I am having trouble understanding the concept of 'frequency' of a signal in an electrical sense in a wire vs the concept of 'frequency' in the electromagnetic sense in a fibre optic cable. Are these the same thing? The electricity travelling through a wire is not the same physical concept as the electromagnetic wave travelling through a fibre optic cable, right?

Do we treat them the same when we calculate the Shannon limit? I.e. can I look at the range of optical frequencies (wavelengths) that can be transmitted through an optic fibre and compare it to the range of electrical frequencies that can be transmitted through a wire, and compare them?

How does generating pulses of light compare to encoding signals with voltage changes?

  • \$\begingroup\$ Frequency it is independent of the carying media. RF transmision in free air calculations for example is exacty the same with light propagation. The λ makes the difference \$\endgroup\$ – GR Tech Apr 8 '14 at 11:38

Strictly speaking, to James Clerk Maxwell they're all the same thing. Discarding bizarre quantum witchcraft, Maxwell's equations applied rigourously work from DC to cosmic rays, it's just we normally use approximations when dealing with various slices of the spectrum.

Broadly, there are three frequency ranges we deal with in electronics:

  1. Low frequency: where we can assume the wavelengths of the signals are much larger than both the structures we use to transmit them and the devices we use to process them. In this case, the signal generally stays confined to the conductor, and we also don't have to worry about reflections in circuits.

  2. Radio frequency and microwave: There's essentially two cases here. From a few MHz up to around 1 GHz or so, the wavelengths are on the same scale as the lengths of conductor we use to ferry them around. In this case, we have to start to worry about reflections and applying our transmission line equations. Interestingly, the signals involved on the higher end of the scale don't travel in the conductor - for example, a signal on coaxial cable is predominately travelling as a wave in the gap between the core and the sheath. This is why the choice of cladding material can change a cable's velocity factor. Above 1GHz or so (microwave scale), things start to get more annoying, because your wavelength starts to approach the scale of the devices you use to process the signal. This usually requires serious FEM modeling to address.

  3. Optical frequency: Once the frequencies get high enough, the wavelengths are so small that we can start using optical approximations like Snell's law. On this scale, transmission is actually easier than RF calculations-wise, but the engineering that goes into device engineering is much harder.

All three cases above agree with Maxwell's equations, they just use their own simplifications where necessary.

EDIT: I completely missed the second part of your question. I suppose, yes, theoretically you could modulate light up to the shannon limit, but the electronics to achieve that don't exist and might never exist due to the physical challenges involved. Most fiber systems use something called Wavelength Division Multiplexing. Essentially, each channel is assigned a wavelength (a color) and then pulsed to encode a digital signal. This way, you can pack many channels onto a single fiber. You might think of this as the fiber equivalent to frequency division multiplexing. Previously, only one wavelength was used, but the invention of a black magic device called an Erbium Doped Fiber Amplifier made WDM possible.

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  • \$\begingroup\$ Thanks. So you are saying that in an optic fibre data is represented by pulses of light, the duration of each pulse lasts for many periods of the light wave at a fixed frequency (and many frequencies are combined as different channels). Why isn't communication in a wire done in a similar way, i.e. stop and start an AC signal to generate 'pulses'? \$\endgroup\$ – Flash Apr 8 '14 at 8:24
  • \$\begingroup\$ Because we can directly synthesize the waveform we need, either as a square wave train or as a root-raised-cosine pulse (that is, we can turn these things on and off within a single cycle). We can't do that with light, at least not yet. You're right though - if you wanted, you could take a 100MHz sine wave and switch it on and off to transmit a message. There's a couple reasons you wouldn't want to: you'd have to deal with transmission line behaviour (see point #2) and it'd probably generate radio waves that you'd have to shield from escaping. \$\endgroup\$ – Peter Apr 8 '14 at 8:35
  • \$\begingroup\$ Ok, so to achieve the Shannon limit we would have to invent a way of generating the electromagnetic signals we need (like a square light wave) which currently can't be done? But if I am calculating the theoretical maximum capacity of a fibre cable, the bandwidth of optical frequencies is still the right number to use, right? Or am I grossly overestimating the theoretical capacity? \$\endgroup\$ – Flash Apr 8 '14 at 8:39
  • \$\begingroup\$ I believe that is correct if we consider only classical E&M. However, there's quantum effects in the real world, and I must admit I don't know enough about QM to quantify them. That Physics SE post you linked to in the other comment does a good job of going over that. \$\endgroup\$ – Peter Apr 8 '14 at 8:47
  • \$\begingroup\$ @Andrew/freyyr: We've already reached the Shannon limit for optical networking. WDM frequencies are in the 1500nm range, with a channel spacing of 12.5, 25, 50 or 100 GHz. Guess which spacing is required for reliably transporting a 10Gbit Ethernet signal? Moreover, 40Gbit Ethernet is actually 4 wavelengths because we can't pule a 1500nm laser that fast. \$\endgroup\$ – RJR Apr 10 '14 at 3:37

The frequency in the wire is the rate at which the electrons are being pushed back and forth. The frequency in an optical communications sense has two meanings. First is the frequency of the light, which in all normal systems does not change. Second is the frequency at which the light is modulated on/off. As for electromagnetic analysis in the optical case, it is the frequency of the light that counts. The light is treated as an electromagnetic wave to analyze diffraction and reflection and absorption.

Current in a wire is often called a phenomena of the sea of free electrons in the metal. In an optical cable I suppose you could think of the sea of photons as the medium for carrying information.

There can be more than one frequency or color of light and they are separated by filters or little diffraction gratings. Similarly there can be more than one frequency in a wire that are separated by filters in a receiver.

For the Shannon limit, not the same. Max frequency for the wire, max switching rate for the optical cable. (Given a theoretical limit of detection, like photon detection, they can be thought of as the same. Otherwise it is the max modulation rate. But I just went over my head, so I'm sure a communications expert will step in.)

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  • \$\begingroup\$ For the Shannon limit, not the same. Are you saying that it is theoretically impossible to achieve the Shannon limit if it's calculated using the bandwidth of light waves that travel through it? Or is it just that we don't have the technology to generate the right electromagnetic signals? I think this question physics.stackexchange.com/questions/56240/… implies you can calculate the Shannon limit like that? \$\endgroup\$ – Flash Apr 8 '14 at 8:27
  • \$\begingroup\$ You're correct - it's just EM, so the shannon limit applies if you can modulate light fast enough. The thing is, at the moment, we have no way of doing so with current tech. There's some interesting ideas around this though, particularily in the intermediate terahertz region where there's been some success using weird materials to do stuff like frequency shifting. That stuff's way above my head though. \$\endgroup\$ – Peter Apr 8 '14 at 8:40

It is worth pointing out that in almost every way, optic transmission of encoded signals depends on first developing an electrically encoded signal to drive the output of an LED or laser diode. The big advantage of fiber optics come from the basic property of light that, provided that two waves are at different wavelengths, they do not interact with each other at all.

It is often convenient to think of voltage as a pressure source - in this way, you can see that there is only really one instantaneous point of information to be tracked, and that is whatever the current voltage or 'pressure' in the line is at any given time. From there, computational wizardry and a decent synch signal can cram rather quite a bit of usable information into that single point of information. Eventually, though, we hit whatever the current state-of-the-art limit there is on how fast we can pull that signal apart meaningfully.

If we want to pass more, then, fiber-optic multiplexers come into play. Each input on the multiplexer passes a separate, coherent beam into a single line of high quality glass. A demultiplexer on the other side breaks it back apart into the various signals, upon which computational wizardry can then be performed in reasonable blocks. Each channel spits out an exact replica of the electrical signal the created the original light, and no two channels interfered with each other. And, we could cover much, much longer distances before needing a repeater of some sort.

It's easy to get confused. The frequency of the electric signal became the frequency of the intensity of the light signal. That light signal occurs at a wavelength that has nothing to do with the frequency of its on-state, or its intensity.

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  • \$\begingroup\$ It wasn't so much wavelength discrimination that was the issue (filters are relatively easy), it was the amplification problem for links greater than about 100km. Before the EDFA (a type of laser-pumped wideband optical amplifier) was invented, the only way to amplify a signal was to demux the signal into different wavelengths, amplify each channel, then recombine them. For land-based fiber, this is at best expensive and impractical. For undersea fiber, this is essentially impossible. The EDFA reduced the whole works to one component, with no need to demux the wavelengths. \$\endgroup\$ – Peter Apr 8 '14 at 8:23
  • \$\begingroup\$ @freyyr You just blew my mind a little. I just had a little optics zen moment. Thank you for so concisely framing the issue. So whats the next boundary right now? I think someone's working on measuring and encoding angular momentum, right? Or is that over and done with? \$\endgroup\$ – Sean Boddy Apr 8 '14 at 9:06
  • \$\begingroup\$ That is to say, optical angular momentum. \$\endgroup\$ – Sean Boddy Apr 8 '14 at 9:28
  • \$\begingroup\$ I'm honestly not all that up to speed with the next-gen fiber R&D. I just know a bit about the current and past state of affairs because I had a microwaves and fiber class taught by an ex-Bell Labs guy who did a lot of research in that field (he's since moved on to deep black magic terahertz communications stuff). \$\endgroup\$ – Peter Apr 8 '14 at 9:44
  • \$\begingroup\$ Yeah, that stuff is so far over my head I confuse it for open sky. I still get cranky running simulations for power inverters. Still, exciting times ahead and all that. \$\endgroup\$ – Sean Boddy Apr 8 '14 at 10:08

For a rather lucid and up-to-date discussion of the practical limits of information transfer through an optical fiber, with examples, see http://spectrum.ieee.org/semiconductors/optoelectronics/is-kecks-law-coming-to-an-end?

This article includes discussion of something called the nonlinear Shannon limit, which is an extension of Shannon's early work in information theory. This article is aimed at a general audience with technical interest, and does not get bogged down in heavy math.

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1) The frequency of light is the same thing as the frequency of an electrical signal

2) The physical concept of light is not the same as the physical concept of electricity. Light has no rest-mass. Electrons have mass even when stationary. The electric field around a wire, which has a frequency and a wavelength and transmits power, is the near field, and is bound to the electrons, which are bound to the wire. However, the two are related (there is a unified field theory), and Frequency is one of the things that is not just related, but the same.

3) Yes, you can calculate a Shannon limit for optical fibre the same as you can for radio or telephone regulations. You can measure a noise floor (set by losses) and frequency limits (set by material characteristics) and an amplitude limit (set by when the fibre gets so hot that it melts). You could do the same for a piece of wire. But normally Shannon limits are set by regulation or by inter-operability. When you don't have that, you're designing to a price/function trade-off, not to a Shannon limit.

4) For Shannon limits, modulating light is the same as modulating voltage. You might want to do that by switching the voltage On and Off, or 'generating pulses of light'. If you've got more bandwidth and dynamic range (and more money), you might want to try a more complex encoding scheme, with different frequency and different phase, polarization and amplitude.

Once you start talking about single photons (which you probably will when you get the noise floor down far enough), the behaviour of your system will be harder to describe, but the end result is the same.

NOTE: I've responded to your question about optic cables. As I've tried to make clear, the limits of Optical Cable are not the limits of the electronics attached to it. And here is a nice discussion from a couple of years ago about the actual upper power limit of actual fibre optic cable: http://www.iotpe.com/IJTPE/IJTPE-2010/IJTPE-Issue5-Vol2-No4-Dec2010/15-IJTPE-Issue5-Vol2-No4-Dec2010-pp85-91.pdf. The technology has moved on since then, and there are several companies offering photonic power over fibre.

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  • \$\begingroup\$ The noise floor that limits the theoretical data-carrying capacity of optical fiber is not set by losses. It's set by the statistical nature of the photodetection process and it's called "shot noise". \$\endgroup\$ – The Photon Apr 8 '14 at 16:45
  • \$\begingroup\$ Also, the amplitude limit is not due to melting the optical fiber. It is due to nonlinear behavior of the fiber that causes interactions between the signals at different wavelengths. IIRC the limiting effect is typically stimulated raman scattering. \$\endgroup\$ – The Photon Apr 8 '14 at 16:52

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