Why is there a minimum spacing?

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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!

 

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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?

 

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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.

 

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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

 

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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

 

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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!

 

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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.

 

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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

 

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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

 

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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

 

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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;
}
}

 

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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

 

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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

 

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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.

 

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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;
> }
>}
>

 

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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

 

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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

 

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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

 

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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

 

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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

 

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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

 

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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

 

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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

 

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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.

 

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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 :/