[SI-LIST] Re: Circle bus topology; Circular Firing Squad?
- From: "Ihsan Erdin" <erdinih@xxxxxxxxx>
- To: weirsi@xxxxxxxxxx
- Date: Mon, 30 Jul 2007 22:01:19 -0400
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
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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
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