The whole functionality of computers (and software in general) is based on binary system. Everything is "translated" in low-level as 0 and 1 bits.

When coming inside the field of networks, where data are transmitted through physical means, the range 0-2 volts represents bit 0, while the range of 3-5 volts represents bit 1. So, in theory, the digital signal consists of 2 possible values 0 and 5 volts that correspond to bits 0 and 1.

After investigating some network protocols (such as Ethernet) that are using carrier sense, I found out that the channel is "listened" and if no other node is not transmitting, the transmission begins. However, there could be nodes transmitting 0 bits during this specific moment, that could not be detected. For this reason, the carrier sense is being executed for a sequence of 96 bits, and if both bits were 0 then, this is accepted as a fact that no one is transmitting.


If (for example) the selected signals were 5 volts for 0 bit, and 10 volts for 1 bit, then there could be a significant performance boost in those protocols, that could complete the carrier sense in only 1 bit duration (1/96 of the current needed time). Is that true, and if yes, then why wasn't that considered from the beginning as a solution ?

The only reason I know, behind the choice of 0 volts, is that 0 volts correspond to ground, which is a reliable absolute scale.

  • This is not always the case for all signaling types. In fact, it looks like it might be rare. One reason why it would be rare is that if the signaling is not symmetrical around 0 V or a common voltage, you are basically transmitting power. For very short distances that is not a problem (e.g., on a motherboard), but between two devices with different power supplies it's a potentially huge problem (no pun intended). 1000Base-T is actually trellis coded and symmetrical around common, and balanced. See: en.wikipedia.org/wiki/Gigabit_Ethernet#1000BASE-T Jul 13, 2015 at 18:17

1 Answer 1


It seems like your real question is "Why is the interframe gap in Ethernet so long?" The answer to that question has nothing to do with the voltages used to transmit symbols. This answer on SuperUser quotes a Google hit which I will re-quote here:

The sole reason for the 9.6 microsecond interframe gap is to allow the station that last transmitted to cycle its circuitry from transmit mode to receive mode.

Here's the original link: http://www.wildpackets.com/resources/compendium/ethernet/interframe_gap Regarding the persistence of the interframe gap in modern, switched, full-duplex Ethernet connections: https://www.safaribooksonline.com/library/view/ethernet-the-definitive/1565926609/ch04.html

Providing the interframe gap ensures that the interfaces at each end of the link can keep up with the full frame rate of the link.

  • The initial question was slightly different, asking if a different approach in digital signalling could render interframe gap almost useless. However, reading your resources, I realised the signalling structure is more complex. I will have to read more. Thanks !
    – Dimos
    Jul 13, 2015 at 18:47

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