# Re: [802.3_100GCU] IEEE P802.3bj

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]
Sent: Thursday, August 27, 2015 10:44 PM
To: STDS-802-3-100GCU@xxxxxxxxxxxxxxxxx
Subject: [802.3_100GCU] IEEE P802.3bj

We have a question about 802.3bj standard(802.3bj-2014.pdf). The question is about linear fit pulse height.

In section 92.8.3.5.2 of 802.3bj-2014, it states that peak value of p(k) shall be greater than 0.45Vf after the transmit equalizer coefficients have been set to the “preset” values. Where p(k) is linear fit pulse and Vf is steady state voltage.

When calculating p(k),  there are two parameters, Np and Dp, specified in standard. The specified value are Np=14 and Dp=2. The relation between p(k) and Vf can be expressed as p(k)=C*Vf, and we are curious about the variation of C as Np and Dp 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 DP=32, and NP=256, the results are shown in Fig.1. The corresponding Vf=0.434 Volt and pulse peak=0.2 Volt. That is p(k)=0.461Vf., which is close to what is given in [ref1].

Fig.1 Dp=32, and Np=256

When we set DP=2, and NP=14 as what specified in standard, we get results shown in Fig.2. With Dp=2, Np=14, corresponding results are Vf=0.396 Volt and pulse peak=0.2 Volt. That is p(k)=0.505 Vf.

Fig.2 Dp=2, and Np=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 Dp=2 and Np=14 results in is p(k)=0.505 Vf , while setting Dp=32 and Np=256 results in is p(k)=0.461 Vf. That is Dp=2 and Np=14 specified in 802.3bj-2014 may underestimate effective loss and seem not consist with original calculation in [ref1].

Ref1: COM and TX Specifications, Charles Moore
http://www.ieee802.org/3/bj/public/jul13/moore_3bj_02a_0713.pdf

Could you help to clarify our question?  Thank you for your kindly help.

Best Regards,
--------------------------------------------------------------------------------------------
Chih-Ching Yu 游志青
System Development Div.II

Data Center Networking BU.

MediaTek Inc.
TEL:+886-3-5670766 Ext:26682
E-mail:CC.Yu@xxxxxxxxxxxx