[SI-LIST] Re: Circle bus topology; Circular Firing Squad?
- From: Kevin Ko <wy_k@xxxxxxxxx>
- To: vinu@xxxxxxxxx, olaney@xxxxxxxx
- Date: Mon, 30 Jul 2007 18:32:21 -0700 (PDT)
Assuming that we are still talking about electromagnetic waves, there is no
reflection occurring when they pass each other unperturbed even if they are of
different phase or magnitude when they meet half way. There goes Newton 3rd law
of motion...
Thanks,
-Kevin
--- Vinu Arumugham <vinu@xxxxxxxxx> 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.
> >>>>
> >>>>
> >>>>
> >>> --
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