Why is there a minimum spacing?

Sorry, should have said, I'm "Henry".

James Knott said

Pop it into the box at the bottom of Google Groups Advanced Search - minus the angle brackets.

Ah, well, Rich Seifert has joined the thread and says otherwise.

That may give you data corruption but it shouldn't trigger the collision detector unless the level is ridiculous.

James Knott said

At the risk of getting controversial, where do standards fit in? :)

Rich Seifert said

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.

James Knott said

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?

James Knott said

One's the sound of one hand clapping, the other's a clash of symbols :)

*/

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.

A bit of Ohms law and Thevenin's equivalent, will show those to be measuring the same thing.

William P. N. Smith said

That is the thing I find odd - could you explain why, please?

glen herrmannsfeldt said

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?

Rich Seifert said

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!

(snip on transceiver spacing calculation)

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

(snip)

With the Fourier series containing odd harmonics proportional to 1/n.

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

(snip)

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.

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.

-- glen

(snip)

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

glen herrmannsfeldt said

With Manchester encoding, even a stream of zeros is actually a square wave.

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

glen herrmannsfeldt said

See below :)

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.

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 -- glen

glen herrmannsfeldt said

....

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.

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 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.

Yes, so far I am only considering the fundamental. The third harmonic is probably also important, but I have to start somewhere.

-- glen