Exercise 1.5: SPC (5, 4) and BEC Model

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Code words of the  $\rm SPC \ (5, 4)$

For this exercise it is required:

• The  Single Parity-check Code with parameters  $k = 4$  and  $n = 5$    ⇒   $\rm SPC \ (5, 4)$  adds to the information bits  $u_{1}$, ... ,  $u_{4}$  adds a check bit  $p$  so that an even number of ones occurs in each codeword  $\underline{x}$ :

$$x_1 \oplus x_2 \oplus x_3 \oplus x_4 \oplus x_5 = 0 \hspace{0.05cm},$$

$$ u_1 \oplus u_2 \oplus u_3 \oplus u_4 \oplus p = 0 \hspace{0.05cm}.$$

• The  Binary Erasure Channel  (BEC) - with binary input values  $x_{i} \in \{0, \ 1\}$  and ternary output  $y_{i} \in \{0, 1, \rm E\}$  leads with probability  $\lambda = 0.1$  to an extinction (English:   Erasure), abbreviated as  $\rm E$.
• Furthermore, ${\rm Pr}(y_{i} = x_{i}) = 1 - \lambda = 0.9$ holds. A real transmission error is excluded:

$$ {\rm Pr} \big[(x_i = 0)\cap (y_i = 1)\big] = {\rm Pr} \big[(x_i = 1)\cap (y_i = 0)\big] = 0\hspace{0.05cm}.$$

The relationship between the information word  $\underline{u}$  and the codeword  $\underline{x}$  is given by the table. From the receive word  $\underline{y}$  the vector  $\underline{v}$  of information bits at the sink is formed by maximum likelihood decision, which should match the information word  $\underline{u}$  as much as possible.

The following nomenclature applies:

$$\underline{u} \ \in \ \{\underline{u}_0, \underline{u}_1,\hspace{0.15cm} \text{...} \hspace{0.2cm}, \underline{u}_{15}\} \hspace{0.05cm},$$

$$ \underline{v} \ \in \ \{\underline{v}_0, \underline{v}_1, \hspace{0.15cm}\text{...} \hspace{0.2cm}, \underline{v}_{15}, \underline{\rm E}\} \hspace{0.05cm}.$$

The result  $\underline{v} =\underline{\rm E} = {\rm (E, E, E, E)}$  indicates that decoding of the code word is not possible due to too many erasures.

Hints:

• This exercise belongs to the chapter  Examples of binary block codes.
• Reference is also made to the chapter  Channel Models and Decision Structures.
• The check bits of  $u_{0}$,  $u_{4}$  and  $u_{13}$  are to be determined in the subtask (1)'.

Questions

1

What is the check bit  $p$ for each of the following information words  $\underline{u}$  ?

$\underline{u} = \underline{u_{0}}\text{:}\hspace{0.4cm}p \ = \ $

$\underline{u} = \underline{u_{4}}\text{:}\hspace{0.4cm}p \ = \ $

$\underline{u} = \underline{u_{13}}\text{:}\hspace{0.25cm}p \ = \ $

2

Let  $ \underline{y} = (0, 0, 0, {\rm E})$. What information word was sent?

	$ \underline{u}_{0}$,
	$ \underline{u}_{4}$,
	$ \underline{u}_{13}$.

3

let  $ \underline{y} = (0, {\rm E}, 0, 0, 1)$. What information word was sent?

	$\underline{u}_{0}$,
	$\underline{u}_{4}$,
	$\underline{u}_{13}$.

4

With what probability does  $\underline{y}$  match the codeword  $\underline{x}$ ?

$\ {\rm Pr} (\underline{y} = \underline{x}) \ = \ $ $\ \%$

5

With what probability do the two vectors  $\underline{u}$  and  $\underline{v}$  match?

$\ {\rm Pr} (\underline{v} = \underline{u}) \ = \ $ $\ \%$

6

What is the probability of a detected error?

$\ {\rm Pr} (\underline{\upsilon} = {\rm {\underline{ E}}}) \ = \ $ $\ \%$

Solution

Solution

(1)  Das Prüfbit $p$ wird beim Single Parity–check Code so bestimmt, dass die Summe aller Einsen im Codewort $\underline{x} = (u_{1}, u_{2}, ... , u_{4}, p)$ geradzahlig ist.
Beispielsweise erhält man:

$$\underline{u}_0 \hspace{-0.1cm}\ = \ \hspace{-0.1cm} (0, 0, 0, 0) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \underline{x}_0 = (0, 0, 0, 0, 0)\hspace{0.3cm} \Rightarrow \hspace{0.3cm} p \hspace{0.15cm} \underline{= 0} \hspace{0.05cm},$$

$$ \underline{u}_4 \hspace{-0.1cm}\ = \ \hspace{-0.1cm} (0, 1, 0, 0) \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \underline{x}_4 = (0, 1, 0, 0, 1)\hspace{0.3cm} \Rightarrow \hspace{0.3cm} p \hspace{0.15cm} \underline{= 1} \hspace{0.05cm},$$

$$\underline{u}_{13} \hspace{-0.1cm}\ = \ \hspace{-0.1cm} (1, 1, 0, 1) \hspace{0.15cm} \Rightarrow \hspace{0.15cm} \underline{x}_{13} = (1, 1, 0, 1, 1)\hspace{0.3cm} \Rightarrow \hspace{0.3cm} p \hspace{0.15cm} \underline{= 1} \hspace{0.05cm}.$$

(2)  Richtig ist die Antwort 1:

• Because of the fact that the number of ones must be even, the erased check bit $p = 0$. Thus, $\underline{u}_{0}$ was sent.

(3)  Correct is answer 2:

• According to the same considerations as in the last subtask, we arrive at $\underline{y} = (0, {\rm E}, 0, 0, 1)$ to the result.

$$\underline{x} = \underline{x}_{4} = (0, 1, 0, 0, 1) ⇒ \underline{u}_{4} = (0, 1, 0, 0).$$

(4)  The event $\underline{y} = \underline{x}$ occurs only if none of the $n = 5$ code bits is cleared by the BEC channel:

$${\rm Pr}(\underline{y} = \underline{x}) = (1 - \lambda)^5 = 0.9^5 \hspace{0.15cm} \underline{= 59.1\%} \hspace{0.05cm}.$$

(5)  The event $v = u$ then occurs,

• if all code bits are transmitted correctly   ⇒   ${\rm Pr}(\underline{y} = \underline{x})$,
• but also when only one codebit is erased. According to the binomial distribution, there are 5 possibilities for this:

$${\rm Pr}(\underline{v} = \underline{u}) \hspace{-0.1cm}\ = \ \hspace{-0.1cm} {\rm Pr}(\underline{y} = \underline{x}) + 5 \cdot (1 - \lambda)^4 \cdot \lambda = 0.591 + 5 \cdot 0.656^4 \cdot 0.1 \hspace{0.15cm} \underline{= 91.9 \%} \hspace{0.05cm}.$$

(6)  Due to the BEC model, the corruption of a codeword $\underline{x}$ is per se impossible, since none of the bits from $0 → 1$ or from $1 → 0$ can be corrupted. Rather:

$${\rm Pr}(\underline{v} = \underline{u}) + {\rm Pr}(\underline{v} = {\rm\underline{ E}}) = 1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm Pr}(\underline{v} = {\rm\underline{ E}}) = 1 - {\rm Pr}(\underline{v} = \underline{u}) \hspace{0.15cm} \underline{= 8.1\%} \hspace{0.05cm}.$$