[SI-LIST] Re: Circle bus topology; Circular Firing Squad?
- From: olaney@xxxxxxxx
- To: erdinih@xxxxxxxxx
- Date: Mon, 30 Jul 2007 20:21:41 -0700
Ihsan:
Which better fits Ockham's razor: an interpretation that requires
splitting the signal into parts so that one may bounce, and the other
travel unimpeded, or one that simply passes everything without
interaction? Why suppose the signals bounce only in the case where it is
impossible to distinguish that from superposition, yet discard that model
in favor of superposition when you can tell the difference? One
interpretation gets into arbitary mental gymnastics, the simpler model
works every time without fail.
Orin
On Mon, 30 Jul 2007 22:01:19 -0400 "Ihsan Erdin" <erdinih@xxxxxxxxx>
writes:
> Steve,
>
> The wave propagation is simply the transfer of the energy in space.
> For the special case a line symmetrically driven at both ends, one
> can
> use the model of an unterminated transmission line driven from one
> side only and no one can tell the difference. This is based on the
> fundamental electromagnetic principle: image theory.
>
> For the uneven drivers of your example, I can rightfully argue that
> the equal frequency components "bounced" and cancelled out while the
> residual part kept on propagating. The idea of waves passing through
> each other is simply a matter of perception; not a rocksolid
> physical
> reality which ridicules the idea of waves bouncing in the middle.
> Both
> cases have equal footing and at the end it all boils down to the
> choice of modeling.
>
> The billiard ball example was an interesting attempt but not quite
> equivalent. At the collision the balls will have to come to a
> momentary full stop before accelerating in the reverse direction.
> This
> is not symmetrical to the case where they (might) pass through each
> other at constant speed.
>
> Best regards,
>
> Ihsan
>
> On 7/30/07, steve weir <weirsi@xxxxxxxxxx> wrote:
> > Vinu but for the discussion at hand:
> >
> > First: The driver is back terminated in the example so both
> wavefronts
> > are completely absorbed and the characteristic impedance is the
> > effective impedance of the line everywhere. Energy propagating
> forward
> > or backwards in the line does not change the impedance.
> >
> > Second: At the point in time where the apparent reflection
> occurs, no
> > wavefront has reached an impedance discontinuity. And in fact as
> stated
> > above, if the source matches perfectly, never will. There are no
> > reflections in this system at all. Each wavefront launches, goes
> its
> > merry way around the path and gets identically absorbed back at
> the
> > driver.
> >
> > To an observer monitoring the line two equal and opposite wave
> fronts
> > will indeed appear to bounce like a perfectly elastic mechanical
> > collision. So let's ask ourselves which is the illusion: the
> apparent
> > 100% reflection, or the continuous propagation of each front.
> Several
> > useful experiments have been offered to resolve the issue. In
> each we
> > send two wavefronts which are not identical and monitor the
> behavior.
> > What do we find? We find that rather than each waveform
> reflecting
> > identically as predicted by the reflection model, the difference
> > continues to propagate forward. IE, the observation EXACTLY
> matches the
> > wave propagation model, while it does not match an unmodified
> reflection
> > model. In order to fix the reflection model we have to
> artificially
> > create a short to the odd mode at the same point where we have an
> open
> > to the even mode INCIDENT waveforms.
> >
> > Best Regards,
> >
> >
> > Steve.
> > Vinu Arumugham wrote:
> > > "There is only one impedance at any given point on the line, and
> for
> > > constant line parameters, the impedance is constant throughout."
> > > Yes, that's the characteristic impedance of the line.
> > >
> > > The input impedance of an unterminated line can vary from zero
> to
> > > infinity depending on the frequency of the driving signal. In
> other
> > > words, the line driver "sees" a high or low impedance that is a
> > > function of the magnitude and phase of the reflected wavefront.
> The
> > > same thing happens when wavefronts meet in a loop. The effective
> > > impedance seen by each wavefront is a function of the magnitude
> and
> > > phase of the other wavefront. So, why is this interpretation
> > > "nonsensical"?
> > >
> > > Thanks,
> > > Vinu
> > >
> > > olaney@xxxxxxxx wrote:
> > >> 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
> > >> <mailto: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
> > >>>> (408) 884-3985 Business
> > >>>> (707) 780-1951 Fax
> > >>>>
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> > >>>> (401) 284-1840 Fax
> > >>>>
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> > >>>>
> > >>>> http://www.teraspeed.com
> > >>>> This e-mail contains proprietary and confidential
> intellectual
> > >>>> property of Teraspeed Consulting Group LLC
> > >>>>
> > >>>>
> > >>>
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> > --
> > Steve Weir
> > Teraspeed Consulting Group LLC
> > 121 North River Drive
> > Narragansett, RI 02882
> >
> > California office
> > (408) 884-3985 Business
> > (707) 780-1951 Fax
> >
> > Main office
> > (401) 284-1827 Business
> > (401) 284-1840 Fax
> >
> > Oregon office
> > (503) 430-1065 Business
> > (503) 430-1285 Fax
> >
> > http://www.teraspeed.com
> > This e-mail contains proprietary and confidential intellectual
> property of Teraspeed Consulting Group LLC
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