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# Data communication line codes and constrained sequences - PowerPoint PPT Presentation

Data communication line codes and constrained sequences. A.J. Han Vinck Revised December 09, 2010. A-synchronous arrivals: Line codes. Binary data transmitted using special signal form Important aspects: Enough transitions in signal to gain timing information

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### Data communicationline codes and constrained sequences

A.J. Han Vinck

Revised December 09, 2010

A-synchronous arrivals: Line codes

• Binary data transmitted using special signal form

• Important aspects:

• Enough transitions in signal to gain timing information

• Constraints to match sequence to the physical channel

• Applications:

• Optical transmission

• Magnetic and Optical recording

• Classical line transmission

A.J. Han Vinck

• maximum run of same symbols

• changes in sequence used to synchronize

• minimum run of same symbols

• used to improve transmision efficiency

Example: ( minimum = 2, maximum = 5 )

00111000001111001100111

A.J. Han Vinck

Line codes for clock recovery: Manchester code

1 0 1 1 0 0 0 0 1 0 1

+5V

0

Basic frequency

1 0 1 1 0 0

+5V

0

-5V

• Longest run = 2, but time to transmit twice as large! Efficiency = ½

Detection is simple:

Transition 1  0 => 1

0  1 => 0

A.J. Han Vinck

Line codes for clock recovery: Manchester code

1 0 1 1 0 0 0 0 1 0 1

+5V

0

Basic frequency

+5V

0

-5V

• Longest run = 2

Detection is simple:

Transition 1  0 => 1

0  1 => 0

Same time to transmit, but

basic frequency factor of 2 higher!

A.J. Han Vinck

4B/5B guarantees: at least one transition per block to allow the clock signal to be recovered.

Efficiency decreased by 25%.

4B5B is used in the following standard: 100BASE-TX standard defined by IEEE 802.3u

A.J. Han Vinck

Definition: between 2 „ones“: maximum of k zeros; minimum of d zeros

ex:0100010010001101

• Conversion from (d,k) into RLL 0100010010001101

1000011100001001

or

0111100011110110

the „ones“ indicate the position of a transition in the RLL sequence

Note, from a (d,k) sequence we can derive 2 RLL sequences depending on the „starting state“ (0 or 1)

A.J. Han Vinck

RLL sequences have minimum run of d+1 and maximum of k+1

ex: the Manchester code has (d,k) = (0,1)

but efficiency ½, i.e. sequence is twice as long

efficiency can be improved to 0.69, i.e. about 1.4 times as long!

Problems: generation of (d,k) sequences

how efficient can this be done

A.J. Han Vinck

• Calculate N(n),

the maximum number of sequences with constraint and length n,

• The „capacity is then defined as

• C := maximum # of information bits/channel use

A.J. Han Vinck

• the following Markov source generates all (d,k) sequences

Very important for practical applications: how far is a construction away from optimal?

0 0 0 0 0

0 1 2 d d+1 k

1 1 1

A.J. Han Vinck

Ex: Modified Frequency Modulation MFM (d,k) = (1,3)

• Use markov state diagram to describe sequence generation

1/01

0/10 1/01

A 0/00 B

Decoding rule:

0 x0

1 01

Efficiency = ½

Maximum C(1,3) = 0.55

A.J. Han Vinck

0 0 0

S1

S2

S3

S4

1 1 1

# sequences in state S1 at time n = # sequences in state S1 at time n-2

+ # sequences in state S1 at time n-3

+ # sequences in state S1 at time n-4

A solution can be of the form: # sequences in state S1 at time n  cZn

Then, we obtain the relation: Zn =Zn-2 +Zn-3 +Zn-4 and

For a solution  we calculate C(d,k) => 1/n log2 c n => log2 = 0.55

A.J. Han Vinck

0

1

S1

S2

1

# sequences in state S1 at time n = # sequences in state S1 at time n-2

+ # sequences in state S1 at time n-1

# sequences in state S1 at time n  cZn. Then, Zn =Zn-1 +Zn-2

For a solution  we calculate C(d,k) => 1/n log2 c n => log2 ~ 0.69

A.J. Han Vinck

A.J. Han Vinck

Ex:CD (EFM) uses d=2, k=10

Note: DvD uses EFM+

A.J. Han Vinck

in recording and transmissionkeep basic basic frequency or minimum symbol length!

• no constraint: L bits take time  L/1

1

0

1

0

d = 1

d = 2

1

0

A.J. Han Vinck

• no constraint: L bits take time  L/1

1

0

• with constraint: L bits take time ‘ L/C(d,k), where C(d,k) < 1

• i.e. it takes more symbols with the contraint included. The best performance is indicated by C(d,k)

1

0

• same spacing: (d+1) units of time ‘= /(d+1)

SURPRISE:‘L/C(d,k) =  L/[(d+1) C(d,k)] <  L for [(d+1) C(d,k)] > 1

A.J. Han Vinck

• 00 00 or 11 2 d = 1

• 01 000 or 111 3

• 10 0000 or 1111 4

• 11 00000 or 11111 5 R = 2/3.5 = 1/1.75 ( = less than C(d,k) )

• Withconstraint L bitstake

‘L/R =  L/(d+1) R = 0.875  L <  L

A.J. Han Vinck

Example: 4 symbols 00,01,10,11

encode sequence 00, 01, 10, 00, 00...

as

00, 111, 0000, 11, 00,...,

where after every symbol we change polarity.

The code is uniquely decodable!

A.J. Han Vinck

Whathappened?thesymboldurationchanged !!

Thus, we can pack more sequences in the same time.

• no constraint (11,01,10,00)

1

0

• with constraint

1

0

• same minimum spacing, but different run lengths

A.J. Han Vinck

Main idea: (like in coded modulation, Ungerböck)

uncoded

RLL- coded; d = 1

0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 1

0 0 0 0 0 0 1 1

0 0 1 0

0 0 0 0 1 1 0 0

2k





> 2k

1 1 1 1

1 1 1 1 1 1 1 1

0 1 1 1 0 0 1 1

T

0 0 0 0 1 1 1 0

0 0 0 1 1 1 0 0

T

- Same minimum transition time,

same information rate k/T= 1/

d=0 d=1 d=2 d=3

• k

• 1 .69

• 2 .88 .41

• 3 .95 .55 .29

• 4 .98 .62 .41 .22

• 5 .99 .65 .46 .32

A.J. Han Vinck

(d,k) RLL

• 10  1000 0000 or 1111

• 11  0100 0111 or 1000

• 011  000100 000111 or 111000

• 010  001000 001111 or 110000

• 000  100100 111000 or 000111

• 0011  00100100 00111000 or 11000111

• 0010  00001000 00001111 or 11110000

Homework: show that this code does not violate the constraint!

A.J. Han Vinck

Remark:

- for every L information bits, we produce exactly 2L channel digits

Thus, the efficiency = ½ (Maximum 0.51 )

- Sequence uniquely decodable!

Instead of using L positions to store L bits of information

we need only L/R*(d+1) = L/1.5 positions or 1/3 less!

A.J. Han Vinck

• 000 00000 R = 3/5 = 0.6 (Maximum = 0.69)

• 001 00001

• 010 00010

• 011 00100 R*(d+1) = 1.2

• 100 00101

• 101 01000

• 110 01001

• 111 01010

Problem:

• What is maximum rate?

• How to encode?

A.J. Han Vinck

• For large d:

exact timing

detection of transition

erroneous shift of transition is called peak shift

(in recording)

A.J. Han Vinck

Equal weight codes: # ones is the same for every code word

example: 0011, 0101, 0110, 1001, 1010, 1100

DC-balanced: try to balance the number of ones and zeros

example: transmit 000 or 111, 010 or 101

001 or 110, 100 or 011

only 1-bit redundancy, for every length

Pulse Position modulation: weight 1 code words

example: 001, 010, 100

A.J. Han Vinck