Aufgaben:Exercise 3.12Z: Ring and Feedback

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Ring and feedback in the state transition diagram

In order to determine the path weighting enumerator function  $T(X)$  of a convolutional code from the state transition diagram, it is necessary to reduce the diagram until it can be represented by a single connection from the initial state to the final state.

In the course of this diagram reduction can occur:

• serial and parallel transitions,
• a ring according to the sketch above,
• a feedback according to the sketch below.

For these two graphs, find the correspondences  $E(X, \, U)$  and  $F(X, \, U)$  depending on the given functions  $A(X, \, U), \ B(X, \ U), \ C(X, \, U), \ D(X, \, U)$ .

Hints:

• This exercise belongs to the chapter  "Distance characteristics and error probability barriers".
• This exercise is intended to prove some of the statements on the  "Rules for manipulating the state transition diagram"  page.
• Applied these rules in the  "Exercise 3.12"  and the  "Exercise 3.13".

Questions

1

Which of the listed transitions are possible with the ring?

	$S_1 → S_2 → S_3$,
	$S_1 → S_2 → S_2 → S_2 → S_3$,
	$S_1 → S_2 → S_1 → S_2 → S_3$.

2

What is the substitution  $E(X, \, U)$  of a ring?

	$E(X, \, U) = [A(X, \, U) + B(X, \, U)] \ / \ [1 \, -C(X, \, U)]$,
	$E(X, \, U) = A(X, \, U) \cdot B(X, \, U) \ / \ [1 \, -C(X, \, U)]$,
	$E(X, \, U) = A(X, \, U) \cdot C(X, \, U) \ / \ [1 \, -B(X, \, U)]$.

3

Which of the listed transitions are possible with feedback?

	$S_1 → S_2 → S_3 → S_4$,
	$S_1 → S_2 → S_3 → S_2 → S_4$,
	$S_1 → S_2 → S_3 → S_2 → S_3 → S_4$,
	$S_1 → S_2 → S_3 → S_2 → S_3 → S_2 → S_3 → S_4$.

4

What is the substitution  $F(X, \, U)$  of a feedback?

	$F(X, \, U) = A(X, \, U) \cdot B(X, \, U) \cdot C(X, \, U) \ / \ [1 \, -C(X, \, U) \cdot D(X, \, U)]$
	$F(X, \, U) = A(X, \, U) \cdot B(X, \, U) \ / \ [1 \, -C(X, \, U) + D(X, \, U)]$.

Solution

Solution

(1)  Correct are solutions 1 and 2:

• In general terms, one first goes from $S_1$ to $S_2$, remains $j$–times in the state $S_2 \ (j = 0, \ 1, \, 2, \ \text{ ...})$, and finally continues from $S_2$ to $S_3$.

(2)  Correct is the solution suggestion 2:

• In accordance with the explanations for the subtask (1), one obtains for the substitution of the ring

$$E \hspace{-0.15cm} \ = \ \hspace{-0.15cm} A \cdot B + A \cdot C \cdot B + A \cdot C^2 \cdot B + A \cdot C^3 \cdot B + \text{ ...} \hspace{0.1cm}=A \cdot B \cdot [1 + C + C^2+ C^3 +\text{ ...}\hspace{0.1cm}] \hspace{0.05cm}.$$

• The parenthesis expression gives $1/(1 \, –C)$.

$$E(X, U) = \frac{A(X, U) \cdot B(X, U)}{1- C(X, U)} \hspace{0.05cm}.$$

(3)  Correct are the solutions 1, 3 and 4:

• one goes first from $S_1$ to $S_2 \ \Rightarrow \ A(X, \, U)$,
• then from $S_2$ to $S_3 \ \Rightarrow \ C(X, \, U)$,
• then $j$–times back to $S_2$ and again to $S_3 \ (j = 0, \ 1, \ 2, \ \text{ ...} \ ) \ \Rightarrow \ E(X, \, U)$,
• finally from $S_3$ to $S_4 \ \Rightarrow \ B(X, \, U)$,

(4)  Thus, the correct solution is suggested solution 1:

• According to the sample solution to subtask (3) applies:

$$F(X, U) = A(X, U) \cdot C(X, U) \cdot E(X, U) \cdot B(X, U)\hspace{0.05cm}$$

• Here $E(X, \, U)$ describes the path "$j$–times" back to $S_2$ and again to $S_3 \ (j =0, \ 1, \ 2, \ \text{ ...})$:

$$E(X, U) = 1 + D \cdot C + (1 + D)^2 + (1 + D)^3 + \text{ ...} \hspace{0.1cm}= \frac{1}{1-C \hspace{0.05cm} D} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} F(X, U) = \frac{A(X, U) \cdot B(X, U)\cdot C(X, U)}{1- C(X, U) \cdot D(X, U)} \hspace{0.05cm}.$$