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

  • From: "Ihsan Erdin" <erdinih@xxxxxxxxx>
  • To: Arpad_Muranyi@xxxxxxxxxx
  • Date: Mon, 30 Jul 2007 16:21:32 -0400

Considering the colored quarks of quantum chromodynamics, not a
far-fetched idea...

Sorry, couldn't resist the temptation.

Ihsan

On 7/30/07, Muranyi, Arpad <Arpad_Muranyi@xxxxxxxxxx> wrote:
> Great idea!  How about coloring the
> electrical waves...   :-)
> Arpad
> ====================================
>
> ________________________________
>
> From: Vinu Arumugham [mailto:vinu@xxxxxxxxx]
> Sent: Monday, July 30, 2007 12:59 PM
> To: Muranyi, Arpad
> Cc: si-list@xxxxxxxxxxxxx
> Subject: Re: [SI-LIST] Re: Circle bus topology; Circular Firing Squad?
>
>
> Hopefully, they can be a different color without affecting their mechanical 
> properties...?
>
> Or, one can perform a thought experiment with two water wavefronts, each of a 
> different color, approaching from either end of a trench and colliding.
>
> Thanks,
> Vinu
>
> Muranyi, Arpad wrote:
>
>         This discussion (argument) makes me think of an analogy.
>
>         Imagine two identical balls in space approaching each other
>         along the same and perfectly straight path.  At some point
>         they collide with a perfectly elastic collision.  Based on
>         the laws of physics they reverse their direction and begin
>         to travel in the opposite direction they came.
>
>         Hmmm.  Or does the ball that came from the right continue
>         on the left and the one that came from the left continue
>         on the right?  I can't tell, they are identical...  :-)
>
>         Arpad
>         
> =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
>         
> =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
>         =3D=3D=3D=3D=3D=3D=3D=3D=3D
>
>         -----Original Message-----
>         From: si-list-bounce@xxxxxxxxxxxxx 
> [mailto:si-list-bounce@xxxxxxxxxxxxx] =
>         On Behalf Of steve weir
>         Sent: Monday, July 30, 2007 11:25 AM
>         To: Vinu Arumugham
>         Cc: olaney@xxxxxxxx; ron@xxxxxxxxxxx; si-list@xxxxxxxxxxxxx
>         Subject: [SI-LIST] Re: Circle bus topology; Circular Firing Squad?
>
>         Vinu, I disagree.  The 'even' mode, the common to reference of each=20
>         waveform encounters the illusion of an open at the intersection.  
> The=20
>         'odd' mode, ie the difference to reference between each encounters 
> the=20
>         illusion of a short.  While this approach yields the correct results, 
> I=20
>         find it more cumbersome than the real behavior which is two 
> wavefronts=20
>         that pass through each other as they propagate in opposite 
> directions=20
>         down the same transmission line.
>
>         As previously discussed in this thread, we can take two drivers at 
> the=20
>         opposite end of the same line and drive them as needed to 
> demonstrate=20
>         any of the possible behaviors.
>
>         Best Regards,
>
>
>         Steve.
>
>         Vinu Arumugham wrote:
>
>
>                 When identical wavefronts are sent through the two branches 
> of the=20
>                 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=20
>                 virtual open circuit. One wavefront encounters a high 
> impedance and=20
>                 the other a low impedance compared to the line impedance. 
> The=20
>                 subsequent reflections of opposite polarity can be described 
> as=20
>                 producing the illusion of the wavefronts flowing through 
> rather than=20
>                 being reflected at that point.
>
>                 In other words, it seems to me that both the reflection and=20
>                 reinforcement descriptions are perfectly valid and each is as 
> real or=20
>                 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> <mailto:weirsi@xxxxxxxxxx>  =
>
>
>         writes:
>
>
>                          =20
>
>
>                                 Ron, yes if the signals exactly match then 
> Ron's description of the=20
>                                 apparent open end matches the illusion.  It 
> is an illusion just the=20
>
>                                 same.  This is where Orin's proposed 
> experiment can provide insight. =
>
>
>
>
>
>                                 =20
>                                 Any difference between the two wavefronts is 
> not accounted for by=20
>                                 the=20
>                                 open end model.  That odd mode if you will 
> encounters the illusion=20
>                                 of a=20
>                                 dead short at the same juncture where the 
> even mode Ron and you=20
>                                 describe=20
>                                 encounters the illusion of an open.  Account 
> for both the even and=20
>                                 odd=20
>                                 signal modes and you will get the right 
> answer from the illusion=20
>                                 just as=20
>                                 you will if you follow the formal, exact, and 
> I think simpler view:=20
>                                 that=20
>                                 the two wavefronts continue to propagate 
> until they are absorbed.=20
>
>                                 Steve.
>                                 ron@xxxxxxxxxxx wrote:
>                                    =20
>
>
>                                         Consider for a moment a 50 ohm source 
> driving two equal length 100=20
>                                              =20
>
>
>                                 ohm=20
>                                    =20
>
>
>                                         lines unterminated(open circuit)
>                                         TDR will show the open circuit at the 
> end of the lines just as if=20
>                                              =20
>                                         there were one 50 ohm open ended line.
>
>                                         Next consider what will happen if you 
> connect the open ended lines=20
>                                              =20
>                                         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=20
>                                              =20
>
>
>                                 and a=20
>                                    =20
>
>
>                                         TDR if you disagree.
>
>
>                                              =20
>
>
>                                 --=20
>                                 Steve Weir
>                                 Teraspeed Consulting Group LLC=20
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>         Steve Weir
>         Teraspeed Consulting Group LLC=20
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>         California office
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