15
\$\begingroup\$

I am not an electrician, or a student of the field. I am a Network Engineer with a curiosity bug and that has recently taken me to exploring cabling and Twisted Pair specifically. I say this to plead that the answers be 'dumbed down' so that I can understand it ^_^.

I have just finally understood the reason why 100BASE-TX and 10BASE-T uses two wires (one pair) for TX and another two wires (another pair) for RX. I understand that across each pair, one wire transmits the original signal, and the other wire transmits the exact inverse.

I also just finally understood why the wires are twisted within the pair. Effectively, to allow ambient sources of Electromagnetic Interference (EMI) to affect both wire pairs equally, rather than one disproportionately to the other.

What lead me to understand it was this image, posted on ResearchGate.net on this Post by Dr. Ismat Aldmour: Twisted Pair and EMI

I'll post his explination here as well, to avoid the risk of link rot:

I had to explain this to my students in networking once by drawing something similar to the attached Figure. In Figure 1, for the case of parallel pair, the interference causes the red wire (the closer to interference source) more pick-up voltage (induced) per unit length (1 mV as example) while less induced (0,5mV) in the blue wire. Total difference at the destination is 3mV. While in the twisted pair case (Figure 2), total difference is 0V at destination because parts (twists) of red and blue wires alternatively subjected to same level of interference and thereby total difference at destination is 0V. I drew this figure for this question hoping to use it in lecturing as well. This is especially useful when teaching networking to non-electrical engineering students who don't recognize terms of impedance, differential mode noise terms, ...etc . By the way, interference in twisted pairs comes mainly from signaling on other pairs running together within the same cable which can have many of them. Thanks. @AlDmour.

With the image and explination, I understand how the six, even twists cause for both wires in the pair to be equally affected by ambient EMI, and the net delta interference to end up at +0. My question is, What happens if there is an odd number of twists in the wire?

For instance, if one more half twist is added to the Figure 2 image above, the interference delta on the Red wire would be +1mV, and the interference delta on the Blue wire would be +0.5mV.

How does the receiving end compensate for that, and/or detect the EMI and determine what mV on each pair it can ignore?

\$\endgroup\$
  • \$\begingroup\$ Back in the day on long runs of mulitiple telephone wires (on poles) they used to have a method of swapping 1 pair over between each pair of poles (now and then) so that in effect you got a twist and therefore cancelation of cross talk to a much more significant degree. It wasn't perfect but it was much more preferable that listening to Mrs Prendergast down the street rabbiting on all day. \$\endgroup\$ – Andy aka Nov 23 '15 at 20:20
  • 2
    \$\begingroup\$ The difference will be tiny (and twisted-pair is not perfect anyhow). If your application is such that you are concerned about a tiny difference, you should use shielded cable. \$\endgroup\$ – Tut Nov 23 '15 at 21:27
  • \$\begingroup\$ Thanks for all the answers, they were all super useful. I think I made the mistake of considering the world (of signaling) around the 2D image that helped me understand it, not realizing the noise source isn't always "above" and "below" the wire, but can be all over the place. I do have one additional (unrelated) question, but I'll create a new thread... \$\endgroup\$ – Eddie Nov 24 '15 at 0:19
  • \$\begingroup\$ If you've got an odd number of 180 degree twists the polarity comes out reversed and nothing works! But serioulsly a 3m ethernet patch cable has over 100 twists, half a twist out lets in less than 1 percent of the interferance an untwisted cable of the same length would seesheileded cable would see. \$\endgroup\$ – Jasen Nov 24 '15 at 9:57
20
\$\begingroup\$

A even number of twists is better, but I am not aware of practical cable situations where this is worth the trouble: there are other sources of interference which are probably more important that the small difference it would make.

Another way to look at it: the amount of magnetic interference is proportional to the area between the two wires. With a perfect even number of twists the area is effectively zero. With an odd number of twists it is essentially one twist area. That is still a vast improvement over no twist at all :)

|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ Totally follow what you mean, how any odd number of twists represents 'one twist area'. But doesn't that still skew the signal (even if just by a small mount)? How does the receiving end extract the original signal from the skewed signal? \$\endgroup\$ – Eddie Nov 23 '15 at 18:41
  • 1
    \$\begingroup\$ @Eddie, at least in Ethernet, they use an encoding where the polarity can be inverted without affecting the received data. For example '0110110011' and '1001001100' both decode to the same message byte. \$\endgroup\$ – The Photon Nov 23 '15 at 19:19
  • 2
    \$\begingroup\$ @Eddie things are never ideal, for instance not all twists cover the exeact same amount of area, and the interferring field is not uniform, so the reciver must always cope with some amount of noise. As long as it is well below the signal level ("the eye must be open") noise is no problem. \$\endgroup\$ – Wouter van Ooijen Nov 23 '15 at 20:11
  • 1
    \$\begingroup\$ Ethernet has specs for how much untwisted wire can be present at a termination. For slower speeds it's more than one twist worth anyway. \$\endgroup\$ – Ben Jackson Nov 23 '15 at 22:41
11
\$\begingroup\$

The an even or odd number of twists is for all intents and purposes arbitrary.

What is more important is the number of Twists per Inch (TPI). The higher this number, the more that noise cancellation will be achieved.

Why? well simply put any noise sources (magnetic fields, etc.) will usually vary over the length of the cable. If you can twist a cable more times, it means that each cable will more closely experience the same noise at any given point.


To visualize it, in the diagram you posted, in a more varying field, imagine that the wire at the top experiences noise on each twist are: 1mv 1mv 0.5mv 2mv 3mv 1mv or some other arbitrarily picked numbers. Then the one on the bottom sees: 2mv 1mv 3mv 0.1mv 1mv 2mv or whatever. Now they don't match up any more so the even/odd thing ceases to matter. Now if you were to double the number of twists, but not change the noise levels, you would see that each wire now experiences the same noise.

So really you would want two twists at any point that the noise sources change. In reality these change continuously, and every environment you use the cable in is different. At which point it basically ceases to matter if there are odd or even twists as the two could never be guaranteed to experience exactly the same noise, just close to the same.

|improve this answer|||||
\$\endgroup\$
3
\$\begingroup\$

Even or odd is not significant for the cable lengths in question. What is more significant is the number of twists per unit of length (and this is also the reason why the specifications limit the amount you are allowed to un-twist when mounting). Instead, the number of twists are even so that no switch of signal polarity occurs along the cable.

Make the following Gedankenexpeirment (or do it with a real cable): If the cable does not run straight, but mostly bends back to itself so that both jacks it connects are relatively close to each other - what do you expect to happen if you rotate one of the jacks/devices by 180 degrees (or both by 90 degrees in opposing directions)? Nothing, of course. And yet, this rotation effectively changed the number of twists by one!

|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ "Instead, the number of twists are even so that no switch of signal polarity occurs along the cable." - I'm not sure what you mean by this - surely no switch of polarity occurs because the individual wires within the cable are colour-coded. \$\endgroup\$ – peterG Nov 23 '15 at 22:02
  • \$\begingroup\$ @peterG -- if you have an odd number of twists, you'll wind up rectifying it when you go to put plugs on the cable :) (either by removing a twist or adding one -- it doesn't matter in the end) \$\endgroup\$ – ThreePhaseEel Nov 23 '15 at 23:17
  • \$\begingroup\$ @ThreePhaseEel You threw me for a minute with 'rectifying' but yes I think I see what you mean! \$\endgroup\$ – peterG Nov 24 '15 at 0:33
3
\$\begingroup\$

The most common installation of Cat-5 cabling for network communications is to 10Base-T standards.

This means 2 pairs, typically blue and green, will be carrying data. Blue has 72 turns per meter, and green has 65 turns per meter.

Over short distances, none of this matters. You could have ribbons wrapped around fluorescent lights connecting your network cards, if you stay under 10 meters. (Source: personal test just to see if I could do it. It was slower than 10Mbit because TCP had to error correct, but bits got through and eventually transferred files. Also, it wasn't tightly wound around the fluorescent tube, probably wrapped around 4 times per meter.)

Worst case scenario for in-code Cat-5 wiring of 10Base-T Ethernet is to have 3 100 meter segments using amplifiers between each segment. (Code says longest any length of Cat-5 for 10Base-T is 100m, and no more than 2 amplifiers between the 100m segments before you need a repeater.) Good luck finding an amplifier instead of a repeater, though: every switch and most dumb hubs produced today will repeat.

With this worst case scenario, you can string up an office building without any data loss, including noise from computers, fluorescent lights, the HVAC system, random aluminum plates, random iron girders, the electrical system, grounded objects like copper pipes for the sprinkler system and plumbing, etc. Naturally, if you're in anything noisier than an office building, such as a manufacturing floor that uses high voltage equipment, you'll want shielded twisted pair.

That's 300 meters without data loss, with at least 65tpm x 300m = 19500 twists on your green pair. There's not much difference between 19500 and 19499 twists in this worst case scenario, where twist-per-meter really does start to matter.

So, in the worst case scenario, you're better off planning your cabling route carefully to avoid high voltage, power lines, noisy EM emitters (lights), and grounded conductors than worrying about whether you have an even or odd number of twists.

And, a bit of trivia: You always have an odd number of twists. Each RJ-45 jack is assembled alternating between tip and ring, and the tip is always the left-most pin, regardless of if you're using A or B standard, so both pass-through and crossover cables always have an odd number of twists. Turning the cable over doesn't change the number of twists that each pair has, either. Even if you have a flat ribbon, there is one 180 twist per pair.

|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ There's not much difference between 19500 and 19499 twists in this worst case scenario <-- well said! \$\endgroup\$ – Eddie Nov 24 '15 at 0:18
2
\$\begingroup\$

The amount of signal picked up is proportional to the area of the "loop". If you have an untwisted pair of length \$L\$ and separation \$d\$, the area is \$Ld\$. If you now have \$2N+1\$ half-turns in the wire, the first \$2N\$ pick-ups cancel in pairs (as you understood) and you are left with a reduced pickup of \$\frac{1}{2N+1}\$ of the original.

For a large number of turns \$N\$, that is very significant. The receivers are "robust" to small amounts of noise because the actual signal of interest is quite large. Further, you can add different circuits to increase robustness to noise. For example, you can add a "Schmitt trigger": this detects when the input reaches a certain trigger level (say 1 V) for the rising edge, then changes the level at which it will trigger again (say 0.8 V) on the falling edge. A small (100 mV) "jiggle" on top of the input signal will not be sufficient to cause an additional trigger: you will trigger once on the rising edge, and once on the falling edge.

There are many other sophisticated tricks for "clock recovery" that can help to clean up the signal. Twisting the wires is just one (very important - because cheap and effective) step.

|improve this answer|||||
\$\endgroup\$
2
\$\begingroup\$

10Base-T and 100-Base-TX are digital protocols that operate at +-2.5V and +-1V/0V, respectively. Additionally, there is a tolerance on the order of +-5-10% for the signal levels.

Assuming this cable is laid in a normal environment, the noise accumulated in a single twist is tiny because: 1) the twists are small, and 2) the wires are close together.

Taken together, the voltage bias from a single, unbalanced, odd twist is insignificant.

|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ Great answer. Could you expand on the +-2.5V? That intrigues me. \$\endgroup\$ – Eddie Nov 24 '15 at 17:09
2
\$\begingroup\$

Others answered the question well. Except:"How does the receiving end compensate for that, and/or detect the EMI and determine what mV on each pair it can ignore?"

The receiver looks at the difference in voltage between the two wires in a pair. It ignores, to a very good approximation, the signal that's common to both wires. As long as the difference due to the original signal is larger than the difference in the induced noise, it recovers the original data. This magic is known as common mode rejection, and it's the reason plain old telephone service, and really long microphone cables, work, despite induced 60Hz hum thousands of times larger than the signal.

|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ Could you elaborate on As long as the difference due to the original signal is larger than the difference in the induced noise, it recovers the original data? Is the difference in the original signal what came from sending a + and - version of the same signal down two different wires? As for induced noise, how can there be a difference if due to the twisting, ambient noise affects both wires approximately the same way? \$\endgroup\$ – Eddie Nov 24 '15 at 23:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.