This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

ADC12DJ3200: JMODE11 Long Transport test Pattern

Part Number: ADC12DJ3200

hi,

I would like to use the Long Transport Test Pattern, however I cannot follow the calculation of the data.

so I'm using JMODE16 , K=20

I get:

                  Frame0     Frame1

DA0 (I0) = 0x0003     0x0002

DA1 (I1) = 0x0002     0x0005

DA2 (Q0) = 0x0004    0x0002

DA3 (Q2)  = 0x0004    0x0004

I'm not sure if this is correct or not.

so I would expect  in DA0 & Frame0:

 0x0001

as AI has CID=0 => CID+1= 0x0001

or how to calculate the corresponding data??

many thanks in advance

  • Hi,

    We are looking at the issue and will get back to you.

    Regards,

    Neeraj

  • Hi Patrick,

    Kindly note that this mode has N=15 and N'=16, where the sample forms the 15 MSBits and the LSBit is a control bit. For each link, we have L=1, M=2, F=2, S=1

    If you apply the formula for the long transport test pattern, it would result in the following. The pattern (in time from left to right), as mapped to the JMODE16 table with a K=10 would be:

    Lane ID Frame 0 CID0 S0 Frame 0 CID1 S0 Frame 1 CID0 S0 Frame 1 CID1 S0 Frame 2 through 9 CID0 S0 Frame 2 through 9 CID1 S0
    A0 0000_0000_0000_001(1) 0000_0000_0000_010(0) 0000_0000_0000_001(0) 0000_0000_0000_001(1) 1000_0000_0000_000(0) 1000_0000_0000_000(0)

    Please note that I have put the control bit in parantheses. The final values on lane A0 are:

    A0I0 = 0x3, A0Q0 = 0x4, A0I1 = 0x2, A0Q1 = 0x3, and the rest of the 8 samples for I and Q as 0x8000.

    The same values would be seen on samples from lane B0.

    Regards,

    Ameet

  • Kindly note that I have listed the example for JMODE16 (as mentioned in the details of your query), and not for JMODE11 (as mentioned in the subject line). The same calculation approach will apply.

  • thanks for you fast response