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

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).

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

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

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

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

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

Encoding Example 0 1 0 1 1 0 1 1 1 α+ α3X+ α5X2 α α3 α5 Binary Information Message 0 1 0 1 1 0 1 1 1 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=(1 0 0 0 0 1 0 1 1 1 0 1 0 1 0 1 1 0 1 1 1)

Encoding Example 0 1 0 1 1 0 1 1 1 α+ α3X+ α5X2 α α3 α5 Binary Information Message 0 1 0 1 1 0 1 1 1 Symbol Mapping Non-Binary Information Message α α3 α5 α+ α3X+ α5X2 Assume u=(α α3 α5) Input Register Contents 0 0 0 0 (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)

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