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

• From: olaney@xxxxxxxx
• To: vinu@xxxxxxxxx
• Date: Wed, 1 Aug 2007 11:19:15 -0700

```Vinu,
With all due respect, the evidence has already been stated: Ockham's
razor, fundamental physics, the tagging experiment...  Ultimately, I
don't care whether you cling to your collision model or not.  However,
superposition is what they teach in engineering school, and it explains
everything simply and completely.  It's you against the world.
Orin

On Wed, 01 Aug 2007 10:45:01 -0700 Vinu Arumugham <vinu@xxxxxxxxx>
writes:
Orin,

Since the reflected and propagated waves have identical properties and
are indistinguishable, how can you be sure it is one and not the other?

Thanks,
Vinu

olaney@xxxxxxxx wrote:
Vinu:
Steve is correct.  We all agree that the single special case 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 that, so why
belabor it?  However, the collision theory supposes that somehow the
physics is different just for this special case.  If you 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 with a special
case at all?  It's a needless complication.

Furthermore, the fundamental physics does not support a collision model.
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 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
> 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

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