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laser communication experiment

Discussion in 'Mathematics and Physics' started by PG1995, Dec 21, 2013.

  1. PG1995

    PG1995 Active Member

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    Hi

    Please note that I'm asking most of the following questions in the context of laser or FSO communication which is explained here.

    Q1: Suppose two laser beams being used for the purpose of communication cross each other vertically. Would there be any interference? Would the information contained in those laser beam be affected?

    Q2: We know that radio communication is affected by electromagnetic interference. I think this interference comes into play at the receiving end such as antenna where both information radio signal and interference radio signal is picked up by an antenna and this combined reception of both signals produces a new noisy signal at the receiving circuit. Do I have it right?

    Q3: What kind of laser is mostly used for optic fiber communication? I believe it's a laser diode which provides continuous laser beam and not a pulsed one, right?

    Q4: Which laser is most popular for optic fiber communication in terms of wavelength? Is it red one? Can they also use infrared laser in optic fiber communication? Ultraviolet lasers also exist so why not use them because they can provide higher data rate than visible light lasers?

    Thank you for the help.

    Regards
    PG
     
  2. JimB

    JimB Super Moderator Most Helpful Member

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    I am not a laser expert, but answering in general terms...

    Q1 Laser light is just another form of electromagnetic radiation in a narrow beam.
    Two beams of "ordinary" electromagnetic radiation (ie Radio) do not affect each other, so why should a laser beam.

    Q2 Assuming a perfect receiver* (which does not exist), the only way for the "electromagnetic interference" to affect the wanted radio signal is for it to have some component at the same frequency as the wanted signal.
    So, yes you have it right.

    If we were to consider a real world receiver* which is not perfect, then unwanted signals at other frequencies than the wanted signal can be a problem.

    Q3 I think a laser diode. Basically continuous, but the light will be pulsed by the modulating signal.

    Q4 Most popular? I do not know.
    The rate at which you can modulate the laser light will give the bandwidth limitation, rather than the "carrier frequency" of the laser light.

    JimB

    * Note to remove any possible ambiguity.
    When I wrote this I was thinking about a radio receiver rather than anything to do with lasers.
    The OP spoke of "radio communication", so it was in that context that I replied.
     
    Last edited: Dec 21, 2013
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  3. ronsimpson

    ronsimpson Well-Known Member Most Helpful Member

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    What color of light has to do with what colors of light the fiber can handle. Some types of light do not pass well through fiber.
    Also the price of lasers. If a blue is 10x more money then a red then the answer is easy.
    The efficiency of a laser is important. If 100 watts into a blue laser does the same job as 10 watts into a red laser then the answer is easy.

    The bottom line is cost/mile.

    The 'color' is not a problem to data rate until the data approaches the color frequency. (not there yet)
     
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  4. dave

    Dave New Member

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  5. rumpfy

    rumpfy Active Member

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    Lets go to optical communication first.
    Optical transmission requires a low loss medium and in fibre, the optical energy is guided through a core material of pure silica glass. Surrounding the core is another layer of silica glass which has a higher refractive index compared to the core, and by total internal reflection, the wave is caused to travel through the core. In modern fibre, the core diameter is around 8 to 10 micron and this fibre is known as single mode fibre. The cladding is around 130 micron from memory. In older fibre designs, the core diameter was around 50 micron with a cladding of 130 micron. This fibre would support multiple modes of propogation and was known as graded index fibre. Because of the multiplicity of modes, the light energy would travel along different paths and due to the time difference of different waves travelling along the fibre, the time difference would destroy the fast rise time of the light pulses. This pulse spreading would reduce the bandwidth of the fibre channel. The single mode fibre does not have this problem but it has a chromatic dispersion problem, so that different wavelengths of light in the original light source would have different transit times so there is still a pulse spreading problem with single mode fibre. This problem is overcome by having a very high degree of spectral purity in the lasers.
    The attenuation of the laser light is wavelength dependent. Initially, the usual led's were used at say 850 nanometre. But there was a low loss transmission window at 1310 nm, and many (most) of the optical systems used today operate at this wavelength. There is a further window at 1550 nanometre, and this window is invariably used for long distance undersea cable systems. With the 1310 nm window, unrepeatered lengths of 40 to 50 kM can be achieved. This is fine for terrestrial systems. For the 1550 nm window, the maximum unrepeatered lengths are around 150 km; but what is being done is that a repeater is actually a laser pumped amplifier using a spliced piece of special fibre. The 'amplifier' is pumped from the terminal end at some wavelength and amplifies the required signal.
    The system does not use continuous lasers, the lasers are pulsed at the bit rate of the system. It is really like morse code.
    For Q1, one beam would have to modulate the other, but in a guided media system, it is hard to see how this can happen. It is possible to couple fibres together using special gratings which clamp over the cladding.
    For Q2, attenuation is generally not important in optical communication systems; it is the reduction in bandwidth due to chromatic and time dispersion effects.
    Q3 is answered.
    Q4 is answered too, but with industrial type systems like the optical link in a audio amp or such, the wavelength can be anything that works because the link lengths are quite short. There is an advantage in using cheap 850 nm led's with large diameter optical fibre (even plastic fibre), because a lot of the light energy can be coupled into the core of the fibre. Visible light is in the 400 to 700 nm wavelength so optical communication is well into the infra red band.
     
    Last edited: Jan 13, 2014
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  6. ericgibbs

    ericgibbs Well-Known Member Most Helpful Member

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    hi PG,
    As a side topic
    In laser communications that are operated through the medium of 'air', particular frequencies are used for the laser, which match the lower absorption level 'windows' through, air at those particular frequencies.
    E
     
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  7. PG1995

    PG1995 Active Member

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    Thank you, JimB, ronsimpson, rumpfy, and Eric for your help.

    rumpfy said, "The system does not use continuous lasers, the lasers are pulsed at the bit rate of the system. It is really like morse code."

    In my humble opinion they use laser diodes which are not pulsed lasers. The continuous laser light is pulsed by the modulating signal as JimB also mentioned in his post above. What do you say? Please let me know.

    Helpful Links:
    1: http://en.wikipedia.org/wiki/Optical_modulator
    2: http://en.wikipedia.org/wiki/Electro-optic_modulator
     
  8. rumpfy

    rumpfy Active Member

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    PG, you may well be right.
    When the optical communication systems were first introduced to Australia, the intended bit rate was 144 Mbit but the first systems were actually running at 565 Mbit. These systems were pulsed lasers. however, the comment I made about systems requiring a high spectral purity, could imply that the laser light is filtered through a grating and the light then modulated for high bit rates. Siemens have announced some years ago that they had developed 10 Gbit systems, and it could be true that modulation is used in the high bitrate systems. I think 2.048 Gbit systems are now being used routinely.
    Dont forget that the 'State of the Art' is changing rapidly, and what is true one year is wrong the next. The first fibres in commercial use were multi mode step index, only to be followed by multi mode graded index within a couple of years, and then to be followed by single mode step index types.
     
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  9. steveB

    steveB Well-Known Member Most Helpful Member

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    PG,

    rumpfy provided a nice summary of some important and relevant facts about fiber optic communications. Overall, your questions are probably answered here, but I can take a run through the questions to give just another viewpoint, in case this helps you. I am likely repeating some of the information given above.

    Generally, in linear media, you do not expect laser beams to interfere (EDIT: on rereading my response I'm realizing we are both using the word "interfere" to mean cross coupling with each other. However, a typical use of the word "interference" in laser terminology is related to "interference patterns" which are the addition of fields which creates constructive and destructive interference. Obviously, laser beams can interfere in this way in linear media, but I assume this is not what you are asking.). Separate signals will be independent based on direction, polarization, spacial mode pattern and frequency, and will not interfere, in principle. However, the real world does not perfectly obey our idealization, and there can be cross coupling of signals. Separation by polarization and spacial modes is tricky and there is often cross coupling due to fiber imperfections. However, frequency and direction tend to be very good ways to separate different signals. The only way for these separate signals to interact is through one of the many types of non-linear processes that are well known. To keep these effects insignificant one just needs to make sure that the signal power is low enough to not induce significant nonlinearity. However, nonlinearity can happen at relativly low power because the fiber confines the light to a very narrow cross sectional area, which makes the light intensity very very high, relative to the actual power level.

    EDIT: I just realized that I should have mentioned that Rayleigh backscattering can cause a signal going in one direction to be reflected into the backward direction. This is a distributed reflection, but it can cause interference. Since Rayleigh backscattering is so prevalent in fibers, liquids and gases, it is worth mentioning separately.

    That is certainly one way, and probably the most significant way in practice.

    As far as "pulsed" versus "not-pulsed", laser diodes are usually run in continuous mode, but they can be pulsed in principle for different applications. The continuous laser is then modulated. You could say that the modulation is a form of pulsing too, but this is just a terminology issue. The idea of "pulsing" a laser is usually meant to describe the trick to get lasing in a material that does not allow continuous lasing very easily (such as a 3-level system)

    Yes, laser diodes are the most practical lasers used in communications. They are specially designed for this use, and come in varying degrees of sophistication in design, depending on the application.

    It all depends on the fiber type. There are plastic fibers designed to be used with red light (or visible light). However, most glass based fibers (silica based being the most practical and common) work best in the infrared. The first fiber optic systems used 800-900 nm infrared because the sources and detectors were available at that wavelength and were cheap and reliable. Also, due to too much water content in the fiber, longer wavelengths would have had higher loss anyway. But, intrinsically, once water impurities are minimized, the longer wavelengths are better, but eventually molecular absorption (which can't be removed) takes over and the fiber becomes opaque if wavelength is too long.

    By the way, ultraviolet light would have too much Rayleigh scattering.

    So, the 800-900 nm range is not optimum for silica fiber. Silica has two interesting bands for communications. The lowest loss wavelength is 1550 nm, and the wavelength of zero material dispersion is near 1300 nm. Fiber design can also affect dispersion, so it is possible to design dispersion-shifted fibers that have zero dispersion at 1550 nm and lowest loss at 1550 nm, with 0.2 dB/km loss. But, these are just details. Basically, 1310 nm was the earlier technology that became practical with 0.4 db/km loss and essentially zero dispersion in single-mode fibers. Then, good sources were developed at 1550 nm. At that point, very narrow linewidth lasers were developed to minimize the dispersion problem. So, depending on the application, those are the better wavelengths to use, particularly over long distance. However, 800-900 nm can still be used in cheap multi-mode fiber links over short distance.

    In the last couple of decades we've seen the development of erbium doped fiber amplifiers and Raman amplifiers, both which work great at 1550 nm.
     
    Last edited: Dec 23, 2013
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  10. PG1995

    PG1995 Active Member

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    Hi

    Could you please help me with these queries related to optical fibers? Thank you.

    Regards
    PG
     

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  11. rumpfy

    rumpfy Active Member

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    Q1; I dont really know exactly what is the answer, BUT I do know who will be able to tell us. My guess is that there is a power loss associated with the light reflection at the interface between the core and the cladding. With higher order modes there are more reflections, so these higher order modes get attenuated more.
    Q2 and 3; The attenuation is given as a logarithm to either the base 10 OR to the base e.
    A decibel is the power ratio using the base 10. A 'NEPER' is the power ratio using the base e.
    EQ 4.69/4.70, they are calculating in neper and the conversion of neper to dB is the ratio 4.34: ie 10 log e = 4.34.
    The factor 0.23 is 1/4.34
    Again, I hope, but I am sure, our 'teacher' will check all this for us.
     
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  12. steveB

    steveB Well-Known Member Most Helpful Member

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    PG,

    rumpfy's answers are good. Certainly Q2 and Q3 are correct and seem complete to me. Q1 is actually a question that could be answered in very great detail. You can keep things simple by considered a "ray" model for the fiber modes, and in that case, talking about reflections at the core cladding interface is a good way to picture it.

    For Q1, you can also go in great depth about field based modes which are discrete field modes with independent propagation constants. These modes are orthogonal and do not cross couple in an ideal world. But due to fiber imperfections, stresses and bending, the modes can couple. Each mode will be described by it's own independent loss per unit length which can be viewed as a constant number to keep things simple. That loss is made of material loss (absorption and Rayleigh scattering) and bending losses (micro bending and macro bending), as well as imperfections of various types. Each mode can have different values for each of the components of the loss and hence the total loss can and typically is different for each mode. So, the very complex loss behavior of the total power in a multimode fiber can be simplified conceptually if one realizes that the total power is distributed across a finite number of guided modes each with it's own loss. The model can be made a little more complicated by considering mode coupling where the power in one mode can also cross couples over to the other modes.

    Basically, the real answer to Q1 is very complex, and requires an electromagnetic field approach to the fiber modes and a detailed understanding of the various effects in fibers including, micro bending loss, macrobending loss, scattering, absorption, mode coupling, guided modes, radiation modes etc. The ray picture of light propagation can simplify the understanding of this to some extent but can also make the behavior seem mysterious, and is not altogether comprehensive enough to include all observed effects.
     
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  13. PG1995

    PG1995 Active Member

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    Thank you, rumpfy, Steve.

    Q1: I find it quite weird that attenuation decreases as the length of the fiber increases because intuitively it should have been that attenuation increases as the length increases.

    Q2: I'm still confused. Are you saying that the "log" in eq. 4.69 is a natural log. Please have a look here.

    Thanks a lot.

    Regards
    PG
     

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  14. steveB

    steveB Well-Known Member Most Helpful Member

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    Q1

    Why should attenuation increase with length?

    The modes with higher loss will lose their power faster and eventually won't have much power further down the line. This means that the only modes with power are the ones that had lower attenuation. It's kind of like "survival of the fittest". Only the modes with lowest lost can propagate the long distance, so further down the line the attenuation is lower because the surviving modes have the lowest attenuation.

    Q2

    Eqn. 4.69 uses base 10 log. But, 4.68 is using natural log/exponent. For natural logs we have units of Nepers per unit length and for base 10 logs we use dB per unit length, with that factor of 10 needed for the Decibel. The factor of 4.34 (0.23) allows conversion between the two units.

    Note that eqn. 4.69 is nothing mysterious. It is just the definition of dB and then you divide by the total length to get attenuation as loss per unit length.
     
    Last edited: Jan 13, 2014
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  15. PG1995

    PG1995 Active Member

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    Thank you, Steve.

    Q1: I think I was picturing it wrongly. Attenuation, A, is defined to be dB/km. It means that if loss during first kilometer length is 2x and then during second kilometer length is x, then combined loss for 2km is 3x, and average loss is 3x/2=1.5x/km. I hope I have the basic idea correct.

    Q2: Eq. 3.11 for attenuation, A, was derived systematically. But is this really correct that eq. 4.69 is just the definition of attenuation constant? I don't see it being derived from anything.

    I understand eq. 4.70 now.

    Regards
    PG
     

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  16. steveB

    steveB Well-Known Member Most Helpful Member

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    Isn't 3.11 and 4.69 the same equation? Why are you ok with 3.11 and not with 4.69? This is just the definition of decibels. I'm sure you've seen it before? Yes? No?
     
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  17. PG1995

    PG1995 Active Member

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    Hi

    I do understand the definition of decibels. The definition of decibel is dB=10log(X2/X1). Eq. 3.11 is derived systematically but on the other hand eq. 4.69 is just thrown into without any derivation. They are the same equations but in one equation we have "A" and in the other there is attenuation constant, alpha. I think I'm just confused and will get around it soon. Thanks.

    Regards
    PG
     
  18. steveB

    steveB Well-Known Member Most Helpful Member

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    OK, it may be that one reference treats it better, but in fairness to the second reference, the basic formula is shown in several places (eqns. 4.59 and 4.65).

    Keep in mind that it is often best to use several different sources to get the best understanding of a new or difficult topic. This is true of text books and even more true of journal and conference papers in the literature.
     
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  19. rumpfy

    rumpfy Active Member

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    Thank you steve.
     
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  20. PG1995

    PG1995 Active Member

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    Thank you, Steve.

    Both references are from a same book, Fiber-Optic Communications Technology by Mynbaev, Scheiner . The top excerpt was taken from chapter #3 and the bottom one was from chapter #4. The material from chapter #4 had reference to a section from chapter #3 so I included both.

    In eq. 4.62, attenuation, A, has unit of 1/km. The attenuation coefficient alpha also has unit of 1/km. Therefore, I think, we can say that "A" in eq. 4.62 is just the definition of "alpha", or A=alpha. Thanks.

    Regards
    PG
     
  21. PG1995

    PG1995 Active Member

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    Hi

    Does it mean that when a single ray enters into a multimode fiber, it spreads into multiple rays where each new ray follows a different path? Could you please let me know? Thanks.

    Regards
    PG
     

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