this is program for LDPC code but just for R=1/2 :
H = dvbs2ldpc(1/2);
hEnc = comm.LDPCEncoder(H);
hMod = comm.PSKModulator(4, 'BitInput',true);
hChan = comm.AWGNChannel(... 'NoiseMethod','Signal to noise ratio (SNR)','SNR',1);
hDemod = comm.PSKDemodulator(4, 'BitOutput',true,... 'DecisionMethod','Approximate log-likelihood ratio', ... 'Variance', 1/10^(hChan.SNR/10));
hDec = comm.LDPCDecoder;
hError = comm.ErrorRate;
for counter = 1:10
data = logical(randi([0 1], 32400, 1));
encodedData = step(hEnc, data);
modSignal = step(hMod, encodedData);
receivedSignal = step(hChan, modSignal);
demodSignal = step(hDemod, receivedSignal);
receivedBits = step(hDec, demodSignal);
errorStats = step(hError, data, receivedBits);
end
fprintf('Error rate = %1.2f\nNumber of errors = %d\n', ... errorStats(1), errorStats(2))
To have other performance I have to change, H = dvbs2ldpc(1/2); by
H = dvbs2ldpc(9/10);
but MATLAB shows me this error
Error using LDPCEncoder/step
Input must be a column vector of length K.
it is impossible to determine ldpc code by rate 9/10 because the max(rate)=8/9
https://in.mathworks.com/matlabcentral/answers/260747-ldpc-h-matrix-with-diffrent-rate
LDPC has 2 mothers codes 1/3 ( like turbo code ) and 1/5 if I use the second base graph.
LDPC has 2 BG : it is according to the rat that I want to use it and the TBS.
I have a graph : in this graph I know if I want to use BG1 or BG2.
and to construct the code rate that I want to use it.
we need the rate matching and the circulat buffer to read the data from this buffer.
in TS 38.214 there is 15 possibles rate that I can use . it is according to a parameter called CQI ( channel quality indicator)
QC_LDPC codes can support flexible code-rates based on puncturing or
parity-check matrix extension
Hello,
In LDPC code 22|46 are the message and parity bits column respectively.
For different code rate the parity bits has to be punctured.
The initial message 2 columns are punctured. Hence the mother code rate defined will be of 22/66(which will be 1/3).
So for 22|24 the code rate will be 22+24 = 46 total columns of encoded bits(N).
k/N-k = 22/ 46-22 (91.67%) code rate.
So to obtain varying code rate above 1/3, we have to puncture the parity bits.
Following this , i have doubt in 5G NR the E value calculated will be max of size 9880. If the code rate less than 85%, how it will get recovered or we have to puncture the remaining parity bits below 85% code rate. For example if we have code rate of 1/3, this will give 25344 code block size. Then from this if only 9880 recovered, then how the remaining bits will get recovered. If it will get punctured then how it will be processed in LDPC decoder?
Thanking you in advance
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