Exercise 2.09: Reed–Solomon Parameters
Adjacent is an incomplete list of possible Reed–Solomon codes known to be based on a Galois field ${\rm GF}(q) = {\rm GF}(2^m)$.
The parameter $m$ specifies with how many bits a Reed–Solomon code symbol is represented. It is valid:
- $m = 4$ (red font),
- $m = 5$ (blue font),
- $m = 6$ (green font).
A Reed–Solomon code is generally denoted as follows: ${\rm RSC}\ (n, \ k, \ d_{\rm min})_q$.
The parameters have the following meaning:
- The parameter $n$ specifies the number of symbols of a code word $\underline{c}$ ⇒ length of the code.
- Tthe parameter $k$ specifies the number of symbols of an information block $\underline{u}$ ⇒ dimension of the code.
- The parameter $d_{\rm min}$ denotes the minimum distance between two code words
$($for all Reed–Solomon codes equal $n-k+1)$. - The parameter $q$ gives an indication of the use of the Galois field ${\rm GF}(q)$.
To the right, there is the binary representation of the same code:
- In this realization of a Reed–Solomon code, each information and code symbol is represented by $m$ bits.
- For example, it can be seen from the first row that the minimum distance in terms of bits is also $d_{\rm min} = 5$ if in ${\rm GF}(2^m)$ the minimum distance is $d_{\rm min} = 5$.
- This code can correct up to $t = 2$ bit errors $($or symbol errors$)$ and detect up to $e = 4$ bit errors $($or symbol errors$)$.
Hints:
- This exercise belongs to the chapter "Definition and Properties of Reed-Solomon Codes".
- However, reference is also made to the chapter "Extension Field".
Questions
Solution
- $$n = q -1 = 2^m -1 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} m = 4 {\rm :}\hspace{0.2cm} n \hspace{0.15cm}\underline {= 15} \hspace{0.05cm}, \hspace{0.4cm}m = 5 {\rm :}\hspace{0.2cm} n \hspace{0.15cm}\underline {= 31} \hspace{0.05cm},\hspace{0.4cm} m = 6 {\rm :}\hspace{0.2cm} n \hspace{0.15cm}\underline {= 63} \hspace{0.05cm}. $$
(2) To be able to correct $t$ symbol errors, the minimum distance must be $d_{\rm min} = 2t + 1$.
- The Reed–Solomon code is a so-called "Maximum Distance Separable $\rm (MDS)$ code".
- For this applies:
- $$d_{\rm min} = n-k+1 = 2t+1 \hspace{0.3cm} \Rightarrow \hspace{0.3cm}k = n -2t = 2^m - ( 2t+1) \hspace{0.05cm}. $$
- This gives for the
- $\rm RSC \ 1$ $($with $m = 4, \ t = 4) \text{:} \hspace{0.2cm} k = 2^4 - (2 \cdot 4 + 1) \ \underline{= 7}$,
- $\rm RSC \ 2$ $($with $m = 5, \ t = 8) \text{:} \hspace{0.2cm} k = 2^5 - (2 \cdot 8 + 1) \ \underline{= 15}$.
(3) The denotation of a Reed–Solomon code is ${\rm RSC} \, (n, \, k, \, d_{\rm min})_q$ with $q = 2^m = n + 1$.
- Correct are the solutions 1 and 4:
- $\rm RSC \ 1 \Rightarrow RSC \, (15, \, 7, \, 9)_{16}$,
- $\rm RSC \ 2 \Rightarrow RSC \, (31, \, 15, \, 17)_{32}$.
(4) If $d_{\rm min}$ denotes the minimum distance of a block code, it can be used to detect $e = d_{\rm min} - 1$ symbol errors and to correct $t = e/2$ symbol errors:
- ${\rm RSC} \ 1 \text{:} \hspace{0.2cm} d_{\rm min} = \ \ 9, \ t = 4, \ \underline{e = 8}$,
- ${\rm RSC} \ 2 \text{:} \hspace{0.2cm} d_{\rm min} = 17, \ t = 8, \ \underline{e = 16}$.
(5) Correct are the two middle solutions 2 and 3:
- For ${\rm RSC} \ 1 \, (m = 4)$: $n = 15$ code symbols from $\rm GF(2^5)$ correspond to $60$ bits and $k = 7$ information symbols are exactly $28$ bits:
- $\rm RSC \ 1 \Rightarrow RSC \, (15, \, 7, \, 9)_{16} \Rightarrow RSC \, (60, \, 28, \, 9)_2$,
- $\rm RSC \ 2 \Rightarrow RSC \, (31, \, 15, \, 17)_{32} \Rightarrow RSC \, (155, \, 75, \, 17)_2$.
- For the minimum distance on bit level ⇒ $\rm GF(2)$, $d_{\rm min} = 9$ resp. $d_{\rm min} = 17$ result in the same values as on symbol level $($see "theory section"$)$.