Why is there a minimum spacing?

(snip)

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 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)

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.

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

glen herrmannsfeldt said

I am not a programmer so I can't read this very easily. I'm not sure what you're doing. I can guess most of it but I'm lost with the index manipulation which I've marked. What's all that about?

(snip)

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

(snip, I wrote)

Round trip on the cable is about 42 bits, so 42 zeros in a row is enough to do it.

-- glen

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

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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

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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

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

(snip regarding transceiver power)

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

glen herrmannsfeldt said

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'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 :)

DHP said

I meant j of course.

glen herrmannsfeldt said

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.

Absolutely essential :) Otherwise a cosine multiplier is going to sweep over your result as you change s, resulting in spurious nulls.

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.

That's the one to worry about :/

glen herrmannsfeldt said

Not many networks are dedicated exclusively to large pure black bitmaps.

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!

Rich Seifert said

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.

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.

glen herrmannsfeldt said

Well, put it this way then, not many networks are dedicated to pure never-ending jam.

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!!!

In article , Rich Seifert 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??

No, repeaters are allowed between segments.

The 500m limit is mostly due to cable attenuation and the ability to properly do collision detect over the length of the cable. Collisions are detected by the DC voltage on the cable. The threshold must be more than the maximum for a single nearby transmitter, and, for receive more detection, less than two at the far end.

The slot time allows for repeaters and FOIRL (fiber) links, at least traditionally.

-- glen