Jeff wrote in
:

Owen Duffy wrote:
Jeff wrote in
:

Owen Duffy wrote:
Jeff wrote in news:hm5dj4$4ls$1
@speranza.aioe.org:

I am sure that in case B it is as for resistors in parallel, ie 2
50ohm cables in parallel give you 25ohms Zo.
Jeff, you you offer more explanation that just that your are
"sure".

If you can't explain it, it speaks of whether you are sure.

Owen
Well having just tried it for real on a network analyser, and
simulated it on Ansoft designer I am now convinced rather than being
sure!!

That is not an explanation at all.

Your confirmation might just be confirmation of a wrong
interpretation of the B configuration.

Owen

and you just might be trolling.

And you just might be wrong on both counts.

I will offer you an explanation of why the ARRL position is IMHO correct.

The configuration is a pair of identical coaxial cables, shields tied
together at each end, and the inner conductors used as a two wire
transmission line (http://www.vk1od.net/transmissionline/stcm/Fig02.png).

Question is what is Zo of the combination?

for each of the coaxial lines, its intinsic Zo defines the ratio of V/I
for a travelling wave at any point. If the two inner conductors are
driven by a differential signal, then I in one conductor equals -I in the
other (defined by differential mode), and V conductor to conductor is
twice V conductor to shield, so V'=2*(I*Zo), and
Zo'=V'/I=2*(I*Zo)/I=2*Zo.

Owen