ve2pid wrote:
In the ARRL's Antenna Book 21st ed page 24-21, we see that if we
connect the two shields of the coax cables together, we obtain
'Shielded parallel Lines' . In that case, the resultant impedance is
simply the sum of the characteristic impedances of each coax.

So, there is quite a difference between the two independent coax I
mentioned in my first message (we connect the shield to the inner
conductor at each of its ends) (A) and the 'Shielded Parallel Lines'
case (B).

I am trying to understand why and it is the reason I posted my first
message...

In (A), the Z=276*log(2S/D) applies, so the Zo of each coax does not
matter.. but in (B), Z=Zo1+Zo2, so the value of each Zo matters.


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

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

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!!

Jeff

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

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.

Jeff

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

Owen Duffy wrote in
:

....
....
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.

I should have expected that this is not obvious to you Jeff, and expand that
to:

....
in the other (defined by differential mode), and V conductor to
conductor is (by definition of differential mode) V'=(I*Zo-(-I*Zo))=2*I*Zo,
and Zo'=V'/I=2*(I*Zo)/I=2*Zo.

Owen