On Mar 17, 1:06*am, Cecil Moore wrote:
Keith Dysart wrote:
Pr.g(t) = -50 + 50 *cos(2wt)
* * * * = -50 -50
* * * * = -100
which would appear to be the 100 watts needed to
heat the source resistor.

Which, contrary to your previous assertions, agrees
with what I have said previously. This is the
destructive interference energy stored in the
feedline 90 degrees earlier and being returned
as constructive interference to the source
resistor. It's impossible to sweep it under the
rug when the source power is zero, huh?

The intriguing question is:
1. Do you just stop reading as soon as you find a snippet
that aligns with your claim?
2. Do you keep reading, but do not understand because you
are so gleeful at finding a snippet that aligns with
your claim?
3. Or do you read, understand, but choose to disingenuously
ignore that which follows the snippet that aligns with
your claim?

Regardless, if you use the snippet above to support
your claim, you have effectively modified your claim.
You are now saying that the reflected energy is
dissipated in the source resistor only at particular
times, such as when the source voltage is 0, and that
you are not interested in what happens during the rest
of the cycle.

For example, if you were to do the same analysis,
except do it for t such that wt equals 100 degrees,
instead of 90, you would find that you need more terms
than just Pr.g to make the energy flows balance.

When wt equals 100 degrees
Ps = -28.170
Prs = 65.798
Pg = -93.96
so, as expected
Ps = Prs + Pg

And
Pr.g = -96.985
Pf.g = 3.015
Ooooppppss, no way to make those add to 65.798.

But
Ps = Prs + Pf.g + Pr.g
since
Pg = Pf.g + Pr.g

So all is well with world.

All power comes from the source. Since power is
being delivered to the source resistor at a time
when the source is delivering zero power, it must
have been previously been stored in the reactance
of the transmission line.

This is true. But it is, of course, Pg(t) that
describes the energy flow in and out of the line,
not Pr.g(t).

So are you now agreeing that it is not the energy
in the reflected wave that accounts for the
difference in the heating of the source resistor
but, rather, the energy stored and returned from
the line, i.e. Pg(t)?

Note also, that when wt is 100 degrees, not only is
the source resistor being heated by energy being
returned from the line, the source is also absorbing
some of the energy being returned from the line
(Ps = -28.170).

...Keith