Why is there a minimum spacing?
henry
Archived from groups: comp.dcom.lans.ethernet (More info?)
I wonder if anyone can give a definitive answer as to why there is a
minimum spacing specified on (some) ethernet cable. The thick stuff
with markers every 2.5m for example which is 1 bit of delay at 10 MHz.
There is some mention of it on various web sites but the reasons for
it are not stated. Maximum lengths etc are simple enough to
understand: you need to be sure that collisions are not late. The only
reason I can think of for specifying a minimum distance is to maximise
the effect of a collision when two MAUs start transmitting at the same
time. Only I can't see that it would. They won't actually start
together. If they're waiting for the line to become free, the last
data going past them will make sure one starts after the other. So the
second will start up at the eaxct moment the first's one's data
arrives. So it will experience a zero timedifference collision. The
first one will have a two bit difference. Even if there's an advantage
in that  which I don't understand it assumes exactly one 2.5m
section of cable. But the 2.5m is only a minimum: the spec doesn't
require exact multiplesof 2.5m over hundreds of metres! So I'm racking
my brains as to why it was ever specified at all.
I wonder if anyone can give a definitive answer as to why there is a
minimum spacing specified on (some) ethernet cable. The thick stuff
with markers every 2.5m for example which is 1 bit of delay at 10 MHz.
There is some mention of it on various web sites but the reasons for
it are not stated. Maximum lengths etc are simple enough to
understand: you need to be sure that collisions are not late. The only
reason I can think of for specifying a minimum distance is to maximise
the effect of a collision when two MAUs start transmitting at the same
time. Only I can't see that it would. They won't actually start
together. If they're waiting for the line to become free, the last
data going past them will make sure one starts after the other. So the
second will start up at the eaxct moment the first's one's data
arrives. So it will experience a zero timedifference collision. The
first one will have a two bit difference. Even if there's an advantage
in that  which I don't understand it assumes exactly one 2.5m
section of cable. But the 2.5m is only a minimum: the spec doesn't
require exact multiplesof 2.5m over hundreds of metres! So I'm racking
my brains as to why it was ever specified at all.
64
answers
Last reply
More about minimum spacing

Archived from groups: comp.dcom.lans.ethernet (More info?)
Henry <me@privacy.net> wrote:
>I wonder if anyone can give a definitive answer as to why there is a
>minimum spacing specified on (some) ethernet cable. The thick stuff
>with markers every 2.5m for example which is 1 bit of delay at 10 MHz.
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. 
Archived from groups: comp.dcom.lans.ethernet (More info?)
Henry wrote:
> I wonder if anyone can give a definitive answer as to why there is a
> minimum spacing specified on (some) ethernet cable. The thick stuff
> with markers every 2.5m for example which is 1 bit of delay at 10 MHz.
>
IIRC, the idea is that each connection to the cable causes an impedance
discontinuity. Spreading them out minimizes the problems caused by those
connections. 
Archived from groups: comp.dcom.lans.ethernet (More info?)
> I wonder if anyone can give a definitive answer as to why there is a
> minimum spacing specified on (some) ethernet cable. The thick stuff
> with markers every 2.5m for example which is 1 bit of delay at 10 MHz.
According to the Spurgeon book, the spacing is a guideline to help
avoid signal reflections resulting from too many transceiver taps being
clumped together.
He goes on to say that maintaining an even 2.5m spacing isn't critical:
when joining cable sections you can ignore the marks, and if two taps
happen to land close together when cables are joined, that's okay too.
/chris 
Archived from groups: comp.dcom.lans.ethernet (More info?)
William P. N. Smith <> said
>Henry <me@privacy.net> wrote:
>>I wonder if anyone can give a definitive answer as to why there is a
>>minimum spacing specified on (some) ethernet cable. The thick stuff
>>with markers every 2.5m for example which is 1 bit of delay at 10 MHz.
Oops, just realized there's a factor of 10 missing there. My attempted
guess at the reasoning was wrong... The mystery deepens.
>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
1980something which explains it, sort of, though it's a bit woolly 
not Rich's explanation but the thinking behind it. 
Archived from groups: comp.dcom.lans.ethernet (More info?)
Henry wrote:
> Thanks. I've found some stuff from Rich Seifert going back to
> 1980something which explains it, sort of, though it's a bit woolly 
> not Rich's explanation but the thinking behind it.
Why not post the link here? 
Archived from groups: comp.dcom.lans.ethernet (More info?)
In article <jlmai1pi08p6do3jo0p0veditvn4ra7v8m@4ax.com>,
Henry <me@privacy.net> wrote:
> William P. N. Smith <> said
>
> >Henry <me@privacy.net> wrote:
> >>I wonder if anyone can give a definitive answer as to why there is a
> >>minimum spacing specified on (some) ethernet cable. The thick stuff
> >>with markers every 2.5m for example which is 1 bit of delay at 10 MHz.
>
> Oops, just realized there's a factor of 10 missing there. My attempted
> guess at the reasoning was wrong... The mystery deepens.
>
As you realized, one bittime at 10 Mb/s is 100 ns, which corresponds to
23.5 m of coaxial cable.
> >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
> 1980something 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 outofphase 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 rootmeansquared 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 randomlydistributed 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 worstcase, figuring that it would surely
show up *somewhere*, and that one installer would have no idea what the
problem was.
By the way, that cablespacing 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) 3955700 Los Gatos, CA 95033
(408) 2280803 FAX
Send replies to: usenet at richseifert dot com 
Archived from groups: comp.dcom.lans.ethernet (More info?)
James Knott <james.knott@rogers.com> said
>Henry wrote:
>
>> Thanks. I've found some stuff from Rich Seifert going back to
>> 1980something which explains it, sort of, though it's a bit woolly 
>> not Rich's explanation but the thinking behind it.
>
>Why not post the link here?
I was going to do exactly that but I tried to retrace my search and
couldn't find it I may have been mixing two posts in my head.
Still, this is short and simple:
<richya023060042802011315570001@nntp.ix.netcom.com>
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>
The fact it talks about nonalignment suggests someone must have
thought there were significant and potentially troublesome components
in the waveform up to 100s of MHz. I can believe that, since the small
mismatches in resistive impedance become very large mismatches,
effectively a short circuit, for very fast edges which encounter a
capacitive tap. However, then it doesn't make sense to insist on using
exact multiples as this will tend to create standing waves.
But at least I know it's nothing to do with collision detection. 
Archived from groups: comp.dcom.lans.ethernet (More info?)
Henry wrote:
> James Knott <james.knott@rogers.com> said
>
>>Henry wrote:
>>
>>> Thanks. I've found some stuff from Rich Seifert going back to
>>> 1980something which explains it, sort of, though it's a bit woolly 
>>> not Rich's explanation but the thinking behind it.
>>
>>Why not post the link here?
>
> I was going to do exactly that but I tried to retrace my search and
> couldn't find it I may have been mixing two posts in my head.
>
> Still, this is short and simple:
> <richya023060042802011315570001@nntp.ix.netcom.com>
That doesn't take me anywhere.
>
> 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.
>
> The fact it talks about nonalignment suggests someone must have
> thought there were significant and potentially troublesome components
> in the waveform up to 100s of MHz. I can believe that, since the small
> mismatches in resistive impedance become very large mismatches,
> effectively a short circuit, for very fast edges which encounter a
> capacitive tap. However, then it doesn't make sense to insist on using
> exact multiples as this will tend to create standing waves.
>
> But at least I know it's nothing to do with collision detection.
Only in that impedance discontinuities will create standing waves, which can
interfere with the signal. 
Archived from groups: comp.dcom.lans.ethernet (More info?)
Rich Seifert wrote:
(snip)
> As you realized, one bittime 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
>>>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
>>1980something 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 outofphase 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 rootmeansquared 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 randomlydistributed 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 worstcase, figuring that it would surely
> show up *somewhere*, and that one installer would have no idea what the
> problem was.
>
> By the way, that cablespacing 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) 3955700 Los Gatos, CA 95033
> (408) 2280803 FAX
>
> Send replies to: usenet at richseifert dot com 
Archived from groups: comp.dcom.lans.ethernet (More info?)
googlegroups@marget.com said
>> I wonder if anyone can give a definitive answer as to why there is a
>> minimum spacing specified on (some) ethernet cable. The thick stuff
>> with markers every 2.5m for example which is 1 bit of delay at 10 MHz.
>
>According to the Spurgeon book, the spacing is a guideline to help
>avoid signal reflections resulting from too many transceiver taps being
>clumped together.
>
>He goes on to say that maintaining an even 2.5m spacing isn't critical:
I can't see why it would matter even an itsywitsy little bit. But
that's where people seem to have different theories.
> when joining cable sections you can ignore the marks, and if two taps
>happen to land close together when cables are joined, that's okay too. 
Archived from groups: comp.dcom.lans.ethernet (More info?)
DHP wrote:
>>He goes on to say that maintaining an even 2.5m spacing isn't critical:
>
> I can't see why it would matter even an itsywitsy little bit. But
> that's where people seem to have different theories.
In theory, practice follows theory. In practice, it doesn't. ;) 
Archived from groups: comp.dcom.lans.ethernet (More info?)
Rich Seifert <usenet@richseifert.com.invalid> said
>In article <jlmai1pi08p6do3jo0p0veditvn4ra7v8m@4ax.com>,
> Henry <me@privacy.net> wrote:
>
>> William P. N. Smith <> said
>>
>> >Henry <me@privacy.net> wrote:
>> >>I wonder if anyone can give a definitive answer as to why there is a
>> >>minimum spacing specified on (some) ethernet cable. The thick stuff
>> >>with markers every 2.5m for example which is 1 bit of delay at 10 MHz.
>>
>> Oops, just realized there's a factor of 10 missing there. My attempted
>> guess at the reasoning was wrong... The mystery deepens.
>>
>
>As you realized, one bittime at 10 Mb/s is 100 ns, which corresponds to
>23.5 m of coaxial cable.
>
>> >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
>> 1980something 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 outofphase 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 rootmeansquared 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 randomlydistributed 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 worstcase, figuring that it would surely
>show up *somewhere*, and that one installer would have no idea what the
>problem was.
>
>By the way, that cablespacing 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.
Hi Rich, thanks for all that.
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 outofband noise.
Thanks for your time in answering this, it seems to crop up regularly
 though the google archive seems to peak in the early 90's
Oh yeah, my Masters is even older than yours! 
Archived from groups: comp.dcom.lans.ethernet (More info?)
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 outofband noise.
>
Actually, I did all of the analysis in the timedomain, rather than the
frequencydomain, although of course they are fully interchangeable.
I started where a communications systems designer SHOULD startwith a
requirement for a maximum biterror rate (which translates into a
frameloss rate). For the specified BER of 10^9 (worstcase), using
Manchester encoding, the minimum signaltonoise ratio turns out to be
14 db, which is a factor of 5:1. You then take the worstcase minimum
transmit level and attenuate it by the maximum amount possible
(worstcase cables, longest specified lengths) to calculate the minimum
received signal level. The allowable noise at that point must be no more
than onefifth of the minimum received signal to achieve the desired BER.
(I could recreate 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 nowobsolete 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 worstcase.
Sure, most environments were much more benign than we assumed for the
design criteria; those environments would experience a much better BER
than worstcase. But even the worst environment would behave acceptably.
When you are planning for millions of networks, and tensofmillions of
installed devices, even 99.9% assurance means a lot of angry customers.

Rich Seifert Networks and Communications Consulting
21885 Bear Creek Way
(408) 3955700 Los Gatos, CA 95033
(408) 2280803 FAX
Send replies to: usenet at richseifert dot com 

Archived from groups: comp.dcom.lans.ethernet (More info?)
James Knott <james.knott@rogers.com> said
>Henry wrote:
>
>> James Knott <james.knott@rogers.com> said
>>
>>>Henry wrote:
>>>
>>>> Thanks. I've found some stuff from Rich Seifert going back to
>>>> 1980something which explains it, sort of, though it's a bit woolly 
>>>> not Rich's explanation but the thinking behind it.
>>>
>>>Why not post the link here?
>>
>> I was going to do exactly that but I tried to retrace my search and
>> couldn't find it I may have been mixing two posts in my head.
>>
>> Still, this is short and simple:
>> <richya023060042802011315570001@nntp.ix.netcom.com>
>
>That doesn't take me anywhere.
Pop it into the box at the bottom of Google Groups Advanced Search 
minus the angle brackets.
>> 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.
>> The fact it talks about nonalignment suggests someone must have
>> thought there were significant and potentially troublesome components
>> in the waveform up to 100s of MHz. I can believe that, since the small
>> mismatches in resistive impedance become very large mismatches,
>> effectively a short circuit, for very fast edges which encounter a
>> capacitive tap. However, then it doesn't make sense to insist on using
>> exact multiples as this will tend to create standing waves.
>>
>> But at least I know it's nothing to do with collision detection.
>
>Only in that impedance discontinuities will create standing waves, which can
>interfere with the signal.
That may give you data corruption but it shouldn't trigger the
collision detector unless the level is ridiculous. 
Archived from groups: comp.dcom.lans.ethernet (More info?)
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? 
Archived from groups: comp.dcom.lans.ethernet (More info?)
DHP wrote:
(snip)
>>>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 
Archived from groups: comp.dcom.lans.ethernet (More info?)
In article <Za2dnZTNBNErT7jeRVnqQ@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.

Rich Seifert Networks and Communications Consulting
21885 Bear Creek Way
(408) 3955700 Los Gatos, CA 95033
(408) 2280803 FAX
Send replies to: usenet at richseifert dot com 
Archived from groups: comp.dcom.lans.ethernet (More info?)
James Knott <james.knott@rogers.com> said
>DHP wrote:
>
>>>He goes on to say that maintaining an even 2.5m spacing isn't critical:
>>
>> I can't see why it would matter even an itsywitsy little bit. But
>> that's where people seem to have different theories.
>
>In theory, practice follows theory. In practice, it doesn't. ;)
At the risk of getting controversial, where do standards fit in? 
Archived from groups: comp.dcom.lans.ethernet (More info?)
Rich Seifert <usenet@richseifert.com.invalid> said
>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 outofband noise.
>>
>
>Actually, I did all of the analysis in the timedomain, rather than the
>frequencydomain, although of course they are fully interchangeable.
>I started where a communications systems designer SHOULD startwith a
>requirement for a maximum biterror rate (which translates into a
>frameloss rate). For the specified BER of 10^9 (worstcase), using
>Manchester encoding, the minimum signaltonoise ratio turns out to be
>14 db, which is a factor of 5:1. You then take the worstcase minimum
>transmit level and attenuate it by the maximum amount possible
>(worstcase cables, longest specified lengths) to calculate the minimum
>received signal level. The allowable noise at that point must be no more
>than onefifth of the minimum received signal to achieve the desired BER.
>(I could recreate 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 nowobsolete 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 worstcase.
>Sure, most environments were much more benign than we assumed for the
>design criteria; those environments would experience a much better BER
>than worstcase. But even the worst environment would behave acceptably.
>When you are planning for millions of networks, and tensofmillions 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. 
Archived from groups: comp.dcom.lans.ethernet (More info?)
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
reflectionbased bit errors. 2M and 3M (for instance) are both worse
than 2.5M. 
Archived from groups: comp.dcom.lans.ethernet (More info?)
DHP wrote:
(snip regarding tap spacing on thick ethernet)
> 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 
Archived from groups: comp.dcom.lans.ethernet (More info?)
James Knott <james.knott@rogers.com> said
>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
similarlysized 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 quasirandom noise which occasionally causes a
data bit to be misread. 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? 
Archived from groups: comp.dcom.lans.ethernet (More info?)
James Knott <james.knott@rogers.com> said
>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

Archived from groups: comp.dcom.lans.ethernet (More info?)
Henry <me@privacy.net> wrote:
>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! 
Archived from groups: comp.dcom.lans.ethernet (More info?)
William P. N. Smith <> said
>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
>reflectionbased 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? 
Archived from groups: comp.dcom.lans.ethernet (More info?)
DHP <me@privacy.net> wrote:
>William P. N. Smith <> said
>>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
>>reflectionbased 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?
Well, Rich said:
/*
I did extensive simulations of the resulting reflections from
transceivers at various spacings, and empirically determined that 2.5
m was "good enough."
[...]
The idea is not just a *minimum* 2.5 m spacing; it is that
transceivers are only placed at the 2.5 m markings.
*/
I'd have to imagine that the 2.5m is somewhere between two distances
that will cause problems in the worst case, perhaps 2m and 3m. 
Archived from groups: comp.dcom.lans.ethernet (More info?)
glen herrmannsfeldt <gah@ugcs.caltech.edu> said
>Rich Seifert wrote:
>
>(snip)
>
>> As you realized, one bittime 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 narrowband signal, a sinewave  can be applied to
a Manchesterencoded bit stream.
In any case, the reflection from a capacitive tap is (approximately)
the timederivative of the origina. With a fast risetime 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?
>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
>>>1980something 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 outofphase 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 rootmeansquared 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 randomlydistributed 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 worstcase, figuring that it would surely
>> show up *somewhere*, and that one installer would have no idea what the
>> problem was.
>>
>> By the way, that cablespacing 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) 3955700 Los Gatos, CA 95033
>> (408) 2280803 FAX
>>
>> Send replies to: usenet at richseifert dot com 
Archived from groups: comp.dcom.lans.ethernet (More info?)
DHP wrote:
(snip)
(I wrote)
>>Constructive interference would result from a half wavelength spacing,
> I do not understand how the concept of constructive interference 
> which applies to a narrowband signal, a sinewave  can be applied to
> a Manchesterencoded 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.
> In any case, the reflection from a capacitive tap is (approximately)
> the timederivative of the origina. With a fast risetime 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 
Archived from groups: comp.dcom.lans.ethernet (More info?)
Rich Seifert <usenet@richseifert.com.invalid> said
>In article <Za2dnZTNBNErT7jeRVnqQ@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! 
Archived from groups: comp.dcom.lans.ethernet (More info?)
DHP wrote:
> 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!
A bit of Ohms law and Thevenin's equivalent, will show those to be measuring
the same thing. 
Archived from groups: comp.dcom.lans.ethernet (More info?)
William P. N. Smith wrote:
(snip on transceiver spacing calculation)
> I'd have to imagine that the 2.5m is somewhere between two distances
> that will cause problems in the worst case, perhaps 2m and 3m.
That could be, but there is also a desire to minimize the spacing
to make it easier for network engineers installing taps.
So, 3m is worse from that point of view, if not from the signal
point of view.
 glen 
Archived from groups: comp.dcom.lans.ethernet (More info?)
glen herrmannsfeldt <gah@ugcs.caltech.edu> said
>DHP wrote:
>
>(snip)
>(I wrote)
>
>>>Constructive interference would result from a half wavelength spacing,
>
>> I do not understand how the concept of constructive interference 
>> which applies to a narrowband signal, a sinewave  can be applied to
>> a Manchesterencoded 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.
With Manchester encoding, even a stream of zeros is actually a square
wave.
>> In any case, the reflection from a capacitive tap is (approximately)
>> the timederivative of the origina. With a fast risetime 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.
However, the reflections from a capacitive tap are not the same shape
as the original signal. I can't see how thay can be said to add
constructively or otherwise to the original signal. So we'd be looking
at two or more reflections a whole bit apart, which is 23.4m.
Which is avoided by using multiples of 2.5m, I suppose. Maybe my
arithmetic is not exact and the figures are tweaked so that the
nearest multiples are 1.25m on each side. Even so, 1.25m is only
5.3ns, so this can't be an issue unless the "dangerous" parts of the
reflection are very narrow.
Maybe if the signal had a rise time of <3ns it would be an issue.
You've got me thinking, now. Not always a good thing 
Archived from groups: comp.dcom.lans.ethernet (More info?)
DHP wrote:
> glen herrmannsfeldt <gah@ugcs.caltech.edu> said
>>DHP wrote:
(snip)
>>>I do not understand how the concept of constructive interference 
>>>which applies to a narrowband signal, a sinewave  can be applied to
>>>a Manchesterencoded 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.
> With Manchester encoding, even a stream of zeros is
> actually a square wave.
With the Fourier series containing odd harmonics
proportional to 1/n.
>>>In any case, the reflection from a capacitive tap is (approximately)
>>>the timederivative of the origina. With a fast risetime 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.
> However, the reflections from a capacitive tap are not the same shape
> as the original signal. I can't see how thay can be said to add
> constructively or otherwise to the original signal. So we'd be looking
> at two or more reflections a whole bit apart, which is 23.4m.
Say the cable has taps spaced 11.75m the whole length, and a signal
starts at one end transmitting all zeros. Some signal will
reflect off the first tap back to the transmitter. A similar
signal will reflect off the next tap and arrive at the transmitter
100ns later. Off the third tap will arrive 100ns later, etc.
Since the original is periodic with period 100ns all the reflections
will be of similar shape, though slightly decreasing amplitude.
Now, say you take a cable and mark off 11.75m regions, numbered from
the end. If you place a tap at all the 2.5m marks that are in odd
numbered regions they will tend to add more than subtract. That will
be about the limit of 100 taps on a 500m cable, maybe the worst case.
The third harmonic should be about 1/3 the amplitude, maybe less if
the square wave isn't perfect. Still worth worrying about but much
less likely to cause problems.
 glen
> Which is avoided by using multiples of 2.5m, I suppose. Maybe my
> arithmetic is not exact and the figures are tweaked so that the
> nearest multiples are 1.25m on each side. Even so, 1.25m is only
> 5.3ns, so this can't be an issue unless the "dangerous" parts of the
> reflection are very narrow.
>
> Maybe if the signal had a rise time of <3ns it would be an issue.
>
> You've got me thinking, now. Not always a good thing 
Archived from groups: comp.dcom.lans.ethernet (More info?)
DHP wrote:
(snip)
> Which is avoided by using multiples of 2.5m, I suppose. Maybe my
> arithmetic is not exact and the figures are tweaked so that the
> nearest multiples are 1.25m on each side. Even so, 1.25m is only
> 5.3ns, so this can't be an issue unless the "dangerous" parts of the
> reflection are very narrow.
Using some simple assumptions it does seem that 2.5m is much better.
First, I assume that for whatever spacing is used taps are only
placed where a period 11.75m cosine is positive.
I then compute the phase shift of the signal reflected off all
pairs of such taps and add them up. If you have awk or gawk
(there is a windows version of gawk around) you can run it and
see. The sum for 2.5m is about one fourth that for 2.4m or 2.6m.
There is also a minimum near 2.0m about twice that for 2.5m.
I didn't put any attenuation into the sum, though.
I don't know if this is at all related to what Rich did, but
it is nice to see 2.5m come out low.
# compare the effect of ethernet taps at different spacing
BEGIN {
# s=11.75;
for(s=1;s<30;s=s+0.1) {
n=int(500/s);
m=x=0;
delete z;
for(i=1;i<=n*4;i++) {
z=cos(i*s*2*3.14159/11.75);
}
for(i=1;i<=n;i++) {
if(z<0) continue;
for(j=1;j<i;j++) {
if(z[j]<0) continue;
x += z;
}
}
printf "%3d %5.2f %5.2f\n", m,s,x;
}
} 
Archived from groups: comp.dcom.lans.ethernet (More info?)
glen herrmannsfeldt wrote:
(snip)
> Say the cable has taps spaced 11.75m the whole length, and a signal
> starts at one end transmitting all zeros. Some signal will
> reflect off the first tap back to the transmitter. A similar
> signal will reflect off the next tap and arrive at the transmitter
> 100ns later. Off the third tap will arrive 100ns later, etc.
> Since the original is periodic with period 100ns all the reflections
> will be of similar shape, though slightly decreasing amplitude.
After doing a few calculations I realized that this determines
the back reflection which may be different than the accumulated
forward reflections. That is, ones that reflect an even number
of times. Still, it should be that 11.75m is bad.
> Now, say you take a cable and mark off 11.75m regions, numbered from
> the end. If you place a tap at all the 2.5m marks that are in odd
> numbered regions they will tend to add more than subtract. That will
> be about the limit of 100 taps on a 500m cable, maybe the worst case.
I will try a few more calculations assuming that taps are only
placed where they add constructively to the first harmonic within
the specified spacing.
> The third harmonic should be about 1/3 the amplitude, maybe less if
> the square wave isn't perfect. Still worth worrying about but much
> less likely to cause problems.
 glen 
Archived from groups: comp.dcom.lans.ethernet (More info?)
glen herrmannsfeldt <gah@ugcs.caltech.edu> said
>DHP wrote:
>
>> glen herrmannsfeldt <gah@ugcs.caltech.edu> said
>>>DHP wrote:
>
>(snip)
>
>>>>I do not understand how the concept of constructive interference 
>>>>which applies to a narrowband signal, a sinewave  can be applied to
>>>>a Manchesterencoded 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.
>
>> With Manchester encoding, even a stream of zeros is
> > actually a square wave.
>
>With the Fourier series containing odd harmonics
>proportional to 1/n.
See below
>>>>In any case, the reflection from a capacitive tap is (approximately)
>>>>the timederivative of the origina. With a fast risetime 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.
>
>> However, the reflections from a capacitive tap are not the same shape
>> as the original signal. I can't see how thay can be said to add
>> constructively or otherwise to the original signal. So we'd be looking
>> at two or more reflections a whole bit apart, which is 23.4m.
>
>Say the cable has taps spaced 11.75m the whole length, and a signal
>starts at one end transmitting all zeros. Some signal will
>reflect off the first tap back to the transmitter. A similar
>signal will reflect off the next tap and arrive at the transmitter
>100ns later. Off the third tap will arrive 100ns later, etc.
>Since the original is periodic with period 100ns all the reflections
>will be of similar shape, though slightly decreasing amplitude.
Oh yea, sorry, I was forgetting the twoway trip... or rather I'd
thought about it and got it wrong Too early in the morning.
>Now, say you take a cable and mark off 11.75m regions, numbered from
>the end. If you place a tap at all the 2.5m marks that are in odd
>numbered regions they will tend to add more than subtract. That will
>be about the limit of 100 taps on a 500m cable, maybe the worst case.
>The third harmonic should be about 1/3 the amplitude, maybe less if
>the square wave isn't perfect. Still worth worrying about but much
>less likely to cause problems.
30MHz has a wavelength of 7.8m doesn't it? Placing taps with an
accuracy of much better than 1m doesn't make much sense unless you're
talking about much higher frequency components to make things happen
over << 1m.
With capacitive taps the 1/f signal spectrum gives a flat one in the
reflections  the spectrum of a series of spikes. So Nature strikes
back.
> glen
>
>
>
>> Which is avoided by using multiples of 2.5m, I suppose. Maybe my
>> arithmetic is not exact and the figures are tweaked so that the
>> nearest multiples are 1.25m on each side. Even so, 1.25m is only
>> 5.3ns, so this can't be an issue unless the "dangerous" parts of the
>> reflection are very narrow.
>>
>> Maybe if the signal had a rise time of <3ns it would be an issue.
>>
>> You've got me thinking, now. Not always a good thing 
Archived from groups: comp.dcom.lans.ethernet (More info?)
glen herrmannsfeldt <gah@ugcs.caltech.edu> said
....
>After doing a few calculations I realized that this determines
>the back reflection which may be different than the accumulated
>forward reflections. That is, ones that reflect an even number
>of times. Still, it should be that 11.75m is bad.
>
>> Now, say you take a cable and mark off 11.75m regions, numbered from
>> the end. If you place a tap at all the 2.5m marks that are in odd
>> numbered regions they will tend to add more than subtract. That will
>> be about the limit of 100 taps on a 500m cable, maybe the worst case.
>
>I will try a few more calculations assuming that taps are only
>placed where they add constructively to the first harmonic within
>the specified spacing.
If you're talking about frequencies ~10MHz, moving a tap by 1.25m is
only going to change the phaseof one reflection by 20 degrees  (and
you can't even move them all in the same direction or you're just
moving the whole cluster!)
If you consider a wide bandwidth system though, the reflection is
actually a forest of spikes. So you can interleave them for
(presumably) minimum impact or superimpose them for worst case.
It seems to me that there's no advantage in moving taps around within
the cluster, but it would be useful to ensure that "forests" arriving
from different clusters always interleave. Insisting on regular
tapping points will acheive this as it creates a predicable set of
time slots for reflections even if not all of them are filled. The
next thing is to ensure the timeslot patterns interleave  by setting
the spacing appropriately. I can see it works, sort of, over a few 10s
of m but whether it can be maintained over a 500m LAN I don't know. 
Archived from groups: comp.dcom.lans.ethernet (More info?)
glen herrmannsfeldt <gah@ugcs.caltech.edu> said
....
>Using some simple assumptions it does seem that 2.5m is much better.
>First, I assume that for whatever spacing is used taps are only
>placed where a period 11.75m cosine is positive.
>I then compute the phase shift of the signal reflected off all
>pairs of such taps and add them up. If you have awk or gawk
>(there is a windows version of gawk around) you can run it and
>see. The sum for 2.5m is about one fourth that for 2.4m or 2.6m.
>There is also a minimum near 2.0m about twice that for 2.5m.
>I didn't put any attenuation into the sum, though.
>
>I don't know if this is at all related to what Rich did, but
>it is nice to see 2.5m come out low.
Ethernet signals have a wide spectrum. What happens with a different
wavelength with the taps in the same place?
># compare the effect of ethernet taps at different spacing
>BEGIN {
># s=11.75;
>for(s=1;s<30;s=s+0.1) {
> n=int(500/s);
> m=x=0;
> delete z;
> for(i=1;i<=n*4;i++) {
> z=cos(i*s*2*3.14159/11.75);
> }
> for(i=1;i<=n;i++) {
> if(z<0) continue;
> for(j=1;j<i;j++) {
> if(z[j]<0) continue;
> x += z;
> }
> }
> printf "%3d %5.2f %5.2f\n", m,s,x;
> }
>}
> 
Archived from groups: comp.dcom.lans.ethernet (More info?)
DHP wrote:
> glen herrmannsfeldt said
>>Using some simple assumptions it does seem that 2.5m is much better.
>>First, I assume that for whatever spacing is used taps are only
>>placed where a period 11.75m cosine is positive.
>>I then compute the phase shift of the signal reflected off all
>>pairs of such taps and add them up. If you have awk or gawk
>>(there is a windows version of gawk around) you can run it and
>>see. The sum for 2.5m is about one fourth that for 2.4m or 2.6m.
>>There is also a minimum near 2.0m about twice that for 2.5m.
>>I didn't put any attenuation into the sum, though.
OK, in words there are n possible taps. For taps i and j, with i>j, and
where an 11.75m period cosine is positive, sum over the phase shift from
the beginning to tap i, back to tap j, and then to the end.
So it is i (to tap i), ij (back to tap j) and nj (to the end).
>>I don't know if this is at all related to what Rich did, but
>>it is nice to see 2.5m come out low.
I did some more tests and it might be that the 2.5m dip is a
rounding artifact. It is there with 199 possible taps over
a 500m cable, but gone with 200 possible taps. The 2m dip
seems to stay, though.
> Ethernet signals have a wide spectrum. What happens with a different
> wavelength with the taps in the same place?
Yes, so far I am only considering the fundamental. The third harmonic
is probably also important, but I have to start somewhere.
 glen 
Archived from groups: comp.dcom.lans.ethernet (More info?)
In article <icmdnYytQqXorbveRVn2w@rogers.com>,
James Knott <james.knott@rogers.com> wrote:
> DHP wrote:
>
> > 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!
>
> A bit of Ohms law and Thevenin's equivalent, will show those to be measuring
> the same thing.
They are not at all the same thing. The voltage level on the coaxial
cable (created by the combined currentsourcing of all active
transmitters) is quite independent of the current drain from the
transceiver power supply, or even the current draw of that particular
transceiver's output driver.

Rich Seifert Networks and Communications Consulting
21885 Bear Creek Way
(408) 3955700 Los Gatos, CA 95033
(408) 2280803 FAX
Send replies to: usenet at richseifert dot com 
Archived from groups: comp.dcom.lans.ethernet (More info?)
In article <7t6dnYVUvoh3HbvenZ2dnUVZ_tGdnZ2d@comcast.com>,
glen herrmannsfeldt <gah@ugcs.caltech.edu> wrote:
> William P. N. Smith wrote:
>
> (snip on transceiver spacing calculation)
>
> > I'd have to imagine that the 2.5m is somewhere between two distances
> > that will cause problems in the worst case, perhaps 2m and 3m.
>
> That could be, but there is also a desire to minimize the spacing
> to make it easier for network engineers installing taps.
That was the ultimate deciding factor. Larger spacings were generally
better, but you don't want to have to unnecessarily coil up lots of
cable in the ceiling between taps.
There was nothing "magical" about the 2.5 m result; it's not like 2 m or
3 m were particularly bad.

Rich Seifert Networks and Communications Consulting
21885 Bear Creek Way
(408) 3955700 Los Gatos, CA 95033
(408) 2280803 FAX
Send replies to: usenet at richseifert dot com 
Archived from groups: comp.dcom.lans.ethernet (More info?)
In article <cvtci1pats9jpsfk29l92r1ilm48qh2cbl@4ax.com>,
DHP <me@privacy.net> wrote:
> With Manchester encoding, even a stream of zeros is actually a square
> wave.
>
It *could* be a square wave, but we intentionally slewrate limited the
signal impressed on the coaxial cable; it has a 25 ns nominal rise/fall
time. This both reduces the effect of tap reflections and reduces the
EMI generated by the signal.
Let's see if I remember the numbers correctly:
The nominal voltage resulting from a single transmitter on the coaxial
cable is ~2V pp. The current in the capacitive tap (which gets
reflected into the coaxial cable) can be calculated as:
I = C dv/dt
dv/dt is 2V / 25 ns, or 80 MV/s (that's MegaVolts per second)
C is 4 pf worstcase, so the current is 4 pf * 80 MV/s = 320 uA.
The impedance seen by the capacitor is 25 ohms (it effectively sees two
50 ohm cables in parallel, one in each direction away from the tap).
Thus the "noise" voltage generated by a single tap is 25 * 320 uA = 8 mV
8 milliVolts by itself would not be a problem, however, if we had the
worstcase situation of all 100 transceivers lumped together, we would
have 800 mV of signal, which would blow away our required 5:1 signal to
noise ratio. So the idea is to make sure that as few of these 8 mV
spikes (they only last for 25 nS, while the voltage on the cable is in
transition) add up in phase. Also, by creating a minimum cable length of
250 m for those 100 taps (i.e., by spacing them by at least 2.5 m), we
are sure to get a fair amount of attenuation, at least for the taps that
are farthest away.
The simulations simply tried a zillion variations on numbers of
transceivers, placement along the cable, spacing requirements, and data
patterns to determine if there were any pathological situations where
the noise exceeded the budget allowance. By the way, this took a HUGE
amount of computer power, at least by the standards of the time. I
managed to distribute the simulation runs across dozens of VAXen (780s,
the only ones in existence at the time) all around the world, using
DEC's private network. I used the idle compute power of just about every
machine in those time zones where the normal work day was over. It was a
rather ambitious task for its day.

Rich Seifert Networks and Communications Consulting
21885 Bear Creek Way
(408) 3955700 Los Gatos, CA 95033
(408) 2280803 FAX
Send replies to: usenet at richseifert dot com 
Archived from groups: comp.dcom.lans.ethernet (More info?)
Rich Seifert wrote:
(snip regarding transceiver power)
> They are not at all the same thing. The voltage level on the coaxial
> cable (created by the combined currentsourcing of all active
> transmitters) is quite independent of the current drain from the
> transceiver power supply, or even the current draw of that particular
> transceiver's output driver.
I used to have a machine with automatic switching between
AUI and BNC outputs. It required a minimum power draw on
the AUI connector to switch, though I believe a jumper was
available in case it didn't.
 glen 
Archived from groups: comp.dcom.lans.ethernet (More info?)
glen herrmannsfeldt <gah@ugcs.caltech.edu> said
>DHP wrote:
>
>> glen herrmannsfeldt said
>
>>>Using some simple assumptions it does seem that 2.5m is much better.
>>>First, I assume that for whatever spacing is used taps are only
>>>placed where a period 11.75m cosine is positive.
>
>>>I then compute the phase shift of the signal reflected off all
>>>pairs of such taps and add them up. If you have awk or gawk
>>>(there is a windows version of gawk around) you can run it and
>>>see. The sum for 2.5m is about one fourth that for 2.4m or 2.6m.
>>>There is also a minimum near 2.0m about twice that for 2.5m.
>>>I didn't put any attenuation into the sum, though.
>
>OK, in words there are n possible taps. For taps i and j, with i>j, and
>where an 11.75m period cosine is positive, sum over the phase shift from
>the beginning to tap i, back to tap j, and then to the end.
>So it is i (to tap i), ij (back to tap j) and nj (to the end).
Sorry, I must be missing something. From the sound of it, you're
looking at a double reflection  a signal travelling the whole length
of the cable and arriving with noise coming in the same direction. The
model Rich is talking about is where tap i sends to tap j but tap k a
bit further down the line provides an unwanted echo. It's travelling
in the opposite direction of course, but k doesn't know that, k just
sees it as noise.
I'm also not sure about your summation. Adding cosines of phase just
gives you the componet which is in phase with the original signal. You
should also do a summation of sines to get the quadrature term. The
poor old receiver doesn't know which is which, it just sees the total
amplitude which, of course is (c^2+s^2)^.5.
>>>I don't know if this is at all related to what Rich did, but
>>>it is nice to see 2.5m come out low.
>
>I did some more tests and it might be that the 2.5m dip is a
>rounding artifact. It is there with 199 possible taps over
>a 500m cable, but gone with 200 possible taps. The 2m dip
>seems to stay, though.
>
>> Ethernet signals have a wide spectrum. What happens with a different
>> wavelength with the taps in the same place?
>
>Yes, so far I am only considering the fundamental. The third harmonic
>is probably also important, but I have to start somewhere.
I'm talking about sidebands, not harmonics. These will be frequencies
close to the 10MHz carrier. Well, close in the sense of being a broad
spectrum covering at least 5 MHz and almost certainly a lot more.
So you need two cosine functions, one to find the quadrants for the
10MHz cosine to get the clusters in the right place, and another for
the actual phase. Well, lots of them actually, an integral would do
nicely, but I'll settle for spot frequencies 
Archived from groups: comp.dcom.lans.ethernet (More info?)
DHP wrote:
> glen herrmannsfeldt <gah@ugcs.caltech.edu> said
(snip)
> Sorry, I must be missing something. From the sound of it, you're
> looking at a double reflection  a signal travelling the whole length
> of the cable and arriving with noise coming in the same direction. The
> model Rich is talking about is where tap i sends to tap j but tap k a
> bit further down the line provides an unwanted echo. It's travelling
> in the opposite direction of course, but k doesn't know that, k just
> sees it as noise.
The one I thought of first was for a near end tap with the reflections
off all the rest of the taps. A far tap, though, won't see single
reflections off many taps. In my case there are 4851 or so pairs of
taps contributing.
> I'm also not sure about your summation. Adding cosines of phase just
> gives you the componet which is in phase with the original signal. You
> should also do a summation of sines to get the quadrature term. The
> poor old receiver doesn't know which is which, it just sees the total
> amplitude which, of course is (c^2+s^2)^.5.
I thought I was doing pretty well to get as far as I did with
only a small amount of work, but yes, it needs that, too.
(snip)
>>>Ethernet signals have a wide spectrum. What happens with a different
>>>wavelength with the taps in the same place?
>>Yes, so far I am only considering the fundamental. The third harmonic
>>is probably also important, but I have to start somewhere.
> I'm talking about sidebands, not harmonics. These will be frequencies
> close to the 10MHz carrier. Well, close in the sense of being a broad
> spectrum covering at least 5 MHz and almost certainly a lot more.
I believe those will fall off pretty fast. There will be a 5MHz
fundamental for alternating 1 and 0, especially from the preamble,
with the same amplitude as the 10MHz from all 0's or all 1's.
Three bit repeats should have 3.33MHz and 6.66MHz at maybe one third
or so the amplitude. The ones near 10MHz need longer bit patterns,
and will have much smaller amplitudes.
> So you need two cosine functions, one to find the quadrants for the
> 10MHz cosine to get the clusters in the right place, and another for
> the actual phase. Well, lots of them actually, an integral would do
> nicely, but I'll settle for spot frequencies
Maybe Rich will say more, but for now I believe that the 5MHz and
10MHz will be much higher amplitude. If we have the tap pattern
that is worst case for that, I don't believe that any other will
be any worse. I just had the thought of someone running a cable
down a hall of rooms spaced 11.75m apart with one or two taps
to each room.
 glen 
Archived from groups: comp.dcom.lans.ethernet (More info?)
In article <yYGdndBliaPjhLreRVnoQ@comcast.com>,
glen herrmannsfeldt <gah@ugcs.caltech.edu> wrote:
> DHP wrote:
>
> > glen herrmannsfeldt <gah@ugcs.caltech.edu> said
>
>
> > I'm also not sure about your summation. Adding cosines of phase just
> > gives you the componet which is in phase with the original signal. You
> > should also do a summation of sines to get the quadrature term. The
> > poor old receiver doesn't know which is which, it just sees the total
> > amplitude which, of course is (c^2+s^2)^.5.
>
>
> >>>Ethernet signals have a wide spectrum. What happens with a different
> >>>wavelength with the taps in the same place?
>
> >>Yes, so far I am only considering the fundamental. The third harmonic
> >>is probably also important, but I have to start somewhere.
>
> > I'm talking about sidebands, not harmonics. These will be frequencies
> > close to the 10MHz carrier. Well, close in the sense of being a broad
> > spectrum covering at least 5 MHz and almost certainly a lot more.
>
> I believe those will fall off pretty fast. There will be a 5MHz
> fundamental for alternating 1 and 0, especially from the preamble,
> with the same amplitude as the 10MHz from all 0's or all 1's.
> Three bit repeats should have 3.33MHz and 6.66MHz at maybe one third
> or so the amplitude. The ones near 10MHz need longer bit patterns,
> and will have much smaller amplitudes.
>
> > So you need two cosine functions, one to find the quadrants for the
> > 10MHz cosine to get the clusters in the right place, and another for
> > the actual phase. Well, lots of them actually, an integral would do
> > nicely, but I'll settle for spot frequencies
>
Now you see why it is easier to work on this problem in the time domain,
rather than the frequency domain! All you need to worry about is signal
amplitude, noise margin, and slew rates.

Rich Seifert Networks and Communications Consulting
21885 Bear Creek Way
(408) 3955700 Los Gatos, CA 95033
(408) 2280803 FAX
Send replies to: usenet at richseifert dot com 
Archived from groups: comp.dcom.lans.ethernet (More info?)
DHP <me@privacy.net> said
> It's travelling
>in the opposite direction of course, but k doesn't know that, k just
>sees it as noise.
I meant j of course. 
Archived from groups: comp.dcom.lans.ethernet (More info?)
glen herrmannsfeldt <gah@ugcs.caltech.edu> said
>DHP wrote:
>
>> glen herrmannsfeldt <gah@ugcs.caltech.edu> said
>
>(snip)
>
>> Sorry, I must be missing something. From the sound of it, you're
>> looking at a double reflection  a signal travelling the whole length
>> of the cable and arriving with noise coming in the same direction. The
>> model Rich is talking about is where tap i sends to tap j but tap k a
>> bit further down the line provides an unwanted echo. It's travelling
>> in the opposite direction of course, but k doesn't know that, k just
>> sees it as noise.
>
>The one I thought of first was for a near end tap with the reflections
>off all the rest of the taps. A far tap, though, won't see single
>reflections off many taps. In my case there are 4851 or so pairs of
>taps contributing.
Yes, but we only need to consider systems that work. If the near tap
sees better than 14dB s/n on single reflections, the far one is going
to see better than 28dB on double ones so we can (probably!) ignore
it.
>> I'm also not sure about your summation. Adding cosines of phase just
>> gives you the componet which is in phase with the original signal. You
>> should also do a summation of sines to get the quadrature term. The
>> poor old receiver doesn't know which is which, it just sees the total
>> amplitude which, of course is (c^2+s^2)^.5.
>
>I thought I was doing pretty well to get as far as I did with
>only a small amount of work, but yes, it needs that, too.
Absolutely essential Otherwise a cosine multiplier is going to
sweep over your result as you change s, resulting in spurious nulls.
>(snip)
>
>>>>Ethernet signals have a wide spectrum. What happens with a different
>>>>wavelength with the taps in the same place?
>
>>>Yes, so far I am only considering the fundamental. The third harmonic
>>>is probably also important, but I have to start somewhere.
>
>> I'm talking about sidebands, not harmonics. These will be frequencies
>> close to the 10MHz carrier. Well, close in the sense of being a broad
>> spectrum covering at least 5 MHz and almost certainly a lot more.
>
>I believe those will fall off pretty fast. There will be a 5MHz
>fundamental for alternating 1 and 0, especially from the preamble,
>with the same amplitude as the 10MHz from all 0's or all 1's.
>Three bit repeats should have 3.33MHz and 6.66MHz at maybe one third
>or so the amplitude. The ones near 10MHz need longer bit patterns,
>and will have much smaller amplitudes.
Actually the fundamental at 10MHz is totally suppressed and *all* the
energy is put into the sidebands. 10Mbps of random data needs a
minimum bandwidth of 5MHz (in both sidebands). A slowly changing
sequence just brings the sidebands in, it doesn't reduce their
amplitude.
>> So you need two cosine functions, one to find the quadrants for the
>> 10MHz cosine to get the clusters in the right place, and another for
>> the actual phase. Well, lots of them actually, an integral would do
>> nicely, but I'll settle for spot frequencies
>
>Maybe Rich will say more, but for now I believe that the 5MHz and
>10MHz will be much higher amplitude. If we have the tap pattern
>that is worst case for that, I don't believe that any other will
>be any worse. I just had the thought of someone running a cable
>down a hall of rooms spaced 11.75m apart with one or two taps
>to each room.
That's the one to worry about
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