🚀 go-pugleaf

In <7380082@hpnmdla.sr.hp.com> stanb@hpnmdla.sr.hp.com (Stan Bischof) writes:

>In sci.electronics, vreeburg@fys.ruu.nl (Jurriaan Vreeburg) writes:

>>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

>Yep.

>>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!!! 

>Nope. Capacitance is doubled, inductance halved, Z halved.
>      Don't forget that Z=1/jwC for cap but jwL for ind.

>Don't believe me? Try connecting a 50 ohm cable to two 50 ohm cables
>      in parallel. Launch pulse and see if you have a reflection at the 
>      junction. Easy to do with any oscilloscope and a pulse generator.

I had some doubt before posting my previous answer to the problem, but imagagine:


                50 ohm
           =============
25 ohm  //               \\   25 ohm
========  A             B  ============
        \\               //            |
          ===============             ---
              50 ohm                  | |
                                      | | 25 ohm
                                      | |
                                      ---
                                       |
                                      GND

A signal arriving at junction A sees an impedance of 25 ohms. So at that point no
reflections occur. But what is the impedance which the signal travelling through
the lower cable sees arriving at junction B? It looks to me that it surely isn't
50 ohm, so at that point reflections are produced. 

One other porblem I have, is that all mailers just replace a coax cable by
an inductance and a capacitance:

        --- L ---
                 |
 Z -->           C
                 |

           1              L
Z = jwL + ---  <> root ( --- )  !!!!!
          jwC             C

So I don't think that one can 'simply' put the inductances and the capacitances
in parallel.

Any comments?

-- 
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