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

• From: Scott McMorrow <scott@xxxxxxxxxxxxx>
• Date: Mon, 30 Jul 2007 15:11:47 -0400

```Linear superposition says that they pass by each other.  You can say
that this "appears" like a reflection, but it is not.  At the driver
end, when the two counter rotating waves meet with current I , they can
be terminated by a current equal to -2I, developed through a termination
impedance of Z/2.
The beauty of the loop topology is in it's natural perfect balance.  The
two traveling waves always return at exactly the same time.

Scott McMorrow
Teraspeed Consulting Group LLC
121 North River Drive
Narragansett, RI 02882
(401) 284-1840 Fax

http://www.teraspeed.com

Teraspeed® is the registered service mark of
Teraspeed Consulting Group LLC

> 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...  :-)
>
> =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> =
>>>
> 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
>>>> 121 North River Drive=20
>>>> Narragansett, RI 02882=20
>>>>
>>>> California office
>>>> (707) 780-1951 Fax
>>>>
>>>> Main office
>>>> (401) 284-1840 Fax=20
>>>>
>>>> Oregon office
>>>> (503) 430-1285 Fax
>>>>
>>>> http://www.teraspeed.com
>>>> This e-mail contains proprietary and confidential intellectual=20
>>>> property of Teraspeed Consulting Group LLC
>>>>
>>>>    =20
>>>>
>>> =
>>>
> -------------------------------------------------------------------------=
>
>
>>> -----------------------------
>>>  =20
>>>
>>>> Teraspeed(R) is the registered service mark of Teraspeed Consulting=20
>>>> Group LLC
>>>>
>>>>
>>>>
>>>>    =20
>>>>
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> --=20
> Steve Weir
> Teraspeed Consulting Group LLC=20
> 121 North River Drive=20
> Narragansett, RI 02882=20
>
> California office
> (707) 780-1951 Fax
>
> Main office
> (401) 284-1840 Fax=20
>
> Oregon office
> (503) 430-1285 Fax
>
> http://www.teraspeed.com
> This e-mail contains proprietary and confidential intellectual property =
> of Teraspeed Consulting Group LLC
> -------------------------------------------------------------------------=
> -----------------------------
> Teraspeed(R) is the registered service mark of Teraspeed Consulting =
> Group LLC
>
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```