[SI-LIST] Re: Circle bus topology; Circular Firing Squad?

  • From: "Dmitriev-Zdorov, Vladimir" <vladimir_dmitriev-zdorov@xxxxxxxxxx>
  • To: <si-list@xxxxxxxxxxxxx>
  • Date: Thu, 2 Aug 2007 09:01:59 -0700

We may try another thing. Take a section of the loss less line of a
certain length / delay. Split it along into two narrower conductor while
keeping the ends at side A and B connected. If by separation or other
reason we exclude coupling, here we have a circle with 'circular'
impedance twice the original (sqrt(2L/(C/2)) and the same propagation
factor and hence the same wavelength: sqrt(2L * (C/2)).

For the outer circuit, the only current that is observable is a
'coherent' component, something like a 'common mode'. Nothing forbids
the existing of the 'differential' component too, that circulates within
the loop. The impedances are different for the 'common' and
'differential' components. But, of course, the differential one does not
affect the outer circuitry (as far as we keep traces connected at A and
B) and therefore we don't need it.

Similarly, the line segment can be split along into arbitrary number of
topologically parallel (but uncoupled) filaments not necessary of same
thickness, and here we could introduce a many circulating currents for
each loop. This is still a valid model but we don't need all this
details if the goal is only to find out how the wave propagates from A
to B.

The concept of loop currents is one of the basics in circuit theory. It
represents any branch current as a superposition of the loop currents
circulating in the contours that include this branch. However, we do not
introduce more than one loop current to keep the system non-singular.
With mode than one current for every loop the solution isn't unique

>Here's another simple example. Connect two 1V voltage sources, one to=20
>each terminal of a 1 ohm resistor. One can say the potential difference

>across the resistor is 0, so by Ohm's law, current is 0. Or one can=20
>ignore one voltage source at a time and say we have 1A going through
the=20
>resistor from each source. The currents are in opposite direction=20
>therefore net current is 0. One can label either approach to be=20
>cumbersome or intuitive but I don't think either approach can claim to=20
>solely represent reality.
>
>Thanks,
>Vinu


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