[SI-LIST] Re: Circle bus topology; Circular Firing Squad?
- From: olaney@xxxxxxxx
- To: vinu@xxxxxxxxx
- Date: Mon, 30 Jul 2007 17:53:08 -0700
Vinu:
The impedance seen by a network analyzer is a composite of the *entire*
path at each measured frequency. Being a composite, it is a mixture of
the constant impedance transmission line and the terminating impedances,
which can be anywhere in the complex plane. That is why the fixed
frequency, steady state, sinusoidal wave input impedance can be radically
different from that of the transmission line itself: It doesn't mean
that the actual line impedance changes in the slightest. To discern what
is actuallly happening at individual points on the line, and
independently from waveshape, the time domain approach is more useful.
The duality is, VNA: entire path at each specific frequency in the sweep
range, TDR: entire range of frequencies (within the bandwidth of the
step) at each specific point along the path length. The surge, or
instantaneous impedance is what we are discussing here.
Orin
On Mon, 30 Jul 2007 15:50:43 -0700 Vinu Arumugham <vinu@xxxxxxxxx>
writes:
> "There is only one impedance at any given point on the line, and for
> constant line parameters, the impedance is constant throughout."
> Yes, that's the characteristic impedance of the line.
> The input impedance of an unterminated line can vary from zero to
> infinity depending on the frequency of the driving signal. In other
> words, the line driver "sees" a high or low impedance that is a
> function
> of the magnitude and phase of the reflected wavefront. The same
> thing
> happens when wavefronts meet in a loop. The effective impedance seen
> by
> each wavefront is a function of the magnitude and phase of the other
>
> wavefront. So, why is this interpretation "nonsensical"?
>
> Thanks,
> Vinu
>
> olaney@xxxxxxxx wrote:
> > If you suppose that the waves meet and rebound like billiard
> balls,
> > that would be incorrect. Each passes through the other as if it
> was
> > the only wave on the transmission line. Only a real open circuit
> (or
> > other impedance discontinuity) can cause reflection. Though
> identical
> > wavefronts might create the illusion of a "virtual open circuit"
> to
> > the viewer, that is not the physical reality. The simultaneous
> "high
> > impedance / low impedance" interpretation is nonsensical. There
> is
> > only one impedance at any given point on the line, and for
> constant
> > line parameters, the impedance is constant throughout. Especially
>
> > note that the impedance of a linear xmsn line has nothing to do
> with
> > the shape or direction of the waves that happen to be traveling on
>
> > it. To suppose otherwise wrenches the laws of physics. Sorry if
> I
> > have to be blunt. Wavefronts passing through each other is the
> > bedrock reality, all else is armwaving.
> >
> > Orin Laney, PE, NCE
> >
> > On Mon, 30 Jul 2007 10:47:33 -0700 Vinu Arumugham <vinu@xxxxxxxxx
> > <mailto:vinu@xxxxxxxxx>> writes:
> >
> > When identical wavefronts are sent through the two branches of
> the
> > 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
> > virtual open circuit. One wavefront encounters a high
> impedance
> > and the other a low impedance compared to the line impedance.
> The
> > subsequent reflections of opposite polarity can be described
> as
> > producing the illusion of the wavefronts flowing through
> rather
> > than being reflected at that point.
> >
> > In other words, it seems to me that both the reflection and
> > reinforcement descriptions are perfectly valid and each is as
> real
> > or 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> writes:
> >>
> >>> Ron, yes if the signals exactly match then Ron's description
> of the
> >>> apparent open end matches the illusion. It is an illusion
> just the
> >>>
> >>> same. This is where Orin's proposed experiment can provide
> insight.
> >>>
> >>> Any difference between the two wavefronts is not accounted
> for by
> >>> the
> >>> open end model. That odd mode if you will encounters the
> illusion
> >>> of a
> >>> dead short at the same juncture where the even mode Ron and
> you
> >>> describe
> >>> encounters the illusion of an open. Account for both the
> even and
> >>> odd
> >>> signal modes and you will get the right answer from the
> illusion
> >>> just as
> >>> you will if you follow the formal, exact, and I think
> simpler view:
> >>> that
> >>> the two wavefronts continue to propagate until they are
> absorbed.
> >>>
> >>> Steve.
> >>> ron@xxxxxxxxxxx wrote:
> >>>
> >>>> Consider for a moment a 50 ohm source driving two equal
> length 100
> >>>>
> >>> ohm
> >>>
> >>>> lines unterminated(open circuit)
> >>>> TDR will show the open circuit at the end of the lines just
> as if
> >>>>
> >>>> there were one 50 ohm open ended line.
> >>>>
> >>>> Next consider what will happen if you connect the open
> ended lines
> >>>>
> >>>> 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
> >>>>
> >>> and a
> >>>
> >>>> TDR if you disagree.
> >>>>
> >>>>
> >>>>
> >>> --
> >>> Steve Weir
> >>> Teraspeed Consulting Group LLC
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