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Old March 22nd 08, 12:27 AM posted to rec.radio.amateur.antenna
Keith Dysart[_2_] Keith Dysart[_2_] is offline
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First recorded activity by RadioBanter: May 2007
Posts: 492
Default The Rest of the Story

On Mar 21, 12:37*pm, Cecil Moore wrote:
Keith Dysart wrote:
But you need to clearly state your limitations and stop
flip flopping.


What you are calling "flip flopping" is me correcting
my errors. Once I correct an error, I don't flip-flop
back.


Actually the flip-flopping I was referring to was the
constant changes in your view of the limitations that
apply to your claim.

I am surprised, this being 2008, that I could actually be
offering a new way to study the question, but if you insist,
I accept the accolade.


I'm sure you are not the first, just the first to think
there is anything valid to be learned by considering
instantaneous power to be important. Everyone except
you discarded that notion a long time ago.

Analysis has shown that when examined with fine granularity,
that for the circuit of Fig 1-1, the energy in the reflected
wave is not always dissipated in the source resistor.


Yes, yes, yes, now you are starting to get it.


Is this a flop or a flip? Are you now agreeing that the
energy in the reflected wave is only dissipated in the
source resistor for those instances when Vs is 0?

When
interference is present, the energy in the reflected
wave is NOT dissipated in the source resistor. Those
facts will be covered in Part 2 & 3 of my web article.

Ptot = P1 + P2 + 2*SQRT(P1*P2)cos(A)


Which of the two needs the 'cos' term?


Ps(t) = Prs(t) + Pg(t)
or
Pg(t) = Pf.g(t) + Pr.g(t)


In fact neither do.


For instantaneous values of voltage, the phase angle is
either 0 or 180 degrees so the cosine term is either +1
or -1. The math is perfectly consistent.

[gratuitous insult snipped]

Non-the-less do feel free to offer corrected expression that include
the 'cos(A)' term.


I did and you ignored it.


I could not find them in the archive. Could you kindly provide
them again, showing where the 'cos(A)' term fits in the
equations:
Ps(t) = Prs(t) + Pg(t)
and
Pg(t) = Pf.g(t) + Pr.g(t)

There is no negative sign in the
power equation yet you come up with negative signs.

[gratuitous insult snipped]

Negative signs also arise when one rearranges equations.

The math holds as it is. But I invite you to offer an alternative
analysis that includes cos(A) terms. We can see how it holds up.

[gratuitous insult snipped]

...Keith