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Why do unshielded twisted pair cables need a different pitch/twist rate for every pair of wires in a cable? Most of my reading on UTP says that the pairs in a cable must have different pitch or else the benefits of UTP will be lost which is the noise cancelling.

From what I understood, twisting the cables essentially ensures that both wires will be near to the noise which will be cancelled since it has differential output. I don't understand why each pairs must have different pitch, too.

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    \$\begingroup\$ This quality is not unique to UTP, but important in STP, too, as explained in the answers. \$\endgroup\$
    – tobalt
    Mar 21 at 8:13

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But I don't understand why each pairs must have different pitch, too.

The different twist pitches are to avoid one pair cross-talking with another pair. If they were the same pitch, they would produce cross-talk. It's got nothing to do with susceptibility to external local noise sources.

enter image description here

  • Two closely coupled twisted pairs are shown above.

  • The upper pair has 4 twists per unit length and,

  • The lower pair has 5 twists per unit length.

  • For the lower pair, its blue wire is initially closest to red (at the far left)

  • Then, moving right, the blue wire becomes closest to neither

  • Then it's closest to blue,

  • Then blue again and,

  • Finally it's closest to red.

The net effect is that neither pair receives an accumulation of cross-talk from its neighbour.

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    \$\begingroup\$ Oh, so if they were the same pitch, they would lay next to each other continuously for the whole length of the wire? That is interesting. It is really a mechanical thing. Sort of like having threaded rods lying next to each other. If they have the same pitch then they will interlock. But if the pitches are different, they can't interlock. \$\endgroup\$
    – mkeith
    Mar 19 at 17:18
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    \$\begingroup\$ @mkeith a good analogy. \$\endgroup\$
    – Andy aka
    Mar 19 at 17:18
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    \$\begingroup\$ "Oh, so if they were the same pitch, they would lay next to each other continuously for the whole length of the wire?" – I don't think that's quite true. But suppose you have two twisted pairs with the same pitch right next to each other, and suppose that one pair has wires numbered A1 and A2, and the other has B1 and B2. Then if there's a point in the cable where A1 is next to B1, then it's going to be followed by a point where A2 is next to B2, then A1 will be next to B1 again, then A2 will be next to B2 again, and so on throughout the whole length of the cable. \$\endgroup\$ Mar 20 at 0:32
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    \$\begingroup\$ But if the pitches of the pairs are different from each other, then A1 will be next to B1 and to B2 equally often. \$\endgroup\$ Mar 20 at 0:34
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    \$\begingroup\$ @TannerSwett right, better to say that if they had the same pitch they'd be "in phase" with each other along their length as far as induced current goes. But with different pitches, each wire of one pair spends about an equal amount of time near either wire of the other pair, so the result is almost-total cancellation. \$\endgroup\$
    – hobbs
    Mar 21 at 3:46
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The purpose for using different twist rates is to reduce crosstalk between pairs.

There are broadly two mechanisms of interference pickup, capacitive amd magnetic, which is more significant will depend on the frequency and impedance of the system.

Twisted pairs mitigate both, capacitive pickup is mitigated by alternating which wire is closer to the interference source, so the two wires pick up the same amount of interference which can then be cancelled by a differential receiver.

magnetic pickup is mitigated by constantly changing the direction of the loop area, so when placed in a uniform magnetic field the magnetic pickup will cancel out.

In both cases though, we have a problem if we place two or more twisted pairs of the same twist rate close to each other. The interference between the two pairs will not cancel because the twists will line up with each other.

Varying the twist rates means the twists will not remain lined up over a long cable and hence mitigates crosstalk.

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    \$\begingroup\$ Hi, can you please elaborate on "magnetic pickup is mitigated by constantly changing the direction of the loop area, so when placed in a uniform magnetic field the magnetic pickup will cancel out.". I'm having a hard time visualizing this. \$\endgroup\$
    – hontou_
    Mar 20 at 14:36
  • \$\begingroup\$ If we place a flat loop of wire in a uniform time-varying magnetic field it will pick up interference. If we rotate the loop of wire then the interference pickup will change, if we start with the loop in the position with maximum pickup, then after rotating it 90 degrees there will be zero pickup. After a further rotation of 90 degrees the pickup will again be at maximum but with opposite polarity. \$\endgroup\$ Mar 20 at 14:43
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    \$\begingroup\$ By twisting the pair we end up with an equal proportion of the pair performing "positive" and "negative" magnetic pickup at any given time, so the net pickup is (ideally) zero. \$\endgroup\$ Mar 20 at 14:48
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As others have noted, it's about crosstalk.

A simpler way to look at it is simply to consider phase: differing twist rates ensure that each pair of pairs have roughly equal lengths in-phase and anti-phase.

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