View source for Aufgaben:Exercise 4.3Z: Conversions of L-value and S-value

{{quiz-Header|Buchseite=Channel_Coding/Soft-in_Soft-Out_Decoder}}

[[File:P_ID3093__KC_Z_4_3neu_v1.png|right|right|frame|Function&nbsp; $y = \tanh {(x)}$&nbsp; <br>in tabular form.]]
We assume a binary random variable&nbsp; $x &#8712; \{+1, \, -1\}$&nbsp; with following probabilities:
:$${\rm Pr}(x =+1) = p\hspace{0.05cm},$$
:$${\rm Pr}(x =-1) = q = 1-p\hspace{0.05cm}.$$

The&nbsp; "reliability"&nbsp; of symbol&nbsp; $x$&nbsp; can be expressed
* by the&nbsp; L&ndash;value&nbsp; $($log likelihood ratio$)$&nbsp; according to the definition
:$$L(x) = {\rm ln} \hspace{0.2cm}  \frac{p}{q} = {\rm ln} \hspace{0.2cm} \frac{p}{1 - p}\hspace{0.05cm}
 \hspace{0.05cm},$$
* by the so called&nbsp; "S&ndash;value":
:$$S(x) = p- q \hspace{0.05cm}.$$

We have created the term&nbsp; "S&ndash;value"&nbsp; in order to be able to formulate the following questions more succinctly.&nbsp; In the literature,&nbsp; one sometimes finds the term&nbsp; "soft bit"&nbsp; for this.

As will be shown in subtask&nbsp; '''(1)''',&nbsp; $L(x)$&nbsp; and&nbsp; $S(x)$&nbsp; can be converted into each other.

Subsequently,&nbsp; these functions shall be used to calculate the following quantities,&nbsp; always assuming a code length&nbsp; $n = 3$:
* the extrinsic&nbsp; L&ndash;value for the third symbol&nbsp; &#8658; &nbsp; $L_{\rm E}(x_3)$,

* the a-posteriori&nbsp; L&ndash;value for the third symbol &nbsp; &#8658; &nbsp; $L_{\rm APP}(x_3)$.


The calculation should be done for the following codes:
* the repetition Code&nbsp; [[Channel_Coding/Examples_of_Binary_Block_Codes#Repetition_Codes|$\text{RC (3, 1, 3)}$]]&nbsp; with the constraint&nbsp; $\sign {(x_1)} = \sign {(x_2)} = \sign {(x_3)}$,

* the single parity&ndash;check code &nbsp; &#8658; &nbsp; [[Channel_Coding/Examples_of_Binary_Block_Codes#Single_Parity-check_Codes|$\text{SPC (3, 2, 2)}$]]&nbsp; with the constraint&nbsp; $x_1 \cdot x_2 \cdot x_3 = +1$.







Hints:
* This exercise belongs to the chapter&nbsp; [[Channel_Coding/Soft-in_Soft-Out_Decoder| "Soft&ndash;in Soft&ndash;out Decoder"]].

* Reference is made in particular to the section&nbsp; [[Channel_Coding/Soft-in_Soft-Out_Decoder#Reliability_information_-_Log_Likelihood_Ratio| "Reliability Information &ndash; Log Likelihood Ratio"]].

* To solve,&nbsp; you need the&nbsp; "hyperbolic tangent"&nbsp; according to the following definition&nbsp; $($this function is given above in tabular form$)$:
:$$y = {\rm tanh}(x) = \frac{{\rm e}^{+x/2} - {\rm e}^{-x/2}}{{\rm e}^{+x/2} + {\rm e}^{-x/2}} 
= \frac{1 - {\rm e}^{-x}}{1 + {\rm e}^{-x}} 
 \hspace{0.05cm}.$$



===Questions===
<quiz display=simple>
{What is the relationship between&nbsp; S&ndash;value and&nbsp; L&ndash;value?
|type="[]"}
- $S(x) = \tanh {(L(x))}$,
+ $S(x) = \tanh {(L(x)/2)}$,
+ $L(x) = 2 \cdot \tanh^{-1}{(S(x))}$.

{The&nbsp; $\text{RC (3, 1, 3)}$&nbsp; is considered.&nbsp; For the&nbsp; a-priori&nbsp; L&ndash;values, &nbsp; let be&nbsp; $\underline{L}_{\rm A} = (+2, -1, +3)$.&nbsp; What is the&nbsp; <u> extrinsicL&ndash;value</u>&nbsp; for the symbol&nbsp; $x_3$?
|type="{}"}
$L_{\rm E}(x_3) \ = \ ${ 1 3% }

{What is the&nbsp; <u>a-posteriori L&ndash;value</u>&nbsp; for the symbol&nbsp; $x_3$?
|type="{}"}
$L_{\rm APP}(x_3) \ = \ ${ 4 3% }

{What is the&nbsp; <u>extrinsic L&ndash;value</u>&nbsp;  at&nbsp; $\text{SPC (3, 2, 2)}$?&nbsp; It is still valid: &nbsp; $\underline{L}_{\rm A} = (+2, -1, +3)$.
|type="{}"}
$L_{\rm E}(x_3) \ = \ ${ -0.757256--0.713144 }

{The a-priori probabilities are now&nbsp; $0.3, \ 0.8, \ 0.9$.&nbsp; What is the&nbsp; <u>extrinsic&nbsp; L&ndash;value</u>&nbsp; for the&nbsp; <u>repetition code</u>?
|type="{}"}
$L_{\rm E}(x_3) \ = \ ${ 0.535 3% }

{What&nbsp; <u>extrinsic L&ndash;value</u>&nbsp; results for the&nbsp; <u>single parity&ndash;check code</u>&nbsp; given the same conditions as in subtask&nbsp; '''(5)'''?
|type="{}"}
$L_{\rm E}(x_3) \ = \ ${ 0.49 3% }
</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''&nbsp; For the binary random variable&nbsp; $x &#8712; \{+1, -1\}$&nbsp; with probabilities 
* $p = {\rm Pr}(x = +1)$,&nbsp; and

* $p = {\rm Pr}(x=-1) = 1-p$


the following definitions apply:
:$$L(x) =  {\rm ln} \hspace{0.2cm} \frac{p}{q} = {\rm ln} \hspace{0.2cm} \frac{p}{1 - p}\hspace{0.05cm}
 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} 
-\infty \le L(x) \le +\infty
\hspace{0.05cm},$$
:$$S(x) = p- q = 2 \cdot p - 1\hspace{0.05cm}
\hspace{0.3cm} \Rightarrow \hspace{0.3cm} 
-1 \le S(x) \le +1
 \hspace{0.05cm}.$$

*Based on the S&ndash;value,&nbsp; we get because of&nbsp; $p + q = 1$:
:$$S(x) = p- q = \frac{p- q}{p+  q} =  \frac{1- q/p}{1+ q/p}
 \hspace{0.05cm}.$$

*Simultaneously&nbsp; $q/p = {\rm e}^{-L(x)}$&nbsp; holds.&nbsp; From this follows:
:$$S(x) = \frac{1- {\rm e}^{-L(x)}}{1+ {\rm e}^{-L(x)}}
 \hspace{0.05cm}.$$
[[File:P_ID3097__KC_Z_4_3c_v2.png|right|frame|Relationship between probability,&nbsp; L&ndash;value and S&ndash;value]]

*Multiplying the numerator and denominator by&nbsp; ${\rm e}^{-L(x)/2}$,&nbsp; we finally get:
:$$S(x) = \frac{{\rm e}^{+L(x)/2}- {\rm e}^{-L(x)/2}}{{\rm e}^{+L(x)/2}+ {\rm e}^{-L(x)/2}}
= {\rm tanh}\big [L(x)/2. \big]
 \hspace{0.05cm}.$$

*The inverse function results in
:$$L(x) = 2 \cdot {\rm tanh}^{-1}[S(x)]
 \hspace{0.05cm}.$$

*Thus,&nbsp; the&nbsp; <u>proposed solutions 2 and 3</u>&nbsp; are correct.&nbsp; The table shows the&nbsp; L&ndash;value&nbsp; and the&nbsp;  S&ndash;value&nbsp; for some probabilities&nbsp; $p = {\rm Pr}(x=+1)$.


'''(2)'''&nbsp; The&nbsp; "extrinsic L&ndash;value"&nbsp; for symbol&nbsp; $x_3$&nbsp; considers only the&nbsp; "a-priori L&ndash;values"&nbsp; $L_{\rm A}(x_1)$&nbsp; and&nbsp; $L_{\rm A}(x_2)$,&nbsp; but not&nbsp; $L_{\rm A}(x_3)$. 
*For the&nbsp; $\text{(3, 1)}$&nbsp; repetition code,&nbsp; this results in: 
:$$L_{\rm E}(x_3) = L_{\rm A}(x_1) + L_{\rm A}(x_2) = 2 + (-1)
\hspace{0.15cm} \underline{= +1}\hspace{0.05cm}.$$


'''(3)'''&nbsp; Thus,&nbsp; for the&nbsp; "a-posteriori L&ndash;value",&nbsp; we obtain:
:$$L_{\rm APP}(x_3) = L_{\rm A}(x_3) + L_{\rm E}(x_3) = 3 + 1
\hspace{0.15cm} \underline{= +4}\hspace{0.05cm}.$$


'''(4)'''&nbsp; In the single parity&ndash;check code,&nbsp; the corresponding calculation rule is:
:$$L_{\rm E}(x_3) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 2 \cdot {\rm tanh}^{-1} \hspace{0.05cm}
 \left [  {\rm tanh}(x_1/2) \cdot {\rm tanh}(x_2/2)  \right ] =  2 \cdot {\rm tanh}^{-1} \hspace{0.05cm}
 \left [  {\rm tanh}(+1) \cdot {\rm tanh}(-0.5)  \right ] =
2 \cdot {\rm tanh}^{-1} \hspace{0.05cm}
 \left [  0.7616 \cdot (-0.4621)  \right ] $$
:$$\Rightarrow \hspace{0.3cm}L_{\rm E}(x_3) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 2 \cdot {\rm tanh}^{-1} \hspace{0.05cm}
 \left [  -0.3519  \right ] =-2 \cdot 0.3676\hspace{0.15cm} \underline{= -0.7352}\hspace{0.05cm}.$$

*The result&nbsp; ${\rm tanh}^{-1} (-0.3519) = 0.3676$&nbsp; was taken from the table on the information page.




'''(5)'''&nbsp; For the repetition code of length&nbsp; $n = 3$,&nbsp; the same holds as in subtask&nbsp; '''(3)''': 
:$$L_{\rm E}(x_3) = L_{\rm A}(x_1) + L_{\rm A}(x_2) = -0.847 +1.382
\hspace{0.15cm} \underline{= +0.535}\hspace{0.05cm}.$$

*The L&ndash;values corresponding to the table for subtask&nbsp; '''(1)'''&nbsp; were used here,&nbsp; for example ${\rm Pr}(x_1 = +1) = 0.3$ &nbsp; &rArr; &nbsp; $L_{\rm A}(x_1) = -0.847$.



'''(6)'''&nbsp; Since here instead of the&nbsp; "a-priori L-values"&nbsp; the&nbsp; "a-priori probabilities"&nbsp; are given,&nbsp; <br>one comes faster to success in comparison with the subtask&nbsp; '''(4)'''&nbsp; on the detour over the&nbsp; "extrinsic S-value".

*We denote the extrinsic probability for the third symbol here by&nbsp; $P_{\rm E}(x_3)$.&nbsp; For this holds:
:$$P_{\rm E}(x_3 = +1) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} P_{\rm A}(x_1 = +1) \cdot P_{\rm A}(x_2 = -1) + 
P_{\rm A}(x_1 = -1) \cdot P_{\rm A}(x_2 = +1) = 0.3 \cdot (1-0.8) + (1-0.3) \cdot 0.8 = 0.62\hspace{0.05cm}.$$

*This results in for the further variables:
:$$S_{\rm E}(x_3) = P_{\rm E}(x_3 = +1) - P_{\rm E}(x_3 = - 1) =  0.62 -0.38 = 0.24\hspace{0.05cm},$$
:$$L_{\rm E}(x_3) = 2 \cdot {\rm tanh}^{-1} \hspace{0.05cm}
 \left [  S_{\rm E}(x_3) \right ] 
= 2 \cdot {\rm tanh}^{-1} \hspace{0.05cm}
 (0.24) = 2 \cdot 0.245
\hspace{0.15cm} \underline{= +0.49}\hspace{0.05cm}$$
{{ML-Fuß}}



[[Category:Channel Coding: Exercises|^4.1 Soft–in Soft–out Decoder^]]