Keith Dysart wrote:
Cecil Moore wrote:

Rs Vg Pfor--25w Vl
+----/\/\/-----+----------------------+
| 50 ohm ~23w--Pref |
| 0.04w / 1.92w
Vs 90 degrees \ Rl
100 cos(wt) 1.96w 50 ohm line / 1 ohm
| \ load
| |
+--------------+----------------------+
gnd

When you find (as you will), that Pfor and Vfor
do not change when the reflected wave returns,
it should be difficult to assert that the
reflected wave is re-reflected at Vg.

Not difficult at all. I've added the powers to the
diagram above. The reflected power is ~23 watts.
The power dissipated in the source resistor is
~0.04 watts. The reflected power is obviously *NOT*
being dissipated in the source resistor so where
is it going instead? The answer is destructive
interference at the source resistor is redistributing
the reflected energy back toward the load as an
equal magnitude of constructive interference.
I call that a reflection. The redistribution of
energy due to interference is a well known and
well understood phenomenon in optical physics but has
been virtually ignored in the field of RF engineering.

It is not a conventional reflection. It is wave
cancellation in action. It is obeying the following
equation:

Ptot = P1 + P2 - 2*SQRT(P1*P2) where
-2*SQRT(P1*P2) is the destructive interference term.

If the reflected wave was re-reflected at Vg,
there would necessarily be a change to Vfor
and Pfor.

Nope, not true. The phase angle
between the voltages determines which direction the
energy flows. If the above example is changed to
1/2WL instead of 1/4WL, the forward and reflected
voltages and powers remain the same but the interference
at the source resistor changes from destructive to
constructive and the source resistor heats up.
--
73, Cecil http://www.w5dxp.com