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i'm doing networking and trying to understand if the twisting in twisted pair reduce attenuation. I know that the cable can't be longer than 100 meters and the book says this

From Microsoft Windows Networking Essentials

Twisted-pair cable is the most commonly used cable type in networks today. It comes in multiple categories with different speed capabilities. A twisted-pair cable used in a network includes four pairs of copper wire. Each wire in the pair is twisted around each other, and the four pairs within a cable are then twisted around the other pairs. The four twisted pairs are then wrapped in a polyethylene or polyvinyl jacket. The number of twists per meter in these cables is different for different categories of cables. Twists in the cable help minimize both cross talk and EMI. Additionally, the number of twists per meter determines the speed and frequency capabilities of the cable. Higher speeds and frequencies allow the cable to carry larger amounts of data. However, all the twisted-pair categories have a maximum distance of 100 meters. In other words, the cable can’t be longer than 100 meters between any two components. It is possible to extend this distance by using a repeater. The repeater amplifies the signal, allowing you to run the cable another 100 meters.

So this means that if i have higher frequency i have less attenuation? I mean i thought attenuation is the distance the data goes not how much data there is, so i think the answer to my question is no, but am i right or wrong?

Thank you

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    \$\begingroup\$ Wherever you read that, throw that book away and find a better one. \$\endgroup\$ – The Photon Apr 19 '16 at 14:46
  • \$\begingroup\$ Where do you have that quote from? Things are quite complicated and stuff like attenuation depends on the frequency, and the speed is always 1GBit for 1GBit networks. Did you know that the twist rate of most twisted pair cables is different for different pairs? \$\endgroup\$ – PlasmaHH Apr 19 '16 at 14:46
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    \$\begingroup\$ It's not attenuation that's improved. It's noise cancellation. At large distances the wire will pick up significant amount of noise from the outside world. Twisted pair simply ensures that both the + and - wires experience the same noise. \$\endgroup\$ – slebetman Apr 19 '16 at 14:50
  • \$\begingroup\$ @ThePhoton Based on two sentences without context? That seems like an unwarranted knee-jerk reaction. Not to mention the fact that the passage quoted is correct at its detail level (it is simplified, of course). See also another comment on here quoting a bit more context. \$\endgroup\$ – marcelm Apr 19 '16 at 17:47
  • \$\begingroup\$ Sorry i wasn't here, anyways the book is microsoft windows networking essential, i've edited the full paragraph in the text \$\endgroup\$ – VUBEDDU Apr 19 '16 at 17:54
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No.

The main effects on the characteristic impedance and the propagation velocity are from the diameter of the individual wires, the separation between the wires, and the dielectric constant of the material between the wires.

The main effects of the attenuation are from the lossiness of the dielectric, the conductivity of the wires, and the diameter of the wires.

The main limiter of the bandwidth is from the attenuation, which increases as the frequency of the signal increases (mainly due to the skin effect)

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  • \$\begingroup\$ Please forgive me. I'm not qualified to argue with your answer. I am trying to understand it. Part of my confusion with this thread comes from Wikipedia where it is stated: "Providing the interfering source remains uniform, or nearly so, over the distance of a single twist, the induced noise will remain common-mode. Differential signaling also reduces electromagnetic radiation from the cable, along with the associated attenuation allowing for greater distance between exchanges." ... \$\endgroup\$ – user109983 May 13 '16 at 13:27
  • \$\begingroup\$ ... You have suggested that the diameter of the wire is a factor with attenuation. If cables are twisted together, do you not end up with one larger de facto cable with an effectively increased diameter? \$\endgroup\$ – user109983 May 13 '16 at 13:27
  • \$\begingroup\$ @Edward, there is insulation between the wires. \$\endgroup\$ – The Photon May 13 '16 at 14:50
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Perhaps the manual meaning is: with more twists per meter the cable can carry higher frequency and higher frequency means larger amounts of data. But I would say the precision of twists is important, the two wires have to be as closer as possible to reduce the free space window to minimum, this ensures low inductance -> higher frequency possible. Larger number of twists just reduces noise susceptibility.

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No. If anything, twisting increases capacitance which increases loss (attenuation) at higher frequencies.

The reason for twisting is to better cancel interference and noise from nearby sources (including the three other pair in the network cable itself).

The more twists per meter, the slightly higher the capacitance between the wires in the pair. And higher capacitance is BAD for trying to send high-speed data. That book was written by (or perhaps edited by) someone who didn't understand the fundamental electronic principles.

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  • \$\begingroup\$ So does CAT5 cable have more twists per meter? \$\endgroup\$ – Tim Spriggs Apr 19 '16 at 17:03
  • \$\begingroup\$ More twists than what? The four different pair inside a typical Cat5 cable have slightly DIFFERENT twist rates. The reason is to reduce cross-talk between adjacent pairs. If the pairs were twisted at exactly the same rate, they would be more vulnerable to cross-talk because of their very close proximity. \$\endgroup\$ – Richard Crowley Apr 19 '16 at 17:19
  • \$\begingroup\$ Not the book's fault - it was misinterpreted and quoted out of context - "Twisted-Pair Cable... Each pair has a specific number of twists per meter, with different pairs having a different number of twists. Even though the pairs are right next to each other in the same cable, these twists prevent signals from crossing over to each other. Additionally, the number of twists per meter determines the frequency capabilities of the cable. Higher frequencies allow the cable to transmit more data.". \$\endgroup\$ – Bruce Abbott Apr 19 '16 at 17:35
  • \$\begingroup\$ I've added the full paragraph with the header being "Understanding twisted pair" the reason i'm asking if it's to reduce attenuation it's because it is a question i have for school and i'm trying to understand if i understood correctly that, according to the textbook, at least, it does not reduce attenuation. \$\endgroup\$ – VUBEDDU Apr 19 '16 at 17:59

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