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 Bounds".

• This exercise is intended to prove some of the statements on the  "Rules for manipulating the state transition diagram"  section.

• Applied these rules in  $\text{Exercise 3.12}$  and  $\text{Exercise 3.13}$.

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 the  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 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 the  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  $S_3 \ (j =0, \ 1, \ 2, \ \text{ ...})$:

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