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In article <1992Oct8.120225.6247@fys.ruu.nl> vreeburg@fys.ruu.nl (Jurriaan Vreeburg) writes:
>From: vreeburg@fys.ruu.nl (Jurriaan Vreeburg)
>Subject: Re: 25 ohm coax cable
>Keywords: coax cable
>Date: Thu, 8 Oct 1992 12:02:25 GMT
>In <bvr5pd.14o@wang.com> garyf@wang.com (Gary Field) writes:
>
>>dhina@dxcern.cern.ch (Mourad Dhina) writes:
>
>>>Hello,
>>>i am trying to realize a circuit i found in an electronics magazine.
>>>one of the components it uses is a semi-rigid coax cable of
>>>25 ohms impedance. this is the first time i hear of this (i only
>>>knew 50, 75 and 93 ohms cables, which are standard). anybody
>>>knows where to find this?
>>>Mourad
>
>>Put two pieces of 50 Ohm cable in parallel. It should be exactly the same.
>>Gary
>
>After reading Gary's answer I had some doubt of its correctness. It implies
>that by putting two 50 Ohm coax cables in series, one is able to create a 100 Ohm
>coax cable, which is certainly not true.
>
>I think that putting two cables in parallel doesn't change the impedance, since
>the impedance is defined as:
>
>                   L
>Z = square_root ( --- )
>                   C
>L: inductance per unit length
>C: capacitance per unit length
>
>What happens if we put two cables in parallel: both the inductance and the capacitance
>increase by a factor of two -> Z doesn't change!!! Putting two cables in series doesn't
>change the impedance as well, since L & C are defined per unit length, which don't
>change at all in that case.
>
>Note that the above formula only holds for cables which are much longer than the wave-
>lengths of the signals it caries.
>
>-- 
>Jurriaan Vreeburg                             email: vreeburg@fys.ruu.nl
>Department of Physics and Astronomy           tel.:  (+31)-(0)30-534566
>P.O. Box 80.000
>3508 TA  Utrecht  Holland
You are partially right. The impedance of the cable Z=(L/C) E 1/2 is the 
result of driving the piecwise (per unit lenght) TERMINATED at the end, 
which means:

              o-----1/2L----+------1/2L-----o-|
                            |                 |
    Zin --->                C                 Zo= 50 ohm
    50 ohm                  |                 |
              o-------------+---------------o-|

so, for calculation, the 50 ohm Zo is involved in order to find Zin 50 ohm. 
This 50 ohm can be another portion of the calbe or just a resistor. That's 
why parallel and series are different.
 





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