Difference between revisions of "Aufgaben:Exercise 4.3Z: Conversions of L-value and S-value"
Line 8: | Line 8: | ||
The "reliability" of symbol $x$ can be expressed | The "reliability" of symbol $x$ can be expressed | ||
* by the L–value $($log likelihood ratio$)$ according to the definition | * by the L–value $($log likelihood ratio$)$ according to the definition | ||
− | :$$L(x) = {\rm ln} \hspace{0.2cm} \frac{p}{q} = \frac{p}{1 - p}\hspace{0.05cm} | + | :$$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},$$ | \hspace{0.05cm},$$ | ||
* by the so called "S–value": | * by the so called "S–value": | ||
Line 77: | Line 77: | ||
===Solution=== | ===Solution=== | ||
{{ML-Kopf}} | {{ML-Kopf}} | ||
− | '''(1)''' For the binary random variable $x ∈ \{+1, -1\}$ with probabilities | + | '''(1)''' For the binary random variable $x ∈ \{+1, -1\}$ with probabilities |
− | * $p = {\rm Pr}(x = +1)$, and | + | * $p = {\rm Pr}(x = +1)$, and |
+ | |||
* $p = {\rm Pr}(x=-1) = 1-p$ | * $p = {\rm Pr}(x=-1) = 1-p$ | ||
the following definitions apply: | the following definitions apply: | ||
− | :$$L(x) = {\rm ln} \hspace{0.2cm} | + | :$$L(x) = {\rm ln} \hspace{0.2cm} {\rm ln} \hspace{0.2cm} \frac{p}{q} = \frac{p}{1 - p}\hspace{0.05cm} |
\hspace{0.3cm} \Rightarrow \hspace{0.3cm} | \hspace{0.3cm} \Rightarrow \hspace{0.3cm} | ||
-\infty \le L(x) \le +\infty | -\infty \le L(x) \le +\infty | ||
Line 92: | Line 93: | ||
\hspace{0.05cm}.$$ | \hspace{0.05cm}.$$ | ||
− | *Based on the | + | *Based on the S–value, we get because of $p + q = 1$: |
:$$S(x) = p- q = \frac{p- q}{p+ q} = \frac{1- q/p}{1+ q/p} | :$$S(x) = p- q = \frac{p- q}{p+ q} = \frac{1- q/p}{1+ q/p} | ||
\hspace{0.05cm}.$$ | \hspace{0.05cm}.$$ | ||
− | *Simultaneously $q/p = {\rm e}^{-L(x)}$ holds. From this follows: | + | *Simultaneously $q/p = {\rm e}^{-L(x)}$ holds. From this follows: |
:$$S(x) = \frac{1- {\rm e}^{-L(x)}}{1+ {\rm e}^{-L(x)}} | :$$S(x) = \frac{1- {\rm e}^{-L(x)}}{1+ {\rm e}^{-L(x)}} | ||
\hspace{0.05cm}.$$ | \hspace{0.05cm}.$$ | ||
+ | [[File:P_ID3097__KC_Z_4_3c_v2.png|right|frame|Relationship between probability, L–value and S–value]] | ||
− | *Multiplying the numerator and denominator by ${\rm e}^{-L(x)/2}$, we finally get: | + | *Multiplying the numerator and denominator by ${\rm e}^{-L(x)/2}$, we finally get: |
− | |||
:$$S(x) = \frac{{\rm e}^{+L(x)/2}- {\rm e}^{-L(x)/2}}{{\rm e}^{+L(x)/2}+ {\rm e}^{-L(x)/2}} | :$$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] | = {\rm tanh}\big [L(x)/2. \big] | ||
Line 109: | Line 110: | ||
:$$L(x) = 2 \cdot {\rm tanh}^{-1}[S(x)] | :$$L(x) = 2 \cdot {\rm tanh}^{-1}[S(x)] | ||
\hspace{0.05cm}.$$ | \hspace{0.05cm}.$$ | ||
− | + | ||
− | Thus, the <u>proposed solutions 2 and 3</u> are correct. The table shows the | + | *Thus, the <u>proposed solutions 2 and 3</u> are correct. The table shows the L–value and the S–value for some probabilities $p = {\rm Pr}(x=+1)$. |
− | '''(2)''' The extrinsic | + | '''(2)''' The "extrinsic L–value" for symbol $x_3$ considers only the "apriori L–values" $L_{\rm A}(x_1)$ and $L_{\rm A}(x_2)$, but not $L_{\rm A}(x_3)$. |
− | *For the (3, 1) | + | *For the $\text{(3, 1)}$ repetition code, this results in: |
:$$L_{\rm E}(x_3) = L_{\rm A}(x_1) + L_{\rm A}(x_2) = 2 + (-1) | :$$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}.$$ | \hspace{0.15cm} \underline{= +1}\hspace{0.05cm}.$$ | ||
− | '''(3)''' Thus, for the | + | '''(3)''' Thus, for the "a-posteriori L–value", we obtain: |
:$$L_{\rm APP}(x_3) = L_{\rm A}(x_3) + L_{\rm E}(x_3) = 3 + 1 | :$$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}.$$ | \hspace{0.15cm} \underline{= +4}\hspace{0.05cm}.$$ | ||
− | '''(4)''' In the | + | '''(4)''' In the single parity–check code, the corresponding calculation rule is: |
:$$L_{\rm E}(x_3) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 2 \cdot {\rm tanh}^{-1} \hspace{0.05cm} | :$$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}(x_1/2) \cdot {\rm tanh}(x_2/2) \right ] = 2 \cdot {\rm tanh}^{-1} \hspace{0.05cm} | ||
Line 133: | Line 134: | ||
\left [ -0.3519 \right ] =-2 \cdot 0.3676\hspace{0.15cm} \underline{= -0.7352}\hspace{0.05cm}.$$ | \left [ -0.3519 \right ] =-2 \cdot 0.3676\hspace{0.15cm} \underline{= -0.7352}\hspace{0.05cm}.$$ | ||
− | The result ${\rm tanh}^{-1} (-0.3519) = 0.3676$ was taken from the table on the information page. | + | *The result ${\rm tanh}^{-1} (-0.3519) = 0.3676$ was taken from the table on the information page. |
+ | |||
− | '''(5)''' For the repetition code of length $n = 3$, the same holds as in subtask (3): | + | '''(5)''' For the repetition code of length $n = 3$, the same holds as in subtask '''(3)''': |
:$$L_{\rm E}(x_3) = L_{\rm A}(x_1) + L_{\rm A}(x_2) = -0.847 +1.382 | :$$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}.$$ | \hspace{0.15cm} \underline{= +0.535}\hspace{0.05cm}.$$ | ||
− | The | + | *The L–values corresponding to the table for subtask '''(1)''' were used here, for example ${\rm Pr}(x_1 = +1) = 0.3$ ⇒ $L_{\rm A}(x_1) = -0.847$. |
+ | |||
− | '''(6)''' Since here instead of the | + | '''(6)''' Since here instead of the "a-priori L-values" the "a-priori probabilities" are given, <br>one comes faster to success in comparison with the subtask '''(4)''' on the detour over the "extrinsic S-value". |
− | *We denote the extrinsic probability for the third symbol here by $P_{\rm E}(x_3)$. For this holds: | + | *We denote the extrinsic probability for the third symbol here by $P_{\rm E}(x_3)$. 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 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}.$$ | P_{\rm A}(x_1 = -1) \cdot P_{\rm A}(x_2 = +1) = 0.3 \cdot (1-0.8) + (1-0.3) \cdot 0.8 = 0.62\hspace{0.05cm}.$$ |
Revision as of 19:40, 29 November 2022
We assume a binary random variable $x ∈ \{+1, \, -1\}$ with following probabilities:
- $${\rm Pr}(x =+1) = p\hspace{0.05cm},$$
- $${\rm Pr}(x =-1) = q = 1-p\hspace{0.05cm}.$$
The "reliability" of symbol $x$ can be expressed
- by the L–value $($log likelihood ratio$)$ 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 "S–value":
- $$S(x) = p- q \hspace{0.05cm}.$$
We have created the term "S–value" in order to be able to formulate the following questions more succinctly. In the literature, one sometimes finds the term "soft bit" for this.
As will be shown in subtask (1), $L(x)$ and $S(x)$ can be converted into each other.
Subsequently, these functions shall be used to calculate the following quantities, always assuming a code length $n = 3$:
- the extrinsic L–value for the third symbol ⇒ $L_{\rm E}(x_3)$,
- the a-posteriori L–value for the third symbol ⇒ $L_{\rm APP}(x_3)$.
The calculation should be done for the following codes:
- the repetition Code $\text{RC (3, 1, 3)}$ with the constraint $\sign {(x_1)} = \sign {(x_2)} = \sign {(x_3)}$,
- the single parity–check code ⇒ $\text{SPC (3, 2, 2)}$ with the constraint $x_1 \cdot x_2 \cdot x_3 = +1$.
Hints:
- This exercise belongs to the chapter "Soft–in Soft–out Decoder".
- Reference is made in particular to the section "Reliability Information – Log Likelihood Ratio".
- To solve, you need the "hyperbolic tangent" according to the following definition $($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
Solution
- $p = {\rm Pr}(x = +1)$, and
- $p = {\rm Pr}(x=-1) = 1-p$
the following definitions apply:
- $$L(x) = {\rm ln} \hspace{0.2cm} {\rm ln} \hspace{0.2cm} \frac{p}{q} = \frac{p}{1 - p}\hspace{0.05cm} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} -\infty \le L(x) \le +\infty \hspace{0.05cm},$$
- $$S(x) = p- q = 2 \cdot p - 1\hspace{0.05cm} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} -1 \le S(x) \le +1 \hspace{0.05cm}.$$
- Based on the S–value, we get because of $p + q = 1$:
- $$S(x) = p- q = \frac{p- q}{p+ q} = \frac{1- q/p}{1+ q/p} \hspace{0.05cm}.$$
- Simultaneously $q/p = {\rm e}^{-L(x)}$ holds. From this follows:
- $$S(x) = \frac{1- {\rm e}^{-L(x)}}{1+ {\rm e}^{-L(x)}} \hspace{0.05cm}.$$
- Multiplying the numerator and denominator by ${\rm e}^{-L(x)/2}$, 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, the proposed solutions 2 and 3 are correct. The table shows the L–value and the S–value for some probabilities $p = {\rm Pr}(x=+1)$.
(2) The "extrinsic L–value" for symbol $x_3$ considers only the "apriori L–values" $L_{\rm A}(x_1)$ and $L_{\rm A}(x_2)$, but not $L_{\rm A}(x_3)$.
- For the $\text{(3, 1)}$ repetition code, 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) Thus, for the "a-posteriori L–value", 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) In the single parity–check code, 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 ${\rm tanh}^{-1} (-0.3519) = 0.3676$ was taken from the table on the information page.
(5) For the repetition code of length $n = 3$, the same holds as in subtask (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–values corresponding to the table for subtask (1) were used here, for example ${\rm Pr}(x_1 = +1) = 0.3$ ⇒ $L_{\rm A}(x_1) = -0.847$.
(6) Since here instead of the "a-priori L-values" the "a-priori probabilities" are given,
one comes faster to success in comparison with the subtask (4) on the detour over the "extrinsic S-value".
- We denote the extrinsic probability for the third symbol here by $P_{\rm E}(x_3)$. 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:
- $$S_{\rm E}(x_3) = P_{\rm E}(x_3 = +1) - P_{\rm E}(x_3 = - 1) = 0.62 -0.38 = 0.24\hspace{0.05cm},$$
- $$L_{\rm E}(x_3) = 2 \cdot {\rm tanh}^{-1} \hspace{0.05cm} \left [ S_{\rm E}(x_3) \right ] = 2 \cdot {\rm tanh}^{-1} \hspace{0.05cm} (0.24) = 2 \cdot 0.245 \hspace{0.15cm} \underline{= +0.49}\hspace{0.05cm}$$