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*To*: STDS-802-3-10GMMF@xxxxxxxxxxxxxxxxx*Subject*: Re: [10GMMF] task 4 Oct. 4th call -- Polarization*From*: Vipul Bhatt <Vipul_Bhatt@xxxxxxxx>*Date*: Mon, 11 Oct 2004 14:12:06 -0700*Importance*: Normal*In-Reply-To*: <6BD8B3E713606246B10066BF3C312FFC0D0A97@blnse201.eu.infineon.com>*Reply-To*: Vipul_Bhatt@xxxxxxxx*Sender*: owner-stds-802-3-10gmmf@xxxxxxxx

Thank you, Stefano. This is helpful. If I have any questions later, I will get back to you. Vipul > -----Original Message----- > From: owner-stds-802-3-10gmmf@IEEE.ORG > [mailto:owner-stds-802-3-10gmmf@IEEE.ORG]On Behalf Of > Bottacchi.external@INFINEON.COM > Sent: Monday, October 11, 2004 6:07 AM > To: STDS-802-3-10GMMF@listserv.ieee.org > Subject: Re: [10GMMF] task 4 Oct. 4th call -- Polarization > Importance: High > > > Vipul, > > thank you for your pertinent comments. Let me try to explain why > polarization works in multimode fiber when offset is taken into account. > I completely agree with you on the fundamental mode solution for the > axial symmetric fiber. > > Let us continue to assume the fiber has rotational symmetry and the > vector wave equation reduces to the well known Helmoltz scalar equation. > According to weakly guiding approximation WGA, all modes belonging to > the same mode group exhibit the same propagation constant and therefore > have the same group velocity (this is strictly true within the WGA, but > it is not relevant for the discussion). > > One important statement is that the whole mode solution (geometrical > mode structure) refers to the assumed polarization of the electric > field. It is customary to assume the x-axis oriented along the linear > polarization of the electric field. At this point, if mode excitation > would be still axial symmetric (OFL), any polarization effect would be > observed. "Everything" would be the same independently from any > azimuthal position. > > What is important now is the result of the overlapping integral at the > launching section to calculate the amount of intensity transferred to > each mode group. If the excitation is eccentric, defined through some > offset coordinate, the coupling coefficients will depend on the relative > orientation of the linear polarization (reference x-axis) and the > (r,phi) coordinate of the offset center. > > This is the important point. For a given mode group assembled by several > modes with the same propagation constant, the amount of intensity > transferred between two offset fibers depends on the relative > orientation of the offset respect to the light polarization. In > addition, at the same radial coordinate region belong different mode > groups with different group velocity. > > Assuming the profile is not optimized we experience a polarization > induced DMD which leads to pulse distortion after few hundreds of meters > at the 10GbE scale. > > I hope to have been clearer in this picture description. This is not in > contradiction with well known mode theory. It just consider one > different situation where polarization, offset and profile optimization > play the role in getting pulse distortion. > > Please, I apologize if something was missing or uncorrected. Looking > forward in receiving your comments, > > Best regards > > Stefano > > -----Original Message----- > From: owner-stds-802-3-10gmmf@IEEE.ORG > [mailto:owner-stds-802-3-10gmmf@IEEE.ORG] On Behalf Of Vipul Bhatt > Sent: Donnerstag, 7. Oktober 2004 20:34 > To: STDS-802-3-10GMMF@listserv.ieee.org > Subject: Re: [10GMMF] task 4 Oct. 4th call -- Polarization > > > I appreciate the work done by Joerg and his team, and the polarization > experiment report seems well organized, but...polarization penalty for > multimode links? I hope someone can educate me, because the way I see > it, polarization shouldn't make any difference to DMD. > > I am starting with the assumption that only the input polarization was > changed, and everything else was unchanged -- i.e., the excitation > region didn't shift, the input field intensity pattern remained the > same, etc. > > DMD changes only if the power distribution among mode groups changes. A > mode group consists of one or more LP modes having nearly identical > propagation constant. My contention is that a change in input > polarization not only leaves power in each mode group intact, but it > also leaves power in each mode intact. > > We normally characterize a mode as LP(m,n), where m is the azimuthal > number and n is the radial number. But actually, each LP(m,n) mode > consists of two (if m=0) or four (if m>0) distinguishable modes having > the same propagation constant -- distinguishable because of two possible > polarizations and two possible azimuthal orientations, sine and cosine. > For example, mode group 6 consists of LP(5,1), LP(3,2) and LP(1,3), and > has a total of 12 distinguishable modes. > > Power in each LP mode is the sum of the powers carried by its > distinguishable modes. These distinguishable modes share the power, the > proportion being decided by how the input polarization aligns with the > polarization axes of the distinguishable modes. When the input > polarization direction is changed, the proportion of power among these > distinguishable modes changes, but the total power in an LP mode remains > the same. Therefore, DMD should not be affected by changes in input > polarization. > > Please tell me what I am missing. Thank you. > > Vipul > > Vipul_Bhatt@ieee.org > +1-650-941-6290 > > -----Original Message----- > From: owner-stds-802-3-10gmmf@IEEE.ORG > [mailto:owner-stds-802-3-10gmmf@IEEE.ORG]On Behalf Of yusun@IEEE.ORG > Sent: Wednesday, October 06, 2004 9:04 AM > To: STDS-802-3-10GMMF@listserv.ieee.org > Subject: [10GMMF] the minitues of task 4 Oct. 4th call > > > Dear colleagues, > > I attached the minutes of tele-conference of task 4 group on Oct. 4th. I > would appreciate your comments. The proposed next task 4 meeting is Oct > 11th, 2004 (Monday), 11am east coast, 8 am west coast and 4 pm UK. > > Best regards, > > Yu

**References**:**Re: [10GMMF] task 4 Oct. 4th call -- Polarization***From:*Bottacchi.external

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