Aufgaben:Exercise 4.09: Recursive Systematic Convolutional Codes - Revision history

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	[[Category:Channel Coding: Exercises\|^4.3 About the Turbo Codes^]]		[[Category:Channel Coding: Exercises\|^4.3 About the Turbo Codes^]]
	[[de:Aufgaben:4.~~09_Wiederholung_zu_den_RSC-~~Codes]]		[[de:Aufgaben:Aufgabe 4.09: Recursive Systematic Convolutional Codes]]

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@@ Line 78: / Line 78: @@
 *This gives the result corresponding to <u>solution proposition 1</u>:
-:$$\underline{p}= (\hspace{0.05cm}1\hspace{0.05cm},
+:$$\underline{p}= (\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm}\text{...}  \hspace{0.05cm})= \underline{g}\hspace{0.05cm}.$$
 *For any RSC code,&nbsp; the impulse response&nbsp; $\underline{g}$&nbsp; is infinite in length and becomes periodic at some point,&nbsp; in this example with period&nbsp; $P = 3$&nbsp; and&nbsp; "$0, \, 1, \, 1$".
@@ Line 103: / Line 94: @@
 '''(3)'''&nbsp; Correct are&nbsp; <u>the proposed solutions 1 and 2</u>:
 *Between the impulse response&nbsp; $\underline{g}$&nbsp; and the&nbsp; $D$&ndash;transfer function&nbsp; $\mathbf{G}(D)$&nbsp; there is the relation according to the first proposed solution:
-:$$\underline{g}= (\hspace{-0.05cm}1\hspace{-0.05cm},
+:$$\underline{g}= (\hspace{-0.05cm}1\hspace{-0.05cm},\hspace{-0.07cm}1\hspace{-0.07cm},\hspace{-0.07cm}1\hspace{-0.07cm},\hspace{-0.07cm}0\hspace{-0.07cm},\hspace{-0.07cm}1\hspace{-0.07cm},\hspace{-0.07cm}1\hspace{-0.07cm},\hspace{-0.07cm}0\hspace{-0.07cm},\hspace{-0.07cm}1\hspace{-0.07cm},\hspace{-0.07cm}1\hspace{-0.07cm},... ) \hspace{0.3cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet \hspace{0.3cm}G(D) =  1\hspace{-0.05cm}+\hspace{-0.05cm} D\hspace{-0.05cm} +\hspace{-0.05cm} D^2\hspace{-0.05cm} +\hspace{-0.05cm} D^4 \hspace{-0.05cm}+\hspace{-0.05cm} D^5 \hspace{-0.05cm}+\hspace{-0.05cm} D^7 \hspace{-0.05cm}+\hspace{-0.05cm} D^8 +   \text{...} .$$*Let us now examine the second proposal::$$G(D) = \frac{1+  D^2}{1+ D  + D^2} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}G(D) \cdot [1+ D  + D^2] = 1+  D^2 \hspace{0.05cm}.$$
 *The following calculation shows that this equation is indeed true:

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	The transfer function matrix of an RSC convolutional code can be specified as follows:		The transfer function matrix of an RSC convolutional code can be specified as follows:
	:$${\boldsymbol{\rm G}}(D) = \left [ 1\hspace{0.05cm},\hspace{0.3cm} G^{(2)}(D)/G^{(1)}(D) \right ]		:$${\boldsymbol{\rm G}}(D) = \left [ 1\hspace{0.05cm},\hspace{0.3cm} G^{(2)}(D)/G^{(1)}(D) \right ]\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	Otherwise,  exactly the same conditions apply here as in Exercise 4.8.  We refer again to the following theory pages:		Otherwise,  exactly the same conditions apply here as in Exercise 4.8.  We refer again to the following theory pages:
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	\hspace{-0.07cm}1\hspace{-0.07cm},		\hspace{-0.07cm}1\hspace{-0.07cm},
	... ) \hspace{0.3cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet \hspace{0.3cm}		... ) \hspace{0.3cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet \hspace{0.3cm}
	G(D) = 1\hspace{-0.05cm}+\hspace{-0.05cm} D\hspace{-0.05cm} +\hspace{-0.05cm} D^2\hspace{-0.05cm} +\hspace{-0.05cm} D^4 \hspace{-0.05cm}+\hspace{-0.05cm} D^5 \hspace{-0.05cm}+\hspace{-0.05cm} D^7 \hspace{-0.05cm}+\hspace{-0.05cm} D^8 + \text{...} .$$		G(D) = 1\hspace{-0.05cm}+\hspace{-0.05cm} D\hspace{-0.05cm} +\hspace{-0.05cm} D^2\hspace{-0.05cm} +\hspace{-0.05cm} D^4 \hspace{-0.05cm}+\hspace{-0.05cm} D^5 \hspace{-0.05cm}+\hspace{-0.05cm} D^7 \hspace{-0.05cm}+\hspace{-0.05cm} D^8 + \text{...} .$$*Let us now examine the second proposal::$$G(D) = \frac{1+ D^2}{1+ D + D^2} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}

	*Let us now examine the second proposal:
	:$$G(D) = \frac{1+ D^2}{1+ D + D^2} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}
	G(D) \cdot [1+ D + D^2] = 1+ D^2 \hspace{0.05cm}.$$		G(D) \cdot [1+ D + D^2] = 1+ D^2 \hspace{0.05cm}.$$

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	:$$+ \hspace{0.8cm}D+ D^2+D^3 \hspace{1.05cm}+ D^5 + D^6 \hspace{1.05cm} +D^8 + D^9 \hspace{1.25cm} +\hspace{0.05cm} \text{...} $$		:$$+ \hspace{0.8cm}D+ D^2+D^3 \hspace{1.05cm}+ D^5 + D^6 \hspace{1.05cm} +D^8 + D^9 \hspace{1.25cm} +\hspace{0.05cm} \text{...} $$
	:$$+ \hspace{1.63cm} D^2+D^3+ D^4 \hspace{1.05cm}+ D^6 +D^7 \hspace{1.05cm}+ D^9 + D^{10} \hspace{0.12cm}+ \hspace{0.05cm} \text{...}$$		:$$+ \hspace{1.63cm} D^2+D^3+ D^4 \hspace{1.05cm}+ D^6 +D^7 \hspace{1.05cm}+ D^9 + D^{10} \hspace{0.12cm}+ \hspace{0.05cm} \text{...}$$
	:$$=\underline{1\hspace{0.72		:$$=\underline{1\hspace{0.72cm}+ D^2} \hspace{0.05cm}.$$
	cm}+ D^2} \hspace{0.05cm}.$$

	*But since equation  '''(2)'''  is true,  the last equation must be false.		*But since equation  '''(2)'''  is true,  the last equation must be false.
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	'''(4)'''  Correct is only the  <u>proposed solution 1</u>:		'''(4)'''  Correct is only the  <u>proposed solution 1</u>:
	*From   $\underline{u} = (1, \, 1, \, 1)$   follows   $U(D) = 1 + D + D^2$.  Thus also holds:		*From   $\underline{u} = (1, \, 1, \, 1)$   follows   $U(D) = 1 + D + D^2$.  Thus also holds:
	:$$P(D) = U(D) \cdot G(D) = (1+D+D^2) \cdot \frac{1+D^2}{1+D+D^2}= 1+D^2\hspace{0.3cm}		:$$P(D) = U(D) \cdot G(D) = (1+D+D^2) \cdot \frac{1+D^2}{1+D+D^2}= 1+D^2\hspace{0.3cm}\Rightarrow\hspace{0.3cm} \underline{p}= (\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}\text{...}\hspace{0.05cm})\hspace{0.05cm}. $$
	\Rightarrow\hspace{0.3cm} \underline{p}= (\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}\text{...}\hspace{0.05cm})\hspace{0.05cm}. $$
	* If the variables   $u_i$   and   $g_i$   were real-valued,  the  $($discrete$)$  convolution   $\underline{p} = \underline{u} * \underline{g} $  would always lead to a broadening.		* If the variables   $u_i$   and   $g_i$   were real-valued,  the  $($discrete$)$  convolution   $\underline{p} = \underline{u} * \underline{g} $  would always lead to a broadening.
	[[File:EN_KC_A_4_9d_v1.png\|right\|frame\|Clarification of  $\underline{p} = (1, \, 1, \, 1, \, 0, \, ...)^{\rm T} \cdot \mathbf{G}$]]		[[File:EN_KC_A_4_9d_v1.png\|right\|frame\|Clarification of  $\underline{p} = (1, \, 1, \, 1, \, 0, \, ...)^{\rm T} \cdot \mathbf{G}$]]

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	[[Category:Channel Coding: Exercises\|^4.3 About the Turbo Codes^]]		[[Category:Channel Coding: Exercises\|^4.3 About the Turbo Codes^]]
			[[de:Aufgaben:4.09_Wiederholung_zu_den_RSC-Codes]]

Guenter at 16:06, 13 March 2023

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	[[File:EN_KC_A_4_9b_v1.png\|right\|frame\|Clarification of  $\underline{p} = (1, \, 1, \, 0, \, 1)^{\rm T} \cdot \mathbf{G}$ ~~'''Korrektur'''~~]]		[[File:EN_KC_A_4_9b_v1.png\|right\|frame\|Clarification of  $\underline{p} = (1, \, 1, \, 0, \, 1)^{\rm T} \cdot \mathbf{G}$]]
	'''(2)'''  The graph shows the solution of this exercise according to the equation  $\underline{p} = \underline{u}^{\rm T} \cdot \mathbf{G}$.		'''(2)'''  The graph shows the solution of this exercise according to the equation  $\underline{p} = \underline{u}^{\rm T} \cdot \mathbf{G}$.
	* Here the generator matrix  $\mathbf{G}$  is infinitely extended downward and to the right.		* Here the generator matrix  $\mathbf{G}$  is infinitely extended downward and to the right.

Guenter at 16:01, 19 December 2022

2022-12-19T16:01:47Z

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	[[File:~~P_ID3054__KC_A_4_9b_v1~~.png\|right\|frame\|Clarification of  $\underline{p} = (1, \, 1, \, 0, \, 1)^{\rm T} \cdot \mathbf{G}$ '''Korrektur''']]		[[File:EN_KC_A_4_9b_v1.png\|right\|frame\|Clarification of  $\underline{p} = (1, \, 1, \, 0, \, 1)^{\rm T} \cdot \mathbf{G}$ '''Korrektur''']]
	'''(2)'''  The graph shows the solution of this exercise according to the equation  $\underline{p} = \underline{u}^{\rm T} \cdot \mathbf{G}$.		'''(2)'''  The graph shows the solution of this exercise according to the equation  $\underline{p} = \underline{u}^{\rm T} \cdot \mathbf{G}$.
	* Here the generator matrix  $\mathbf{G}$  is infinitely extended downward and to the right.		* Here the generator matrix  $\mathbf{G}$  is infinitely extended downward and to the right.

Guenter at 15:40, 14 December 2022

2022-12-14T15:40:38Z

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	*Between the impulse response  $\underline{g}$  and the  $D$–transfer function  $\mathbf{G}(D)$  there is the relation according to the first proposed solution:		*Between the impulse response  $\underline{g}$  and the  $D$–transfer function  $\mathbf{G}(D)$  there is the relation according to the first proposed solution:
	:$$\underline{g}= (\hspace{-0.05cm}1\hspace{-0.05cm},		:$$\underline{g}= (\hspace{-0.05cm}1\hspace{-0.05cm},
	\hspace{-0.~~05cm~~}1\hspace{-0.~~05cm~~},		\hspace{-0.07cm}1\hspace{-0.07cm},
	\hspace{-0.~~05cm~~}1\hspace{-0.~~05cm~~},		\hspace{-0.07cm}1\hspace{-0.07cm},
	\hspace{-0.~~05cm~~}0\hspace{-0.~~05cm~~},		\hspace{-0.07cm}0\hspace{-0.07cm},
	\hspace{-0.~~05cm~~}1\hspace{-0.~~05cm~~},		\hspace{-0.07cm}1\hspace{-0.07cm},
	\hspace{-0.~~05cm~~}1\hspace{-0.~~05cm~~},		\hspace{-0.07cm}1\hspace{-0.07cm},
	\hspace{-0.~~05cm~~}0\hspace{-0.~~05cm~~},		\hspace{-0.07cm}0\hspace{-0.07cm},
	\hspace{-0.~~05cm~~}1\hspace{-0.~~05cm~~},		\hspace{-0.07cm}1\hspace{-0.07cm},
	\hspace{-0.~~05cm~~}1\hspace{-0.~~05cm~~},		\hspace{-0.07cm}1\hspace{-0.07cm},
	... ) \hspace{0.3cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet \hspace{0.3cm}		... ) \hspace{0.3cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet \hspace{0.3cm}
	G(D) = 1\hspace{-0.05cm}+\hspace{-0.05cm} D\hspace{-0.05cm} +\hspace{-0.05cm} D^2\hspace{-0.05cm} +\hspace{-0.05cm} D^4 \hspace{-0.05cm}+\hspace{-0.05cm} D^5 \hspace{-0.05cm}+\hspace{-0.05cm} D^7 \hspace{-0.05cm}+\hspace{-0.05cm} D^8 + \text{...} .$$		G(D) = 1\hspace{-0.05cm}+\hspace{-0.05cm} D\hspace{-0.05cm} +\hspace{-0.05cm} D^2\hspace{-0.05cm} +\hspace{-0.05cm} D^4 \hspace{-0.05cm}+\hspace{-0.05cm} D^5 \hspace{-0.05cm}+\hspace{-0.05cm} D^7 \hspace{-0.05cm}+\hspace{-0.05cm} D^8 + \text{...} .$$
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	cm}+ D^2} \hspace{0.05cm}.$$		cm}+ D^2} \hspace{0.05cm}.$$

	*But since equation (2) is true, the last equation must be false.		*But since equation  '''(2)'''  is true,  the last equation must be false.



	'''(4)'''  Correct is only the <u>proposed solution 1</u>:		'''(4)'''  Correct is only the  <u>proposed solution 1</u>:
	*From $\underline{u} = (1, \, 1, \, 1)$ follows $U(D) = 1 + D + D^2$. Thus also holds:		*From   $\underline{u} = (1, \, 1, \, 1)$   follows   $U(D) = 1 + D + D^2$.  Thus also holds:
	:$$P(D) = U(D) \cdot G(D) = (1+D+D^2) \cdot \frac{1+D^2}{1+D+D^2}= 1+D^2\hspace{0.3cm}		:$$P(D) = U(D) \cdot G(D) = (1+D+D^2) \cdot \frac{1+D^2}{1+D+D^2}= 1+D^2\hspace{0.3cm}
	\Rightarrow\hspace{0.3cm} \underline{p}= (\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}\text{...}\hspace{0.05cm})\hspace{0.05cm}. $$		\Rightarrow\hspace{0.3cm} \underline{p}= (\hspace{0.05cm}1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}\text{...}\hspace{0.05cm})\hspace{0.05cm}. $$
	* If the variables  $u_i$  and  $g_i$  were real-valued, the (discrete) convolution  $\underline{p} = \underline{u} * \underline{g}$  would always lead to a broadening  ⇒   $\underline{p}$  in this case would be broader than  $\underline{u}$  and also broader than  ~~$\underline{g}$.~~		* If the variables   $u_i$   and   $g_i$   were real-valued,  the  $($discrete$)$  convolution   $\underline{p} = \underline{u} * \underline{g} $  would always lead to a broadening.
	* On the ~~other hand, for~~  $~~u_i ∈ {\rm GF}~~(2)$~~  and  $g_i ∈ {\rm GF}(2)~~$~~  it may (but need not~~) ~~happen that even with unbounded  $\underline{u}$  or for unbounded  $\underline{g}~~$  ~~the~~ convolution ~~product~~  $\underline{p} = \underline{u} * \underline{g}$  ~~is bounded~~.		[[File:EN_KC_A_4_9d_v1.png\|right\|frame\|Clarification of  $\underline{p} = (1, \, 1, \, 1, \, 0, \, ...)^{\rm T} \cdot \mathbf{G}$]]
	[[File:EN_KC_A_4_9d_v1.png\|right\|frame\|~~Verdeutlichung von~~  $\underline{p} = (1, \, 1, \, 1, \, 0, \, ...)^{\rm T} \cdot \mathbf{G}$]]		*In this case   $\underline{p}$  would be broader than   $\underline{u}$   and also broader than   $\underline{g}$.
	*~~The result is finally checked according to the equation~~ $\underline{p} = \underline{u}^{\~~rm T} \cdot \mathbf~~{G}$ ~~is checked~~.

			* On the other hand,  for   $u_i ∈ {\rm GF}(2)$   and   $g_i ∈ {\rm GF}(2)$   it may  $($but need not$)$  happen that even with unlimited   $\underline{u}$   or for unlimited   $\underline{g}$   the convolution product   $\underline{p} = \underline{u} * \underline{g}$  is limited.

			*The result is finally checked according to the equation  $\underline{p} = \underline{u}^{\rm T} \cdot \mathbf{G}$  is checked.  See graph on the right.

	'''(5)'''  Following a similar procedure as in [[Aufgaben:Exercise_4.08:_Repetition_to_the_Convolutional_Codes\|"Exercise A4.8, (4)"]], one can see:
	*The free distance is again determined by the path  $S_0 → S_0 → S_1 → S_2 → S_0 → S_0 → \hspace{0.05cm}\text{...}\hspace{0.05cm}$.
	*But the associated ~~code~~ sequence  $\underline{x}$  is now  " $00 \ 11 \ 10 \ 11 \ 00 \ ... $".		'''(5)'''  Following a similar procedure as in  [[Aufgaben:Exercise_4.08:_Repetition_to_the_Convolutional_Codes\|$\text{Exercise A4.8}$,  (4)]],  one can see:
	*This gives the free distance to  $d_{\rm F} \ \underline{= 5}$.		*The free distance is again determined by the path
	*In contrast, in the non-recursive code (~~exercise~~ 4.8), the free distance was  $d_{\rm F} = 3$.		:$$S_0 → S_0 → S_1 → S_2 → S_0 → S_0 → \hspace{0.05cm}\text{...}\hspace{0.05cm}.$$
			*But the associated encoded sequence  $\underline{x}$  is now  " $00 \ 11 \ 10 \ 11 \ 00 \ ... $".

			*This gives the free distance to  $d_{\rm F} \ \underline{= 5}$.

			*In contrast,  in the non-recursive code  $($Exercise 4.8$)$,  the free distance was  $d_{\rm F} = 3$.
	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 14:12, 14 December 2022

2022-12-14T14:12:35Z

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	'''(3)'''  Correct are  <u>the proposed solutions 1 and 2</u>:		'''(3)'''  Correct are  <u>the proposed solutions 1 and 2</u>:
	*Between the impulse response  $\underline{g}$  and the  $D$–transfer function  $\mathbf{G}(D)$  there is the relation according to the first proposed solution:		*Between the impulse response  $\underline{g}$  and the  $D$–transfer function  $\mathbf{G}(D)$  there is the relation according to the first proposed solution:
	$$\underline{g}= (\hspace{-0.05cm}1\hspace{-0.05cm},		:$$\underline{g}= (\hspace{-0.05cm}1\hspace{-0.05cm},
	\hspace{-0.05cm}1\hspace{-0.05cm},		\hspace{-0.05cm}1\hspace{-0.05cm},
	\hspace{-0.05cm}1\hspace{-0.05cm},		\hspace{-0.05cm}1\hspace{-0.05cm},
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	\hspace{-0.05cm}1\hspace{-0.05cm},		\hspace{-0.05cm}1\hspace{-0.05cm},
	\hspace{-0.05cm}1\hspace{-0.05cm},		\hspace{-0.05cm}1\hspace{-0.05cm},
	... ) \hspace{0.3cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\\hspace{0.3cm}		... ) \hspace{0.3cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet \hspace{0.3cm}
	G(D) = 1\hspace{-0.05cm}+\hspace{-0.05cm} D\hspace{-0.05cm} +\hspace{-0.05cm} D^2\hspace{-0.05cm} +\hspace{-0.05cm} D^4 \hspace{-0.05cm}+\hspace{-0.05cm} D^5 \hspace{-0.05cm}+\hspace{-0.05cm} D^7 \hspace{-0.05cm}+\hspace{-0.05cm} D^8 + ~~\hspace{0.05cm}~~ \text{...~~} \hspace{0.05cm~~}.$$		G(D) = 1\hspace{-0.05cm}+\hspace{-0.05cm} D\hspace{-0.05cm} +\hspace{-0.05cm} D^2\hspace{-0.05cm} +\hspace{-0.05cm} D^4 \hspace{-0.05cm}+\hspace{-0.05cm} D^5 \hspace{-0.05cm}+\hspace{-0.05cm} D^7 \hspace{-0.05cm}+\hspace{-0.05cm} D^8 + \text{...} .$$

	*Let us now examine the second proposal:		*Let us now examine the second proposal:

Guenter at 14:10, 14 December 2022

2022-12-14T14:10:36Z

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  Tracing the transitions in the state diagram for the sequence  $\underline{u} = (1, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0)$  at the input, we get the path		'''(1)'''  Tracing the transitions in the state diagram for the sequence   $\underline{u} = (1, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0)$   at the input,  we get the path
	:$$S_0 → S_1 → S_3 → S_2 → S_1 → S_3 → S_2 → S_1 → S_3 → \hspace{0.05cm}\text{...} \hspace{0.05cm}$$		:$$S_0 → S_1 → S_3 → S_2 → S_1 → S_3 → S_2 → S_1 → S_3 → \hspace{0.05cm}\text{...} \hspace{0.05cm}$$

	*For each transition, the first code symbol $x_i^{(1)}$ is equal to the information bit $u_i$ and the code symbol $x_i^{(2)}$ indicates the parity-check bit $p_i$.		*For each transition,  the first code symbol  $x_i^{(1)}$  is equal to the information bit  $u_i$  and the code symbol  $x_i^{(2)}$  indicates the parity-check bit  $p_i$.

	*This gives the result corresponding to <u>solution proposition 1</u>:		*This gives the result corresponding to <u>solution proposition 1</u>:
	:$$\underline{p}= (\hspace{0.05cm}1\hspace{0.05cm},		:$$\underline{p}= (\hspace{0.05cm}1\hspace{0.05cm},
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	= \underline{g}\hspace{0.05cm}.$$		= \underline{g}\hspace{0.05cm}.$$

	*For any RSC code, the impulse response $\underline{g}$ is infinite in length and becomes periodic at some point, in this example with period  $P = 3$  and  "$0, \, 1, \, 1$".		*For any RSC code,  the impulse response  $\underline{g}$  is infinite in length and becomes periodic at some point,  in this example with period  $P = 3$  and  "$0, \, 1, \, 1$".



	[[File:P_ID3054__KC_A_4_9b_v1.png\|right\|frame\|Clarification of  $\underline{p} = (1, \, 1, \, 0, \, 1)^{\rm T} \cdot \mathbf{G}$]]		[[File:P_ID3054__KC_A_4_9b_v1.png\|right\|frame\|Clarification of  $\underline{p} = (1, \, 1, \, 0, \, 1)^{\rm T} \cdot \mathbf{G}$ '''Korrektur''']]
	'''(2)'''  The graph shows the solution of this exercise according to the equation  $\underline{p} = \underline{u}^{\rm T} \cdot \mathbf{G}$.		'''(2)'''  The graph shows the solution of this exercise according to the equation  $\underline{p} = \underline{u}^{\rm T} \cdot \mathbf{G}$.
	* Here the generator matrix  $\mathbf{G}$  is infinitely extended downward and to the right.		* Here the generator matrix  $\mathbf{G}$  is infinitely extended downward and to the right.
	*The correct solution is <u>proposal 2</u>.
	~~<br clear=all>~~		*The correct solution is  <u>proposal 2</u>.
	'''(3)'''  Correct are <u>the proposed solutions 1 and 2</u>:
	*Between the impulse response $\underline{g}$ and the $D$–transfer function $\mathbf{G}(D)$ there is the relation according to the first proposed solution:
	:$$\underline{g}= (\hspace{-0.05cm}1\hspace{-0.05cm},
			'''(3)'''  Correct are  <u>the proposed solutions 1 and 2</u>:
			*Between the impulse response  $\underline{g}$  and the  $D$–transfer function  $\mathbf{G}(D)$  there is the relation according to the first proposed solution:
			$$\underline{g}= (\hspace{-0.05cm}1\hspace{-0.05cm},
	\hspace{-0.05cm}1\hspace{-0.05cm},		\hspace{-0.05cm}1\hspace{-0.05cm},
	\hspace{-0.05cm}1\hspace{-0.05cm},		\hspace{-0.05cm}1\hspace{-0.05cm},
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	\hspace{-0.05cm}1\hspace{-0.05cm},		\hspace{-0.05cm}1\hspace{-0.05cm},
	\hspace{-0.05cm}1\hspace{-0.05cm},		\hspace{-0.05cm}1\hspace{-0.05cm},
	... ) \~~quad~~ \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\~~quad~~		... ) \hspace{0.3cm} \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\\hspace{0.3cm}
	G(D) = 1\hspace{-0.05cm}+\hspace{-0.05cm} D\hspace{-0.05cm} +\hspace{-0.05cm} D^2\hspace{-0.05cm} +\hspace{-0.05cm} D^4 \hspace{-0.05cm}+\hspace{-0.05cm} D^5 \hspace{-0.05cm}+\hspace{-0.05cm} D^7 \hspace{-0.05cm}+\hspace{-0.05cm} D^8 + \hspace{0.05cm} \text{...} \hspace{0.05cm}.$$		G(D) = 1\hspace{-0.05cm}+\hspace{-0.05cm} D\hspace{-0.05cm} +\hspace{-0.05cm} D^2\hspace{-0.05cm} +\hspace{-0.05cm} D^4 \hspace{-0.05cm}+\hspace{-0.05cm} D^5 \hspace{-0.05cm}+\hspace{-0.05cm} D^7 \hspace{-0.05cm}+\hspace{-0.05cm} D^8 + \hspace{0.05cm} \text{...} \hspace{0.05cm}.$$

Guenter at 13:55, 14 December 2022

2022-12-14T13:55:38Z

← Older revision		Revision as of 15:55, 14 December 2022
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	* The following vectorial quantities are used in the questions for this exercise:		* The following vectorial quantities are used in the questions for this exercise:
	** the information sequence:  $\hspace{0.2cm} \underline{u} = (u_1, \, u_2, \text{...} \hspace{0.05cm} )$,

	** the parity-check sequence:  $\hspace{0.2cm} \underline{p} = (p_1, \, p_2, \text{...} \hspace{0.05cm})$,		:* the information sequence:  $\hspace{0.2cm} \underline{u} = (u_1, \, u_2, \text{...} \hspace{0.05cm} )$,
			:* the parity-check sequence:  $\hspace{0.2cm} \underline{p} = (p_1, \, p_2, \text{...} \hspace{0.05cm})$,
	** the impulse response:  $\hspace{0.2cm} \underline{g} = (g_1, \, g_2, \text{...} \hspace{0.05cm} ); \hspace{0.2cm}$ this is equal to the parity sequence $\underline{p}$  for  $\underline{u} = (1, \, 0, \, 0, \text{...} \hspace{0.05cm} )$.		:* the impulse response:  $\hspace{0.2cm} \underline{g} = (g_1, \, g_2, \text{...} \hspace{0.05cm} ); \hspace{0.2cm}$ this is equal to the parity sequence  $\underline{p}$  for  $\underline{u} = (1, \, 0, \, 0, \text{...} \hspace{0.05cm} )$.


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	{What is the impulse response  $\underline{g}$ ?		{What is the impulse response  $\underline{g}$ ?
	\|type="()"}		\|type="()"}
	+  $\underline{g} = (1, \, 1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 1, \, 1, \text{...} \hspace{0.05cm})$ ~~holds~~.		+  $\underline{g} = (1, \, 1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 1, \, 1, \text{...} \hspace{0.05cm})$.
	-  $\underline{g} = (1, \, 0, \, 1, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \text{...} \hspace{0.05cm})$ ~~holds~~.		-  $\underline{g} = (1, \, 0, \, 1, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \text{...} \hspace{0.05cm})$.

	{It holds  $\underline{u} = (1, \, 1, \, 0, \, 1)$. Which statements hold for the parity sequence  $\underline{p}$ ?		{It holds   $\underline{u} = (1, \, 1, \, 0, \, 1)$.  Which statements hold for the parity sequence  $\underline{p}$ ?
	\|type="[]"}		\|type="[]"}
	-  $\underline{p} = (1, \, 0, \, 1, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \text{...} \hspace{0.05cm})$ ~~holds~~.		-  $\underline{p} = (1, \, 0, \, 1, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \text{...} \hspace{0.05cm})$.
	+  $\underline{p} = (1, \, 0, \, 0, \, 0, \, 0, \, 1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 1, \text{...} \hspace{0.05cm})$ ~~holds~~.		+  $\underline{p} = (1, \, 0, \, 0, \, 0, \, 0, \, 1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 1, \text{...} \hspace{0.05cm})$.
	- With limited information sequence  $\underline{u}$  ~~is always also~~  $\underline{p}$  limited.		- With limited information sequence  $\underline{u}$   ⇒   $\underline{p}$  is also limited.

	{What is the  $D$–transfer function  $G(D)$?		{What is the  $D$–transfer function  $G(D)$?
	\|type="[]"}		\|type="[]"}
	+  $G(D) = 1 + D + D^2 + D^4 + D^5 + D^7 + D^8 + \text{...} \hspace{0.05cm}$ ~~holds~~.		+  $G(D) = 1 + D + D^2 + D^4 + D^5 + D^7 + D^8 + \text{...} \hspace{0.05cm}$.
	+  $G(D) = (1 + D^2)/(1 + D + D^2)$ ~~holds~~.		+  $G(D) = (1 + D^2)/(1 + D + D^2)$.
	-  $G(D) = (1 + D + D^2)/(1 + D^2)$ ~~holds~~.		-  $G(D) = (1 + D + D^2)/(1 + D^2)$.

	{Now let  $\underline{u} = (1, \, 1, \, 1)$ ~~hold~~. Which statements hold for the parity sequence  $\underline{p}$?		{Now let  $\underline{u} = (1, \, 1, \, 1)$.  Which statements hold for the parity sequence  $\underline{p}$?
	\|type="[]"}		\|type="[]"}
	+  $\underline{p} = (1, \, 0, \, 1, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \text{...} \hspace{0.05cm})$ ~~holds~~.		+  $\underline{p} = (1, \, 0, \, 1, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \text{...} \hspace{0.05cm})$.
	-  $\underline{p} = (1, \, 0, \, 0, \, 0, \, 0, \, 1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 1, \text{...} \hspace{0.05cm})$ ~~holds~~.		-  $\underline{p} = (1, \, 0, \, 0, \, 0, \, 0, \, 1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 1, \text{...} \hspace{0.05cm})$.
	- With ~~unlimited impulse response~~  $\underline{g}$  ~~is always also~~  $\underline{p}$  ~~unlimited~~.		- With limited information sequence  $\underline{u}$   ⇒   $\underline{p}$  is also limited.

	{What is the free distance  $d_{\rm F}$  of this RSC encoder?		{What is the free distance  $d_{\rm F}$  of this RSC encoder?