Aufgaben:Exercise 4.4: Extrinsic L-values at SPC - Revision history

Hwang at 15:25, 23 January 2023

2023-01-23T15:25:52Z

Guenter at 12:02, 3 December 2022

2022-12-03T12:02:27Z

Guenter at 11:21, 3 December 2022

2022-12-03T11:21:47Z

Guenter at 11:16, 3 December 2022

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Guenter at 16:46, 30 November 2022

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Hwang at 19:14, 1 November 2022

2022-11-01T19:14:45Z

Noah at 17:44, 29 October 2022

2022-10-29T17:44:55Z

Noah at 17:44, 29 October 2022

2022-10-29T17:44:44Z

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Noah at 15:54, 29 October 2022

2022-10-29T15:54:43Z

Noah at 11:40, 29 October 2022

2022-10-29T11:40:19Z

@@ Line 24: / Line 24: @@
 p_4 = 0.6  \hspace{0.05cm}.$$
-From this the apriori log likelihood ratios result to:
+From this the a-priori log likelihood ratios result to:
 :$$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}

@@ Line 79: / Line 79: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; For the apriori $L$ values of the first two bits of the code word:
+'''(1)'''&nbsp; For the a-priori log likelihood ratios of the first two bits of the code word:
 :$$L_{\rm A}(i = 1) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm ln} \hspace{0.1cm} \left [ \frac{1-p_1}{p_1} \right ] =  {\rm ln} \hspace{0.1cm} 4 \hspace{0.15cm}\underline{= +1.386}
   \hspace{0.05cm},$$
@@ Line 85: / Line 85: @@
   \hspace{0.05cm}.$$
-The values can be read from the fourth column of the table attached to the information page.
+*The values can be read from the fourth column of the table attached to the information page.
-'''(2)'''&nbsp; To calculate the extrinsic $L$ value over the $i$th bit, only the information about the other three bits $(j &ne; i)$ may be used. With the given equation holds:
 :$$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}.$$
-*For the product, we obtain according to the third column of [[Aufgaben:Exercise_4.4:_Extrinsic_L-values_at_SPC|"table"]]:
+*For the product,&nbsp; we obtain according to the third column of the&nbsp; [[Aufgaben:Exercise_4.4:_Extrinsic_L-values_at_SPC|$\rm table$]]:
 :$$\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
@@ Line 99: / Line 100: @@
   \hspace{0.05cm}.$$
-*In terms of bit 2, one obtains accordingly:
+*In terms of bit 2,&nbsp; one obtains accordingly:
 :$$\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
@@ Line 107: / Line 108: @@
-'''(3)'''&nbsp; For the apriori $L$ value holds:
 :$$L_j = L_{\rm A}(j) = {\rm ln} \hspace{0.1cm} \left [ \frac{{\rm Pr}(x_j = 0)}{{\rm Pr}(x_j = 1)}
 \right ] = {\rm ln} \hspace{0.1cm} \left [ \frac{1-p_j}{p_j}
@@ Line 119: / Line 121: @@
 = \frac{{\rm e}^{L_j}-1}{{\rm e}^{L_j} +1}\hspace{0.05cm} .$$
-*Multiplying the numerator and denominator by ${\rm e}^{-L_j/2}$, we get:
+*Multiplying the numerator and denominator by&nbsp; ${\rm e}^{-L_j/2}$,&nbsp; 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 <u>all proposed solutions</u> are correct.
+*Thus&nbsp; <u>all proposed solutions</u> are correct.
-*The function <i>hyperbolic tangent</i> can be found, for example, in tabular form in collections of formulas or in the last column of the table given in front.
-'''(4)'''&nbsp; We first calculate $L_{\rm E}(i = 3)$ in the same way as in subtask (2):
+'''(4)'''&nbsp; We first calculate&nbsp; $L_{\rm E}(i = 3)$&nbsp; in the same way as in subtask&nbsp; '''(2)''':
 :$$\prod\limits_{j =1, \hspace{0.05cm}2,\hspace{0.05cm} 4} \hspace{0.05cm}(1-2p_j) =
 (+0.6) \cdot (-0.8) \cdot (-0.2) = +0.096
@@ Line 133: / Line 137: @@
   \hspace{0.05cm}.$$
-*We calculate the extrinsic $L$ value with respect to the last bit according to the equation
+*We calculate the extrinsic L&ndash;value with respect to the last bit according to the equation
 :$$L_{\rm E}(i = 4) = {\rm ln} \hspace{0.2cm}  \frac{1 + \pi}{1 - \pi}\hspace{0.05cm}, \hspace{0.3cm}
-{\rm mit} \hspace{0.3cm} \pi = {\rm tanh}(L_1/2) \cdot {\rm tanh}(L_2/2) \cdot {\rm tanh}(L_3/2)
+{\rm with} \hspace{0.3cm} \pi = {\rm tanh}(L_1/2) \cdot {\rm tanh}(L_2/2) \cdot {\rm tanh}(L_3/2)
   \hspace{0.05cm}.$$
-*This results in accordance with the above [[Aufgaben:Exercise_4.4:_Extrinsic_L-values_at_SPC|"table"]]:
+*This results in accordance with the above&nbsp; [[Aufgaben:Exercise_4.4:_Extrinsic_L-values_at_SPC|$\rm table$]]:
 :$$p_1 = 0.2 \hspace{0.2cm}\Rightarrow \hspace{0.2cm}
 L_1 = +1.386 \hspace{0.2cm}\Rightarrow \hspace{0.2cm}

@@ Line 24: / Line 24: @@
 p_4 = 0.6  \hspace{0.05cm}.$$
-From this the apriori L&ndash;values result to:
+From this the apriori log likelihood ratios result to:
 :$$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}
@@ Line 37: / Line 37: @@
 *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#Calculation_of_extrinsic_LLRs|"Calculation of the extrinsic&nbsp; L&ndash;values"]].
+*Reference is made in particular to the section&nbsp; [[Channel_Coding/Soft-in_Soft-Out_Decoder#Calculation_of_extrinsic_log_likelihood_ratios|"Calculation of the extrinsic log likelihood ratios"]].
 * In the table are given for&nbsp; $p_i = 0$&nbsp; to&nbsp; $p_i = 1$&nbsp; with step size&nbsp; $0.1$&nbsp; $($column 1$)$:
 ::In column 2: &nbsp; the probability&nbsp; $q_i = {\rm Pr}(x_i = 0) = 1 - p_i$,
 ::in column 3: &nbsp; the values for&nbsp; $1 - 2p_i$,
-::in column 4: &nbsp; the a-priori L&ndash;values&nbsp; $L_i = \ln {\big [(1 - p_i)/p_ i \big ]} = L_{\rm A}(i)$.
+::in column 4: &nbsp; the a-priori log likelihood ratios&nbsp; $L_i = \ln {\big [(1 - p_i)/p_ i \big ]} = L_{\rm A}(i)$.
 * The&nbsp; "hyperbolic tangent"&nbsp; $(\tanh)$&nbsp; of $L_i/2$&nbsp; is identical to&nbsp; $1-2p_i$ &nbsp; &#8658; &nbsp; column 3.
@@ Line 55: / Line 55: @@
 ===Questions===
 <quiz display=simple>
-{It holds&nbsp; $p_1 = 0.2, \ p_2 = 0.9, \ p_3 = 0.3, \ p_4 = 0.6$.&nbsp; From this,&nbsp; calculate the a-priori L&ndash;values of the&nbsp; $\text{SPC (4, 3, 4)}$&nbsp; for bit 1 and bit 2.
+{It holds&nbsp; $p_1 = 0.2, \ p_2 = 0.9, \ p_3 = 0.3, \ p_4 = 0.6$.&nbsp; From this,&nbsp; calculate the a-priori log likelihood ratios of the&nbsp; $\text{SPC (4, 3, 4)}$&nbsp; for bit 1 and bit 2.
 |type="{}"}
 $L_{\rm A}(i = 1) \ = \ ${ 1.386 3% }
 $L_{\rm A}(i = 2) \ = \ ${ -2.26291--2.13109 }
-{What are the extrinsic L&ndash;values for bit 1 and bit 2.
+{What are the extrinsic log likelihood ratios for bit 1 and bit 2.
 |type="{}"}
 $L_{\rm E}(i = 1) \ = \ ${ 0.128 3% }
@@ Line 71: / Line 71: @@
 + It holds&nbsp; $1-2p_j = \tanh {(L_j/2)}$.
-{It is further&nbsp; $p_1 = 0.2, \ p_2 = 0.9, \ p_3, \ p_4 = 0.6$.&nbsp; Calculate the extrinsic L&ndash;values for bit 3 and bit 4.&nbsp; Use different equations for this purpose.
+{It is further&nbsp; $p_1 = 0.2, \ p_2 = 0.9, \ p_3, \ p_4 = 0.6$.&nbsp; Calculate the extrinsic log likelihood ratios for bit 3 and bit 4.&nbsp; Use different equations for this purpose.
 |type="{}"}
 $L_{\rm E}(i = 3) \ = \ ${ 0.193 3% }

@@ Line 2: / Line 2: @@
 [[File:EN_KC_A_4_4.png|right|frame|Suitable auxiliary table]]
-We consider again the&nbsp; [[Channel_Coding/Examples_of_Binary_Block_Codes#Single_Parity-check_Codes|"single parity&ndash;check code"]]. In such a&nbsp; ${\rm SPC} \ (n, \, n-1, \, 2)$&nbsp; the&nbsp; $n$&nbsp; bits of a code word&nbsp; $\underline{x}$&nbsp; come from the&nbsp; $k = n -1$&nbsp; bits from the source sequence&nbsp; $\underline{u}$&nbsp; and only a single check bit&nbsp; $p$&nbsp; is added, such that the number of ones in the code word is even:
+We consider again the&nbsp; [[Channel_Coding/Examples_of_Binary_Block_Codes#Single_Parity-check_Codes|"single parity&ndash;check code"]].&nbsp; In such a&nbsp; ${\rm SPC} \ (n, \, n-1, \, 2)$&nbsp; the&nbsp; $n$&nbsp; bits of a code word&nbsp; $\underline{x}$&nbsp; come from the&nbsp; $k = n -1$&nbsp; bits from the source sequence&nbsp; $\underline{u}$&nbsp; and only a single check bit&nbsp; $p$&nbsp; is added,&nbsp; such that the number of&nbsp; "ones"&nbsp; in the code word&nbsp; $\underline{x}$&nbsp; 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&nbsp; $i$th code bit is formed over all other symbols&nbsp; $(j &ne; i)$&nbsp;. Therefore we write for the code word shorter by one bit:
+The extrinsic information about the&nbsp; $i$th code bit is formed over all other bits&nbsp; $(j &ne; i)$.&nbsp; 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&nbsp; $L$ value over the&nbsp; $i$th code symbol reads with the&nbsp; [[Channel_Coding/Objective_of_Channel_Coding#Important_definitions_for_block_coding|"Hamming weight"]]&nbsp; $w_{\rm H}$&nbsp; of the truncated sequence&nbsp; $\underline{x}^{(-i)}$:
+The extrinsic&nbsp; L&ndash;value over the&nbsp; $i$th code symbol reads with the&nbsp; [[Channel_Coding/Objective_of_Channel_Coding#Important_definitions_for_block_coding|"Hamming weight"]]&nbsp; $w_{\rm H}$&nbsp; of the truncated sequence&nbsp; $\underline{x}^{(-i)}$:
-:$$L_{\rm E}(i) = \frac{{\rm Pr} \left [w_{\rm H}(\underline{x}^{(-i)})\hspace{0.15cm}{\rm ist \hspace{0.15cm} gerade} \hspace{0.05cm} | \hspace{0.05cm}\underline{y} \hspace{0.05cm}\right ]}{{\rm Pr} \left [w_{\rm H}(\underline{x}^{(-i)})\hspace{0.15cm}{\rm ist \hspace{0.15cm} ungerade} \hspace{0.05cm} | \hspace{0.05cm}\underline{y} \hspace{0.05cm}\right ]}
+:$$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}.$$
-*If the probability in the numerator is greater than that in the denominator, then&nbsp; $L_{\rm E}(i) > 0$&nbsp; and thus the a posteriori&ndash;$L$&ndash; value&nbsp; $L_{\rm APP}(i) = L_{\rm A}(i) + L_{\rm E}(i)$&nbsp; magnified, that is tends to be affected in the direction of the symbol&nbsp; $x_i = 0$&nbsp;.
+*If the probability in the numerator is greater than that in the denominator,&nbsp; then&nbsp; $L_{\rm E}(i) > 0$&nbsp; and thus the a-posteriori  L&ndash;value&nbsp; $L_{\rm APP}(i) = L_{\rm A}(i) + L_{\rm E}(i)$&nbsp; magnified,&nbsp; that is tends to be affected in the direction of the symbol&nbsp; $x_i = 0$.
-*If&nbsp; $L_{\rm E}(i) < 0$&nbsp; from the point of view of the other symbols&nbsp; $(j &ne; i)$&nbsp; there is much to be said for&nbsp; $x_i = 1$&nbsp;.
-Only the&nbsp; $\text{SPC (4, 3, 4)}$ is treated, where for the probabilities&nbsp; $p_i = {\rm Pr}(x_i = 1)$&nbsp; holds:
+Only the&nbsp; $\text{SPC (4, 3, 4)}$&nbsp; is treated,&nbsp; where for the probabilities&nbsp; $p_i = {\rm Pr}(x_i = 1)$&nbsp; holds:
 :$$p_1 = 0.2 \hspace{0.05cm}, \hspace{0.3cm}
 p_2 = 0.9 \hspace{0.05cm}, \hspace{0.3cm}
@@ Line 23: / Line 24: @@
 p_4 = 0.6  \hspace{0.05cm}.$$
-From this the apriori&ndash;$L$ values result to:
+From this the apriori L&ndash;values result to:
 :$$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}
@@ Line 33: / Line 34: @@
-Hints:
+* In&nbsp; [[Aufgaben:Exercise_4.4Z:_Supplement_to_Exercise_4.4|$\text{Exercise 4.4Z}$]]&nbsp; it is shown that for the extrinsic&nbsp; L&ndash;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)
@@ Line 53: / Line 55: @@
 ===Questions===
 <quiz display=simple>
-{It holds&nbsp; $p_1 = 0.2, \ p_2 = 0.9, \ p_3 = 0.3, \ p_4 = 0.6$. From this, calculate the apriori $L$ values of the&nbsp; $\text{SPC (4, 3, 4)}$&nbsp; for bit 1 and bit 2.
+{It holds&nbsp; $p_1 = 0.2, \ p_2 = 0.9, \ p_3 = 0.3, \ p_4 = 0.6$.&nbsp; From this,&nbsp; calculate the a-priori L&ndash;values of the&nbsp; $\text{SPC (4, 3, 4)}$&nbsp; for bit 1 and bit 2.
 |type="{}"}
 $L_{\rm A}(i = 1) \ = \ ${ 1.386 3% }
 $L_{\rm A}(i = 2) \ = \ ${ -2.26291--2.13109 }
-{What are the extrinsic $L$ values for bit 1 and bit 2.
+{What are the extrinsic L&ndash;values for bit 1 and bit 2.
 |type="{}"}
 $L_{\rm E}(i = 1) \ = \ ${ 0.128 3% }
@@ Line 65: / Line 67: @@
 {What are the relationships between&nbsp; $p_j$&nbsp; and&nbsp; $L_j = L_{\rm A}(j)$?
 |type="[]"}
-+ Es gilt&nbsp; $p_j = 1/(1 + {\rm e}^ {L_j})$.
++ It holds&nbsp; $p_j = 1/(1 + {\rm e}^ {L_j})$.
-+ Es gilt&nbsp; $1-2p_j = ({\rm e}^ {L_j} - 1) \ / \ ({\rm e}^ {L_j} + 1)$.
++ It holds&nbsp; $1-2p_j = ({\rm e}^ {L_j} - 1) \ / \ ({\rm e}^ {L_j} + 1)$.
-+ Es gilt&nbsp; $1-2p_j = \tanh {(L_j/2)}$.
++ It holds&nbsp; $1-2p_j = \tanh {(L_j/2)}$.
-{It further&nbsp; $p_1 = 0.2, \ p_2 = 0.9, \ p_3$ and $p_4 = 0.6$. Calculate the extrinsic $L$ values for bit 3 and bit 4. <br>Use different equations for this purpose.
+{It is further&nbsp; $p_1 = 0.2, \ p_2 = 0.9, \ p_3, \ p_4 = 0.6$.&nbsp; Calculate the extrinsic L&ndash;values for bit 3 and bit 4.&nbsp; Use different equations for this purpose.
 |type="{}"}
 $L_{\rm E}(i = 3) \ = \ ${ 0.193 3% }

@@ Line 1: / Line 1: @@
 {{quiz-Header|Buchseite=Channel_Coding/Soft-in_Soft-Out_Decoder}}
-[[File:P_ID2993__KC_A_4_4_v1.png|right|frame|Suitable auxiliary table]]
+[[File:EN_KC_A_4_4.png|right|frame|Suitable auxiliary table]]
 We consider again the&nbsp; [[Channel_Coding/Examples_of_Binary_Block_Codes#Single_Parity-check_Codes|"single parity&ndash;check code"]]. In such a&nbsp; ${\rm SPC} \ (n, \, n-1, \, 2)$&nbsp; the&nbsp; $n$&nbsp; bits of a code word&nbsp; $\underline{x}$&nbsp; come from the&nbsp; $k = n -1$&nbsp; bits from the source sequence&nbsp; $\underline{u}$&nbsp; and only a single check bit&nbsp; $p$&nbsp; is added, such that the number of ones in the code word 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 ) =

@@ Line 2: / Line 2: @@
 [[File:P_ID2993__KC_A_4_4_v1.png|right|frame|Geeignete Hilfstabelle]]
-Wir betrachten nochmals den&nbsp; [[Channel_Coding/Beispiele_bin%C3%A4rer_Blockcodes#Single_Parity.E2.80.93check_Codes|Single Parity&ndash;check Code]]. Bei einem solchen&nbsp; ${\rm SPC} \ (n, \, n-1, \, 2)$&nbsp; stammen von den&nbsp; $n$&nbsp; Bit eines Codewortes&nbsp; $\underline{x}$&nbsp; die ersten&nbsp; $k = n -1$&nbsp; Bit von der Quellenfolge&nbsp; $\underline{u}$&nbsp; und es wird nur ein einziges Prüfbit&nbsp; $p$&nbsp; hinzugefügt, und zwar derart, dass die Anzahl der Einsen im Codewort geradzahlig ist:
+Wir betrachten nochmals den&nbsp; [[Channel_Coding/Examples_of_Binary_Block_Codes#Single_Parity-check_Codes|"Single Parity&ndash;check Code"]]. Bei einem solchen&nbsp; ${\rm SPC} \ (n, \, n-1, \, 2)$&nbsp; stammen von den&nbsp; $n$&nbsp; Bit eines Codewortes&nbsp; $\underline{x}$&nbsp; die ersten&nbsp; $k = n -1$&nbsp; Bit von der Quellenfolge&nbsp; $\underline{u}$&nbsp; und es wird nur ein einziges Prüfbit&nbsp; $p$&nbsp; hinzugefügt, und zwar derart, dass die Anzahl der Einsen im Codewort geradzahlig ist:
 :$$\underline{x} = \big ( \hspace{0.03cm}x_1, \hspace{0.03cm} x_2, \hspace{0.05cm} \text{...} \hspace{0.05cm} , x_{n-1}, \hspace{0.03cm} x_n \hspace{0.03cm} \big ) =
 \big ( \hspace{0.03cm}u_1, \hspace{0.03cm} u_2, \hspace{0.05cm} \text{...} \hspace{0.05cm} , u_{k}, \hspace{0.03cm} p \hspace{0.03cm} \big )\hspace{0.03cm}. $$
@@ Line 9: / Line 9: @@
 :$$\underline{x}^{(-i)} = \big ( \hspace{0.03cm}x_1, \hspace{0.05cm} \text{...} \hspace{0.05cm} , \hspace{0.03cm} x_{i-1}, \hspace{0.43cm} x_{i+1},  \hspace{0.05cm} \text{...} \hspace{0.05cm} , x_{n} \hspace{0.03cm} \big )\hspace{0.03cm}. $$
-Der extrinsische&nbsp; $L$&ndash;Wert über das&nbsp; $i$&ndash;te Codesymbol lautet mit dem&nbsp; [[Channel_Coding/Zielsetzung_der_Kanalcodierung#Einige_wichtige_Definitionen_zur_Blockcodierung|Hamming&ndash;Gewicht]]&nbsp; $w_{\rm H}$&nbsp; der verkürzten Folge&nbsp; $\underline{x}^{(-i)}$:
+Der extrinsische&nbsp; $L$&ndash;Wert über das&nbsp; $i$&ndash;te Codesymbol lautet mit dem&nbsp; [[Channel_Coding/Objective_of_Channel_Coding#Important_definitions_for_block_coding|"Hamming&ndash;Gewicht"]]&nbsp; $w_{\rm H}$&nbsp; der verkürzten Folge&nbsp; $\underline{x}^{(-i)}$:
 :$$L_{\rm E}(i) = \frac{{\rm Pr} \left [w_{\rm H}(\underline{x}^{(-i)})\hspace{0.15cm}{\rm ist \hspace{0.15cm} gerade} \hspace{0.05cm} | \hspace{0.05cm}\underline{y} \hspace{0.05cm}\right ]}{{\rm Pr} \left [w_{\rm H}(\underline{x}^{(-i)})\hspace{0.15cm}{\rm ist \hspace{0.15cm} ungerade} \hspace{0.05cm} | \hspace{0.05cm}\underline{y} \hspace{0.05cm}\right ]}
   \hspace{0.05cm}.$$
@@ Line 37: / Line 37: @@
 ''Hinweise:''
-* Die Aufgabe gehört zum Kapitel&nbsp; [[Channel_Coding/Soft%E2%80%93in_Soft%E2%80%93out_Decoder| Soft&ndash;in Soft&ndash;out Decoder]].
+* Die Aufgabe gehört zum Kapitel&nbsp; [[Channel_Coding/Soft-in_Soft-Out_Decoder| "Soft&ndash;in Soft&ndash;out Decoder"]].
-*Bezug genommen wird insbesondere auf die Seite&nbsp; [[Channel_Coding/Soft–in_Soft–out_Decoder#Zur_Berechnung_der_extrinsischen_L.E2.80.93Werte|Zur Berechnung der extrinsischen&nbsp; $L$&ndash;Werte]].
+*Bezug genommen wird insbesondere auf die Seite&nbsp; [[Channel_Coding/Soft-in_Soft-Out_Decoder#Calculation_of_extrinsic_LLRs|"Zur Berechnung der extrinsischen&nbsp; $L$&ndash;Werte"]].
 * In der Tabelle sind für&nbsp; $p_i = 0$&nbsp; bis&nbsp; $p_i = 1$&nbsp; mit Schrittweite $0.1$ (Spalte 1) angegeben:
 ::In Spalte 2:  &nbsp; die Wahrscheinlichkeit&nbsp; $q_i = {\rm Pr}(x_i = 0) = 1 - p_i$,
@@ Line 44: / Line 44: @@
 ::in Spalte 4:  &nbsp; die Apriori&ndash;$L$&ndash;Werte&nbsp; $L_i = \ln {\big [(1 - p_i)/p_ i \big ]} = L_{\rm A}(i)$.
 * Der&nbsp; <i>Tangens Hyperbolicus</i>&nbsp; ($\tanh$)&nbsp;  von $L_i/2$&nbsp; ist identisch mit&nbsp; $1-2p_i$ &nbsp; &#8658; &nbsp; Spalte 3.
-* In der&nbsp; [[Aufgaben:Aufgabe_4.4Z:_Ergänzung_zur_Aufgabe_4.4|Aufgabe 4.4Z]]&nbsp; wird gezeigt, dass für den extrinsischen&nbsp; $L$&ndash;Wert auch geschrieben werden kann:
+* In der&nbsp; [[Aufgaben:Exercise_4.4Z:_Supplement_to_Exercise_4.4|"Aufgabe 4.4Z"]]&nbsp; wird gezeigt, dass für den extrinsischen&nbsp; $L$&ndash;Wert auch geschrieben werden kann:
 :$$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)