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

Guenter at 15:11, 22 December 2022

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Guenter at 21:04, 12 December 2022

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Guenter at 20:59, 12 December 2022

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@@ Line 75: / Line 75: @@
 ===Solution===
 {{ML-Kopf}}
-[[File:P_ID3060__KC_A_4_10c_v3.png|right|frame|Polynomial division for subtask&nbsp;'''(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&nbsp;'''(3)''': $G(D) = (1 + D + D^3) \ / \ (1 + D^2 + D^3)$]]
 '''(1)'''&nbsp; The code parameters are&nbsp; $k = 1$&nbsp; and&nbsp; $n = 3$ &nbsp; &#8658; &nbsp; code rate&nbsp; $\underline{R = 1/3}$.
 *The memory is&nbsp; $\underline{m = 3}$.

@@ Line 80: / Line 80: @@
 *The influence lengths result in&nbsp; $\nu = 1, \ \nu_2 = 4$&nbsp; and&nbsp; $\nu_3 = 4$ &nbsp; &#8658; &nbsp; Total influence length&nbsp; $\underline{\nu = 9}$.
@@ Line 85: / Line 86: @@
-The upper graph illustrates the polynomial division $(1 + D + D^3) \ / \ (1 + D^2 + D^3)$. For explanation:
+'''(3)'''&nbsp; Correct are the&nbsp; <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&nbsp; $(1 + D + D^3) \ / \ (1 + D^2 + D^3)$.&nbsp; 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&nbsp; $D^8 + D^9 = D^7 \cdot (D + D^2)$.&nbsp; Thus also holds:
 *After summarizing:
 :$$G(D) = 1 + D + D^2 + D^3 + D^6 + D^8+ D^9+ D^{10} + D^{13} + \hspace{0.05cm}\text{ ... }\hspace{0.05cm} \hspace{0.05cm}. $$
@@ Line 112: / Line 113: @@
 * The impulse response continues to infinity &nbsp; &#8658; &nbsp; Solution proposal 3 is also correct.
@@ Line 127: / Line 129: @@
 \hspace{0.05cm}. $$
-In the state transition diagram (right), the impulse response $\underline{g}$ is highlighted in yellow. The impulse response results as the parity-check sequence&nbsp; $\underline{p}$&nbsp; for the information sequence&nbsp; $\underline{u} = (1, \, 0, \, 0, \, 0, \, 0, \, \text{ ... })$.
+*In the state transition diagram&nbsp; $($right$)$,&nbsp; the impulse response&nbsp; $\underline{g}$&nbsp; is highlighted in yellow.&nbsp;
 * The transitions in the diagram are labeled "$u_i\hspace{0.05cm}|\hspace{0.05cm}\underline{x}_i$", which is equivalent to "$u_i\hspace{0.05cm}|\hspace{0.05cm}u_i \hspace{0.05cm}p_i$".
-*The parity-check sequence&nbsp; $\underline{p} \ (=$ impulse response&nbsp; $\underline{g})$&nbsp; thus results from the respective second coder output symbol.
+*The impulse response results as the parity-check sequence &nbsp; $\underline{p}$ &nbsp; for the information sequence &nbsp; $\underline{u} = (1, \, 0, \, 0, \, 0, \, 0, \, \text{ ... })$.
 * $\underline{g}$ is represented by the following states:
 :$$S_0 &#8594; [S_1 &#8594; S_2 &#8594; S_5 &#8594; S_3 &#8594; S_7 &#8594; S_6 &#8594; S_4 ] &#8594; [S_1 &#8594; \ ... \ &#8594; S_4] &#8594; \ \text{ ... } $$
-'''(5)'''&nbsp; The following graphic shows the solution using the generator matrix $\mathbf{G}$. It holds $\underline{u} = (0, \, 1, \, 1, \, 0, \, 0, \, \text{ ... } )$.
+'''(5)'''&nbsp; The following graph shows the solution using the generator matrix&nbsp; $\mathbf{G}$.&nbsp; It holds&nbsp; $\underline{u} = (0, \, 1, \, 1, \, 0, \, 0, \, \text{ ... } )$.
 [[File:EN_KC_A_4_10e_v1.png|right|frame|$\underline{p} = (0, \, 1, \, 1, \, \text{ ... } ) \cdot \mathbf{G}$]]
-It can be seen that <u>solutions 1, 2 and 3</u> are correct:
+[[File:EN_KC_A_4_10f_v1.png|right|frame|$\underline{p} = (0, \, 1, \, 0, \, 0, \, 0, \, 0, \, 0, \, 0, \, 1, \, 0, \, 0, \, \text{ ... }) \cdot \mathbf{G}$]]
-In the transition diagram, the states $S_0 &#8594; S_0 &#8594; S_1 &#8594; S_3 &#8594; S_7 &#8594; S_6 &#8594; S_4 &#8594; S_1$ are passed through first. This is followed (infinitely often) by the periodic portion $S_1 &#8594; S_2 &#8594; S_5 &#8594; S_3 &#8594; S_7 &#8594; S_6 &#8594; S_4 &#8594; S_1$.
+* The Hamming weight of the&nbsp; $($limited$)$&nbsp; input sequence is actually&nbsp; $w_{\rm H}(\underline{u}) = 2$.
-[[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)'''&nbsp; The last graph shows the solution for&nbsp;
-Correct are the <u>proposed solutions 3 and 4</u>:
+*From position 10&nbsp; the output sequence is&nbsp; $\underline{p} \equiv\underline{0}$ &nbsp; <br>&#8658; &nbsp; proposals 1 and 2 therefore do not apply.
 <br clear=all>
-''Further notes:''
+<u>Further notes:</u>
-* For a turbo code, especially those input sequences $\underline{u}$ whose $D$&ndash;transforms are representable as $U(D) = f(D) \cdot [1 + D^{P}]$ are extremely unfavorable.
+# For turbo codes,&nbsp; especially those input sequences&nbsp; $\underline{u}$&nbsp; whose&nbsp; $D$&ndash;transforms are representable as&nbsp; $U(D) = f(D) \cdot [1 + D^{P}]$&nbsp; 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&nbsp; [[Channel_Coding/The_Basics_of_Turbo_Codes#Performance_of_the_turbo_codes|"Performance of Turbo Codes"]] page in the theory section.
-*$P$&nbsp; here indicates the period of the impulse response&nbsp; $\underline{g}$.
+#$P$&nbsp; indicates here the period of the impulse response&nbsp; $\underline{g}$.&nbsp; In our example&nbsp; $f(D) = D$&nbsp; and&nbsp; $P = 7$.
 {{ML-Fuß}}

@@ Line 75: / Line 75: @@
 ===Solution===
 {{ML-Kopf}}
-[[File:P_ID3060__KC_A_4_10c_v3.png|right|frame|Polynomial division for subtask&nbsp;''(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&nbsp;'''(3)''': $G(D) = (1 + D + D^3) \ / \ (1 + D^2 + D^3)$]]
 '''(1)'''&nbsp; The code parameters are&nbsp; $k = 1$&nbsp; and&nbsp; $n = 3$ &nbsp; &#8658; &nbsp; code rate&nbsp; $\underline{R = 1/3}$.
 *The memory is&nbsp; $\underline{m = 3}$.
@@ Line 82: / Line 82: @@
-'''(2)'''&nbsp; As the comparison of the [[Channel_Coding/Algebraic_and_Polynomial_Description#Filter_structure_with_fractional.E2.80.93rational_transfer_function|"recursive filter"]]&nbsp; on the data page with the&nbsp; [[Channel_Coding/Algebraic_and_Polynomial_Description#Filter_structure_with_fractional.E2.80.93rational_transfer_function|"filter structure"]]&nbsp; in the theory section for fractional&ndash;rational&nbsp; $G(D)$,&nbsp; the&nbsp; <u>proposed solution 1</u> is correct.
+'''(2)'''&nbsp; As the comparison of the&nbsp; "recursive filter"&nbsp; on the data page with the&nbsp; [[Channel_Coding/Algebraic_and_Polynomial_Description#Filter_structure_with_fractional.E2.80.93rational_transfer_function|"filter structure"]]&nbsp; in the theory section for fractional&ndash;rational&nbsp; $G(D)$,&nbsp; the&nbsp; <u>proposed solution 1</u> is correct.

← Older revision		Revision as of 17:46, 14 December 2022
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−	'''(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.

← Older revision		Revision as of 17:42, 14 December 2022
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−	'''(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.

@@ Line 76: / Line 76: @@
 {{ML-Kopf}}
 [[File:P_ID3060__KC_A_4_10c_v3.png|right|frame|Polynomial division for subtask&nbsp;''(3)''': $G(D) = (1 + D + D^3) \ / \ (1 + D^2 + D^3)$]]
-'''(1)'''&nbsp; The code parameters are $k = 1$ and $n = 3$ &nbsp; &#8658; &nbsp; Code rate $\underline{R = 1/3}$.
+'''(1)'''&nbsp; The code parameters are&nbsp; $k = 1$&nbsp; and&nbsp; $n = 3$ &nbsp; &#8658; &nbsp; code rate&nbsp; $\underline{R = 1/3}$.
-*The memory is $\underline{m = 3}$.
+*The memory is&nbsp; $\underline{m = 3}$.
 *The influence lengths result in $\nu = 1, \ \nu_2 = 4$ and $\nu_3 = 4$ &nbsp;&#8658;&nbsp; Total influence length $\underline{\nu = 9}$.
-'''(2)'''&nbsp; 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&ndash;rational $G(D)$, the <u>proposed solution 1</u> is correct.
+'''(2)'''&nbsp; As the comparison of the [[Channel_Coding/Algebraic_and_Polynomial_Description#Filter_structure_with_fractional.E2.80.93rational_transfer_function|"recursive filter"]]&nbsp; on the data page with the&nbsp; [[Aufgaben:Exercise_4. 10:_Turbo_Encoder_for_UMTS_and_LTE|"filter structure"]]&nbsp; in the theory section for fractional&ndash;rational&nbsp; $G(D)$,&nbsp; the&nbsp; <u>proposed solution 1</u> is correct.

@@ Line 22: / Line 22: @@
 ** the&nbsp; [[Channel_Coding/Code_Description_with_State_and_Trellis_Diagram|"Code Description with State and Trellis Diagram"]].
-* For further guidance on how to do this, see&nbsp; [[Aufgaben:Exercise_4.08:_Repetition_to_the_Convolutional_Codes|$\text{Exercise 4.8}$]]&nbsp; and&nbsp; [[Tasks:Exercise_4.09:_Recursive_Systematic_Convolutional_Codes|$\text{Exercise 4.9}$]].
+* For further guidance on how to do this, see&nbsp; [[Aufgaben:Exercise_4.08:_Repetition_to_the_Convolutional_Codes|$\text{Exercise 4.8}$]]&nbsp; and&nbsp; [[Aufgaben:Exercise_4.09:_Recursive_Systematic_Convolutional_Codes|$\text{Exercise 4.9}$]].
 * The information sequence&nbsp; $\underline{u}$&nbsp; is partially specified by its&nbsp; $D$&ndash;transform for easier description in the subtasks.&nbsp; For example:

@@ Line 1: / Line 1: @@
 {{quiz-Header|Buchseite=Channel_Coding/The_Basics_of_Turbo_Codes}}
-[[File:EN_KC_A_4_10_v1.png|right|frame|Turbocoder for UMTS and LTE]]
+[[File:EN_KC_A_4_10_v1.png|right|frame|Turbo encoder for UMTS and LTE]]
-The mobile communications standards&nbsp; [[Mobile_Communications/Characteristics_of_UMTS|"UMTS"]]&nbsp; and&nbsp; [[Mobile_Communications/General_Information_on_the_LTE_Mobile_Communications_Standard|"LTE"]]&nbsp; each use a turbo code that is largely identical to the encoder described in the&nbsp; [[Channel_Coding/The_Basics_of_Turbo_Codes|"The Basics of Turbo Codes"]]&nbsp; chapter.
+The mobile communications standards&nbsp; [[Mobile_Communications/Characteristics_of_UMTS|$\rm UMTS$]]&nbsp; and&nbsp; [[Mobile_Communications/General_Information_on_the_LTE_Mobile_Communications_Standard|$\rm LTE$]]&nbsp; each use a turbo code that is largely identical to the encoder described in the&nbsp; [[Channel_Coding/The_Basics_of_Turbo_Codes|"The Basics of Turbo Codes"]]&nbsp; chapter.
-* The&nbsp; $1/n$ convolutional code is systematic, meaning that the code sequence&nbsp; $\underline{x}$&nbsp; includes the information sequence&nbsp; $\underline{u}$&nbsp; as a component.
+* The&nbsp; $1/n$ convolutional code is systematic, meaning that the encoded sequence&nbsp; $\underline{x}$&nbsp; includes the information sequence&nbsp; $\underline{u}$&nbsp; as a component.
 * The parity-check sequences&nbsp; $\underline{p}_1$&nbsp; and&nbsp; $\underline{p}_2$&nbsp; are based on the same transfer function:
 :$$G_1(D) = G_2(D) = G(D).$$
-* $\underline{p}_1$&nbsp; and&nbsp; $\underline{p}_2$&nbsp; however, use different input sequences&nbsp; $\underline{u}$&nbsp; and&nbsp; $\underline{u}_{\pi}$, respectively. Here,&nbsp; ${\rm \Pi}$&nbsp; marks the interleaver, for UMTS and LTE mostly a&nbsp; $S$ random interleaver.
+* $\underline{p}_1$&nbsp; and&nbsp; $\underline{p}_2$&nbsp; however,&nbsp; use different input sequences&nbsp; $\underline{u}$&nbsp; and&nbsp; $\underline{u}_{\pi}$,&nbsp; respectively.&nbsp; Here,&nbsp; ${\rm \Pi}$&nbsp; marks the interleaver,&nbsp; for UMTS and LTE mostly a&nbsp; $S$&ndash;random interleaver.
 [[File:P_ID3052__KC_A_4_10b_v2.png|left|frame|Given filter structure]]
-<br><br><br><br><br><br>The main difference compared to the description in the theory part results from a different transfer function&nbsp; $G(D)$ given by the recursive filter structure drawn on the left.
+<br><br><br><br><br><br>The main difference compared to the description in the theory part results from a different transfer function&nbsp; $G(D)$&nbsp; given by the recursive filter structure drawn on the left.
 <br clear=all>
@@ Line 16: / Line 17: @@
 Hints:
 * The exercise belongs to the chapter&nbsp; [[Channel_Coding/The_Basics_of_Turbo_Codes| "Basics of Turbo Codes"]].
 * Knowledge is expected about
 ** the&nbsp; [[Channel_Coding/Algebraic_and_Polynomial_Description|"Algebraic and Polynomial Description of Convolutional Codes"]],
 ** the&nbsp; [[Channel_Coding/Code_Description_with_State_and_Trellis_Diagram|"Code Description with State and Trellis Diagram"]].
-* For further guidance on how to do this, see the&nbsp; [[Aufgaben:Exercise_4.08:_Repetition_to_the_Convolutional_Codes|"Exercise 4.8"]] and the&nbsp; [[Tasks:Exercise_4.09:_Recursive_Systematic_Convolutional_Codes|"Exercise 4.9"]].
-* The information sequence&nbsp; $\underline{u}$&nbsp; is partially specified by its&nbsp; $D$&ndash;transform for easier description in the subtasks. For example:
+* For further guidance on how to do this, see&nbsp; [[Aufgaben:Exercise_4.08:_Repetition_to_the_Convolutional_Codes|$\text{Exercise 4.8}$]]&nbsp; and&nbsp; [[Tasks:Exercise_4.09:_Recursive_Systematic_Convolutional_Codes|$\text{Exercise 4.9}$]].
 :$$\underline{u}= (\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 1\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm}0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm},\hspace{0.05cm} 0\hspace{0.05cm}\hspace{0.05cm} \text{...}\hspace{0.05cm}) \quad \circ\!\!-\!\!\!-^{\hspace{-0.25cm}D}\!\!\!-\!\!\bullet\quad
 U(D) = D+  D^2\hspace{0.05cm},$$
@@ Line 31: / Line 35: @@
 ===Questions===
 <quiz display=simple>
-{What are the characteristics of the considered turbo code (memory&nbsp; $m$,&nbsp; influence length&nbsp; $\nu$,&nbsp; rate $R$)?
+{What are the characteristics of the considered turbo code&nbsp; $($memory&nbsp; $m$,&nbsp; influence length&nbsp; $\nu$,&nbsp; rate $R)$?
 |type="{}"}
 $ m \hspace{0.2cm} = \ ${ 3 3% }
@@ Line 37: / Line 41: @@
 $R \hspace{0.2cm} = \ ${ 0.333 3% }
-{What are the (identical) transfer functions&nbsp; $G_1(D) = G_2(D) = G(D)$?
+{What are the&nbsp; (identical)&nbsp; transfer functions&nbsp; $G_1(D) = G_2(D) = G(D)$?
 |type="()"}
 + &nbsp; $G(D) = (1 + D + D^3)/(1 + D^2 + D^3)$.
 - &nbsp; $G(D) = (1 + D^2 + D^3)/(1 + D + D^3)$.
-{What is the impulse response&nbsp; $\underline{g}$&nbsp;?
+{What is the impulse response&nbsp; $\underline{g}$?
 |type="[]"}
 -&nbsp; $\underline{g} = (1, \, 1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 1, \, 1, \, 0, \, 1, \, 1, \hspace{0.05cm} \text{...}\hspace{0.05cm})$
@@ Line 50: / Line 54: @@
 {Are there periodic components within the impulse response&nbsp; $\underline{g}$&nbsp;?
 |type="()"}
-+ Yes, with the period&nbsp; $P = 7$.
++ Yes,&nbsp; with period&nbsp; $P = 7$.
-- Yes, with the period&nbsp; $P = 8$.
+- Yes,&nbsp; with period&nbsp; $P = 8$.
 - No.