Exercise 4.5: On the Extrinsic L-values again

Tabelle nach dem ersten $L_{\rm E}(i)$–Ansatz

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

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

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

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

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

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.

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

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 "For calculation of extrinsic L values".
Only the second solution approach is treated here.
For the first solution approach we refer to the "Exercise 4.5Z" .

Questions

Musterlösung

Solution

Ergebnisse für $\underline{L}=(+1.0, +0.4, –1.0)$

(1) Entsprechend dem zweiten $L_{\rm E}(i)$–Ansatz gilt:

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

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

In gleicher Weise erhält man:

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

(2) Die Aposteriori–$L$–Werte zu Beginn der ersten Iteration $(I = 1)$ ergeben sich aus der Summe der bisherigen $L$–Werte (für $I = 0$) und den unter (1) berechneten extrinsischen Werten:

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

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

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

(3) Wie aus obiger Tabelle hervorgeht, sind die Lösungsvorschläge 1 und 2 richtig im Gegensatz zur Antwort 3:

Mit jeder neuen Iteration werden die Beträge von $L(1), \ L(2)$ und $L(3)$ signifikant größer.

Ergebnisse für $\underline{L}=(+0.6, +1.0, –0.4)$

(4) Wie aus nebenstehender Tabelle hervorgeht, sind die Antworten 1 und 3 richtig:

Die Entscheidung fällt also für das Codewort $\underline{x}_0 = (+1, +1, +1)$.
Ab $I = 1$ wäre dies auch die Entscheidung von Hard Decision.

Ergebnisse für $\underline{L}=(+0.6, +1.0, –0.8)$

(5) Richtig sind die Antworten 2 und 3:

Wegen $|L(3)| > |L(1)|$ gilt bereits ab $I = 1$: $L_1 < 0 \hspace{0.05cm},\hspace{0.2cm} L_2 > 0 \hspace{0.05cm},\hspace{0.2cm} L_3 < 0 \hspace{0.05cm}.$
Ab dieser Iterationsschleife liefert Hard Decision das Codewort $\underline{x}_2 = (-1, +1, -1)$.

Ergebnisse für $\underline{L}=(+0.6, +1.0, –0.6)$

(6) Richtig sind der Lösungsvorschlag 3:

Die nebenstehende Tabelle zeigt, dass unter der Voraussetzung $|L(1)| = |L(3)|$ ab der Iterationsschleife $I = 1$ alle extrinsischen $L$–Werte Null sind.
Damit bleiben die Aposteriori–$L$–Werte auch für $I > 1$ konstant gleich $\underline{L} = (0., +0.4, 0.)$, was keinem Codewort zugeordnet werden kann.

	Hard Decision nach $I = 1$ führt zum Codewort $\underline{x}_1 = (+1, -1, -1)$.
	Daran ändert sich auch nach weiteren Iterationen nichts.
	Weitere Iterationen erhöhen die Zuverlässigkeit für $\underline{x}_1$ nicht.

	Die iterative Decodierung führt zum Ergebnis $\underline{x}_0 = (+1, +1, +1)$.
	Die iterative Decodierung führt zum Ergebnis $\underline{x}_2 = (-1, +1, -1)$.
	Dieses Ergebnis liefert auch Hard Decision ab $I = 1$.

	Die iterative Decodierung führt zum Ergebnis $\underline{x}_0 = (+1, +1, +1)$.
	Die iterative Decodierung führt zum Ergebnis $\underline{x}_2 = (-1, +1, -1)$.
	Die iterative Decodierung führt hier nicht zum Ziel.