On 6/24/2011 7:41 PM, Cecil Moore wrote:
On Jun 24, 2:20 pm, John wrote:
If the reflected wave is re-reflected, it must be by an impedance
other than the virtual impedance generated by the reflected wave
itself. If the reflected wave is being used to generate a virtual
impedance, it cannot at the same time be being re-reflected.

I disagree. There are 100W supplied by the source and 100W consumed by
the load. There are 200W in the 291.4 ohm line. 100W of that is just
"passing through". The other 100W is circulating, that is, stored energy
which was put there by the start-up transient. If it is circulating,
then it must be reflected from each end of the 291.4 ohm line.

Let's assume that the 100 watts is just "passing through". It would
change from 70.7 volts in the 50 ohm environment to 170.7 volts in the
291.4 ohm environment. The reflected power is 100 watts so the
reflected voltage is also 170.7 volts. Those two voltages would have
to add together to get the forward voltage. The forward voltage is
known to be 241.4 volts. Exactly how do you add two 170.7 volt in-
phase waves to get a total of 241.4 volts?

Here is actually what happens. Since the physical power reflection
coefficient at the Z0-match point is 0.5, only 50 watts of the source
power makes it through the impedance discontinuity. That voltage is
120.7 volts. The same thing applies to the reflected power - only 50
watts is re-reflected by the 0.5 power reflection coefficient. So the
re-reflected voltage is also 120.7 volts. Adding those two voltages
together yields 241.4 volts which we know is the correct forward
voltage.

Now you are going to ask how two 50 watt waves can add up to 200 watts
forward power. That's just the nature of constructive interference
since power is proportional to voltage squared. If we add two 50 watt
waves in phase, we get a 200 watt wave. Where did the extra 100 watts
come from? Why, from the 100 watts of destructive interference toward
the source that eliminated the reflections toward the source.
--
73, Cecil, w5dxp.com

If the reflections toward the source is eliminated, how is it that it
appears to be 50 ohms at that point rather than 291.4 ohms?

John