"Dr. Slick" wrote in message
om...
"Tarmo Tammaru" wrote in message
...
Let's start all over on this.

1. You have a transmission line of Zo= 1 - j1
2. Zo* = 1 + j1
3 Short the end of the line
4. By the classic formula RC= (0 - (1 - j1))/(0 + (1 - j1))
= -(1 - j1)/(1 - j1) = -1
5. By Besser's formula RC = (0 - (1 + j1))/(0 + (1 - j1))
which is (-1 -j1)/(1 - j1)
6. The above is SQRT(2)/_-135 / SQRT(2) /_-45 = 1/_-90
7. Note that RC has angle -90, not 180.
8. Reflected voltage is now (V+) x (1/_-90). This is in quadrature with
V+,
not in opposition with.
9. This won't meet the boundary condition that ( V+) + (V-) = 0
10. You can use a different phase angle for Zo if you like, but the only
Zo
phase angle that will give you RC = 1/_180 is 0.

Tam/WB2TT


You are correct. But please note that the rho is 1 in both cases.

And i believe that the conjugate equation is correct, that the
reflected voltage will have a phase shift of twice the phase of the
Zo, which in this case would be 45 degrees.

And you bring up an excellent point Tam, that the only Zo that
will give you a RC=-1 (or 1 /_ 180 ), with an ideal short as a Zload,
would be a purely REAL Zo, with no reactance.

So again, the "normal" equation fails in this respect. I'll state
once again that the normal equation assumes a purely real Zo.

Besser and Kurokawa and the ARRL are all correct.

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Read this again. The *normal* equation works. It is the ARRL/Besser equation
that does not.

Tam

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Slick