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 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

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

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) 1 1 1 1 NRZ (non-return to zero) Clock

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 1 1 1 1 NRZI (non-return to zero intverted) Clock

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 1 1 1 1 Manchester Clock

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 0000 11110 0001 01001 0010 10100 0011 10101 0100 01010 0101 01011 0110 01110 1111 01111 1000 10010 1001 10011 1010 10110 1011 10111 1100 11010 1101 11011 1110 11100 1111 11101

Overview Encoding Framing Error detection & Correction

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?

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

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.

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)

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

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

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)

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!)

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

Overview Two-dimensional Parity Checksum Cyclic Redundancy Check Hamming Codes

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 0010110 1 1110101 1 1001011 1000110 1

How Many Errors Can you Detect? All 1-bit errors Example: 0110100 1 1011010 0000110 odd number of 1’s 1 error bit 1110101 1 1001011 1000110 1

How Many Errors Can you Detect? All 2-bit errors Example: 0110100 1 1011010 0000111 1 error bits 1110101 1 1001011 1000110 1 odd number of 1’s on columns

How Many Errors Can you Detect? All 3-bit errors Example: 0110100 1 1011010 0000111 1 error bits 1100101 1 1001011 1000110 1 odd number of 1’s on column

How Many Errors Can you Detect? Most 4-bit errors Example of 4-bit error that is not detected: 0110100 1 1011010 0000111 1 error bits 1100100 1 1001011 1000110 1 How many errors can you correct?

Overview Two-dimensional Parity Checksum Cyclic Redundancy Check Hamming Codes

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

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 -15+16 = 1 + 16 = 00010000 00000000 1 + 1 00000001

Overview Two-dimensional Parity Checksum Cyclic Redundancy Check Hamming Codes

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)

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

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 a b a b 1

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)

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 101 111 100 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! 101) 11001101 101 110 111 1 R’(x) Note: an error is not detected iff C(x) divides T’(x) – T(x)

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

Overview Two-dimensional Parity Checksum Cyclic Redundancy Check Hamming Codes

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

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)

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

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… 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

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) b1 b2 b3 b4 b5 b6 b7 Position 1 10 11 100 101 110 111 Check:b1 x Check:b2 Check:b4 b1 b2 1 b4 Position 10 11 100 101 110 111 Check:b1 Check:b2 Check:b4

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