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

  • From: Vinu Arumugham <vinu@xxxxxxxxx>
  • To: olaney@xxxxxxxx
  • Date: Wed, 01 Aug 2007 10:45:01 -0700

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 
> <http://www.glenbrook.k12.il.us/GBSSCI/PHYS/CLASS/newtlaws/u2l2a.html#top>, 
> 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 
> <mailto: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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