basic circuit analysis which shows that the
source is continuously generating 40 watts, ...

Of course, but during the power-on transient period, 68 watts is NOT
delivered to the load. Please don't make me waste my time calculating
the forward and reflected power during the power-on transient period.
When you actually do those calculations, you will agree with me that,
during the power-on transient phase, 68 watts of power has been stored
in the feedline and remains there until the power-off transient phase.
It is *EXACTLY* the amount of power required by the forward and
reflected waves.

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.

Which must necessarily include the energy stored in the feedline during
the power-on transient condition because it is *still there during
steady-state*. Ignoring the energy stored in the feedline during the
power-on transient phase is both irrational and illogical. It could
even be bad for your mental health.

What exactly is my premise, please?

from my earlier posting:

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 sorry, Roy, but that is just BS! Either you admit there is enough
energy to support the steady-state forward and reflected waves or you
don't.

My analysis doesn't need to consider the stored energy at all.

And that is exactly why your analysis is wrong. You have, once again,
been seduced by the steady-state model and are spreading old wives'
tales as a result. Hopefully, you don't really want to do that.

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

I think you mean 68 joules.

68 joules/sec in your original example. 68 joules in my one-second-
long feedline example.

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

The 68 joules were stored in the feedline during the power-on
transient phase. They are still there during steady-state. An
S-parameter analysis yields the correct results because it
includes the reflected power, |a2|^2, as an energy source,
something you deny. Wonder what the S-parameter analysis folk
know that you don't choose to admit?

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?

It never encounters the source resistor as it is re-reflected by
wave cancellation, not by an impedance discontinuity. I have a QEX
article coming soon that will explain the details. Stand by. With
an unprejudiced open mind, you might actually learn something.

What's (s22*a2)^2?

Same as (1-rho^2)*Pref2 in ham terms. Reflected power from the
load that is re-reflected back toward the load by an impedance
discontinuity.

What's s11*a1?

Same as Vfor1*rho in ham terms. Forward voltage that is reflected
back toward the source by an impedance discontinuity.

I'll leave you to the folks who regard this kind of gobbledegook as
convincing evidence.

Do you realize what intellectual shape you would be in if you had
adopted that attitude when you were one year old? :-)
--
73, Cecil http://www.qsl.net/w5dxp


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