当前位置:网站首页>UCI and data multiplexing are transmitted on Pusch (Part VI) -- LDPC coding
UCI and data multiplexing are transmitted on Pusch (Part VI) -- LDPC coding
2022-07-03 09:40:00 【Communication pawn】
ldpc The coding is for the coding of uplink and downlink data , It's also 5G The most important kind of coding .
1 TB Block add CRC
This is and UCI POLAR One difference ,UCI Yes for each. cb Block add crc.
% 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
** This will be true of TB Block on crc The addition of , Add crc The process has been in the last article
《UCI And data reuse in pusch Up transfer ( The fifth part )---polar code 》 Shared ,
There is no retelling here .**
2 graph choice
% LDPC base graph selection
if A <= 292 || (A <= 3824 && R <= 0.67) || R <= 0.25
bgn = 2;
else
bgn = 1;
end
3 cb Block segmentation and cb Block add crc
Add crc The process has been in the last article
《UCI And data reuse in pusch Up transfer ( The fifth part )—polar code 》 Shared ,
There is no retelling here .
% Get the maximum code block size
if bgn == 1
Kcb = 8448;
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));
Bd = B+C*L;
end
% Obtain the number of bits per code block (excluding CB-CRC bits)
cbz = ceil(B/C);
% Get number of bits in each code block (excluding filler bits)
Kd = ceil(Bd/C);
% 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));
% 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 UL_SCH Channel code of ( This is the most difficult part of coding, which needs to be further clarified )
% Get lifting set number
ZSets = {
[2 4 8 16 32 64 128 256],... % Set 1
[3 6 12 24 48 96 192 384],... % Set 2
[5 10 20 40 80 160 320],... % Set 3
[7 14 28 56 112 224],... % Set 4
[9 18 36 72 144 288],... % Set 5
[11 22 44 88 176 352],... % Set 6
[13 26 52 104 208],... % Set 7
[15 30 60 120 240]}; % Set 8
for setIdx = 1:8 % LDPC lifting size set index
if any(Zc==ZSets{
setIdx})
break;
end
end
% Pre-stored H types for each BGN and setIdx pair
% How did this come from? I don't know , Need to take a further look
Htype = {
3, 3, 3, 3, 3, 3, 2, 3;
4, 4, 4, 1, 4, 4, 4, 1};
% Get the matrix with base graph number 'bgn' and set number 'setIdx'.
% The element of matrix V in the following is H_BG(i,j)*V(i,j), where
% H_BG(i,j) and V(i,j) are defined in TS 38.212 5.3.2; if V(i,j) is not
% defined in Table 5.3.2-2 or Table 5.3.2-3, the elements are -1.
% % This is in the agreement v_ij
switch bgn
case 1
switch setIdx
case 1
V = bgs.BG1S1;
case 2
V = bgs.BG1S2;
case 3
V = bgs.BG1S3;
case 4
V = bgs.BG1S4;
case 5
V = bgs.BG1S5;
case 6
V = bgs.BG1S6;
case 7
V = bgs.BG1S7;
otherwise % 8
V = bgs.BG1S8;
end
Nplus2Zc = Zc*(66+2);
otherwise % bgn = 2
switch setIdx
case 1
V = bgs.BG2S1;
case 2
V = bgs.BG2S2;
case 3
V = bgs.BG2S3;
case 4
V = bgs.BG2S4;
case 5
V = bgs.BG2S5;
case 6
V = bgs.BG2S6;
case 7
V = bgs.BG2S7;
otherwise % 8
V = bgs.BG2S8;
end
Nplus2Zc = Zc*(50+2);% Why ask this and ?
end
P = nr5g.internal.ldpc.calcShiftValues(V,Zc);% according to V and Zc obtain P P=mod(V,Zc)
% The element of matrix P in the following is P(i,j) in TS 38.212 5.3.2
% when V(i,j) are defined in Table 5.3.2-2 or Table 5.3.2-3. If not
% defined, the elements are -1.
P = zeros(size(V));
for i = 1:size(V,1)
for j = 1:size(V,2)
if V(i,j) == -1
P(i,j) = -1;
else
P(i,j) = mod(V(i,j),Z);
end
end
end
P1 = P(:,1:(size(P,2) - size(P,1)));% Take it P In front of (52-42) perhaps (67-45) That's ok , Why take this ?
C = size(infoBits,2); % Get layers
codewords = zeros(Nplus2Zc,C);
% Get shift values matrix
P = nr5g.internal.ldpc.calcShiftValues(V,Zc);
P1 = P(:,1:(size(P,2) - size(P,1)));
C = size(infoBits,2);
codewords = zeros(Nplus2Zc,C);
for r = 1:C
% Per code-block processing only
infoVec = reshape(infoBits(:,r),Zc,[]);% The information bit( In the agreement C_k) become Zc That's ok ,K/Zc Column .
d = blockMultiply(P1,infoVec);
d0 = d(:,1:4);
% Solve 4 equations for Htype * m = d0
switch Htype{
bgn,setIdx}
case 1
% H1 = [ 1 0 -1 -1;
% -1 0 0 -1;
% 0 -1 0 0;
% 1 -1 -1 0 ];
m1 = sum(d0, 2);
m2 = d0(:,1) + m1([2:end 1]);
m3 = d0(:,2) + m2;
m4 = d0(:,3) + m1 + m3;
case 2
% H2 = [ 0 0 -1 -1;
% 105 0 0 -1;
% -1 -1 0 0;
% 0 -1 -1 0 ];
m1 = sum(d0, 2);
shift = mod(105, Zc);
if shift > 0
m1 = m1([(end-shift+1):end 1:(end-shift)]);
end
m2 = d0(:,1) + m1;
m4 = d0(:,4) + m1;
m3 = d0(:,3) + m4;
case 3
% H3 = [ 1 0 -1 -1;
% 0 0 0 -1;
% -1 -1 0 0;
% 1 -1 -1 0 ];
m1 = sum(d0, 2);
m2 = d0(:,1) + m1([2:end 1]);
m3 = d0(:,2) + m1 + m2;
m4 = d0(:,3) + m3;
otherwise % == 4
% H4 = [ 0 0 -1 -1;
% -1 0 0 -1;
% 1 -1 0 0;
% 0 -1 -1 0 ];
m1 = sum(d0, 2);
m1 = m1([end 1:(end-1)]);
m2 = d0(:,1) + m1;
m3 = d0(:,2) + m2;
m4 = d0(:,4) + m1;
end
% Get other parity bits
P3 = P(5:end,size(P1,2)+(1:4));
p = blockMultiply(P3, [m1 m2 m3 m4]) + d(:,5:end);
% Form codeword and assign to output
codewords(:,r) = [infoBits(:,r); mod([m1; m2; m3; m4; p(:)],2)];
end
end
5 Rate matching
% Get code block soft buffer size
if ~isempty(Nref)
fcnName = 'nrRateMatchLDPC';
validateattributes(Nref, {
'numeric'}, ...
{
'scalar','integer','positive'},fcnName,'NREF');
Ncb = min(N,Nref);
else % No limit on buffer size
Ncb = N;
end
% Get starting position in circular buffer
if bgn == 1
if rv == 0
k0 = 0;
elseif rv == 1
k0 = floor(17*Ncb/N)*Zc;
elseif rv == 2
k0 = floor(33*Ncb/N)*Zc;
else % rv is equal to 3
k0 = floor(56*Ncb/N)*Zc;
end
else
if rv == 0
k0 = 0;
elseif rv == 1
k0 = floor(13*Ncb/N)*Zc;
elseif rv == 2
k0 = floor(25*Ncb/N)*Zc;
else % rv is equal to 3
k0 = floor(43*Ncb/N)*Zc;
end
end
% Get rate matching output for all scheduled code blocks and perform
% code block concatenation according to Section 5.4.2 and 5.5
out = [];
for r = 0:C-1
if r <= C-mod(outlen/(nlayers*Qm),C)-1
E = nlayers*Qm*floor(outlen/(nlayers*Qm*C));
else
E = nlayers*Qm*ceil(outlen/(nlayers*Qm*C));
end
out = [out; cbsRateMatch(in(:,r+1),E,k0,Ncb,Qm)]; %#ok<AGROW>
end
end
function e = cbsRateMatch(d,E,k0,Ncb,Qm)
% Rate match a single code block segment as per TS 38.212 Section 5.4.2
% Bit selection, Section 5.4.2.1
k = 0;
j = 0;
e = zeros(E,1,class(d));
while k < E
if d(mod(k0+j,Ncb)+1) ~= -1 % Filler bits
e(k+1) = d(mod(k0+j,Ncb)+1);
k = k+1;
end
j = j+1;
end
% Bit interleaving, Section 5.4.2.2
e = reshape(e,E/Qm,Qm);
e = e.';
e = e(:);
end
边栏推荐
- Directory and switching operation in file system
- About the configuration of vs2008+rade CATIA v5r22
- Epoll read / write mode in LT and et modes
- Definition and use of enum in C language
- [male nanny style] teach you to open the first wechat applet
- Leetcode daily question (985. sum of even numbers after queries)
- [solution to the new version of Flink without bat startup file]
- unbuntu(debian)下TFTP服务器搭建及测试
- 307. Range Sum Query - Mutable
- Flink CDC practice (including practical steps and screenshots)
猜你喜欢
[combinatorics] Introduction to Combinatorics (combinatorial thought 2: mathematical induction | mathematical induction promotion | multiple induction thought)
Directory and switching operation in file system
【男保姆式】教你打开第一个微信小程序
MySQL environment variable configuration
Convert IP address to int
[csdn] C1 analyse des questions de formation Partie III Bar _ JS Foundation
全球KYC服务商ADVANCE.AI 活体检测产品通过ISO国际安全认证 产品能力再上一新台阶
Leetcode daily question (2212. maximum points in an archery competition)
[successful graduation] [1] - visit [student management information system]
UCI and data multiplexing are transmitted on Pusch - determine the bit number of harqack, csi1 and csi2 (Part II)
随机推荐
LeetCode每日一题(1162. As Far from Land as Possible)
Basic knowledge of database design
Flink学习笔记(九)状态编程
LeetCode每日一题(1362. Closest Divisors)
1922. Count Good Numbers
LeetCode每日一题(968. Binary Tree Cameras)
【顺利毕业】[1]-游览 [学生管理信息系统]
QT sub window is blocked, and the main window cannot be clicked after the sub window pops up
Make the most basic root file system of Jetson nano and mount NFS file system on the server
Shell logic case
Directory and switching operation in file system
Powerdesign reverse wizard such as SQL and generates name and comment
Leetcode daily question (2090. K radius subarray averages)
Solve editor MD uploads pictures and cannot get the picture address
专利查询网站
数字身份验证服务商ADVANCE.AI顺利加入深跨协 推进跨境电商行业可持续性发展
Successful graduation [3]- blog system update...
LeetCode每日一题(516. Longest Palindromic Subsequence)
Flask+supervisor installation realizes background process resident
Epollet lessons