Question

6. Suppose we want an error-correcting code that will allow all single-bit errors to be corrected for memory words of length 10. a) How many parity bits are necessary?

Answer 1

Answer 2

Solution : a) Number of Pacity bits needed For 12 length memory word = mtxt1 <= 2² where me length of memoky word and t is the number of Parity bits. mtxt 1 L=2t 12 txt 1 2= 2t 13t&c=2² which implies that a should be s. So, number of parity bits needed fose 12 length memory worked is 5. b) The code word for 100100011010 will ſbe as shown below: since m-12. 12 ttt1 <= 2t 13tkC=2t which implies that t should be greated than equal to 5. choose t=5. find bit positions are sums of those numbers that are Power of 2.

1 = 1 2-2 3=1+2 4 - 4 5 = 1t4 6=2+4 7= 1+2+4 88 g= 1+8 10 =2+8 11 - 1+2 +8 12 a 4t8 13 = 1+4+8 14 = 2+4+8 15 - 1+2 +4 +8 16 = 4+4+8 17 - 1+4+4+8 Parity bit 1 checks the patity over 1,3,5, 7.9, 11, 13, 15,17. Therefore, bit lis 1. parity bit 2 checks the parity over 2,3,6, 7,10,11,14,15. Therefore bit 2 is 1.

→ Parity bit 3 checks the parity over 4,5.6, 7,12,13,14,15. There Fare bit 4 is 1. y Parity bit 4 checks the parity over 8, 9, 10, 11, 12, 13, 14, 15. Therefore bit 4 is o. → Parity bit 5 checks the passity over 16, 17. Therefore, bit 5 is 1. 1 1 0 1 0 0 0 1 0 1 0 1 1 0 0 1 0 1 1 - 11 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 + The code word for 100100011010 (assuming even pazity) is 11001000 100011 011.

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