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

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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>>>>>
>>>>>           
>>>>>>     Teraspeed(R) is the registered service mark of Teraspeed Consulting
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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
>> (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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>>     
>
>
>   


-- 
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
------------------------------------------------------------------------------------------------------
Teraspeed(R) is the registered service mark of Teraspeed Consulting Group LLC

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