[SI-LIST] Re: Circle bus topology; Circular Firing Squad?
- From: Vinu Arumugham <vinu@xxxxxxxxx>
- To: Arpad_Muranyi@xxxxxxxxxx
- Date: Mon, 30 Jul 2007 12:59:09 -0700
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> =
>>>
> 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
>>>> (408) 884-3985 Business
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>>>>
>>>> =20
>>>>
>>> =
>>>
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> --=20
> Steve Weir
> Teraspeed Consulting Group LLC=20
> 121 North River Drive=20
> Narragansett, RI 02882=20
>
> California office
> (408) 884-3985 Business
> (707) 780-1951 Fax
>
> Main office
> (401) 284-1827 Business=20
> (401) 284-1840 Fax=20
>
> Oregon office
> (503) 430-1065 Business
> (503) 430-1285 Fax
>
> http://www.teraspeed.com
> This e-mail contains proprietary and confidential intellectual property =
> of Teraspeed Consulting Group LLC
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> Teraspeed(R) is the registered service mark of Teraspeed Consulting =
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