Error Detection and Correction - doc.ic.ac.ukpjm/nac/lectures_2008/lecture_error.pdf · 5 A Simple...

1

Error Detection and Correction

2

Handling Errors (1): Ignore Them!

data in→

data out→

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..

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..

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.

.

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. .

. ..

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Tx -.

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DATA

frame

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Rx

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..

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.

..

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.

.

.

. .

. ..

..

..Some errors are not significant

audio

video

3

Handling Errors (2): Do something

data in→

data out→

.

.Let’s meet at1pm tomorrow

.

.

Tx -.

.

DATA

frame

.

.

Rx.

. Let’s meet at3pm tomorrow

..Most errors are significant — must be detected

..If only detect errors→ Reverse Error Correction (REC)

must ask for retransmission — ARQ protocol

unsuited to real-time applications

unsuited to simplex communication

..If detect and correct errors→ Forward Error Correction (FEC)

4

A Simple REC System

data in→

. .

Let’s meet at1pm tomorrow

.

.

Tx

-

-

�

.

.


..


.

.

NAK.

.

Rx

..Send data twice

..Rx delivers data if two copies identical and generates ACK-frame

..Rx sends NAK-frame if two copies differ

5

A Simple FEC System

data in→

data out→

..


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Tx

-

-

-

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.


..


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.


.

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Rx.

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..Send data thrice

..Rx delivers the majority

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Defeating Error Correction

data in→

data out→

..


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Tx

-

-

-

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..


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Rx.

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..More severe errors may not be detected

7

The Basis for Error Handling

..addredundancy to data to formcodewords

..handle two types of errors:

random errors→ bit error rate (BER)

error bursts → up toB bits in error

Codewords should have enough redundancy to detectprobableerrors

8

Example of valid codewords

binary decimalsymbolb2 b1 b0 d0 represented0 0 0 0 A0 0 1 1 Invalid0 1 0 2 Invalid0 1 1 3 Invalid1 0 0 4 Invalid1 0 1 5 Invalid1 1 0 6 Invalid1 1 1 7 B

?Send’A’ . .

0..

0..

0 -

.

.

Tx..

Channel withBER = 0.1

.

.

Rx -..

?..

?..

?

6

binary decimalp(if ’A’b2 b1 b0 d0 is sent)0 0 0 0 0.7290 0 1 1 0.0810 1 0 2 0.0810 1 1 3 0.0091 0 0 4 0.0811 0 1 5 0.0091 1 0 6 0.0091 1 1 7 0.001

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Coordination Failure with Communication Failure

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H1

m1 -

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H2

...

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H1

mn−1�

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H2

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H1

mn -

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H2

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Hamming Distance

- b0

0 1- b0

6

b1

00 01

10 11

- b0

6

b1

>b2

000 001

010 011

100 101

110 111

..HD is number of bits between codewords

e.g.0002 is HD of two from0112, 1012, and1102

..Error detection/correction based on HD between valid codewords

11

Hamming Distance of Two

000 001

010 011

100 101

110 111

000 001

010 011

100 101

110 111

..One bit error will change valid codeword to invalid codeword

..Two bit error will change valid codeword into different valid codeword

12

Hamming Distance of Three

000 001

010 011

100 101

110 111

..One bit error will change valid codeword to invalid codeword

invalid code HD of one from only one valid codeword

..Two bit error will change valid codeword to invalid codeword

13

Examples of Hamming Distance

binary decimal symbol HD HD

b2 b1 b0 d0 representedfrom ’A’ from ’B’

0 0 0 0 A 0 3

0 0 1 1 Invalid 1 2

0 1 0 2 Invalid 1 2

0 1 1 3 Invalid 2 1

1 0 0 4 Invalid 1 2

1 0 1 5 Invalid 2 1

1 1 0 6 Invalid 2 1

1 1 1 7 B 3 0

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Using Hamming Distance

..To reliably detect up toe bits in error

bits in error must be less than HD

e = HD − 1

HD = e + 1

..To reliably detect and correct up toe bits in error

bits in error must be less than halfway of HD

e = ⌊HD − 1

2⌋

HD = 2e + 1

15

Parity Bits

codeword

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r.

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m6.

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m5.

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m4.

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m3.

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m2.

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m1.

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m0

r = m0 +2 . . . +2 m6

ASCII ‘X’ ( 10110002)

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1.

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1.

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0.

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1.

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1.

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0.

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0.

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0

modulo arithmeticno carry

0 +2 0 = 0

1 +2 0 = 1

0 +2 1 = 1

1 +2 1 = 0

..this defineseven parity

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Parity Bits

codeword

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.

r.

.

m6.

.

m5.

.

m4.

.

m3.

.

m2.

.

m1.

.

m0

r = m0 +2 . . . +2 m6

ASCII ‘X’ ( 10110002)

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1.

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1.

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0.

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1.

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1.

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0.

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0.

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0

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1.

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1.

.

0.

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0.

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1.

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0.

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0.

.

0

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Parity Bits

codeword

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r.

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m6.

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m5.

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m4.

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m3.

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m2.

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m1.

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m0

r = m0 +2 . . . +2 m6

ASCII ‘X’ ( 10110002)

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1.

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1.

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0.

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1.

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1.

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0.

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0.

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0

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1.

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1.

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0.

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1.

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1.

.

0.

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1.

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0

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Parity Bits

codeword

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r.

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m6.

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m5.

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m4.

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m3.

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m2.

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m1.

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m0

r = m0 +2 . . . +2 m6

ASCII ‘X’ ( 10110002)

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1.

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1.

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0.

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1.

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1.

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0.

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0.

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0

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0.

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1.

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0.

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1.

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1.

.

0.

.

0.

.

0

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Odd and Even Parity

000 001

010 011

100 101

110 111 even parityr = m0 +2 . . . +2 mn

..hardware: XOR returns 0

..by eye: even number of one bits

000 001

010 011

100 101

110 111 odd parityr = m0 +2 . . . +2 mn + 1

..hardware: XOR returns 1

..by eye: odd number of one bits

..Don’t need to know which bit is redundant bit

..Implements HD=2→ reliably detects one bit errors

20

Worksheet: Hamming Distance and Parity

1. Draw arcs between the codewords HD=1

1011 1100 1001

1111 1010 1000

0111 0000

2. If 1000 is to be a valid codeword, shade codewords which maintain HD=2.

3. What type of parity do codewords have?

4. If 1010 was received, with single bit, what codewords might have been sent?

5. How bit errors can parity check detect, and how many can it correct?

21

Block Sum Check

r m6 m5 m4 m3 m2 m1 m0

w0 0 1 0 0 1 0 0 0

w1 0 1 1 0 0 1 0 1

w2 0 1 1 0 1 1 0 0

w3 0 1 1 0 1 1 0 0

w4 0 1 1 0 1 1 1 1

BCC 0 1 0 0 0 0 1 0

..Parity only detects one bit errors

..Improve by adding BCC at end, doing column-wise parity

22

Block Sum Check: Random Errors

r m6 m5 m4 m3 m2 m1 m0

w0 0 1 0 0 1 0 0 0

w1 0 1 0 0 1 1 0 1

w2 0 1 1 0 1 1 0 0

w3 0 1 0 0 0 1 0 0

w4 0 1 1 0 1 1 1 1

BCC 0 1 0 0 0 0 1 0

..BCC defeated by four bits in error

..HD=4

detect three bit errors

detect and correct one bit errors

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Block Sum Check: Burst Errors

r m6 m5 m4 m3 m2 m1 m0

w0 0 1 0 0 1 0 0 0

w1 0 1 1 0 0 1 0 1

w2 0 1 1 0 1 1 1 1

w3 0 1 1 0 1 1 1 1

w4 0 1 1 0 1 1 1 1

BCC 0 1 0 0 0 0 1 0

..Smallest burst with four bit square is ten bits

remember, not all bits in a burst need be in error, just the first and last

..BCC copes with 9 bit bursts

24

Hamming Code

b12 b11 b10 b9 b8 b7 b6 b5 b4 b3 b2 b1

20

21

22

23

m8 m7 m6 m5 r4 m4 m3 m2 r3 m1 r2 r1

‘X’ 0 1 0 1 ? 1 0 0 ? 0 ? ?

..Each bit positionbn wheren = 2x used for redundancy bit

..Parity generated on all bitsbm wherem hasx bit set

25

Numbering bits means each has unique identifier

b12 1 1 0 0

b11 1 0 1 1

b10 1 0 1 0

b9 1 0 0 1

b8 1 0 0 0

b7 0 1 1 1

b6 0 1 1 0

b5 0 1 0 1

b4 0 1 0 0

b3 0 0 1 1

b2 0 0 1 0

b1 0 0 0 1

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Hamming Code: Compute20 Value

b12 b11 b10 b9 b8 b7 b6 b5 b4 b3 b2 b1

20 1 1 1 0 0 ?

21

22

23

m8 m7 m6 m5 r4 m4 m3 m2 r3 m1 r2 r1

‘X’ 0 1 0 1 ? 1 0 0 ? 0 ? ?

..The value ofr1 is the even parity for allbn wheren hasb1 = 1

..Thusr1 = 1

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b12 b11 b10 b9 b8 b7 b6 b5 b4 b3 b2 b1

20

21 1 0 1 0 0 ?

22

23

m8 m7 m6 m5 r4 m4 m3 m2 r3 m1 r2 r1

‘X’ 0 1 0 1 ? 1 0 0 ? 0 ? 1


..Thusr2 = 0

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b12 b11 b10 b9 b8 b7 b6 b5 b4 b3 b2 b1

20

21

22 0 1 0 0 ?

23

m8 m7 m6 m5 r4 m4 m3 m2 r3 m1 r2 r1

‘X’ 0 1 0 1 ? 1 0 0 ? 0 0 1


..Thusr3 = 1

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Hamming Code

b12 b11 b10 b9 b8 b7 b6 b5 b4 b3 b2 b1

20 1 1 1 0 0 1

21 1 0 1 0 0 0

22 0 1 0 0 1

23 0 1 0 1 0

m8 m7 m6 m5 r4 m4 m3 m2 r3 m1 r2 r1

‘X’ 0 1 0 1 0 1 0 0 1 0 0 1

..Each bit positionbn wheren = 2x use for redundancy bit

..Parity generated on all bitsbm wherem hasx bit set

30

Hamming Code: Error Correction

b12 b11 b10 b9 b8 b7 b6 b5 b4 b3 b2 b1

20 1 1 1 0 0 1

21 1 0 1 1 0 0

22 0 1 1 0 1

23 0 1 0 1 0

m8 m7 m6 m5 r4 m4 m3 m2 r3 m1 r2 r1

‘X’ 0 1 0 1 0 1 1 0 1 0 0 1

..Parity checks failing ‘points’ to error

..HD=3→ detect and correct one bit errors

31

Worksheet: Hamming Code

m8 m7 m6 m5 r4 m4 m3 m2 r3 m1 r2 r1

1. Write message11101110 into the message bit boxes above.

2. Compute using even parity the bitsr4r3r2r1

3. Invert message bitm6, and write down in the orderr4r3r2r1 a1 if rn parity is

now incorrect, or a 0 if correct.

4. Invert the bit position in the codeword which the number represents.

32

Cyclic Redundancy Check

If we take a numbern and divide by divisord, result is

..integer quotientq

..integer remainderr

..obeys the equationn = qd + r

If we add an errore to n, thenn + e = q′d + r′

..e < d → r′ 6= r

..r′ = r → e must be an integer multiple ofd

33

Examples of CRC Checking

n d q r e n + e r′ OK?

245 161 1 84 0 245 84 Yes

1 246 85 No

2 247 86 No

160 405 83 No

161 406 84 Yes

162 407 85 No

321 566 83 No

322 567 84 Yes

323 568 85 No

..First error pattern not detected= 16110 = 101000012

..Second error pattern not detected= 32210 = 1010000102

34

Properties of CRC

name polynomial binary divisor

CRC-16 X16 + X15 + X2 + 1 11000000000000101

CRC-CCITT X16 + X12 + X5 + 1 10001000000100001

Choosing values for polynomials need care to achieve the best performance. Those

above detect:

..error bursts of 16 bits

..all error bursts with an odd number of bits

..99.997% of 17 bit errors

..99.998% of 18 or longer bit errors

35

Error Bursts

message1+redundant1 =.

.

n1

5 .

.

n1

4 .

.

n1

3 .

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n1

2 .

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n1

1


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n2

5 .

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n2

4 .

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n2

3 .

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n2

2 .

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n2

1


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n3

5 .

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n3

4 .

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n3

3 .

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n3

2 .

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n3

1


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n4

5 .

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n4

4 .

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n4

3 .

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n4

2 .

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n4

1

- . . ..

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n1

3 .

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n4

2 .

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n3

2 .

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n2

2 .

.

n1

2 .

.

n4

1 .

.

n3

1 .

.

n2

1 .

.

n1

1

transmitted data interleaves codewords

..If basic error checking can detectB bit bursts, interleavingN codewords

allowsNB bursts to be detected

Error Detection and Correction - doc.ic.ac.ukpjm/nac/lectures_2008/lecture_error.pdf · 5 A Simple...

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