[TowerTalk] re: coax as balanced line

[John Brosnahan]

At 02:36 PM 8/1/97 -0500, you wrote:
>I was wondering about the possibility or practicality of doing this
>where a long run would exist from the shack to the antenna. Cost wise it
>would be cheaper than hardline and should avoid some of the problems of
>ladderline: rain, snow, ect. Would this work at vhf, uhf, or just hf?
>Would this setup have the same losses as ladderline or somewhere between
>that & coax? Has anyone tried using 4 lines for 50 ohms or the 2 line
>100 ohm setup, if so were you using RG-8 or RG-58 size cable & how were
>you terminating it? Thanks for your ideas.
>73' Steve KG8TX


Steve, this issue was brought up earlier this year and here is my
offline response to the following comment that was made during the
previous discussion:

>>     Using two runs of coaxial cable as a balanced feeder will result in an 
>>     additional 3-dB loss.  


This is BS  (Bogus Speculation)!

I have used a balanced RG-213 feedline on my 8/8/8/8 10M array
that got such good EME echoes when it was up.  It was my contention 
that the loss would stay the same--simple thought problem.  If you think 
in terms of coax with 3 dB loss over the run from rig to antenna and then 
use two of them in a balanced feeder it was my contention that you 
could think in terms of half the power going into each one and each 
one would have 3 dB loss.

So if a single 50 ohm coax had 10 watts going in and had 3 dB loss
then the antenna would receive 5 watts.  If the balanced pair of
coaxes each had 5 watts going in and each had 3 dB loss then you would
have 2.5 watts at the far end of EACH coax--combined back into 5 watts.
Maybe not the best analogy but it gave me the answer that made most
sense.  I wanted to believe that with all of the extra copper that maybe the
loss would be less, but I couldn't get a logical handle on it that made sense
and I certainly didn't believe that the losses would be greater than with a 
single coax as the original questioner had posed.

But if you want a mathematical answer rather than an answer based
on logic here is what you need:


If you believe in the RGLC circuit model for the T-line, then the following
might convince you that the loss stays the same.

Let R, G, L, C be the resistance, conductance, inductance, and capacitance
of the original line (per unit length).

Then your new line has parameters R', G', L', C' where:

         R'=2R, G'=G/2, L'=2L, C'=C/2

Characteristic impedance of old line:  Zo=sqrt(L/C)
Characteristic impedance of new line:  Zo'=sqrt(L'/C')=sqrt(2L/(C/2))=2*Zo

Using low-loss approximation, Jordan and Balmain, p219
(not necessary to use approximation, but easy):

Loss per unit length (nepers) of old line: alpha = 0.5*(R/Zo + G*Zo)
Loss per unit length (nepers) of new line: alpha'= 0.5*(R'/Zo' + G'*Zo')
                                                 = 0.5*(2R/(2Zo) + (G/2)*2Zo)
                                                 = 0.5*(R/Zo + G*Zo)

i.e. the loss of the new line is the same as that of the old line.  The
factors of 2 (which combined to make the characteristic impedance go up)
cancel out in the loss factor...

Regarding any questions on increased copper losses - note that if the 50 and
100 ohm lines are carrying equal power, then the current on the 100 ohm
line is smaller by 1/sqrt(2) and the voltage on the 100 ohm line is bigger
by sqrt(2).  That's why the increased copper loss of the 100 ohm line
doesn't increase the loss, the current is just enough smaller to keep the
I**2*R loss the same.  That's also why the decreased conductance of the 100
ohm line (by virtue of the two conductors being separated by twice as much
dielectric) doesn't help - because the voltage is sqrt(2) higher, so that
V**2*G loss stays the same...


Hope this answers your question adequately.

73  John  W0UN




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