Exercise 4.5: On the Extrinsic L-values again

Table according to the first $L_{\rm E}(i)$ approach

We assume as in "theory section" from single parity–check code $\rm SPC \, (3, \, 2, \, 2)$ . The possible code words are $\underline{x} \hspace{-0.01cm}\in \hspace{-0.01cm} \{ \underline{x}_0,\hspace{0.05cm} \underline{x}_1,\hspace{0.05cm} \underline{x}_2,\hspace{0.05cm} \underline{x}_3\}$ with

$$\underline{x}_0 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} (0\hspace{-0.03cm},\hspace{0.05cm}0\hspace{-0.03cm},\hspace{0.05cm}0)\hspace{0.35cm}{\rm bzw. } \hspace{0.35cm} \underline{x}_0 \hspace{-0.05cm}=\hspace{-0.05cm} (+1\hspace{-0.03cm},\hspace{-0.05cm}+1\hspace{-0.03cm},\hspace{-0.05cm}+1)\hspace{0.05cm},$$

$$\underline{x}_1 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} (0\hspace{-0.03cm},\hspace{0.05cm}1\hspace{-0.03cm},\hspace{0.05cm}1)\hspace{0.35cm}{\rm bzw. } \hspace{0.35cm} \underline{x}_1 \hspace{-0.05cm}=\hspace{-0.05cm} (+1\hspace{-0.03cm},\hspace{-0.05cm}-1\hspace{-0.03cm},\hspace{-0.05cm}-1)\hspace{0.05cm},$$

$$\underline{x}_2 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} (1\hspace{-0.03cm},\hspace{0.05cm}0\hspace{-0.03cm},\hspace{0.05cm}1)\hspace{0.35cm}{\rm bzw. } \hspace{0.35cm} \underline{x}_2 \hspace{-0.05cm}=\hspace{-0.05cm} (-1\hspace{-0.03cm},\hspace{-0.05cm}+1\hspace{-0.03cm},\hspace{-0.05cm}-1)\hspace{0.05cm},$$

$$\underline{x}_3 \hspace{-0.15cm} \ = \ \hspace{-0.15cm} (1\hspace{-0.03cm},\hspace{0.05cm}1\hspace{-0.03cm},\hspace{0.05cm}0)\hspace{0.35cm}{\rm bzw. } \hspace{0.35cm} \underline{x}_3 \hspace{-0.05cm}=\hspace{-0.05cm} (-1\hspace{-0.03cm},\hspace{-0.05cm}-1\hspace{-0.03cm},\hspace{-0.05cm}+1)\hspace{0.05cm}.$$

In the exercise we mostly use the second (bipolar) representation of the code symbols: $x_i ∈ \{+1, -1\}$.

It is not that the $\rm SPC \, (3, \, 2, \, 2)$ would be of much practical interest, since, for example, in hard decision because of $d_{\rm min} = 2$ only one error can be detected and none can be corrected. However, the code is well suited for practice and demonstration purposes because of the manageable effort involved.
With iterative symbol-wise decoding one can also correct one error. In the present code, the extrinsic $L$–values $\underline{L}_{\rm E} = \big (L_{\rm E}(1), \ L_{\rm E}(2), \ L_{\rm E}(3)\big )$ must be calculated according to the following equation.

$$L_{\rm E}(i) = {\rm ln} \hspace{0.15cm}\frac{{\rm Pr} \left [w_{\rm H}(\underline{x}^{(-i)})\hspace{0.15cm}{\rm is \hspace{0.15cm} even} \hspace{0.05cm} \right ]}{{\rm Pr} \left [w_{\rm H}(\underline{x}^{(-i)})\hspace{0.15cm}{\rm is \hspace{0.15cm} odd} \hspace{0.05cm} \hspace{0.05cm}\right ]}.$$

Here $\underline{x}^{(-1)}$ denotes all symbols except $x_i$ and is thus a vector of length $n - 1 = 2$.

As the first $L_{\rm E}(i)$ approach we refer to the approach corresponding to the equations

$$L_{\rm E}(1) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}2 \cdot {\rm tanh}^{-1} \left [{\rm tanh}(L_2/2) \cdot {\rm tanh}(L_3/2) \right ] \hspace{0.05cm},$$

$$L_{\rm E}(2) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}2 \cdot {\rm tanh}^{-1} \left [{\rm tanh}(L_1/2) \cdot {\rm tanh}(L_3/2) \right ] \hspace{0.05cm},$$

$$L_{\rm E}(3) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}2 \cdot {\rm tanh}^{-1} \left [{\rm tanh}(L_1/2) \cdot {\rm tanh}(L_2/2) \right ] \hspace{0.05cm}.$$

(1) This $L_{\rm E}(i)$ approach underlies the results table above (red entries), assuming the following a posteriori $L$ values:

$$\underline {L}_{\rm APP} = (+1.0\hspace{0.05cm},\hspace{0.05cm}+0.4\hspace{0.05cm},\hspace{0.05cm}-1.0) \hspace{0.5cm}{\rm kurz\hspace{-0.1cm}:}\hspace{0.25cm} L_1 = +1.0\hspace{0.05cm},\hspace{0.05cm} L_2 = +0.4\hspace{0.05cm},\hspace{0.05cm} L_3 = -1.0\hspace{0.05cm}.$$

(2) The extrinsic $L$–values for the zeroth iteration result in (derivation in "Exercise 4.5Z"):

$$L_{\rm E}(1) = -0.1829, \ L_{\rm E}(2) = -0.4337, \ L_{\rm E}(3) = +0.1829.$$

(3) The a posteriori values at the beginning of the first iteration are thus

$$\underline{L}^{(I=1)} = \underline{L}^{(I=0)} + \underline{L}_{\hspace{0.02cm}\rm E}^{(I=0)} = (+0.8171\hspace{0.05cm},\hspace{0.05cm}-0.0337\hspace{0.05cm},\hspace{0.05cm}-0.8171) \hspace{0.05cm} . $$

(4) From this, the new extrinsic values for the iteration loop $I = 1$ are as follows:

$$L_{\rm E}(1) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}2 \cdot {\rm tanh}^{-1} \big [{\rm tanh}(-0.0337/2) \cdot {\rm tanh}(-0.8171/2) \big ] = 0.0130 = -L_{\rm E}(3)\hspace{0.05cm},$$

$$L_{\rm E}(2) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}2 \cdot {\rm tanh}^{-1} \big [{\rm tanh}(+0.8171/2) \cdot {\rm tanh}(-0.8171/2) \big ] = - 0.3023\hspace{0.05cm}.$$

Further, one can see from the above table:

A hard decision according to the signs before the first iteration $(I = 0)$ fails, since $(+1, +1, -1)$ is not a valid $\rm SPC \, (3, \, 2, \, 2)$–code word.
But already after $I = 1$ iterations, a hard decision yields a valid code word, namely $\underline{x}_2 = (+1, -1, -1)$. Also in later graphs, the rows with correct HD decisions for the first time are highlighted in blue.
Hard decisions after further iterations $(I ≥ 2)$ each lead to the same code word $\underline{x}_2$. This statement is not only valid for this example, but in general.

Besides, in this exercise we consider a second $L_{\rm E}(i)$ approach, which is given here for the example of the first symbol $(i = 1)$:

$${\rm sign} \big[L_{\rm E}(1)\big] \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm sign} \big[L_{\rm E}(2)\big] \cdot {\rm sign} \big[L_{\rm E}(3)\big]\hspace{0.05cm},\hspace{0.8cm} |L_{\rm E}(1)| \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm Min} \left ( |L_{\rm E}(2)|\hspace{0.05cm}, \hspace{0.05cm}|L_{\rm E}(3)| \right ) \hspace{0.05cm}.$$

This second approach is based on the assumption that the reliability of $L_{\rm E}(i)$ is essentially determined by the most unreliable neighbor symbol. The better (larger) input–LLR is completely disregarded. – Let us consider two examples for this:

(1) For $L_2 = 1.0$ and $L_3 = 5.0$ we get for example

after the first approach: $L_{\rm E}(1) =2 \cdot {\rm tanh}^{-1} \big [{\rm tanh}(0.5) \cdot {\rm tanh}(2.5) \big ] =2 \cdot {\rm tanh}^{-1}(0.4559) = 0.984\hspace{0.05cm},$
according to the second approach: $|L_{\rm E}(1)| \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm Min} \big ( 1.0\hspace{0.05cm}, \hspace{0.05cm}5.0 \big ) = 1.000 \hspace{0.05cm}.$

(2) On the other hand one obtains for $L_2 = L_3 = 1.0$

according to the first approach: $L_{\rm E}(1) =2 \cdot {\rm tanh}^{-1} \big [{\rm tanh}(0.5) \cdot {\rm tanh}(0.5) \big ] =2 \cdot {\rm tanh}^{-1}(0.2135) = 0.433\hspace{0.05cm},$
according to the second approach: $|L_{\rm E}(1)| \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm Min} \big ( 1.0\hspace{0.05cm}, \hspace{0.05cm}1.0 \big ) = 1.000 \hspace{0.05cm}.$

One can see the clear discrepancy between the two approaches. The second approach (approximation) is clearly more positive than the first (correct) approach. However, it is actually only important that the iterations lead to the desired decoding result.

Hints:

The exercise belongs to the chapter "Soft–in Soft–out Decoder".
Referred to in particular "Calculation of extrinsic LLRs".
Only the second solution approach is treated here.
For the first solution approach we refer to the "Exercise 4.5Z" .

Questions

Solution

Results for $\underline{L}=(+1.0, +0.4, –1.0)$

(1) According to the second $L_{\rm E}(i)$ approach holds:

$${\rm sign} [L_{\rm E}(1)] \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm sign} [L_{\rm E}(2)] \cdot {\rm sign} [L_{\rm E}(3)] = -1 \hspace{0.05cm},$$

$$|L_{\rm E}(1)| \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm Min} \left ( |L_{\rm E}(2)|\hspace{0.05cm}, \hspace{0.05cm}|L_{\rm E}(3)| \right ) = {\rm Min} \left ( 0.4\hspace{0.05cm}, \hspace{0.05cm}1.0 \right ) = 0.4\hspace{0.3cm} \Rightarrow \hspace{0.3cm}L_{\rm E}(1) \hspace{0.15cm} \underline{-0.4}\hspace{0.05cm}.$$

In the same way you get:

$$L_{\rm E}(2) \hspace{0.15cm} \underline{-1.0}\hspace{0.05cm}, \hspace{0.3cm} L_{\rm E}(3) \hspace{0.15cm} \underline{+0.4}\hspace{0.05cm}.$$

(2) The a posteriori $L$ values at the beginning of the first iteration $(I = 1)$ are the sum of the previous $L$–values (for $I = 0$) and the extrinsic values calculated in (1):

$$L_1 = L_{\rm APP}(1) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}1.0 + (-0.4)\hspace{0.15cm} \underline{=+0.6}\hspace{0.05cm},$$

$$L_2 = L_{\rm APP}(2) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 0.4 + (-1.0)\hspace{0.15cm} \underline{=-0.6}\hspace{0.05cm},$$

$$L_3 = L_{\rm APP}(3) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} (-1.0) + 0.4\hspace{0.15cm} \underline{=-0.6}\hspace{0.05cm}.$$

(3) As can be seen from the above table, solutions 1 and 2 are correct in contrast to answer 3:

With each new iteration, the amounts of $L(1), \ L(2)$ and $L(3)$ become significantly larger.

Results for $\underline{L}=(+0.6, +1.0, –0.4)$

(4) As can be seen from the adjacent table, answers 1 and 3 are correct:

So the decision is made for the code word $\underline{x}_0 = (+1, +1, +1)$.
From $I = 1$ this would also be the decision of hard decision.

Results for $\underline{L}=(+0.6, +1.0, –0.8)$

(5) Correct are answers 2 and 3:

Because $|L(3)| > |L(1)|$ the following is valid from $I = 1$: $L_1 < 0 \hspace{0.05cm},\hspace{0.2cm} L_2 > 0 \hspace{0.05cm},\hspace{0.2cm} L_3 < 0 \hspace{0.05cm}.$
From this iteration loop, hard decision returns the code word $\underline{x}_2 = (-1, +1, -1)$.

Results for $\underline{L}=(+0.6, +1.0, –0.6)$

(6) Correct is the proposed solution 3:

The adjacent table shows that under the condition $|L(1)| = |L(3)|$, starting from the iteration loop $I = 1$, all extrinsic $L$–values are zero.
Thus, the a posteriori $L$ values remain constantly equal to $\underline{L} = (0., +0.4, 0.)$ even for $I > 1$, which cannot be assigned to any code word.

	Hard decision after $I = 1$ leads to the code word $\underline{x}_1 = (+1, -1, -1)$.
	This does not change after further iterations.
	Further iterations do not increase the reliability for $\underline{x}_1$ .

	The iterative decoding leads to the result $\underline{x}_0 = (+1, +1, +1)$.
	The iterative decoding leads to the result $\underline{x}_2 = (-1, +1, -1)$.
	This result also returns hard decision $I = 1$.

	The iterative decoding produces the result $\underline{x}_0 = (+1, +1, +1)$.
	The iterative decoding leads to the result $\underline{x}_2 = (-1, +1, -1)$.
	This result also returns hard decision $I = 1$.

	Iterative decoding leads to the result $\underline{x}_0 = (+1, +1, +1)$.
	The iterative decoding leads to the result $\underline{x}_2 = (-1, +1, -1)$.
	The iterative decoding does not lead to the result here.