Tutorial – the sum-product algorithm for decoding of LDPC codes
Use a small parity-check matrix H and make sure that the sum-product algorithm works
appropriately before simulating communication systems with larger and more complex LDPC
codes (i.e. codes corresponding to a large matrix H).
Use for example the small matrix H (which we considered during the lecture):
1
1
H
0

0
0
0
1
1
1
0
1
0
0
1
0
1
1
0
0
1
0
1
1
0
1
0
1
0
0
1
0

1
which corresponds to the factor graph:
Start by constructing the codeword c that we want to transmit. For all linear block codes, the
all zero codeword is a valid codeword, hence we assume that we want to transmit the message
(the structure of the parity check matrix H above indicates that the information message
consists of 4 elements and the codeword of 8 elements):
m = [0 0 0 0] which corresponds to the codeword c = mG = [0 0 0 0 0 0 0 0]
With respect to this assumption, we do not need the generator matrix G (which would be the
case if the elements of m would be different).
For example, assume that BPSK modulation is used and that all codeword bits are assigned
values +1 or -1 before transmission over the channel (these values +1 and -1 are related to the
location of the codeword symbols in the signal space after BPSK modulation). After
modulation and with respect to the mapping [0,1] → [-1,+1], the all zero codeword
corresponds to:
c_mod = [-1 -1 -1 -1 -1 -1 -1 -1]
Now, assume that c_mod is sent over a noisy channel which yields the received codeword r at
the receiver:
r = [-2.5467
0.2358 -1.3929 -3.0287 -1.8290 -1.1768 -1.9434 -0.1152]
Given r and the description of the sum-product algorithm (see the documentation on the
course website), the output of the steps 1-5 should approximately be as follows:

Output of Step 1 (see equation (3)):
Assume transmission over an AWGN channel with normalized channel variance N0/2 = 1.
v = -2r / (N0/2)2 = [5.0935 -0.4716 2.7859 6.0574 3.6580 2.3535 3.8867 0.2305]

Output of Step 2 (1st iteration) (see equation (4)):
(The first time the variables send their probability of being in a specific state, the messages
simply correspond to the values v from Step 1, since no messages have been received from
the checks which will be the case for the subsequent iterations.)
From Variable 1 to Check 1 = 5.0935
From Variable 1 to Check 2 = 5.0935
From Variable 2 to Check 3 = -0.4716
From Variable 2 to Check 4 = -0.4716
From Variable 3 to Check 1 = 2.7859
From Variable 3 to Check 3 = 2.7859
From Variable 4 to Check 2 = 6.0574
From Variable 4 to Check 4 = 6.0574
From Variable 5 to Check 1 = 3.6580
From Variable 5 to Check 4 = 3.6580
From Variable 6 to Check 2 = 2.3535
From Variable 6 to Check 3 = 2.3535
From Variable 7 to Check 1 = 3.8867
From Variable 7 to Check 3 = 3.8867
From Variable 8 to Check 2 = 0.2305
From Variable 8 to Check 4 = 0.2305

Output of Step 3 (1st iteration) (see equation (8)):
Satisfaction of Check 1 = 201.7455
Satisfaction of Check 2 = 16.7372
Satisfaction of Check 3 = -12.0177
Satisfaction of Check 4 = -2.4086
Since not all satisfaction values of the checks are > 0, we do not have a valid codeword and
hence the iteration continues.

Output of Step 4 (1st iteration) (see equation (5)):
From Check 1 to Variable 1 = 2.2290
From Check 2 to Variable 1 = 0.1893
From Check 3 to Variable 2 = 1.7387
From Check 4 to Variable 2 = 0.2178
From Check 1 to Variable 3 = 2.9490
From Check 3 to Variable 3 = -0.3715
From Check 2 to Variable 4 = 0.1879
From Check 4 to Variable 4 = -0.0505
From Check 1 to Variable 5 = 2.4284
From Check 4 to Variable 5 = -0.0529
From Check 2 to Variable 6 = 0.2266
From Check 3 to Variable 6 = -0.3980
From Check 1 to Variable 7 = 2.3708
From Check 3 to Variable 7 = -0.3415
From Check 2 to Variable 8 = 2.2689
From Check 4 to Variable 8 = -0.4449

Output of Step 2 (2nd iteration) (see equation (4)):
(This time we take messages from the checks into account.)
From Variable 1 to Check 1 = 5.2828
From Variable 1 to Check 2 = 7.3224
From Variable 2 to Check 3 = -0.2538
From Variable 2 to Check 4 = 1.2671
From Variable 3 to Check 1 = 2.4144
From Variable 3 to Check 3 = 5.7348
From Variable 4 to Check 2 = 6.0069
From Variable 4 to Check 4 = 6.2453
From Variable 5 to Check 1 = 3.6051
From Variable 5 to Check 4 = 6.0865
From Variable 6 to Check 2 = 1.9556
From Variable 6 to Check 3 = 2.5801
From Variable 7 to Check 1 = 3.5452
From Variable 7 to Check 3 = 6.2575
From Variable 8 to Check 2 = -0.2144
From Variable 8 to Check 4 = 2.4994

Output of Step 3 (2nd iteration) (see equation (8)):
Satisfaction of Check 1 = 163.0174
Satisfaction of Check 2 = -18.4419
Satisfaction of Check 3 = -23.4991
Satisfaction of Check 4 = 120.3822
Since not all satisfaction values of the checks are > 0, we do not have a valid codeword and
hence the iteration continues.

Output of Step 4 (2nd iteration) (see equation (5)):
From Check 1 to Variable 1 = 1.9331
From Check 2 to Variable 1 = -0.1602
From Check 3 to Variable 2 = 2.5148
From Check 4 to Variable 2 = 2.4497
From Check 1 to Variable 3 = 2.7959
From Check 3 to Variable 3 = -0.2169
From Check 2 to Variable 4 = -0.1608
From Check 4 to Variable 4 = 1.0285
From Check 1 to Variable 5 = 2.0957
From Check 4 to Variable 5 = 1.0293
From Check 2 to Variable 6 = -0.2131
From Check 3 to Variable 6 = -0.2512
From Check 1 to Variable 7 = 2.1092
From Check 3 to Variable 7 = -0.2163
From Check 2 to Variable 8 = 1.9342
From Check 4 to Variable 8 = 1.2535

Output of Step 2 (3rd iteration):
From Variable 1 to Check 1 = 4.9333
From Variable 1 to Check 2 = 7.0265
From Variable 2 to Check 3 = 1.9781
From Variable 2 to Check 4 = 2.0432
From Variable 3 to Check 1 = 2.5690
From Variable 3 to Check 3 = 5.5817
From Variable 4 to Check 2 = 7.0859
From Variable 4 to Check 4 = 5.8966
From Variable 5 to Check 1 = 4.6874
From Variable 5 to Check 4 = 5.7537
From Variable 6 to Check 2 = 2.1023
From Variable 6 to Check 3 = 2.1405
From Variable 7 to Check 1 = 3.6704
From Variable 7 to Check 3 = 5.9960
From Variable 8 to Check 2 = 1.4840
From Variable 8 to Check 4 = 2.1647

Output of Step 3 (3rd iteration) (see equation (8)):
Satisfaction of Check 1 = 218.0386
Satisfaction of Check 2 = 155.3315
Satisfaction of Check 3 = 141.7045
Satisfaction of Check 4 = 150.0543
All satisfaction values of the checks are > 0, we have found a valid codeword! The next step
is to obtain the binary representation of the codeword and the message.

Output of Step 5:
We form the resulting output values as (see equation (6))
v’ = [6.8664
4.4929
5.3648
6.9251
6.7831 1.8893
5.7796 3.4181]
and the hard decisions with respect to these values give us the received codeword (see
equation (7))
c = [0 0 0 0 0 0 0 0].
Finally, given the assumption that the first part of the codeword corresponds to the message
m, we get m as
m = [0 0 0 0].
Comparing the received message sequence to the transmitted message, we observe that no
error occurred (however, recall that the algorithm can converge to a codeword that is valid but
not equal to the transmitted codeword, and thus the received message may contain errors
although the sum-product algorithm stops iterating due to a fulfilled stopping criterion).
HINT:
The sum-product algorithm consists of an inverse hyperbolic tangent (i.e. tanh-1) computation.
Since tanh-1(x) approaches +/- infinity as x goes to +/- 1, there is a potential risk of error in
MATLAB due to limitations of the numerical precision. Hence, in order to avoid that
MATLAB returns +/- infinity when making the tanh-1 computation, it is suitable to include a
limitation similar to the pseudo code (note: tanh-1(0.9999) ≈ 5):
if (-0.9999 < x) & (x < 0.9999)
% perform the atanh computation as specified in the
% sum-product algorithm
else
atanh_computation = 5*sign(tanh-1(x));
end