Exercise 3.12: Trellis Diagram for Two Precursors

Trellis diagram for two precursors

We assume the basic pulse values $g_0\ne 0$ , $g_{\rm –1}\ne 0$ and $g_{\rm –2}\ne 0$ :

This means that the decision on the symbol $a_{\rm \nu}$ is also influenced by the subsequent coefficients $a_{\rm \nu +1}$ and $a_{\rm \nu +2}$ .

Thus, for each time point $\nu$ , exactly eight metrics $\varepsilon_{\rm \nu}$ have to be determined, from which the minimum accumulated metrics ${\it \Gamma}_{\rm \nu}(00)$ , ${\it \Gamma}_{\rm \nu}(01)$ , ${\it \Gamma}_{\rm \nu}(10)$ and ${\it \Gamma}_{\rm \nu}(11)$ can be calculated.

For example, ${\it \Gamma}_{\rm \nu}(01)$ provides information about the symbol $a_{\rm \nu}$ under the assumption that $a_{\rm \nu +1} = 0$ and $a_{\rm \nu +2} = 1$ will be.

Here, the minimum accumulated metric ${\it \Gamma}_{\rm \nu}(01)$ is the smaller value obtained from the comparison of

$\big[{\it \Gamma}_{\nu-1}(00) + \varepsilon_{\nu}(001)\big] \hspace{0.15cm}{\rm and} \hspace{0.15cm}\big[{\it \Gamma}_{\nu-1}(10) + \varepsilon_{\nu}(101)\big].$

To calculate the minimum accumulated metric ${\it \Gamma}_2(10)$ in subtasks (1) and (2), assume the following numerical values:

unipolar amplitude coefficients: $a_{\rm \nu} ∈ \{0, 1\}$ ,

basic pulse values $g_0 = 0.5$ , $g_{\rm –1} = 0.3$ , $g_{\rm –2} = 0.2$ ,

applied noisy detection sample: $d_2 = 0.2$ ,

minimum accumulated metric at time $\nu = 1$ :

${\it \Gamma}_{1}(00) = 0.0,\hspace{0.2cm}{\it \Gamma}_{1}(01) = 0.2, \hspace{0.2cm} {\it \Gamma}_{1}(10) = 0.6,\hspace{0.2cm}{\it \Gamma}_{1}(11) = 1.2 \hspace{0.05cm}.$

The graph shows the simplified trellis diagram for time points $\nu = 1$ to $\nu = 8$ .

Blue branches come from either ${\it \Gamma}_{\rm \nu –1}(00)$ or ${\it \Gamma}_{\rm \nu –1}(01)$ and denote a hypothetical " $0$ ".

In contrast, all red branches – starting from the ${\it \Gamma}_{\rm \nu –1}(10)$ or ${\it \Gamma}_{\rm \nu –1}(11)$ states – indicate the symbol " $1$ ".

Notes:

The exercise belongs to the chapter "Viterbi Receiver".

All quantities here are to be understood normalized.

Also, assume unipolar and equal probability amplitude coefficients: ${\rm Pr} (a_\nu = 0) = {\rm Pr} (a_\nu = 1)= 0.5.$

Questions

Solution

(1) The first metric is calculated as follows:

$\varepsilon_{2}(010) = [d_0 - 0 \cdot g_0 - 1 \cdot g_{-1}- 0 \cdot g_{-2}]^2= [0.2 -0.3]^2\hspace{0.15cm}\underline {=0.01} \hspace{0.05cm}.$

Correspondingly, for the other metrics:

$\varepsilon_{2}(011) \ = \ [0.2 -0.3- 0.2]^2\hspace{0.15cm}\underline {=0.09}\hspace{0.05cm},$

$\varepsilon_{2}(110) \ = \ [0.2 -0.5- 0.3]^2\hspace{0.15cm}\underline {=0.36}\hspace{0.05cm},$

$\varepsilon_{2}(111) \ = \ [0.2 -0.5- 0.3-0.2]^2\hspace{0.15cm}\underline {=0.64} \hspace{0.05cm}.$

(2) The task is to find the minimum value of each of two comparison values:

${\it \Gamma}_{2}(10) \ = \ {\rm Min}\left[{\it \Gamma}_{1}(01) + \varepsilon_{2}(010), \hspace{0.2cm}{\it \Gamma}_{1}(11) + \varepsilon_{2}(110)\right] = {\rm Min}\left[0.2+ 0.01, 1.2 + 0.36\right]\hspace{0.15cm}\underline {= 0.21} \hspace{0.05cm},$

${\it \Gamma}_{2}(11) \ = \ {\rm Min}\left[{\it \Gamma}_{1}(01) + \varepsilon_{2}(011), \hspace{0.2cm}{\it \Gamma}_{1}(11) + \varepsilon_{2}(111)\right] = {\rm Min}\left[0.2+ 0.09, 1.2 + 0.64\right]\hspace{0.15cm}\underline {= 0.29} \hspace{0.05cm}.$

(3) The first and last solutions are correct:

The sequence " $1011010$ " can be recognized from the continuous path: "red – blue – red – red – blue – red – blue".

On the other hand, no final statement can be made about the symbol $a_8$ at time $\nu = 8$ :

Only under the hypothesis $a_9 = 1$ and $a_{\rm 10} = 1$ one would decide for $a_8 = 0$ , under other hypotheses for $a_8 = 1$ .

	The first seven symbols are " $1011010$ ".
	The first seven symbols are " $1101101$ ".
	The last symbol $a_8 = 1$ is safe.
	No definite statement can be made about the symbol $a_8$ .