Ian McCrum Room 5B18
Tel: 90 366364 voice mail on 6 th ring
Email: IJ.McCrum@Ulster.ac.uk
Web site: http://www.eej.ulst.ac.uk
http://www.eej.ulst.ac.uk/~ian/modules/COM342/COM342_L4C.ppt
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• k parity bits are added to n data bits giving a new word
(k+n) bits long
• The word is numbered from bit
1 to (k+n) for more convenience when working things out
• The parity bits are inserted throughout the word – at positions given by powers of two. I.e bits 1,2,4,8, 16 etc
• Each parity bit is calculated from a subset of the data bits, these overlap
• Comparing the parity bits received with those calculated at the receiver will give all zeroes if there is no discrepancy
(which is why we number from one)
• If the comparison is non-zero, e.g 101 then bit 5 is in error
– this is very handy, easy to correct in hardware http://www.eej.ulst.ac.uk/~ian/modules/COM342/COM342_L4C.ppt
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• In general the total number of bits that can be accommodated by the method is 2 k -1
• So (n+k) = 2 k -1
• If k is 3 we can accommodate 7 bits (4 data)
• If k is 4 we can accommodate up to 15 bits (11 data)
• If k is 5 we can accommodate up to 31 bits (26 data)
• There are better codes, hamming only fixes single bit errors (c.f convolutional codes although we won’t cover them here).
• If a 100 character block needs protected, it is better to use hamming on the 100 bits vertically through the block, so all bit 0s are protected, then all bit 1s etc., this extends each 100 bits to 107 bits since 7 hamming bits can protect 127 data bits. This is good for bursts of noise of a few bits… http://www.eej.ulst.ac.uk/~ian/modules/COM342/COM342_L4C.ppt
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[0010 0011] msb…lsb (b7..b0) b1 : b2 : b3 : b4 : b5 : b6 : b7 : b8 : b9 : b10: b11:b12
• We reserve the bit positions that are powers of two to hold the special parity bits, usually called check bits
P1 : P2 : 1 : P4 : 1 : 0 : 0 : P8 : 0 : 1 : 0 : 0
3,5,7,9,11
3,6,7,10,11
5,6,7,12
9,10,11,12
•
We apply parity to subsets of the original 8 data bits, each parity bit tests 4 or 5 data bits, there is a pattern as to which!
http://www.eej.ulst.ac.uk/~ian/modules/COM342/COM342_L4C.ppt
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P1 : P2 : 1 : P4 : 1 : 0 : 0 : P8 : 0 : 1 : 0 : 0
3,5,7,9,11
3,6,7,10,11
5,6,7,12
9,10,11,12
P1 must be 0 to make {0, 1,1,0,0,0,} even
P2 must be 0 to make {0, 1,0,0,1,0,} even
P4 must be 1 to make {1, 1,0,0,0 } even
P8 must be 1 to make {1, 0,1,0,0 } even
0 : 0 : 1 : 1 : 1 : 0 : 0 : 1 : 0 : 1 : 0 : 0 = 00 1 1 100 1 0100
• The new 12 bit word can now be sent… http://www.eej.ulst.ac.uk/~ian/modules/COM342/COM342_L4C.ppt
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• The parity bits are labelled P1, P2, P4 and P8
• Data bits are in positions 3,5,6,7,9,10,11,12
• For each data bit, write its position as powers of 2 so position 10 is 8 +
2
• We use P2 and P8 to “cover” 10
• 3=>P1 and P2, 5=>P1 and P4, 6=>P4 & P2
• 7=>P1 & P2 & P4, 9=>1001 => P8,P4,P2,P1
• Hence you can draw the arrows and work out the parity bits.
• A better way is to arrange a 2D table, see the website for a good link showing this ( http://candle.ctit.utwente.nl/wp5/tel-sys/exercises ) or the answers to the tutorial questions later in this presentation.
http://www.eej.ulst.ac.uk/~ian/modules/COM342/COM342_L4C.ppt
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00 1 1 100 1 0100 bit positions… 123456789012
• 00 1 1 100 1 0100 arrives into the receiver
• The receiver calculates four parity bits, it uses the original grouping but includes the parity bits themselves.
• So C1 = {b1, b3, b5, b7, b9, b11 } = 0 11000
• So C2 = {b2, b3, b6, b7, b10, b11 } = 0 10010
• So C4 = {b4, b5, b6, b7, b12 } = 1 1000
• So C8 = {b8, b9, b10, b11, b12 } = 1 0100
• These should all be zero 0000 (even parity)
• The next slide puts an error in bit 5 http://www.eej.ulst.ac.uk/~ian/modules/COM342/COM342_L4C.ppt
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00 1 1 0 00 1 0100 bit positions… 123456789012
• So C1 = {b1, b3, b5 , b7, b9, b11 } = 0 1 0 000
• So C2 = {b2, b3, b6, b7, b10, b11 } = 0 10010
• So C4 = {b4, b5 , b6, b7, b12 } = 1 0 000
• So C8 = {b8, b9, b10, b11, b12 } = 1 0100
• Now C1 and C4 are set, The checkbits arranged as a binary number C8 C4 C2 C1 are 0101
• This is five in binary, so invert bit 5 to correct the code
• 00 1 1 0 00 1 0100 = > 00 1 1 1 00 1 0100 http://www.eej.ulst.ac.uk/~ian/modules/COM342/COM342_L4C.ppt
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• [1] A 12 bit Hamming code containing 8 bits of data and 4 parity bits is received, what was the original?
• [1](a)
0000 1110 1010
• [1](b)
1011 1000 0110
• [1](c) 1011 1111 0100
• [2] Given the 8 bit data word 0101 1011 give the 12 bit that can correct a single bit error
• [3] Given a 11 bit data word, generate the 15 bit hamming code word http://www.eej.ulst.ac.uk/~ian/modules/COM342/COM342_L4C.ppt
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Answers to tutorial questions from Lecture 4C
NB C1={1,3,5,7,9,11}, C2={2,3,6,7,10,11}, C4={4,5,6,7,12}, C8={8,9,10,11,12}
And these groups should have an even number of ones, if not then set that C bit
00 0 0 111 0 1010 10 1 1 100 0 0110 10 1 1 111 1 0100
C1=0 _0_1_1_1_1_ C1= 1 _1_1_0_0_1_ C1=1 _1_1_1_0_0_
C2= _ 0 0__11__01_ C2= _ 0 1__00__11_ C2= _ 0 1__11__10_
C4= ___ 0 111____0 C4= ___ 1 100____0 C4= ___ 1 111____0
C8= _______ 0 1010
So C8421 = 0110
I.e bit 6 is bad, invert it giving
00 0 0 1 0 1 0 1010
C8= _______ 0 0110
So C8421 = 0010
I.e bit 2 is bad, invert it giving
1 1 1 1 100 0 0110
C8= _______ 1 0100
So C8421 = 0000
I.e No bits are bad so code is
101111110100 http://www.eej.ulst.ac.uk/~ian/modules/COM342/COM342_L4C.ppt
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Answers to lecture 4C page 2
•[2] Given the 8 bit data word 0101 1011 give the 12 bit that can correct a single bit error
Code is :P1 :P2 :b3 :P4 :b5 :b6 :b7 :P8 :b9 :b10:b11:b12: i.e : ? : ? : 1 : ? : 1 : 0 : 1 : ? : 1 : 0 : 1 : 0 :
Take b3 :[ 1 ]:[ 1 ]: : (3 = 1+2)
Take b5 :[ 1 ]: : :[ 1 ]: (5 = 1+4)
Take b6 : :[ 0 ]: :[ 0 ]: (6 = 2+4)
Take b7 :[ 1 ]:[ 1 ]: :[ 1 ]: (7 = 1+2+4)
Take b9 :[ 1 ]: : : : : : :[ 1 ]: (9 = 1+8)
Take b10: :[ 0 ]: : : : : :[ 0 ]: (10= 2+8)
Take b11:[ 1 ]:[ 1 ]: : : : : :[ 1 ]: (11= 1+2+8)
Take b12: : : :[ 0 ]: : : :[ 0 ]: (12= 4+8)
Parity Pn[1] [1] [0] [0] (make column even)
12 bits = 1 1 1 0 1 0 1 0 1 0 1 0 i.e [1110010101010] http://www.eej.ulst.ac.uk/~ian/modules/COM342/COM342_L4C.ppt
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Answers to Lecture 4C Tutorials Page 3
•[3] Given a 11 bit data word, generate the 15 bit hamming code word, I pick an arbitrary 001 1010 11011
Code is : P1 : P2 :b3 : P4 :b5 :b6 :b7 : P8 :b9 :b10:b11:b12:b13:b14:b15: P16 :b17: i.e : ? : ?
: 1 : ?
: 1 : 0 : 1 : ?
: 1 : 0 : 1 : 0 : 1 : 1 : 0 : ? : 0 :
Take b3 :[1]:[1]: : (3 = 1+2)
Take b5 :[1]: : :[1]: (5 = 1+4)
Take b6 : :[0]: :[0]: (6 = 2+4)
Take b7 :[1]:[1]: :[1]: (7 = 1+2+4)
Take b9 :[1]: : : : : : :[1]: (9 = 1+8)
Take b10: :[0]: : : : : :[0]: (10= 2+8)
Take b11:[1]:[1]: : : : : :[1]: (11= 1+2+8)
Take b12: : : :[0]: : : :[0]: (12= 4+8)
Take b13:[1]: : :[1]: : : :[1]: (13= 1+4+8)
Take b14: :[1]: :[1]: : : :[1]: (14= 2+4+8)
Take b15:[0]:[0]: :[0]: : : :[0]: (15= 1+2+4+8)
Take b17:[0]: : : : : : : : : : : : : : :[0]: (17= 1+16)
Take b18: :[0]: : : : : : : : : : : : : :[0]: (18= 2+16)
Parity Pn[0] [0] [0] [0] [0] (make column even)
12 bits = 0 0 1 0 1 0 1 0 1 0 1 0 1 1 0 0 0 i.e 00101010101011000 http://www.eej.ulst.ac.uk/~ian/modules/COM342/COM342_L4C.ppt
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