EECS 122: Encoding, Framing, Error Detection and Recovery

1
EECS 122 Encoding, Framing, Error Detection and
Recovery

• Computer Science Division
• Department of Electrical Engineering and Computer
  Sciences
• University of California, Berkeley
• Berkeley, CA 94720-1776

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Overview

• Encoding
• Framing
• Error detection correction

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Encoding

• Goal send bits from one node to another node on
  the same physical media
• This service is provided by the physical layer
• Problem specify a robust and efficient encoding
  scheme to achieve this goal

Signal
Adaptor
Adaptor
Adaptor convert bits into physical signal and
physical signal back into bits
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Assumptions

• Use two discrete signals, high and low, to encode
  0 and 1
• The transmission is synchronous, i.e., there is a
  clock used to sample the signal
• In general, the duration of one bit is equal to
  one or two clock ticks

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Non-Return to Zero (NRZ)

• 1 ? high signal 0 ? low signal

• Disadvantages when there is a long sequence of
  1s or 0s
• Sensitive to clock skew, i.e., difficult to do
  clock recovery
• Difficult to interpret 0s and 1s (baseline
  wander)

0
0
1
0
1
0
1
1
0
Clock
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Non-Return to Zero Inverted (NRZI)

• 1 ? make transition 0 ? stay at the same level
• Solve previous problems for long sequences of
  1s, but not for 0s

0
0
1
0
1
0
1
1
0
Clock
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Manchester

• 1 ? high-to-low transition 0 ? low-to-high
  transition
• Addresses clock recovery and baseline wander
  problems
• Disadvantage needs a clock that is twice as fast
  as the transmission rate

0
0
1
0
1
0
1
1
0
Clock
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4-bit/5-bit

• Goal address inefficiency of Manchester
  encoding, while avoiding long periods of low or
  high signals
• Solution
• Use 5 bits to encode every sequence of four bits
  such that no 5 bit code has more than one leading
  0 and two trailing 0s
• Use NRZI to encode the 5 bit codes

4-bit 5-bit
4-bit 5-bit

• 0000 11110
• 0001 01001
• 0010 10100
• 0011 10101
• 0100 01010
• 0101 01011
• 0110 01110
• 0111 01111

• 1000 10010
• 1001 10011
• 1010 10110
• 1011 10111
• 1100 11010
• 1101 11011
• 1110 11100
• 1111 11101

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Overview

• Encoding
• Framing
• Error detection Correction

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Framing

• Goal send a block of bits (frames) between nodes
  connected on the same physical media
• This service is provided by the data link layer
• Use a special byte (bit sequence) to mark the
  beginning (and the end) of the frame
• Problem what happens if this sequence appears in
  the data payload?

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Byte-Oriented Protocols Sentinel Approach
8
8
Text (Data)
STX
ETX

• STX start of text
• ETX end of text
• Problem what if ETX appears in the data portion
  of the frame?
• Solution
• If ETX appears in the data, introduce a special
  character DLE (Data Link Escape) before it
• If DLE appears in the text, introduce another DLE
  character before it
• Protocol examples
• BISYNC, PPP, DDCMP

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Byte-Oriented Protocols Byte Counting Approach

• Sender
• Insert the length of the data (in bytes) at the
  beginning of the frame, i.e., in the frame header
• Receiver
• Extract this length and decrement it every time a
  byte is read when this counter becomes zero, we
  are done

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Bit-Oriented Protocols
8
8
Start sequence
End sequence
Text (Data)

• Both start and end sequence can be the same
• E.g., 01111110 in HDLC (High-level Data Link
  Protocol)
• Sender inserts a 0 after five consecutive 1s
• Receiver when it sees five 1s makes decision on
  the next two bits
• if next bit 0 (this is a stuffed bit), remove it
• if next bit 1 (sixth 1 in a row), look at the
  next bit
• If 0 this is end-of-frame (receiver has seen
  01111110)
• If 1 this is an error, discard the frame
  (receiver has seen 01111111)

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Clock-Based Framing (SONET)

• SONET (Synchronous Optical NETwork)
• Example SONET ST-1 51.84 Mbps

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Clock-Based Framing (SONET)

• First two bytes of each frame contain a special
  bit pattern that allows to determine where the
  frame starts
• No bit-stuffing is used
• Receiver looks for the special bit pattern every
  810 bytes
• Size of frame 9x90 810 bytes

Data (payload)
overhead
9 rows
SONET STS-1 Frame
90 columns
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Clock-Based Framing (SONET)

• Details
• Overhead bytes are encoded using NRZ
• To avoid long sequences of 0s or 1s the payload
  is XOR-ed with a special 127-bit patter with many
  transitions from 1 to 0
• Duration of a frame is 51.84 usec (51.84 Mbps for
  STS-1)

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High Level View

• Goal transmit correct information
• Problem bits can get corrupted
• Electrical interference, thermal noise
• Solution
• Detect errors
• Recover from errors
• Correct errors
• Retransmission (already done this!)

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Overview

• Encoding
• Framing
• Error detection Correction

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Error Detection (and Correction)

• Problem detect bit errors in packets (frames)
• Solution add extra bits to each packet
• Goals
• Reduce overhead, i.e., reduce the number of added
  bits
• Increase the number and the type of bit error
  patterns that can be detected
• Examples
• Two-dimensional parity
• Checksum
• Cyclic Redundancy Check (CRC)

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Overview

• Encoding
• Framing
• Error detection Correction
• Two-dimensional parity
• Checksum
• Cyclic Redundancy Check (CRC)

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Two-dimensional Parity

• Add one extra bit to a 7-bit code such that the
  number of 1s in the resulting 8 bits is even (or
  odd for odd parity)
• Add a parity byte for the packet
• Example five 7-bit character packet, even parity

0110100
1
1011010
0
0010110
1
1110101
1
1001011
0
1000110
1
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How Many Errors Can you Detect?

• All 1-bit errors
• Example

0110100
1
1011010
0
0000110
1
error bit
1110101
1
1001011
0
1000110
1
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How Many Errors Can you Detect?

• All 2-bit errors
• Example

0110100
1
1011010
0
0000111
1
error bits
1110101
1
1001011
0
1000110
1
odd number of 1s on columns
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How Many Errors Can you Detect?

• All 3-bit errors
• Example

0110100
1
1011010
0
0000111
1
error bits
1100101
1
1001011
0
1000110
1
odd number of 1s on column
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How Many Errors Can you Detect?

• Most 4-bit errors
• Example of 4-bit error that is not detected

0110100
1
1011010
0
0000111
1
error bits
1100100
1
1001011
0
1000110
1
How many errors can you correct?
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Overview

• Encoding
• Framing
• Error detection Correction
• Two-dimensional parity
• Checksum
• Cyclic Redundancy Check (CRC)

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Checksum

• Sender add all words of a packet and append the
  result (checksum) to the packet
• Receiver add all words of a packet and compare
  the result with the checksum
• Can detect all 1-bit errors
• Example Internet checksum
• Use 1s complement addition

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1s Complement Revisited

• Negative number x is x with all bits inverted
• When two numbers are added, the carry-on is added
  to the result
• Example -15 16 assume 8-bit representation

15 00001111 ? -15 11110000

16 00010000
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Overview

• Encoding
• Framing
• Error detection Correction
• Two-dimensional parity
• Checksum
• Cyclic Redundancy Check (CRC)

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Cyclic Redundancy Check (CRC)

• Represent a (n1)-bit message as an n-degree
  polynomial M(x)
• E.g., 10101101 ? M(x) x7 x5 x3 x2 x0
• Choose k-degree polynomial C(x) as divisor
• Compute reminder R(x) of M(x)xk / C(x), i.e.,
  compute A(x) such that
• M(x)xk A(x)C(x) R(x), where degree(R(x)) lt
  k
• Let
• T(x) M(x)xk R(x) A(x)C(x)
• Then
• T(x) is divisible by C(x)
• First n coefficients of T(x) represent M(x)

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Cyclic Redundancy Check (CRC)

• Sender
• Compute and send T(x), i.e., the coefficients of
  T(x)
• Receiver
• Let T(x) be the (nk)-degree polynomial
  generated from the received message
• If C(x) divides T(x) ? no errors otherwise
  errors
• Note all computations are modulo 2

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Arithmetic Modulo 2

• Like binary arithmetic but without
  borrowing/carrying from/to adjacent bits
• Examples
• Addition and subtraction in binary arithmetic
  modulo 2 is equivalent to XOR

101 010 111
101 001 100
1011 0111 1100
1011 - 0111 1100
101 - 010 111
101 - 001 100
a b a b
0 0 0
0 1 1
1 0 1
1 1 0
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Some Polynomial Arithmetic Modulo 2 Properties

• If C(x) divides B(x), then degree(B(x)) gt
  degree(C(x))
• Subtracting/adding C(x) from/to B(x) modulo 2 is
  equivalent to performing an XOR on each pair of
  matching coefficients of C(x) and B(x)
• E.g.

B(x) x7 x5 x3 x2 x0
(10101101)
C(x) x3 x1
x0 (00001011)
B(x) - C(x) x7 x5 x2 x1
(10100110)
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Example (Sender Operation)

• Send packet 110111, i.e., M(X) x5x4x2x1
• Choose C(x) 101 (k 2)
• M(x)xK (x5x4x2x1)x2 x7x6x3x2 ?
  11011100
• Compute the reminder R(x) of M(x)xk / C(x)
• Compute T(x) M(x)xk - R(x) ? 11011100 xor 1
  11011101
• Send T(x)

101) 11011100 101 111
101 101 101
100 101
1
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Example (Receiver Operation)

• Assume T(x) 11011101
• C(x) divides T(x) ? no errors
• Assume T(x) 11001101
• Reminder R(x) 1 ? error!

• Note an error is not detected iff C(x) divides
  T(x) T(x)

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CRC Properties

• Detect all single-bit errors if coefficients of
  xk and x0 of C(x) are one
• Detect all double-bit errors, if C(x) has a
  factor with at least three terms
• Detect all burst of errors smaller than k bits
• Detect all number of odd errors, if C(x) contains
  factor (x1)