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Hi Adam, We have a question about 802.3bj standard(802.3bj2014.pdf). The question is about linear fit pulse height. In section 92.8.3.5.2 of 802.3bj2014, it states that peak value of p(k) shall be greater than 0.45V_{f} after the transmit equalizer coefficients have
been set to the “preset” values. Where p(k) is linear fit pulse and V_{f} is steady state voltage. When calculating p(k), there are two parameters, N_{p} and D_{p}, specified in standard. The specified value are N_{p}=14 and D_{p}=2.
The relation between p(k) and V_{f} can be expressed as p(k)=C*V_{f}, and we are curious about the variation of C as N_{p} and D_{p} are changed. We find a reference document [ref1] written by Charles Moore, and try to repeat
results of test case in [ref1]. The selected test case is highlighted in the Table I.
Table I. As we set D_{P}=32, and N_{P}=256, the results are shown in Fig.1. The corresponding V_{f}=0.434 Volt and pulse peak=0.2 Volt. That is p(k)=0.461V_{f}., which is close to what is given
in [ref1]. Fig.1 D_{p}=32, and N_{p}=256 When we set D_{P}=2, and N_{P}=14 as what specified in standard, we get results shown in Fig.2. With D_{p}=2, N_{p}=14, corresponding results are V_{f}=0.396 Volt and pulse peak=0.2
Volt. That is p(k)=0.505 V_{f. } Fig.2 D_{p}=2, and N_{p}=14 In summary, standard specify p(k)>=0.45Vf. p(k)>=0.45Vf seems coming from the results calculated in [ref1] except package length is changed from 35mm to 30mm. However, for the test case, setting D_{p}=2 and
N_{p}=14 results in is p(k)=0.505 V_{f} , while setting D_{p}=32 and N_{p}=256 results in
is p(k)=0.461 V_{f. }That is D_{p}=2 and N_{p}=14 specified in 802.3bj2014 may underestimate effective loss and seem not consist with original calculation in [ref1]. Ref1: COM and TX Specifications, Charles Moore Could you help to clarify our question? Thank you for your kindly help. Best Regards, MediaTek Inc.
