[SI-LIST] Re: Circle bus topology; Circular Firing Squad?
- From: jkeeble@xxxxxxxxxxxx
- To: si-list@xxxxxxxxxxxxx
- Date: Wed, 01 Aug 2007 10:05:38 +1000
Steve is right; the 'principle of superposition' applies, as
this is a linear system. Things that can be said about such
systems include
- events proceed without regard to other events
- total behaviour is the arithmetic sum of the parts
Regards
Jon
> Vinu we all agree that the special equal amplitude case
> yields the same results whether we choose a continuous
> propagation model, or a reflection model where each
> incident wavefront sees a virtual open. So let's please
> put that aside and concentrate on the unequal case.
>
> Do we all agree that whatever the actual behavior,
> propagated or reflected that at the point of the
> intersection there will be one and only one
> characteristic impedance? If not why?
>
> Do we agree that this characteristic impedance will be
> encountered by both wavefronts?
> If not why?
>
> If we do agree on both points, then what impedance of the
> line can account for the behavior you depict where 50% of
> the L=>R wave reflects, 50% propagates, but 100% of the
> R=>L reflects? Solving for the L=>R the impedance at the
> intersection is 3*Zchar, while for R=>L that same point
> is an open. Then at an immeasurably tiny distance from
> the intersection on each side the impedance magically
> changes back to Zchar in both directions. But neither of
> those discontinuities generate reflections of their own.
> I hope you have an elegant explanation for how such
> apparent behavior is reasonable.
>
> The wave propagation model doesn't suffer from such
> complications. What is headed L=>R just continues on its
> merry way, just like what is headed R=>L.
>
> Best Regards,
>
>
> Steve.
>
>
> Vinu Arumugham wrote:
> > Here is some ASCII art with "colored" waves. The LLL and
> > RRR represent portions of a T-line charged by a
> > wavefront. <,> represent wavefronts traveling in
> indicated direction. >
> > Assume 50 ohm lines, 20mA current changes resulting in
> 1V wavefronts. >
> > Incident waves of equal magnitude:
> > LLLLLLLLLLLL> <RRRRRRRRRRR
> >
> > Reflection model :
> > <LLLLLLLLLLLRRRRRRRRRR>
> > LLLLLLLLLLLLRRRRRRRRRRR
> >
> > Wave propagation model:
> > <LLLLLLLLLLLRRRRRRRRRR>
> > RRRRRRRRRRRLLLLLLLLLLLL
> >
> > Unequal incident waves, left is 2V, 40mA, right is 20mA,
> > 1V: LLLLLLLLLLLL>
> > LLLLLLLLLLLL> <RRRRRRRRRRR
> >
> > Reflection model:
> > <LLLLLLLLLLLLLLLLLLLLLL>
> > LLLLLLLLLLLLRRRRRRRRRR>
> > LLLLLLLLLLLLRRRRRRRRRRR
> >
> > For the unequal wave reflection model above, at the
> > point where the wavefronts meet, no charge can flow
> > (I=0) from the right wavefront to the left T-line
> > because the lines are charged to the same potential.
> I=0 means the right wavefront sees an open circuit and
> > reflects. When this point reaches 2V, the 40mA left
> > incident wavefront can charge both the T-lines with
> > 20mA each, sending a 1V wavefront into the right T-line
> , and a 1V reflected wavefront going back on the left
> > T-line. So the left incident wavefront can be described
> > as having encountered a high impedance (>50 ohm and
> > <open circuit) whereby a 2V incident wave produced a 1V
> reflected wave. >
> > Wave propagation model:
> > <RRRRRRRRRRLLLLLLLLLLL>
> > LLLLLLLLLLLLLLLLLLLLLLL>
> > LLLLLLLLLLLLRRRRRRRRRRR
> >
> > Thanks,
> > Vinu
> >
> >
> > steve weir wrote:
> >> Ihsan, I've presented two methods that both correctly
> predict the >> results: One based on modeling the
> intersection as an open to the >> even mode, while short
> to the odd mode, and the other on what I think >> is far
> simpler: continuous propagation of each of the original
> wave >> fronts. Use whichever model makes your day
> simpler, but for my money >> I'll stick with the latter.
> I prefer the view that discontinuities >> and resulting
> reflections in quasi uniform, infinite length, ie >>
> terminated transmissions are the result of physical
> variations in the >> channel, not patterns of energy I
> happen to launch into them. >>
> >> Consider for example +1.0V step from the left, and a
> +0.5V step from >> the right. After they meet, the
> voltage moving rightward continues >> to rise by +1.0V
> from its previous value, and the voltage moving >>
> leftward continues to rise by +0.5V from its previous
> value. The >> waves just linearly superimpose.
> >>
> >> Regards,
> >>
> >>
> >> Steve.
> >>
> >>
> >> Ihsan Erdin wrote:
> >>> 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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