#4 1 Victor S. Frost Dan F. Servey Distinguished Professor Electrical Engineering and Computer...

#4 1

Victor S. FrostDan F. Servey Distinguished Professor Electrical Engineering and Computer

ScienceUniversity of Kansas2335 Irving Hill Dr.

Lawrence, Kansas 66045Phone: (785) 864-4833 FAX:(785) 864-

7789 e-mail: [email protected]

http://www.ittc.ku.edu/

How to cope with last hop impairments?Part 1

Error Detection and Correction#4

All material copyright 2006Victor S. Frost, All Rights Reserved

http://www.ittc.ku.edu/

#4 2

How to cope with last hop impairments?

• Techniques for coping with noise

–Forward error detection/correction coding

– Automatic Repeat reQuest (ARQ)– Co-existance or modifications to end-to-end protocols

• Techniques for coping with multipath fading fading mitigation techniques, e.g., – Equalizers– Diversity– RAKE receivers– OFDM

#4 3

Techniques for coping with noise

• Forward error detection/correction coding– Detection of bit error is needed to enable

reliable communications– Requires the addition of bits (redundancy) to

packets• If the time between packet error is long then

detection and request retransmission (see ARQ)• If the time between packet error is short then often

it “pays” to add enough bits to correct errored bits– Wireless– Fiber links

• Real time applications do not have time for retransmissions

#4 4

Error Detection

EncodeEncodein.data.framein.data.frame

k bitsk bits

out.data.frameout.data.frameor code word cor code word cjj

n bitsn bitsn>kn>k

r = n -k =redundant bits added tor = n -k =redundant bits added to detect errorsdetect errors

#4 5

Error Detection

Example: n=2k•Repeat the frame•k redundant bits•Code word has 2k bits

Redundancy Ratio =

nk

#4 6

Error Detection

• Example: Simple Parity Check• Even Parity

– number of ‘1’ per block is even

• Odd Parity– number of ‘1’ per block is odd

• Redundancy ratio = k+1/k

#4 7

Error Detection

• Performance criteria for error detection codes is the probability of an undetected error,

Let p = probability of a bit error

and assume random errors.

then

Prob[1 bit error in a block of n bits]

= p(1) = np(1- p)n-1

for m errors

p(m) = n!

(n- m)!m!pm(1 p)n-m

Pe

#4 8

Error Detection

Pe = P(2) + P(4) + P(6) ...

Note P(m) 0 for M > 3.

Pe = n(n - 1)

np2(1 p)n-2

Example: Find the Example: Find the time between undetectedtime between undetectederrors for:errors for:

n = 1000n = 1000p = 1 in millionp = 1 in millionRate = 100 Mb/sRate = 100 Mb/s

Pe = 5 x 10-7

Time to transmit a block=10s

Time between undetected errors=10s

5 x 10-7

= 20 seconds

For simple parity For simple parity check:check:

#4 9

Hamming distance

• Hamming distance=the number of bits two code words differ– XOR the

code words– Hamming

distance = d=3

CodeWord

1

1 0 0 0 1 0 0 1

CodeWord

2

1 0 1 1 0 0 0 1

0 0 1 1 1 0 0 0

• For d=3 it takes 3 bit errors to map one code word into another

• In general it take d bit errors to map one code word into another

#4 10

A Geometric Interpretation ofError correcting Codes

message. received the todistance Hammingin closest''

wordcode theselectsdecoder distance minimum The

cc=d Note

2.=d is c and c

between distance Hamming then the

001100 =c

and 101101 = c

Let

jiij

ijji

j

i

#4 11

Hamming distance

• There are 2n possible code words

• Only 2k are valid

• Among all the 2k valid code words ci and cj have the minimum

Hamming distance d• Then the complete code is defined by this Hamming distance

d• To detect m errors you need a code with a Hamming distance

of d = m+1, clearly m errors can not map a valid code word into another valid code word

• To correct m errors you need a code with a Hamming distance of d = 2m + 1– Decoding algorithm

• Let r be the received code word• Calculate the Hamming distance between r and each valid code word• Declare the valid code word, e.g., ci with that has the minimum

Hamming to be correct– So m errors will result in a received code word that is “closer” to ci

then all other code words

#4 12

Example of Hamming Distances

d cd1=00 c1=0000d2=01 c2=0101d3=10 c3=1010d4=11 c4=1111

r d1 d2 d3 d4

0000 0 2 2 40001 1 1 2 30010 1 3 1 30011 2 2 2 20100 1 1 2 30101 2 0 4 20110 2 2 2 20111 3 2 3 11000 1 2 1 31001 2 2 2 21010 2 4 0 21011 3 3 1 11100 2 2 2 21101 3 1 2 11110 3 3 1 11111 4 2 2 0

Data Code word

All possible received

Code words

#4 13

Block CodesRepresentation and manipulation of message

blocks

,1,0,0)(0,1,0,0,1 01001100 Example,

)d...., ,d ,d ,(d=d

klength ofvector

binary aby drepresente is messagebit -k

k321

#4 14

Block Codes

,0),1,0,0,1,1(0,1,0,0,1 00100110011

Example,

)c...., ,c ,c ,(c=c

nlength ofvector

binary aby drepresente is wordcodebit -n

n321

#4 15

Block CodesTransformation between message and

code vectors.

Let c1 d1, c2 d2 , c3 d3,.. ck dk

ck+1 = p11d1 p21d2 p31d3 p41d4 ... pk1dk

ck+2 = p12d1 p22d2 p32d3 p42d4... pk2dk

.

.

cn = p1rd1 p2rd2 p3rd3 p4rd4 ... pkrdk

Note 0 0 0,1 0 1, 0 1 1, 11 0

#4 16

Block codes: Example

• c1= d1

• c2= d2

• c3= d3

• c4= d1 + d3

• c5= d2 + d3

• c6= d1 + d2

• Let d= (0, 0, 1)

– C = (0, 0, 1, 1, 1, 0)

#4 17

Block Codes

In matrix form c =

d G

where

G =

1 0 0 0 0 p11 p12 p13 p14 p15 p1r

0 1 0 0 0 p21 p22 p23 p24 p25 p1r

0 0 1 0 0 p31 p32 p33 p34 p35 p1r

0 0 0 1 0 p41 p42 p43 p44 p45 p1r

0 0 0 0 1 pk1 pk2 pk3 pk4 pk5 pkr

G = [IkP] where I k = k x k idenity matrix

G = Code Generator Matrix

#4 18

Block Codes

• G Defines a Linear Systematic (n,k) Code

k Message k Message BitsBits

r Code r Code BitsBits

Code Word of n bitsCode Word of n bits

#4 19

Block Codes: Example

011100 =

011100

101010

110001

100 =G d = c

000000 =

011100

101010

110001

]000[ =G d = c

wordscode unique 8 are There

011100

101010

110001

=G

011

101

110

= P

22

11

#4 20

Block Codes

• Problem: given a received message determine if an error has occurred.

Model for the recieved messager =

c

e

wheree = error vector

e i = 1 means that a error occured in the i th

bit of the code word

e = (001010) Errors in: bit 3 & 5

#4 21

Block Codes

• Define a Parity Check Matrix H

H = [P tIr ] =

p11 . . . pk1 1 0 0 0 0 0

. . . . . 0 1 0 0 0 0

. . . . . 0 0 1 0 0 0

. . . . . 0 0 0 1 0 0

. . . . . 0 0 0 0 1 0

p1r . . . pkr 0 0 0 0 0 1

#4 22

Block codes

one.another into wordcode

one mapcan errorsenough that note error, detectable no then 0=s if Therefore,

Syndrome = s

occured. haserror an then 0Hr if So

occured. haserror an and 0e then 0He if and

He = HeHc = )Hec( = Hr sThen

0=Hc shown that becan it

words,code valid2 allFor

t

t

ttttt

t

k

#4 23

Block Codes: Example

G =

1 0 0 0 1 1

0 1 0 1 0 1

0 0 1 1 1 0

P =

0 1 1

1 0 1

1 1 0

H =

0 1 1 1 0 0

1 0 1 0 1 0

1 1 0 0 0 1

H t =

0 1 1

1 0 1

1 1 0

1 0 0

0 1 0

0 0 1

r = 1 1 1 1 1 1

s = 1 1 1 1 1 1

0 1 1

1 0 1

1 1 0

1 0 0

0 1 0

0 0 1

= 1 1 1 0

An error has been detected .

#4 24

Binary Cyclic Codes

• A subclass of linear block codes are the binary cyclic codes.

• If ck is a code word in a binary cyclic code then a lateral (or cyclic) shift of it is also a code word.

If c1 c2 c3 c4 c5 is a code word

then c5 c1 c2 c3 c4 and

c4 c5 c1 c2 c3 can also be code words .

#4 25

Binary Cyclic Codes

• Advantages:– Ease of syndrome calculation– Simple and efficient coding/decoding

structure

• Cyclic codes restrict the form of the G matrix

• The cyclic nature of these codes indicates that there is a underlying pattern.

#4 26

Binary Cyclic Codes

1or 1)(

:Example

1....1 =)(

g(x) PolynomialGenerator

2151621516

23

131

r

xxxxxxxg

xgxggxxg rr

rrr

ggi i = 0, 1= 0, 1

#4 27

Polynomial Coding

• Code has binary generating polynomial of degree n–k

• k information bits define polynomial of degree k – 1

• Define the codeword polynomial of degree n – 1

b(x) = xn-ki(x) + r(x)

n bits k bits n-k bits

g(x) = xn-k + gn-k-1xn-k-1 + … + g2x2 + g1x + 1

i(x) = ik-1xk-1 + ik-2xk-2 + … + i2x2 + i1x + i0

Modified from: Leon-Garcia & Widjaja: Communication Networks

#4 28

Binary Cyclic Codes: Construction of G from g(x)

• 1) Use g(x) to form the kth row of G.• 2) Use kth row to form (k-1) row by a shift

left, or xg(x).• 3) If (k-1) row in not in standard form

then add kth row to shifted row, i.e.., (k-1) row becomes xg(x) + g(x).

• 4) Continue to form rows from the row below.

• 5) If the (k-j) row not in standard form the add g(x) to the shifted row.

#4 29

1101000

0110100

1110010

1010001

.

1010001

1101000

0111001.

01110013:4

.

1110010

1101000

0011010.

00110102:3

.

01101001:2

1101000:1

347

1110=g(x)

:3323

G

DoneokForm

oknotForm

GofrowkStep

okForm

oknotForm

GofrowkStep

okForm

GofrowkStep

GofkStep

rkn

xxxxxxx

Example

th

#4 30

Binary Cyclic Codes

Codes for this example are :

c 1 0 0 0 1

1 0 0 0 1 0 1

0 1 0 0 1 1 1

0 0 1 0 1 1 0

0 0 0 1 0 1 1

0 0 0 1 0 1 1

c 2 0 0 1 0

1 0 0 0 1 0 1

0 1 0 0 1 1 1

0 0 1 0 1 1 0

0 0 0 1 0 1 1

0 0 1 0 1 1 0

Note that c 1 and

c 2 are cyclic shifts of each other .

#4 31

Binary Cyclic Codes:Standard generator polynomials

1)(12

1)(

1)(16

231112

51216

21516

xxxxxxgCRC

xxxxgCCITTCRC

xxxxgCRC

#4 32

Convolutional Codes*

*Based on: http://complextoreal.com/chapters/convo.pdf

Example:ui = input bitsvi = output bits

1 input bit 3 output bitsRate 1/3 code

Calculation of output bits

When next input bit u2 arrives contents of the register shift to right by 1 and a new set of 3 outputs calculated

= x2 + x + 1 (1,1,1) = x + 1 (0,1,1) = x2 + 1 (1,0,1)

Calculation of output bits:

Generator polynomials

#4 33


• The properties of the code are governed by– (n, k, m)– Generator polynomials


#4 34

Convolutional Codes*• Specification of convolution codes

– (n, k, m)• k = number of input bits• n = number of output bits• m =

– For k = 1: m = maximum number of input bits used to calculate an output bit

– k > 1 later– In the example (3, 1, 3) code – Note a coded bit is a function of several past input

bits, the length of the memory is represented by the Constraint Length = L = k(m-1)

– In the example 1(3-1) = L = 2– Note then that the number of states the code can be

is 2L


#4 35


• For k>1 the form of the coder can be represented as


...uk ... u1 ...U-mk ... u...U-0 ... u-k+1

1 2 m

v1

+ + +

v2 vn

#4 36

Convolutional Codes*• Example:

• Note that the output bits depend on the current k inputs plus k(m-1) previous input bits so L = k(m-1)

• For k>1 m= L/k + 1• For this example L = 6, k = 3

m=3


#4 37


• Systematic convolution codes


#4 38


• Example:– n=2, k=1, m=4

(2,1, 4)– L = 3– Number of states =

8000…111


#4 39


• Find the output sequence for in 1 0 input, assume encoder starts in state 000

• Output sequence = – From first 1 11110111 – From shifting the input “0”0000– Output 111101110000


#4 40

Convolutional Codes*• Find the output sequence for in 0 input,

assume encoder starts in state 000• Output sequence = 00000000

• Note “response” to a “0” given an initial state of 000 is 00000000

• Note “response” to a “1” given an initial state of 000 is 11110111

• The “response” to a “1” in given an initial state of 000 is called the impulse response.

• Now an output to any input can be found by convolving the input with the impulse response


#4 41



Shift and add

#4 42

Convolutional Codes*• Graphical view of encoding process Trellis


Sta

tes

Input bit

Output bitsAfter L bits (L=3 here) trellis fully populated, i.e., all possible transitions are included. The Trellis repeats after that.

#4 43

Convolutional Codes*Finding the output for 10110


0(01

)

Output 1111011101

#4 44


• Approaches– Compare received sequence to all

permissible sequences pick “closest” (Hamming Distance) Hard decision

– Pick sequence with highest correlation Soft decision

• Problem given received message length of s bits decode without “checking” all 2s codewords– Sequential decoding Fano– Maximum likelihood Viterbi


#4 45


• Sequential decoding– Move forward and backward in trellis– Keeps track of results and can backtrack

• Continue example (2,1,4) code– Input 1011 000

• Last three bits used to flush out the last bit these are called flush bits or end tail bits

– With no errors 11110111010111– Assume received bits are

01110111010111


#4 46


• Sequential decoding– Based on channel stats set an “error

threshold” = 3– Decoder starts n state 00 looks at first

2 bits 01• Note starting in state 00 only possible

outputs are 00 and 11• Knows there is an error but not the

position of the error


#4 47


• Randomly assumes that a 0 was transmitted and selects the path corresponding to 00 in the trellis Set error count = 1


Start

#4 48


• Now at point 2: Incoming bits are 11 so assume a 1 transmitted


Start

#4 49


• Now at point 3: Incoming bits are 0 1 so another error increment error count; error count = 2 randomly assume a 0 transmitted


Start

#4 50


• Now at point 4: Incoming bits are 11 so another error increment error count; error count = 3; too many so back track


Start

#4 51


• Now at point 3: Incoming bits are 01 so another error, now another error and go to point 2


Start

#4 52

Convolutional Codes*• Now at point 2: But point 2 was correct (taking other path will

cause error) so backtrack to point 1 and assume a 1 was sent. Now from point 1 assuming a 1 was sent the forward path through the trellis will not encounter any errors


Start

#4 53


• Correct path through the Trellis 1011000


0

1

1

1

0

0

0

#4 54


• Viterbi decoding– Assumes infrequent errors– Random error i.e., Prob(consecutive bits

errors)< Prob(single bit error)– For a given received sequence length

• Computes metric (e.,g Hamming for hard decision) for each path

• Makes decision based on metric• All paths through Trellis are followed until two

paths converge• The path with the higher metric survives (called

the survivor)• Metrics are added and path with largest is the

winner


#4 55


• Continue example (2,1,4) code– Input 1011 000

• Last three bits used to flush out the last bit these are called flush bits or end tail bits

– With no errors 11110111010111– Assume received bits are 01

11 01 11 01 01 11 (7 – tuples)

• Decoder starts in state 000


#4 56


• Step 1: Two paths are possible, compute metric for each path an go on…


#4 57


• Step 2: Now there are 4 paths,


=M(0111, 1111)

=M(0111, 0000)

=M(0111, 0011)

=M(0111, 1100)

Metric=M(X, Y)= number bits the same

#4 58


• Step 3: Now have all 8 states covered


#4 59


• Step 4: Now there are multiple paths and the metric is calculated for each path, the largest survive


#4 60


• Step 4: After discarding


#4 61


• Step 5: Continue note some ties, keep both paths if equal


#4 62


• Step 6


#4 63

Convolutional Codes*• Step 7: Used up all the input and have a winner: states visited

000 100 101 110 011 001 000 1011 000


M=8

M=13

#4 64

Convolutional Codes*• Punctured codes

– Take a rate 1/n code and not send some of the bits

– Example • this is a 2/3 code• Do not send the circled output


#4 65

Convolution Codes

• For an interactive example see• http://

www.ece.drexel.edu/commweb/cov/cov_html.html#mainstart

http://www.ece.drexel.edu/commweb/cov/cov_html.html#mainstart



#4 66

Errors

• Random Codes usually assume that bit errors are statistically independent

• Burst However fading and other effects induce burst errors, or a string of consecutive errors.

• Interleaving make burst errors look like random errors

#4 67

Interleaving

Write into Write into MemoryMemory

Read outRead outof of MemoryMemory

#4 68

Interleaving

• Increases delay, must read in and buffer entire block of data before it can be sent.

• Used in CD players– Specifications: correct 90 ms of

errors– at 44 kb/s & 16 bits/sample

•4000 consecutive bit errors

#4 69

References

• Leon-Garcia & Widjaja: Communication Networks, McGraw Hill, 2004

• http://complextoreal.com/chapters/convo.pdf

• S.H. Lin, D.J. Castello: 'Error Correction Coding',. McGraw Hill, 1983

http://complextoreal.com/chapters/convo.pdf



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#4 1 Victor S. Frost Dan F. Servey Distinguished Professor Electrical Engineering and Computer...

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