Aufgaben:Exercise 3.1Z: Convolution Codes of Rate 1/2 - Revision history

Guenter at 17:26, 16 November 2022

2022-11-16T17:26:26Z

Guenter at 17:26, 16 November 2022

2022-11-16T17:26:04Z

Guenter at 16:54, 6 November 2022

2022-11-06T16:54:33Z

Guenter at 16:36, 6 November 2022

2022-11-06T16:36:26Z

Guenter at 15:12, 27 October 2022

2022-10-27T15:12:24Z

Noah at 18:14, 21 September 2022

2022-09-21T18:14:49Z

Show changes

Noah at 19:46, 19 September 2022

2022-09-19T19:46:15Z

Noah: Noah moved page Aufgaben:Erercise 3.1Z: Convolution Codes of Rate 1/2 to Aufgaben:Exercise 3.1Z: Convolution Codes of Rate 1/2: Erercise - > Exercise

2022-09-19T19:45:25Z

Noah moved page Erercise 3.1Z: Convolution Codes of Rate 1/2 to Exercise 3.1Z: Convolution Codes of Rate 1/2: Erercise - > Exercise

Guenter at 08:59, 12 September 2022

2022-09-12T08:59:28Z

Guenter: Guenter moved page Aufgaben:Aufgabe 3.1Z: Faltungscodes der Rate 1/2 to Aufgaben:Erercise 3.1Z: Convolution Codes of Rate 1/2

2022-09-12T08:57:42Z

Guenter moved page Aufgabe 3.1Z: Faltungscodes der Rate 1/2 to Erercise 3.1Z: Convolution Codes of Rate 1/2

@@ Line 1: / Line 1: @@
 {{quiz-Header|Buchseite=Channel_Coding/Basics_of_Convolutional_Coding}}
-[[File:EN_KC_Z_3_1_jetztaber.png|right|frame|Convolutional codes of rate &nbsp;$1/2$ '''Korrektur''']]
+[[File:EN_KC_Z_3_1_jetztaber.png|right|frame|Convolutional codes of rate &nbsp;$1/2$]]
 The graphic shows two convolutional encoders of rate&nbsp; $R = 1/2$:&nbsp;
 *At the input there is the information sequence&nbsp; $\underline {u} = (u_1, u_2, \ \text{...} \ , u_i, \ \text{...})$.&nbsp;

@@ Line 1: / Line 1: @@
 {{quiz-Header|Buchseite=Channel_Coding/Basics_of_Convolutional_Coding}}
-[[File:P_ID2589__KC_Z_3_1.png|right|frame|Convolutional codes of rate &nbsp;$1/2$ '''Korrektur''']]
+[[File:EN_KC_Z_3_1_jetztaber.png|right|frame|Convolutional codes of rate &nbsp;$1/2$ '''Korrektur''']]
 The graphic shows two convolutional encoders of rate&nbsp; $R = 1/2$:&nbsp;
 *At the input there is the information sequence&nbsp; $\underline {u} = (u_1, u_2, \ \text{...} \ , u_i, \ \text{...})$.&nbsp;

@@ Line 62: / Line 62: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; For both encoders, $k = 1$ and $n = 2$.
+'''(1)'''&nbsp; For both encoders,&nbsp; $k = 1$&nbsp; and&nbsp; $n = 2$.
 *The memory $m$ and the influence length $\nu$ are different &nbsp; &#8658; &nbsp; <u>Answers 3 and 4</u>.
-[[File:EN_KC_A_3_1.png|right|frame|Equivalent coder representations]]
+'''(2)'''&nbsp; The shift register of encoder &nbsp;$\rm A$&nbsp; does contain two memory cells.&nbsp; However,&nbsp; since&nbsp; $x_i^{(1)} = u_i$&nbsp; and&nbsp; $x_i^{(2)} = u_i + u_{i-1}$&nbsp; is influenced only by the immediately preceding bit&nbsp; $u_{i-1}$&nbsp; besides the current information bit&nbsp; $u_i$,
-'''(2)'''&nbsp; The shift register of encoder &nbsp;$\rm A$&nbsp; does contain two memory cells.
+[[File:EN_KC_A_3_1.png|right|frame|Equivalent encoder representations]]
-The graphic shows the two coders in another representation, whereby the "memory cells" are highlighted in yellow.
+*Thus,&nbsp; the&nbsp; <u>proposed solution 2</u>&nbsp; is correct.
@@ Line 84: / Line 86: @@
 :$$x_i^{(1)} = u_{i} + u_{i-1}+ u_{i-2} \hspace{0.05cm}.$$
-Considering the preassignment ($u_0 = u_{-1} = 0$), we obtain with the above data:
+*Considering the preassignment&nbsp;($(u_0 = u_{-1} = 0)$,&nbsp; we obtain with the above data:
 :$$x_1^{(1)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} u_{1} + u_{0}+ u_{-1} = 1+0+0 = 1 \hspace{0.05cm},\hspace{1cm}x_2^{(1)} = u_{2} + u_{1}+ u_{0} = 0+1+0 = 1\hspace{0.05cm},$$
 :$$x_3^{(1)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} u_{3} + u_{2}+ u_{1} \hspace{0.25cm}= 1+0+1 = 0 \hspace{0.05cm},\hspace{1cm}x_4^{(1)} = u_{4} + u_{3}+ u_{2} = 1+1+0 = 0\hspace{0.05cm},$$
@@ Line 90: / Line 92: @@
 :$$x_7^{(1)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm} x_8^{(1)} = \text{...} \hspace{0.05cm}= 0 \hspace{0.05cm}.$$
-*The <u>proposed solution 1</u> is therefore correct.
+*The&nbsp; <u>proposed solution 1</u>&nbsp; is correct.&nbsp; The second solution suggestion &nbsp; &#8658; &nbsp; $\underline {x}^{(1)} = \underline {u}$&nbsp; would only be valid for a systematic code&nbsp; $($which is not present here$)$.
+'''(4)'''&nbsp; Analogous to subtask&nbsp; '''(3)'''&nbsp; &rArr; &nbsp; $x_i^{(2)} = u_i + u_{i&ndash;2}$:
 :$$x_1^{(2)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm}   1+0 = 1 \hspace{0.05cm},\hspace{0.2cm}x_2^{(2)} = 0+0 = 0\hspace{0.05cm},
 \hspace{0.2cm}x_3^{(3)} = 1+1 = 0\hspace{0.05cm},\hspace{0.2cm}x_4^{(2)} =  1+0 = 1 \hspace{0.05cm},$$
@@ Line 102: / Line 102: @@
 x_7^{(2)} = x_8^{(2)} = \text{...} \hspace{0.05cm}= 0 \hspace{0.05cm}.$$
-*The correct solution is therefore <u>proposed solution 2</u>.
+*The correct solution is the&nbsp; <u>proposed solution 2</u>.
-'''(5)'''&nbsp; For the (entire) code sequence, one can formally write:
+'''(5)'''&nbsp; For the&nbsp; $($entire$)$&nbsp; code sequence,&nbsp; one can formally write:
 :$$\underline{\it x} =  \big( \hspace{0.05cm}\underline{\it x}_1\hspace{0.05cm}, \hspace{0.05cm} \underline{\it x}_2\hspace{0.05cm}, \hspace{0.05cm}\text{...}\hspace{0.05cm}  \underline{\it x}_i \hspace{0.05cm}, \text{...} \hspace{0.05cm} \big )\hspace{0.05cm}, \hspace{0.3cm}  \underline{\it x}_i = \big( x_i^{(1)}\hspace{0.05cm}, x_i^{(2)} \big)
 \hspace{0.4cm}\Rightarrow \hspace{0.4cm} \underline{\it x} =  \big( \hspace{0.05cm}x_1^{(1)}\hspace{0.01cm},\hspace{0.05cm} x_2^{(1)}\hspace{0.01cm},\hspace{0.05cm} x_1^{(2)}\hspace{0.01cm},\hspace{0.05cm} x_2^{(2)}\hspace{0.01cm}, \hspace{0.05cm} \text{...} \hspace{0.05cm} \big )\hspace{0.05cm}. $$
-A comparison with the solutions of exercises (3) and (4) shows the correctness of <u>proposed solution 1</u>.
+*A comparison with the solutions of subtasks&nbsp; '''(3)'''&nbsp; and&nbsp; '''(4)'''&nbsp; shows the correctness of the&nbsp; <u>proposed solution 1</u>.
 {{ML-Fuß}}
 [[Category:Channel Coding: Exercises|^3.1 Basics of Convolutional Coding^]]

@@ Line 1: / Line 1: @@
 {{quiz-Header|Buchseite=Channel_Coding/Basics_of_Convolutional_Coding}}
-[[File:P_ID2589__KC_Z_3_1.png|right|frame|Two convolutional codes of rate &nbsp;$1/2$]]
+[[File:P_ID2589__KC_Z_3_1.png|right|frame|Convolutional codes of rate &nbsp;$1/2$ '''Korrektur''']]
-The graphic shows two convolutional encoders of rate&nbsp; $R = 1/2$. At the input there is the information sequence&nbsp; $\underline {u} = (u_1, u_2, \ \text{...} \ , u_i, \ \text{...})$&nbsp;. From this, modulo 2 operations generate the two sequences
+The graphic shows two convolutional encoders of rate&nbsp; $R = 1/2$:&nbsp;
 :$$\underline{\it x}^{(1)} \hspace{-0.15cm} \ = \ \hspace{-0.15cm}   \big( \hspace{0.05cm}x_1^{(1)}\hspace{0.05cm},\hspace{0.05cm} x_2^{(1)}\hspace{0.05cm},\hspace{0.05cm} \text{...} \hspace{0.05cm},\hspace{0.05cm}
   x_i^{(1)} \hspace{0.05cm},\text{...} \hspace{0.05cm} \big )\hspace{0.05cm},$$
@@ Line 8: / Line 11: @@
   x_i^{(2)} \hspace{0.05cm}, \text{...} \hspace{0.05cm} \big )$$
-where&nbsp; $x_i^{(j)}$&nbsp; with&nbsp; $j = 1$&nbsp; resp.&nbsp; $j = 2$&nbsp; except from&nbsp; $u_i$&nbsp; also from the previous information bits&nbsp; $u_{i-1}, \ \text{...} \ , u_{i-m}$&nbsp; may depend. One refers&nbsp; $m$&nbsp; as the memory and&nbsp; $\nu = m + 1$&nbsp; as the influence length of the code or the encoder. The considered coders &nbsp;$\rm A$&nbsp; and &nbsp;$\rm B$&nbsp; differ with respect to these quantities.
+:where&nbsp; $x_i^{(j)}$&nbsp; with&nbsp; $j = 1$&nbsp; resp.&nbsp; $j = 2$&nbsp;  may depend except from&nbsp; $u_i$&nbsp; also from the previous information bits&nbsp; $u_{i-1}, \ \text{...} \ , u_{i-m}$.
 *Not shown in the diagram is the multiplexing of the two subsequences&nbsp; $\underline {x}^{(1)}$&nbsp; and&nbsp; $\underline {x}^{(2)}$&nbsp; to the resulting code sequence&nbsp;
 :$$\underline {x} = (x_1^{(1)}, x_1^{(2)}, x_2^{(1)}, x_2^{(2)}, \ \text{...}).$$
-*In subtasks '''(3)''' to '''(5)''' you are to determine the respective start of the sequences&nbsp; $\underline {x}^{(1)}, \underline{x}^{(2)}$&nbsp; and&nbsp; $\underline{x}$&nbsp; assuming the information sequence&nbsp; $\underline{u} = (1, 0, 1, 1, 0, 0, \ \text{. ..})$&nbsp; is to be assumed.
+*In subtasks&nbsp; '''(3)'''&nbsp; to&nbsp; '''(5)'''&nbsp; you are to determine the start of the sequences&nbsp; $\underline {x}^{(1)}, \underline{x}^{(2)}$&nbsp; and&nbsp; $\underline{x}$&nbsp; assuming the information sequence&nbsp;
@@ Line 26: / Line 32: @@
 ===Questions===
 <quiz display=simple>
-{In which code parameters do coder &nbsp;$\rm A$&nbsp; and coder &nbsp;$\rm B$ differ?
+{In which code parameters do encoder &nbsp;$\rm A$&nbsp; and encoder &nbsp;$\rm B$ differ?
 |type="[]"}
 - $k$: &nbsp; &nbsp; Number of information bits processed per coding step,
 - $n$: &nbsp; &nbsp; Number of code bits output per coding step,
-+ $m$: &nbsp; memory order of the code or encoder,
++ $m$: &nbsp; memory order of the code,
 + $\nu$: &nbsp; &nbsp; influence length of the code.
-{Which encoder exhibits the memory&nbsp; $m = 2$&nbsp;?
+{Which encoder exhibits the memory&nbsp; $m = 2$?
 |type="[]"}
 - Encoder &nbsp;$\rm A$,
-+ Encoder &nbsp;$\rm B$.
++ encoder &nbsp;$\rm B$.
 {What is the partial code sequence&nbsp; $\underline {x}^{(1)}$&nbsp; of encoder &nbsp;$\rm B$&nbsp; for&nbsp; $\underline {u} = (1, 0, 1, 1, 0, 0, \ \text{...})$?

← Older revision		Revision as of 15:12, 27 October 2022
Line 60:		Line 60:


−	[[File:~~P_ID2599__KC_Z_3_1b~~.png\|right\|frame\|Equivalent coder representations]]	+	[[File:EN_KC_A_3_1.png\|right\|frame\|Equivalent coder representations]]
	'''(2)'''  The shift register of encoder  $\rm A$  does contain two memory cells.		'''(2)'''  The shift register of encoder  $\rm A$  does contain two memory cells.

← Older revision		Revision as of 19:46, 19 September 2022
Line 1:		Line 1:
−	{{quiz-Header\|Buchseite=~~Kanalcodierung~~/~~Grundlagen der Faltungscodierung~~}}	+	{{quiz-Header\|Buchseite=Channel_Coding/Basics_of_Convolutional_Coding}}

	[[File:P_ID2589__KC_Z_3_1.png\|right\|frame\|Zwei Faltungscodes der Rate  $1/2$]]		[[File:P_ID2589__KC_Z_3_1.png\|right\|frame\|Zwei Faltungscodes der Rate  $1/2$]]

← Older revision		Revision as of 08:59, 12 September 2022
Line 108:		Line 108:


−	[[Category:Channel Coding: Exercises\|^3.1 ~~Grundlagen der Faltungscodierung~~^]]	+	[[Category:Channel Coding: Exercises\|^3.1 Basics of Convolutional Coding^]]

← Older revision	Revision as of 08:57, 12 September 2022
(No difference)