On Apr 20, 9:13 am, Cecil Moore wrote:
Keith Dysart wrote:
But a key realization is that the behaviour on the line
can not depend on arcane details about the construction of the
generator.

Agreed, but the behavior on the line also doesn't depend
upon where the forward and reverse traveling wave came from.
All that is important is their existence. The reverse
traveling wave might come from a separate source and the
line doesn't know the difference.

All true, and as long as you think of voltage and current waves
you won't get into trouble.

So when such statements are made, it is instructive
to explore the situation with a different generator. If the outcome
is different, then there is something wrong with the theory.

No, if the outcome is different, something different is
happening while abiding by the wave reflection model rules.
Why are you surprised that changes occur when something is
changed?

The key point is that the line conditions did not change, so the
same reflections must be occuring and yet your explanation
claims that sometimes nothing is reflected and sometimes
all is reflected.

Note
that the standard explanations only require that the impedance
of the source be known.

If the "standard explanations" were correct, this would
have been put to bed a long time ago.

And it was. But there are always a few stragglers.

But how does it know, since the output (source) impedances are
the same for both cases?

Conclusion: The output source impedance is NOT the impedance
encountered by the reflected wave. That's essentially what
Walter Maxwell's paper says.

Did you read the same paper I read? I recall that the claim was a
conjugate match of the effective impedances AFTER tuning. The
systems under discussion here have not been tuned for maximum
power transfer.

I think you have made this mistakte before. In my expression P=VI,
V and I are simultaneous instantaneous values and P is the
instantaneous power. If you want average you need to integrate
and divide.

No, if you want to average sinusoidals , use RMS values, which
is the result of said integrating and dividing. Somebody else
already did all the work.

This is true, but reflection is an instantaneous behaviour.
Simplifying
the analysis to RMS or average values, while often effective, bears
many risks in misleading the understanding, as demonstrated here.

Your expression applies only to sine waves and V and I are peak
voltages.

I is a peak voltage??????? You are more confused than
I ever realized. :-)

Oh darn. A typo. I know it will be quoted over and over in subsequent
posts. So be it.

Well, actually two typos. It should read "V and I are RMS voltages
and currents".

Interestingly, the expression P=V*I*cos(theta) for sine
waves is always derived by starting with Pinst=Vinst*Iinst,

Interestingly, you forgot to say you were talking about
instantaneous power until now.

My apologies. I did not mean to confuse you. Since reflection is
an instantaneous phenomenon, I just tend to think of it that way.

True. But there IS energy flowing at every point that is not a voltage
minimum or maximum. At these points there is NEVER any energy
flowing.

Every half cycle in every EM traveling wave, the instantaneous
power is zero. Would you therefore argue that traveling waves
cannot transfer any energy? Please get real.

Of course not. But when the instaneous power is 0 for all instants
then no energy can be flowing.

Given two parallel water pipes carrying 100 gallons/minute
in opposite directions. I can see you arguing that there
is no NET flow of water. But I cannot see you arguing that
there is no flow of water at all when the pressure of just
one stream can knock you off your feet.

Given one pipe of water, I also would not argue there are 10**9
gallons flowing per minute in one direction and 10**9 gallons
flowing per minute in the other, but the net is 0 so all is well.

That is analagous to the claim being made for forward and
reverse energy carrying travelling waves. Are you arguing
that for the water?

Well that does leave you with a bit of a conundrum since on
an open ended line, zero energy is flowing at every quarter
wavelength point back from the load.

The conundrum is all yours. Every half cycle, every EM
traveling wave is at a zero instantaneous power point.
If EM traveling waves can transfer energy while the
instantaneous power is zero every half cycle, superposing
two of them doesn't present a conundrum at all.

Except at those points where for every instant the instantaneous
energy transfer is zero.

At such a point a real instantaneous wattmeter would always
indicate zero.

A real instantaneous wattmeter would always indicate zero
at the zero crossing of every EM traveling wave in the
universe so exactly how does the light from Alpha Centauri
ever make it to Earth when the instantaneous power is zero
every half cycle? Please get real.

There is quite a difference between the instanteous power being
occasionally zero and being zero for all instances. Real enough?

....Keith