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

• From: olaney@xxxxxxxx
• To: vinu@xxxxxxxxx
• Date: Mon, 30 Jul 2007 11:43:09 -0700

```If you suppose that the waves meet and rebound like billiard balls, that
would be incorrect.  Each passes through the other as if it was the only
wave on the transmission line.  Only a real open circuit (or other
impedance discontinuity) can cause reflection.  Though identical
wavefronts might create the illusion of a "virtual open circuit" to the
viewer, that is not the physical reality.  The simultaneous "high
impedance / low impedance" interpretation is nonsensical.  There is only
one impedance at any given point on the line, and for constant line
parameters, the impedance is constant throughout.  Especially note that
the impedance of a linear xmsn line has nothing to do with the shape or
direction of the waves that happen to be traveling on it.  To suppose
otherwise wrenches the laws of physics.  Sorry if I have to be blunt.
Wavefronts passing through each other is the bedrock reality, all else is
armwaving.
Orin Laney, PE, NCE

On Mon, 30 Jul 2007 10:47:33 -0700 Vinu Arumugham <vinu@xxxxxxxxx>
writes:
When identical wavefronts are sent through the two branches of the loop
and meet at the far end, each wavefront can be described as being
reflected by the virtual open circuit.
When one wavefront is "marked", the wavefronts do not encounter a virtual
open circuit. One wavefront encounters a high impedance and the other a
low impedance compared to the line impedance. The subsequent reflections
of opposite polarity can be described as producing the illusion of the
wavefronts flowing through rather than being reflected at that point.

In other words, it seems to me that both the reflection and reinforcement
descriptions are perfectly valid and each is as real or illusory as the
other.

Thanks,
Vinu

olaney@xxxxxxxx wrote:
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
121 North River Drive
Narragansett, RI 02882

California office
(707) 780-1951 Fax

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