OK here is my next question 
One of the other parity codes in P25 TDMA known as “ISCH” is a similar
“binary code” which the specs decribe as follows:
The Information field in the ISCH is 40 bits encoded with a
(40, 9,16) binary code…The (40, 9,16) code derived from a
(40,10,16) binary code with a generator matrix, g,
1 0 0 0 1 0 0 0 0 0 0 1 0 1 1 0 1 1 0 0 1 1 1 0 0 0 1 1 0 1 1 0 1 1 0 1
0 1 1 1
0 0 1 0 0 0 0 0 0 0 0 1 1 1 0 1 1 1 1 1 1 1 0 1 0 1 0 0 1 1 1 1 0 1 1 0
0 1 0 0
0 0 0 1 0 0 0 0 0 0 0 0 1 1 1 1 0 1 0 0 1 0 1 1 0 0 0 1 0 1 1 1 0 1 0 1
1 0 0 0
0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 1 1 1 1 0 1 1 0 1 0 0 0 1 1 0 0 0
1 1 1 0
0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1
1 1 1 1
0 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 1 0 0 0 1 1 0 1 1 0 0 1 1 0 1 1 0 1 1 1
0 0 1 0
0 0 0 0 0 0 0 0 1 0 0 1 1 1 0 1 1 0 1 0 0 0 1 1 1 0 1 0 0 0 0 1 0 1 1 1
0 0 0 1
0 0 0 0 0 0 0 0 0 1 0 1 1 0 0 0 1 1 0 0 1 0 1 1 1 0 1 0 1 0 1 0 0 1 0 0
1 1 1 0
0 0 0 0 0 0 0 0 0 0 1 1 0 1 0 0 0 0 1 1 1 1 0 1 1 0 0 0 0 1 0 1 1 0 0 1
0 1 1 1 Looking at this generator matrix it’s not clear how to even
convert it into the parity check
matrix form that is needed by the cldpc module. In particular it does
not have an identity
matrix on the left hand side (but there is something vaguely resembling
a diagonal row
of ones in the leftmost 9x11 chunk of the matrix ‘g’…
So, the question for this time around is whether the above generator
matrix is from a
type of code that ldpc will work OK with? If so, how would I remap g
above into the proper
form of parity check matrix? FWIW, the above generator matrix g can be
used to form
proper P25 codewords by taking the dot product using numpy.dot() as in
my previous
question [which was about the 4x8 generator matrix]…
Thx again
Best
Max