[[File:P_ID2589__KC_Z_3_1.png|right|frame|Two convolutional codes of rate $1/2$]]
[[File:EN_KC_Z_3_1_jetztaber.png|right|frame|Convolutional codes of rate $1/2$]]
The graphic shows two convolutional encoders of rate $R = 1/2$. At the input there is the information sequence $\underline {u} = (u_1, u_2, \ \text{...} \ , u_i, \ \text{...})$ . From this, modulo 2 operations generate the two sequences
The graphic shows two convolutional encoders of rate $R = 1/2$:
where $x_i^{(j)}$ with $j = 1$ resp. $j = 2$ except from $u_i$ also from the previous information bits $u_{i-1}, \ \text{...} \ , u_{i-m}$ may depend. One refers $m$ as the memory and $\nu = m + 1$ as the influence length of the code or the encoder. The considered coders $\rm A$ and $\rm B$ differ with respect to these quantities.
*From this, modulo-2 operations generate the two sequences
:where $x_i^{(j)}$ with $j = 1$ resp. $j = 2$ may depend except from $u_i$ also from the previous information bits $u_{i-1}, \ \text{...} \ , u_{i-m}$.
*One refers $m$ as the "memory" and $\nu = m + 1$ as the "influence length" of the code $($or of the encoder$)$.
* The considered encoders $\rm A$ and $\rm B$ differ with respect to these quantities.
Hints:
<u>Hints:</u>
*The exercise refers to the chapter [[Channel_Coding/Basics_of_Convolutional_Coding| "Basics of Convolutional Coding"]].
*The exercise refers to the chapter [[Channel_Coding/Basics_of_Convolutional_Coding| "Basics of Convolutional Coding"]].
*Not shown in the diagram is the multiplexing of the two subsequences $\underline {x}^{(1)}$ and $\underline {x}^{(2)}$ to the resulting code sequence
*Not shown in the diagram is the multiplexing of the two subsequences $\underline {x}^{(1)}$ and $\underline {x}^{(2)}$ to the resulting code sequence
*In subtasks '''(3)''' to '''(5)''' you are to determine the respective start of the sequences $\underline {x}^{(1)}, \underline{x}^{(2)}$ and $\underline{x}$ assuming the information sequence $\underline{u} = (1, 0, 1, 1, 0, 0, \ \text{. ..})$ is to be assumed.
*In subtasks '''(3)''' to '''(5)''' you are to determine the start of the sequences $\underline {x}^{(1)}, \underline{x}^{(2)}$ and $\underline{x}$ assuming the information sequence
'''(2)''' The shift register of encoder $\rm A$ does contain two memory cells. However, since $x_i^{(1)} = u_i$ and $x_i^{(2)} = u_i + u_{i-1}$ is influenced only by the immediately preceding bit $u_{i-1}$ besides the current information bit $u_i$,
'''(2)''' The shift register of encoder $\rm A$ does contain two memory cells.
However, since $x_i^{(1)} = u_i$ and $x_i^{(2)} = u_i + u_{i-1}$ is influenced only by the immediately preceding bit $u_{i-1}$ besides the current information bit $u_i$, is
*the memory $m = 1$, and
*the influence length $\nu = m + 1 = 2$.
The graphic shows the two encoders in another representation, whereby the "memory cells" are highlighted in yellow.
*For the encoder $\rm A$ one recognizes only one memory ⇒ $m = 1$.
*In contrast, for the encoder $\rm B$ actually $m = 2$ and $\nu = 3$.
The graphic shows the two coders in another representation, whereby the "memory cells" are highlighted in yellow.
*Thus, the <u>proposed solution 2</u> is correct.
*For the encoder $\rm A$ one recognizes only one such memory ⇒ $m = 1$.
*In contrast, for the encoder $\rm B$ actually $m = 2$ and $\nu = 3$.
*The <u>proposed solution 2</u> is therefore correct.
*The <u>proposed solution 1</u> is therefore correct.
*The <u>proposed solution 1</u> is correct. The second solution suggestion ⇒ $\underline {x}^{(1)} = \underline {u}$ would only be valid for a systematic code $($which is not present here$)$.
*The second solution suggestion ⇒ $\underline {x}^{(1)} = $\underline {u}$ would only be valid for a systematic code (which is not present here).
'''(4)''' Analogous to subtask (3), $x_i^{(2)} = u_i + u_{i–2}$ is obtained:
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 '''(3)''' and '''(4)''' shows the correctness of the <u>proposed solution 1</u>.
{{ML-Fuß}}
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[[Category:Channel Coding: Exercises|^3.1 Basics of Convolutional Coding^]]
[[Category:Channel Coding: Exercises|^3.1 Basics of Convolutional Coding^]]
[[de:Aufgaben:Aufgabe 3.1Z: Faltungscodes der Rate 1/2]]
In subtasks (3) to (5) you are to determine the start of the sequences $\underline {x}^{(1)}, \underline{x}^{(2)}$ and $\underline{x}$ assuming the information sequence
The memory $m$ and the influence length $\nu$ are different ⇒ Answers 3 and 4.
(2) The shift register of encoder $\rm A$ does contain two memory cells. However, since $x_i^{(1)} = u_i$ and $x_i^{(2)} = u_i + u_{i-1}$ is influenced only by the immediately preceding bit $u_{i-1}$ besides the current information bit $u_i$,
Equivalent encoder representations
the memory is $m = 1$, and
the influence length is $\nu = m + 1 = 2$.
The graphic shows the two encoders in another representation, whereby the "memory cells" are highlighted in yellow.
For the encoder $\rm A$ one recognizes only one memory ⇒ $m = 1$.
In contrast, for the encoder $\rm B$ actually $m = 2$ and $\nu = 3$.
Thus, the proposed solution 2 is correct.
(3) For the upper output of encoder $\rm B$ applies in general:
The proposed solution 1 is correct. The second solution suggestion ⇒ $\underline {x}^{(1)} = \underline {u}$ would only be valid for a systematic code $($which is not present here$)$.
