Moving this video to kulpims's personal queue. It failed to receive enough votes to get sifted up to the front page within 2 days.

Very interesting video, but a pity he went in to all that detail to end with a figure as vague as "1 terabyte in a few hundredths of a second".

If "a few" is 5 we'd get 20 Tbytes/s, if a few is 50 we'd get 2 Tbytes/s.

From wikipedia it seems optical fibre bandwidth is also limited by distance, due to dispersion, but current experimental research results are approaching 100 Terabit/s over a single channel.
(26Tbit/s in 2011, 73.7 Tbit/s in 2013)
http://en.wikipedia.org/wiki/Fiber-optic_communication#Parameters

100 Terabit/s is 12.5 Terabyte/s, which is 1 Terabyte in 0.08 seconds.

So while it still doesn't tell us what the theoretical limit is, for our currently achievable maximum speeds "a few" is 8.

http://what-if.xkcd.com/31/

Lol, I remember studying the theoretical maximum bandwidth of a telephone modem at university. I can barely remember the maths/physics of it, but I remember the result which, interestingly, turns out to be less than 56k!

Christ, I feel old now.

Fibre can go a pretty long distance before it affects the signal though...

Fibre is comprised mainly of silicone, the more pure the fibre, the less dispersion issues occur at or around 1550nm (one of the main wavelengths used for long distance transmission, as we can easily and cheaply amplify this using ebrium doped segments and some pumps!)

Any impurities in the fibre will absorb the 1550 at a greater rate than other wavelengths, causing linear distortions in the received carrier along greater distances. This is called Brillouin scattering.

In the context of the above video, consider a paralell cable sending data over 100m. If one of those lines is 98m, then every bit that is sent down that line, will be out of order.

Same deal with Brillouin scattering, only on the optical level. Thats one of the main issues we gotta deal with at distance, however it only ever occurs at or around 1550nm, and only ever when you are driving that carrier at high powers (i.e. launching into the fibre directy from an ebrium doped amplifier at +15 dBm)

Theres some fancy ways of getting around that, but its not cheap.

Anywhere from say around 1260 to 1675nm is the typical bandwidth window we use today.

So, say 415nm of available bandwidth.
If we want that in frequency to figure out the theoritcal bits/sec value from the shannon-hartley theory, then we just take the inverse of the wavelength and times it by the speed of light.

7.2239e+14 hz is the available spectrum.

...thats 7.2239e+5 terahertz....

Assume typical signal to noise on fibre carrier of +6dB (haha, not a chance in hell it would be this good across this much bandwidth, but whatever..)

For a single fibre you would be looking at an average peak bandwidth of around 20280051221451.9 mbps.

Thats 19,340,564 Terabits per second, or 18,887.3 Petabits per second.

You can fudge that +/- a couple of million Tbps based on what the actual SnR would be, but thats your average figure.....thats a lot of Terabits.

On one fibre.

Source: Im a telecoms engineer

are these absolute maximums, or maximums before some clever ?quadrature phase shift? encoding scheme makes it pack even more information than is being mentioned

You are thinking about QAM, Quadrature Amplitude Modulation. Thats an interesting question because QAM essentially produces the same results that the prof talks about in this video. By using interesting ways to change the beat and phase of a single carrier, one can represent a whole array of numbers greater than just a 1 or a zero with a single pulse, case in point.

In QAM, lets just use the easy example of QAM, QPSK (4QAM), where there are 4 possible binary positions for any given 'carrier' signal at a known frequency.

By shifting both phase and amplitude, you can get a 0, 1, 2 or 3, where each position represents a power of 2, up to a total value of 16 unique numbers.

Rather than just a 0 or a 1, you can have 0 through to 15. However doing this requires both a timeslot, and a known carrier window.

The fastest the QAM transmitter can encode onto a carrier is limited by the nyquist rate, that is, less than half the frequency which the receiver can sample at its fastest rate (on the remote end). As you increase the speed of the encoding, you also increase the error rate, and introduce more noise into the base carrier signal, in turn, reducing your effective available bandwidth.

So it then becomes a balancing act, do I want to encode faster, or do I want to increase my constellation density? The obvious answer is the one we went with, increase in constellation density.

There are much more dense variants, I think the highest ive heard of was 1024 QAM, where a single carrier of 8MHz wide could represent 1024 bits (1,050,625 unique values for a given 'pulse' within a carrier).

I actually had a lot more typed out here, but the maths that goes into this gets very ugly, and you have to account for noise products that are introduced as you increase both your transmission speed, and your receiver sensitivty, thus lowering your SNR, reducing your effective bandwidth for a given QAM scheme.

So rather than bore you with the details, the Shannon Hartley theorem is the hard wired physical limitation.

Think of it as an asymptote, that QAM is one method of trying to milk the available space of.

You can send encoded pulses very fast, but you are limited by nyquist, and your receivers ability to determine noise from signal.

The faster you encode, the more noise, the less effective bandwidth....and so begins the ritule of increasing constellation density, and receivers that can decode them....etc....

There is also the aspect of having carriers too close to one another that you must consider. If you do not have enough of a dead band between your receivers cut off for top end, and the NEXT carrier alongs cutoff for deadband at its LOW end, you can induce what is known as a heterodyne. These are nasty, especially so when talking about fibre, as the wavelengths used can cause a WIDE BAND noise product that results in your effective RF noise floor to jump SUBSTANTIALLY, destroying your entire network in the process.

So not only can you not have a contiguous RF bandwidth of carriers, one directly after another...if you try and get them close, you end up ruining everyones day.

I am sure there will be newer more fancy ways to fill that spectrum with useable numbers, but I seriously doubt they will ever go faster than the limit I proposed earlier (unless they can get better SNR, again that was just a stab in the dark).

It gives you a good idea of how it works though.

If you want to read more on this, I suggest checking wikipedia for the following;

Shannon Hartley theorem.
Nyquist Rate
Quadrature Phase Shift Key
Quadrature Amplitude Modulation
Fibre Optic Communication Wavelengths
Stimulated Brillouin Scattering
Ebrium Doped Fibre Amplifiers

Sorry...I went too far with that

I take it you enjoy your work?

Nah, doesn't even interest me

@charliem great answer thanks a lot for the education

> 1024 QAM
thats crazy

Im actually reading that the next constellation density in research is 16,384 QAM...if you thought 1024 was crazy >.>