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Hi CC The question is what is V_a. Vf follows from convolution.
… Rich From: CC Yu (游志青) [mailto:cc.yu@xxxxxxxxxxxx]
Hi Richard, Two conditions in my simulation are as follows.
1.
D_p=2 and N_p=14
è p1(k)=0.2V and Vf1=0.396V
è p1(k)=0.505Vf1
2.
D_p=32 and N_p=256
è p2(k)=0.2V and Vf2=0.434V
è p2(k)=0.461Vf2 The value of p1(k) is equal to p2(k) In your simulation, but how about your Vf1 and Vf2? Thank you for your help. Best Regards, MediaTek Inc.
From: Mellitz, Richard [mailto:richard.mellitz@xxxxxxxxx]
Hi CC, Starting with V_a=0.434 and using the Quadra+8p5in… and the COM 30mm package + loads I get:
A SBR the peak of 189.609mv just from convolving the impulse response. Vf is compute to be 413.6mv ( don’t fo get the filter). So this the reference to compare to.
If convolve the IR with the PRBS9 pattern and fit for D_p =32 and N_p= 256 or D_p=2 and DP=14 I get max(p(K))= 188.60mV for both. Sigma_e is different but that is expected/ That’s pretty close. The ratio is 0.456 which is passing. Oh BTW, that ratio in independent of V_a. So all looks OK to me. … Rich From: CC Yu (游志青) [mailto:cc.yu@xxxxxxxxxxxx]
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.
