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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:



Questions

1

Which of the listed transitions are possible with the ring?

S1S2S3,
S1S2S2S2S3,
S1S2S1S2S3.

2

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

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

3

Which of the listed transitions are possible with feedback?

S1S2S3S4,
S1S2S3S2S4,
S1S2S3S2S3S4,
S1S2S3S2S3S2S3S4.

4

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

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


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 = AB+ACB+AC2B+AC3B+ ...=AB[1+C+C2+C3+ ...].
  • The parenthesis expression gives  1/(1C).
E(X,U)=A(X,U)B(X,U)1C(X,U).


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

  • One goes first from  S1  to  S2  A(X,U),
  • then from  S2  to  S3  C(X,U),
  • then  j–times back to  S2  and again to  S3 (j=0, 1, 2,  ... )  E(X,U),
  • finally from  S3  to  S4  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)C(X,U)E(X,U)B(X,U)
  • Here  E(X,U)  describes the path  "j–times"  back to  S2  and again to  S3 (j=0, 1, 2,  ...):
E(X,U)=1+DC+(1+D)2+(1+D)3+ ...=11CDF(X,U)=A(X,U)B(X,U)C(X,U)1C(X,U)D(X,U).