| Author |
Message |
Rich Seifert
Guest
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 Posted:
Mon Sep 12, 2005 11:36 pm Post subject:
Re: Why is there a minimum spacing? |
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In article <re6bi1pu62s7ndelmdqgnoq77c7lil7fou@4ax.com>,
DHP <me@privacy.net> wrote:
| Quote: |
Could I now quiz you a bit more? 4pF on a 50 ohm system gives a
characteristic time of some 200ps or a frequency of about 800MHz. So
I'm guessing (having forgotten the theory ages ago), without doing a
phasor diagram, that you'd get a reflection coefficient ~f/800 for
each component. But at the same time, you only need to worry about
reflections that interfere constructively, i.e. over about half a
wavelength = 117m/f.
So if the allowable reflection is 5%, the number of taps in 117/f m of
cable is 5/100 * 800/f, which is about 1 tap per 2.5m, though there
should be the odd fudge factor to upset the convenient result. Anyway
I can see the point of having a lowish average density of taps! Would
I be right in thinking that the requirement to place taps at equal
spacing is a result of needing to cater for the higher frequencies?
My thinking is that the allowable density of taps taken over a
fraction of a wavelength brings you down to just a small handful of
taps so you may as well just space them equally rather than worsen the
noise with a cluster? Is that the "real" criterion - to avoid clusters
over short distances? It would seem to assume that NICs are sensitive
to out-of-band noise.
|
Actually, I did all of the analysis in the time-domain, rather than the
frequency-domain, although of course they are fully interchangeable.
I started where a communications systems designer SHOULD start--with a
requirement for a maximum bit-error rate (which translates into a
frame-loss rate). For the specified BER of 10^-9 (worst-case), using
Manchester encoding, the minimum signal-to-noise ratio turns out to be
14 db, which is a factor of 5:1. You then take the worst-case minimum
transmit level and attenuate it by the maximum amount possible
(worst-case cables, longest specified lengths) to calculate the minimum
received signal level. The allowable noise at that point must be no more
than one-fifth of the minimum received signal to achieve the desired BER.
(I could re-create the actual numbers, or even find my old notebooks if
I looked, but my point here is to show methodology, which should apply
to a wide variety of communications systems, rather than show the
specific numbers for a now-obsolete system like coaxial Ethernet.)
I then apportioned the allowable noise among the various contributors:
tap reflections, reflections from cable impedance variations, external
EMI, etc. The tap reflection allowance resulted in the specification for
maximum shunt capacitance and the "2.5 meter" rule. The cable impedance
allowance resulted in the specification for maximum deviation from
nominal impedance (50 +/- 2 ohms), and the rules for concatenating long
lengths from shorter pieces. The EMI allowance resulted in the
specification for transfer impedance of the cable shield (effectively
mandating the quad shield design).
Our motto was always that the system had to work in the worst-case.
Sure, most environments were much more benign than we assumed for the
design criteria; those environments would experience a much better BER
than worst-case. But even the worst environment would behave acceptably.
When you are planning for millions of networks, and tens-of-millions of
installed devices, even 99.9% assurance means a lot of angry customers.
--
Rich Seifert Networks and Communications Consulting
21885 Bear Creek Way
(408) 395-5700 Los Gatos, CA 95033
(408) 228-0803 FAX
Send replies to: usenet at richseifert dot com |
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DHP
Guest
|
| Posted:
Tue Sep 13, 2005 12:03 am Post subject:
Re: Why is there a minimum spacing? |
|
|
Rich Seifert <usenet@richseifert.com.invalid> said
| Quote: | In article <re6bi1pu62s7ndelmdqgnoq77c7lil7fou@4ax.com>,
DHP <me@privacy.net> wrote:
Could I now quiz you a bit more? 4pF on a 50 ohm system gives a
characteristic time of some 200ps or a frequency of about 800MHz. So
I'm guessing (having forgotten the theory ages ago), without doing a
phasor diagram, that you'd get a reflection coefficient ~f/800 for
each component. But at the same time, you only need to worry about
reflections that interfere constructively, i.e. over about half a
wavelength = 117m/f.
So if the allowable reflection is 5%, the number of taps in 117/f m of
cable is 5/100 * 800/f, which is about 1 tap per 2.5m, though there
should be the odd fudge factor to upset the convenient result. Anyway
I can see the point of having a lowish average density of taps! Would
I be right in thinking that the requirement to place taps at equal
spacing is a result of needing to cater for the higher frequencies?
My thinking is that the allowable density of taps taken over a
fraction of a wavelength brings you down to just a small handful of
taps so you may as well just space them equally rather than worsen the
noise with a cluster? Is that the "real" criterion - to avoid clusters
over short distances? It would seem to assume that NICs are sensitive
to out-of-band noise.
Actually, I did all of the analysis in the time-domain, rather than the
frequency-domain, although of course they are fully interchangeable.
I started where a communications systems designer SHOULD start--with a
requirement for a maximum bit-error rate (which translates into a
frame-loss rate). For the specified BER of 10^-9 (worst-case), using
Manchester encoding, the minimum signal-to-noise ratio turns out to be
14 db, which is a factor of 5:1. You then take the worst-case minimum
transmit level and attenuate it by the maximum amount possible
(worst-case cables, longest specified lengths) to calculate the minimum
received signal level. The allowable noise at that point must be no more
than one-fifth of the minimum received signal to achieve the desired BER.
(I could re-create the actual numbers, or even find my old notebooks if
I looked, but my point here is to show methodology, which should apply
to a wide variety of communications systems, rather than show the
specific numbers for a now-obsolete system like coaxial Ethernet.)
I then apportioned the allowable noise among the various contributors:
tap reflections, reflections from cable impedance variations, external
EMI, etc. The tap reflection allowance resulted in the specification for
maximum shunt capacitance and the "2.5 meter" rule. The cable impedance
allowance resulted in the specification for maximum deviation from
nominal impedance (50 +/- 2 ohms), and the rules for concatenating long
lengths from shorter pieces. The EMI allowance resulted in the
specification for transfer impedance of the cable shield (effectively
mandating the quad shield design).
Our motto was always that the system had to work in the worst-case.
Sure, most environments were much more benign than we assumed for the
design criteria; those environments would experience a much better BER
than worst-case. But even the worst environment would behave acceptably.
When you are planning for millions of networks, and tens-of-millions of
installed devices, even 99.9% assurance means a lot of angry customers.
|
I appreciate the design philosophy! I was just trying to get a handle
on why you went for multiples of 2.5m rather than have it as a simple
minimum. |
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James Knott
Guest
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| Posted:
Tue Sep 13, 2005 12:10 am Post subject:
Re: Why is there a minimum spacing? |
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DHP wrote:
| Quote: | That may give you data corruption but it shouldn't trigger the
collision detector unless the level is ridiculous.
|
What's the difference, between two signals colliding and a signal and it's
reflection colliding? |
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DHP
Guest
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| Posted:
Tue Sep 13, 2005 12:55 am Post subject:
Re: Why is there a minimum spacing? |
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|
James Knott <james.knott@rogers.com> said
| Quote: | DHP wrote:
That may give you data corruption but it shouldn't trigger the
collision detector unless the level is ridiculous.
What's the difference, between two signals colliding and a signal and it's
reflection colliding?
|
A collision actually means two units transmitting at once, not two
signals superimposing. So a true collision results in two
similarly-sized signals superimposing. This can be detected almost
instantly by purely electrical means.
The reflections we are talking about here are much smaller than the
original signal so they don't trigger the collision detection
circuitry. They produce quasi-random noise which occasionally causes a
data bit to be mis-read. The corruption won't be detected by this
layer but will be spotted by the next layer when it does a CRC check.
That's what I meant by the level being ridiculous. You can create huge
reflections by not terminating the line. In that case you may very
well get the collision detection circuitry triggering - when the
transmitting unit gets its own data back. You could even get it
happening if it was the only unit on the line. But it's a bit academic
as the data would be so coruupted that CRC would not be able to mend
it.
Is that what you wanted to know? |
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Henry
Guest
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| Posted:
Tue Sep 13, 2005 12:59 am Post subject:
Re: Why is there a minimum spacing? |
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James Knott <james.knott@rogers.com> said
| Quote: | DHP wrote:
That may give you data corruption but it shouldn't trigger the
collision detector unless the level is ridiculous.
What's the difference, between two signals colliding and a signal and it's
reflection colliding?
|
One's the sound of one hand clapping, the other's a clash of symbols
:) |
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William P. N. Smith
Guest
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| Posted:
Tue Sep 13, 2005 2:07 am Post subject:
Re: Why is there a minimum spacing? |
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Henry <me@privacy.net> wrote:
| Quote: | James Knott <james.knott@rogers.com> said
What's the difference, between two signals colliding and a signal and it's
reflection colliding?
One's the sound of one hand clapping, the other's a clash of symbols
|
HA! Thanks, I needed that! Can't explain it to most folks, but
thanks anyway! |
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William P. N. Smith
Guest
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| Posted:
Tue Sep 13, 2005 2:12 am Post subject:
Re: Why is there a minimum spacing? |
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DHP <me@privacy.net> wrote:
| Quote: | on why you went for multiples of 2.5m rather than have it as a simple
minimum.
|
It's not a minimum or maximum, it's the spacing that minimizes
reflection-based bit errors. 2M and 3M (for instance) are both worse
than 2.5M. |
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glen herrmannsfeldt
Guest
|
| Posted:
Tue Sep 13, 2005 2:39 am Post subject:
Re: Why is there a minimum spacing? |
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DHP wrote:
(snip regarding tap spacing on thick ethernet)
| Quote: | I appreciate the design philosophy! I was just trying to get a handle
on why you went for multiples of 2.5m rather than have it as a simple
minimum.
|
Having actually put taps into cables in cable trays and suspended
ceilings, it is sometimes hard to know where the other taps are.
Sometimes I have done it by feel when I could barely see the cable.
(Well, enough to know it was the right one.)
Then again, it is hard to know that there aren't more than 100.
As previously discussed here, random spacing would be even better,
but guaranteeing it is hard, and it seems that making a machine to
mark cables at random spacings is also hard.
-- glen |
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glen herrmannsfeldt
Guest
|
| Posted:
Tue Sep 13, 2005 2:53 am Post subject:
Re: Why is there a minimum spacing? |
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DHP wrote:
(snip)
| Quote: | Unfortunately it seems 802.3 is ambiguous (anyone got the thing?) and
can't make up its mind whether 2.5m is a minimum or whether you're
supposed to tap in ONLY at multiples of it.
sysrick.713115865@starbase.spd.louisville.edu
It's minimum distance, though the old thicknet cables had specific points
marked on the sheath, where a vampire tap could be attached.
Ah, well, Rich Seifert has joined the thread and says otherwise.
|
There is physics, and then there are rules. The rules are set so
that the system will work within the physical limitations.
In many cases the rules are more strict than necessary to make
them simpler. The 2.5m tap rule is simple to state, not too
restrictive for actual use, and allows the system to work.
In many cases you can't see all of the cable, so you couldn't
guarantee a minimum. With cable marked at 2.5m you can be
sure that if you tap at marks you meet the requirement.
For thin ethernet the rule is 0.5m minimum. As BNC cables
commonly come premade in lengths that are not multiples of
0.5m it is good that it isn't required to be multiples.
Also, in the thin ethernet case, the real restriction is
against the lumped impedance effect of many taps close
together as seen from some distance away. A very short cable
with a (relatively) large number of taps will work just fine.
-- glen |
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Rich Seifert
Guest
|
| Posted:
Tue Sep 13, 2005 3:01 am Post subject:
Re: Why is there a minimum spacing? |
|
|
In article <Za2dnZTNBNErT7jeRVn-qQ@rogers.com>,
James Knott <james.knott@rogers.com> wrote:
| Quote: | DHP wrote:
That may give you data corruption but it shouldn't trigger the
collision detector unless the level is ridiculous.
What's the difference, between two signals colliding and a signal and it's
reflection colliding?
|
In coaxial Ethernet (the subject of the original post), collisions are
detected by measuring the average DC voltage on the cable, NOT by
comparison between the transmitted and received signal. A tap reflection
does not change the average DC; thus, while it might cause data
corruption, it will never cause a false collision.
--
Rich Seifert Networks and Communications Consulting
21885 Bear Creek Way
(408) 395-5700 Los Gatos, CA 95033
(408) 228-0803 FAX
Send replies to: usenet at richseifert dot com |
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glen herrmannsfeldt
Guest
|
| Posted:
Tue Sep 13, 2005 3:25 am Post subject:
Re: Why is there a minimum spacing? |
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|
Rich Seifert wrote:
(snip)
| Quote: | As you realized, one bit-time at 10 Mb/s is 100 ns, which corresponds to
23.5 m of coaxial cable.
|
I am not sure how accurate the velocity factor is, but...
Constructive interference would result from a half wavelength spacing,
so 11.75m. A 500m cable could have 43 taps with that spacing,
which could be significant. If you put 44 taps equally distributed
over the same distance they will pretty much cancel each other out.
If you put 43 taps spaced at 11.75m and the velocity factor is
off by 2% they also pretty much cancel out.
The first odd multiple of 11.75m that is close to a multiple of 2.5m
seems to be 82.25m.
It seems to me very unlikely that, unless someone intentionally spaced
them at 11.75m that they would cause problems, but it is nice to have
a rule with a known effect.
-- glen
| Quote: | Our own Rich Seifert certainly can, but IIRC it has to do with keeping
impedance discontinuities caused by taps far enough apart that they
don't reinforce each other.
{google,deja} news is your friend.
Thanks. I've found some stuff from Rich Seifert going back to
1980-something which explains it, sort of, though it's a bit woolly -
not Rich's explanation but the thinking behind it.
The basic problem is that transceiver taps appear to the transmission
line as discrete, lumped capacitive loads; the specification mandates a
maximum of 4 pf, but this is still significant. When the signal
encounters this capacitance, it creates an out-of-phase reflection of a
portion of the energy. To all other devices on the cable, this
reflection appears as asynchronous "noise," i.e., a signal that
interferes with the desired signal.
The situation to be avoided is where all of the transceiver taps are
spaced such that the reflections from each of them add up in phase, thus
combining *algebraically* (i.e., simple summation). The small reflection
from 99 transceivers added up could create enough interference to cause
bit errors. Ideally, one would want the transceivers to be *randomly*
spaced along the cable; this would ensure that the reflections added not
algebraically, but on a root-mean-squared basis, yielding much less
reflected energy. In fact, my original proposal was to do exactly that;
I even had a patent application prepared for a method of manufacturing
cables with randomly-distributed markings for this purpose!
As it turns out, random markings were neither practical (installers
didn't like the idea, and neither did the cable manufacturers) nor
necessary. I did extensive simulations of the resulting reflections from
transceivers at various spacings, and empirically determined that 2.5 m
was "good enough." It was relatively easy to mark the cables with a
uniform 2.5 m marking; as the cable comes flying out of the extruder, it
passes across a roller with a 2.5 m circumference, which places a mark
at every rotation.
The idea is not just a *minimum* 2.5 m spacing; it is that transceivers
are only placed at the 2.5 m markings. However, as another poster noted,
it's not all that critical; if a few transceivers are offset, or even
lumped together, it is unlikely to cause a noticeable problem. I was
just trying to design for the worst-case, figuring that it would surely
show up *somewhere*, and that one installer would have no idea what the
problem was.
By the way, that cable-spacing work, along with the work that defined
the proper lengths to use for concatenating short coaxial cables into
long runs, constituted a major part of my EE master's thesis some 25
years ago.
--
Rich Seifert Networks and Communications Consulting
21885 Bear Creek Way
(408) 395-5700 Los Gatos, CA 95033
(408) 228-0803 FAX
Send replies to: usenet at richseifert dot com |
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DHP
Guest
|
| Posted:
Tue Sep 13, 2005 3:46 am Post subject:
Re: Why is there a minimum spacing? |
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|
William P. N. Smith <> said
| Quote: | DHP <me@privacy.net> wrote:
on why you went for multiples of 2.5m rather than have it as a simple
minimum.
It's not a minimum or maximum, it's the spacing that minimizes
reflection-based bit errors. 2M and 3M (for instance) are both worse
than 2.5M.
|
That is the thing I find odd - could you explain why, please? |
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DHP
Guest
|
| Posted:
Tue Sep 13, 2005 3:57 am Post subject:
Re: Why is there a minimum spacing? |
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|
glen herrmannsfeldt <gah@ugcs.caltech.edu> said
| Quote: | Rich Seifert wrote:
(snip)
As you realized, one bit-time at 10 Mb/s is 100 ns, which corresponds to
23.5 m of coaxial cable.
I am not sure how accurate the velocity factor is, but...
Constructive interference would result from a half wavelength spacing,
|
I do not understand how the concept of constructive interference -
which applies to a narrow-band signal, a sinewave - can be applied to
a Manchester-encoded bit stream.
In any case, the reflection from a capacitive tap is (approximately)
the time-derivative of the origina. With a fast rise-time on the
transmitter, you'll just get a series of short pulses. With random
polarity of data, how can you ensure they cancel rather than add?
| Quote: | so 11.75m. A 500m cable could have 43 taps with that spacing,
which could be significant. If you put 44 taps equally distributed
over the same distance they will pretty much cancel each other out.
If you put 43 taps spaced at 11.75m and the velocity factor is
off by 2% they also pretty much cancel out.
It seems to me very unlikely that, unless someone intentionally spaced
them at 11.75m that they would cause problems, but it is nice to have
a rule with a known effect.
-- glen
Our own Rich Seifert certainly can, but IIRC it has to do with keeping
impedance discontinuities caused by taps far enough apart that they
don't reinforce each other.
{google,deja} news is your friend.
Thanks. I've found some stuff from Rich Seifert going back to
1980-something which explains it, sort of, though it's a bit woolly -
not Rich's explanation but the thinking behind it.
The basic problem is that transceiver taps appear to the transmission
line as discrete, lumped capacitive loads; the specification mandates a
maximum of 4 pf, but this is still significant. When the signal
encounters this capacitance, it creates an out-of-phase reflection of a
portion of the energy. To all other devices on the cable, this
reflection appears as asynchronous "noise," i.e., a signal that
interferes with the desired signal.
The situation to be avoided is where all of the transceiver taps are
spaced such that the reflections from each of them add up in phase, thus
combining *algebraically* (i.e., simple summation). The small reflection
from 99 transceivers added up could create enough interference to cause
bit errors. Ideally, one would want the transceivers to be *randomly*
spaced along the cable; this would ensure that the reflections added not
algebraically, but on a root-mean-squared basis, yielding much less
reflected energy. In fact, my original proposal was to do exactly that;
I even had a patent application prepared for a method of manufacturing
cables with randomly-distributed markings for this purpose!
As it turns out, random markings were neither practical (installers
didn't like the idea, and neither did the cable manufacturers) nor
necessary. I did extensive simulations of the resulting reflections from
transceivers at various spacings, and empirically determined that 2.5 m
was "good enough." It was relatively easy to mark the cables with a
uniform 2.5 m marking; as the cable comes flying out of the extruder, it
passes across a roller with a 2.5 m circumference, which places a mark
at every rotation.
The idea is not just a *minimum* 2.5 m spacing; it is that transceivers
are only placed at the 2.5 m markings. However, as another poster noted,
it's not all that critical; if a few transceivers are offset, or even
lumped together, it is unlikely to cause a noticeable problem. I was
just trying to design for the worst-case, figuring that it would surely
show up *somewhere*, and that one installer would have no idea what the
problem was.
By the way, that cable-spacing work, along with the work that defined
the proper lengths to use for concatenating short coaxial cables into
long runs, constituted a major part of my EE master's thesis some 25
years ago.
--
Rich Seifert Networks and Communications Consulting
21885 Bear Creek Way
(408) 395-5700 Los Gatos, CA 95033
(408) 228-0803 FAX
Send replies to: usenet at richseifert dot com |
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| Back to top |
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DHP
Guest
|
| Posted:
Tue Sep 13, 2005 4:05 am Post subject:
Re: Why is there a minimum spacing? |
|
|
Rich Seifert <usenet@richseifert.com.invalid> said
| Quote: | In article <Za2dnZTNBNErT7jeRVn-qQ@rogers.com>,
James Knott <james.knott@rogers.com> wrote:
DHP wrote:
That may give you data corruption but it shouldn't trigger the
collision detector unless the level is ridiculous.
What's the difference, between two signals colliding and a signal and it's
reflection colliding?
In coaxial Ethernet (the subject of the original post), collisions are
detected by measuring the average DC voltage on the cable, NOT by
comparison between the transmitted and received signal. A tap reflection
does not change the average DC; thus, while it might cause data
corruption, it will never cause a false collision.
|
A *resistive* mismatch will cause a reflection that would alter the
"DC" level as it's a carbon copy of the original signal, just smaller.
But a reflection from a capacitive tap has no "DC" component.
Incidentally, I read somewhere, that it's not the "DC" level on the
line that's measured, but the DC current taken from the power supply,
but I dare say that's as accurate as all the other bits of lore! |
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glen herrmannsfeldt
Guest
|
| Posted:
Tue Sep 13, 2005 5:04 am Post subject:
Re: Why is there a minimum spacing? |
|
|
DHP wrote:
(snip)
(I wrote)
| Quote: | Constructive interference would result from a half wavelength spacing,
I do not understand how the concept of constructive interference -
which applies to a narrow-band signal, a sinewave - can be applied to
a Manchester-encoded bit stream.
|
I don't believe that a scrambler is used, and repetitive bit streams
are fairly common. One could easily imagine the entire cable filled
with all zero bits. Otherwise, yes, for each combination of bits
there should be an appropriate combination of taps where they will
add constructively.
| Quote: | In any case, the reflection from a capacitive tap is (approximately)
the time-derivative of the origina. With a fast rise-time on the
transmitter, you'll just get a series of short pulses. With random
polarity of data, how can you ensure they cancel rather than add?
|
Whatever it looks like it will add in phase to one delayed by
one cycle. For a stream of zero bits that is one half a bit time
down the cable.
-- glen |
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