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

  • From: olaney@xxxxxxxx
  • To: weirsi@xxxxxxxxxx,ron@xxxxxxxxxxx
  • Date: Sun, 29 Jul 2007 01:26:05 -0700

There is a difference, Ron, and my experiment illustrates it.  It is that
rather than bouncing back as a relection on the same trace, the loop
return signals are the result of a round trip without reflection.  Two
open ended lines in parallel will show an impedance profile similar to
that of the loop *only* if the trace lengths are matched.  The fact that
this special case is indistinguishable from a loop at the driving point
is interesting, but does not make it equivalent in terms of the origin of
each return signal.  If you have a means to mark the driving signals so
that they can be distinguished from each other, the difference between
double open ended traces and with the ends shorted together can be
observed.  As you say, try it with a couple of pieces of coax and a TDR
if you disagree.  It'll work best if you use a separate series
termination for each trace rather than a single backmatch resistor for
both so that you can see the return signals separately.  I mentioned
ferrite but a high frequency LC trap on one leg to notch out a specific
frequency might be more convincing.  With two traces, the marked signal
returns on the same trace.  Create a loop by shorting the ends (making
sure that the short maintains the correct path impedance), and the marked
signal returns on the other trace.  With identical traces (or coax) and
identical driving signals, as you propose, the difference is there but
you can't see it.  That does not mean that the cases are equivalent, just
that your experimental setup cannot distinguish between them.  Hence, the
need to mark the signals.  Steve explained it well.  This would make a
good question for the electrical engineering professional licensing exam.

Orin

On Sat, 28 Jul 2007 23:29:35 -0700 steve weir <weirsi@xxxxxxxxxx> writes:
> Ron, yes if the signals exactly match then Ron's description of the 
> apparent open end matches the illusion.  It is an illusion just the 
> 
> same.  This is where Orin's proposed experiment can provide insight. 
>  
> Any difference between the two wavefronts is not accounted for by 
> the 
> open end model.  That odd mode if you will encounters the illusion 
> of a 
> dead short at the same juncture where the even mode Ron and you 
> describe 
> encounters the illusion of an open.  Account for both the even and 
> odd 
> signal modes and you will get the right answer from the illusion 
> just as 
> you will if you follow the formal, exact, and I think simpler view: 
> that 
> the two wavefronts continue to propagate until they are absorbed. 
> 
> Steve.
> ron@xxxxxxxxxxx wrote:
> >
> > Consider for a moment a 50 ohm source driving two equal length 100 
> ohm 
> > lines unterminated(open circuit)
> > TDR will show the open circuit at the end of the lines just as if 
> 
> > there were one 50 ohm open ended line.
> >
> > Next consider what will happen if you connect the open ended lines 
> 
> > together.  No change.  It will still reflect back as an open.
> >
> > Ponder that for a little and try it with a couple pieces of coax 
> and a 
> > TDR if you disagree.
> >
> >
> 
> -- 
> Steve Weir
> Teraspeed Consulting Group LLC 
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