HAMMING CODE PPT

8 Bit Hamming Hammin g Code Correction And Detection

Presented by TE-1 -Shoubhik Pande -Vijay Nair -Viral Nagda -Pritesh Varsolkar

39 44 45 50

Error Detection and Correction 



Improve the performance of communication network  A simple error detection scheme  



a parity bit a single bit error can be detected, but cannot be corrected

 An error-correction code  



generates multiple parity check bits the check bits generate a unique pattern, called a syndrome the specific bit in error can be identified

Hamming Code 

k  parity

bits are added to an n -bit data word to form a new word of  n +k  bits  





Criteria: (2k  – 1 n  + k ) The bit positions are numbered in sequence from 1 to n  + k  Those positions numbered as a power of 2 are reserved for the parity bits The remaining bits are the data bits

Example: 8-bit data word 11000100 

Include 4 parity bits and the 8-bit word



12 bits

2k  1  n  + k , n  = 8  k  = 4 Bit position: 1 2 3 4 5 6 7 8 9 10 11 12 P1 P2 1 P4 1 0 0 P8 0 1 0 0  – 



Calculate the parity bits: even parity    assumption P1 = XOR of bits (3, 5, 7, 9, 11) = 1  1  0  0  0 = 0 P2 = XOR of bits (3, 6, 7, 10, 11) = 1  0  0  1  0 = 0 P4 = XOR of bits (5, 6, 7, 12) = 1  0  0  0 = 1 P8 = XOR of bits (9, 10, 11, 12) = 0  1  0  0 = 1



Store the 12-bit composite word in memory. Bit position: 1 2 3 4 5 6 7 8 9 10 11 12 0 0 1 1 1 0 0 1 0 1 0 0

How to Check? 

When the 12 bits are read from the memory 



Check bits are calculated C1 = XOR of bits (1, 3, 5, 7, 9, 11)=XOR of bits(1, P1) C2 = XOR of bits (2, 3, 6, 7, 10, 11)=XOR of bits(2, P2) C4 = XOR of bits (4, 5, 6, 7, 12)=XOR of bits(4, P4) C8 = XOR of bits (8, 9, 10, 11, 12)=XOR of bits(8, P8)

If no error has occurred Bit position: 1 2 3 4 5 6 7 8 9 10 11 12 0 0 1 1 1 0 0 1 0 1 0 0  C  =

C8C4C2C1 = 0000

When Error Happen !! 

One-bit error  

error in bit 1     



error in bit 5 



C1 = XOR of bits (1, 3, 5, 7, 9, 11) = 1 C2 = XOR of bits (2, 3, 6, 7, 10, 11) = 0 C4 = XOR of bits (4, 5, 6, 7, 12) = 0 C8 = XOR of bits (8, 9, 10, 11, 12) = 0 C8C4C2C1 = 0001 C8C4C2C1 = 0101

Two-bit error  

errors in bits 1 and 5 

C8C4C2C1 = 0100  can not detect !!



12-bit Hamming code word containing 8 bits of data and 4 parity bits is read from memory. 



What was the original 8-bit data word that was written into memory if the 12-bit word read out is 000011101010?

 Answer  Bit position: 1 2 3 4 5 6 7 8 9 10 11 12 P1 P2 0 P4 1 1 1 P8 1 0 1 0    



C1 = XOR of bits (1, 3, 5, 7, 9, 11) = 0 C2 = XOR of bits (2, 3, 6, 7, 10, 11) = 1 C4 = XOR of bits (4, 5, 6, 7, 12) = 1 C8 = XOR of bits (8, 9, 10, 11, 12) = 0

C=0110  Error in bit 6  correct 8-bit data 01011010

How to Determine Check Bits Locations? 



Bit position: 1 2 3 4 5 6 7 8 9 10 11 12 P1 P2 D7 P4 D6 D5 D4 P8 D3 D2 D1 D0 Check bits ( cb ) locations: C1 = XOR of bits (1, 3, 5, 7, 9, 11) C2 = XOR of bits (2, 3, 6, 7, 10, 11) C4 = XOR of bits (4, 5, 6, 7, 12) C8 = XOR of bits (8, 9, 10, 11, 12) . cb  for 2k-1=XOR of bits with 2 k-1 set in their bit # …

Single-Error Correction, Double-Error  Detection 

Hamming code  



Can detect and correct only a single error  Multiple errors may not be detected

Hamming code + a parity bit  

Can detect double errors and correct a single error. The additional parity bit is the XOR of all the other  bits 

E.g.: the previous 12-bit coded word 0 0 1 1 1 0 0 1 0 1 0 0 P13  0 0 1 1 1 0 0 1 0 1 0 0 1 (even parity).

 Applications: 



Hamming code is useful for detecting and correcting errors. VHDL implementation of Hamming code can be part of soft core communication system.