Phone lines to 10/100

Yup. And then a few years down the pike Moore's Law turns the multimillion dollar laboratory curiosity into something that is standard equipment on a child's plaything. When FDDI first shipped would you have ever thought that something ten times as fast would be a $50/node consumer product that ran over something not a whole heck of a lot fancier than phone wire?

Lots of variables to be considered.

Reading quickly through IEEE 802.3-2002 Clause 40, it would seem that using the 4 twisted pair in each direction, each one providing 250 Mb/s, with PAM-5 modulation such that the symbol rate is only 125 Msymbols/sec in each twisted pair, results in an attenuation of up to 11.8 dB per 100 meters. Is that the best the most up to date receivers can do? Maybe not. Maybe greater attenuation can be survived, so the symbol rate can be increased somewhat.

The modulation scheme used in 1000BASE-T is PAM-5. This allows for 2 bits of information to be transferred with each symbol. You send 2 b/symbol * 0.125 Gsymbols/sec * 4 lanes = 1 Gb/s total throughput. Is that the best we can do? Maybe not. Maybe you can send 6 or 8 or 10 bits per symbol, and still be capable of accurate decoding at the other end.

Clause 40 states:

"40.7.5 Noise environment

"The 1000BASE-T noise environment consists of noise from many sources. The primary noise sources that impact the objective BER are NEXT and echo interference, which are reduced to a small residual noise using cancelers. The remaining noise sources, which are secondary sources, are discussed in the following list."

Since crosstalk and echo can apparently be well compenated for, according to Clause 40, is it possible to raise the signal level at the source, to achieve a higher SNR than what is required for 1000BASE-T? Maybe yes, in which case you'll have a better chance of decoding the more delicate symbols required in the more complicated modulation techniques than PAM-5 (e.g. 256-QAM, in which each symbol can represent one of 16 amplitude levels and one of 16 phase variations from center frequency, allowing for the transfer of 8 bits/symbol rather than just

Without measuring cable bandwidth and noise levels in 100 meter segments, and without knowing what the best receivers can accept in terms of SNR today, it's impossible to compute the actual Shannon limit of a 100 meter segment of Cat 5e. My guess is that we aren't even close to it. But at the same time, getting closer costs money and development time.

Bert

(snip)

First order approximation, the bandwidth is half the baud (transition) rate. Each cycle the signal makes two transitions. Response can't fall too fast near that frequency, but that is the number that is used in characterizing the cable, including attenuation and crosstalk.

-- glen

LOL! Just because stupidity abounds does not mean that it is correct or even always occurs.

Certainly there are lots. You don't heard much about them because bad news sells.

Non technical managers _can_ be the very best. The real problem is when managers micromanage their employees. Even when they have great technical strengths (wich sometimes means poor people skills), they usually are missing information. Even when they have it all, it's rather disrespectful of a competent professional to meddle.

LOL! At NASA, certainly.

Is that safe? We're talking about marketroids here! I'm not sure they're competant.

It doesn't matter whether there are many shareholders or only one. All corporations have shares even if not publicly traded. Conscientious owners are very careful what and how they manage. More is not better. Keeping powder dry is an important concept.

-- Robert

We can do better than assume. ADSL points the way to some rough calculations:

Over short distances (3-5 kft), 6 (12?) Mbit/s are possible and lets say that corresponds to your "whopping" 42.1 dB S/N. Then voice-grade telco distribution UTP has a bandwidth of at least 430 kHz.

Many people have trouble getting even 1.5 Mb/s at long distances (15+kft), so the S/N must have dropped to 7 dB or below. At least 3.5 dB/kft. How does that sound?

-- Robert

That part sounds right.

Over long distances, the bandwidth of voice grade cannot be assumed to still be 430 KHz, right? The high frequencies will be attenuated more than the lower frequencies, as a function of cable length.

If the bandwidth is less than you assume, then it follows that the SNR you need will be higher than you state.

Bert