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

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
> >>>>>>>>
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