The higher twist (and generally every other characteristic of twisted pair designed for 10baseT frequencies) makes it a superb cable for VF frequencies.
The most likely problem is that any kinks or other damage to the cable would allow cross-talk from ring current on the phone line to interfere with the 10baseT signal, but obviously that would be rare, intermittent, and of very short duration to the point of virtually never being even noticed (unless you have a couple teenagers using the phone line... :-) .
The most serious problem would be wiring the connectors wrong. If they are connected just straight across, with pairs on 1 & 2,
3 & 4, and so on, the Ethernet segment will have a crossed pair using the tip from one pair and the ring from another. That will guaranteed result in the phone line, and perhaps other sources of noise, interfering with the Ethernet. On relatively typical short runs of CAT5 it will still work though, for
10baseT, and may not ever be discovered. The same cable will not work at all for 100Mbs Ethernet though, so if you move to a higher speed network that would immediately become critical.
You clearly have not Googled that question! But we can talk about it anyway... ;-)
There are 8 pairs of pins on an RJ-45 connector, and at least three different wiring arrangements that are not compatible. (And given the 10BaseT arrangement, you can use the 4,5 and/or the 7/8 pins for something other than 10BaseT on the same cable.)
For example, the same connector and CAT5 cable is commonly used by telecom people for T1 or DSX-1 circuits. A home user might not ever see that, but these days even small business users may have T1 or Fractional T1 services, and could even have a patch panel and therefore have patch cords for it that are wired for T1's.
Those will not work when used for 10/100BaseT Ethernet.
The typical mistake that a non-aware person makes is just wiring the Blue/Orange/Green/Brown pairs straight across to pins 1 through 8 in order. As you've noted above, 1 & 2 is one pair and 3 and 6 are the other for 10BaseT, hence a straight across cable will put that signal on the Ring of the orange pair and the Ring of the Green pair, which is called a "split pair". Since it is not a twisted pair there is a lot of unbalanced induction into the pairs which is not canceled by the common-mode operation as expected when using twisted pair cable. For short runs it will actually work though, at 10Mbps, and might go undetected if the only test made on the cable is for continuity. It won't work at all past a few feet for 100Mbps Ethernet.
Of course if you are are wiring *both* the plugs on the cords and the jacks, it makes no difference at all as long as they are done the same. Hence you can wire them for T1's, for 10BaseT, or just pairs 1 to 4... but then you have to keep all the cords from getting mixed up too! That problem is already bad enough, what with crossover and non-crossover already needed for the
10BaseT cords, so even if you don't have T1's you'd still potentially have 3 types of cords just for 10BaseT.
I'm not going to look that up to be certain, but I don't think it is correct. T1 was certainly designed for telephone lines, but I don't think 10baseT was. Look at the difference in the twist on CAT5 compared to lesser grade twisted pair!
The frequency has little to do with whether it will interfere. The voltage level is what makes a difference. Besides, 10baseT does *not* run at 10Mbps... that is merely the highest bit rate. The actual frequency spectrum runs from DC on up through at least 15 Mhz (three times the 5 Mhz fundamental for 10 Mbps). Which is to say that VF circuits and 10baseT do in fact share the same frequency range for the entire VF range.
10BASE-T was designed to operate on early Cat-3 UTP wiring. This was better than untwisted (quad-type) telephone wire, or "silver satin" modular phone cords. Cat-5 wiring was designed for 100 Mb/s operation.
Actually, the maximum fundamental frequency for 10BASE-T signals is
10 MHz, not 5.
-- Rich Seifert Networks and Communications Consulting 21885 Bear Creek Way (408) 395-5700 Los Gatos, CA 95033 (408) 228-0803 FAX
Makes it a pain in the neck to wire at both ends since all the termination equipment is designed with one jack-one cable paradigm in mind. Also makes it not standard-compliant as you are not supposed to share horizontal cable between applications, but it will work. As long as you don't mess up the pinouts on both ends, of course. Make sure you use pair #1 (blue) for POTS and #2 (orange) and #3 (green) for Ethernet. Creates less confusion that way.
CAT3 is not "telephone" cable though. Twisted pair telephone cable is CAT1, useful up to relatively 1 MHz. (A 1.544 Mbps rate of a T1 has a maximum required frequency bandwidth of 750 KHz, and works just fine on CAT1.)
CAT2 cable was designed for 4Mbps Token Ring and is useful up to
1.5MHz. CAT3 was designed 10Mbps 10baseT, and is useful up to
16 MHz. CAT4 for 20 Mbps (meant for 16Mbps token ring) or
20MHz. CAT5 for 100Mbps 100baseTX and useful up to 100MHz.
The point is still that 10baseT was not designed to work over telephone cable.
Opps. That's true. It uses Manchester encoding, not NRZ or AMI. There is either always 1 transition per bit, and sometimes two. So half the bit rate is the lower frequency limit and the upper frequency limit is the bit rate (one Hz per bit).
The spectrum used by 10baseT is 5 to 10 MHz. Which makes CAT3 cable a relatively conservative choice... and CAT5 is simply overkill, but then again I don't even know if anyone stocks CAT3 cable these days!
If the transition period of the eithernet signal happens at a time when the induced voltage from crosstalk is greater than the threshold for the ethernet receiver, the receiver will decode an incorrect value. The significance that frequency plays is just how often and for how long that effect lasts.
Essentially a 20Hz ring voltage could, for example, wipe out
1/40th of a second worth of 10baseT data if, for example, the entire positive transition of the ring signal is able to block the eithernet receiver.
Granted however that the sensitivity to a 20 Hz interfering signal is going to be significantly less than to a single frequency that falls inside the spectrum of the desired signal. But then the voltage level for the ring voltage is huge by comparison too...
I misspoke about the spectrum of 10baseT ethernet too. You are right that it is 5 - 10 Mhz, not DC to 5 Mhz (which is what 10 Mbps would be using NRZ encoding).
Yes, it's entirely acceptable. When twisted pair ethernet was created, it was intended to be run on existing cat 3 telephone cables, along side phone lines. Also, the frequencies used on telephones are nowhere near those used by ethernet, so interference is unlikely. I've even seen installations, where two 100 Mb ethernet circuits share one cable.
10baseT was designed to run on the same cat 3 cable as was used to wire phone systems. The whole idea was that offices had lots of the stuff.
Where shall I start. Lessee now...
10baseT uses Manchester encoding, where there's one complete cycle per bit, with the phase inverted for a "1" bit in the data. So, if you had a steady "1" or a steady "0", you'd see a continuous 10 MHz square wave, which has a fundamental frequency of 10 MHz, plus harmonics. An alternating "1 0" bit pattern would result in a fundamental frequency of 5 MHz, plus harmonics. Every other bit pattern would fall between those extremes. I doubt you'd see frequencies anywhere near the voice frequencies and DC is impossible, as the lines are transformer coupled. So the highest voice frequency would be in the vicinity of 4 KHz and the lowest ethernet frequency around 5 MHz or over 1000x greater. I'd be surprised, if much voice energy made it through the transformers. As for voltage, it won't make much difference, unless great enough to cause physical damage, considering any voice signal is well removed from the frequencies used by ethernet and likely to be blocked by the transformers.
CAT1 is telephone grade UTP. There was no significant amount of CAT3 cable in a telephone office at the time 10baseT was designed, and in fact there still isn't.
There just is no use for CAT3 cable in a telephone office other than for 10baseT. Runs of T1/DS1 between racks in the same row could use it, but it doesn't have a ground wire, so it isn't used. Any runs of T1/DS1 between rows requires what is called ABAM, which is shielded twisted pair. For voice pairs CAT3 could be used, but CAT1 is sufficient and is less expensive.
CAT3 was designed for data networks, not telephone networks.
Yes. I was thinking of NRZ or AMI as is used in T1/DS1 encoding. The point is still the same though, because the frequency is *not* what interfers. It's the voltage.
The spectrum for 10baseT has 5 and 10 Mhz fundamental components, and typically would have significant 15 and 30 Mhz
3rd harmonic components (which are unnecessary for proper operation).
You are, like I was, thinking of how NRZ works.
With Manchester coding "every other bit pattern" is merely a combination of one or the other of the two extremes. *Every* bit has a transition, which is the 5 Mhz component. Any change from either a 1 to an 0, or from a 0 to a 1, results in a 10 Mhz component by adding a transition between the regular once per bit transition.
That is true, and certainly is a characteristic that has significance; but it doesn't mean that crosstalk from 20 Hz ring current cannot interfere with the the 10baseT receiver. It merely limits the mechanics of what can happen.
Same as for ring current, except that since the levels are significantly lower it is very unlikely that any VF signals actually will cause interference.
There can be and often are high frequency components in the ring current, and since we are talking about an extremely high level signal by comparison, the levels that would interfere can be present. Note that the exact same cable problems can result in problems from "60Hz" power influence too, which might also have significant components at higher frequencies (for both telephone and power cabling it is not uncommon to see spikes on the cable that are in the 1 GHz frequency range).
The significant point though, for this type of problem, is that it would be very intermittant and would probably not produce errors at a rate sufficient to even be noticed.
Actually I do this almost daily. Usually where a temporary second connection is required. Or where is is difficult or impossible to run a second cable. Have never had a problem. Or one reportrd to me from the data types I work for.
The intent is juxtapositioned. CAT3 cable was designed for
10baseT, and allows VF circuits to ride on the two unused pairs.
True, but the typical *telephone* cable was CAT1 quality, not CAT3 (and higher) until 10baseT came along.
We are discussing systems which encode voltage levels, not frequency or phase. By definition "noise" is a unwanted voltage change, and the frequency or phase of the change is insignificant. Only on systems where frequency or phase is used for encoding or modulation would that be defined as noise.
However, frequency relates to the ability to *transport* the signal, even though it is the voltage level which actually causes interference.
If the frequency is such that it will not pass through various parts of the transmission path, there will likely be insufficient voltage at a point (the receiver) where it can cause a problem. But if the voltage is sufficient, it makes no difference what frequency it is.
At the point where it enters the system. But at the point where it actually interferes with the intended signal, frequency is insignificant.
See the various discussion articles on Manchester coding and frequency spectrum. The bit rate is indeed fixed at 10 Mbps, but that does *not* necessarily translate directly to a 10 MHz bandwidth or a 10 MHz fundamental. In the case of Manchester coding only (others are distinctly different) it means a 5 MHz bandwidth that ranges from 5 to 10 MHz. (Other components are generated at 15 and 30 MHz, but are not required at the receiver for proper operation.)
I was thinking about NRZ encoding, not Manchester. Sorry about that.
Virtually *all* essential spectral components are at 5 and 10 MHz. The transmitter may generate a square wave, but the receiver only requires a sine wave, therefore none of the other components have significance at the receiver (or anywhere else).