If the maximum data bandwidth is sent down a fibre optic cable, constantly, maxing out the cable's capacity, how much data is in transit over any given distance?

Perhaps I have an incorrect understanding of how this works, but I assume that the data packets are sent one after the other, in a line down the wire, not one at a time, so if a wire has a 1000GB/second bandwidth, it can have many of these terabyte 'seconds' going down the same wire at the same time.

Is this understanding in any way correct? If so, how much data can a fibre optic cable have in transit over x distance if sending maximum capacity throughout the length of the wire?

  • \$\begingroup\$ You can work it out the length of one bit in metres, lambda=L, from: fL= 3x10^8, where f is the frequency in bit/sec, assuming free-space propagation speed (probably slower in medium) \$\endgroup\$
    – Chu
    Apr 7, 2015 at 13:19
  • \$\begingroup\$ In the optical communication light in the fiber is modulated all the time even if there's nothing to send.<br> To send more data the Wavelength Division Multiplexing is used. The WDM (or DWDM) uses different colors of light (usually IR) to send data in parallel over the same fiber. \$\endgroup\$
    – Alexxx
    Apr 7, 2015 at 13:35
  • \$\begingroup\$ The Photon's answer is correct; it's worth noting that having a large amount of data in flight at once needs to be thought about from a software point of view. A single TCP connection is limited by its "window". \$\endgroup\$
    – pjc50
    Apr 7, 2015 at 15:41

1 Answer 1


Let's say you transmit at 10 Gb/s. Then the bit period is 0.1 ns.

The index of refraction of glass is about 1.5, so the speed of propagation through the fiber is \$c/1.5\$, or about \$200\times10^6\ \mathrm{m}/\mathrm{s}\$.

So each bit occupies about 20 mm of length, and there are about 50 bits "in flight" in every meter of fiber.

If you want to know what happens when you use the "maximum data bandwidth" of the fiber, you'll have to define what you mean by that. In practical systems, the bandwidth isn't limited by the fiber but by the transmitters and receivers at each end. The number of wavelength division multiplexing channels that can be used is limited by nonlinear behavior that occurs in the fiber when the total power (across all channels) gets too high. So even that limit depends on the sensitivity of your receivers (which determines how low you can make the power in each channel).

  • 1
    \$\begingroup\$ Note that this answer assumes you transmit one bit per symbol and as such, the symbolrate = bitrate. There are (optical) modulation techniques that encode multiple bits per symbol. As such, the number of bits per meter increases. \$\endgroup\$
    – RJR
    Apr 7, 2015 at 23:23
  • \$\begingroup\$ @RJR. Good point. I work in the data networking (as opposed to telecommunications) world of fiber optics. In data networking there is not yet any commercial systems available with more than one bit per baud. But we do expect to see them within the next year or two. \$\endgroup\$
    – The Photon
    Apr 7, 2015 at 23:26
  • \$\begingroup\$ For future readers: just an update that these systems have definitely arrived and Terabit backhauls are a thing, with multiple wavelengths being used as multiple carriers, and high-rate fiberoptics doing coherent transmission/reception on these, using plain or modified QAM constellations with >> 1 bit/channel access. The future is awesome. \$\endgroup\$ Jul 28, 2020 at 7:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.