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

  • From: <steven.d.corey@xxxxxxxxxxxxxx>
  • To: <si-list@xxxxxxxxxxxxx>
  • Date: Wed, 1 Aug 2007 11:06:17 -0700

Hi Orin -- I have to say I don't understand why you say that describing
the effect as two waves passing through each other in opposite
directions represents the physics any more faithfully than a weighted
sum of symmetric and anti-symmetric waves reflecting at the point of
symmetry.  Maxwell's equations predict a single E-field vector and a
single H-field vector at each point at each time, and either
decomposition breaks that field into two arbitrary components, neither
any more physical than the other.  Whether one description is needless
or is a complication or is non-sensical as compared to the other is a
highly subjective point, more one of personal opinion than mathematical
or physical rigor.

It's worth noting that waves are not required to describe the field
distributions predicted by Maxwell's equations -- they just happen to be
quite convenient.  Superposition doesn't obligate us to choose a
particular set of basis functions to decompose the problem -- it gives
us the freedom to choose any set that we please, out of an infinite
reservoir, as long as they span the problem space.  Characteristic
impedance is a construct that is convenient for analyzing certain
geometries and excitation schemes, but is by no means a necessity for
physically describing a system.

In terms of straight physics, at the time when the two waves reach the
point of symmetry and either reflect or pass through, depending on your
preference, the system could care less where they came from.  It has no
interest in the history of the signals, whether they came from the left,
right, top, or bottom, or what they saw along the way.  It really only
cares about the resultant field distribution at that point in time
(including the derivatives, of course).  For all it knows, they could
have been placed there by a magical initial condition generator.  And
when it sees that distribution, it responds by kicking out a wave in
either direction.

We regularly play any number of mathematical games to simplify problem
solving, or lend insight into the problem, often without realizing it.
Exhibit A is the venerable Fourier transform, whose basis is a set of
sinusoids that have been oscillating since time immemorial and must
continue forever if the solution is to be complete.  Clearly nonphysical
(although perhaps metaphysical), yet none of us complains because it
simplifies the path to the right solution, and that in turn keeps the
bills paid.

  -- Steve

------------------------------------------------------------------------
-
Steven D. Corey, Ph.D.
Principal Engineer
Tektronix - Enabling Innovation

http://www.tdasystems.com
http://www.tektronix.com

email: steven.corey@xxxxxxxxxxxxx
phone: (503) 627-6816
fax:   (503) 627-2260
------------------------------------------------------------------------
-
=20

>-----Original Message-----
>From: si-list-bounce@xxxxxxxxxxxxx=20
>[mailto:si-list-bounce@xxxxxxxxxxxxx] On Behalf Of olaney@xxxxxxxx
>Sent: Tuesday, July 31, 2007 6:04 PM
>To: vinu@xxxxxxxxx
>Cc: weirsi@xxxxxxxxxx; erdinih@xxxxxxxxx; ron@xxxxxxxxxxx;=20
>si-list@xxxxxxxxxxxxx
>Subject: [SI-LIST] Re: Circle bus topology; Circular Firing Squad?
>
>Vinu:
>Steve is correct.  We all agree that the single special case=20
>of identical
>waves (or some fraction of both waves that can be said to be that way)
>approaching from both directions looks and acts in a way that can be
>interpreted as if there was wave collision.  We all agree on=20
>that, so why
>belabor it?  However, the collision theory supposes that somehow the
>physics is different just for this special case.  If you=20
>acknowledge that
>non-identical signals pass through each other, then why not accept the
>exact and correct results if you assume that everything, different or
>identical, passes through without interaction?  Why bother=20
>with a special
>case at all?  It's a needless complication.
>Furthermore, the fundamental physics does not support a=20
>collision model.=20
>Here's a quote:
>
>"As discussed in Lesson 2, some forces result from contact interactions
>(normal, frictional, tensional, and applied forces are examples of
>contact forces) and other forces are the result of action-at-a-distance
>interactions (gravitational, electrical, and magnetic forces)."
>http://www.glenbrook.k12.il.us/GBSSCI/PHYS/CLASS/newtlaws/u2l4a.html
>
>Applying a contact model to an action-at-a-distance interaction is a
>conceptual error.  Gravity does not rebound as the planets pass each
>other, and broadcast waves slip through each other=20
>effortlessly.  The sea
>of electrons in metal supports electrical waves that do the same.
>
>Orin
>
>On Tue, 31 Jul 2007 15:20:23 -0700 Vinu Arumugham <vinu@xxxxxxxxx>
>writes:
>> Here is some ASCII art with "colored" waves. The LLL and RRR=20
>> represent=20
>> portions of a T-line charged by a wavefront. <,> represent=20
>> wavefronts=20
>> traveling in indicated direction.
>> Assume 50 ohm lines, 20mA current changes resulting in 1V=20
>> wavefronts.
>>=20
>> Incident waves of equal magnitude:
>> LLLLLLLLLLLL>   <RRRRRRRRRRR
>>=20
>> Reflection model :
>> <LLLLLLLLLLLRRRRRRRRRR>
>> LLLLLLLLLLLLRRRRRRRRRRR
>>=20
>> Wave propagation model:
>> <LLLLLLLLLLLRRRRRRRRRR>
>> RRRRRRRRRRRLLLLLLLLLLLL
>>=20
>> Unequal incident waves, left is 2V, 40mA, right is 20mA, 1V:
>> LLLLLLLLLLLL>
>> LLLLLLLLLLLL>   <RRRRRRRRRRR
>>=20
>> Reflection model:
>> <LLLLLLLLLLLLLLLLLLLLLL>
>> LLLLLLLLLLLLRRRRRRRRRR>
>> LLLLLLLLLLLLRRRRRRRRRRR
>>=20
>> For the unequal wave reflection model above, at the point where the=20
>> wavefronts meet, no charge can flow (I=3D0) from the right wavefront=20
>> to=20
>> the left T-line because the lines are charged to the same potential.=20
>> I=3D0=20
>> means the right wavefront sees an open circuit and reflects. When=20
>> this=20
>> point reaches 2V, the 40mA left incident wavefront can charge both=20
>> the=20
>> T-lines with 20mA each, sending a 1V wavefront into the right=20
>> T-line,=20
>> and a 1V reflected wavefront going back on the left T-line. So the=20
>> left=20
>> incident wavefront can be described as having encountered a high=20
>> impedance (>50 ohm and <open circuit) whereby a 2V incident wave=20
>> produced a 1V reflected wave.
>>=20
>> Wave propagation model:
>> <RRRRRRRRRRLLLLLLLLLLL>
>> LLLLLLLLLLLLLLLLLLLLLLL>
>> LLLLLLLLLLLLRRRRRRRRRRR
>>=20
>> Thanks,
>> Vinu
>>=20
>>=20
>> steve weir wrote:
>> > Ihsan, I've presented two methods that both correctly predict the=20
>> > results:  One based on modeling the intersection as an open to the=20
>>=20
>> > even mode, while short to the odd mode, and the other on what I=20
>> think=20
>> > is far simpler:  continuous propagation of each of the original=20
>> wave=20
>> > fronts.  Use whichever model makes your day simpler, but for my=20
>> money=20
>> > I'll stick with the latter.  I prefer the view that=20
>> discontinuities=20
>> > and resulting reflections in quasi uniform, infinite length, ie=20
>> > terminated transmissions are the result of physical variations in=20
>> the=20
>> > 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=20
>> from=20
>> > the right.  After they meet, the voltage moving rightward=20
>> continues to=20
>> > rise by +1.0V from its previous value, and the voltage moving=20
>> leftward=20
>> > continues to rise by +0.5V from its previous value.  The waves=20
>> just=20
>> > linearly superimpose.
>> >
>> > Regards,
>> >
>> >
>> > Steve.
>> >
>> >
>> > Ihsan Erdin wrote:
>> >> Steve,
>> >>
>> >> The wave propagation is simply the transfer of the energy in=20
>> space.
>> >> For the special case a line symmetrically driven at both ends,=20
>> one can
>> >> use the model of an unterminated transmission line driven from=20
>> one
>> >> side only and no one can tell the difference. This is based on=20
>> the
>> >> fundamental electromagnetic principle: image theory.
>> >>
>> >> For the uneven drivers of your example, I can rightfully argue=20
>> that
>> >> the equal frequency components "bounced" and cancelled out while=20
>> the
>> >> residual part kept on propagating. The idea of waves passing=20
>> through
>> >> each other is simply a matter of perception; not a rocksolid=20
>> physical
>> >> reality which ridicules the idea of waves bouncing in the middle.=20
>> 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=20
>> quite
>> >> equivalent. At the collision the balls will have to come to a
>> >> momentary full stop before accelerating in the reverse direction.=20
>> This
>> >> is not symmetrical to the case where they (might) pass through=20
>> each
>> >> other at constant speed.
>> >>
>> >> Best regards,
>> >>
>> >> Ihsan
>
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