"Dr. Slick" wrote in message
om...
"Tarmo Tammaru" wrote in message
...
you settled on (Zl - Z0*)/(Zl + Zo). For Zl = 0 this comes
to -Z0*/Z0,
which, for Zo having phase angle b equates to -1 at angle(-2b).

Agreed. This doesn't prove anything, does it?

How does this allow for the sum of V+ and V- to be 0? That is what you
have
across a short.


This is the correct answer, you just interpret it incorrectly.

The reflection coefficient is defined as the reflected voltage
divided by the incident voltage. Sure, the voltage is zero right at
the short, but there is a reflected voltage wave that moves back
towards the generator.

When you have a short, the phase is flipped 180 degrees, which is
exactly what the -1 means. Notice if you had a Zl=infinity, that the
RC would be +1, which would be full reflections INPHASE with the
generator, or in phase with the incident voltage wave.

It's very simple stuff, but many people don't understand this.

That's the whole point. By the conjugate formula RC is *not* -1. It is -1
with a phase angle. I agree it works for an open circuit, since you can
divide both sides by Zl, and anything divided by infinity is 0



.........................................
P= (1/2)Go|V+|^2e**(-2az)[1 - |CM|^2 +2(Bo/Go)Im(CM)]

Remember, z is distance from the load.

Tam/WB2TT



ok, you've done a nice job of copying the text you sent me.

As I recall, you said you were not familiar with these diagrams and did
not
understand them. What do you want me to do? derive it in a different
way?


I want you to tell me the significance of the fact that the
normalized load impedance Zn=Zr/Zo does NOT have the same angle as Zr

Obviously true


because Zo is complex in the general case. And how this may make this
example incorrect for this discussion.

Slick

I am not sure which example you mean, but A*/A does not have 0 phase angle,
unless the phase angle of A is 0.

PS I scanned in the other stuff. Compressed TIF

Tam