[SI-LIST] Re: Circle bus topology; Circular Firing Squad?
- From: Vinu Arumugham <vinu@xxxxxxxxx>
- To: steve weir <weirsi@xxxxxxxxxx>
- Date: Tue, 31 Jul 2007 15:20:23 -0700
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
>>>>>>>
>>>>>>> Main office
>>>>>>> (401) 284-1827 Business
>>>>>>> (401) 284-1840 Fax
>>>>>>>
>>>>>>> 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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>>>>>>> Consulting
>>>>>>> Group LLC
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>
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>>>>>>
>>>>>
>>>>>
>>> --
>>> Steve Weir
>>> Teraspeed Consulting Group LLC
>>> 121 North River Drive
>>> Narragansett, RI 02882
>>>
>>> California office
>>> (408) 884-3985 Business
>>> (707) 780-1951 Fax
>>>
>>> Main office
>>> (401) 284-1827 Business
>>> (401) 284-1840 Fax
>>>
>>> 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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>
>
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