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

2. Overview • Encoding • Framing • Error detection & correction

3. 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

4. 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

5. NRZ (non-return to zero) Non-Return to Zero (NRZ) • 1  high signal; 0  low signal • Disadvantages: when there is a long sequence of 1’s or 0’s • Sensitive to clock skew, i.e., difficult to do clock recovery • Difficult to interpret 0’s and 1’s (baseline wander) 0 0 1 0 1 0 1 1 0 Clock

6. NRZI (non-return to zero intverted) Non-Return to Zero Inverted (NRZI) • 1  make transition; 0  stay at the same level • Solve previous problems for long sequences of 1’s, but not for 0’s 0 0 1 0 1 0 1 1 0 Clock

7. Manchester 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

8. 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 0’s • Use NRZI to encode the 5 bit codes 4-bit 5-bit 4-bit 5-bit • 1000 10010 • 1001 10011 • 1010 10110 • 1011 10111 • 1100 11010 • 1101 11011 • 1110 11100 • 1111 11101 • 0000 11110 • 0001 01001 • 0010 10100 • 0011 10101 • 0100 01010 • 0101 01011 • 0110 01110 • 0111 01111

9. Overview • Encoding • Framing • Error detection & Correction

10. 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?

11. Byte-Oriented Protocols: Sentinel Approach 8 8 • 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 STX Text (Data) ETX

12. 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

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

14. Clock-Based Framing (SONET) • SONET (Synchronous Optical NETwork) • Example: SONET ST-1: 51.84 Mbps

15. 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 SONET STS-1 Frame 9 rows 90 columns

16. Clock-Based Framing (SONET) • Details: • Overhead bytes are encoded using NRZ • To avoid long sequences of 0’s or 1’s 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)

17. 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!)

18. Overview • Encoding • Framing • Error detection & Correction

19. 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)

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

21. Two-dimensional Parity • Add one extra bit to a 7-bit code such that the number of 1’s 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

22. odd number of 1’s 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

23. 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 1’s on columns

24. 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 1’s on column

25. 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?

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

27. 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 1’s complement addition

28. -15+16 = 1 00000000 1 + 1 00000001 1’s 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

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

30. Cyclic Redundancy Check (CRC) • Represent a (n+1)-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)) < 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)

31. Cyclic Redundancy Check (CRC) • Sender: • Compute and send T(x), i.e., the coefficients of T(x) • Receiver: • Let T’(x) be the (n+k)-degree polynomial generated from the received message • If C(x) divides T’(x)  no errors; otherwise errors • Note: all computations are modulo 2

32. 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 101 - 010 111 101 - 001 100 1011 - 0111 1100

33. Some Polynomial Arithmetic Modulo 2 Properties • If C(x) divides B(x), then degree(B(x)) >= 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)

34. R(x) Example (Sender Operation) • Send packet 110111, i.e., M(X) = x5+x4+x2+x+1 • Choose C(x) = 101 (k = 2) • M(x)*xK = (x5+x4+x2+x+1)*x2 = x7+x6+x3+x2  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

35. 101) 11001101 101 110 101 111 101 101 101 1 R’(x) 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)

36. 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 (x+1)