Aufgaben:Exercise 3.09Z: Viterbi Algorithm again

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Trellis for a rate 1/2 code, memory  $m = 1$

The diagram shows the trellis of the convolutional code according to  "Exercise 3.6", characterized by the following quantities:

• Rate 1/2  ⇒   $k = 1, \ n = 2$,
• memory  $m = 1$,
• transfer function matrix  $\mathbf{G}(D) = (1, \ 1 + D)$,
• length of information sequence:  $L = 4$,
• sequence length including termination:  $L\hspace{0.05cm}' = L + m = 5$.

On the basis of this representation, the Viterbi decoding is to be understood step by step, starting from the following reception sequence:   $\underline{y} = (11, \, 01, \, 01, \, 11, \, 01)$.

Drawn into the trellis are:

• The initial value  ${\it \Gamma}_0(S_0)$  for the Viterbi–algorithm, which is always chosen to  $0$ .
• The two branch metrics for the first decoding step  $(i = 1)$  are obtained with  $\underline{y}_1 = (11)$  as follows:

$${\it \Gamma}_1(S_0) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\it \Gamma}_0(S_0) + d_{\rm H} \big ((00)\hspace{0.05cm},\hspace{0.05cm} (11) \big ) = 2 \hspace{0.05cm},$$

$${\it \Gamma}_1(S_1) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\it \Gamma}_0(S_0) + d_{\rm H} \big ((11)\hspace{0.05cm},\hspace{0.05cm} (11) \big ) = 0 \hspace{0.05cm}.$$

• The branch metrics to step  $i = 2$   ⇒   $\underline{y}_2 = (01)$  are obtained by the following comparisons:

$${\it \Gamma}_2(S_0) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}{\rm min} \left [{\it \Gamma}_{1}(S_0) + d_{\rm H} \big ((00)\hspace{0.05cm},\hspace{0.05cm} (01) \big )\hspace{0.05cm}, \hspace{0.2cm}{\it \Gamma}_{1}(S_1) + d_{\rm H} \big ((01)\hspace{0.05cm},\hspace{0.05cm} (01) \big ) \right ] $$

$$\Rightarrow\hspace{0.3cm} {\it \Gamma}_2(S_0) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm min} \big [ 2+1\hspace{0.05cm},\hspace{0.05cm} 0+0 \big ] = 0\hspace{0.05cm},$$

$${\it \Gamma}_2(S_1) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}{\rm min} \left [{\it \Gamma}_{1}(S_0) + d_{\rm H} \big ((11)\hspace{0.05cm},\hspace{0.05cm} (01) \big )\hspace{0.05cm}, \hspace{0.2cm}{\it \Gamma}_{1}(S_1) + d_{\rm H} \big ((10)\hspace{0.05cm},\hspace{0.05cm} (01) \big ) \right ]$$

$$\Rightarrow\hspace{0.3cm} {\it \Gamma}_2(S_1) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm min} \big [ 2+1\hspace{0.05cm},\hspace{0.05cm} 0+2 \big ] = 2\hspace{0.05cm}.$$

In the same way you are to

• compute the branch metrics at time points  $i = 3, \ i = 4$  and  $i = 5$  (termination), and
• eliminate the less favorable paths to a node  ${\it \Gamma}_i(S_{\mu})$  in each case. In the graph this is indicated by dotted lines for  $i = 2$ .

Then the continuous path from  ${\it \Gamma}_0(S_0)$  to  ${\it \Gamma}_5(S_0)$  is to be found, where the backward direction is recommended.

If one follows the found path in forward direction, one recognizes.

• the most likely code sequence  $\underline{z}$  $($ideally equal  $\underline{x})$  by the labels,
• the most probable information sequence  $\underline{v}$  $($ideally equal  $\underline{u})$  at the colors.

Hints:

• This exercise belongs to the chapter  "Decoding of Convolutional Codes".

Questions

1

Calculate the minimum branch metrics for time  $i = 3$.

${\it \Gamma}_3(S_0) \ = \ $

${\it \Gamma}_3(S_1) \ = \ $

2

Calculate the minimum branch metrics for time  $i = 4$.

${\it \Gamma}_4(S_0) \ = \ $

${\it \Gamma}_4(S_1) \ = \ $

3

Calculate the minimum branch metrics for time  $i = 5$ (end).

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

4

What are the final results of the Viterbi algorithm:

	$\underline{z} = (11, \, 01, \, 00, \, 11, \, 01)$.
	$\underline{z} = (11, \, 01, \, 11, \, 01, \, 00)$.
	$\underline{v} = (1, \, 0, \, 0, \, 1, \, 0)$.
	$\underline{v} = (1, \, 0, \, 1, \, 0, \, 0)$.

5

What decision would have been made without scheduling?

	The same,
	another.

Solution

Solution

(1) 

Trellis with branch metrics

Starting from  ${\it \Gamma}_2(S_0) = 0, \ {\it \Gamma}_2(S_1) = 2$ we get $\underline{y}_3 = (01)$:

$${\it \Gamma}_3(S_0) \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm min} \left [0 + d_{\rm H} \big ((00), (01) \big ), \hspace{0.05cm}2 + d_{\rm H} \big ((01), (01) \big ) \right ] = {\rm min} \left [ 0+1\hspace{0.05cm},\hspace{0.05cm} 2+0 \right ] \hspace{0.15cm}\underline{= 1}\hspace{0.05cm},$$

$${\it \Gamma}_3(S_1) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}{\rm min} \left [0 + d_{\rm H} \big ((11), (01) \big ), \hspace{0.05cm}2 + d_{\rm H} \big ((10), (01) \big ) \right ] {\rm min} \left [ 0+1\hspace{0.05cm},\hspace{0.05cm} 2+2 \right ] \hspace{0.15cm}\underline{= 1}\hspace{0.05cm}.$$

Thus, eliminated are the two subpaths that start from state $S_1$ at time $i = 2$ (i.e., at the third decoding step)   ⇒   Dotted in the graph.

(2)  Analogous to subtask (1), we obtain with  $y_4 = (11)$:

$${\it \Gamma}_4(S_0) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}{\rm min} \left [1 + d_{\rm H} \big ((00), (11) \big ), \hspace{0.05cm}1 + d_{\rm H} \big ((01), (11) \big ) \right ] = {\rm min} \left [ 1+2\hspace{0.05cm},\hspace{0.05cm} 1+1 \right ] \hspace{0.15cm}\underline{= 2}\hspace{0.05cm},$$

$${\it \Gamma}_4(S_1) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}{\rm min} \left [1 + d_{\rm H} \big ((11), (11) \big ), \hspace{0.05cm}1 + d_{\rm H} \big ((10), (11) \big ) \right ] ={\rm min} \left [ 1+0\hspace{0.05cm},\hspace{0.05cm} 1+1 \right ] \hspace{0.15cm}\underline{= 1}\hspace{0.05cm}$$

⇒   Elimination in the fourth decoding step of the two subpaths $S_0 → S_0$ and $S_1 → S_1$.

Path search

(3)  Für $i = 5$   ⇒   "Termination" is obtained with  $\underline{y}_5 = (01)$:

$${\it \Gamma}_5(S_0) \hspace{-0.15cm} \ = \ \hspace{-0.15cm}{\rm min} \left [2 + d_{\rm H} \big ((00), (01) \big ), \hspace{0.05cm}1 + d_{\rm H} \big ((01), (01) \big ) \right ] {\rm min} \left [ 2+1\hspace{0.05cm},\hspace{0.05cm} 1+0 \right ] \hspace{0.15cm}\underline{= 1}\hspace{0.05cm}.$$

To be eliminated here is the subpath $S_0 → S_0$.

(4)  The backward search of the continuous path from ${\it \Gamma}_5(S_0)$ to ${\it \Gamma}_0(S_0)$ yields

$$S_0 ← S_1 ← S_0 ← S_0 ← S_1 ← S_0.$$

In the forward direction, this yields the path $S_0 → S_1 → S_0 → S_0 → S_1 → S_0$ and thus the

• the most likely code sequence $\underline{z} = (11, \, 01, \, 00, \, 11, \, 01)$,
• the most likely information sequence $\underline{v} = (1, \, 0, \, 0, \, 1, \, 0)$.

Thus, the proposed solutions 1 and 3 are correct:

• A comparison with the given received vector $\underline{y} = (11, \, 01, \, 01, \, 11, \, 01)$ shows that the sixth bit was corrupted during transmission.

(5)  Without termination ⇒ final decision at $i = 4$, there would have been two continuous paths:

• from $S_0 → S_1 → S_0 → S_1 → S_0$ (shown in yellow),
• from $S_0 → S_1 → S_0 → S_0 → S_1$ (the ultimately correct path).

The constraint decision at time $i = 4$ would have led here to the second path and thus to the result $\underline{v} = (1, \, 0, \, 0, \, 1)$ because of ${\it \Gamma}_4(S_1) < {\it \Gamma}_4(S_0)$.

• In the considered example, therefore, to the same decision as in subtask (4) with termination bit.
• However, there are many constellations where only the termination bit enables the correct and safe decision.