Aufgaben:Exercise 4.4Z: Supplement to Exercise 4.4 - Revision history

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	[[Category:Channel Coding: Exercises\|^4.1 Soft–in Soft–out Decoder^]]		[[Category:Channel Coding: Exercises\|^4.1 Soft–in Soft–out Decoder^]]
	[[de:Aufgaben:4.~~4Z_Ergänzung_zur_Aufgabe_A4~~.4]]		[[de:Aufgaben:Aufgabe 4.4Z: Ergänzung zur Aufgabe 4.4]]

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	Gallager solved the problem in an analytical way:		Gallager solved the problem in an analytical way:
	:$$ {\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even} \right ]		:$$ {\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even} \right ]\hspace{-0.15cm} \ = \ \hspace{-0.15cm} 1/2 \cdot [1 + \pi]\hspace{0.05cm},$$
	\hspace{-0.15cm} \ = \ \hspace{-0.15cm} 1/2 \cdot [1 + \pi]\hspace{0.05cm},$$		:$${\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} odd}\right ] \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 1/2 \cdot [1 - \pi]\hspace{0.05cm}.$$
	:$${\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} odd}
	\right ] \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 1/2 \cdot [1 - \pi]\hspace{0.05cm}.$$

	Here the following auxiliary variable is used:		Here the following auxiliary variable is used:
	:$$\pi = \prod\limits_{i =1}^{n} \hspace{0.25cm}(1-2p_i)		:$$\pi = \prod\limits_{i =1}^{n} \hspace{0.25cm}(1-2p_i)\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	One applies the equation,  for example,  to calculate the extrinsic  L–values of a  "single parity–check code".		One applies the equation,  for example,  to calculate the extrinsic  L–values of a  "single parity–check code".

	Indeed,  as already pointed out in  [[Aufgaben:Exercise_4.4:_Extrinsic_L-values_at_SPC\| $\text{Exercise 4.4}$]]  the extrinsic log likelihood ratio with Hamming–weight  $w_{\rm H}$  is the truncated sequence  $\underline{x}^{(-i)}$:		Indeed,  as already pointed out in  [[Aufgaben:Exercise_4.4:_Extrinsic_L-values_at_SPC\| $\text{Exercise 4.4}$]]  the extrinsic log likelihood ratio with Hamming–weight  $w_{\rm H}$  is the truncated sequence  $\underline{x}^{(-i)}$:
	:$$L_{\rm E}(i) = {\rm ln} \hspace{0.15cm}\frac{{\rm Pr} \left [w_{\rm H}(\underline{x}^{(-i)})\hspace{0.15cm}{\rm is \hspace{0.15cm} even} \hspace{0.05cm} \| \hspace{0.05cm}\underline{y} \hspace{0.05cm}\right ]}{{\rm Pr} \left [w_{\rm H}(\underline{x}^{(-i)})\hspace{0.15cm}{\rm is \hspace{0.15cm} odd} \hspace{0.05cm} \| \hspace{0.05cm}\underline{y} \hspace{0.05cm}\right ]}		:$$L_{\rm E}(i) = {\rm ln} \hspace{0.15cm}\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}.$$
	\hspace{0.05cm}.$$

	Here it is taken into account that for  $L_{\rm E}(i)$  one may refer only to the other symbols  $(j ≠ i)$ :		Here it is taken into account that for  $L_{\rm E}(i)$  one may refer only to the other symbols  $(j ≠ i)$ :
Line 87:		Line 83:
	[[File:P_ID2996__KC_Z_4_4a_v1.png\|frame\|right\|Derivation  "$w_{\rm H}$ is even"  for code length  $n = 2$.]]		[[File:P_ID2996__KC_Z_4_4a_v1.png\|frame\|right\|Derivation  "$w_{\rm H}$ is even"  for code length  $n = 2$.]]
	'''(1)'''  According to the adjacent table applies:		'''(1)'''  According to the adjacent table applies:
	:$${\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.10cm}{\rm is \hspace{0.10cm} even}\right ] =		:$${\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.10cm}{\rm is \hspace{0.10cm} even}\right ] ={\rm Pr} \left [w_{\rm H} = 0 \right] + {\rm Pr} \left [w_{\rm H} = 2 \right]\hspace{0.05cm}. $$
	{\rm Pr} \left [w_{\rm H} = 0 \right] + {\rm Pr} \left [w_{\rm H} = 2 \right]
	\hspace{0.05cm}. $$

	*With the probabilities		*With the probabilities
Line 96:		Line 90:

	:one obtains:		:one obtains:
	:$${\rm Pr} \left [w_{\rm H}(\underline{x}) = 0\right] \hspace{-0.15cm} \ = \ \hspace{-0.15cm}		:$${\rm Pr} \left [w_{\rm H}(\underline{x}) = 0\right] \hspace{-0.15cm} \ = \ \hspace{-0.15cm}{\rm Pr} \left [(x_1 = 0)\cap (x_2 = 0) \right] = q_1 \cdot q_2 = 0.8 \cdot 0.1= 0.08 \hspace{0.05cm},$$
	{\rm Pr} \left [(x_1 = 0)\cap (x_2 = 0) \right] = q_1 \cdot q_2 = 0.8 \cdot 0.1		:$${\rm Pr} \left [w_{\rm H}(\underline{x}) = 2\right] \hspace{-0.15cm} \ = \ \hspace{-0.15cm}{\rm Pr} \left [(x_1 = 1)\cap (x_2 = 1) \right] = p_1 \cdot p_2 = 0.2 \cdot 0.9 = 0.18$$
	= 0.08 \hspace{0.05cm},$$		:$$\Rightarrow \hspace{0.3cm} {\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even}\right] = 0.8 + 0.18 \hspace{0.15cm} \underline{= 0.26}\hspace{0.05cm}.$$
	:$${\rm Pr} \left [w_{\rm H}(\underline{x}) = 2\right] \hspace{-0.15cm} \ = \ \hspace{-0.15cm}
	{\rm Pr} \left [(x_1 = 1)\cap (x_2 = 1) \right] = p_1 \cdot p_2 = 0.2 \cdot 0.9 = 0.18$$
	:$$\Rightarrow \hspace{0.3cm} {\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even}\right] = 0.8 + 0.18 \hspace{0.15cm} \underline{= 0.26}
	\hspace{0.05cm}.$$

	*The Gallager's equation provides for the same set of parameters:		*The Gallager's equation provides for the same set of parameters:
	:$${\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even}\right] \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 0.5 + 0.5 \cdot \prod\limits_{i =1}^{2} \hspace{0.25cm}(1-2\cdot p_i) = 0.5 + 0.5 \cdot (1 - 2 \cdot 0.2)\cdot (1 - 2 \cdot 0.9) = 0.26		:$${\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even}\right] \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 0.5 + 0.5 \cdot \prod\limits_{i =1}^{2} \hspace{0.25cm}(1-2\cdot p_i) = 0.5 + 0.5 \cdot (1 - 2 \cdot 0.2)\cdot (1 - 2 \cdot 0.9) = 0.26\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	*The equation given by Gallager 1963 was hereby verified for  $n = 2$.		*The equation given by Gallager 1963 was hereby verified for  $n = 2$.
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	*The occurrence probabilities of each combination are given in the last column.  Thus,  the result is:		*The occurrence probabilities of each combination are given in the last column.  Thus,  the result is:
	:$$ {\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even}\right] = 0.056 + 0.216 + 0.006 + 0.126 \hspace{0.15cm} \underline{= 0.404}		:$$ {\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even}\right] = 0.056 + 0.216 + 0.006 + 0.126 \hspace{0.15cm} \underline{= 0.404}\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	*The red rows provide the complementary event:		*The red rows provide the complementary event:
	:$$ {\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} odd}\right] = 0.024 + 0.504 + 0.014 + 0.054= 0.596		:$$ {\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} odd}\right] = 0.024 + 0.504 + 0.014 + 0.054= 0.596\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	*Gallager's equation again gives the exact same result,  although it should be noted that this equation is valid for all  $n$  and all arbitrary probabilities:		*Gallager's equation again gives the exact same result,  although it should be noted that this equation is valid for all  $n$  and all arbitrary probabilities:
	:$${\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even}\right] \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 0.5 + 0.5 \cdot \prod\limits_{i =1}^{3} \hspace{0.25cm}(1-2\cdot p_i) $$		:$${\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even}\right] \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 0.5 + 0.5 \cdot \prod\limits_{i =1}^{3} \hspace{0.25cm}(1-2\cdot p_i) $$
	:$$\Rightarrow\hspace{0.3cm}{\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even}\right] \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 0.5 + 0.5 \cdot (+0.6) \cdot (-0.8) \cdot (+0.4) = 0.404		:$$\Rightarrow\hspace{0.3cm}{\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even}\right] \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 0.5 + 0.5 \cdot (+0.6) \cdot (-0.8) \cdot (+0.4) = 0.404\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$


	'''(3)'''  According to the specification page applies:		'''(3)'''  According to the specification page applies:
	:$$\pi = \prod\limits_{i =1}^{4} \hspace{0.25cm}(1-2\cdot p_i) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} (1 - 2 \cdot 0.2) \cdot (1 - 2 \cdot 0.9) \cdot (1 - 2 \cdot 0.3) \cdot (1 - 2 \cdot 0.6) $$		:$$\pi = \prod\limits_{i =1}^{4} \hspace{0.25cm}(1-2\cdot p_i) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} (1 - 2 \cdot 0.2) \cdot (1 - 2 \cdot 0.9) \cdot (1 - 2 \cdot 0.3) \cdot (1 - 2 \cdot 0.6) $$
	:$$\Rightarrow\hspace{0.3cm}\pi = \prod\limits_{i =1}^{4} \hspace{0.25cm}(1-2\cdot p_i) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}(+0.6) \cdot (-0.8) \cdot (+0.4) \cdot (-0.2) = 0.0384		:$$\Rightarrow\hspace{0.3cm}\pi = \prod\limits_{i =1}^{4} \hspace{0.25cm}(1-2\cdot p_i) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}(+0.6) \cdot (-0.8) \cdot (+0.4) \cdot (-0.2) = 0.0384\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	*From this can be calculated:		*From this can be calculated:
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	*If you add up the blue and red probabilities on the information page,V you get exactly the values calculated here.  For the quotient we get:		*If you add up the blue and red probabilities on the information page,V you get exactly the values calculated here.  For the quotient we get:
	:$$Q = \frac{{\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even}\right]} { {\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} odd}\right]} = \frac{0.5192}{0.4808}\hspace{0.15cm} \underline{= 1.0799}		:$$Q = \frac{{\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even}\right]} { {\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} odd}\right]} = \frac{0.5192}{0.4808}\hspace{0.15cm} \underline{= 1.0799}\hspace{0.05cm}. $$
	\hspace{0.05cm}. $$



	'''(4)'''  For the  "single parity–check code",  the extrinsic log likelihood ratio with respect to the  $i^{th}$  bit was specified as follows:		'''(4)'''  For the  "single parity–check code",  the extrinsic log likelihood ratio with respect to the  $i^{th}$  bit was specified as follows:
	:$$L_{\rm E}(i) = {\rm ln} \hspace{0.15cm}\frac{{\rm Pr} \left [w_{\rm H}(\underline{x}^{(-i)})\hspace{0.15cm}{\rm is \hspace{0.15cm} even} \hspace{0.05cm} \| \hspace{0.05cm}\underline{y} \hspace{0.05cm}\right ]}{{\rm Pr} \left [w_{\rm H}(\underline{x}^{(-i)})\hspace{0.15cm}{\rm is \hspace{0.15cm} odd} \hspace{0.05cm} \| \hspace{0.05cm}\underline{y} \hspace{0.05cm}\right ]}		:$$L_{\rm E}(i) = {\rm ln} \hspace{0.15cm}\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},$$
	\hspace{0.05cm},$$

	*Or:		*Or:
	:$$L_{\rm E}(i) = {\rm ln} \hspace{0.15cm}\frac{1+\prod_{j \ne i} \hspace{0.25cm}(1-2\cdot p_j)}{1-\prod_{j \ne i} \hspace{0.25cm}(1-2\cdot p_j)}		:$$L_{\rm E}(i) = {\rm ln} \hspace{0.15cm}\frac{1+\prod_{j \ne i} \hspace{0.25cm}(1-2\cdot p_j)}{1-\prod_{j \ne i} \hspace{0.25cm}(1-2\cdot p_j)}\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	*At  $\text{SPC (5, 4, 2}$)   ⇒   $n = 5$,  this product for  $i = 5$  results from the following four factors:		*At  $\text{SPC (5, 4, 2}$)   ⇒   $n = 5$,  this product for  $i = 5$  results from the following four factors:
	:$$\pi = \prod\limits_{j = 1, \hspace{0.05cm}2, \hspace{0.05cm}3, \hspace{0.05cm}4} \hspace{0.05cm}(1-2\cdot p_j) =		:$$\pi = \prod\limits_{j = 1, \hspace{0.05cm}2, \hspace{0.05cm}3, \hspace{0.05cm}4} \hspace{0.05cm}(1-2\cdot p_j) =(1-2\cdot p_1) \cdot (1-2\cdot p_2) \cdot (1-2\cdot p_3) \cdot (1-2\cdot p_4)\hspace{0.05cm}.$$
	(1-2\cdot p_1) \cdot (1-2\cdot p_2) \cdot (1-2\cdot p_3) \cdot (1-2\cdot p_4)
	\hspace{0.05cm}.$$

	The comparison with subtask  '''(3)'''  shows that  $L_{\rm E}(i = 5) = \ln {Q} = \ln {(1.0799)} \ \underline{\approx 0.077}$.		The comparison with subtask  '''(3)'''  shows that  $L_{\rm E}(i = 5) = \ln {Q} = \ln {(1.0799)} \ \underline{\approx 0.077}$.

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← Older revision		Revision as of 14:33, 16 March 2026
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	[[Category:Channel Coding: Exercises\|^4.1 Soft–in Soft–out Decoder^]]		[[Category:Channel Coding: Exercises\|^4.1 Soft–in Soft–out Decoder^]]
			[[de:Aufgaben:4.4Z_Ergänzung_zur_Aufgabe_A4.4]]

Guenter at 14:29, 3 December 2022

2022-12-03T14:29:45Z

← Older revision		Revision as of 16:29, 3 December 2022
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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	[[File:P_ID2996__KC_Z_4_4a_v1.png\|frame\|right\|Derivation "$w_{\rm H}$ is even" for code length $n = 2$.]]		[[File:P_ID2996__KC_Z_4_4a_v1.png\|frame\|right\|Derivation  "$w_{\rm H}$ is even"  for code length  $n = 2$.]]
	'''(1)'''  According to the adjacent table applies:		'''(1)'''  According to the adjacent table applies:
	:$${\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.10cm}{\rm is \hspace{0.10cm} even}\right ] =		:$${\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.10cm}{\rm is \hspace{0.10cm} even}\right ] =
Line 91:		Line 91:
	\hspace{0.05cm}. $$		\hspace{0.05cm}. $$

	With the probabilities		*With the probabilities
	:$$p_1 = {\rm Pr} (x_1 = 1) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 0.2\hspace{0.05cm},\hspace{0.3cm}q_1 = {\rm Pr} (x_1 = 0) = 0.8\hspace{0.05cm},$$		:$$p_1 = {\rm Pr} (x_1 = 1) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 0.2\hspace{0.05cm},\hspace{0.3cm}q_1 = {\rm Pr} (x_1 = 0) = 0.8\hspace{0.05cm},$$
	:$$p_2 = {\rm Pr} (x_2 = 1) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 0.9\hspace{0.05cm},\hspace{0.3cm}q_2 = {\rm Pr} (x_2 = 0) = 0.1$$		:$$p_2 = {\rm Pr} (x_2 = 1) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 0.9\hspace{0.05cm},\hspace{0.3cm}q_2 = {\rm Pr} (x_2 = 0) = 0.1$$

	one obtains:		:one obtains:
	:$${\rm Pr} \left [w_{\rm H}(\underline{x}) = 0\right] \hspace{-0.15cm} \ = \ \hspace{-0.15cm}		:$${\rm Pr} \left [w_{\rm H}(\underline{x}) = 0\right] \hspace{-0.15cm} \ = \ \hspace{-0.15cm}
	{\rm Pr} \left [(x_1 = 0)\cap (x_2 = 0) \right] = q_1 \cdot q_2 = 0.8 \cdot 0.1		{\rm Pr} \left [(x_1 = 0)\cap (x_2 = 0) \right] = q_1 \cdot q_2 = 0.8 \cdot 0.1
Line 104:		Line 104:
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	The Gallager's equation provides for the same set of parameters:		*The Gallager's equation provides for the same set of parameters:
	:$${\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even}\right] \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 0.5 + 0.5 \cdot \prod\limits_{i =1}^{2} \hspace{0.25cm}(1-2\cdot p_i) = 0.5 + 0.5 \cdot (1 - 2 \cdot 0.2)\cdot (1 - 2 \cdot 0.9) = 0.26		:$${\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even}\right] \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 0.5 + 0.5 \cdot \prod\limits_{i =1}^{2} \hspace{0.25cm}(1-2\cdot p_i) = 0.5 + 0.5 \cdot (1 - 2 \cdot 0.2)\cdot (1 - 2 \cdot 0.9) = 0.26
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	The equation given by Gallager 1963 was hereby verified for $n = 2$.		*The equation given by Gallager 1963 was hereby verified for  $n = 2$.


	[[File:P_ID2997__KC_Z_4_4b_v1.png\|right\|frame\|Derivation "$w_{\rm H}$ is even" for code length $n = 3$.]]
	~~'''(2)'''  In the second table, the four combinations with an even number of ones are marked in blue.~~ The probabilities ~~of occurrence~~ of each combination are given in the last column. Thus, the result is:		'''(2)'''  In the second table,  the four combinations with an even number of  "ones"  are marked in blue.
			[[File:P_ID2997__KC_Z_4_4b_v1.png\|right\|frame\|Derivation "$w_{\rm H}$ is even"  for code length  $n = 3$.]]

			*The occurrence probabilities of each combination are given in the last column.  Thus,  the result is:
	:$$ {\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even}\right] = 0.056 + 0.216 + 0.006 + 0.126 \hspace{0.15cm} \underline{= 0.404}		:$$ {\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even}\right] = 0.056 + 0.216 + 0.006 + 0.126 \hspace{0.15cm} \underline{= 0.404}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	The red rows provide the complementary event:		*The red rows provide the complementary event:
	:$$ {\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} odd}\right] = 0.024 + 0.504 + 0.014 + 0.054= 0.596		:$$ {\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} odd}\right] = 0.024 + 0.504 + 0.014 + 0.054= 0.596
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	~~The~~ Gallager's equation again gives the exact same result, although it should be noted that this equation is valid for all $n$ and all arbitrary probabilities:		*Gallager's equation again gives the exact same result,  although it should be noted that this equation is valid for all  $n$  and all arbitrary probabilities:
	:$${\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even}\right] \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 0.5 + 0.5 \cdot \prod\limits_{i =1}^{3} \hspace{0.25cm}(1-2\cdot p_i) $$		:$${\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even}\right] \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 0.5 + 0.5 \cdot \prod\limits_{i =1}^{3} \hspace{0.25cm}(1-2\cdot p_i) $$
	:$$\Rightarrow\hspace{0.3cm}{\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even}\right] \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 0.5 + 0.5 \cdot (+0.6) \cdot (-0.8) \cdot (+0.4) = 0.404		:$$\Rightarrow\hspace{0.3cm}{\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even}\right] \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 0.5 + 0.5 \cdot (+0.6) \cdot (-0.8) \cdot (+0.4) = 0.404
Line 131:		Line 134:
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	From this can be calculated:		*From this can be calculated:
	:$${\rm Pr}({\rm blue}) = {\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even}\right] \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 0.5 + 0.5 \cdot \pi = 0.5 + 0.5 \cdot 0.0384\hspace{0.15cm} \underline{= 0.5192}\hspace{0.05cm},$$		:$${\rm Pr}({\rm blue}) = {\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even}\right] \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 0.5 + 0.5 \cdot \pi = 0.5 + 0.5 \cdot 0.0384\hspace{0.15cm} \underline{= 0.5192}\hspace{0.05cm},$$
	:$${\rm Pr}({\rm red}) = {\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} odd}\right] \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 0.5 - 0.5 \cdot \pi = 0.5 - 0.5 \cdot 0.0384\hspace{0.15cm} \underline{= 0.4808}\hspace{0.05cm}. $$		:$${\rm Pr}({\rm red}) = {\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} odd}\right] \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 0.5 - 0.5 \cdot \pi = 0.5 - 0.5 \cdot 0.0384\hspace{0.15cm} \underline{= 0.4808}\hspace{0.05cm}. $$

	If you add up the blue and red probabilities on the information page, you get exactly the values calculated here.		*If you add up the blue and red probabilities on the information page,V you get exactly the values calculated here.  For the quotient we get:

	For the quotient we get:
	:$$Q = \frac{{\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even}\right]} { {\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} odd}\right]} = \frac{0.5192}{0.4808}\hspace{0.15cm} \underline{= 1.0799}		:$$Q = \frac{{\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even}\right]} { {\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} odd}\right]} = \frac{0.5192}{0.4808}\hspace{0.15cm} \underline{= 1.0799}
	\hspace{0.05cm}. $$		\hspace{0.05cm}. $$


	'''(4)'''  For the ~~<i>~~single parity–check code~~</i>~~, the extrinsic ~~$L$ value~~ with respect to the $i$th bit was specified as follows:
			'''(4)'''  For the  "single parity–check code",  the extrinsic log likelihood ratio with respect to the  $i^{th}$  bit was specified as follows:
	:$$L_{\rm E}(i) = {\rm ln} \hspace{0.15cm}\frac{{\rm Pr} \left [w_{\rm H}(\underline{x}^{(-i)})\hspace{0.15cm}{\rm is \hspace{0.15cm} even} \hspace{0.05cm} \| \hspace{0.05cm}\underline{y} \hspace{0.05cm}\right ]}{{\rm Pr} \left [w_{\rm H}(\underline{x}^{(-i)})\hspace{0.15cm}{\rm is \hspace{0.15cm} odd} \hspace{0.05cm} \| \hspace{0.05cm}\underline{y} \hspace{0.05cm}\right ]}		:$$L_{\rm E}(i) = {\rm ln} \hspace{0.15cm}\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},$$		\hspace{0.05cm},$$

	or:		*Or:
	:$$L_{\rm E}(i) = {\rm ln} \hspace{0.15cm}\frac{1+\prod_{j \ne i} \hspace{0.25cm}(1-2\cdot p_j)}{1-\prod_{j \ne i} \hspace{0.25cm}(1-2\cdot p_j)}		:$$L_{\rm E}(i) = {\rm ln} \hspace{0.15cm}\frac{1+\prod_{j \ne i} \hspace{0.25cm}(1-2\cdot p_j)}{1-\prod_{j \ne i} \hspace{0.25cm}(1-2\cdot p_j)}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	At  $\text{SPC (5, 4, 2}$)   ⇒   $n = 5$, this product for $i = 5$ results from the following four factors:		*At  $\text{SPC (5, 4, 2}$)   ⇒   $n = 5$,  this product for  $i = 5$  results from the following four factors:
	:$$\pi = \prod\limits_{j = 1, \hspace{0.05cm}2, \hspace{0.05cm}3, \hspace{0.05cm}4} \hspace{0.05cm}(1-2\cdot p_j) =		:$$\pi = \prod\limits_{j = 1, \hspace{0.05cm}2, \hspace{0.05cm}3, \hspace{0.05cm}4} \hspace{0.05cm}(1-2\cdot p_j) =
	(1-2\cdot p_1) \cdot (1-2\cdot p_2) \cdot (1-2\cdot p_3) \cdot (1-2\cdot p_4)		(1-2\cdot p_1) \cdot (1-2\cdot p_2) \cdot (1-2\cdot p_3) \cdot (1-2\cdot p_4)
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	The comparison with ~~the~~ subtask '''(3)''' shows that $L_{\rm E}(i = 5) = \ln {Q} = \ln {(1.0799)} \ \underline{\approx 0.077}$.		The comparison with subtask  '''(3)'''  shows that  $L_{\rm E}(i = 5) = \ln {Q} = \ln {(1.0799)} \ \underline{\approx 0.077}$.


	'''(5)'''  Correct is <u>proposed solution 3</u> because the result for $L_{\rm E}(i = 5)$ is independent of $p_5$.		'''(5)'''  Correct is the  <u>proposed solution 3</u>  because the result for  $L_{\rm E}(i = 5)$  is independent of  $p_5$.
	{{ML-Fuß}}		{{ML-Fuß}}


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

Guenter at 14:09, 3 December 2022

2022-12-03T14:09:07Z

← Older revision		Revision as of 16:09, 3 December 2022
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	*This exercise belongs to the chapter  [[Channel_Coding/Soft%E2%80%93in_Soft%E2%80%93out_Decoder\|"Soft–in Soft–out Decoder"]].		*This exercise belongs to the chapter  [[Channel_Coding/Soft%E2%80%93in_Soft%E2%80%93out_Decoder\|"Soft–in Soft–out Decoder"]].

	*Reference is made in particular to the section  [[Channel_Coding/Soft-in_Soft-Out_Decoder#~~Calculation_of_extrinsic_LLRs~~\|"For calculating the extrinsic log likelihood ratios"]].		*Reference is made in particular to the section  [[Channel_Coding/Soft-in_Soft-Out_Decoder#Calculation_of_extrinsic_log_likelihood_ratios\|"For calculating the extrinsic log likelihood ratios"]].

	*The exercise is intended as a supplement to  [[Aufgaben:Exercise_4.4:_Extrinsic_L-values_at_SPC\| $\text{Exercise 4.4}$]] .		*The exercise is intended as a supplement to  [[Aufgaben:Exercise_4.4:_Extrinsic_L-values_at_SPC\| $\text{Exercise 4.4}$]] .

Guenter at 14:07, 3 December 2022

2022-12-03T14:07:39Z

← Older revision		Revision as of 16:07, 3 December 2022
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	One applies the equation,  for example,  to calculate the extrinsic  L–values of a  "single parity–check code".		One applies the equation,  for example,  to calculate the extrinsic  L–values of a  "single parity–check code".

	Indeed,  as already pointed out in  [[Aufgaben:Exercise_4.4:_Extrinsic_L-values_at_SPC\| $\text{Exercise A4.4}$]]  the extrinsic log likelihood ratio with Hamming–weight  $w_{\rm H}$  is the truncated sequence  $\underline{x}^{(-i)}$:		Indeed,  as already pointed out in  [[Aufgaben:Exercise_4.4:_Extrinsic_L-values_at_SPC\| $\text{Exercise 4.4}$]]  the extrinsic log likelihood ratio with Hamming–weight  $w_{\rm H}$  is the truncated sequence  $\underline{x}^{(-i)}$:
	:$$L_{\rm E}(i) = {\rm ln} \hspace{0.15cm}\frac{{\rm Pr} \left [w_{\rm H}(\underline{x}^{(-i)})\hspace{0.15cm}{\rm is \hspace{0.15cm} even} \hspace{0.05cm} \| \hspace{0.05cm}\underline{y} \hspace{0.05cm}\right ]}{{\rm Pr} \left [w_{\rm H}(\underline{x}^{(-i)})\hspace{0.15cm}{\rm is \hspace{0.15cm} odd} \hspace{0.05cm} \| \hspace{0.05cm}\underline{y} \hspace{0.05cm}\right ]}		:$$L_{\rm E}(i) = {\rm ln} \hspace{0.15cm}\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}.$$		\hspace{0.05cm}.$$

Guenter at 14:06, 3 December 2022

2022-12-03T14:06:32Z

← Older revision		Revision as of 16:06, 3 December 2022
Line 1:		Line 1:
	{{quiz-Header\|Buchseite=Channel_Coding/Soft-in_Soft-Out_Decoder}}		{{quiz-Header\|Buchseite=Channel_Coding/Soft-in_Soft-Out_Decoder}}

	[[File:P_ID2994__KC_Z_4_4_v3.png \|right\|frame\|Hamming weights ~~and~~ sequence probabilities ]]		[[File:P_ID2994__KC_Z_4_4_v3.png \|right\|frame\|Hamming weights  $w_{\rm H}(\underline{x})$, <br>sequence probabilities  ${\rm Pr}(\underline{x})$ ]]
	The information theorist  [https://en.wikipedia.org/wiki/Robert_G._Gallager "Robert G. Gallager"]  already in 1963 ~~dealt~~ with the following problem:		The information theorist  [https://en.wikipedia.org/wiki/Robert_G._Gallager $\text{Robert G. Gallager}$]  dealt already in 1963 with the following problem:
	* Given a random vector  $\underline{x} = (x_1, \, x_2, \hspace{-0.04cm} \text{ ...} \hspace{0.08cm} , x_n)$  with  $n$  binary elements  $x_i ∈ \{0, \, 1\}$.		* Given a random vector  $\underline{x} = (x_1, \, x_2, \hspace{-0.04cm} \text{ ...} \hspace{0.08cm} , x_n)$  with  $n$  binary elements  $x_i ∈ \{0, \, 1\}$.

	* Known are all probabilities  $p_i = {\rm Pr}(x_i = 1)$  and  $q_i = {\rm Pr}(x_i = 0) = 1 - p_i$  with index  $i = 1, \hspace{-0.04cm}\text{ ...} \hspace{0.08cm} ,\ n$.		* Known are all probabilities  $p_i = {\rm Pr}(x_i = 1)$  and  $q_i = {\rm Pr}(x_i = 0) = 1 - p_i$  with index  $i = 1, \hspace{-0.04cm}\text{ ...} \hspace{0.08cm} ,\ n$.
	* What we are looking for is the probability that the number of ones in this vector is even.
			* What we are looking for is the probability that the number of  "ones"  in this vector is even.

	* Or expressed using the  [[Channel_Coding/Objective_of_Channel_Coding#Important_definitions_for_block_coding\|"Hamming weight"]]:   What is the probability  ${\rm Pr}[w_{\rm H}(\underline{x}) {\rm \ is \ even}]$?		* Or expressed using the  [[Channel_Coding/Objective_of_Channel_Coding#Important_definitions_for_block_coding\|"Hamming weight"]]:   What is the probability  ${\rm Pr}[w_{\rm H}(\underline{x}) {\rm \ is \ even}]$?

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	* For the row highlighted in green   ⇒   $\underline{x} = (1, \, 0, \, 0, \, 1)$  holds  $w_{\rm H}(\underline{x}) = 2$  and		* For the row highlighted in green   ⇒   $\underline{x} = (1, \, 0, \, 0, \, 1)$  holds  $w_{\rm H}(\underline{x}) = 2$  and
	:$${\rm Pr}(\underline{x}) = p_1 \cdot q_2 \cdot q_3 \cdot p_4 = 0.0084.$$		:$${\rm Pr}(\underline{x}) = p_1 \cdot q_2 \cdot q_3 \cdot p_4 = 0.0084.$$
	* Blue font means "$w_{\rm H}(\underline{x})$ is even". Red font stands for "$w_{\rm H}(\underline{x})$ is odd."		* Blue font means  "$w_{\rm H}(\underline{x})$  is even".  Red font stands for  "$w_{\rm H}(\underline{x})$  is odd."
	*The probability  ${\rm Pr}[w_{\rm H}(\underline{x}) {\rm \ is \ even}]$  is the sum of the blue numbers in the last column.
			*The probability  ${\rm Pr}[w_{\rm H}(\underline{x}) {\rm \ is \ even}]$  is the sum of the blue numbers in the last column.

	*The sum of the red numbers gives  ${\rm Pr}[w_{\rm H}(\underline{x}) {\rm \ is \ odd}] = 1 - {\rm Pr}[w_{\rm H}(\underline{x}) {\rm \ is \ even}]$.		*The sum of the red numbers gives  ${\rm Pr}[w_{\rm H}(\underline{x}) {\rm \ is \ odd}] = 1 - {\rm Pr}[w_{\rm H}(\underline{x}) {\rm \ is \ even}]$.

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	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	One applies the equation, for example, to calculate the extrinsic  $L$–values of a  ~~<i>~~single parity–check code~~</i> ~~.		One applies the equation,  for example,  to calculate the extrinsic  L–values of a  "single parity–check code".

	Indeed, as already pointed out in  [[Aufgaben:Exercise_4.4:_Extrinsic_L-values_at_SPC\| "Exercise A4.4"]]  the extrinsic ~~$L$–value~~ with Hamming–weight  $w_{\rm H}$  is the truncated sequence  $\underline{x}^{(-i)}$:		Indeed,  as already pointed out in  [[Aufgaben:Exercise_4.4:_Extrinsic_L-values_at_SPC\| $\text{Exercise A4.4}$]]  the extrinsic log likelihood ratio with Hamming–weight  $w_{\rm H}$  is the truncated sequence  $\underline{x}^{(-i)}$:
	:$$L_{\rm E}(i) = {\rm ln} \hspace{0.15cm}\frac{{\rm Pr} \left [w_{\rm H}(\underline{x}^{(-i)})\hspace{0.15cm}{\rm is \hspace{0.15cm} even} \hspace{0.05cm} \| \hspace{0.05cm}\underline{y} \hspace{0.05cm}\right ]}{{\rm Pr} \left [w_{\rm H}(\underline{x}^{(-i)})\hspace{0.15cm}{\rm is \hspace{0.15cm} odd} \hspace{0.05cm} \| \hspace{0.05cm}\underline{y} \hspace{0.05cm}\right ]}		:$$L_{\rm E}(i) = {\rm ln} \hspace{0.15cm}\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}.$$		\hspace{0.05cm}.$$
Line 41:		Line 46:


			<u>Hints:</u>
			*This exercise belongs to the chapter  [[Channel_Coding/Soft%E2%80%93in_Soft%E2%80%93out_Decoder\|"Soft–in Soft–out Decoder"]].

			*Reference is made in particular to the section  [[Channel_Coding/Soft-in_Soft-Out_Decoder#Calculation_of_extrinsic_LLRs\|"For calculating the extrinsic log likelihood ratios"]].

			*The exercise is intended as a supplement to  [[Aufgaben:Exercise_4.4:_Extrinsic_L-values_at_SPC\| $\text{Exercise 4.4}$]] .
	~~Hints:~~
	*This exercise belongs to the chapter  [[Channel_Coding/Soft%E2%80%93in_Soft%E2%80%93out_Decoder\|"Soft–in Soft–out Decoder"]].
	*Reference is made in particular to the page  [[Channel_Coding/Soft-in_Soft-Out_Decoder#Calculation_of_extrinsic_LLRs\|"For calculating the extrinsic  $L$ values"]].
	*The exercise is intended as a supplement to  [[Aufgaben:Exercise_4.4:_Extrinsic_L-values_at_SPC\| "Exercise 4.4"]] .


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	===Questions===		===Questions===
	<quiz display=simple>		<quiz display=simple>
	{We consider the vector  $\underline{x} = (x_1, \, x_2) \ \Rightarrow \ n = 2$  with  $x_i ∈ \{0, \, 1\}$  and  $p_1 = 0.2, \ p_2 = 0.9$. <br>What is the probability that  $\underline{x}$  contains an even number of ones?		{We consider the vector  $\underline{x} = (x_1, \, x_2) \ \Rightarrow \ n = 2$  with  $x_i ∈ \{0, \, 1\}$  and  $p_1 = 0.2, \ p_2 = 0.9$. <br>What is the probability that   $\underline{x}$   contains an even number of  "ones" ?
	\|type="{}"}		\|type="{}"}
	${\rm Pr}\big [w_{\rm H}(\underline{x}) \ {\rm is \ even}\big ] \ = \ ${ 0.26 3% }		${\rm Pr}\big [w_{\rm H}(\underline{x}) \ {\rm is \ even}\big ] \ = \ ${ 0.26 3% }

	{~~Berechnen Sie die gleiche Wahrscheinlichkeit für~~  $\underline{x} = (x_1, \, x_2, \, x_3) \ \Rightarrow \ n = 3$  ~~und~~  $p_1 = 0.2, \ p_2 = 0.9, \ p_3 = 0.3$.		{Compute the same probability for  $\underline{x} = (x_1, \, x_2, \, x_3) \ \Rightarrow \ n = 3$  and  $p_1 = 0.2, \ p_2 = 0.9, \ p_3 = 0.3$.
	\|type="{}"}		\|type="{}"}
	${\rm Pr}\big [w_{\rm H}(\underline{x}) \ {\rm is \ even}\big ] \ = \ ${ 0.404 3% }		${\rm Pr}\big [w_{\rm H}(\underline{x}) \ {\rm is \ even}\big ] \ = \ ${ 0.404 3% }

	{Now let  $n = 4$  and  $p_1 = 0.2, \ p_2 = 0.9, \ p_3 = 0.3, \ p_4 = 0.6$. Calculate the following quantities according to Gallager's equation:		{Now let be  $n = 4$  and  $p_1 = 0.2, \ p_2 = 0.9, \ p_3 = 0.3, \ p_4 = 0.6$.  Calculate the following quantities according to Gallager's equation:
	\|type="{}"}		\|type="{}"}
	${\rm Pr(blue) = Pr}\big [w_{\rm H}(\underline{x}) \ {\rm is \ even}\big ] \hspace{0.33cm} = \ ${ 0.5192 3% }		${\rm Pr(blue) = Pr}\big [w_{\rm H}(\underline{x}) \ {\rm is \ even}\big ] \hspace{0.33cm} = \ ${ 0.5192 3% }
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	$\text{Quotient }Q = {\rm Pr(blau)/Pr(rot)} \hspace{0.4cm} = \ ${ 1.0799 3% }		$\text{Quotient }Q = {\rm Pr(blau)/Pr(rot)} \hspace{0.4cm} = \ ${ 1.0799 3% }

	{What is the extrinsic~~  $L$ value~~ for the symbol  $i = 5$  at  $\text{SPC (5, 4, 2)}$  with  $p_1 = 0.2, \ p_2 = 0.9, \ p_3 = 0.3, \ p_4 = 0.6, \ p_5 = 0.9$?		{What is the extrinsic log likelihood ratio for the symbol  $i = 5$  at  $\text{SPC (5, 4, 2)}$  with  $p_1 = 0.2, \ p_2 = 0.9, \ p_3 = 0.3, \ p_4 = 0.6, \ p_5 = 0.9$?
	\|type="{}"}		\|type="{}"}
	$L_{\rm E}(i = 5) \ = \ ${ 0.077 3% }		$L_{\rm E}(i = 5) \ = \ ${ 0.077 3% }

	{How does  $L_{\rm E}(i = 5)$ change if we assume  $p_5 = 0.1$  instead?		{How does  $L_{\rm E}(i = 5)$  change if we assume  $p_5 = 0.1$  instead?
	\|type="()"}		\|type="()"}
	- $L_{\rm E}(i = 5)$  becomes larger.		- $L_{\rm E}(i = 5)$  becomes larger.
	- $L_{\rm E}(i = 5)$  becomes smaller.		- $L_{\rm E}(i = 5)$  becomes smaller.
	+ $L_{\rm E}(i = 5)$  is not changed compared to subtask '''(4)'''.		+ $L_{\rm E}(i = 5)$  is not changed compared to subtask  '''(4)'''.
	</quiz>		</quiz>

Noah at 22:09, 29 October 2022

2022-10-29T22:09:48Z

← Older revision		Revision as of 00:09, 30 October 2022
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	Indeed, as already pointed out in  [[Aufgaben:Exercise_4.4:_Extrinsic_L-values_at_SPC\| "Exercise A4.4"]]  the extrinsic $L$–value with Hamming–weight  $w_{\rm H}$  is the truncated sequence  $\underline{x}^{(-i)}$:		Indeed, as already pointed out in  [[Aufgaben:Exercise_4.4:_Extrinsic_L-values_at_SPC\| "Exercise A4.4"]]  the extrinsic $L$–value with Hamming–weight  $w_{\rm H}$  is the truncated sequence  $\underline{x}^{(-i)}$:
	:$$L_{\rm E}(i) = {\rm ln} \hspace{0.15cm}\frac{{\rm Pr} \left [w_{\rm H}(\underline{x}^{(-i)})\hspace{0.15cm}{\rm ~~ist~~ \hspace{0.15cm} ~~gerade~~} \hspace{0.05cm} \| \hspace{0.05cm}\underline{y} \hspace{0.05cm}\right ]}{{\rm Pr} \left [w_{\rm H}(\underline{x}^{(-i)})\hspace{0.15cm}{\rm ~~ist~~ \hspace{0.15cm} ~~ungerade~~} \hspace{0.05cm} \| \hspace{0.05cm}\underline{y} \hspace{0.05cm}\right ]}		:$$L_{\rm E}(i) = {\rm ln} \hspace{0.15cm}\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}.$$		\hspace{0.05cm}.$$

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	Hints:		Hints:
	*This exercise belongs to the chapter  [[Channel_Coding/Soft%E2%80%93in_Soft%E2%80%93out_Decoder\|"Soft–in Soft–out Decoder"]].		*This exercise belongs to the chapter  [[Channel_Coding/Soft%E2%80%93in_Soft%E2%80%93out_Decoder\|"Soft–in Soft–out Decoder"]].
	*Reference is made in particular to the page  [[Channel_Coding/Soft-in_Soft-Out_Decoder#Calculation_of_extrinsic_LLRs\|"For calculating the extrinsic  $L$~~–~~values"]].		*Reference is made in particular to the page  [[Channel_Coding/Soft-in_Soft-Out_Decoder#Calculation_of_extrinsic_LLRs\|"For calculating the extrinsic  $L$ values"]].
	*The exercise is intended as a supplement to  [[Aufgaben:Exercise_4.4:_Extrinsic_L-values_at_SPC\| "Exercise 4.4"]] .		*The exercise is intended as a supplement to  [[Aufgaben:Exercise_4.4:_Extrinsic_L-values_at_SPC\| "Exercise 4.4"]] .

Noah at 22:09, 29 October 2022

2022-10-29T22:09:25Z

← Older revision		Revision as of 00:09, 30 October 2022
Line 6:		Line 6:
	* Known are all probabilities  $p_i = {\rm Pr}(x_i = 1)$  and  $q_i = {\rm Pr}(x_i = 0) = 1 - p_i$  with index  $i = 1, \hspace{-0.04cm}\text{ ...} \hspace{0.08cm} ,\ n$.		* Known are all probabilities  $p_i = {\rm Pr}(x_i = 1)$  and  $q_i = {\rm Pr}(x_i = 0) = 1 - p_i$  with index  $i = 1, \hspace{-0.04cm}\text{ ...} \hspace{0.08cm} ,\ n$.
	* What we are looking for is the probability that the number of ones in this vector is even.		* What we are looking for is the probability that the number of ones in this vector is even.
	* Or expressed using the  [[Channel_Coding/Objective_of_Channel_Coding#Important_definitions_for_block_coding\|"Hamming weight"]]:   What is the probability  ${\rm Pr}[w_{\rm H}(\underline{x}) {\rm \ is \even}]$?		* Or expressed using the  [[Channel_Coding/Objective_of_Channel_Coding#Important_definitions_for_block_coding\|"Hamming weight"]]:   What is the probability  ${\rm Pr}[w_{\rm H}(\underline{x}) {\rm \ is \ even}]$?


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	:$${\rm Pr}(\underline{x}) = p_1 \cdot q_2 \cdot q_3 \cdot p_4 = 0.0084.$$		:$${\rm Pr}(\underline{x}) = p_1 \cdot q_2 \cdot q_3 \cdot p_4 = 0.0084.$$
	* Blue font means "$w_{\rm H}(\underline{x})$ is even". Red font stands for "$w_{\rm H}(\underline{x})$ is odd."		* Blue font means "$w_{\rm H}(\underline{x})$ is even". Red font stands for "$w_{\rm H}(\underline{x})$ is odd."
	*The probability  ${\rm Pr}[w_{\rm H}(\underline{x}) {\rm \is \even}]$  is the sum of the blue numbers in the last column.		*The probability  ${\rm Pr}[w_{\rm H}(\underline{x}) {\rm \ is \ even}]$  is the sum of the blue numbers in the last column.
	*The sum of the red numbers gives  ${\rm Pr}[w_{\rm H}(\underline{x}) {\rm \ is \ odd}] = 1 - {\rm Pr}[w_{\rm H}(\underline{x}) {\rm \ is \ even}]$.		*The sum of the red numbers gives  ${\rm Pr}[w_{\rm H}(\underline{x}) {\rm \ is \ odd}] = 1 - {\rm Pr}[w_{\rm H}(\underline{x}) {\rm \ is \ even}]$.


	Gallager ~~hat das Problem~~ in ~~analytischer Weise gelöst~~:		Gallager solved the problem in an analytical way:
	:$$ {\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm ~~ist~~ \hspace{0.15cm} ~~gerade~~} \right ]		:$$ {\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} even} \right ]
	\hspace{-0.15cm} \ = \ \hspace{-0.15cm} 1/2 \cdot [1 + \pi]\hspace{0.05cm},$$		\hspace{-0.15cm} \ = \ \hspace{-0.15cm} 1/2 \cdot [1 + \pi]\hspace{0.05cm},$$
	:$${\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm ~~ist~~ \hspace{0.15cm} ~~ungerade~~}		:$${\rm Pr} \left [w_{\rm H}(\underline{x})\hspace{0.15cm}{\rm is \hspace{0.15cm} odd}
	\right ] \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 1/2 \cdot [1 - \pi]\hspace{0.05cm}.$$		\right ] \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 1/2 \cdot [1 - \pi]\hspace{0.05cm}.$$

	~~Hierbei ist die folgende Hilfsgröße verwendet~~:		Here the following auxiliary variable is used:
	:$$\pi = \prod\limits_{i =1}^{n} \hspace{0.25cm}(1-2p_i)		:$$\pi = \prod\limits_{i =1}^{n} \hspace{0.25cm}(1-2p_i)
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	~~Die Gleichung wendet man zum Beispiel an~~, ~~um die extrinsischen~~  $L$–~~Werte eines~~  <i>~~Single Parity~~–check ~~Codes~~</i>  ~~zu berechnen~~.		One applies the equation, for example, to calculate the extrinsic  $L$–values of a  <i>single parity–check code</i> .

	~~Wie bereits~~ in ~~der~~  [[Aufgaben:Exercise_4.4:_Extrinsic_L-values_at_SPC\| "~~Aufgabe~~ A4.4"]]  ~~dargelegt, lautet nämlich der extrinsische~~ $L$–~~Wert mit dem~~ Hamming–~~Gewicht~~  $w_{\rm H}$  ~~der verkürzten Folge~~  $\underline{x}^{(-i)}$:		Indeed, as already pointed out in  [[Aufgaben:Exercise_4.4:_Extrinsic_L-values_at_SPC\| "Exercise A4.4"]]  the extrinsic $L$–value with Hamming–weight  $w_{\rm H}$  is the truncated sequence  $\underline{x}^{(-i)}$:
	:$$L_{\rm E}(i) = {\rm ln} \hspace{0.15cm}\frac{{\rm Pr} \left [w_{\rm H}(\underline{x}^{(-i)})\hspace{0.15cm}{\rm ist \hspace{0.15cm} gerade} \hspace{0.05cm} \| \hspace{0.05cm}\underline{y} \hspace{0.05cm}\right ]}{{\rm Pr} \left [w_{\rm H}(\underline{x}^{(-i)})\hspace{0.15cm}{\rm ist \hspace{0.15cm} ungerade} \hspace{0.05cm} \| \hspace{0.05cm}\underline{y} \hspace{0.05cm}\right ]}		:$$L_{\rm E}(i) = {\rm ln} \hspace{0.15cm}\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}.$$		\hspace{0.05cm}.$$

	~~Hierbei ist berücksichtigt, dass man für~~  $L_{\rm E}(i)$  ~~nur die anderen Symbole~~  $(j ≠ i)$  ~~heranziehen darf~~:		Here it is taken into account that for  $L_{\rm E}(i)$  one may refer only to the other symbols  $(j ≠ i)$ :
	:$$\underline{x}^{(-i)} = \big ( \hspace{0.03cm}x_1, \hspace{-0.04cm} \text{ ...} \hspace{0.08cm} , \hspace{0.03cm} x_{i-1}, \hspace{0.43cm} x_{i+1}, \hspace{-0.04cm} \text{ ...} \hspace{0.08cm} , x_{n} \hspace{0.03cm} \big )\hspace{0.03cm}. $$		:$$\underline{x}^{(-i)} = \big ( \hspace{0.03cm}x_1, \hspace{-0.04cm} \text{ ...} \hspace{0.08cm} , \hspace{0.03cm} x_{i-1}, \hspace{0.43cm} x_{i+1}, \hspace{-0.04cm} \text{ ...} \hspace{0.08cm} , x_{n} \hspace{0.03cm} \big )\hspace{0.03cm}. $$

Noah at 22:06, 29 October 2022

2022-10-29T22:06:50Z

← Older revision		Revision as of 00:06, 30 October 2022
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	* Known are all probabilities  $p_i = {\rm Pr}(x_i = 1)$  and  $q_i = {\rm Pr}(x_i = 0) = 1 - p_i$  with index  $i = 1, \hspace{-0.04cm}\text{ ...} \hspace{0.08cm} ,\ n$.		* Known are all probabilities  $p_i = {\rm Pr}(x_i = 1)$  and  $q_i = {\rm Pr}(x_i = 0) = 1 - p_i$  with index  $i = 1, \hspace{-0.04cm}\text{ ...} \hspace{0.08cm} ,\ n$.
	* What we are looking for is the probability that the number of ones in this vector is even.		* What we are looking for is the probability that the number of ones in this vector is even.
	* Or expressed using the  [[Channel_Coding/Objective_of_Channel_Coding#Important_definitions_for_block_coding\|"Hamming weight"]]:   What is the probability  ${\rm Pr}[w_{\rm H}(\underline{x}) {\rm \ is \ even}]$?		* Or expressed using the  [[Channel_Coding/Objective_of_Channel_Coding#Important_definitions_for_block_coding\|"Hamming weight"]]:   What is the probability  ${\rm Pr}[w_{\rm H}(\underline{x}) {\rm \ is \even}]$?

← Older revision		Revision as of 16:09, 3 December 2022
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	*This exercise belongs to the chapter  [[Channel_Coding/Soft%E2%80%93in_Soft%E2%80%93out_Decoder\|"Soft–in Soft–out Decoder"]].		*This exercise belongs to the chapter  [[Channel_Coding/Soft%E2%80%93in_Soft%E2%80%93out_Decoder\|"Soft–in Soft–out Decoder"]].

	*Reference is made in particular to the section  [[Channel_Coding/Soft-in_Soft-Out_Decoder#~~Calculation_of_extrinsic_LLRs~~\|"For calculating the extrinsic log likelihood ratios"]].		*Reference is made in particular to the section  [[Channel_Coding/Soft-in_Soft-Out_Decoder#Calculation_of_extrinsic_log_likelihood_ratios\|"For calculating the extrinsic log likelihood ratios"]].

	*The exercise is intended as a supplement to  [[Aufgaben:Exercise_4.4:_Extrinsic_L-values_at_SPC\| $\text{Exercise 4.4}$]] .		*The exercise is intended as a supplement to  [[Aufgaben:Exercise_4.4:_Extrinsic_L-values_at_SPC\| $\text{Exercise 4.4}$]] .