1. V. Non-Binary Codes: Introduction to Reed Solomon Codes

2. Reed Solomon Codes Reed Solomon Codes are: Linear Block Codes
Cyclic Codes
Non-Binary Codes (Symbols are made up of m-bit sequences)
Used in channel coding for a wide range of applications, including
Storage devices (tapes, compact disks, DVDs, bar-codes)
Wireless communication (cellular telephones, microwave links, satellites, …)
Digital television
High-speed modems (ADSL, xDSL).

3. Reed Solomon Codes
For any positive integer m≥3, there exists a non-binary Reed Solomon code such that:
Code Length: n = 2m-1
No. of information symbols: k = 2m-1-2t
No. of parity check symbols: n-k = 2t
Symbol Error correcting capability: t

4. Generator Polynomial
Example (7,3) double-symbol correcting Reed-Solomon Code over GF(23)

5. Encoding in Systematic Form
Three Steps:
Multiply the non-binary message polynomial u(X) by Xn-k
Dividing Xn-ku(X) by g(X) to obtain the remainder b(X)
Forming the codeword b(X)+Xn-ku(X)
Encoding Circuit is a Division Circuit
Gate g0 g1 g2
gn-k-1 + + .. b1 b2 + b0
bn-k-1 +
Xn-ku(X)
Codeword
Information Symbols
Parity Check Symbols

6. Encoding Circuit Example: Implementation
Encoding Circuit of (7,3) RS Cyclic Code with
g(X)=α3+α1X+ α0X2+ α3X3+X4
Gate α3 α1 α0 α3 x x x x + + + b0 b1 b2 b3 +
Xn-ku(X)
Codeword
Information Symbols
Parity Check Symbols

7. Bits to Symbols Mapping
Transmitter Side
Receiver Side
Received Non-Binary Data
Decoded Non-Binary Data
Decoded binary Data
Binary Data
Non-Binary Data
Encoded Non-Binary Data
Symbol
Mapping
Channel
Encoder
Channel
Decoder
Bit
Mapping
Information message is usually binary
A symbol mapping is usually necessary for the encoding process
Remember the polynomial representation of symbols in GF(2m)
Symbol
ai(α)
Mapping
0 0 0 α0
1 0 0 α1
α 1
0 1 0 α2
0 0 1 α3
1+α
1 1 0 α4
α +α2
0 1 1 α5
1+α +α2
1 1 1 α6
1 +α2
1 0 1
The coefficients of the polynomial representation may be used for the mapping

8. Encoding Example 0 1 0 1 1 0 1 1 1 α+ α3X+ α5X2 α α3 α5
Binary Information
Message
Symbol
Mapping
Non-Binary
Information Message α α3 α5
α+ α3X+ α5X2
Encoding Mathematics:
X4u(X)= αX4+ α3X5+ α5X6.
Dividing by g(X). The remainder b(X)= α0+ α2X+ α4X2+ α6X3
v(X)=b(X)+X4u(X)= α0+ α2X+ α4X2+ α6X3+ αX4+ α3X5+ α5X6
V=(α0 α2 α4 α6 α α3 α5)
In Binary: V=( )

9. Encoding Example 0 1 0 1 1 0 1 1 1 α+ α3X+ α5X2 α α3 α5
Binary Information
Message
Symbol
Mapping
Non-Binary
Information Message α α3 α5
α+ α3X+ α5X2
Assume u=(α α3 α5)
Input
Register Contents
(Initial State) α5
α α6 α5 α (1st Shift) α3
α3 0 α2 α2 (2nd Shift) α
α0 α2 α4 α6 (3rd Shift)
α α3 α5
V=(α0 α2 α4 α6 α α3 α5)

10. Syndrome Computation Valid Codewords V(X) are divisible by g(X)
α, α2, α3, …, α2t are roots of g(X)
Example in (7,3) RS Code:
r(X) = α0+ α2X+ α4X2+ α0X3+ α6X4+ α3X5+ α5X6
r is not a codeword