Exercise 3.12Z: Ring and Feedback
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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 Exercise 3.12 and Exercise 3.13.
Questions
Solution
(1) Correct are the solutions 1 and 2:
- In general terms, one first goes from S1 to S2, remains j–times in the state S2 (j=0, 1,2, ...), and finally continues from S2 to S3.
(2) Correct is the solution suggestion 2:
- In accordance with the explanations for subtask (1), one obtains for the substitution of the ring:
- E = A⋅B+A⋅C⋅B+A⋅C2⋅B+A⋅C3⋅B+ ...=A⋅B⋅[1+C+C2+C3+ ...].
- 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}.