In networks capacity_of_link = bandwidth*propagation_delay

My doubt is how can a same wire carry multiple signals? Is is because the wire is very long (still the wire can be either high or low). I can understand if parallel wires are the reason for increase in bandwidth and thus capacity increase.

  • \$\begingroup\$ "Capacity of link" is a strange phrase to use for that. What that equation actually gives you is the delay of the link in terms of symbols. For example, if you know that the time delay is 10 us, you can multiply this by the bandwidth of, say, 1 Gb/s to get the data delay of 10,000 bits. \$\endgroup\$
    – Dave Tweed
    Jan 16, 2015 at 12:30

2 Answers 2


the wire can be either high or low

This assumption is wrong and is the source of your doubt. At high speed it's perfectly possible for one end to be high and the other low. A transition ("edge") from high/low takes time to propagate along a wire. It doesn't propagate at the speed of light either, but at a rate determined by the capacitance of the cable.

At high ethernet speeds it's perfectly possible to have multiple bits in flight along a cable.


Ethernet standards use signal modulation to increase bandwidth without much frequency increase.

For instance 1000Base-T use PAM-5 modulation, it allows to transmit 2 bits at a time:

  • 00 : -1V
  • 01 : -0.5V
  • Control : 0V
  • 10 : +0.5V
  • 11 : +1V

And for 10GBase-T a PAM-16 modulation is used, allowing around 4 bits to be transmitted at the same time.


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