Exercise 4.4: Extrinsic L-values at SPC

Suitable auxiliary table

We consider again the "single parity–check code". In such a ${\rm SPC} \ (n, \, n-1, \, 2)$ the $n$ bits of a code word $\underline{x}$ come from the $k = n -1$ bits from the source sequence $\underline{u}$ and only a single check bit $p$ is added, such that the number of "ones" in the code word $\underline{x}$ is even:

$$\underline{x} = \big ( \hspace{0.03cm}x_1, \hspace{0.03cm} x_2, \hspace{0.05cm} \text{...} \hspace{0.05cm} , x_{n-1}, \hspace{0.03cm} x_n \hspace{0.03cm} \big ) = \big ( \hspace{0.03cm}u_1, \hspace{0.03cm} u_2, \hspace{0.05cm} \text{...} \hspace{0.05cm} , u_{k}, \hspace{0.03cm} p \hspace{0.03cm} \big )\hspace{0.03cm}. $$

The extrinsic information about the $i$th code bit is formed over all other bits $(j ≠ i)$. Therefore we write for the code word shorter by one bit:

$$\underline{x}^{(-i)} = \big ( \hspace{0.03cm}x_1, \hspace{0.05cm} \text{...} \hspace{0.05cm} , \hspace{0.03cm} x_{i-1}, \hspace{0.43cm} x_{i+1}, \hspace{0.05cm} \text{...} \hspace{0.05cm} , x_{n} \hspace{0.03cm} \big )\hspace{0.03cm}. $$

The extrinsic L–value over the $i$th code symbol reads with the "Hamming weight" $w_{\rm H}$ of the truncated sequence $\underline{x}^{(-i)}$:

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

If the probability in the numerator is greater than that in the denominator, then $L_{\rm E}(i) > 0$ and thus the a-posteriori L–value $L_{\rm APP}(i) = L_{\rm A}(i) + L_{\rm E}(i)$ magnified, that is tends to be affected in the direction of the symbol $x_i = 0$.

If $L_{\rm E}(i) < 0$ then there is much to be said for $x_i = 1$ from the point of view of the other symbols $(j ≠ i)$.

Only the $\text{SPC (4, 3, 4)}$ is treated, where for the probabilities $p_i = {\rm Pr}(x_i = 1)$ holds:

$$p_1 = 0.2 \hspace{0.05cm}, \hspace{0.3cm} p_2 = 0.9 \hspace{0.05cm}, \hspace{0.3cm} p_3 = 0.3 \hspace{0.05cm}, \hspace{0.3cm} p_4 = 0.6 \hspace{0.05cm}.$$

From this the a-priori log likelihood ratios result to:

$$L_{\rm A}(i) = {\rm ln} \hspace{0.1cm} \left [ \frac{{\rm Pr}(x_i = 0)}{{\rm Pr}(x_i = 1)} \right ] = {\rm ln} \hspace{0.1cm} \left [ \frac{1-p_i}{p_i} \right ] \hspace{0.05cm}.$$

Hints:

This exercise belongs to the chapter "Soft–in Soft–out Decoder".

Reference is made in particular to the section "Calculation of the extrinsic log likelihood ratios".

In the table are given for $p_i = 0$ to $p_i = 1$ with step size $0.1$ $($column 1$)$:

In column 2: the probability $q_i = {\rm Pr}(x_i = 0) = 1 - p_i$,

in column 3: the values for $1 - 2p_i$,

in column 4: the a-priori log likelihood ratios $L_i = \ln {\big [(1 - p_i)/p_ i \big ]} = L_{\rm A}(i)$.

The "hyperbolic tangent" $(\tanh)$ of $L_i/2$ is identical to $1-2p_i$ ⇒ column 3.

In $\text{Exercise 4.4Z}$ it is shown that for the extrinsic L–value can also be written:

$$L_{\rm E}(i) = {\rm ln} \hspace{0.2cm} \frac{1 + \pi}{1 - \pi}\hspace{0.05cm}, \hspace{0.3cm} {\rm mit} \hspace{0.3cm} \pi = \prod\limits_{j \ne i}^{n} \hspace{0.25cm}(1-2p_j) \hspace{0.05cm}.$$

Questions

$L_{\rm A}(i = 1) \ = \ $

$L_{\rm A}(i = 2) \ = \ $

$L_{\rm E}(i = 1) \ = \ $

$L_{\rm E}(i = 2) \ = \ $

	It holds $p_j = 1/(1 + {\rm e}^ {L_j})$.
	It holds $1-2p_j = ({\rm e}^ {L_j} - 1) \ / \ ({\rm e}^ {L_j} + 1)$.
	It holds $1-2p_j = \tanh {(L_j/2)}$.

$L_{\rm E}(i = 3) \ = \ $

$L_{\rm E}(i = 4) \ = \ $

Solution

(1) For the a-priori log likelihood ratios of the first two bits of the code word:

$$L_{\rm A}(i = 1) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm ln} \hspace{0.1cm} \left [ \frac{1-p_1}{p_1} \right ] = {\rm ln} \hspace{0.1cm} 4 \hspace{0.15cm}\underline{= +1.386} \hspace{0.05cm},$$

$$L_{\rm A}(i = 2) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm ln} \hspace{0.1cm} \left [ \frac{1-p_2}{p_2} \right ] = {\rm ln} \hspace{0.1cm} 1/9 \hspace{0.15cm}\underline{= -2.197} \hspace{0.05cm}.$$

The values can be read from the fourth column of the table attached to the information page.

(2) To calculate the extrinsic L–value over the $i$th bit, only the information about the other three bits $(j ≠ i)$ may be used. With the given equation holds:

$$L_{\rm E}(i = 1) = {\rm ln} \hspace{0.2cm} \frac{1 + \prod\limits_{j \ne 1} \hspace{0.25cm}(1-2p_j)}{1 - \prod\limits_{j \ne 1} \hspace{0.25cm}(1-2p_j)} \hspace{0.05cm}.$$

For the product, we obtain according to the third column of the $\rm table$:

$$\prod\limits_{j =2, \hspace{0.05cm}3,\hspace{0.05cm} 4} \hspace{0.05cm}(1-2p_j) = (-0.8) \cdot (+0.4) \cdot (-0.2) = 0.064 \hspace{0.05cm}\hspace{0.3cm} \Rightarrow \hspace{0.3cm}L_{\rm E}(i = 1) = {\rm ln} \hspace{0.2cm} \frac{1 + 0.064}{1 - 0.064} = {\rm ln} \hspace{0.1cm} (1.137)\hspace{0.15cm}\underline{= +0.128} \hspace{0.05cm}.$$

In terms of bit 2, one obtains accordingly:

$$\prod\limits_{j =1, \hspace{0.05cm}3,\hspace{0.05cm} 4} \hspace{0.05cm}(1-2p_j) = (+0.6) \cdot (+0.4) \cdot (-0.2) = -0.048 \hspace{0.05cm}\hspace{0.3cm} \Rightarrow \hspace{0.3cm}L_{\rm E}(i = 2) = {\rm ln} \hspace{0.2cm} \frac{1 -0.048}{1 +0.048} = {\rm ln} \hspace{0.1cm} (0.908)\hspace{0.15cm}\underline{= -0.096} \hspace{0.05cm}.$$

(3) For the a-priori L– value holds:

$$L_j = L_{\rm A}(j) = {\rm ln} \hspace{0.1cm} \left [ \frac{{\rm Pr}(x_j = 0)}{{\rm Pr}(x_j = 1)} \right ] = {\rm ln} \hspace{0.1cm} \left [ \frac{1-p_j}{p_j} \right ]\hspace{0.3cm} \Rightarrow \hspace{0.3cm} 1-p_j = p_j \cdot {\rm e}^{L_j} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} p_j = \frac{1}{1+{\rm e}^{L_j} } \hspace{0.05cm} .$$

Thus also applies:

$$1- 2 \cdot p_j = 1 - \frac{2}{1+{\rm e}^{L_j} } = \frac{1+{\rm e}^{L_j}-2}{1+{\rm e}^{L_j} } = \frac{{\rm e}^{L_j}-1}{{\rm e}^{L_j} +1}\hspace{0.05cm} .$$

Multiplying the numerator and denominator by ${\rm e}^{-L_j/2}$, we get:

$$1- 2 \cdot p_j = \frac{{\rm e}^{L_j/2}-{\rm e}^{-L_j/2}}{{\rm e}^{L_j/2}+{\rm e}^{-L_j/2}}={\rm tanh} (L_j/2) \hspace{0.05cm} .$$

Thus all proposed solutions are correct.

The function "hyperbolic tangent" can be found, for example, in tabular form in formula collections or in the last column of the table given in front.

(4) We first calculate $L_{\rm E}(i = 3)$ in the same way as in subtask (2):

$$\prod\limits_{j =1, \hspace{0.05cm}2,\hspace{0.05cm} 4} \hspace{0.05cm}(1-2p_j) = (+0.6) \cdot (-0.8) \cdot (-0.2) = +0.096 \hspace{0.05cm}\hspace{0.3cm} \Rightarrow \hspace{0.3cm}L_{\rm E}(i = 3) = {\rm ln} \hspace{0.2cm} \frac{1 +0.096}{1 -0.096} = {\rm ln} \hspace{0.1cm} (1.212)\hspace{0.15cm}\underline{= +0.193} \hspace{0.05cm}.$$

We calculate the extrinsic L–value with respect to the last bit according to the equation

$$L_{\rm E}(i = 4) = {\rm ln} \hspace{0.2cm} \frac{1 + \pi}{1 - \pi}\hspace{0.05cm}, \hspace{0.3cm} {\rm with} \hspace{0.3cm} \pi = {\rm tanh}(L_1/2) \cdot {\rm tanh}(L_2/2) \cdot {\rm tanh}(L_3/2) \hspace{0.05cm}.$$

This results in accordance with the above $\rm table$:

$$p_1 = 0.2 \hspace{0.2cm}\Rightarrow \hspace{0.2cm} L_1 = +1.386 \hspace{0.2cm}\Rightarrow \hspace{0.2cm} L_1/2 = +0.693 \hspace{0.2cm} \Rightarrow \hspace{0.2cm} {\rm tanh}(L_1/2) = \frac{{\rm e}^{+0.693}-{\rm e}^{-0.693}}{{\rm e}^{+0.693}+{\rm e}^{-0.693}} = 0.6 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm identisch \hspace{0.15cm}mit\hspace{0.15cm} }1-2\cdot p_1\hspace{0.05cm},$$

$$p_2 = 0.9 \hspace{0.2cm}\Rightarrow \hspace{0.2cm} L_2 = -2.197 \hspace{0.2cm}\Rightarrow \hspace{0.2cm} L_2/2 = -1.099\hspace{0.2cm} \Rightarrow \hspace{0.2cm} {\rm tanh}(L_2/2) = \frac{{\rm e}^{-1.099}-{\rm e}^{+1.099}}{{\rm e}^{-1.099}+{\rm e}^{+1.099}} = -0.8 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm identisch \hspace{0.15cm}mit\hspace{0.15cm} }1-2\cdot p_2\hspace{0.05cm},$$

$$p_3 = 0.3 \hspace{0.2cm}\Rightarrow \hspace{0.2cm} L_3 = 0.847 \hspace{0.2cm}\Rightarrow \hspace{0.2cm} L_3/2 = +0.419 \hspace{0.2cm} \Rightarrow \hspace{0.2cm} {\rm tanh}(L_3/2) = \frac{{\rm e}^{+0.419}-{\rm e}^{-0.419}}{{\rm e}^{+0.419}+{\rm e}^{-0.419}} = 0.4 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm identisch \hspace{0.15cm}mit\hspace{0.15cm} }1-2\cdot p_3\hspace{0.05cm}.$$

The final result is thus:

$$\pi = (+0.6) \cdot (-0.8) \cdot (+0.4) = -0.192 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} L_{\rm E}(i = 4) = {\rm ln} \hspace{0.2cm} \frac{1 -0.192}{1 +0.192}\hspace{0.15cm}\underline{= -0.389} \hspace{0.05cm}.$$