Keith Dysart wrote:
My issue was that you seemed, in that sentence, to be saying
that the reflected energy was dissipated in the source
resistor. But earlier you had stated that was not your claim.

My earlier claim was that the average power in a reflected
wave is dissipated in the source resistor when the forward
wave is 90 degrees out of phase with the reflected wave at
the source resistor. In that earlier claim, I didn't care
to discuss instantaneous power and thus excluded instantaneous
power from that claim.

For instantaneous values, it will be helpful to change the
example while leaving the conditions at the source resistor
unchanged. Here's the earlier example:

Rs Vg Vl
+----/\/\/-----+----------------------+
| 50 ohm |
| 1/8 WL |
Vs 45 degrees 12.5 ohm
100v RMS 50 ohm line Load
| |
| |
+--------------+----------------------+
gnd

Here's the present example:

Rs Vg Vl
+----/\/\/-----+----------------------+
| 50 ohm |
| 1 WL |
Vs 360 degrees 23.5+j44.1
100v RMS 50 ohm line ohm Load
| |
| |
+--------------+----------------------+
gnd

If I haven't made some stupid mistake, the conditions at
the source resistor are identical in both examples. But
in the second example, it is obvious that energy can
be stored in the transmission line during part of a cycle
(thus avoiding dissipation at that instant in time) and be
delivered back to the source resistor during another part
of the cycle (to be dissipated at a later instant in time).
That is the nature of interference energy and is exactly
equal to the difference between the two powers that you
calculated. You neglected to take into account the ability
of the network reactance to temporarily store energy and
dissipate it later in time.

All of the reflected energy is dissipated in the source
resistor, just not at the time you thought it should be.

While performing this analysis, you will discover the ability
of superposing waves to redistribute energy (even in the
complete absence of ordinary reflections). Not possible in
the above special case examples, but under other conditions,
the redistribution of energy can attain perpetual steady-state
status (even in the complete absence of ordinary reflections).

There are two mechanisms involved in the re-routing of
steady-state reflected energy back toward the load.

1. Ordinary reflection of single EM waves governed by the
rules of the wave reflection model.

2. A redistribution of reflected energy back toward the
load as a result of superposition of waves accompanied
by an average level of steady-state interference.
--
73, Cecil http://www.w5dxp.com