1 / 42

EECS 122: EE122: Error Detection and Reliable Transmission

EECS 122: EE122: Error Detection and Reliable Transmission. Computer Science Division Department of Electrical Engineering and Computer Sciences University of California, Berkeley Berkeley, CA 94720-1776. Overview. Encoding Framing Error detection & correction. Encoding.

rmorena
Download Presentation

EECS 122: EE122: Error Detection and Reliable Transmission

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. EECS 122: EE122: Error Detection and Reliable Transmission 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 • 1111 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. 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) • Hamming Codes

  19. Overview • Two-dimensional Parity • Checksum • Cyclic Redundancy Check • Hamming Codes

  20. 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

  21. 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

  22. 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

  23. 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

  24. 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?

  25. Overview • Two-dimensional Parity • Checksum • Cyclic Redundancy Check • Hamming Codes

  26. 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

  27. -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

  28. Overview • Two-dimensional Parity • Checksum • Cyclic Redundancy Check • Hamming Codes

  29. 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)

  30. 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

  31. 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

  32. 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)

  33. R(x) Example (Sender Operation) • Send packet 110111; choose C(x) = 101 (k = 2) • M(x)*xK  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

  34. 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)

  35. 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 number of odd errors, if C(x) contains factor (x+1) • Detect all burst of errors smaller than k bits

  36. Overview • Two-dimensional Parity • Checksum • Cyclic Redundancy Check • Hamming Codes

  37. Code words • Combination of the n payload bits and the k check bits as being a n+k bit code word • For any error correcting scheme, not all n+k bit strings will be valid code words • Errors can be detected if and only if the received string is not a valid code word • Example: even parity check only detects an odd number of bit errors

  38. Hamming Distance • Given code words A and B, the Hamming distance between them is the number of bits in A that need to be flipped to turn it into B • E.g., H(011101,000000) = 4 • If all code words are at least d Hamming distance apart, then up to d-1 bit errors can be detected Richard W. Hamming (1915-1998)

  39. Error Correction • If all the code words are at least a Hamming distance of 2d+1 apart then up to d bit errors can be corrected • Just pick the codeword closest to the one received! • How many bits are required to correct d errors when there are n bits in the payload? • Example: d=1: Suppose n=3. Then any payload can be transformed into 3 other payload strings (e.g., 000 into 001, 010 or 100). • Need at least two extra bits to differentiate between 4 possibilities • In general need at least k ≥ log2(n+1) bits to correct one error • A scheme that is optimal is called a perfect parity code

  40. Perfect Parity Codes • Consider a codeword of n+k bits • b1 b2 b3 b4 b5 b6 b7 b8 b9 b10 b11… • Parity bits are in positions 20, 21, 22 ,23 ,24… • b1b2b3 b4b5 b6 b7 b8b9 b10 b11… • A parity bit in position 2h, checks all data bits bp such that if you write out p in binary, the hth place in p’s binary representation is an one

  41. Example: (7,4)-Parity Code • n=4, k=3 • Corrects one error • log2(1+n) = 2.32  k = 3, perfect parity code • data payload = 1010 • For each error there is a unique combination of checks that fail • E.g., 3rd bit is in error,:1000  both b2 and b4 fail (single case in which only b2 and b4 fail)

  42. Summary • Encoding – specify how bits are transmitted on the physical media • Challenge – achieve • Efficiency – ideally, bit rate = clock rate • Robust – avoid de-synchronization between sender and receiver when there is a large sequence of 1’s or 0’s • Framing – specify how blocks of data are transmitted • Challenge • Decide when a frame starts/ends • Differentiate between the true frame delimiters and delimiters appearing in the payload data

More Related