L4139 - PowerPoint PPT Presentation

Title: L4139


1
COM347J1Networks and Data Communications
Lecture 4 Data Compression, Error Detection and
Error Correction

• Ian McCrum Room 5D03B
• Tel 90 366364 voice mail on 6th ring
• Email IJ.McCrum_at_Ulster.ac.uk
• Web site http//www.eej.ulst.ac.uk

2
The Encoding and compression of data

• Introduction
• Information Content of a message stream
• simple coding methods
• Huffman coding
• compression techniques

3
REDUNDANCY

• Consider that you were in receipt of the
  following telegram
• RONMIE (ROCKTT) OSULLIVON 146 CREAK
• It is possible due to the inherent redundancy of
  natural language to perform a reconstruction
  leading to the message on the next slide.

4
REDUNDANCY

• Consider that you were in receipt of the
  following telegram
• RONMIE (ROCKTT) OSULLIVON 146 CREAK
• It is possible due to the inherent redundancy of
  natural language to perform a reconstruction
  leading to the message below.
• RONNIE (ROCKET) OSULLIVAN 146 BREAK
• but what about the numbers in the message?

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Redundancy

• Redundancy arises due to the correlation of
  letters occurring in natural language, consider
  the word
• YACH ( if T is sent it will carry no information)
• Is it possible for a coding schema to produce an
  Ideal code?

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Reduction of Redundancy

• observe the
• Statistical occurrence of symbols
• Repetition of symbols
• employ
• Fano coding, Huffman coding (the most common
  symbols are given shorter codes)
• data compression (e,g code repetition as a
  special case)

7
Packed decimal / half byte compression

• When frames just contain numeric characters
• use binary coded decimal instead of 7 bit ASCII
  or 8 bit EBCDIC as only the four least
  significant bits change with number.
• In ASCII and in same column are used as
  decimal pt and space respectively

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Packed Decimal
STX Cntrl XX 26 3 2 45
ETX BCC
Closing flag Block CC
1st number 26.32
Number of digits following
Control character half byte compression
Opening Flag
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Relative encoding

• Whenever only small differences occur between
  successive values
• send only that difference
• very effective in data logging
• consider level of a river

Relative encoding sign, number and delimiter
STX 1 4
ETX BCC
Relative encoding using signed 8 bit integers
STX 3 -95 11 124 -100
ETX BCC
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Character suppression

• in a stream of digits there are often sequences
  of the same characters, most frequently spaces.
• if a continuous string of three or more chars in
  a sequence it is replaced by Cntrl,char,number
• thus CntrlF25 means 25Fs in a sequence.
• type of run-length encoding

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Character suppression
STX Cntrl sp 45 A B
ETX BCC
Single letters
Closing flag Block CC
number of chars
Char being suppressed
Control character
Opening Flag
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Run length encoding

• Run-length compression where the codeword
  actually contains the number of repetitions.
• A three byte minimum repetition is chosen such
  that all occurrences of repetitions greater or
  equal to 3 will be encoded thus.
• ltchargtltchargtltchargtltngt
• this four byte codeword can represent repetitions
  up to 259
• ltchargt ltchargt
• ltchargtltchargt ltchargtltchargt
• ltchargtltchargtltchargt ltchargtltchargtltchargtlt0gt
• ltchargtltchargtltchargtltchargt ltchargtltchargtltchargtlt1gt
• ltchargtltchargtltchargtltchargtltchargt ltchargtltchargtltchargt
  lt2gt

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Huffman coding

• Instead of representing symbols with a fixed no
  of bits, fewer bits are used for frequently
  occurring symbols and vice versa
• Method Determine the relative frequency of
  symbols. Create an unbalanced tree with unequal
  branches.

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Example of Huffman

• Consider that a group of characters A to H is to
  be transmitted. This comprises
• 9As, 9Bs, 5Cs, 5Ds, 2Es, 2Fs, 2Gs, 2Hs
• Sequence of operations.
• a) Order the symbols in terms of probability
• b) Combine the two least frequently occurring
  symbols
• c) assigning 1(upper) and 0(lower) to each.
• d) This is now considered to be one entity.

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Huffman continued

• Perform the same steps until only two symbols are
  left.
• Determine the codeword by reading from left to
  right. The first bit being read is the least
  significant one.

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Comparison

• If there were N symbols then N codewords would be
  sent. In the case of fixed length binary codes
  this would be represented by 3N bits.
• How does this compare with those required by this
  example of Huffman encoding?

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Therefore there has been a saving of 0.28N bits
in comparison with fixed length binary each of 3
bits. Redundancy it can shown that the ideal
code for this sequence of symbols would take
2.53N bits ie. this is the actual information
content of the stream of codewords. Thus for
fixed length binary codes the
Information content Redundancy 1 -
------------------------- Number of
bits sent or 1 -
2.53N/3.0N 16 for Huffman
1 - 2.53N/2.72N
7
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MNP Class 5 Compression

• is a combination of Huffman and run-length
  encoding.
• The symbol stream is run-length encoded with a
  minimum repetition of 3 bytes and then Huffman
  encoded using a statistically generated table.
• During transmission the statistics for the
  occurrence of each symbol are updated and the
  allocation of codewords are dynamically changed.
• MNP Class 5 compression achieves 21 compression
  on a regular basis. Its major drawback is that
  cannot turn itself off when it offers no gain, so
  that an incompressible file actually expands by
  gt10.

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Error detection and protection

• Introduction
• Error Detection
• recognise that one has happened
• Error Correction
• repair damaged data
• parity and CRC.
• BCC and Hamming,

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Data errors

• Errors can arise due to attenuation of signal
  strength and due to other reasons.
• well shaped signals can become distorted and
  thus misinterpreted.
• Random errors (each occurs with certain
  probability)
• noise in electronics
• distance traveled
• Burst errors (groups of bits in error occur)
• source interference
• faults in equipment

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Error detection

• A sequence of bits (I0 In) is subjected to some
  processing (P) giving rise to a check sequence
  (C0Ck)
• Both are transmitted toward a receiver and incur
  a possibility of corruption.
• Upon reception the bit stream is separated into
  received data (I0r Inr) and received check
  sequence (C0rCkr).
• The received data (I0r Inr) is assumed to be
  correct and the same processing (P) is performed
  on it giving the reconstructed sequence
  (C0rr...Ckrr).
• If received check sequence (C0rCkr) and the
  reconstructed sequence (C0rr...Ckrr) are equal
  then no detectable error has occurred.

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Parity for ASCII codes

• Consider a seven-bit ASCII code to comprise the
  following bits which can be labeled I6, I5, I4,
  I3, I2, I1, I0
• A Parity bit P0 is placed beside the most
  significant bit I6 so that the codeword P0, I6,
  I5, I4, I3, I2, I1, I0 is formed.
• The Parity bit is determined as before so that
  for Odd parity there are an odd number of 1s in
  the codeword.
• and for Even parity there are an even number of
  1s in the codeword.

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Block Sum Check Character
P0 I6 I5 I4 I3 I2 I1 I0 1 0 1 1 1 0 1 0 0 1 1 1
1 0 0 1 1 1 0 1 1 1 0 0 0 0 1 0 0 1 1 0 1 1 1 0 1
0 0 1 1 1 1 1 0 1 1 1 0 1 0 1 1 0 0 0
Codeword 1 Codeword 2 Codeword 3 Codeword
4 Codeword 5 Codeword 6 Block Check Char.
Hey! See me!!
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Block Sum Check Character

• Consider what this method can do
• in terms of detecting errors.
• in terms of correcting errors.
• Can you see where it might be used in practice?
• Where will it cease to work adequately?

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Cyclic Redundancy Check (CRC)

• The CRC is so called because the codes fall into
  a class of cyclic codes each forming new legal
  code which shifted, when added to a sequence of
  bits they increase the redundancy of the
  codeword.
• The data sequence is divided by a standard
  polynomial and the remainder is the check bits or
  CRC.
• Polynomial is of the form
• 1.X4 0.X3 1.X2 0.X1 1
• more usually written X4 X2 1
• and in binary take the form 10101

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The arithmetic is different! But easier

• In decimal 0..9 and 0..9 means 100 different
  additions and 21 different answers (0..20)
• In binary using a half adder or exclusive OR
  there are (0 1) and (0 1) meaning 4 different
  additions and only 2 answers.
• Thus 0 ? 0 0, 0 ? 1 1, 1 ? 0 1 and
  1 ? 1 0
• ? being the symbol for exclusive OR.
• think of a half adder being an adder without a
  carry.

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To perform CRC determination

• Get data to be protected, ok 11011
• Choose polynomial ok X4 X2 1
• append to data the number of bits indicated by
  the maximum order of the polynomial (4) giving
  110110000
• divide this number by the polynomial thus
• 110110000 / 10101
• Take the remainder and send after the original
  data.
• Upon reception check received CRC with
  reconstructed CRC to determine error conditions.

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Use the polynomial x4 x2 1 to generate CRC
11101 10101 110110000
10101 11100 10101
10010 10101 11100
10101 1001
Thus the remainder is 1001and codeword 110111001
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Does 111010010 contain an error, generated by
using the same polynomial as before.
11000 10101 111010000
10101 10000 10101
10100 10101 010

Thus the remainder is 0010 and codeword 111010010
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Or divide rx data and crc by generating
polynomial and remainder should be zero
11010 10101 111010010
10101 10000 10101
10101 10101 000

Thus the remainder is 000 and codeword 11101 was
rx ok!
33
Hamming Codes
11 10 9 8 7 6 5 4 3 2 1
position in codeword I6 I5 I4 C3 I3 I2 I1 C2 I0
C1 C0 information and checks
Given an ASCII code 1001010 what is the Hamming
Code?
11 10 9 8 7 6 5 4 3 2 1 I6
I5 I4 C3 I3 I2 I1 C2 I0 C1 C0 1 0 0 x 1 0 1 x 0
x x
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How to determine the values of C3C2C1C0
C3 C2 C1 C0 11 1 0 1 1 7 0 1 1 1 5 0 1
0 1 1 0 0 1
I6 I5 I4 C3 I3 I2 I1 C2 I0 C1 C0 1 0 0 1 1 0 1 0
0 0 1
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How does this detect an error?
I6 I5 I4 C3 I3 I2 I1 C2 I0 C1 C0 1 0 0 1 1 1 1 0
0 0 1
Bit in error
C3 C2 C1 C0 11 1 0 1 1 8 1 0 0 0 7 0 1 1
1 6 0 1 1 0 5 0 1 0 1 1 0 0 0 1 0 1 1 0
Therefore 6th bit was received in error
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Summary

• Hamming codes have their redundant bits in the
  positions which are powers of 2 ie 1,2,4,8 etc
• They can detect and correct single errors.
• They can indicate multiple error conditions but
  cannot correct.
• Used for random errors.
• Can you think of how they might be applied to a
  circumstance a burst error could occur? Assume
  that the burst is shorter that 8 bits and there
  are 256 bytes to be transmitted.