# Why is there a minimum spacing?

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Last response: in Networking

henry

September 12, 2005 2:27:18 PM

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 time-difference 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 time-difference 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.

More about : minimum spacing

Anonymous

September 12, 2005 2:27:19 PM

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.

Anonymous

September 12, 2005 2:27:19 PM

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.

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Anonymous

September 12, 2005 2:27:19 PM

> 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

henry

September 12, 2005 4:33:54 PM

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

1980-something which explains it, sort of, though it's a bit woolly -

not Rich's explanation but the thinking behind it.

Anonymous

September 12, 2005 4:33:55 PM

Henry wrote:

> 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.

Why not post the link here?

Anonymous

September 12, 2005 4:33:55 PM

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 bit-time 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

> 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

henry

September 12, 2005 5:54:54 PM

James Knott <james.knott@rogers.com> said

>Henry wrote:

>

>> 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.

>

>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:

<rich-ya023060042802011315570001@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 non-alignment 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.

Anonymous

September 12, 2005 5:54:55 PM

Henry wrote:

> James Knott <james.knott@rogers.com> said

>

>>Henry wrote:

>>

>>> 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.

>>

>>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:

> <rich-ya023060042802011315570001@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 non-alignment 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.

Anonymous

September 12, 2005 7:25:15 PM

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,

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

>>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

Anonymous

September 12, 2005 7:29:53 PM

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 itsy-witsy 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.

Anonymous

September 12, 2005 7:29:54 PM

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 itsy-witsy little bit. But

> that's where people seem to have different theories.

In theory, practice follows theory. In practice, it doesn't. ;-)

Anonymous

September 12, 2005 9:11:10 PM

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 bit-time 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

>> 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.

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 out-of-band 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!

Anonymous

September 12, 2005 9:11:11 PM

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.

--

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

Anonymous

September 12, 2005 9:39:25 PM

Anonymous

September 12, 2005 10:50:09 PM

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

>>>> 1980-something 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:

>> <rich-ya023060042802011315570001@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 non-alignment 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.

Anonymous

September 12, 2005 10:50:10 PM

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?

Anonymous

September 12, 2005 10:50:10 PM

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

Anonymous

September 12, 2005 10:50:11 PM

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.

--

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

Anonymous

September 12, 2005 10:52:34 PM

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 itsy-witsy 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?

Anonymous

September 13, 2005 12:03:37 AM

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 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.

Anonymous

September 13, 2005 12:03:38 AM

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.

Anonymous

September 13, 2005 12:03:38 AM

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

Anonymous

September 13, 2005 12:55:43 AM

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

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?

henry

September 13, 2005 12:59:51 AM

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

Anonymous

September 13, 2005 12:59:52 AM

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!

Anonymous

September 13, 2005 3:46:33 AM

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

>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?

Anonymous

September 13, 2005 3:46:34 AM

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

>>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?

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.

Anonymous

September 13, 2005 3:57:35 AM

glen herrmannsfeldt <gah@ugcs.caltech.edu> said

>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?

>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

Anonymous

September 13, 2005 3:57:36 AM

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 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.

> 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

Anonymous

September 13, 2005 4:05:48 AM

Rich Seifert <usenet@richseifert.com.invalid> said

>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!

Anonymous

September 13, 2005 4:05:49 AM

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.

Anonymous

September 13, 2005 4:34:40 AM

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

Anonymous

September 13, 2005 12:22:37 PM

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 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.

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 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.

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

Anonymous

September 13, 2005 12:22:38 PM

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 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.

> 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 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.

> 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

Anonymous

September 13, 2005 12:22:38 PM

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[i+(i-j)+n-j];

}

}

printf "%3d %5.2f %5.2f\n", m,s,x;

}

}

Anonymous

September 13, 2005 12:22:39 PM

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

Anonymous

September 13, 2005 1:15:33 PM

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 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.

>

>> 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 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.

>

>> 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 two-way 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

Anonymous

September 13, 2005 1:39:51 PM

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 time-slot 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.

Anonymous

September 13, 2005 2:24:10 PM

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[i+(i-j)+n-j];

> }

> }

> printf "%3d %5.2f %5.2f\n", m,s,x;

> }

>}

>

Anonymous

September 13, 2005 2:24:11 PM

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), i-j (back to tap j) and n-j (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

Anonymous

September 13, 2005 7:40:18 PM

In article <icmdnYytQqXorbveRVn-2w@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 current-sourcing 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) 395-5700 Los Gatos, CA 95033

(408) 228-0803 FAX

Send replies to: usenet at richseifert dot com

Anonymous

September 13, 2005 7:42:23 PM

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) 395-5700 Los Gatos, CA 95033

(408) 228-0803 FAX

Send replies to: usenet at richseifert dot com

Anonymous

September 13, 2005 8:02:06 PM

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 slew-rate 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 p-p. 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 worst-case, 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

worst-case 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) 395-5700 Los Gatos, CA 95033

(408) 228-0803 FAX

Send replies to: usenet at richseifert dot com

Anonymous

September 13, 2005 8:26:05 PM

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 current-sourcing 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

Anonymous

September 13, 2005 10:26:20 PM

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), i-j (back to tap j) and n-j (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

Anonymous

September 13, 2005 10:26:21 PM

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

Anonymous

September 13, 2005 10:26:22 PM

In article <yYGdndBliaPjhLreRVn-oQ@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) 395-5700 Los Gatos, CA 95033

(408) 228-0803 FAX

Send replies to: usenet at richseifert dot com

Anonymous

September 13, 2005 10:56:43 PM

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.

Anonymous

September 14, 2005 12:28:03 AM

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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