Aufgaben:Exercise 3.1: Analysis of a Convolutional Encoder - Revision history

Guenter at 16:08, 6 November 2022

2022-11-06T16:08:45Z

Guenter at 15:59, 6 November 2022

2022-11-06T15:59:16Z

Guenter at 15:55, 6 November 2022

2022-11-06T15:55:33Z

Noah at 19:52, 19 September 2022

2022-09-19T19:52:20Z

Noah at 19:50, 19 September 2022

2022-09-19T19:50:41Z

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Noah: Noah moved page Aufgaben:Erercise 3.1: Analysis of a Convolutional Encoder to Aufgaben:Exercise 3.1: Analysis of a Convolutional Encoder: Erercise - > Exercise

2022-09-19T19:45:16Z

Noah moved page Erercise 3.1: Analysis of a Convolutional Encoder to Exercise 3.1: Analysis of a Convolutional Encoder: Erercise - > Exercise

Guenter at 08:58, 12 September 2022

2022-09-12T08:58:59Z

Guenter: Guenter moved page Aufgaben:Aufgabe 3.1: Analyse eines Faltungscodierers to Aufgaben:Erercise 3.1: Analysis of a Convolutional Encoder

2022-09-12T08:56:33Z

Guenter moved page Aufgabe 3.1: Analyse eines Faltungscodierers to Erercise 3.1: Analysis of a Convolutional Encoder

Javier: Text replacement - "Category:Aufgaben zu Kanalcodierung" to "Category:Channel Coding: Exercises"

2021-03-23T12:50:53Z

Text replacement - "Category:Aufgaben zu Kanalcodierung" to "Category:Channel Coding: Exercises"

Javier: Text replacement - "[[Kanalcodierung" to "[[Channel_Coding"

2020-07-09T12:58:09Z

Text replacement - "[[Kanalcodierung" to "[[Channel_Coding"

@@ Line 68: / Line 68: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; At each clock&,nbsp; $\underline {k = 3}$&nbsp; new information bits are processed and&nbsp; $\underline {n = 4}$&nbsp; code bits are output.
+'''(1)'''&nbsp; At each clock,&nbsp; $\underline {k = 3}$&nbsp; new information bits are processed and&nbsp; $\underline {n = 4}$&nbsp; code bits are output.
@@ Line 75: / Line 75: @@
 \nu_2  \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 1\hspace{0.05cm},\hspace{0.2cm}\nu_3 = 2\hspace{0.05cm}.$$
-*The memory order&nbsp; $m$&nbsp; (or memory for short)&nbsp; of the encoder is given by the longest shift register &nbsp; &#8658; &nbsp; $\underline {m = 2}$.
+*The memory order&nbsp; $m$&nbsp; $($or memory for short$)$&nbsp; of the encoder is given by the longest shift register &nbsp; &#8658; &nbsp; $\underline {m = 2}$.
-*The total influence length&nbsp; (or influence length for short)&nbsp; is equal to the number of memory elements of the entire encoder circuit &nbsp; &#8658; &nbsp; $\underline {\nu = 3}$.
+*The total influence length&nbsp; $($or influence length for short$)$&nbsp; is equal to the number of memory elements of the entire encoder circuit &nbsp; &#8658; &nbsp; $\underline {\nu = 3}$.
 [[File:P_ID2623__KC_A_3_1_neu.png|right|frame|Predefined convolutional encoder]]
-'''(3)'''&nbsp; In general, for the $n = 4$ code bits to step $i$:
+'''(3)'''&nbsp; In general,&nbsp; for the&nbsp; $n = 4$&nbsp; code bits of step&nbsp; $i$:
 :$$x_i^{(1)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} u_{i}^{(1)}  \hspace{0.05cm},$$
 :$$x_i^{(2)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} u_{i}^{(1)} + u_{i}^{(2)} + u_{i-1}^{(2)} + u_{i-1}^{(3)} \hspace{0.05cm},$$
@@ Line 87: / Line 87: @@
 :$$x_i^{(4)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} u_{i}^{(1)} + u_{i}^{(2)} + u_{i}^{(3)}+ u_{i-2}^{(3)}\hspace{0.05cm}.$$
-*In the first coding step, all memory elements are assigned zeros.
+*In the first coding step,&nbsp; all memory elements are assigned zeros.
 *Therefore for $i = 1$ all bits with the indices $i \, &ndash;1$ or $i \, &ndash;2$ can be omitted.
-*In accordance with the specification, let further hold: $u_1^{(1)} = 0$, $u_1^{(2)} = 1$, $u_1^{(3)} = 1$.
+*Therefore for&nbsp; $i = 1$&nbsp; all bits with the indices &nbsp; $i \, &ndash;1$ &nbsp; or &nbsp; $i \, &ndash;2$ &nbsp; can be omitted.
 *This is obtained by modulo 2 addition:
 :$$x_1^{(1)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} u_{1}^{(1)} \hspace{0.15cm}\underline {=0}   \hspace{0.05cm},$$
 :$$x_1^{(2)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} u_{1}^{(1)} + u_{1}^{(2)} = 0+1 \hspace{0.15cm}\underline {=1} \hspace{0.05cm},$$
@@ Line 97: / Line 100: @@
-'''(4)'''&nbsp; In the coding step $i = 3$ the information bits are:
+'''(4)'''&nbsp; In the coding step&nbsp; $i = 3$&nbsp; the information bits are:
 :$$u_i^{(1)} \hspace{-0.2cm} \ = \ \hspace{-0.2cm}1\hspace{0.05cm},\hspace{0.2cm}u_{i-1}^{(1)} = 1\hspace{0.05cm},\hspace{0.2cm}u_{i-2}^{(1)} = 0\hspace{0.05cm},$$
 :$$u_i^{(2)} \hspace{-0.2cm} \ = \ \hspace{-0.2cm}0\hspace{0.05cm},\hspace{0.2cm}u_{i-1}^{(2)} = 1\hspace{0.05cm},\hspace{0.2cm}u_{i-2}^{(2)} = 1\hspace{0.05cm},$$
 :$$u_i^{(3)} \hspace{-0.2cm} \ = \ \hspace{-0.2cm}1\hspace{0.05cm},\hspace{0.2cm}u_{i-1}^{(3)} = 0\hspace{0.05cm},\hspace{0.2cm}u_{i-2}^{(3)} = 1\hspace{0.05cm},$$
-which results in the following codebits:
+:which results in the following code bits:
 :$$x_3^{(1)} \hspace{0.15cm}\underline {=1} \hspace{0.05cm},\hspace{0.4cm}x_3^{(2)} = 1+0+1+0\hspace{0.15cm}\underline {=0}\hspace{0.05cm},\hspace{0.4cm}x_3^{(3)}= 0+1+1+1 \hspace{0.15cm}\underline {=1} \hspace{0.05cm},\hspace{0.4cm}x_3^{(4)} = 1+0+1+1\hspace{0.15cm}\underline {=1}\hspace{0.05cm}.$$
-The codebits in the coding step $i = 2$ have already been mentioned on the information page. Here still follows the derivation:
+&rArr; &nbsp; The code bits in the coding step&nbsp; $i = 2$&nbsp; have already been mentioned on the information page.&nbsp; Here still follows the derivation:
 :$$x_2^{(1)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} u_{2}^{(1)} = 1 \hspace{0.05cm},$$
 :$$x_2^{(2)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} u_{2}^{(1)} + u_{2}^{(2)} + u_{1}^{(2)} + u_{1}^{(3)} = 1+1+1+1 = 0 \hspace{0.05cm},$$
@@ Line 111: / Line 114: @@
 :$$x_2^{(4)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} u_{2}^{(1)} + u_{2}^{(2)} + u_{2}^{(3)}+ u_{0}^{(3)}= 1+1+0+(0) = 0\hspace{0.05cm}.$$
-Thus, the code sequence $\underline {x}$ (after multiplexing) starts with $(0100, 1000, 1011, \ \text{...})$. The grouping was done here only for clarity.
+*Thus,&nbsp; the code sequence&nbsp; $\underline {x}$&nbsp; $($after multiplexing$)$&nbsp; starts with&nbsp; $(0100, 1000, 1011, \ \text{...})$.
 {{ML-Fuß}}
 [[Category:Channel Coding: Exercises|^3.1 Basics of Convolutional Coding^]]

@@ Line 68: / Line 68: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; At each clock $\underline {k = 3}$ new information bits are processed and $\underline {n = 4}$ code bits are output.
+'''(1)'''&nbsp; At each clock&,nbsp; $\underline {k = 3}$&nbsp; new information bits are processed and&nbsp; $\underline {n = 4}$&nbsp; code bits are output.
-'''(2)'''&nbsp; We denote here the influence lengths of the information sequences&nbsp; $\underline{u}^{(j)}$&nbsp; by&nbsp; $\nu_j$, where&nbsp; $1 &#8804; j &#8804; k = 3$&nbsp; is to be set. Then holds:
+'''(2)'''&nbsp; We denote here the influence lengths of the information sequences&nbsp; $\underline{u}^{(j)}$&nbsp; by&nbsp; $\nu_j$,&nbsp; where&nbsp; $1 &#8804; j &#8804; k = 3$&nbsp; is to be set.&nbsp; Then holds:
 :$$\nu_1  \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 0 \hspace{0.15cm}{\rm (for \hspace{0.15cm}the \hspace{0.15cm}first \hspace{0.15cm}sequence\hspace{0.15cm}no \hspace{0.15cm}shift \hspace{0.15cm}register \hspace{0.15cm}is \hspace{0.15cm}necessary)}\hspace{0.05cm},\hspace{0.2cm}
 \nu_2  \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 1\hspace{0.05cm},\hspace{0.2cm}\nu_3 = 2\hspace{0.05cm}.$$
-*The memory order $m$ (or memory for short) of the encoder is given by the longest shift register &nbsp;&#8658;&nbsp; $\underline {m = 2}$.
+*The memory order&nbsp; $m$&nbsp; (or memory for short)&nbsp; of the encoder is given by the longest shift register &nbsp; &#8658; &nbsp; $\underline {m = 2}$.
 *The total influence length (or influence length for short) is equal to the number of memory elements of the entire encoder circuit &nbsp; &#8658; &nbsp; $\underline {\nu = 3}$.

@@ Line 10: / Line 10: @@
 :$$\underline{\it u}^{(3)} \hspace{-0.15cm} \ =  \hspace{-0.15cm}   \big( 1,\hspace{0.05cm} 0,\hspace{0.05cm} 1,\hspace{0.05cm}\text{...}  \big )\hspace{0.05cm}.$$
-The bits present at the coder at time&nbsp; $i$&nbsp; are denoted by&nbsp; $u_i^{\rm (1)}$,&nbsp; $u_i^{\rm (2)}$&nbsp; and&nbsp; $u_i^{\rm (3)}$&nbsp;. For example&nbsp; $u_1^{\rm (1)} = 0$,&nbsp; $u_2^{\rm (2)} = 1$&nbsp; and&nbsp; $u_3^{\rm (3)} = 1$.
+The bits present at the coder at time&nbsp; $i$&nbsp; are denoted by&nbsp; $u_i^{\rm (1)}$,&nbsp; $u_i^{\rm (2)}$&nbsp; and&nbsp; $u_i^{\rm (3)}$&nbsp;. For example&nbsp; $u_1^{\rm (1)} = 0$,&nbsp; $u_2^{\rm (2)} = 1$,&nbsp;  $u_3^{\rm (3)} = 1$.
-The purpose of this exercise is to determine:
+The purpose of this exercise is to determine
 * the number&nbsp; $k$&nbsp; of information bits processed per coding step,
 * the number&nbsp; $n$&nbsp; of code bits output per coding step,
@@ Line 31: / Line 30: @@
 Hints:
 * This exercise refers to the chapter&nbsp; [[Channel_Coding/Basics_of_Convolutional_Coding| "Basics of Convolutional Coding"]].
 * For completeness, the code bits to the clock step&nbsp; $i = 2$&nbsp; are also given here:
 :$$x_2^{(1)} = 1 \hspace{0.05cm},\hspace{0.4cm}x_2^{(2)} = 0 \hspace{0.05cm},\hspace{0.4cm}
 x_2^{(3)} = 0 ,\hspace{0.4cm}x_2^{(4)} = 0 \hspace{0.05cm}.$$
-* However, this last specification is not needed to solve the exercise.
+* However,&nbsp; this last specification is not needed to solve the exercise.
@@ Line 46: / Line 46: @@
 $n \ = \ ${ 4 3% }
-{What are the memory order&nbsp; $m$&nbsp; and total influence length&nbsp; $\nu$?
+{What are the memory order&nbsp; $m$&nbsp; and the total influence length&nbsp; $\nu$?
 |type="{}"}
 $m \ = \ ${ 2 3% }
 $\nu \ = \ ${ 3 3% }
-{What are the four codebits in the first coding step&nbsp; $(i = 1)$?
+{What are the code bits in the first coding step&nbsp; $(i = 1)$?
 |type="{}"}
 $x_1^{(1)} \ = \ ${ 0 3% }

@@ Line 72: / Line 72: @@
 '''(2)'''&nbsp; We denote here the influence lengths of the information sequences&nbsp; $\underline{u}^{(j)}$&nbsp; by&nbsp; $\nu_j$, where&nbsp; $1 &#8804; j &#8804; k = 3$&nbsp; is to be set. Then holds:
-:$$\nu_1  \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 0 \hspace{0.15cm}{\rm (f\ddot{u}r \hspace{0.15cm}die \hspace{0.15cm}erste \hspace{0.15cm}Sequenz \hspace{0.15cm}ist \hspace{0.15cm}kein \hspace{0.15cm}Schieberegister \hspace{0.15cm}n\ddot{o}tig)}\hspace{0.05cm},\hspace{0.2cm}
+:$$\nu_1  \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 0 \hspace{0.15cm}{\rm (for \hspace{0.15cm}the \hspace{0.15cm}first \hspace{0.15cm}sequence\hspace{0.15cm}no \hspace{0.15cm}shift \hspace{0.15cm}register \hspace{0.15cm}is \hspace{0.15cm}necessary)}\hspace{0.05cm},\hspace{0.2cm}
 \nu_2  \hspace{-0.15cm} \ = \ \hspace{-0.15cm} 1\hspace{0.05cm},\hspace{0.2cm}\nu_3 = 2\hspace{0.05cm}.$$

← Older revision		Revision as of 08:58, 12 September 2022
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−	[[Category:Channel Coding: Exercises\|^3.1 ~~Grundlagen der Faltungscodierung~~^]]	+	[[Category:Channel Coding: Exercises\|^3.1 Basics of Convolutional Coding^]]

← Older revision	Revision as of 19:45, 19 September 2022
(No difference)

← Older revision	Revision as of 08:56, 12 September 2022
(No difference)

← Older revision		Revision as of 12:50, 23 March 2021
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−	[[Category:~~Aufgaben zu Kanalcodierung~~\|^3.1 Grundlagen der Faltungscodierung^]]	+	[[Category:Channel Coding: Exercises\|^3.1 Grundlagen der Faltungscodierung^]]

@@ Line 30: / Line 30: @@
 ''Hinweise:''
-* Die Aufgabe bezieht sich auf das Kapitel&nbsp; [[Kanalcodierung/Grundlagen_der_Faltungscodierung| Grundlagen der Faltungscodierung]].
+* Die Aufgabe bezieht sich auf das Kapitel&nbsp; [[Channel_Coding/Grundlagen_der_Faltungscodierung| Grundlagen der Faltungscodierung]].
 * Der Vollständigkeit halber werden hier auch die Codebits zum Taktschritt&nbsp; $i = 2$&nbsp; angegeben:
 :$$x_2^{(1)} = 1  \hspace{0.05cm},\hspace{0.4cm}x_2^{(2)} = 0 \hspace{0.05cm},\hspace{0.4cm}