You have Cat6, 23 AWG, rated for 250Mhz frequency. You then have Cat6a, 23 AWG, rated for 500Mhz. Both are unshielded.

How is Cat6a able to transmit data at a higher frequency then its counter part?

They are both the same gauge wire, same copper, but one is rated faster. How?

  • \$\begingroup\$ Careful control of the physical construction of the cable. Any more detail than that and you'd have to ask someone more knowledgeable than I. \$\endgroup\$ – Hearth Feb 11 at 3:40
  • \$\begingroup\$ Turns per inch and dielectric properties (of the insulation) play a huge role at such frequencies. \$\endgroup\$ – Sparky256 Feb 11 at 4:22
  • \$\begingroup\$ A search on Google has turned up only basic construction demands for CAT6 & CAT6a, including special RJ-45 connectors and rules about connector attachment. It is likely that the chemistry of the insulation is proprietary, thus not available to the public-yet. Polypropylene would be a good choice, but new aerogel based insulation maybe in use here. I have no clear answer so I am not posting one. \$\endgroup\$ – Sparky256 Feb 11 at 4:35
  • \$\begingroup\$ I could understand if it was based on a special type of shielding material that ensures the frequencies integrity, but I am starting to wonder if its some sort of "marketing" ploy they throw out there to make a sale. \$\endgroup\$ – Ryan F Feb 11 at 9:12
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    \$\begingroup\$ Most of the high frequency attenuation in controlled impedance cabling is caused by power dissipation in the dielectric material. In other words, the type of plastic will have a controlling effect on the high frequency attenuation. \$\endgroup\$ – mkeith Feb 12 at 5:23

There are multiple factors that affect the behavior of transmission lines, including their top transmission frequency. To start of simple, the gauge of the wires will determine the maximum current rating, and the insulation between the wires will determine the maximum voltage rating.

From here, the next topic to visit is Telegrapher's Equations. These equations break down how in every transmission line there is line resistance and inductance along the wires, and conductance and capacitance between the wire pair.

In lossless transmission lines, the transmission speed is infinite. However, lossless transmission lines only exist in theory, as all conductors will present some resistance and all insulators will present some conductance.

The problem is these lossless lines will degrade the signals the further they travel. In the case of sending a square pulse for example, you might remember a square pulse can be simulated out of multiple sin waves through the Fourier Series, each with a multiple frequency.

Each of these sin waves will travel at a different speed down the transmission line, as the lossy transmission line model is dependent on frequency. Thus, as the wave travels down the line, the pulse will slowly become deformed, first around the corner edges, and then until it is no longer recognizable. I tried finding an image of this concept but was unsuccessful, invite anyone else to post a link

Thus the transmission speed is limited by the parameters of the lossy transmission line. The line works up until the individual pulses are no longer distinguishable as 1's or 0's at the other end.

Dispersion In Transmission Line

  • \$\begingroup\$ Good start but there are other aspects that are highly relevant such as crosstalk, shielding, length matching, twist rate control, common mode to differential mode conversion. \$\endgroup\$ – pericynthion Feb 12 at 4:11
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    \$\begingroup\$ Though that is a very interesting read A.S. It is not the answer I seek. I'm not seeking WHY their is frequency, but HOW it is determined. Perhaps I an dense, and missed it. I have 10 meters of patch cord, both are same AWG, but one is 250 MHZ rated, the other is 500 MHZ rated. WHY. Is it based on how the wire is twisted as it is being manufactured (To help offset EMF)? I don't ask this to get an answer per-se, If anyone can at least point me to the answer, i'd be happy to read it myself. If it is in fact an east answer i'd love to hear it. \$\endgroup\$ – Ryan F Feb 12 at 10:03

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