• State TI Thinks Resolved
  • Locked Locked
  • Replies 2 replies
  • Answers 1 answer
  • Subscribers 102 subscribers
  • Views 714 views
  • Users 0 members are here
Support feedback
Options
Options
Similar topics

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.

BCP Rate Dematcher LTE-Harq Output Size

user4514136
Prodigy 10 points

Hello,

I am facing the problem about rate dematcher harq output size, as we saw BCP is using more memory than we allocate for harq output for LTE.

We are working with TCI6638K2K. We are allocating (3*Transport Block Size + 128*Codeblock Number(for Codeblock allignment) ) Bytes for harq output, we aligned the buffer start addresses to 128 byte, and we are using DDR3 memory. In some cases that we could not figure out, BCP is using 16 or 32 bytes extra memory.

I could not find any harq output size formula in BCP User Guide(sprugz1) and in the TI forum, is there any mathematical formula about LTE-harq output size?

Thanks & Regards

Murat

  • 0
    TI__Guru* 81335 points

    Welcome to the TI E2E forum. I hope you will find many good answers here and in the TI.com documents and in the TI Wiki Pages (for processor issues). Be sure to search those for helpful information and to browse for the questions others may have asked on similar topics (e2e.ti.com). Please read all the links below my signature.

    We will get back to you on the above query shortly. Thank you for your patience.

  • 0 in reply to Raja
    TI__Mastermind 30405 points

    Hi Murat,

    The LTE HARQ output data format is defined in section 4.8.1.2 (LTE - Output Data) under Figure 4-136 (RD Output Format for HARQ for LTE).

    The Rate Dematcher adds zero padding after each individual stream (SYS, PAR1 and PAR2) to align the start of the next stream on a 16 byte (128-bit QWORD) boundary. Given that the Rate Dematcher outputs 8-bit soft values, there can be 16 soft-bits in a 128-bit quad-word. Thus the amount of zero-bytes/padding inserted after each individual stream (Sys, P1, P2) can be 15 bytes at max to make the next stream start at a 128 bit quad-word.

    For eaxmple, if there are 146 bytes in the systematic stream, there will be 14 zero bytes (16 minus (146 Mod 16) = 14)  inserted at the end of systematic output so that the P1 stream starts at a 128-bit quad-word (or 16 byte) boundary, and so on for P1 and P2.

    Therefore, the general calculation for BCP HARQ output block size can be written as:

    RD HARQ_BUFFER_SIZE_IN_BYTES = (CB_SIZE + NUM_PADDING_BYTES) x 3 x NUM_CODE_BLOCKS_IN_TB

    Where NUM_PADDING_BYTES = 16 - (CB_SIZE Mod 16).

    Regards
    -Nitin