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I am a bit confused about this Wikipedia article about Category 5 cables.

It states:

The cable standard provides performance of up to 100 MHz and is suitable for most varieties of Ethernet over twisted pair up to 2.5GBASE-T.

How is it possible to transmit data with a speed of 2.5 gigabit/s when using a frequency of 100 MHz? Wouldn't you need at least 2.5 GHz?

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    \$\begingroup\$ "it uses a frequency of 100 MHz", no a bandwidth of 100 MHz, and not even that is correct. \$\endgroup\$ Aug 3 at 8:18
  • \$\begingroup\$ See en.wikipedia.org/wiki/Line_code \$\endgroup\$
    – NStorm
    Aug 3 at 8:33
  • \$\begingroup\$ @MarcusMüller Ohh, I see. So there's not the signal beeing transmitted at 100 MHz, there are multiple signals beeing transmitted simultaniously over the same pair at different frequencies, correct? \$\endgroup\$ Aug 3 at 10:35
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    \$\begingroup\$ @SteveSh the question asked was how to transmit data using a frequency, so that's what that I commented on, not the cabling standard (which indeed only guarantees any particular behaviour on frequencies, but it does limit the delay spectrum, so that's inherently a guarantee over a bandwidth). \$\endgroup\$ Aug 3 at 15:14
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    \$\begingroup\$ basically the same as this question yesterday: How am I getting 1 Gbps through a CAT3 cable? \$\endgroup\$
    – phuclv
    Aug 4 at 2:18
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No, you don't need 2.5 GHz bandwidth to transmit 2.5 Gbit/s. You can do it of course, but that would be very inefficient.

The data is encoded into symbols.

100Base-T Ethernet uses one pair in one direction and uses MLT-3 encoding to convert 4B5B encoded 125 Mbit/s bit stream into a signal that has 31.25 MHz bandwidth, so it uses 3.2 bits per hertz per channel.

2.5Base-T Ethernet uses all four pairs and complex encoding methods to achieve 6.25 bits per hertz per channel to end up being 100 MHz bandwidth for which the Cat 5e cable is rated for.

The encoding method includes the use of PAM-16 modulation, which means that each symbol on each wire pair is one of 16 voltage levels and thus carries four bits per symbol. It also uses Tomlinson-Harashima precoding (THP) precoding and DSQ128 encoding to achieve the goal of getting 2.5 Gbit/s sent over four pairs with 100 MHz bandwidth per pair, or 400 MHz total bandwidth for the whole link.

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  • \$\begingroup\$ How did you arrive at the number of 6.25 bits per channel per Hz? Shouldn't it be 12.5? Or is 2.5Base-T half duplex? \$\endgroup\$ Aug 3 at 23:39
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    \$\begingroup\$ Somehow I never realized this before, but it looks like it's full duplex using all four pairs in both directions: qdpma.com/GigabitEthernet.html . That sounds wacky, but it's no different from an ordinary telephone line. You just have to filter your own transmissions from reaching your receiver, which is fairly straightforward. \$\endgroup\$ Aug 4 at 1:31
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    \$\begingroup\$ AIUI, the bandwidth limits of CAT5e are part of why 2.5GBase-T even exists, it provides an acceptable improvement over 1GBazse-T that (usually) can use existing 1GBase-T infrastructure, thus allowing some degree of upgrade with limited up-front costs. \$\endgroup\$ Aug 4 at 2:03
  • \$\begingroup\$ @AustinHemmelgarn it’s also for longer runs of cat6 and it is at a sweetspot where internet uplink or wlan might be over 1GE capacity but below 10. (but the uptake was not reflecting that) \$\endgroup\$
    – eckes
    Aug 4 at 7:34
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    \$\begingroup\$ @DeanMacGregor Are you specifically asking how is it possible to have fractional bits per Hz, or why there is? It's a calculated value from total link speed (2.5Gbps) divided by number of wire pairs (4) which gives 625 Mbps for single pair. That is sent over 100 MHz bandwidth in wire. 6.25 bits per channel per Hz. \$\endgroup\$
    – Justme
    Aug 4 at 12:09
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The rate at which you can communicate information through any channel (like a CAT5 link) is determined by its channel capacity

For normal channels like CAT5, this is well approximated by the Shannon-Hartley Theorem. This determines the channel capacity for a given bandwidth in the presence of noise as:

C = Wlog2(1 + S/N)

... where C is the channel capacity in bits/s, W is the bandwidth, and S/N is signal power divided by noise power.

So you can see that the amount of information that you can transmit is proportional to bandwidth, but also increases according to the clarity of the channel and the power of the transmitter.

100 MHz corresponds to 100 Mb/s only when the noise is as powerful as the signal. Cat 5 cable is designed to minimize high-frequency noise, so you can transmit a lot of data at the higher frequencies. It also has 4 pairs that you can use independently.

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    \$\begingroup\$ Did you mean 100 Mbits/s when you wrote "100 bits/s"? \$\endgroup\$
    – Michael
    Aug 5 at 4:11
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    \$\begingroup\$ Yes. Fixed. Thanks. \$\endgroup\$ Aug 5 at 12:28
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Thanks to @Justme and @MarcusMüller for pointing me in the right direction. I found this video by Ben Eater which answers my question quite well.

Basically, the frequency of 100 MHz is not directly related to the bits pre second that get transmitted. I assumed there was one signal (a square wave or something) per pair, which is not the case. Instead, sine wave modulation is used to transmit "symbols", which can be just ones and zeros (that's the case for slow transmissions like 10BaseT), but for gigabit ethernet, a symbol can have more than two states. This is part of the Line Code that @Nstorm mentioned. See this table for which ethernet variant uses which line code. The symbols that get transmitted per second are called the "baud rate". The 100 MHz bandwith limits how "sophisticated" a symbol can be.

Feel free to edit this answer if I understood something wrong.

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    \$\begingroup\$ bps was divorced from Hz forty years ago, using Information Theory back when 300 baud modems were Really Expensive. \$\endgroup\$
    – RonJohn
    Aug 4 at 19:08

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