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Can anyone explain this concept with figures for both non twisted pair cable and twisted pair cable?

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  • \$\begingroup\$ What's the loop area with a twisted pair? How about with flat ladder line? \$\endgroup\$ – Hearth May 4 at 4:05
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The twisting of the two wires in a signal pair does not reduce the coupling of external radiation, it assures that the coupling to each of the two wires is the same because the average distance of each of the wires from the noise source is the same. If you think of a point source of radiated noise somewhere near a long untwisted pair run, the noise source can be anywhere on a circle out at some radius from the pair. There are only two points on the circle where the distance to the noise source, and hence the coupling, is exactly the same for both wires. Everywhere else around the circle, the noise current will be greater in one of the wires. As you travel down a twisted pair over a long enough distance, each wire spends the same amount of time being closer to the external noise, so the induced noise current is the same in both wires.

At the receiver, if the two signals (one on each wire) are subtracted algebraically, the noise coupled into the wires cancels out because (+1) + (-(+1)) = 0.

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  • \$\begingroup\$ If you posted some figure, it would be more helpful. \$\endgroup\$ – Team B.I May 4 at 7:07
  • \$\begingroup\$ @AnalogKid, Exactly! The twisted pair only equalizes the induced voltages; the differential input does the rest. Here is an interesting RG discussion. \$\endgroup\$ – Circuit fantasist May 4 at 9:17

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