Aufgaben:Exercise 3.10: Metric Calculation

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Only partially evaluated trellis

In the  "theory section"  of this chapter, the calculation of the branch metrics  ${\it \Gamma}_i(S_{\mu})$  has been discussed in detail, based on the Hamming distance  $d_{\rm H}(\underline{x}\hspace{0.05cm}', \ \underline{y}_i)$  between the possible codewords  $\underline{x}\hspace{0.05cm}' ∈ \{00, \, 01, \, 10, \, 11\}$  and the 2–bit–words  $\underline{y}_i$  received at time  $i$  is based.

The exercise deals exactly with this topic. In the adjacent graph

• the considered trellis is shown– valid for the code with rate  $R = 1/2$,  memory  $m = 2$  and  $\mathbf{G}(D) = (1 + D + D^2, \ 1 + D^2)$,
• are the received words  $\underline{y}_1 = (01), \hspace{0.05cm}\text{ ...} \hspace{0.05cm} , \ \underline{y}_7 = (11)$  indicated in the rectangles,
• are all branch metrics  ${\it \Gamma}_0(S_{\mu}), \hspace{0.05cm}\text{ ...} \hspace{0.05cm} , \ {\it \Gamma}_4(S_{\mu})$  already entered.

For example, the branch metric  ${\it \Gamma}_4(S_0)$  with  $\underline{y}_4 = (01)$  as the minimum of the two comparison values

• ${\it \Gamma}_3(S_0) + d_{\rm H}((00), \ (01)) = 3 + 1 = 4$, and
• ${\it \Gamma}_3(S_2) + d_{\rm H}((11), \ (01)) = 2 + 1 = 3$.

The surviving branch – here from  ${\it \Gamma}_3(S_2)$  to  ${\it \Gamma}_4(S_0)$  – is drawn solid, the eliminated branch from  ${\it \Gamma}_3(S_0)$  to  ${\it \Gamma}_4(S_0)$  dotted. Red arrows represent the information bit $u_i = 0$, blue arrows $u_i = 1$.

In the subtask (4) the relationship between

• the  ${\it \Gamma}_i(S_{\mu})$ minimization and
• the  ${\it \Lambda}_i(S_{\mu})$ maximization.

shall be worked out. Here, we refer to the nodes  ${\it \Lambda}_i(S_{\mu})$  as metrics, where the metric increment over the predecessor nodes results from the correlation value  $〈\underline{x}_i\hspace{0.05cm}', \, \underline{y}_i 〉$ . For more details on this topic, see the following theory pages:

• "Relationship between Hamming distance and correlation"
• "Viterbi algorithm based on correlation and metrics"
• "Viterbi decision for non–terminated convolutional codes".

Hints:

• The exercise refers to the chapter  "Decoding Convolutional Codes".
• For the time being, the search of surviving paths is not considered. This will be dealt with for the same example in the later  "Exercise 3.11".

Questions

1

What are the branch metrics for time  $i = 5$?

${\it \Gamma}_5(S_0) \ = \ $

${\it \Gamma}_5(S_1) \ = \ $

${\it \Gamma}_5(S_2) \ = \ $

${\it \Gamma}_5(S_3) \ = \ $

2

What are the branch metrics for time  $i = 6$?

${\it \Gamma}_6(S_0) \ = \ $

${\it \Gamma}_6(S_2) \ = \ $

3

What is the final value of this trellis based on  ${\it \Gamma}_i(S_{\mu})$?

	It holds  ${\it \Gamma}_7(S_0) = 3$.
	This final value suggests one error-free transmission.
	This final value suggests three transmission errors.

4

Which statements are true for the  ${\it \Lambda}_i(S_{\mu})$ evaluation?

	The metrics  ${\it \Lambda}_i(S_{\mu})$  provide the same information as  ${\it \Gamma}_i(S_{\mu})$.
	For all nodes,  ${\it \Lambda}_i(S_{\mu}) = 2 \cdot \big [i \, –{\it \Gamma}_i(S_{\mu})\big ]$.
	For the metric increments,  $〈 \underline{x}_i', \, \underline{y}_i 〉 ∈ \{0, \, 1, \, 2\}$.

Solution

Solution

(1)  At all nodes $S_{\mu}$ a decision must be made between the two incoming branches. The branch that led to the (minimum) error metric ${\it \Gamma}_5(S_{\mu})$ is then selected in each case. With $\underline{y}_5 = (01)$ one obtains:

$${\it \Gamma}_5(S_0) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}{\rm min} \left [{\it \Gamma}_{4}(S_0) + d_{\rm H} \big ((00)\hspace{0.05cm},\hspace{0.05cm} (01) \big )\hspace{0.05cm},\hspace{0.2cm}{\it \Gamma}_{4}(S_2) + d_{\rm H} \big ((11)\hspace{0.05cm},\hspace{0.05cm} (01) \big ) \right ] ={\rm min} \left [ 3+1\hspace{0.05cm},\hspace{0.05cm} 2+1 \right ] \hspace{0.15cm}\underline{= 3}\hspace{0.05cm},$$

$${\it \Gamma}_5(S_1) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}{\rm min} \left [{\it \Gamma}_{4}(S_0) + d_{\rm H} \big ((11)\hspace{0.05cm},\hspace{0.05cm} (01) \big )\hspace{0.05cm},\hspace{0.2cm}{\it \Gamma}_{4}(S_2) + d_{\rm H} \big ((00)\hspace{0.05cm},\hspace{0.05cm} (01) \big ) \right ] ={\rm min} \left [ 3+1\hspace{0.05cm},\hspace{0.05cm} 2+1 \right ] \hspace{0.15cm}\underline{= 3}\hspace{0.05cm},$$

$${\it \Gamma}_5(S_2) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}{\rm min} \left [{\it \Gamma}_{4}(S_1) + d_{\rm H} \big ((10)\hspace{0.05cm},\hspace{0.05cm} (01) \big )\hspace{0.05cm},\hspace{0.2cm}{\it \Gamma}_{4}(S_3) + d_{\rm H} \big ((01)\hspace{0.05cm},\hspace{0.05cm} (01) \big ) \right ] = {\rm min} \left [ 3+2\hspace{0.05cm},\hspace{0.05cm} 2+0 \right ] \hspace{0.15cm}\underline{= 2}\hspace{0.05cm},$$

$${\it \Gamma}_5(S_3) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}{\rm min} \left [{\it \Gamma}_{4}(S_1) + d_{\rm H} \big ((01)\hspace{0.05cm},\hspace{0.05cm} (01) \big )\hspace{0.05cm},\hspace{0.2cm}{\it \Gamma}_{4}(S_3) + d_{\rm H} \big ((10)\hspace{0.05cm},\hspace{0.05cm} (01) \big ) \right ] ={\rm min} \left [ 3+0\hspace{0.05cm},\hspace{0.05cm} 2+2 \right ] \hspace{0.15cm}\underline{= 3}\hspace{0.05cm}.$$

The left graph shows the final evaluated ${\it \Gamma}_i(S_{\mu})$ trellis.

Evaluated trellis diagrams

(2)  At time $i = 6$ the termination is already effective and there are only two branch metrics left. For these one obtains with $\underline{y}_6 = (01)$:

$${\it \Gamma}_6(S_0) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}{\rm min} \left [{\it \Gamma}_{5}(S_0) + d_{\rm H} \big ((00)\hspace{0.05cm},\hspace{0.05cm} (01) \big )\hspace{0.05cm},\hspace{0.2cm}{\it \Gamma}_{5}(S_2) + d_{\rm H} \big ((11)\hspace{0.05cm},\hspace{0.05cm} (01) \big ) \right ] ={\rm min} \left [ 3+1\hspace{0.05cm},\hspace{0.05cm} 2+1 \right ] \hspace{0.15cm}\underline{= 3}\hspace{0.05cm},$$

$${\it \Gamma}_6(S_2) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}{\rm min} \left [{\it \Gamma}_{5}(S_1) + d_{\rm H} \big ((10)\hspace{0.05cm},\hspace{0.05cm} (01) \big )\hspace{0.05cm},\hspace{0.2cm}{\it \Gamma}_{5}(S_3) + d_{\rm H} \big ((01)\hspace{0.05cm},\hspace{0.05cm} (01) \big ) \right ] ={\rm min} \left [ 3+2\hspace{0.05cm},\hspace{0.05cm} 3+0 \right ] \hspace{0.15cm}\underline{= 3}\hspace{0.05cm}.$$

(3)  The final value results to

$${\it \Gamma}_7(S_0) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}{\rm min} \left [{\it \Gamma}_{6}(S_0) + d_{\rm H} \big ((00)\hspace{0.05cm},\hspace{0.05cm} (11) \big )\hspace{0.05cm},\hspace{0.2cm}{\it \Gamma}_{6}(S_2) + d_{\rm H} \big ((11)\hspace{0.05cm},\hspace{0.05cm} (11) \big ) \right ] ={\rm min} \left [ 3+2\hspace{0.05cm},\hspace{0.05cm} 3+0 \right ] \hspace{0.15cm}\underline{= 3}\hspace{0.05cm}.$$

In the BSC model, one can infer from ${\it \Gamma}_7(S_{\mu}) = 3$ that three transmission errors occurred   ⇒   solutions 1 and 3.

(4)  Correct are statements 1 and 2:

• Maximizing the branch metrics ${\it \Lambda}_i(S_{\mu})$ according to the right sketch in the above graph gives the same result as minimizing the branch metrics ${\it \Gamma}_i(S_{\mu})$ shown on the left. Also, the surviving and deleted branches are identical in both graphs.
• The given equation is also correct, which is shown here only on the example $i = 7$:

$${\it \Lambda}_7(S_0)) = 2 \cdot \big [i - {\it \Gamma}_7(S_0) \big ] = 2 \cdot \big [7 - 3 \big ] \hspace{0.15cm}\underline{= 8}\hspace{0.05cm}.$$

• The last statement is false. Rather applies  $〈x_i', \, y_i〉 ∈ \{–2, \, 0, \, +2\}$.

Hints: In "Exercise 3.11", path finding is demonstrable for the same example, assuming ${\it \Lambda}_i(S_{\mu})$–metrics as shown in the right graph.