1

Error Detection and Correction

2

Handling Errors (1): Ignore Them!

data in→

data out→

Tx -DATA

frame Rx

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

-

-

�

Let’s meet at3pm tomorrow

Let’s meet at1pm tomorrow

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→

Let’s meet at1pm tomorrow Tx

-

-

-

Let’s meet at3pm tomorrow

Let’s meet at1pm tomorrow

Let’s meet at1pm tomorrow

Rx Let’s meet at1pm tomorrow

Send data thrice

Rx delivers the majority

6

Defeating Error Correction

data in→

data out→

Let’s meet at1pm tomorrow Tx

-

-

-

Let’s meet at3pm tomorrow

Let’s meet at3pm tomorrow

Let’s meet at1pm tomorrow

Rx Let’s meet at3pm tomorrow

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 with

BER = 0.1Rx - ? ? ?

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

9

Coordination Failure with Communication Failure

H1

m1 - H2

...

H1

mn−1� H2

H1

mn - H2

10

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

14

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

codewordr m6 m5 m4 m3 m2 m1 m0

r = m0 +2 . . . +2 m6

ASCII ‘X’ ( 10110002)

1 1 0 1 1 0 0 0

modulo arithmeticno carry

0 +2 0 = 0

1 +2 0 = 1

0 +2 1 = 1

1 +2 1 = 0

this defineseven parity

16

Parity Bits

codewordr m6 m5 m4 m3 m2 m1 m0

r = m0 +2 . . . +2 m6

ASCII ‘X’ ( 10110002)

1 1 0 1 1 0 0 0

1 1 0 0 1 0 0 0

17

Parity Bits

codewordr m6 m5 m4 m3 m2 m1 m0

r = m0 +2 . . . +2 m6

ASCII ‘X’ ( 10110002)

1 1 0 1 1 0 0 0

1 1 0 1 1 0 1 0

18

Parity Bits

codewordr m6 m5 m4 m3 m2 m1 m0

r = m0 +2 . . . +2 m6

ASCII ‘X’ ( 10110002)

1 1 0 1 1 0 0 0

0 1 0 1 1 0 0 0

19

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

23

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

26

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

27

Hamming Code: Compute21 Value

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

The value ofr2 is the even parity for allbn wheren hasb2 = 1

Thusr2 = 0

28

Hamming Code: Compute22 Value

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

The value ofr3 is the even parity for allbn wheren hasb3 = 1

Thusr3 = 1

29

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

CRC uses this idea (but using modulo 2 division)

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 16 bit CRC

name polynomial binary divisor application

CRC-16 X16 + X15 + X2 + 1 11000000000000101 USB

CRC-CCITT X16 + X12 + X5 + 1 10001000000100001Bluetooth

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

Properties of 32 bit CRC

IEEE 802.3 CRC-32 is

x32 +x26 +x23 +x22 +x16 +x12 +x11 +x10 +x8 +x7 +x5 +x4 +x2 +x+1

As a shorthand write as a number representing polynomials (missing off the

last +1).e.g.IEEE 802.3 0x82608EDB, iSCSI 0x8F6E37A0

HD max no message bits

0x82608EDB 0x8F6E37A0 0xBA0DC66B 0xD419CC15 0x80108400

8 91 177 152 58 0

7 171 (177) (152) 81 0

6 268 5,243 16,360 1,060 0

5 2,974 (5,243) (16,360) 65,505 65,505

4 91,607 ≥ 128K 114,663 (65,505) (65,505)

3 ≥ 128K unknown (114,663) (65,505) (65,505)

36

Error Bursts

message1+redundant1 = n1

5n1

4n1

3n1

2n1

1

message2+redundant2 = n2

5n2

4n2

3n2

2n2

1

message3+redundant3 = n3

5n3

4n3

3n3

2n3

1

message4+redundant4 = n4

5n4

4n4

3n4

2n4

1

- . . . n1

3n4

2n3

2n2

2n1

2n4

1n3

1n2

1n1

1

transmitted data interleaves codewords

If basic error checking can detectB bit bursts, interleavingN codewords

allowsNB bursts to be detected