The nonlinear scattering effects in optical fibers are due to the inelastic scattering of a photon to a lower energy photon. The energy difference is absorbed by the molecular vibrations or phonons in the medium. In other words one can state that the energy of a light wave is transferred to another wave, which is at a higher wavelength (lower  energy) such that energy difference appears in form of phonons [1]. The other

wave is known as the Stokes wave. The signal can be considered as pump wave. Of course, high-energy photon at the so-called anti-Stokes frequency can also be created if phonon of right energy and momentum is available.

There are two nonlinear scattering phenomenon in fibers and both are related to vibrational excitation modes of silica [2–6, 31–33]. These phenomenon are known as stimulated Raman scattering (SRS) and stimulated Brillouin scattering (SBS). The fundamental difference is that, the optical phonons participate in SRS while SBS is through acoustic phonons.

To better understand this nonlinear effects you can consult:

http://www.jpier.org/PIER/pier74/23.07051102.Singh.GS.pdf

or: 

http://www.ijrte.org/attachments/File/v2i1/A0525032113.pdf

or

http://www.rp-photonics.com/brillouin_scattering.html

Hello Steve!

I would like to bring a complementary view to Gerson's summary to capture qualitatively the relationship between the Brillouin gain spectrum and the acoustic wave decay time.

Let's start with the fact that Brillouin scattering comes from scattering of the incident lightwave by the acoustic phonons (in other words, we would say the acoustic waves).  These acoustic phonons do not have an infinite lifetime and will be themselves scattered by impurties in the material provoking a decay of the acoustic wave.  The acoustic wave decay is assumed to be exponential which has the consequence that the spectrum of the acoustic wave is a Lorentzian distribution whose linewidth is the inverse of the decay time.  The spectrum of the acoustic wave is the so called Brillouin gain spectrum. The spectrum of the acoustic wave will define the scattered lightwave shape and in particular its linewidth.  In linear regime, the scattered lightwave intensity will be the convolution between the Brillouin gain and the incident lightwave spectrum.

You can find more details and further references in chapter 2 and 3 of my thesis which is in free access on my Research gate account. 

Should have have other questions or comments, please, do not hesitate to ask me again.  It will be more than a pleasure.

Best regards,

Fabien.

Thank you Mr Gerson.

Actually the first part of my question was explained by Mr. Fabien. It was really very helpful. I have read many papers and i didn't find an such simple explanation about phonon decay and the  physical relation between it and SBS gain.

Concerning the spatial resolution (the second part of of the question) I am trying to find the answer on your PHD. Thank you, it is helpful too.

Best regards.

Hello Steve!

I can probably give you some hints on the spatial resolution part of your question.

First of all, the spatial resolution depends on the pulse length used by any distributed sensor (Rayleigh, Brillouin or Raman based).  It also depends on the bandwidth of the detection electronic circuit but we can keep it aside in the current discussion.  Due to the physics  of the interaction, it can be understood that the measured information is the result of the integration over the pulse length.  As a rule of thumb, it can be said that a 10ns pulse corresponds to a 1m spatial resolution.  When you want to measure accurately small details (short temperature or strain events), your need to reduce the pulse length.

Reducing the pulse length has an important consequence: if you narrow a pulse in the time domain, you will expand its linewidth in the frequency domain.  A pulse of 1000ns (100m spatial resolution) has a linewidth close to the one of a CW signal.  If we consider a Brillouin sensor, it means that it is negligible compared to the Brillouin gain linewidth.  It also means that the measured  gain linewidth will be equal to the Brillouin gain linewidth which is about 50MHz or larger.  If we reduce the pulsewidth, the pulse spectrum will become wider.  For a 1m spatial resolution, the pulse spectrum width iwill be equal or larger to the Brillouin spectrum.  Each spectrum component interacts with the acoustic phonons and each scattered component of the pulse has a spectrum equals to the Brillouin spectrum,  It can be pictured as a convolution of the pulse spectrum with the Brillouin spectrum. It has two consequences:

Both affect the peak frequency measurement resolution and hence temperature and strain measurement resolution. That is the reason why the improvement of the spatial resolution can be detrimental to the overall measurement accuracy.  This is also how the spatial resolution and the acoustic phonon lifetime "relates".

I hope this helps.  Of course you will find details and a quantitative discussion in my thesis as well as in the associated references.

Again, feel free to ask for clarifications.

Best regards,

Fabien.

Thank you for your help, I'll read your thesis completely, it is a simple way to explain SBS. If I find any question I'll not be shy to ask.

Thank you very much.

My regards

Can we say that limiting the spatial resolution to 1 m in the conventional systems (without using RZ codes or other techniques) leads to miss-interaction between acoustic wave and the spectrum component, because the pulse width becomes10 ns smaller than the acoustic life time or  the pulse spectrum width will be equal or larger to the Brillouin spectrum?

So this is the reason of limiting the spatial resolution to 1 m. Ofcourse this will lead to lower SNR as you said but we suppose that we are working at short distances (less than 1 km) and spatial resolution is more important than SNR.

Best regards.

Hello Steve,

It is really a scattered light spectrum broadening.

What the physics tells us is that once the pump lightwave spectrum becomes equal or larger than the acoustic phonon bandwidth (i.e. when the pulse narrows down to 10ns), the scattered light spectrum is much broader (>200MHz) making difficult the detection of small frequency changes, which are the purpose of the Brillouin sensor.  A 1 MHz frequency change corresponds in a standard telecom fiber to a temperature variation of 1°C  and / or a strain variation of  20microstrain.  A spectrum of 200MHz or more makes difficult the detection of a small peak frequency change, in particular if the SNR is poor.  Actually both SNR and spectrum bandwidth interplay to affect the frequency resolution.

The "10ns limit" should not be viewed as a threshold that tells above it is good and below it is bad.  Itis a smooth transition from high frequency resolution to low frequency resolution and the cause is the Brillouin gain spectrum (and hence the acoustic phonon lifetime which is its time domain equivalent).

I hope that clarifies a little bit more your doubts.  Do not hesitate for more questions.

Best regards,

Fabien.

Thank you very much, it was very helpful

Best regards.