"Dr. Slick" wrote in message
om...
"Tarmo Tammaru" wrote in message
...
Why didn't you do the gamma for a shorted line using your formula? I
think
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.

..........................................
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?


They also mention that the normalized load impedance Zn=Zr/Zo does
NOT have the same angle as Zr because Zo is complex in the general
case.

"They anticipate people being concerned about |Gamma| 1 and later
come up
with a formula for time average power."

Really? Could you tell us more about that? Could you email me
more pages, especially the one that has Eq. 5.2b?



Slick

I can't scan the whole chapter, and part of the next. You should be able to
get your library to borrow the book for you. Or, look at some other college
textbook under lossy lines. 5.2B is straightforward enough:

Zo=(1/Yo)=SQRT[(R+jwL)/(G+jwC)]=Ro+jXo

Tam/WB2TT