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Mike, Where does 56 byte packets come from? I thought that 64 octets was the minimum MAC frame size. According to my calculations, using an 800 octet MAC frame only makes 0.07 dB of difference to the 10Log(Qo/Qi) value. In my opinion, we are better off using the smallest MAC frame size to base the calculation on. I have added a slide on this to my draft presentation for Thursday (assuming the MMF Ad Hoc chair accepts the presentation request). Regards, Pete Anslow | Senior Standards Advisor From: Mike Dudek [mailto:mike.dudek@xxxxxxxxxx] Sending to both reflectors again. Haven’t we missed something here. If we are “trashing” the FEC frame when we have uncorrectable errors so that we don’t have an MTTFPA problem won’t that create close to 10 Ethernet packet (frame) errors with 56Byte packets with minimum IPG whenever we have an FEC frame with uncorrectable errors (The only time we expect to get any errors at the MAC service interface) Ie although the scrambler doesn’t multiply the Ethernet FER this “trashing” will. The multiplication factor is however very much a function of the packet size. Varying from this max of 10 for the shortest packets with minimum IPG, down to approx. 1.4 for a 1.5Kbyte packets with minimum IPG. What is the appropriate Ethernet packet size for doing this comparison. This type of issue was covered in the referenced papers but the “trashing” will definitely change the details. ? (Without this FEC for a given BER the Ethernet frame error rate gets smaller for smaller packets). With the FEC the Ethernet frame error rate is getting larger for smaller packet size due to this multiplication. Note that the 10GBASE-T precedent of using 800 Octet frames would result in a much higher allowed BER than if we use the 56Byte packet comparison. Mike Dudek QLogic Corporation Senior Manager Signal Integrity 26650 Aliso Viejo Parkway Aliso Viejo CA 92656 949 389 6269 - office. From: John Petrilla [mailto:john.petrilla@xxxxxxxxxxxxx] Hello Pete Thanks for taking on this analysis. For now I have a question regarding use of the formula proposed by Roy Cideciyan. In the referenced contribution, the term, ‘i’, is defined as t+1. As I understand it, t = (528-514)/2 = 7. In the calculation below, Regards, John From: Anslow, Peter [mailto:panslow@xxxxxxxxx] John, (sent to both the Cu and Optx reflectors) To return to the BER/FER requirements discussion for 100GBASE-SR4. Following Matt’s argument that the interPacketGap (IPG) is nominally 96 bits and is followed by a 7 bit preamble and rarely shrinks to the minimum of 8 bits, it will be a rare occurrence for errors in one Ethernet frame to cause errors in the following frame due to descrambler error multiplication. As discussed previously, if the errors are randomly distributed, then a BER at the MAC/PLS service interface of 1E-12 is equivalent to an Ethernet FER of 5.12E-10 for 512 bit (64 octet) frames. So, what FEC FER is this equivalent to? According to slide 7 of brown_3bj_02_0912.pdf the Ethernet FER = FEC FER * (1+512/5280) = FEC FER * 1.097 But it seems to me that this formula is pessimistic in that it assumes that there are many errors in an errored FEC frame. If we consider the case of random errors in the FEC frame, then there would only be 1 error in most FEC frames and the appropriate formula would be Ethernet FER = FEC FER * (512/5280), i.e. the Ethernet FER would be 1/10 of the FEC FER. For the case of random errors at the FEC decoder input, we would expect 8 errors per FEC frame most of the time. (According to my calculation, the probability of there being 9 errors is a factor of 33 smaller and 10 errors 1200 smaller). So mostly we have 8 errors distributed across 10 Ethernet frames. If we make the conservative assumption that Ethernet FER = FEC FER, then I think that this will also cover the optimistic assumption made earlier that the descrambler error multiplication with a reduced IPG never causes additional Ethernet frame errors. This would mean a FEC FER of 5.12E-10. Slide 5 of cideciyan_3bj_01a_0912.pdf (second to bottom line) quotes the formula to calculate the FEC FER from the input 10-bit symbol error ratio. If the symbol error ratio is Ps, then this is: FEC FER = sum over i=9 to 5280 [ 5280 C i * Ps^i * (1-Ps)^(5280-i)] Where 5280 C i is the number of combinations of i from 5280 And Ps = 1 – (1 – BER)^10 which is approximately BER * 10 for small BER Evaluating this for a FEC FER of 5.12E-10 gives an input BER of 5.1E-5 which is a Q of 3.89 and 10Log(Qo/Qi) = 2.58 dB If no one points out a flaw with this analysis, then I will capture it in a slideset for the forthcoming MMF Ad Hoc. Regards, Pete Anslow | Senior Standards Advisor From: Matt Brown [mailto:mbrown@xxxxxxx] Hi Pete, That’s an accurate summary of the intent of the frame error rate requirement. From: Anslow, Peter [mailto:panslow@xxxxxxxxx] Matt, (sent to both the Cu and Optx reflectors) Thanks. Yes, I had read through both of the presentations that you reference, but couldn’t get a figure of a FER of 1.7E-10 for 512 bit Ethernet frames from either of them. So, if I understand you correctly, the idea is that when you have a FER of 1.7E-10 for 512 bit Ethernet frames at the MAC/PLS service interface, the burst error environment of CR4/KR4/KP4 together with the fact that the FEC will ensure that there are at least 8 errors per 5280 bit FEC block for CR4/KR4 or 16 errors per 5280 bit FEC block for KP4 is accounted for by assuming that there are no more than 3 bit errors per 512 bit Ethernet frame. Regards, Pete Anslow | Senior Standards Advisor From: Matt Brown [mailto:mbrown@xxxxxxx] Hi Pete, The target MAC frame error ratio was discussed in the following presentations: http://www.ieee802.org/3/bj/public/sep12/brown_3bj_02_0912.pdf http://www.ieee802.org/3/bj/public/sep12/cideciyan_3bj_01a_0912.pdf Clause 91 refers to the FEC codeword (not frame). The presentations refer to FEC uncorrectable codeword ratio (UCR) and the MAC frame error ratio (FER). For consistency it may be helpful to use those terms. The choice of a target FER 1.7E-10 was a compromise between the two presentations. I agree with your observation that for a random error rate of 1E-12 at the PMD will result in a frame error rate of 5.12E-12 at the MAC layer if you ignore the error multiplication of the PCS layer. However, when considering the effect of 3x error multiplication of the scrambling process no correction factor is required. A frame error occurs only if an error occurs in the octets from the SFD to the last octet of the FCS. This leaves minimum 8 bits IPG plus 7 bits preamble between frames where errors can occur without causing a frame error (in the worst case). On average, the IPG must be 12 octets so the worst case is infrequent frequent even during a burst of back to back packets. This leaves a minimum of 64 bits between back to back frames where errors can occur without causing a frame error. The errors caused by the scrambler can be spread by no more than 57 bits so a single bit error at the PMD cannot cause more than one frame error. The appropriate target FER should be 5.12E-10 for 64-octet MAC frames for the case where there is no error propagation with a literal interpretation of the error rate target. If we disagree on this point, the difference is so small it should not matter either way. If this target is acceptable for the random error case, I propose that it should also be acceptable for the case where there is error propagation and/or there is an FEC; where the literal BER may be higher but has no worse effect on FER (the observable result). From: Anslow, Peter [mailto:panslow@xxxxxxxxx] Mike, Yes, the discussion we were having below was concerned with FEC frames (5280 bits long) and the text in P802.3bj D 1.2 which says: “For a complete Physical Layer, this specification is considered to be satisfied by a frame error ratio less than 1.7 × 10–10 for 64 octet frames with minimum inter-packet gap.” Is talking about Ethernet frames. That is why I was careful to use the term “FEC frame error ratio” in my original post. However, I am struggling to understand where the value of 1.7E-10 comes from. I had thought that it was supposed to be the FER for a 64-octet frame assuming random errors and a BER of 1E-12. That would be = 1-(1-P_bit_error)^512 = 5.12E-10. The text copied above doesn’t say where this FER is evaluated (I think that this needs fixing, I will try to remember to comment). The value of 5.12E-10 would be appropriate to the MAC/PLS service interface where the objective of 1E-12 BER applies. If we assume that the intent of the text copied above is that the FER has to be met after the FEC but before the descrambler, then the error multiplication of a factor of three must be considered and dividing 5.12E-10 by 3 gives 1.71E-10. If this is the case, then I am missing something because the FER before the descrambler and the FER after the descrambler do not differ by a factor of 3, although the BER’s do. The minimum IPG is 8 bits. This means that an error in an Ethernet frame at the descrambler input will not cause errors in the following frame unless the original error occurs in the last 58 - 8 = 50 bits. For randomly distributed errors, this means that the probability of an error in one frame affecting the next is about 0.1 which would indicate a FER at the descrambler input of about 5.12E-10/1.1 = 4.65E-10. If one assumed that the errors were not randomly distributed, but occurred in groups of 8 for KR4/CR4 FEC and 16 for KP4 FEC, then the ratio of the FER before the descrambler to the FER after the descrambler would still be less than 2. (Error multiplication can only affect the next frame). And the number would be different for Clause 94 than for Clauses 92 and 93. What am I missing? Regards, Pete Anslow | Senior Standards Advisor From: Mike Dudek [mailto:mike.dudek@xxxxxxxxxx] I’ve added some folk from the .bj task force FEC ad-hoc to this thread and changed the title. As stated in a separate e-mail I think that it would be good if 802.3bj and 802.3bm came to a common understanding of the error ratio requirements. First I’m wondering whether we may be having some confusion between FEC frames and Ethernet Packets (aka frames). The 802.3bj has created a requirement (in draft 1.2) that is stated in terms of frame error rate, but in context I think it means packet error rate, and I think that may be the most meaningful for a system. Obviously if it is the Packet Error Rate that really matters to the system then we also have to consider how shorter packets get distributed within FEC frames, (or FEC frames are distributed in long packets). I think the thread below is talking about the FEC frame error rate which I think would be just an intermediate calculation point. Mike Dudek QLogic Corporation Senior Manager Signal Integrity 26650 Aliso Viejo Parkway Aliso Viejo CA 92656 949 389 6269 - office. From: Anslow, Peter [mailto:panslow@xxxxxxxxx] John, a) No, I wouldn’t. The probability of a frame containing 5280 bits having one or more errors in it is one minus the probability that all of the bits are correct. The probability that they are all correct is the probability that any one bit is correct raised to the power of the number of bits in the frame. Hence the formula. If the BER is low, then this is dominated by the case where there is only one error per frame and it evaluates to be very close to 1E-12 * 5280 = 5.28E-9. When the BER becomes lower, say 1E-5, the effect of more than one error starts to be seen and it evaluates to 5.14E-2 rather than 5.28E-2. b) A very simple minded analysis might say that the process in a) can be reversed by taking the Frame Error Ratio (FER) and applying the formula: BER = 1-(1-FER)^(1/5280) which evaluates to 1E-12 for FER = 5.28E-9 and is approximately FER/5280 unless the BER is high. However, this conversion assumes that the errored frames have random errors (since it is the reversal of the calculation in a). However, as you point out, the effect of FEC is to remove the errors from all frames with 7 or less symbol errors. This means that any frame that contains an error contains at least 8 errors. (8 is actually the dominant case, 9 errors in a frame is much less probable). Consequently, the formula to go from the FER to the BER is approximately BER = FER*8/5280. Hope that makes my meaning clear, Pete Anslow | Senior Standards Advisor From: John Petrilla [mailto:john.petrilla@xxxxxxxxxxxxx] Hello Pete Thanks a) Regarding item 6) below, why would you not use binomial statistics? b) Further down, you write, “Because the errors with FEC occur in groups of 8 or more per FEC frame, the BER in this case will be at least 8E-12.” I’m not sure what you mean. Could you clarify? Regards, John From: Anslow, Peter [mailto:panslow@xxxxxxxxx] John, Your proposal with changes / comments in red. 1) For the 100 m MMF objective, definition will be based on the assumption that RS-FEC, RS(528, 514), defined in Clause 91 for 100GBASE-CR4 or 100GBASE-KR4 (hereinafter referred to as KR4 FEC) is available. 2) All of the error correction capability of KR4 FEC is allocated to the link supporting the 100 m MMF objective. 3) The incoming BER for the MMF PMD (including any errors generated by CAUI-4 if present) will be equal to or better than 10^-12 and the target corrected BER for the link output will be equal to or better than 10^-12. 4) KR4 FEC uses 528 symbols of 10 bits/symbol yielding a frame size of 5280 bits. 5) There are 514 data symbols and 14 parity symbols providing the ability to correct (528-514)/2 = 7 corrupted symbols. 6) The Frame Error Ratio, FER, for operation without FEC for a BER = 1E-12 using P_frame_error = 1-(1-P_bit_error)^5280 is 5.28E-9. 7) In a optical link, assume bit errors are noise generated, independent and random. Further, since there will be no DFEs, error multiplication is not expected before the descrambler. 8) The worst case that can be corrected is 7 10-bit symbols in error. Now, I get stuck because I think we need to agree on what the error criterion at the FEC output is. I think that John has assumed that the FEC Frame error ratio (FER) is 5.28E-9 i.e. the same FER as you would get if you had randomly distributed errors and a BER of 1E-12 or in other words, the same FER as you would expect if you were operating without FEC. Because the errors with FEC occur in groups of 8 or more per FEC frame, the BER in this case will be at least 8E-12. Before going further and calculating what input BER corresponds to a FEC FER of 5.28E-9, I think we need to get agreement that this is the criterion that we are going to use as it is not strictly in agreement with the objective: “Support a BER better than or equal to 10-12 at the MAC/PLS service interface” Regards, Pete Anslow | Senior Standards Advisor From: John Petrilla [mailto:john.petrilla@xxxxxxxxxxxxx] Hello Jonathan Thanks for proposing the meeting and especially for the 8:30 AM Pacific start time. Perhaps we can make some progress on FEC details prior to the meeting. Along that line I’ll propose the following strawman, so that we may have common values to use for Q and BER in our various analyses. 1) For the 100 m MMF objective, definition will be based on the assumption that RS-FEC, RS(528, 514), defined in Clause 91 for 100GBASE-CR4 or 100GBASE-KR4 (hereinafter referred to as KR4 FEC) is available. 2) All of the error correction capability of KR4 FEC is allocated to the link supporting the 100 m MMF objective. 3) The incoming BER for the MMF PMD will be equal to or better than 10^-12 and the target corrected BER for the link output will be equal to or better than 10^-12. 4) KR4 FEC uses 528 symbols of 10 bits/symbol yielding a frame size of 5280 bits. 5) There are 514 data symbols and 14 parity symbols providing the ability to correct (528-514)/2 = 7 corrupted symbols. 6) The Frame Error Rate, FER, for operation without FEC for a BER = 1E-12 using binomial statistics (probability density function) is 5.30E-9. 7) In a optical link, assume bit errors are noise generated, independent and random. Further, since there will be no DFEs, error multiplication is not expected. 8) The worst case that can be corrected is 7 bit errors for 7 symbols with 1 bit error/symbol. 9) The case equivalent to operation without KR4 FEC is where 8 symbols are corrupted, since for only 7, all errors are corrected. 10) For operation with KR4 FEC, a BER = 6.90E-5 yields an FER of 5.30E-9 (Qi = 3.8119) to match the FER for a BER = 1E-12 (Q = 7.034) without FEC and 10Log(Qo/Qi) = 2.66 dB. 11) The link model supporting the 100 m MMF objective will use a value of 3.8119 for Q. If there are objection or counter proposals, let’s try to resolve them before the ad hoc meeting. Regards, John From: Jonathan King [mailto:jonathan.king@xxxxxxxxxxx] Dear all, I’d like to propose we meet for an MMF Ad Hoc conference call (via Webex) on 29th November, 8.30am Pacific Standard time (4.30pm GMT) . Call duration will be up to 1.5 hours, but can end earlier if meaningful discussion ends. If there aren’t too many objections I’ll confirm the date/time in a couple of days, and send out call details. Our main goal is to progress development of baseline proposals for the 100m reach and 20m reach MMF objectives, and hopefully remove a few more TBDs from the proposed Tx and Rx spec tables. Please send presentation requests to me by close of business on 28th November- (I’ll send out the agenda that day). Best wishes jonathan |