UCI and data multiplexing are transmitted on Pusch - Part I

One . total

UCI Include harq-ack,csi,sr. these uci Information can be obtained through PUCCH transmission , It can also be done through pusch transmission （ There is a question about this ,sr Is in the pusch Before dispatching ,ue Need to send sr, But when sr When the dispatching base station cannot receive , By re connecting , then msg3 Carry the bsr Instead of , normal sr It can't go through pusch Sent , This is to be determined .）,csi Pass periodically pucch transmission , Aperiodic passage pusch transmission , Semi persistent line passing PUCCH perhaps dci Activation of pusch.CSI The report consists of two parts ：CSI port1 and CSI port2.CSI port1 There is a fixed load , Used to identify CSI port2 Number of information bits ,CSI port1 Must be in CSI port2 Complete transmission before transmission .CSI and harq-ack Can be in PUSCH Upper transmission can be multiplexed with data , It can also be transmitted separately （ This scene I understand is PUCCH There are not enough resources on , To be determined ）. This time, there is UC_SCH Data reuse .

Case description

The introduction strategy of this article is based on matlab The built-in use case introduces the Protocol , Sort out the agreement , I wrote it myself directly , But on the company computer , Don't rewrite , Use the example of thread to sort . Because it will be empty to sort out the agreement directly , So it's better to sort out the agreement by case , This is my way of learning . This case ack =1 ,csi1=10,csi2=10, Choosing the example of comparative envoys can better introduce , Of course, the agreement introduced later is not just ; Count this case , Other information will also be introduced .

## Cell parameters entered in the use case

Inside NSizeGrid It refers to the community prb Count , This parameter is determined by two factors , Bandwidth and subcarrier spacing , bandwidth /（ Subcarrier spacing 12）= 10M/(15K12)=55, Reserve a part on both sides, that is 52 了 ,
## Use case input ue Level parameter

tcr Is the target bit rate

1. According to the data a The length of A Calculation ,crc The length of , And then calculate crc Put it in the data Back , become b

2.graph The choice is 1 still 2, Later, you need to use .

The calculation process ：

   if A <= 292 || (A <= 3824 && R <= 0.67) || R <= 0.25
      bgn = 2;
    else
      bgn = 1;
    end

    % Get transport block size after CRC attachment according to 38.212
    % 6.2.1 and 7.2.1, and assign CRC polynomial to CRC field of output
    % structure info
    if A > 3824
      L        = 24;
      info.CRC = '24A';
    else
      L        = 16;
      info.CRC = '16';
    end

    % Get the length of transport block after CRC attachment
    B = A + L;

3., First calculate for each cb Size of the block data allocation container ,B‘=B+C*L , Then find out K’=B’/C,

Find another Zc The multiple is greater than the nearest K As the size of the container , Put the data b, according to K Put it into cr, Every cr Add your own crc, From here on, the following are based on each cb Block handled by itself .

    % Get the maximum code block size
    if bgn == 1  % This means graph
      Kcb = 8448;  % This refers to every cb The size of the block 
    else
      Kcb = 3840;
    end

    % Get number of code blocks and length of CB-CRC coded block
    if B <= Kcb
      L = 0;
      C = 1;
      Bd = B;
    else
      L = 24; % Length of the CRC bits attached to each code block
      C = ceil(B/(Kcb-L));  % This is divided into how many cb block 
      Bd = B+C*L;
    end

    % Obtain the number of bits per code block (excluding CB-CRC bits)
    cbz = ceil(B/C);% This is exclusion crc Every cb The inclusion of the block bit The amount of 

    % Get number of bits in each code block (excluding filler bits)
    Kd = ceil(Bd/C);% This is each cb Blocks exclude padding bit, contain crc Of bit Count 

    % Find the minimum value of Z in all sets of lifting sizes in 38.212
    % Table 5.3.2-1, denoted as Zc, such that Kb*Zc>=Kd
    if bgn == 1
      Kb = 22;
    else
      if B > 640
        Kb = 10;
      elseif B > 560
        Kb = 9;
      elseif B > 192
        Kb = 8;
      else
        Kb = 6;
      end
    end
    Zlist = [2:16 18:2:32 36:4:64 72:8:128 144:16:256 288:32:384];
    Zc =  min(Zlist(Kb*Zlist >= Kd));% To calculate the ZC

    % Get number of bits (including <NULL> filler bits) to be input to the LDPC
    % encoder
    if bgn == 1
      K = 22*Zc;  % Why must this be Zc Multiple , This is a problem ？
    else
      K = 10*Zc;
    end

4. Get including padding bit Count bit Count

  % Get number of bits (including filler bits) to be encoded by LDPC
    % encoder
    if bgInfo.BGN == 1
        N = 66*cbInfo.Zc;
    else
        N = 50*cbInfo.Zc;
    end

5. Calculation UCI and data Location and size that can be placed

according to type type , Select the table below , Yes type-1 Come on ,dmrs The location of

How to arrange the specific position ？

Calculation UCI Can be placed bit Subcarriers , Calculate the subcarrier position that the array can place

UCI It can only be done in the non dmr Put on the symbol , Data can be placed on data symbols , It can also be placed where there is no pilot station on the pilot symbol .

  % Calculate bit capacity of PUSCH (excluding DM-RS and PT-RS)
    E = sum(rinfo.MULSCH);
    G = nlqm*E;
    % Get the number of resource elements available for UCI transmission on
    % PUSCH from l_0
    if ~isempty(l0)
        s2 = sum(rinfo.MUCI(l0(1):end)); %s2 Means uci Where you can put it 
    else
        s2 = 0;
    end