Re: [802.3_100GCU] IEEE P802.3bj

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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,
--------------------------------------------------------------------------------------------
Chih-Ching Yu 游志青
System Development Div.II系???二?
Data Center Networking BU. ??中心?路事?部

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

From: Mellitz, Richard [mailto:richard.mellitz@xxxxxxxxx]
Sent: Tuesday, September 01, 2015 6:16 AM
To: CC Yu (游志青); STDS-802-3-100GCU@xxxxxxxxxxxxxxxxx
Subject: 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

Hi Adam,

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 0.45V_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 802.3bj-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.

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