Aufgaben:Exercise 1.11: Syndrome Decoding

• Return to book

Syndrome and coset leaders
$($incomplete list$)$  of  $\rm HC \ (7, \, 4, \, 3)$

To decode a  $(7, \, 4, \, 3)$  Hamming code,  which is defined by its parity-check matrix

$${ \boldsymbol{\rm H}}_{\rm } = \begin{pmatrix} 1 &1 &0 &1 &1 &0 &0\\ 0 &1 &1 &1 &0 &1 &0\\ 1 &0 &1 &1 &0 &0 &1 \end{pmatrix} \hspace{0.05cm},$$

"syndrome decoding"  is also suitable.  Since all Hamming codes are perfect,  the result is as good as with the  $($in the general case$)$  more complicated maximum likelihood decoding.

For  "syndrome decoding"  one proceeds as follows:

• From the received vector  $\underline{y}$   one forms the syndrome  $($it holds  $m = n - k)$:

$$\underline{s} = \underline{y} \cdot { \boldsymbol{\rm H}}^{\rm T} \in {\rm GF}(2^m) \hspace{0.05cm}.$$

• In  "BSC channel"  the received word   $\underline{y} = \underline{x} \, ({\rm codeword}) + \underline{e} \, ({\rm error\:vector})$   is also an element of  ${\rm GF}(2^n)$,  and it holds because of   $ \underline{x} · \boldsymbol {{\rm H} }^{\rm T} = \underline{0}$:

$$\underline{s} = \underline{e} \cdot { \boldsymbol{\rm H}}^{\rm T} \hspace{0.05cm}.$$

• Many error patterns  $\underline{e}$  lead to the same syndrome  $\underline{s}$.  One now groups those error patterns with the same syndrome  $\underline{s}_{\mu}$  to the coset  ${\it \Psi}_{\mu}$.

• The coset leader  $\underline{e}_{\mu}$  is the fault vector that has the lowest Hamming weight within the class  ${\it \Psi}_{\mu}$  and is accordingly the most probable.  If there are several of these,  one chooses one of them arbitrarily.

The above graph shows the incomplete list of coset leaders  $\underline{e}_{\mu}$  for each  $\underline{s}_{\mu}$. The most likely error vectors

• $\underline{e}_{3}$  with syndrome  $\underline{s}_{3} = (0, 1, 1)$,
• $\underline{e}_{5}$  with syndrome  $\underline{s}_{5} = (1, 0, 1)$,
• $\underline{e}_{6}$  with syndrome  $\underline{s}_{6} = (1, 1, 0)$,
• $\underline{e}_{7}$  with syndrome  $\underline{s}_{7} = (1, 1, 1)$

are to be determined in the subtasks  (4)  and  (5).

Hints:

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

• Underlying is a Hamming code with parameters  $n = 7$,  $k = 4$   ⇒   $m = 3$.  All code words have the following format:

$$\underline{x} = ( x_1, x_2, x_3, x_4, x_5, x_6, x_7) = ( u_1, u_2, u_3, u_4, p_1, p_2, p_{3}) \hspace{0.05cm}.$$

• The parity-check equations are illustrated on the information sheet for  "Exercise 1.11Z"  which considers the same constellation as in the present exercise.

• In the last subtask  (6),  use the BSC parameter  $ \varepsilon = 0.1$.

Questions

1

How many received words  $(N_{0})$  lead to the syndrome  $\underline{s} = \underline{s}_{0} = (0, 0, 0)$?

$N_{0} \ = \ $

2

How many received words  $(N_{7})$  lead to the syndrome  $\underline{s} = \underline{s}_{7} = (1, 1, 1)$?

$N_{7} \ = \ $

3

What are the characteristics of all coset leaders  $\underline{e}_{\mu}$?

	The last three bits of  $\underline{e}_{\mu}$  are identical to  $\underline{s}_{\mu}$ .
	All  $\underline{e}_{\mu}$  contain a single  "$1$"  each.
	All  $\underline{e}_{\mu}$  include at most one  "$1$".

4

What syndrome  $\underline{s}_{\mu}$  does the error vector  $(1, 0, 0, 0, 0, 0, 0)$  lead to?

$\underline{e} = (1, 0, 0, 0, 0, 0, 0) \text{:} \hspace{0.4cm} {\rm Index} \ \mu \ = \ $

5

Calculate the syndrome  $\underline{s}_{\mu}$  $($Input:  Index $\mu)$  for

$\ \underline{e} = (0, 1, 0, 0, 0, 0, 0) \text{:} \hspace{0.4cm} {\rm Index} \ \mu \ = \ $

$\ \underline{e} = (0, 0, 1, 0, 0, 0, 0) \text{:} \hspace{0.4cm} {\rm Index} \ \mu \ = \ $

$\ \underline{e} = (0, 0, 0, 1, 0, 0, 0) \text{:} \hspace{0.4cm} {\rm Index} \ \mu \ = \ $

6

What is the block error probability for the BSC model with falsification probability  $\varepsilon = 0.1$?

${\rm Pr(block\:error)} \ = \ $ $\ \%$

Solution

Solution

(1)  There are a total of  $2^7 = 128$  different code words  $\underline{x}$  and according to the BSC model also  $2^7$  different received words  $y$  and as many error vectors  $\underline{e}$.

• With  $m = 3$  parity bits,  there are  $2^3 = 8$  distinct values for the syndrome,

$$\underline{s} \hspace{0.05cm} \in \hspace{0.05cm} \{ \underline{s}_0, \underline{s}_1,\hspace{0.05cm} \text{...} \hspace{0.05cm} , \underline{s}_7\} = \{ \underline{s}_{\mu} \}\hspace{0.05cm}, \hspace{0.2cm} \mu = 0, \hspace{0.05cm} \text{...} \hspace{0.05cm} , 7 \hspace{0.05cm},$$

and just as many  "cosets".

• Since in the Hamming code,  which is perfect,  all error vectors belong to one of the eight cosets  ${\it \Psi}_{\mu}$  and,  moreover,  the number of all vectors in all cosets is the same  ("Why should it be different?" Is that enough proof for you?),  we get:

$$ N_0 = \frac{2^n}{2^m} = 2^k \hspace{0.15cm} \underline{= 16} \hspace{0.05cm}.$$

• The cosets  ${\it \Psi}_{0}$  include for example - see sample solution to  "Exercise 1.11Z" - the vectors

$$\underline{e} = (1, 1, 0, 0, 0, 0, 1),$$

$$\underline{e} = (1, 1, 1, 1, 1, 1, 1).$$

(2)  According to the comments of the last partial result,  the following applies equally:  $N_{7} \ \underline{= 16}$.

(3)  Correct is  answer 3:

• The coset leader  $\underline{e}_{\mu}$  is the error vector  $\underline{e}$  with the lowest "Hamming weight"  $w_{\rm H}(\underline{e})$  that leads to the syndrome  $\underline{s}_{\mu}$.

• The (7, 4, 3) Hamming code considered here is perfect.  That is,  all eight coset leaders therefore contain.

• either no  "one"  $(\underline{e}_{0}$   ⇒   no correction is required$)$,  or
• exactly a single  "one"  $(\underline{e}_{1}, \hspace{0.05cm} \text{...} \hspace{0.05cm} , \underline{e}_{7}$   ⇒   an information or parity bit must be corrected$)$.

(4)  It holds  $\underline{s} = \underline{e} · \boldsymbol{\rm H}^{\rm T}$:

$$\underline{s} = \begin{pmatrix} 1 &0 &0 &0 &0 &0 &0 \end{pmatrix} \cdot \begin{pmatrix} 1 &0 &1\\ 1 &1 &0\\ 0 &1 &1\\ 1 &1 &1\\ 1 &0 &0\\ 0 &1 &0\\ 0 &0 &1 \end{pmatrix} = \begin{pmatrix} 1 &0 &1 \end{pmatrix} = \underline{s}_5 \hspace{0.3cm} \Rightarrow\hspace{0.3cm} \hspace{0.15cm} \underline{ \mu= 5} \hspace{0.05cm}.$$

All coset leaders of the  $(7, \, 4, \, 3)$  Hamming code Korrektur

(5)  A comparison with the solution to the last subtask shows that  $(0, 1, 0, 0, 0, 0, 0) \cdot \boldsymbol{\rm H}^{\rm T}$  as a syndrome gives the second row of the transposed matrix:

$$\begin{pmatrix} 0 &1 &0 &0 &0 \hspace{0.2cm}\text{... }\end{pmatrix} \cdot { \boldsymbol{\rm H}}^{\rm T} = \begin{pmatrix} 1 &1 &0 \end{pmatrix} = \underline{s}_6$$

$$\Rightarrow\hspace{0.45cm} \underline{ \mu= 6} \hspace{0.05cm}, \hspace{0.5cm} \underline{e}_6 = \begin{pmatrix} 0 &1 &0 &0 &0 &0 &0 \end{pmatrix}\hspace{0.05cm}.$$

$$\begin{pmatrix} 0 &0 &1 &0 &0 \hspace{0.2cm}\text{... }\end{pmatrix} \cdot { \boldsymbol{\rm H}}^{\rm T} = \begin{pmatrix} 0 &1 &1 \end{pmatrix} = \underline{s}_3$$

$$\Rightarrow\hspace{0.45cm} \underline{ \mu= 3} \hspace{0.05cm}, \hspace{0.5cm} \underline{e}_3 = \begin{pmatrix} 0 &0 &1 &0 &0 &0 &0 \end{pmatrix}\hspace{0.05cm}.$$

$$\begin{pmatrix} 0 &0 &0 &1 &0 \hspace{0.2cm}\text{... } \end{pmatrix} \cdot { \boldsymbol{\rm H}}^{\rm T} = \begin{pmatrix} 1 &1 &1 \end{pmatrix} = \underline{s}_7$$

$$\Rightarrow\hspace{0.45cm} \underline{ \mu= 7} \hspace{0.05cm}, \hspace{0.5cm} \underline{e}_7 = \begin{pmatrix} 0 &0 &0 &1 &0 &0 &0 \end{pmatrix}\hspace{0.05cm}.$$

The adjacent graph summarizes again the results of the subtasks  (4)  and  (5).

(6)  For the  $(7, \, 4, \, 3)$  Hamming code under consideration,  the correct information word is decided if at most one bit within the code word is corrupted during transmission.  It follows:

$${ \rm Pr(block\:errors)} \hspace{-0.15cm}\ = \ \hspace{-0.15cm} { \rm Pr(\ge 2\hspace{0.15cm}bit\hspace{0.15cm}errors\hspace{0.15cm})} =1 - { \rm Pr(0\hspace{0.15cm}bit\hspace{0.15cm}errors)} - { \rm Pr(1\hspace{0.15cm}bit\hspace{0.15cm}errors)} =1 - 0.9^7 - 7 \cdot 0.1 \cdot 0.9^7 \hspace{0.15cm} \underline{ = 0.15} \hspace{0.05cm}.$$

• Uncoded transmission of a block with  $n = k = 4$  bits would result in the same BSC channel:

$${ \rm Pr(block\:errors)}= 1 - 0.9^4 \approx 0.344 \hspace{0.05cm}.$$

• The comparison is not entirely fair,  however,  since with  $n = 4$  a smaller falsification probability  $\varepsilon$  would have to be applied than with  $n = 7$  $($smaller symbol rate   ⇒   smaller bandwidth   ⇒   smaller noise power$)$.