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A fiber optic receiver needs to convert the incoming optical signal to an electrical signal so it can be sampled. How is this achieved in practice?

I believe a photodiode is used, but would a photodiode not have the problem that the frequency of the incoming optical signal is too high? Such high frequencies cannot propagate through the conductors.

This is what Wikipedia says:

Coherent receivers use a local oscillator laser in combination with a pair of hybrid couplers and four photodetectors per polarization, followed by high speed ADCs and digital signal processing to recover data modulated with QPSK, QAM, or OFDM.

If I am not mistaken, this implies that you interfere the incoming signal with one of a plurality of coherent optical reference signals such that each of said reference signals is equal to one possible received signal. This way you can detect whether the input signal is one of the expected values given the interference pattern. Another possibility is to have each reference signal be such that it destructively interferes with the incoming signal. But wouldn’t that limit the modulation schemes you can use? Can you use frequency modulation over fiber optics?

Any references, block diagrams etc. will be greatly appreciated.

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    \$\begingroup\$ At most data rates and using AM, the output of the photodiode is amplified and directly converted to data. \$\endgroup\$ – Andy aka Jul 5 at 16:38
  • \$\begingroup\$ @Andyaka Yes, I can see how this can work for ASK. \$\endgroup\$ – user110971 Jul 5 at 16:41
  • \$\begingroup\$ If the frequency of the signal is too high to propagate through normal conductors, how do you think they got into the fiber in the first place? \$\endgroup\$ – Elliot Alderson Jul 5 at 16:56
  • \$\begingroup\$ @ElliotAlderson The low frequency signal (GHz) in the metal conductor turns into a high frequency (THz) optical signal with a diode. \$\endgroup\$ – user110971 Jul 5 at 16:58
  • \$\begingroup\$ Does the fiber carry a higher data rate than the conductor(s) leading up to it? If so, how does that happen? How does the extra data get into the fiber if it didn't come from plain old conductors in the first place? Is there magic inside the "diode"? Be sure you don't confuse the modulated data rate with the carrier frequency. \$\endgroup\$ – Elliot Alderson Jul 5 at 17:08
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I believe that your "can't be carried by wires" notion stems from the fact that fiber optics can carry signals at data rates that would be wildly uneconomical to carry with wires -- but it could still, in theory, be done.

The frequencies involved in the modulated data cannot be higher than can be carried on a wire, because the signals are generated electronically before being modulated.

So yes, some pretty impressive high-speed photodiodes are used, and then some pretty impressive high-speed electronics.

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  • \$\begingroup\$ What I meant is that optical signals in the THz range can’t be carried by conductors. You need to shift the frequency down by some method. I am asking what said method is. \$\endgroup\$ – user110971 Jul 5 at 16:53
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    \$\begingroup\$ "signals in the THz range" if that's really your question, put it in the question. Thinking in terms of the frequency of the light is probably not helpful at all -- think in terms of photons and wavelength. Photons smack into the junction of a photodiode, shake electrons loose, and those electrons get detected. It's a quantum-mechanical process involving photons as particles -- not a continuous-time, continuous-wave, or continuous-anything process involving THz signals. \$\endgroup\$ – TimWescott Jul 5 at 16:59
  • \$\begingroup\$ It is in the question. That’s why I said optical signal and that’s what the entire Wikipedia quote discusses. Sorry if it was not clear. I’m specifically asking about the optics between the fiber optics cable and the photodiode. \$\endgroup\$ – user110971 Jul 5 at 17:04
  • \$\begingroup\$ Well, "optics" is the key word. It's photon handling until one of them collides with an electron in a PN junction, and then it's electronics, at the modulation speed, not at the speed of oscillation within a photon. \$\endgroup\$ – TimWescott Jul 5 at 17:24
  • \$\begingroup\$ Direct sampling of the optical signal works only with ASK modulation. If you have a coherent system wherein there is additional optical modulation, e.g. an interferometer, after the transmitting diode in order to vary the phase as well as the amplitude of the optical signal, it’s not so straight forward. You need to have some optical system to handle this before the photodiode. Such is the case for QAM, QPSK, and OFDM. My question pertains to said optical system. See the Wikipedia link for more information. \$\endgroup\$ – user110971 Jul 5 at 17:31

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