Aufgaben:Exercise 3.11Z: Metric and Accumutated Metric

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Calculation of the minimum total error quantities

For the maximum likelihood constellation with bipolar amplitude coefficient   $a_{\rm \nu} ∈ \{+1, –1\}$  discussed in   "Exercise 3.11",  behandelte Maximum–Likelihood–Konstellation mit bipolaren Amplitudenkoeffizient  $a_{\rm \nu} ∈ \{+1, –1\}$  the error quantities   $\varepsilon_{\rm \nu}(i)$  and the minimum total error quantities   ${\it \Gamma}_{\rm \nu}(–1)$ and ${\it \Gamma}_{\rm \nu}(+1)$  are to be determined.

The basic pulse is given by the two values   $g_0$  and  $g_{\rm –1}$.  These, as well as the detection samples   $d_0$  and  $d_1$,  can be taken from the following calculations for the error quantities   $\varepsilon_{\rm \nu}(i)$  at the times   $\nu = 0$  and  $\nu = 1$.  Note that the symbol  $a_0 = 0$  is always sent before the actual message   $(a_1$, $a_2$, $a_3)$. 

For the time  $\nu = 0$  holds:

$$\varepsilon_{0}(+1) \ = \ \big[-0.4- 0.4\big]^2=0.64 \hspace{0.05cm},$$

$$\varepsilon_{0}(-1) \ = \ \big[-0.4+ 0.4\big]^2=0.00 \hspace{0.05cm}.$$

From this it could be concluded already at time   $\nu = 0$  that with high probability   $a_1 = -\hspace{-0.05cm}1$. 

For time   $\nu = 1$,  the following error quantities result:

$$\varepsilon_{1}(+1, +1) \ = \ \big[-0.8- 0.6 -0.4\big]^2=3.24 \hspace{0.05cm},$$

$$\varepsilon_{1}(+1, -1) \ = \ \big[-0.8- 0.6 +0.4\big]^2=1.00 \hspace{0.05cm},$$

$$\varepsilon_{1}(-1, +1) \ = \ \big[-0.8+ 0.6 -0.4\big]^2=0.36 \hspace{0.05cm},$$

$$ \varepsilon_{1}(-1, -1) \ = \ \big[-0.8+ 0.6 +0.4\big]^2=0.04 \hspace{0.05cm}.$$

The minimum total error quantities   ${\it \Gamma}_{\rm \nu}(-\hspace{-0.07cm}1)$  and  ${\it \Gamma}_{\rm \nu}(+1)$ that can be calculated with these six error quantities are already plotted in the graph. The other detection samples are   $d_{2}=0.1 \hspace{0.05cm},\hspace{0.1cm} d_{3}=0.5 \hspace{0.05cm}.$

Notes:

• The exercise belongs to the chapter   "Viterbi Receiver".

• All quantities are to be understood normalized here.
• Also assume bipolar and equal probability amplitude coefficients:   ${\rm Pr} (a_\nu = -\hspace{-0.05cm}1) = {\rm Pr} (a_\nu = +1)= 0.5.$
• The topic is also covered in the interactive applet   "Properties of the Viterbi Receiver".

Questions

1

What detection samples   $d_0$  and  $d_1$  were assumed here?

$d_0 \ = \ $

$d_1\ = \ $

2

Which basic pulse values were assumed here?

$g_0\ = \ $

$g_{-1} \ = \ $

3

Which of the listed detection samples are possible for   $\nu ≥ 1$? 

	$±0.2,$
	$±0.4,$
	$±0.6,$
	$±1.0.$

4

Give the minimum total error quantities for time   $\nu = 2$   $(d_2 = 0.1)$.

${\it \Gamma}_2(+1)\ = \ $

${\it \Gamma}_2(-\hspace{-0.05cm}1)\ = \ $

5

Calculate the minimum total error quantities for time   $\nu = 3$  $(d_3 = 0.5)$.

${\it \Gamma}_3(+1) \ = \ $

${\it \Gamma}_3(-\hspace{-0.05cm}1) \ = \ $

Solution

Solution

(1)  From the equations on the information section one can see $d_0 = \underline{–0.4}$ and $d_1 = \underline {–0.8}$.

(2)  The error quantities $\varepsilon_0(i)$ include the basic pulse value $g_{\rm –1}$, which is used to establish the relationship between the amplitude coefficient $a_1$ and the detection sample $d_0$ ($g_0$ is not included in these equations).

• One can see $g_{\rm –1}\ \underline {= 0.4}$.
• From the equations for $\nu = 1$, the main value $g_0 \ \underline {= 0.6}$ can be read.

(3)  The correct solutions are 1 and 4:

• The possible useful samples are $\pm g_0 \pm g_{\rm –1} = \pm 0.6 \pm0.4$, i.e. $\underline {±0.2}$ and $\underline {±1.0}$.
• In contrast, unipolar signaling   ⇒   $a_\nu \in \{0, \hspace{0.05cm} 1\}$ would result in values of $0, \ 0.4, \ 0.6$ and $1$.
• The relationship between bipolar values $b_i$ and unipolar equivalents $u_i$ is generally:   $b_i = 2 \cdot u_i - 1 \hspace{0.05cm}.$

(4)  The error quantities are obtained for $\nu = 2$ considering the result from (3) as follows:

$$\varepsilon_{2}(+1, +1) \ = \ [0.1 - 1.0]^2=0.81,\hspace{0.2cm} \varepsilon_{2}(-1, +1) = [0.1 +0.2]^2=0.09 \hspace{0.05cm},$$

$$\varepsilon_{2}(+1, -1) \ = \ [0.1 -0.2]^2=0.01,\hspace{0.2cm} \varepsilon_{2}(-1, -1) = [0.1 +1.0]^2=1.21 \hspace{0.05cm}.$$

Thus, the minimum total error quantities are:

$${\it \Gamma}_{2}(+1) \ = \ {\rm Min}\left[{\it \Gamma}_{1}(+1) + \varepsilon_{2}(+1, +1), \hspace{0.2cm}{\it \Gamma}_{1}(-1) + \varepsilon_{2}(-1, +1)\right] = {\rm Min}\left[0.36 + 0.81, 0.04 + 0.09\right]\hspace{0.15cm}\underline {= 0.13} \hspace{0.05cm},$$

$${\it \Gamma}_{2}(-1) \ = \ {\rm Min}\left[{\it \Gamma}_{1}(+1) + \varepsilon_{2}(+1, -1), \hspace{0.2cm}{\it \Gamma}_{1}(-1) + \varepsilon_{2}(-1, -1)\right] = {\rm Min}\left[0.36 + 0.01, 0.04 + 1.21\right]\hspace{0.15cm}\underline {= 0.37} \hspace{0.05cm}.$$

Calculation of the minimum total error quantities

In the adjacent trellis diagram, the state "$1$" is to be interpreted as "$+1$" and "$0$" as "$–1$".

Then holds:

• ${\it \Gamma}_2(+1) = 0.13$ is the minimum total error quantity under the hypothesis that the following symbol will be $a_3 = +1$.
• Under this assumption, $a_2 = \ –1$ is more likely than $a_2 = +1$, as shown in the trellis diagram (the incoming path is blue).
• A realistic alternative to the combination "$a_2 = \ –1, a_3 = +1$" is "$a_2 = +1, a_3 = \ –1$", which lead to the minimum total error quantity ${\it \Gamma}_2(–1) = 0.37$. Here, the incoming path is red.

(5)  For time $\nu = 3$, the following equations hold:

$$\varepsilon_{3}(+1, +1) \ = \ [0.5 - 1.0]^2=0.25,\hspace{0.2cm} \varepsilon_{3}(-1, +1) = [0.5 +0.2]^2=0.49 \hspace{0.05cm},$$

$$\varepsilon_{3}(+1, -1) \ = \ [0.5 -0.2]^2=0.09,\hspace{0.2cm} \varepsilon_{3}(-1, -1) = [0.5 +1.0]^2=2.25 \hspace{0.05cm}.$$

$$\Rightarrow \hspace{0.3cm}{\it \Gamma}_{3}(+1) \ = \ {\rm Min}\left[0.13 + 0.25, 0.37 + 0.49\right]\hspace{0.15cm}\underline {= 0.38} \hspace{0.05cm},\hspace{0.8cm} {\it \Gamma}_{3}(-1) \ = \ {\rm Min}\left[0.13 + 0.09, 0.37 + 2.25\right]\hspace{0.15cm}\underline {= 0.22} \hspace{0.05cm}.$$

• In both equations, the first term in each case is the smaller, with ${\it \Gamma}_2(+1) = 0.13$ included in each case.
• Therefore, the Viterbi receiver will certainly output $a_3 = +1$, no matter what information it will still get at later times ($\nu > 3$).

If we follow the continuous path in the trellis diagram, the other amplitude coefficients are also fixed by fixing $a_3 = +1$:

$$a_1 = a_2 = \ –1.$$