[SI-LIST] Re: Circle bus topology; Circular Firing Squad?
- From: "Muranyi, Arpad" <Arpad_Muranyi@xxxxxxxxxx>
- To: <si-list@xxxxxxxxxxxxx>
- Date: Mon, 30 Jul 2007 13:01:51 -0700
Great idea! How about coloring the
electrical waves... :-)
Arpad
====================================
________________________________
From: Vinu Arumugham [mailto:vinu@xxxxxxxxx]
Sent: Monday, July 30, 2007 12:59 PM
To: Muranyi, Arpad
Cc: si-list@xxxxxxxxxxxxx
Subject: Re: [SI-LIST] Re: Circle bus topology; Circular Firing Squad?
Hopefully, they can be a different color without affecting their mechanical
properties...?
Or, one can perform a thought experiment with two water wavefronts, each of a
different color, approaching from either end of a trench and colliding.
Thanks,
Vinu
Muranyi, Arpad wrote:
This discussion (argument) makes me think of an analogy.
Imagine two identical balls in space approaching each other
along the same and perfectly straight path. At some point
they collide with a perfectly elastic collision. Based on
the laws of physics they reverse their direction and begin
to travel in the opposite direction they came.
Hmmm. Or does the ball that came from the right continue
on the left and the one that came from the left continue
on the right? I can't tell, they are identical... :-)
Arpad
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D
-----Original Message-----
From: si-list-bounce@xxxxxxxxxxxxx
[mailto:si-list-bounce@xxxxxxxxxxxxx] =
On Behalf Of steve weir
Sent: Monday, July 30, 2007 11:25 AM
To: Vinu Arumugham
Cc: olaney@xxxxxxxx; ron@xxxxxxxxxxx; si-list@xxxxxxxxxxxxx
Subject: [SI-LIST] Re: Circle bus topology; Circular Firing Squad?
Vinu, I disagree. The 'even' mode, the common to reference of each=20
waveform encounters the illusion of an open at the intersection. The=20
'odd' mode, ie the difference to reference between each encounters
the=20
illusion of a short. While this approach yields the correct results,
I=20
find it more cumbersome than the real behavior which is two
wavefronts=20
that pass through each other as they propagate in opposite directions=20
down the same transmission line.
As previously discussed in this thread, we can take two drivers at
the=20
opposite end of the same line and drive them as needed to demonstrate=20
any of the possible behaviors.
Best Regards,
Steve.
Vinu Arumugham wrote:
When identical wavefronts are sent through the two branches of
the=20
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=20
virtual open circuit. One wavefront encounters a high impedance
and=20
the other a low impedance compared to the line impedance. The=20
subsequent reflections of opposite polarity can be described
as=20
producing the illusion of the wavefronts flowing through rather
than=20
being reflected at that point.
In other words, it seems to me that both the reflection and=20
reinforcement descriptions are perfectly valid and each is as
real or=20
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> <mailto:weirsi@xxxxxxxxxx> =
writes:
=20
Ron, yes if the signals exactly match then
Ron's description of the=20
apparent open end matches the illusion. It is
an illusion just the=20
same. This is where Orin's proposed experiment
can provide insight. =
=20
Any difference between the two wavefronts is
not accounted for by=20
the=20
open end model. That odd mode if you will
encounters the illusion=20
of a=20
dead short at the same juncture where the even
mode Ron and you=20
describe=20
encounters the illusion of an open. Account
for both the even and=20
odd=20
signal modes and you will get the right answer
from the illusion=20
just as=20
you will if you follow the formal, exact, and I
think simpler view:=20
that=20
the two wavefronts continue to propagate until
they are absorbed.=20
Steve.
ron@xxxxxxxxxxx wrote:
=20
Consider for a moment a 50 ohm source
driving two equal length 100=20
=20
ohm=20
=20
lines unterminated(open circuit)
TDR will show the open circuit at the
end of the lines just as if=20
=20
there were one 50 ohm open ended line.
Next consider what will happen if you
connect the open ended lines=20
=20
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=20
=20
and a=20
=20
TDR if you disagree.
=20
--=20
Steve Weir
Teraspeed Consulting Group LLC=20
121 North River Drive=20
Narragansett, RI 02882=20
California office
(408) 884-3985 Business
(707) 780-1951 Fax
Main office
(401) 284-1827 Business=20
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Steve Weir
Teraspeed Consulting Group LLC=20
121 North River Drive=20
Narragansett, RI 02882=20
California office
(408) 884-3985 Business
(707) 780-1951 Fax
Main office
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http://www.teraspeed.com
This e-mail contains proprietary and confidential intellectual property
=
of Teraspeed Consulting Group LLC
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