Aufgaben:Exercise 5.1: Error Distance Distribution

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Error distance distributions

Any digital channel model can be described in the same way by

• the error sequence  $〈e_{\rm \nu}〉$, and
• the error distance sequence  $〈a_{\rm \nu \hspace{0.05cm}'}〉$.

As an example, we consider the sequences:

$$<\hspace{-0.1cm}e_{\nu} \hspace{-0.1cm}> \ = \ < \hspace{-0.1cm}0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, \text{...} \hspace{-0.1cm}> \hspace{0.05cm},$$

$$< \hspace{-0.1cm}a_{\nu\hspace{0.05cm} '} \hspace{-0.15cm}> \ = \ <\hspace{-0.1cm}2, 3, 1, 4, 2, 5, 1, 1, 3, 4, 1, 2, \text{...} \hspace{-0.1cm}> \hspace{0.05cm}.$$

One can see from this, for example:

• The error distance  $a_2 = 3$  means that there are two error-free symbols between the first and the second error.
• In contrast,  $a_3 = 1$  indicates that the second error is immediately followed by a third.

The different indices  $(\nu$  and  $\nu\hspace{0.05cm} '$, each starting with  $1$)  are necessary because there is no synchrony between the error distance sequence and the error sequence.

In the graph, for two different models  $M_1$  and  $M_2$,  the error distance distribution (EDD) is given as

$$V_a(k) = {\rm Pr}(a \ge k) = 1 - \sum_{\kappa = 1}^{k} {\rm Pr}(a = \kappa)\hspace{0.05cm}$$

This table is to be evaluated in this exercise.

Note:

• The exercise belongs to the chapter  "Parameters of Digital Channel Models".

Questions

1

What are the following error values  $(0$  or  $1)$?

$e_{\rm 16} \ = \ $

$e_{\rm 17} \ = \ $

$e_{\rm 18} \ = \ $

2

What is the value of  $V_a(k = 1)$ for both models?

$V_a(k = 1) \ = \ $

3

For model  $M_1$,  determine the probabilities of the error distances.

${\rm Pr}(a = 1) \ = \ $

${\rm Pr}(a = 2) \ = \ $

${\rm Pr}(a = 3) \ = \ $

${\rm Pr}(a = 4) \ = \ $

${\rm Pr}(a = 5) \ = \ $

4

What is the maximum possible error distance for model  $M_1$?

$k_{\rm max} \ = \ $

5

Calculate the mean error distance for model  $M_1$. 

${\rm E}\big[a \big] \ = \ $

6

For model  $M_1$,  what is the mean error probability  $p_{\rm M} = {\rm E}[e]$?

$p_{\rm M} \ = \ $

7

Which statements are true for the model  $M_2$  with certainty?

	Two errors cannot directly follow each other.
	The most frequent error distance is  $a = 6$.
	The average error probability is  $p_{\rm M} = 0.25$.

Solution

Solution

(1)  Evaluation of the error distance sequence indicates errors at $\nu = 2, 5, 6, 10, 12, 17, 18, 19, 22, 26, 27$ and $29$.

• It follows:   $e_{\rm 16} \ \underline {= 0}$,     $e_{\rm 17} \ \underline {= 1}$,     $e_{\rm 18} \ \underline {= 1}$.

(2)  From the definition equation follows already

$$V_a(k = 1) = {\rm Pr}(a \ge 1)\hspace{0.15cm}\underline {= 1} \hspace{0.05cm}.$$

(3)  ${\rm Pr}(a = k) = V_a(k) \, –V_a(k+1)$ holds. From this we obtain for the individual probabilities:

$${\rm Pr}(a = 1)\hspace{-0.1cm} \ = \ \hspace{-0.1cm}V_a(1) - V_a(2) = 1 - 0.7\hspace{0.15cm}\underline {= 0.3}\hspace{0.05cm},$$

$${\rm Pr}(a = 2)\hspace{-0.1cm} \ = \ \hspace{-0.1cm}V_a(2) - V_a(3) = 0.7 - 0.45 \hspace{0.15cm}\underline {= 0.25}\hspace{0.05cm},$$

$${\rm Pr}(a = 3)\hspace{-0.1cm} \ = \ \hspace{-0.1cm}V_a(3) - V_a(4) = 0.45 - 0.25 \hspace{0.15cm}\underline {= 0.2}\hspace{0.05cm},$$

$${\rm Pr}(a = 4)\hspace{-0.1cm} \ = \ \hspace{-0.1cm}V_a(4) - V_a(5) = 0.25 - 0.10 \hspace{0.15cm}\underline {= 0.15}\hspace{0.05cm},$$

$${\rm Pr}(a = 5)\hspace{-0.1cm} \ = \ \hspace{-0.1cm}V_a(5) - V_a(6) = 0.10 - 0 \hspace{0.15cm}\underline {= 0.10}\hspace{0.05cm}.$$

(4)  From $V_a(k=6) = {\rm Pr}(a ≥ 6) = 0$, it follows directly for the maximum error distance $k_{\rm max} \ \underline {= 5}$.

(5)  Using the probabilities calculated in (3), the expected value we are looking for is:

$${\rm E}\big[a \big] = \sum_{k = 1}^{5} k \cdot {\rm Pr}(a = k) = 1 \cdot 0.3 +2 \cdot 0.25 +3 \cdot 0.2 +4 \cdot 0.15 +5 \cdot 0.1\hspace{0.15cm}\underline { = 2.5} \hspace{0.05cm}.$$

(6)  The mean error probability is the inverse of the mean error distance:   $p_{\rm M} \ \underline {= 0.4}$.

(7)  With certainty, only statement 1 is true:

• The first statement is true because ${\rm Pr}(a = 1) = V_a(1) \, – V_a(2) = 0$.
• The second statement is not certain because $V_a(6)$ gives only the sum of the probabilities ${\rm Pr}(a ≥ 6)$, but not ${\rm Pr}(a = 6)$ alone.
• Only with the additional specification $V_a(7) = 0$ would statement 2 be true.
• Likewise, for the expected value ${\rm E}[a]$, no definite statement is possible due to missing information. With $V_a(7) = 0$ the result would be:

$${\rm E}[a] = 2 \cdot 0.1 +3 \cdot 0.2 +4 \cdot 0.2 +5 \cdot 0.2 +6 \cdot 0.3= 4.4$$

• Without this specification, only the statement ${\rm E}[a] ≥ 4.4$ is possible. But this means that the condition $p_{\rm M} < 1/4.4 < 0.227$ is valid for the mean error probability.
• The statement 3 is therefore also not true with certainty.