On Thu, 17 Apr 2008 18:43:41 GMT
Cecil Moore wrote:

Roger Sparks wrote:
Yes, in nature the returning power will affect the
frequency of the source.

If the reflected wave is a sine wave coherent with
the forward wave, why would the frequency change
when they are superposed?


I grant that the case for true frequency change was overstated. There is a one time phase change, causing a frequency "bump" for one cycle. This phase change is a one time event, occuring for each successive and diminishing reflected wave.

GND--Vs--x--Rs--------45deg 50 ohm----short
100v 50ohm

This circuit has an impedance of 73.5 + J44.1 ohms
at the point x. Energy sinking into the source occurs
during the cycle. It is not an illusion.

I got a different impedance value but for purposes
of discussion, all that matters is that the impedance
is NOT 50 ohms.

Right, in my haste, I picked up the impedance for a 12.5 ohm load. Sorry. The impedance should be 50 + 50j ohms.

Let's assume your impedance is correct. That means
the ratio of reflected power to forward power at
point 'x' is 0.1452 in the 50 ohm environment.
Why would reflected energy flowing back through
the source be considered "energy sinking". Why
cannot the reflected energy flow back through the
source unimpeded by the 0+j0 impedance and be
reflected by the short to ground on the other side?
That same thing happens all the time to standing
wave current on the outside of the coax braid.
Why do the laws of physics change inside a source?

It is the arbitrariness that I object to. As you say, you chose 50 ohms to calculate the reflection factor. So we have a 50 ohm transmission line zero length long, reflecting from a short circuit. This does nothing for us. We still have the source absorbing power, but now we have now added a mechanism for how the absorbed power behaves within the source.

In this example, we are delivering 100 watts to Rs.
With a power reflection coefficient of 0.1452, the
forward power has to be 117 watts making the reflected
power equal to 17 watts at point 'x'. That reflected
power of 17 watts is not absorbed by the source. It
flows back through the source, unimpeded by the 0+j0
impedance, and is reflected by the short to ground.
It joins the 100 watts being generated by the source
to become the forward wave of 117 watts. No absorbing
of energy required - just good old distributed network
physics.

Once this 17 watt wave is tracked back through the
source, reflected, and superposed with the 100 watts
being generated by the source, all will become clear
without any power absorption by the source being
necessary. Here's a diagram of what's happening.

100w 100w
GND--------Vs----------Rs--------45deg----------short
17w-- 117w-- 50w--
--17w --17w --50w

All energy flows balance perfectly and there is no
absorption by the source.
--

This is much too arbitrary for me.
--
73, Roger, W7WKB