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

• From: steve weir <weirsi@xxxxxxxxxx>
• To: ron@xxxxxxxxxxx
• Date: Fri, 03 Aug 2007 14:15:53 -0700

```Ron, I like experiments.  Try this:  Instead of traces just get two 1
meter pieces of coax and a coupler with a very low VSWR.  TDR each of
the cables.  How long is it?  Join them with the coupler?  How long is
it now?

Here's another to think about.  Get a power splitter with a very low
VSWR value and a length of matching coax, say 1 meter long, and a couple
of terminators.

Experiment #1:  TDR to 0dB splitter port.  Terminator on -6dB port A.
Coax on -6dB port B.  Terminator at the far end of the coax.  What does
the TDR report?
Experiment #2:  Same as #1 but reverse which end of the coax is on port
B and which is terminated.  What does the TDR report?
Experiment #3:  Remove the terminator from the end of the coax.  What
does the TDR report?
Experiment #4:  Connect the coax between port A and port B of the power
splitter.  What does the TDR report?

I hope that we all agree that in Experiments 1 and 2 if the matching is
perfect the TDR reports an infinite length line.  Similarly, Experiment
3 reports a 1 meter line, and Experiment 2 a 0.5 meter line.  We can
explain all of these results with either the reflection or the
propagation models.  As much as it offends you to think that the signal
does not propagate all the way down the coax, why then do you think the
first three experiments behave as they do?  In each case we used the
identical piece of coax.  The crowd that I belong to look at that coax,
see that it is unchanged from Experiment 1 to Experiment 4 and therefore
take the view that the TDR suffers the illusion of a reflection.

Now connect a pulse generator to a power splitter 0dB port.  Connect a
fast scope through a short piece of coax to the -6dB A port.  Connect
one end of the 1 meter coax to the -6dB B port.  Do the same thing with
a second pulse generator at the opposite end.  Set the pulse generators
to orthogonal pulse rates.

Experiment #5:  With the left side pulse generator set to 0 amplitude,
fire a train of widely spaced pulses from the right side pulse
generator.  What does each scope channel report?

Experiment #6:  Repeat #5 but with the right side pulse generator set to
0 amplitude and the left side generator issuing pulses.  What does each
scope channel report?

Experiment #7:  Now allow each pulse generator to drive but at rates
that are orthogonal to each other.  What does each scope channel report?

Does the continuous propagation model correctly predict the results of
each experiment or not?  If it does, why is it so much "foolishness"?
Do images from one pulse generator not arrive at the opposite scope port
superimposed on the local pulses Tflight_1meter after launch?

If that doesn't strike your fancy, try the LASER pointer experiments.

In the greater ultimate TRUTH, it might be that wavefronts reflect in
such a way as to mimic continuous propagation.  Or it could be that EM
waves really do propagate independently of one another.  Or it may be
that we are all the puppets of the ElectroMagnetic Pixies who manipulate
all our perceptions to their whimsical desires.  The problem is that
without any difference in the predicted outcome we can't discern the
ultimate TRUTH.  What we can do is use the simplest solution that
predicts the right results, or a more complicated one if it suits us.  I
vote simple.

Regards,

Steve.
ronald miller wrote:
> Hi Guys
> Since all this theoretical garbage is still trailing on I would like to
> suggest a simple analogy to make it clear.
>
> Consider a 50 ohm microstrip on a board terminated in an open
> Split the line lengthwise with a cut down the center -
> each side will have approximately 100 ohm impedance if the other side is
> removed.
> TDR the two connected together at the source and you will again see 50 ohms.
> Put a small solder ball at the end connecting the two lines
> TDR the two again and you will again see 50 ohms.
>
> The coupling factor can be igored for this analysis.
>
> The idea that you have a circle is foolishness.  It is one line split
> into two forward paths and we know how they act.
>
> Now, if  you add different loads at different points along the two lines
> you have a more complex situation
> but can again analyze it without regard for the circle.
>
> Ron Miller
>
> David Banas wrote:
>
>
>> I don't think it's correct to think of the electron and photon
>> "collisions" as similar, since electrons are Fermions and photons are
>> Bosons. Whereas Fermions are prohibited by the laws of quantum mechanics
>>
> >from occupying the same state and, therefore, must repel each other when
>
>> they get close, Bosons are actually "encouraged" by quantum mechanics to
>> occupy identical states. In fact, the more Bosons that are in a
>> particular state, the more likely it is that a new one will join them.
>> (This is why lasers work.)
>>
>> -db
>>
>>
>>
>>
>>
>>> -----Original Message-----
>>> From: si-list-bounce@xxxxxxxxxxxxx
>>>
>>>
>>>
>> [mailto:si-list-bounce@xxxxxxxxxxxxx]
>>
>>
>>
>>> On Behalf Of Muranyi, Arpad
>>> Sent: Wednesday, August 01, 2007 11:13 AM
>>> To: si-list@xxxxxxxxxxxxx
>>> Subject: [SI-LIST] Re: Circle bus topology; Circular Firing Squad?
>>> =20
>>> This is why we need to use coloring dye...  :-)
>>> =20
>>> Seriously, I wonder about the particle and
>>> wave duality of electrons.  Remember, the
>>> drift velocity of electrons is not the same
>>> as the propagation velocity of the EM waves.
>>> =20
>>> So what happens at the "collision" point (i.e.
>>> the electrical midpoint of the loop)?  Do=3D20
>>> the electrons hit each other and bounce back
>>> like balls with the drift velocity speeds,
>>> while at the same time the waves go through
>>> each other (at speeds close to c) without
>>> changing directions?
>>> =20
>>> Another analogy comes to my mind, when two laser
>>> beams on different path cross each other.  After
>>> the cross point we are still going to see their
>>> original colors without any mixing.  However,
>>> if you did the same with two water streams,
>>> the cross point will result in a splash in all
>>> different directions...
>>> =20
>>> Is there someone out there with a good physics=3D20
>>> background who could shed some light on this?
>>> =20
>>> Thanks,
>>> =20
>>>
>>>
>>>
>>>
>> =3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D=
>> 3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D
>> =3D3
>>
>>
>>
>>> D=3D
>>>
>>>
>>>
>>>
>> =3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D=
>> 3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D=3D3D
>> =3D3
>>
>>
>>
>>> D=3D
>>> =3D3D
>>> =3D20
>>> =20
>>> -----Original Message-----
>>> From: si-list-bounce@xxxxxxxxxxxxx
>>>
>>>
>>>
>> [mailto:si-list-bounce@xxxxxxxxxxxxx]
>>
>>
>>
>>> On Behalf Of Vinu Arumugham
>>> Sent: Wednesday, August 01, 2007 10:45 AM
>>> To: olaney@xxxxxxxx
>>> Cc: weirsi@xxxxxxxxxx; erdinih@xxxxxxxxx; ron@xxxxxxxxxxx;
>>> si-list@xxxxxxxxxxxxx
>>> Subject: [SI-LIST] Re: Circle bus topology; Circular Firing Squad?
>>> =20
>>> Orin,
>>> Since the reflected and propagated waves have identical properties
>>>
>>>
>>>
>> and=3D20
>>
>>
>>
>>> are indistinguishable, how can you be sure it is one and not the
>>>
>>>
>>>
>> other?
>>
>>
>>
>>> =20
>>> Thanks,
>>> Vinu
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>>
>>
>>
>
>

--
Steve Weir
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