[SI-LIST] Re: Circle bus topology; Circular Firing Squad?
- From: steve weir <weirsi@xxxxxxxxxx>
- To: Ihsan Erdin <erdinih@xxxxxxxxx>
- Date: Mon, 30 Jul 2007 20:44:29 -0700
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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>>>>
>> --
>> 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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>> Teraspeed(R) is the registered service mark of Teraspeed 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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