Temperature and fiber

That's just the temperature coefficient of optical path length in quartz, about 10**-5/K. (Almost all types of glass are near that value--TC_opl = dn/dT + n*CTE--quartz has a high dn/dT but low CTE, whereas other glasses have lower dn/dT and higher CTE.) Most of the time, if you have a fibre application for which that matters, you'll have a much worse time with etalon fringes and scatter than with temperature coefficient.

Cheers,

Phil Hobbs

Bonsoir Phil,

Thanks for this interesting reply.

This view is a little new for me, let me try this reasoning to see if I understand correctly.

I wished to consider the coefficient a 30 ps/km/K.

If we can suppose that dn is relatively less than dL, the speed is yet c =3D 3 x 10e8 m/s. So the 30 ps are given by dL =3D 3 x 10e8 x 30 x

10e-12 =3D 9 x 10e-3 m.

With L =3D 1 km, the coefficient 1/L x dL/dT is 9 x 10e-6, around 10 ppm/ K.

Is it globally correct, or can we improve that? Best regards, Michelot

Every kind of common glass I know about has a TC_opl right around there. Some are a bit more, some a bit less. There are some near zero TCN (i.e. dn/dT) glasses sold for special uses, e.g. Nd:glass lasers, but nothing that is commonly made into fibre. You might be able to get a special run. If you're trying to preserve timing coherence across some big gizmo e.g. a laser fusion system, and really need that low tempco because there's no way to control the temperature everywhere, this might be fine.

If you're building a fibre interferometer or other sensor, you also have to worry about facet reflections and the piezooptic (stress-optic) effects e.g. bend birefringence.

Cheers,

Phil Hobbs

Bonjour Phil,

Thanks for that valuable information.

My study is more ordinary. I just wanted illustrate the impact of diurnal and nocturnal temperature on fibres, in relation to the wander phase noise (slow variations < 10 Hz)

With a delay of 30 ps. km-1.K-1 (even 40 or 50 seems more common),

1000 km on fibres, 20 C of variation, we have a wander of 0,6 to 1 ns. It's not negligible when the specification can recommand around 5 ns during a day. Such a wander can only be fitered (at the electrical level) by a dual memory, with a write timing different on the read timing.

You're right Michelot to be not satisfy in what you read. The good unity is rather "ps.km-1.K-1" or, "ps/(km.K)" or "(ps/km)K-1".

Best regards, Michelot

1000 km? If this is a long-distance application, a lot of those are going to be underground, so diurnal variations would be small for that bit. If you have a big coil in the lab--and 1000 km is a pretty big coil--it's easy to temperature control.

Cheers,

Phil Hobbs

1000 km? If this is a long-distance application, a lot of those are going to be underground, so diurnal variations would be small for that bit. If you have a big coil in the lab--and 1000 km is a pretty big coil--it's easy to temperature control.

Cheers,

Phil Hobbs

Bonjour Phil,

You're right, the variation considered is a little bit too high.

Probably, it would be also the case for submarine cables.

Glurps! I have to be precise.

Between 2 OEO equipments (Optical input, Electrical treatments, Optical output) we can have a trail length of 1000 km. This electrical trail is supported by several physical optical sections and several OOO equipments (all-optic equipments). For 1000 km, we can have e.g.

13 optical sections of 80 km length, with G.653 fiber.

The all-optic equipments can be wavelength multiplexers or switches, optical amplifiers. Probably you know that because it is always subjects of research. For example, remote all-optic wavelength conversions would not seem really solved.

And the electrical trail of 1000 km is necessary (it would be a maximum value) to regenerate the electrical signal (by reamplification, reshaping and resynchronization).

Best regards, Michelot

..remotly configurable (and not locally)...

Best regards, Michelot

Given that you're going to have clock-recovery circuits on all those lines, why does such a slow variation--maximum slope (1/f)(df/dt) of 1 part in 10^13 per second--worry you at all? I'd be much more likely to lose sleep over double Rayleigh scatter and spontaneous emission jitter.

It sounds almost as though you're expecting the whole system to be globally synchronous--is that so? That would be pretty unusual in the telecom infrastructure, but I suppose if you rented a pipe the entire way, you could do anything you wanted with it.

Cheers,

Phil Hobbs

Bonjour Phil,

Right, we have to recover the timing at the interface between the optical server layer and the electrical client layer. This timing is used to recover the data and, possibly, to synchronize the equipment when it needs to treate several signals.

If you're alone, there is no problem. If you have several signals with independant timing coming from different locations, which have to suffer common treatments, you need a centralized synchronization in the equipment. We are not yet in the issue of a synchronous network

How did you find this value?

This frequency drift of 10^-13 /s corresponds to a phase drift of

10^-4 ns/s2, that gives around 10^6 ns for a day (we have perhaps to divide this by 2). This value has to be compared with the time bit of a STM-256 e.g., that is 25 ps = UI for a wavelength of 40 Gbit/s.

If the equipement needs to exchange data between 2 x STM-256 with independant timing, it has to take in charge the long term wander.

But other factors than temperature can affect the phase of an electrical signal.

Best regards, Michelot

Sorry, I misspoke. That number is actually 1 ns*pi/86400 seconds per day, i.e. d(delta t)/dt, rounded up to the nearest power of 10. My point was that this is so slow that any clock recovery circuit would never know it even existed. You could probably lock a rubidium reference to that.

If you're expecting to be able to have picosecond time stability in a network with 1000 km loops (physical or logical), that'll be hard. What's wrong with a bit of buffering? It isn't going to hurt your latency much on a 1000 km link.

Cheers,

Phil Hobbs

Bonsoir Phil,

It's same for me. I start again.

With a delay of 50 ps.km-1.K-1, a fiber link of 1000 km between 2 OEO equipments, only 2 C of variation, we have a wander of 0,1 =B5s (it represents 500 bytes of STM-256).

I agree, it is very small and it is added to other wander factors, as asynchronous mapping, clock noises, phase drifts, power supply noises... Wander are effects < 10 Hz and jitter beyond.

The problem is not the clock recovery circuit that doesn't filter wander. The clock recovered (C0) is used to recover data (D0). And C0 and D0 recovered have same wander than the received signal.

The problem is when the clock recovered is not choose to synchronize the OEO equipment which has to take in charge multiple sink and source interfaces, when C0 is not a synchronization source. In this case, the wander in the D0 data recovered has to be considered, relatively to the wander of the clock that transfers its timing to the output interfaces. And it causes hitless justifications or slips.

Too many hitless justifications are a measure of the network degradation, and slips are disastrous. When the reading pointer is catching up with the writing pointer (or contrary), because only some bytes are missing, all the frame is lost.

System clock can be rubidium, but not PLL to recover the timing in a received signal (I prefer write timing in this case and not clock).

Best regards, Michelot