Aufgaben:Exercise 4.10: Turbo Encoder for UMTS and LTE - Revision history

Maintenance script: Fix interlanguage link: resolve redirect chain

2026-03-16T15:58:39Z

Fix interlanguage link: resolve redirect chain

← Older revision		Revision as of 17:58, 16 March 2026
Line 155:		Line 155:

	[[Category:Channel Coding: Exercises\|^4.3 About the Turbo Codes^]]		[[Category:Channel Coding: Exercises\|^4.3 About the Turbo Codes^]]
	[[de:Aufgaben:4.~~10_UMTS/LTE–Turbocoder~~]]		[[de:Aufgaben:Aufgabe 4.10: Turbocoder für UMTS und LTE]]

Maintenance script: Join 2 split formula blocks

2026-03-16T14:16:42Z

Join 2 split formula blocks

@@ Line 94: / Line 94: @@
 * The $D$&ndash;inverse transform gives the proposed solution 2:
-:$$\underline{g}= (\hspace{0.05cm}1\hspace{0.05cm},
+:$$\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} 0\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} 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} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}\text{...}\hspace{0.05cm})\hspace{0.05cm}. $$
 * The impulse response continues to infinity &nbsp; &#8658; &nbsp; Solution proposal 3 is also correct.
@@ Line 116: / Line 102: @@
 [[File:P_ID3061__KC_A_4_10d_v2.png|right|frame|State transition diagram and impulse response]]
 '''(4)'''&nbsp;  The impulse response can be expressed as follows:
-:$$\underline{g}=  \Big (\hspace{0.03cm}1\hspace{0.03cm},
+:$$\underline{g}=  \Big (\hspace{0.03cm}1\hspace{0.03cm},\big [ \hspace{0.03cm} 1\hspace{0.03cm},\hspace{0.03cm} 1\hspace{0.03cm},\hspace{0.03cm} 1\hspace{0.03cm},\hspace{0.03cm} 0\hspace{0.03cm},\hspace{0.03cm} 0\hspace{0.03cm},\hspace{0.03cm} 1\hspace{0.03cm},\hspace{0.03cm} 0\hspace{0.03cm} \big ]_{\rm per}\Big ) \hspace{0.15cm}\Rightarrow \hspace{0.15cm} \underline{P = 7}\hspace{0.05cm}. $$
 *In the state transition diagram&nbsp; $($right$)$,&nbsp; the impulse response&nbsp; $\underline{g}$&nbsp; is highlighted in yellow.&nbsp;

Maintenance script: Join split $$ formula lines for correct rendering

2026-03-16T13:48:22Z

Join split $$ formula lines for correct rendering

← Older revision		Revision as of 15:48, 16 March 2026
Line 25:		Line 25:

	* The information sequence  $\underline{u}$  is partially specified by its  $D$–transform for easier description in the subtasks.  For example:		* The information sequence  $\underline{u}$  is partially specified by its  $D$–transform for easier description in the subtasks.  For example:
	:$$\underline{u}= (\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm}\hspace{0.05cm} \text{...}\hspace{0.05cm}) \quad \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\~~quad~~		:$$\underline{u}= (\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} 0\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm}\hspace{0.05cm} \text{...}\hspace{0.05cm}) \quad \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\quadU(D) = D+ D^2\hspace{0.05cm},$$
	U(D) = D+ D^2\hspace{0.05cm},$$		:$$\underline{u}= (\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm}\hspace{0.05cm} \text{...}\hspace{0.05cm}) \quad \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\quadU(D) = D+ D^8\hspace{0.05cm}.$$
	:$$\underline{u}= (\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm}\hspace{0.05cm} \text{...}\hspace{0.05cm}) \quad \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\~~quad~~
	U(D) = D+ D^8\hspace{0.05cm}.$$

Maintenance script: Fix spaces in LaTeX braces to prevent Quiz misparse

2026-03-16T13:21:29Z

Fix spaces in LaTeX braces to prevent Quiz misparse

Maintenance script: Add German interlanguage link

2026-03-16T12:33:01Z

Add German interlanguage link

← Older revision		Revision as of 14:33, 16 March 2026
Line 180:		Line 180:

	[[Category:Channel Coding: Exercises\|^4.3 About the Turbo Codes^]]		[[Category:Channel Coding: Exercises\|^4.3 About the Turbo Codes^]]
			[[de:Aufgaben:4.10_UMTS/LTE–Turbocoder]]

Guenter at 15:11, 22 December 2022

2022-12-22T15:11:30Z

← Older revision		Revision as of 17:11, 22 December 2022
Line 75:		Line 75:
	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	[[File:~~P_ID3060__KC_A_4_10c_v3~~.png\|right\|frame\|Polynomial division for subtask '''(3)''': $G(D) = (1 + D + D^3) \ / \ (1 + D^2 + D^3)$]]		[[File:EN_KC_A_4_10c_v3_neu2.png\|right\|frame\|Polynomial division for subtask '''(3)''': $G(D) = (1 + D + D^3) \ / \ (1 + D^2 + D^3)$]]
	'''(1)'''  The code parameters are  $k = 1$  and  $n = 3$   ⇒   code rate  $\underline{R = 1/3}$.		'''(1)'''  The code parameters are  $k = 1$  and  $n = 3$   ⇒   code rate  $\underline{R = 1/3}$.
	*The memory is  $\underline{m = 3}$.		*The memory is  $\underline{m = 3}$.

Guenter at 14:18, 15 December 2022

2022-12-15T14:18:21Z

← Older revision		Revision as of 16:18, 15 December 2022
Line 80:		Line 80:

	*The influence lengths result in  $\nu = 1, \ \nu_2 = 4$  and  $\nu_3 = 4$   ⇒   Total influence length  $\underline{\nu = 9}$.		*The influence lengths result in  $\nu = 1, \ \nu_2 = 4$  and  $\nu_3 = 4$   ⇒   Total influence length  $\underline{\nu = 9}$.



Line 85:		Line 86:


	~~'''(3)'''  Correct are the <u>proposed solutions 2 and 3</u>:~~

	The upper graph illustrates the polynomial division $(1 + D + D^3) \ / \ (1 + D^2 + D^3)$. For explanation:		'''(3)'''  Correct are the  <u>proposed solutions 2 and 3</u>:
	*Cancelled is the representation with the remainder $D^8 + D^9 = D^7 \cdot (D + D^2)$.
	*Thus also holds:		The upper graph illustrates the polynomial division  $(1 + D + D^3) \ / \ (1 + D^2 + D^3)$.  For explanation:
	:$$(D^8 + D^9) \hspace{0.05cm} /\hspace{0.05cm} (1+ D^2+ D^3 ) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} D^7 \cdot (D+ D^2+ D^3 + D^6) + {\rm ~~rest_2~~}$$		*Cancelled is the representation with the remainder  $D^8 + D^9 = D^7 \cdot (D + D^2)$.  Thus also holds:
			:$$(D^8 + D^9) \hspace{0.05cm} /\hspace{0.05cm} (1+ D^2+ D^3 ) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} D^7 \cdot (D+ D^2+ D^3 + D^6) + {\rm remainder_2}$$
	*After summarizing:		*After summarizing:
	:$$G(D) = 1 + D + D^2 + D^3 + D^6 + D^8+ D^9+ D^{10} + D^{13} + \hspace{0.05cm}\text{ ... }\hspace{0.05cm} \hspace{0.05cm}. $$		:$$G(D) = 1 + D + D^2 + D^3 + D^6 + D^8+ D^9+ D^{10} + D^{13} + \hspace{0.05cm}\text{ ... }\hspace{0.05cm} \hspace{0.05cm}. $$
Line 112:		Line 113:

	* The impulse response continues to infinity   ⇒   Solution proposal 3 is also correct.		* The impulse response continues to infinity   ⇒   Solution proposal 3 is also correct.



Line 127:		Line 129:
	\hspace{0.05cm}. $$		\hspace{0.05cm}. $$

	In the state transition diagram (right), the impulse response $\underline{g}$ is highlighted in yellow. The impulse response results as the parity-check sequence  $\underline{p}$  for the information sequence  $\underline{u} = (1, \, 0, \, 0, \, 0, \, 0, \, \text{ ... })$.		*In the state transition diagram  $($right$)$,  the impulse response  $\underline{g}$  is highlighted in yellow.
	* The transitions in the diagram are labeled "$u_i\hspace{0.05cm}\|\hspace{0.05cm}\underline{x}_i$", which is equivalent to "$u_i\hspace{0.05cm}\|\hspace{0.05cm}u_i \hspace{0.05cm}p_i$".
	*The parity-check sequence  $\underline{p} \ (=$ impulse response  $\underline{g})$  thus results from the respective second coder output symbol.		*The impulse response results as the parity-check sequence   $\underline{p}$   for the information sequence   $\underline{u} = (1, \, 0, \, 0, \, 0, \, 0, \, \text{ ... })$.
	* $\underline{g}$ is represented by the following states:
			* The transitions in the diagram are labeled  "$u_i\hspace{0.05cm}\|\hspace{0.05cm}\underline{x}_i$",  which is equivalent to  "$u_i\hspace{0.05cm}\|\hspace{0.05cm}u_i \hspace{0.05cm}p_i$".

			*The parity-check sequence   $\underline{p} \ (=$ impulse response  $\underline{g})$   thus results from the respective second coder output symbol.

			* The impulse response  $\underline{g}$  is represented by the following states:
	:$$S_0 → [S_1 → S_2 → S_5 → S_3 → S_7 → S_6 → S_4 ] → [S_1 → \ ... \ → S_4] → \ \text{ ... } $$		:$$S_0 → [S_1 → S_2 → S_5 → S_3 → S_7 → S_6 → S_4 ] → [S_1 → \ ... \ → S_4] → \ \text{ ... } $$


	'''(5)'''  The following ~~graphic~~ shows the solution using the generator matrix $\mathbf{G}$. It holds $\underline{u} = (0, \, 1, \, 1, \, 0, \, 0, \, \text{ ... } )$.		'''(5)'''  The following graph shows the solution using the generator matrix  $\mathbf{G}$.  It holds  $\underline{u} = (0, \, 1, \, 1, \, 0, \, 0, \, \text{ ... } )$.

	[[File:EN_KC_A_4_10e_v1.png\|right\|frame\|$\underline{p} = (0, \, 1, \, 1, \, \text{ ... } ) \cdot \mathbf{G}$]]		[[File:EN_KC_A_4_10e_v1.png\|right\|frame\|$\underline{p} = (0, \, 1, \, 1, \, \text{ ... } ) \cdot \mathbf{G}$]]

	~~It can be seen that <u>solutions 1, 2 and 3</u> are correct~~:		[[File:EN_KC_A_4_10f_v1.png\|right\|frame\|$\underline{p} = (0, \, 1, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 1, \, 0, \, 0, \, \text{ ... }) \cdot \mathbf{G}$]]
	* The present parity-check sequence $\underline{p}~~$ has the same period $P~~ = ~~7$ as the impulse response $~~\~~underline{g}$.~~
	* The Hamming weight of the (bounded) input sequence is actually $w_{\~~rm H}(~~\~~underline~~{u}) ~~= 2$.~~
	* Proposition 4 is incorrect. Rather, the semi–infinite output sequence here is $w_{\~~rm H}(~~\~~underline~~{p}~~) → \infty~~$.

			It can be seen that  <u>solutions 1, 2 and 3</u>  are correct:
			* The present parity-check sequence  $\underline{p}$  has the same period  $P = 7$  as the impulse response  $\underline{g}$.

	In the ~~transition diagram, the states $S_0 → S_0 → S_1 → S_3 → S_7 → S_6 → S_4~~ &~~#8594~~; ~~S_1~~$ ~~are passed through first. This is followed~~ (~~infinitely often~~) ~~by the periodic portion~~ $~~S_1~~ &~~#8594~~; ~~S_2~~ &~~#8594~~; ~~S_5 → S_3 → S_7 → S_6 → S_4 → S_1~~$.		* The Hamming weight of the  $($limited$)$  input sequence is actually  $w_{\rm H}(\underline{u}) = 2$.

			* Proposition 4 is incorrect.  Rather,  for the semi–infinite output sequence:   $w_{\rm H}(\underline{p}) → \infty$.

			*In the transition diagram,  the states
			:$$S_0 → S_0 → S_1 → S_3 → S_7 → S_6 → S_4 → S_1$$
			: are passed through first.

			*This is followed  $($infinitely often$)$ by the periodic portion
			:$$S_1 → S_2 → S_5 → S_3 → S_7 → S_6 → S_4 → S_1.$$

	~~'''(6)'''  The last graph shows the solution for $U(D) = D + D^8 \Rightarrow \underline{u} = (0, \, 1, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 1, \, 0, \, 0, \, \text{ ... })$.~~

	~~[[File~~:~~EN_KC_A_4_10f_v1~~.~~png\|right\|frame\|$~~\underline{p} = (0, \, 1, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 1, \, 0, \, 0, \, \text{ ... }) \~~cdot~~ \~~mathbf~~{G}$]]		'''(6)'''  The last graph shows the solution for
			:$$U(D) = D + D^8$$
			:$$ \Rightarrow \hspace{0.3cm}\underline{u} = (0, \, 1, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 1, \, 0, \, 0, \, \text{ ... }).$$
			*Correct are the  <u>proposed solutions 3 and 4</u>:

			*The input sequence  $\underline{u}$  contains two  "ones"  and the output sequence  $\underline{p}$  six  "ones".

	~~Correct are the <u>proposed solutions 3 and 4</u>:~~		*From position 10  the output sequence is  $\underline{p} \equiv\underline{0}$   <br>⇒   proposals 1 and 2 therefore do not apply.
	*The input sequence $\underline{u}$ contains two ones and the output sequence $\underline{p}$ six ones.
	*From position 10 ~~now~~ the output sequence $\underline{p} \equiv\underline{0}$   <br>⇒   proposals 1 and 2 therefore do not apply.
	<br clear=all>		<br clear=all>
	''Further notes:''		<u>Further notes:</u>
	* For a turbo ~~code~~, especially those input sequences $\underline{u}$ whose $D$–transforms are representable as $U(D) = f(D) \cdot [1 + D^{P}]$ are extremely unfavorable.		# For turbo codes,  especially those input sequences  $\underline{u}$  whose  $D$–transforms are representable as  $U(D) = f(D) \cdot [1 + D^{P}]$  are extremely unfavorable.
	*They cause the ~~<i>~~error floor~~</i>~~ as seen on the [[Channel_Coding/The_Basics_of_Turbo_Codes#Performance_of_the_turbo_codes\|"Performance of Turbo Codes"]] page in the theory section.		#They cause the error floor as seen on the  [[Channel_Coding/The_Basics_of_Turbo_Codes#Performance_of_the_turbo_codes\|"Performance of Turbo Codes"]] page in the theory section.
	*$P$  here ~~indicates~~ the period of the impulse response  $\underline{g}$.		#$P$  indicates here the period of the impulse response  $\underline{g}$.  In our example  $f(D) = D$  and  $P = 7$.
	*In our example  $f(D) = D$  and  $P = 7$.
	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 17:50, 14 December 2022

2022-12-14T17:50:33Z

← Older revision		Revision as of 19:50, 14 December 2022
Line 75:		Line 75:
	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	[[File:P_ID3060__KC_A_4_10c_v3.png\|right\|frame\|Polynomial division for subtask ''(3)''': $G(D) = (1 + D + D^3) \ / \ (1 + D^2 + D^3)$]]		[[File:P_ID3060__KC_A_4_10c_v3.png\|right\|frame\|Polynomial division for subtask '''(3)''': $G(D) = (1 + D + D^3) \ / \ (1 + D^2 + D^3)$]]
	'''(1)'''  The code parameters are  $k = 1$  and  $n = 3$   ⇒   code rate  $\underline{R = 1/3}$.		'''(1)'''  The code parameters are  $k = 1$  and  $n = 3$   ⇒   code rate  $\underline{R = 1/3}$.
	*The memory is  $\underline{m = 3}$.		*The memory is  $\underline{m = 3}$.
Line 82:		Line 82:


	'''(2)'''  As the comparison of the ~~[[Channel_Coding/Algebraic_and_Polynomial_Description#Filter_structure_with_fractional.E2.80.93rational_transfer_function\|~~"recursive filter"]]  on the data page with the  [[Channel_Coding/Algebraic_and_Polynomial_Description#Filter_structure_with_fractional.E2.80.93rational_transfer_function\|"filter structure"]]  in the theory section for fractional–rational  $G(D)$,  the  <u>proposed solution 1</u> is correct.		'''(2)'''  As the comparison of the  "recursive filter"  on the data page with the  [[Channel_Coding/Algebraic_and_Polynomial_Description#Filter_structure_with_fractional.E2.80.93rational_transfer_function\|"filter structure"]]  in the theory section for fractional–rational  $G(D)$,  the  <u>proposed solution 1</u> is correct.

Guenter at 17:46, 14 December 2022

2022-12-14T17:46:26Z

← Older revision		Revision as of 19:46, 14 December 2022
Line 82:		Line 82:


	'''(2)'''  As the comparison of the [[Channel_Coding/Algebraic_and_Polynomial_Description#Filter_structure_with_fractional.E2.80.93rational_transfer_function\|"recursive filter"]]  on the data page with the  [[~~Aufgaben:Exercise_4~~.~~10:_Turbo_Encoder_for_UMTS_and_LTE~~\|"filter structure"]]  in the theory section for fractional–rational  $G(D)$,  the  <u>proposed solution 1</u> is correct.		'''(2)'''  As the comparison of the [[Channel_Coding/Algebraic_and_Polynomial_Description#Filter_structure_with_fractional.E2.80.93rational_transfer_function\|"recursive filter"]]  on the data page with the  [[Channel_Coding/Algebraic_and_Polynomial_Description#Filter_structure_with_fractional.E2.80.93rational_transfer_function\|"filter structure"]]  in the theory section for fractional–rational  $G(D)$,  the  <u>proposed solution 1</u> is correct.

Guenter at 17:42, 14 December 2022

2022-12-14T17:42:19Z

← Older revision		Revision as of 19:42, 14 December 2022
Line 82:		Line 82:


	'''(2)'''  As the comparison of the [[Channel_Coding/Algebraic_and_Polynomial_Description#Filter_structure_with_fractional.E2.80.93rational_transfer_function\|"recursive filter"]]  on the data page with the  [[Aufgaben:Exercise_4. 10:_Turbo_Encoder_for_UMTS_and_LTE\|"filter structure"]]  in the theory section for fractional–rational  $G(D)$,  the  <u>proposed solution 1</u> is correct.		'''(2)'''  As the comparison of the [[Channel_Coding/Algebraic_and_Polynomial_Description#Filter_structure_with_fractional.E2.80.93rational_transfer_function\|"recursive filter"]]  on the data page with the  [[Aufgaben:Exercise_4.10:_Turbo_Encoder_for_UMTS_and_LTE\|"filter structure"]]  in the theory section for fractional–rational  $G(D)$,  the  <u>proposed solution 1</u> is correct.

← Older revision		Revision as of 15:21, 16 March 2026
Line 93:		Line 93:
	:$$(D^8 + D^9) \hspace{0.05cm} /\hspace{0.05cm} (1+ D^2+ D^3 ) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} D^7 \cdot (D+ D^2+ D^3 + D^6) + {\rm remainder_2}$$		:$$(D^8 + D^9) \hspace{0.05cm} /\hspace{0.05cm} (1+ D^2+ D^3 ) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} D^7 \cdot (D+ D^2+ D^3 + D^6) + {\rm remainder_2}$$
	*After summarizing:		*After summarizing:
	:$$G(D) = 1 + D + D^2 + D^3 + D^6 + D^8+ D^9+ D^{10} + D^{13} + \hspace{0.05cm}\text{ ... }\hspace{0.05cm} \hspace{0.05cm}. $$		:$$G(D) = 1 + D + D^2 + D^3 + D^6 + D^8+ D^9+ D^{10} + D^{13} + \hspace{0.05cm}\text{...}\hspace{0.05cm} \hspace{0.05cm}. $$

	* The $D$–inverse transform gives the proposed solution 2:		* The $D$–inverse transform gives the proposed solution 2:
Line 110:		Line 110:
	\hspace{0.05cm} 0\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}\text{ ... }\hspace{0.05cm})\hspace{0.05cm}. $$		\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}\text{...}\hspace{0.05cm})\hspace{0.05cm}. $$

	* The impulse response continues to infinity   ⇒   Solution proposal 3 is also correct.		* The impulse response continues to infinity   ⇒   Solution proposal 3 is also correct.
Line 131:		Line 131:
	*In the state transition diagram  $($right$)$,  the impulse response  $\underline{g}$  is highlighted in yellow.		*In the state transition diagram  $($right$)$,  the impulse response  $\underline{g}$  is highlighted in yellow.

	*The impulse response results as the parity-check sequence   $\underline{p}$   for the information sequence   $\underline{u} = (1, \, 0, \, 0, \, 0, \, 0, \, \text{ ... })$.		*The impulse response results as the parity-check sequence   $\underline{p}$   for the information sequence   $\underline{u} = (1, \, 0, \, 0, \, 0, \, 0, \, \text{...})$.

	* The transitions in the diagram are labeled  "$u_i\hspace{0.05cm}\|\hspace{0.05cm}\underline{x}_i$",  which is equivalent to  "$u_i\hspace{0.05cm}\|\hspace{0.05cm}u_i \hspace{0.05cm}p_i$".		* The transitions in the diagram are labeled  "$u_i\hspace{0.05cm}\|\hspace{0.05cm}\underline{x}_i$",  which is equivalent to  "$u_i\hspace{0.05cm}\|\hspace{0.05cm}u_i \hspace{0.05cm}p_i$".
Line 138:		Line 138:

	* The impulse response  $\underline{g}$  is represented by the following states:		* The impulse response  $\underline{g}$  is represented by the following states:
	:$$S_0 → [S_1 → S_2 → S_5 → S_3 → S_7 → S_6 → S_4 ] → [S_1 → \ ... \ → S_4] → \ \text{ ... } $$		:$$S_0 → [S_1 → S_2 → S_5 → S_3 → S_7 → S_6 → S_4 ] → [S_1 → \ ... \ → S_4] → \ \text{...} $$


	'''(5)'''  The following graph shows the solution using the generator matrix  $\mathbf{G}$.  It holds  $\underline{u} = (0, \, 1, \, 1, \, 0, \, 0, \, \text{ ... } )$.		'''(5)'''  The following graph shows the solution using the generator matrix  $\mathbf{G}$.  It holds  $\underline{u} = (0, \, 1, \, 1, \, 0, \, 0, \, \text{...} )$.

	[[File:EN_KC_A_4_10e_v1.png\|right\|frame\|$\underline{p} = (0, \, 1, \, 1, \, \text{ ... } ) \cdot \mathbf{G}$]]		[[File:EN_KC_A_4_10e_v1.png\|right\|frame\|$\underline{p} = (0, \, 1, \, 1, \, \text{...} ) \cdot \mathbf{G}$]]

	[[File:EN_KC_A_4_10f_v1.png\|right\|frame\|$\underline{p} = (0, \, 1, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 1, \, 0, \, 0, \, \text{ ... }) \cdot \mathbf{G}$]]		[[File:EN_KC_A_4_10f_v1.png\|right\|frame\|$\underline{p} = (0, \, 1, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 1, \, 0, \, 0, \, \text{...}) \cdot \mathbf{G}$]]

	It can be seen that  <u>solutions 1, 2 and 3</u>  are correct:		It can be seen that  <u>solutions 1, 2 and 3</u>  are correct:
Line 164:		Line 164:
	'''(6)'''  The last graph shows the solution for		'''(6)'''  The last graph shows the solution for
	:$$U(D) = D + D^8$$		:$$U(D) = D + D^8$$
	:$$ \Rightarrow \hspace{0.3cm}\underline{u} = (0, \, 1, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 1, \, 0, \, 0, \, \text{ ... }).$$		:$$ \Rightarrow \hspace{0.3cm}\underline{u} = (0, \, 1, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 1, \, 0, \, 0, \, \text{...}).$$
	*Correct are the  <u>proposed solutions 3 and 4</u>:		*Correct are the  <u>proposed solutions 3 and 4</u>: