| Author |
Message |
DHP
Guest
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 Posted:
Wed Sep 14, 2005 12:28 am Post subject:
Re: Why is there a minimum spacing? |
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glen herrmannsfeldt <gah@ugcs.caltech.edu> said
| Quote: | 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.
| Quote: | 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.
| Quote: | (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.
| Quote: | 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.
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That's the one to worry about :/ |
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glen herrmannsfeldt
Guest
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| Posted:
Wed Sep 14, 2005 1:51 am Post subject:
Re: Why is there a minimum spacing? |
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DHP wrote:
(snip)
| Quote: | 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.
|
Many systems, such as most modems, use a scrambler to randomize
the bit stream. Otherwise the data may not be all that random.
For random data, yes, the 10MHz is zero, but consider ftping
a file full of zeros. Maybe the background for an uncompressed
image file, for example.
-- glen |
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DHP
Guest
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| Posted:
Wed Sep 14, 2005 2:23 am Post subject:
Re: Why is there a minimum spacing? |
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glen herrmannsfeldt <gah@ugcs.caltech.edu> said
| Quote: | DHP wrote:
(snip)
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.
Many systems, such as most modems, use a scrambler to randomize
the bit stream. Otherwise the data may not be all that random.
For random data, yes, the 10MHz is zero, but consider ftping
a file full of zeros. Maybe the background for an uncompressed
image file, for example.
|
Not many networks are dedicated exclusively to large pure black
bitmaps. |
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glen herrmannsfeldt
Guest
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| Posted:
Wed Sep 14, 2005 3:19 am Post subject:
Re: Why is there a minimum spacing? |
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DHP wrote:
(snip, I wrote)
| Quote: | For random data, yes, the 10MHz is zero, but consider ftping
a file full of zeros. Maybe the background for an uncompressed
image file, for example.
Not many networks are dedicated exclusively to large pure black
bitmaps.
|
Round trip on the cable is about 42 bits, so 42 zeros in a row
is enough to do it.
-- glen |
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Walter Roberson
Guest
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| Posted:
Wed Sep 14, 2005 3:24 am Post subject:
Re: Why is there a minimum spacing? |
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In article <usenet-7C29BA.16020613092005@news.isp.giganews.com>,
Rich Seifert <usenet@richseifert.com.invalid> wrote:
: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.
The 750 was officially introduced in October 1980, 25 years ago next
month. In an earlier posting, you indicated your thesis was
"some 25 years ago". Coincidence??
--
"I will speculate that [...] applications [...] could actually see a
performance boost for most users by going dual-core [...] because it
is running the adware and spyware that [...] are otherwise slowing
down the single CPU that user has today" -- Herb Sutter |
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Rich Seifert
Guest
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| Posted:
Wed Sep 14, 2005 3:40 am Post subject:
Re: Why is there a minimum spacing? |
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In article <icmdnYytQqXorbveRVn-2w@rogers.com>,
James Knott <james.knott@rogers.com> wrote:
| Quote: | 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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Rich Seifert
Guest
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| Posted:
Wed Sep 14, 2005 3:42 am Post subject:
Re: Why is there a minimum spacing? |
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In article <7t6dnYVUvoh3HbvenZ2dnUVZ_tGdnZ2d@comcast.com>,
glen herrmannsfeldt <gah@ugcs.caltech.edu> wrote:
| Quote: | 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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Rich Seifert
Guest
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| Posted:
Wed Sep 14, 2005 4:02 am Post subject:
Re: Why is there a minimum spacing? |
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In article <cvtci1pats9jpsfk29l92r1ilm48qh2cbl@4ax.com>,
DHP <me@privacy.net> wrote:
| Quote: | 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
Guest
|
| Posted:
Wed Sep 14, 2005 4:07 am Post subject:
Re: Why is there a minimum spacing? |
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In article <yYGdndBliaPjhLreRVn-oQ@comcast.com>,
glen herrmannsfeldt <gah@ugcs.caltech.edu> wrote:
| Quote: | 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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glen herrmannsfeldt
Guest
|
| Posted:
Wed Sep 14, 2005 4:26 am Post subject:
Re: Why is there a minimum spacing? |
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|
Rich Seifert wrote:
(snip regarding transceiver power)
| Quote: | 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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DHP
Guest
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| Posted:
Wed Sep 14, 2005 4:52 am Post subject:
Re: Why is there a minimum spacing? |
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Rich Seifert <usenet@richseifert.com.invalid> said
| Quote: | 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.
|
Or, to be perverse, it limits the bandwidth :)
I agree about it being simpler in the time domain though, because we
are really talking about narrow pulses - the derivative of the edges.
| Quote: | 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.
|
Interesting, because with 2.5m taps, a 25ns pulse produces
(practically) non-overlapping reflections. So if you're trying to
break the system you don't gain anything by clustering them. That
means the amplitude is that of a single reflection per half-bit of
line, 11.7m. So the worst possible case is about 42 reflections or
328mV. What's that, about -10dB in a 1V system? But as the noise isn't
random it can *never* go any higher so, ignoring other noise, it
should be possible to set threshol voltages that avoid errors
altogether. But only just.
| Quote: | 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. |
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DHP
Guest
|
| Posted:
Wed Sep 14, 2005 5:06 am Post subject:
Re: Why is there a minimum spacing? |
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glen herrmannsfeldt <gah@ugcs.caltech.edu> said
| Quote: | DHP wrote:
(snip, I wrote)
For random data, yes, the 10MHz is zero, but consider ftping
a file full of zeros. Maybe the background for an uncompressed
image file, for example.
Not many networks are dedicated exclusively to large pure black
bitmaps.
Round trip on the cable is about 42 bits, so 42 zeros in a row
is enough to do it.
|
Well, put it this way then, not many networks are dedicated to pure
never-ending jam. |
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Dougie!
Guest
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| Posted:
Wed Sep 14, 2005 8:20 am Post subject:
Re: Why is there a minimum spacing? |
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|
the 500 meters part of 10BASE5 actually relates to the Slot time of
Ethernet... Or the maximum time taken to realize a cdollision has
occurred... 2165 nanseconds.
With regards to the spacing, when you look at the impedance variation
of a vampire tap and associated capacitive coupling of the MAU, you get
some level of reflection. If these elements are introduced and managed
at the odd integrals of the quarter wavelength, they cancel each other
out. Just like to 17.5, 70.7, and 177 M lengths specified for
mismatches of cable segments when mixing cables from different
maufacturer and lot numbers.
Dougie! |
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DHP
Guest
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| Posted:
Wed Sep 14, 2005 1:44 pm Post subject:
Re: Why is there a minimum spacing? |
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"Dougie!" <dougie.stevenson@gmail.com> said
| Quote: | the 500 meters part of 10BASE5 actually relates to the Slot time of
Ethernet... Or the maximum time taken to realize a cdollision has
occurred... 2165 nanseconds.
With regards to the spacing, when you look at the impedance variation
of a vampire tap and associated capacitive coupling of the MAU, you get
some level of reflection. If these elements are introduced and managed
at the odd integrals of the quarter wavelength, they cancel each other
out. Just like to 17.5, 70.7, and 177 M lengths specified for
mismatches of cable segments when mixing cables from different
maufacturer and lot numbers.
|
Sure. But that's lengths of cable where one may reasonably expect
small mismatches in the resistive impedance and it makes sense to try
to cancel some of the reflections. I'd be interested to know the exact
reasoning, since successive bits need bear no relationship to each
other, so what cancels for a 1xx1 pattern will reinforce for a 1xx0
pattern.
That aside, my query was about the spacing between taps which is
specified as multiples of 2.5m but also referred to as a minimum of
2.5m. Taps produce short pulse reflections, occupying about 5m of
cable and it is much easier to think of a lot of short pulses being
reflected than to try to use frequency domain concepts like wavelength
on what is a wide-band signal.
The multiples vs minimum question keeps cropping up but from what Rich
has said I infer there's no technical benefit of sticking to multiples
rather than minimums, in fact random spacing is probably better.
However, if you do so, you *will* meet a minimum spacing criterion
automatically as well as having a simple and practical way of being
absolutely sure you're doing it right.
According to Rich, who did the work for the standard, it's all about
reflections. I've seen several people earnestly assuring each other
that it's to do with triggering collision detection but that would
appear to be an urban myth. It seems that even the minimum spacing is
a really an "on average" criterion. However, to put that into a
standard, you'd have to say something "a maximum of 7 taps in any
10m, a maximum of10 taps in any 20m and a maximum of 12 taps in any
30m, each maximum being taken over the entire length of the [segment]"
etc. That would be far too complex and I can understand why they
decided on a simple foolproof method. |
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Guest
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| Posted:
Wed Sep 14, 2005 2:18 pm Post subject:
Re: Why is there a minimum spacing? |
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| Quote: | Various illuminating prompts and comments,
the latter particularly by Rich Seifert,
|
Thanks all for your efforts, it is always a pleasure to read such
material.
I have never actually worked with "thick" Ethernet however I had
assumed that the regular spacing was to cause some effect
rather that to prevent some effect. - I hope that makes sense.
Very interesting.
And apparently he is becomming a Lawyer!!! |
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