[SI-LIST] Re: Circle bus topology; Circular Firing Squad?
- From: steve weir <weirsi@xxxxxxxxxx>
- To: Vinu Arumugham <vinu@xxxxxxxxx>
- Date: Tue, 31 Jul 2007 16:23:37 -0700
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
>>>>>>>>
>>>>>>>> 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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>>>>>>
>>>>>>
>>>> --
>>>> 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
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>>>> 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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>
--
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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