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I was reading about the Twisted Pair Cables, and how they implement Balanced Line methodology to cancel out all the noise. All fine and good.

However, I just couldn't grasp this one simple doubt:

When we are twisting two wires (in a pair) close up, which are carrying equal but opposite signals, aren't they going to interfere with each other? Isn't this interference going to add noise?

I couldn't find the answer to this simple doubt anywhere. Perhaps, I am missing something very fundamental.

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Note: The original question appeared to refer to wires in multiple pairs. The OP has advised that he is asking about interference between the two wires within a single pair. The original subject line and text has been edited to reflect this.

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    \$\begingroup\$ Yes. But the signals they are carrying are the same signal with opposite polarity. So the interference just appears as attenuation. \$\endgroup\$ – Brian Drummond Jul 25 '16 at 11:56
  • \$\begingroup\$ @RussellMcMahon it looks like you have altered the question to suit your answer now. It makes my answer look stupid. Please roll-back the edit. If the op feels he wants to change the question then I really do think he should do it rather someone who has posted an answer. \$\endgroup\$ – Andy aka Jul 25 '16 at 14:08
  • \$\begingroup\$ @Andyaka I was not trying to make my answer "look good" or anyone else's "look bad" or stupid or whatever. I adapted the answer to what the OP said he had intended it to be. The potential cost of leaving the answer uncorrected is to mislead more people. Your rep makes it obvious that you are not stupid, and you and I can afford to be thought stupid if anyone feels the need without it having any importance. ||| The OP said: "Sorry. Yes I made a major mistake in the question. What I meant was just 2 wires in one pair." \$\endgroup\$ – Russell McMahon Jul 25 '16 at 14:29
  • \$\begingroup\$ @Andyaka See note at end of question. This can presumably be removed in due course. \$\endgroup\$ – Russell McMahon Jul 25 '16 at 14:32
  • \$\begingroup\$ Copied from other comments as easily missed: [shivams] ... In a given pair, consider the two wires. Aren't these two wires going to interfere with each other? Forget all other pairs. | [ANDY] ... you say "two pairs" - in English that means 4 wires with 2 of the wires forming one pair and the other 2 wires forming the other pair. I do not understand what your comment has to do with your original question. Maybe you misunderstood/miswrote the question? | [shivams] Sorry. Yes I made a major mistake in the question. What I meant was just 2 wires in one pair. \$\endgroup\$ – Russell McMahon Jul 25 '16 at 14:35
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Within the one pair the difference in voltage or current IS the signal. If used as a single pair carrying a signal the signal is the relative difference between the two conductors. IF the pair was used as two circuits with voltages relative to eg ground then you would get 'interference' but that is not how a twisted pair is properly used.

What you do get is series inductance and resistance and parallel capacitance (and possibly interwire leakage but that is usually negligible) which causes losses. This leads to increasing loss with frequency. This can be passively addressed with "loading coils" which flatten the frequency response across a desired range with a consequent high frequency cutoff and low pass filter action. This method was traditionally used with twisted pair telephone cables to get extended range at voice frequencies and a very sharp cutoff above the range of interest. From dim memory 3.2 kHz was considered to be the top of the speechband for residential circuits.

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When we are twisting two pairs of wires close up, which are carrying equal but opposite signals, aren't they going to interfere with each other? Isn't this interference going to add noise?

The problem we're trying to avoid here is creating a loop antenna that will pick up interference from other sources. We can do this reasonably well by running the feed and return conductors very close together but twisting the pair improves this significantly. Each alternate twist forms a small loop but each consecutive loop is out of phase with the previous one and the net effect is to cancel out the interferece.

So, I hear you repeat, "aren't they going to interfere with each other?"

enter image description here

Figure 1. The feed and return currents' magnetic fields do not oppose each other.

There's no reason to. Because the feed and return carry the same current in opposite directions the magnetic field of each conductor goes through the loop in the same way.

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When we are twisting two pairs of wires close up, which are carrying equal but opposite signals, aren't they going to interfere with each other? Isn't this interference going to add noise?

Usually, the two individual pairs are twisted at a different spacing (say one twist per inch versus 1 twist for every 2 inches) to each other thus, over a length of cable, cancellation of one pairs signal onto another pair is fairly good. If you look closely at the pairs below they have different twist spacings: -

enter image description here

Alternatively two pairs can be twisted all together is a diamond, star or quad formation. One pair will use wires at opposite corners of the "diamond" and this will theoretically cancel any cross-talk to the other pair due to symmetry: -

enter image description here

Given that the question has changed, here is my revamped answer. Consider the following diagram showing two conductors with equal and opposite voltages present: -

enter image description here

Lines of equipotential exist between the two wires and at exactly halfway is a zero volts equipotential plane extending to infinity. It should take only a small leap of faith to imagine this plane is in fact a solid but this highly conductive earth plane. This can be regarded as an electric field screen and the problem then reduces to one wire above a ground plane on the left and ditto on the right. Thus, there is no interference between the two cables only a reduction in capacitance.

The fact that twisted pair cable is errrm, twisted does make the visualization more difficult but the truth is there is still effectively a 0V equipotential point between the two conductors.

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  • \$\begingroup\$ Thank you for the answer, but I think you misunderstood the question. My doubt is this: In a given pair, consider the two wires. Aren't these two wires going to interfere with each other? Forget all other pairs. \$\endgroup\$ – shivams Jul 25 '16 at 11:28
  • \$\begingroup\$ In your question you say "two pairs" - in English that means 4 wires with 2 of the wires forming one pair and the other 2 wires forming the other pair. I do not understand what your comment has to do with your original question. Maybe you misunderstood/miswrote the question? \$\endgroup\$ – Andy aka Jul 25 '16 at 12:50
  • \$\begingroup\$ Sorry. Yes I made a major mistake in the question. What I meant was just 2 wires in one pair. \$\endgroup\$ – shivams Jul 25 '16 at 13:04
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    \$\begingroup\$ If driven differentially and balanced then halfway between the two wires is a point in space that MUST be 0 volts. This point can be extended to an infinite plane (ignoring the twists) and this infinite plane can be replaced with a highly conductive infinite earth plane without loss of theory (providing the plane is thin compared to the wire dimensions involved). This is an electric field shield thus no E field interference. The twisting makes it impossible to visualize an earth plane but it will still exist. \$\endgroup\$ – Andy aka Jul 25 '16 at 14:21

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