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

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

As I said, I corrected an error in my thinking. You
are free to consider that to be a flip-flop, I consider
it to be a step forward.

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?

Yes, for instantaneous reflected energy exactly as I
previously stated, but not true for average reflected
energy. 100% of the average reflected energy is
dissipated in the source resistor when the transmission
line is 45 degrees long. That intra-cycle interference
exists, thus delaying the dissipation by 90 degrees,
is irrelevant to where the net energy winds up going.

Your conservation of power principle would have you
demanding that the power sourced by a battery charger
must be instantaneously dissipated. Everyone except
you seems to realize that is an invalid concept. The
dissipation of energy can be delayed by a battery
or a reactance. In the present example, the dissipation
of the instantaneous reflected energy is delayed by
90 degrees by the reactance.

I could not find them in the archive. Could you kindly provide
them again, showing where the 'cos(A)' term fits in the
equations:

Go back and read the part where I said cos(0)=+1 and
cos(180)=-1. There is no such thing as conservation of
power. Your equations assume a conservation of power
principle that doesn't exist in reality. The forward
power is positive power. The reflected power is positive
power. The only negative power is destructive interference
which must be offset by an equal magnitude of constructive
interference. When the instantaneous interference power is
negative, the two voltages are 180 degrees out of phase and
that is your cos(180)=-1. When the instantaneous interference
power is positive, the two voltages are in phase and that
is your cos(0)=+1. Why don't you already know all this
elementary stuff?
--
73, Cecil http://www.w5dxp.com