'''(1)''' Correct is the <u>proposed solution 3</u>:
'''(1)''' Correct is the <u>proposed solution 3</u>:
*The received word $\underline{y}_2 = (0, 1, 0)$ is not a valid code word of the <i>single parity–check code</i> SPC (3, 2). Thus, the first statement is false.
*The received word $\underline{y}_2 = (0, 1, 0)$ is not a valid code word of the single parity–check code $\text{SPC (3, 2)}$. Thus, the first statement is false.
*In addition, since the SPC (3, 2) has only the minimum distance $d_{\rm min} = 2$, no error can be corrected.
*In addition, since the $\text{SPC (3, 2)}$ has only the minimum distance $d_{\rm min} = 2$, no error can be corrected.
'''(2)''' Correct is the <u>proposed solution 2</u>:
*The possible code words at RP (3, 1) are $\underline{x}_0 = (0, 0, 0)$ and $\underline{x}_1 = (1, 1, 1)$.
*The minimum distance of this code is $d_{\rm min} = 3$, so $t = (d_{\rm min} \, - 1)/2 = 1$ error can be corrected.
*In addition to $\underline{y}_0 = (0, 0, 0)$, $\underline{y}_1 = (0, 0, 1), \ \underline{y}_2 = (0, 1, 0)$, and $\underline{y}_4 = (1, 0, 0)$ are also assigned to the decoding result $\underline{x}_0 = (0, 0, 0)$.
'''(2)''' Correct is the <u>proposed solution 2</u>:
*The possible code words at the $\text{RP (3, 1)}$ are $\underline{x}_0 = (0, 0, 0)$ and $\underline{x}_1 = (1, 1, 1)$.
*The minimum distance of this code is $d_{\rm min} = 3$, so $t = (d_{\rm min} \, - 1)/2 = 1$ error can be corrected.
*In addition to $\underline{y}_0 = (0, 0, 0)$, the received words $\underline{y}_1 = (0, 0, 1), \ \underline{y}_2 = (0, 1, 0)$, and $\underline{y}_4 = (1, 0, 0)$ are also assigned to the decoding result $\underline{x}_0 = (0, 0, 0)$.
'''(3)''' According to the BSC model, the conditional probability is that $\underline{y}_2 = (0, 1, 0)$ is received, given that $\underline{x}_0 = (0, 0, 0)$ was sent:
'''(3)''' According to the BSC model, the conditional probability that $\underline{y}_2 = (0, 1, 0)$ is received, given that $\underline{x}_0 = (0, 0, 0)$ was sent:
*The first term $(1 \, –\varepsilon)^2$ indicates the probability that the first and the third bit were transmitted correctly and $\varepsilon$ considers the corruption probability for the second bit.
*The first term $(1 \, –\varepsilon)^2$ indicates the probability that the first and third bit were transmitted correctly. $\varepsilon$ describes the falsification probability for the second bit.
*Correspondingly, for the second possible code word $\underline{x}_1 = (1, 1, 1)$:
*Correspondingly, for the second possible code word $\underline{x}_1 = (1, 1, 1)$:
'''(4)''' The sign of the channel log likelihood ratios $L_{\rm K}(i)$ is
*positive if $y_i = 0$, and
*negative for $y_i = 1$.
'''(4)''' The sign of the channel LLR $L_{\rm K}(i)$ is positive if $y_i = 0$, and negative for $y_i = 1$.
*The absolute value indicates the reliability of $y_i$. In the BSC model, $|L_{\rm K}(i)| = \ln {(1 \, – \varepsilon)/\varepsilon} = 1$ for all $i$. Thus:
The absolute value $|L_{\rm K}(i)|$ indicates the reliability of $y_i$.
[[File:P_ID2988__KC_A_4_3e_v1.png|rightr|frame| Iterative decoding of $(+1, -1, +1)$]]
*In the BSC model, $|L_{\rm K}(i)| = \ln {(1 \, – \varepsilon)/\varepsilon} = 1$ for all $i$. Thus:
'''(5)''' The adjacent table illustrates the iterative bit-wise decoding starting from $\underline{y}_2 = (0, \, 1, \, 0)$.
'''(5)''' The adjacent table illustrates the iterative bit-wise decoding starting from $\underline{y}_2 = (0, \, 1, \, 0)$.
<br clear=all>
These results can be interpreted as follows:
These results can be interpreted as follows:
* The preassignment (iteration $I = 0$) happens according to $\underline{L}_{\rm APP} = \underline{L}_{\rm K}$. A hard decision ⇒ "$\sign {\underline{L}_{\rm APP}(i)}$" would lead to the decoding result $(0, \, 1, \, 0)$. The reliability of this obviously incorrect result is given as $|{\it \Sigma}| = 1$. This value agrees with the "$\ln (S)$" calculated in subtasks (3).
* The preassignment $($iteration $I = 0)$ happens according to $\underline{L}_{\rm APP} = \underline{L}_{\rm K}$. A hard decision ⇒ "$\sign \ {\underline{L}_{\rm APP}(i)}$" would lead to the decoding result $(0, \, 1, \, 0)$.
* After the first iteration $(I = 1)$ all a posteriori LLRs are $L_{\rm APP}(i) = +1$. A hard decision here would yield the (expected) correct result $\underline{x}_{\rm APP} = (0, \, 0, \, 0)$. The probability that this outcome is correct is quantified by $|{\it \Sigma}_{\rm APP}| = 3$:
*The reliability of this obviously incorrect result is given as $|{\it \Sigma}_{\rm APP}| = 1$. This value agrees with $\ln (S)$ calculated in subtask '''(3)'''.
* After the first iteration $(I = 1)$ all a-posteriori log likelihood ratios are $L_{\rm APP}(i) = +1$. A hard decision here would yield the $($expected$)$ correct result $\underline{x}_{\rm APP} = (0, \, 0, \, 0)$.
*The probability that this outcome is correct is quantified by $|{\it \Sigma}_{\rm APP}| = 3$:
* The second iteration confirms the decoding result of the first iteration. The reliability is even quantified here with "$9$". This value can be interpreted as follows:
* The second iteration confirms the decoding result of the first iteration. The reliability is even quantified here with $9$. This value can be interpreted as follows:
*With each further iteration the reliability value and thus the probability ${\rm Pr}(\underline{x}_0 | \underline{y}_2)$ increases drastically ⇒ <u>All proposed solutions</u> are correct.
*With each further iteration the reliability value and thus the probability ${\rm Pr}(\underline{x}_0 | \underline{y}_2)$ increases drastically ⇒ <u>All proposed solutions</u> are correct.
[[File:P_ID2991__KC_A_4_3f_v1.png|right|frame|Iterative decoding of $(–1, –1, +1)$]]
'''(6)''' Correct are <u>the proposed solutions 2 and 3</u>:
'''(6)''' Correct are <u>the proposed solutions 2 and 3</u>:
*For the received vector $\underline{y}_6 = (1, \, 1, \, 0)$, the second table applies.
*For the received vector $\underline{y}_6 = (1, \, 1, \, 0)$, the second table applies.
*The decoder now decides for the sequence $\underline{x}_1 = (1, \, 1, \, 1)$.
*The decoder now decides for the sequence $\underline{x}_1 = (1, \, 1, \, 1)$.
*The case "$\underline{y}_3 = (1, \, 1, \, 0)$ received under the condition $\underline{x}_1 = (1, \, 1, \, 1)$ sent" would correspond exactly to the constellation "$\underline{y}_2 = (1, \, 0, \, 1)$ received and $\underline{x}_0 = (0, \, 0, \, 0)$ sent" considered in the last subtask.
*But since $\underline{x}_0 = (0, \, 0, \, 0)$ was sent, there are now two bit errors with the following consequence:
*The case »$\underline{y}_6 = (1, \, 1, \, 0)$ received under the condition $\underline{x}_1 = (1, \, 1, \, 1)$ sent« would correspond exactly to the constellation »$\underline{y}_2 = (0, \, 1, \, 0)$ received and $\underline{x}_0 = (0, \, 0, \, 0)$ sent« considered in the last subtask.
*But since $\underline{x}_0 = (0, \, 0, \, 0)$ was sent, there are now two bit errors with the following consequence:
:* The iterative decoder decides incorrectly.
:* The iterative decoder decides incorrectly.
:* With each further iteration the wrong decision is declared as more reliable.
:* With each further iteration the wrong decision is declared as more reliable.
Here, $\varepsilon$ denotes the falsification probability of the BSC model.
Except for the last subtask, the following received value is always assumed:
$$\underline{y} = (0, 1, 0) =\underline{y}_2
\hspace{0.05cm}. $$
The chosen indexing of all possible received vectors can be taken from the graphic.
The most considered vector $\underline{y}_2$ is highlighted in red.
Only for subtask (6) applies:
$$\underline{y} = (1, 1, 0) =\underline{y}_6
\hspace{0.05cm}. $$
For decoding purposes, the exercise will examine:
the "syndrome decoding", which follows the concept "hard decision maximum likelihood detection" $\rm (HD-ML)$ for the codes under consideration. $($because soft values are not available at the BSC$)$;
The received word $\underline{y}_2 = (0, 1, 0)$ is not a valid code word of the single parity–check code $\text{SPC (3, 2)}$. Thus, the first statement is false.
In addition, since the $\text{SPC (3, 2)}$ has only the minimum distance $d_{\rm min} = 2$, no error can be corrected.
(2) Correct is the proposed solution 2:
The possible code words at the $\text{RP (3, 1)}$ are $\underline{x}_0 = (0, 0, 0)$ and $\underline{x}_1 = (1, 1, 1)$.
The minimum distance of this code is $d_{\rm min} = 3$, so $t = (d_{\rm min} \, - 1)/2 = 1$ error can be corrected.
In addition to $\underline{y}_0 = (0, 0, 0)$, the received words $\underline{y}_1 = (0, 0, 1), \ \underline{y}_2 = (0, 1, 0)$, and $\underline{y}_4 = (1, 0, 0)$ are also assigned to the decoding result $\underline{x}_0 = (0, 0, 0)$.
(3) According to the BSC model, the conditional probability that $\underline{y}_2 = (0, 1, 0)$ is received, given that $\underline{x}_0 = (0, 0, 0)$ was sent:
The first term $(1 \, –\varepsilon)^2$ indicates the probability that the first and third bit were transmitted correctly. $\varepsilon$ describes the falsification probability for the second bit.
Correspondingly, for the second possible code word $\underline{x}_1 = (1, 1, 1)$:
(5) The adjacent table illustrates the iterative bit-wise decoding starting from $\underline{y}_2 = (0, \, 1, \, 0)$.
These results can be interpreted as follows:
The preassignment $($iteration $I = 0)$ happens according to $\underline{L}_{\rm APP} = \underline{L}_{\rm K}$. A hard decision ⇒ "$\sign \ {\underline{L}_{\rm APP}(i)}$" would lead to the decoding result $(0, \, 1, \, 0)$.
The reliability of this obviously incorrect result is given as $|{\it \Sigma}_{\rm APP}| = 1$. This value agrees with $\ln (S)$ calculated in subtask (3).
After the first iteration $(I = 1)$ all a-posteriori log likelihood ratios are $L_{\rm APP}(i) = +1$. A hard decision here would yield the $($expected$)$ correct result $\underline{x}_{\rm APP} = (0, \, 0, \, 0)$.
The probability that this outcome is correct is quantified by $|{\it \Sigma}_{\rm APP}| = 3$:
The second iteration confirms the decoding result of the first iteration. The reliability is even quantified here with $9$. This value can be interpreted as follows:
With each further iteration the reliability value and thus the probability ${\rm Pr}(\underline{x}_0 | \underline{y}_2)$ increases drastically ⇒ All proposed solutions are correct.
Iterative decoding of $(–1, –1, +1)$
(6) Correct are the proposed solutions 2 and 3:
For the received vector $\underline{y}_6 = (1, \, 1, \, 0)$, the second table applies.
The decoder now decides for the sequence $\underline{x}_1 = (1, \, 1, \, 1)$.
The case »$\underline{y}_6 = (1, \, 1, \, 0)$ received under the condition $\underline{x}_1 = (1, \, 1, \, 1)$ sent« would correspond exactly to the constellation »$\underline{y}_2 = (0, \, 1, \, 0)$ received and $\underline{x}_0 = (0, \, 0, \, 0)$ sent« considered in the last subtask.
But since $\underline{x}_0 = (0, \, 0, \, 0)$ was sent, there are now two bit errors with the following consequence:
The iterative decoder decides incorrectly.
With each further iteration the wrong decision is declared as more reliable.
