On May 13, 5:51*pm, Wimpie wrote:
When a source behaves like a 50 Ohms source, it will not rereflect the
reflected power back to the load. In other words the forward power
generated by the source will not change when changing the load. The
reflected power that is absorbed by the source may result in increase
or reduction of power dissipation of the active device. It all depends
on the change of the integral of V*I (for the active device).

Where is the reflected power that is "absorbed by the source"
dissipated? How can a "reduction of power dissipation" be the result?
The answer is that if the power dissipation of the source decreases
when the reflected power is incident, then destructive interference
energy is being redistributed back toward the load. That destructive
interference energy includes some (or all) of the reflected energy.

The forward wave is associated with EfwdxHfwd power. The reflected
wave is associated with ErefxHref power. The energy in those two waves
traveling in opposite directions at the speed of light in the medium
must be conserved.

For instance, when the load at the end of an ideal 1/4WL feedline is
changed to a short, the forward power is still the same and the
reflected power is equal to the forward power. Yet, the source
dissipation has been reduced to zero even though the reflected power
"is absorbed by the source". In the source resistor, the forward
current is equal in magnitude and 180 degrees out of phase with the
reflected current and the net current superposes to zero. That's the
definition of "total destructive interference" and according to the
conservation of energy principle, the energy involved has to go
somewhere - so it is redistributed in the only possible direction -
back toward the load associated with the "total destructive
interference" that occurs in the 50 ohm source resistor.

The main point to this part of the discussion is to realize that the
forward and reflected wave energy is flowing through the 50 ohm source
resistor because the physical reflection coefficient looking back into
the 50 ohm source from a 50 ohm feedline is zero.
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK