1. Univ. of TehranIntroduction to Computer Network1 An Introduction to Computer Networks University of Tehran Dept. of EE and Computer Engineering By: Dr. Nasser Yazdani Point-to-Point link Lecture 4: Point-to-Point link

2. Univ. of TehranIntroduction to Computer Network2 Concepts: Data signal Links Link functions Encoding Framing Error Detection Error correction or reliable transmission Outline

3. Univ. of TehranIntroduction to Computer Network3 Data Discrete data: an instance is binary. Computer works with discrete data. Discrete is encoded in 0s and 1s. Continuous data: change with time or space. It is converted to discrete data by sampling Data is delivered by signals in the links

4. Univ. of TehranIntroduction to Computer Network4 Link Functions Functions 1. Construct Frame with Error Detection Code 2. Encode bit sequence into analog signal 3. Transmit bit sequence on a physical medium (Modulation) 4. Receive analog signal 5. Convert Analog Signal to Bit Sequence 6. Recover errors through error correction and/or ARQ Adaptor Signal Adaptor: convert bits into physical signal and physical signal back into bits

5. Univ. of TehranIntroduction to Computer Network5 Link Components NRZI

6. Univ. of TehranIntroduction to Computer Network6 Link Properties Function Duplex/Half Duplex One stream, multiple streams Characteristics Bit Error Rate Data Rate (this sometimes mistakenly called bandwidth!) Degradation with distance

7. Univ. of TehranIntroduction to Computer Network7 Example: Optical Links

8. Univ. of TehranIntroduction to Computer Network8 Electromagnetic waves propagating in the light speed. Frequency Wavelength A (periodic) signal can be viewed as a sum of sine waves of different frequencies and strengths. Every signal has an equivalent representation in the frequency domain. » What frequencies are present and what is their strength (energy)Signals

9. Univ. of TehranIntroduction to Computer Network9 Example- The following signal is the sum of sine waves. Signals (cont)

10. Univ. of TehranIntroduction to Computer Network10 Sender changes the nature of the signal in a way that the receiver can recognize. » Similar to radio: AM or FM Digital transmission: encodes the values 0 or 1 in the signal. » It is also possible to encode multi-valued symbols Amplitude modulation: change the strength of the signal, typically between on and off. » Sender and receiver agree on a “rate” » On means 1, Off means 0 Similar: frequency or phase modulation. Can also combine method modulation types.Modulation

11. Univ. of TehranIntroduction to Computer Network11 Modulation The function of transmitting the encoded signal over a link, often by combining it with another (carrier signal) E.g. Frequency Modulation (FM) Combine the signal with a carrier signal in such a way that the instantaneous frequency of the received signal contains the information of the carrier E.g. Frequency Hopping (OFDM) Signal transmitted over multiple frequencies Sequence of frequencies is pseudo random

12. Univ. of TehranIntroduction to Computer Network12 Link rate and Distance Links become slower with distance because of attenuation of the signal Amplifiers and repeaters can help

13. Univ. of TehranIntroduction to Computer Network13 Noise: “random” energy is added to the signal. Attenuation: some of the energy in the signal leaks away. We need repeaters. Dispersion: attenuation and propagation speed are frequency dependent. » Changes the shape of the signal Limits in sending signals

14. Univ. of TehranIntroduction to Computer Network14 Noise A signal s(t) sent over a link is generally Distorted by the physical nature of the medium This distortion may be known and reversible at the receiver Affected by random physical effects Shot noise Fading Multipath Effects Also interference from other links Wireless Crosstalk Dealing with noise is what communications engineers do

15. Univ. of TehranIntroduction to Computer Network15 Noise limits the link rate Suppose there were no noise E.g. Send s(t) always receive s(t+Δ) Take a message of N bits say b 1 b 2 ….b N, and send a pulse of amplitude of size 0.b 1 b 2 ….b N Can send at an arbitrarily high rate This is true even if the link distorts the signal but in a known way In practice the signal always gets distorted in an unpredictable (random) way Receiver tries to estimate the effects but this lowers the effective rate One way to mitigate noise is to jack up the power of the signal Signal to Noise ratio (SNR) measures the extent of the distortion effects

16. Univ. of TehranIntroduction to Computer Network16 Every transmission medium supports transmission in a certain frequency range. This is called channel capacity. » The channel bandwidth is determined by the transmission medium and the nature of the transmitter and receivers A noiseless channel of width H can at most transmit a binary signal at a rate 2 x H. » E.g. a 3000 Hz channel can transmit data at a rate of at most 6000 bits/second Assumes binary amplitude encoding Shannon extended this result by accounting for the effects of noise. More aggressive encoding can increase the channel bandwidth. » Example: modems Channel capacity

17. Univ. of TehranIntroduction to Computer Network17 Bandwidth affects the data rate There is usually a fixed range of frequencies at which the analog wave can traverse a link The physical characteristics of the link might govern this Example: Voice Grade Telephone line 300Hz – 3300Hz The bandwidth is 3000Hz For the same SNR, a higher bandwidth gives a higher rate

18. Univ. of TehranIntroduction to Computer Network18 Signaling bits on a link Multi-level Signaling 1 Levels 2 3 4 0011100100 1 Levels 2 3 4 5 6 7 8 000010101001110 2-bits per symbol 3-bits per symbol Ultimately, what limits the number of bits I can send per symbol?

19. Univ. of TehranIntroduction to Computer Network19 Maximum Capacity/Data Rate Shannon Capacity: Bandwidth of linkSignal-to-Noise ratio For example:  Bandwidth of telephone link from telephone to a typical home is approx 3300Hz – 300Hz = 3kHz  Signal-to-noise ratio is approx 30dB = 10log 10 (S/N)  Therefore, C = 3000*log 2 (1001) ~= 30kb/s Optical fiber has a higher capacity because the bandwidth, B, of a fiber is much greater than for wire; and it is less susceptible to noise, N.

20. Univ. of TehranIntroduction to Computer Network20 Sampling Result (Nyquist) Suppose a signal s(t) has a bandwidth B. Sampling Result: Suppose we sample it (accurately) every T seconds. If T≤ 1/2B then it is possible to reconstruct the s(t) correctly Only one signal with bandwidth B has these sample points There are multiple signals with these sample points for signals with bandwidth greater than B Increasing the bandwidth results in a richer signal space No noise allowed in the sampling result

21. Univ. of TehranIntroduction to Computer Network21 Unshielded twisted pair » Two copper wires twisted - avoid antenna effect » Grouped into cables: multiple pairs with common sheath » Category 3 (voice grade) versus category 5 » 100 Mbps up to 100 m, 1 Mbps up to a few km » Cost: ~ 10cents/foot Coax cables. » One connector is placed inside the other connector » Holds the signal in place and keeps out noise » Gigabit up to a km Copper Wire

22. Univ. of TehranIntroduction to Computer Network22 Copper Wire

23. Univ. of TehranIntroduction to Computer Network23  High frequency electromagnetic waves (>1GHz)  Line of sight terrestrial transmissions and for communications via satellites.  Some atmospheric interference occurs but reliable transmission can be obtained over distances up to 50 Km.  Microwaves is absorbed by rain and does not penetrate obstacles.Microwaves

24. Univ. of TehranIntroduction to Computer Network24  Thin thread of glass or plastic  Lightweight.  Fibers act as wave-guides for light which is usually produced by lasers.  Visible light has frequency around 5*10 15 Hz, which ensures an extremely high bandwidth.  The raw materials are cheap.  Immune to electrical interference.  Difficult to join and tap.  Security advantages Fiber Optic

25. Univ. of TehranIntroduction to Computer Network25 Leased Lines Dedicated link from Telephone Companies DS1 24 digital voice of 64Kbps DS3 28 DS1 STS stands for Synch. Transfer signal LineBandwidth DS1 (T1)1.544 Mbps DS3 (T3)44.736 Mbps STS-1 (OC1)51.840 Mbps STS-3 (OC3)155.250 Mbps STS-12 (OC12)622.080 Mbps STS-48 (OC48)2.48 Gbps STS-12 (OC12)622.080 Mbps STS-192 (OC192)9.95 MGbps

26. Univ. of TehranIntroduction to Computer Network26 Last-Mile Links Connect from home to network service providers xDSl (Digital Subscriber line), runs on local loop on telephone line. ADSL, VDSL CATV- Cabel TV, BW of 6 MHZ, is asymmetric. LineBandwidth POTS (modem)28.8-65Kbps ISDN64-128 Kbps XDSL16Kbps-55.2 Mbps CATV20-40 Mpbs

27. Univ. of TehranIntroduction to Computer Network27 Encoding Goal: send bits from one node to another node on the same physical media Encode binary data onto signals This service is provided by the physical layer Bits flow is between network adaptors. Problem: specify a robust and efficient encoding scheme to achieve this goal How to send clock information? Extract from changes in signals.

28. Univ. of TehranIntroduction to Computer Network28 Encoding (crieria) Criteria for encoding: Bit rate (in limited BW) Recovering time information (in LAN) Error detecting Immunity to noise and interference Complexity and cost of implementation. Bit rates V.s Baud Rates Baud rates is the number of pulses sent over the link or changing in signals.

29. Univ. of TehranIntroduction to Computer Network29 Assumptions We 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 If the amplitude and duration of the signals is large enough, the receiver can do a reasonable job of looking at the distorted signal and estimating what was sent.

30. Univ. of TehranIntroduction to Computer Network30 Non-Return to Zero (NRZ) 1  high signal; 0  low signal 0 01010110 NRZ (non-return to zero) Clock

31. Univ. of TehranIntroduction to Computer Network31 Problem with NRZ Consecutive 0’s or 1’s may create problems. Synchronization problem because of difference in the sender or receiver clocks. The average of signals which is used to distinguish between low or high may move and make the decoding difficult. This is called baseline wander Unable to recover clock

32. Univ. of TehranIntroduction to Computer Network32 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 01010110 Clock NRZI (non-return to zero intverted)

33. Univ. of TehranIntroduction to Computer Network33 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 Efficiency of 50% 0 01010110 Clock Manchester

34. Univ. of TehranIntroduction to Computer Network34 Encodings (cont) Bits NRZ Clock Manchester NRZI 0010111101000010

35. Univ. of TehranIntroduction to Computer Network35 4-bit/5-bit (100Mb/s Ethernet) Goal: address inefficiency of Manchester encoding, while avoiding long periods of low/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 Efficiency is 80% 0000 11110 0001 01001 0010 10100 0011 10101 0100 01010 0101 01011 0110 01110 0111 01111 4-bit 5-bit 1000 10010 1001 10011 1010 10110 1011 10111 1100 11010 1101 11011 1110 11100 1111 11101 4-bit 5-bit  Unused code are used for control. I.e. 11111 is for line idle or 00000 for the line is dead.  FDDI is using this scheme.

36. Univ. of TehranIntroduction to Computer Network36 Framing How to distinguish between data and garbage. Break sequence of bits into a frame Typically implemented by network adaptor Frames Bits AdaptorAdaptor Node B Node A

37. Univ. of TehranIntroduction to Computer Network37 Byte-oriented Approaches Sentinel-based delineate frame with special pattern: 01111110 or STX, ETX, etc characters. e.g., HDLC, SDLC, PPP Problem: special pattern or characters appear in the payload. Character escaping or stuffing: in BISYNC (from IBM) with DLE character. Body 8 16 8 CRC Beginning sequenc e Ending Header

38. Univ. of TehranIntroduction to Computer Network38 Byte-oriented Approaches (cont) Counter-based include payload length in the header e.g., DDCMP protocol from DEC. problem: count field corrupted solution: catch when CRC fails

39. Univ. of TehranIntroduction to Computer Network39 Bit-oriented Approaches Frame is a collection of bits, SDLC (IMB) (synch. Data Link Control), later changed to HDLC (high) by OSI Beginning and end with 01111110 problem: How if the sequence appear in the body. solution: Bit stuffing: add after 5 consecutive 1 a zero. Beginning SequenceHeader Body …….CRCEnding Sequence 8 16 8

40. Univ. of TehranIntroduction to Computer Network40 Clock-Based Framing e.g., SONET: Synchronous Optical Network By Bellcore Use NRZ, but scramble it for enough transition Multiplex multiple low-speed links into a high sp STS-n (STS-1 = 51.84 Mbps), each frame 125 us Overhead Payload 90 columns 9 rows

41. Univ. of TehranIntroduction to Computer Network41 Error Detection To send extra information to find error in the frame. The simplest form is sending two copies, inefficient. Sending the sum of values (?) in the frame, checksum. In Internet, consider 16 bits sequences and then use one-complement to find the result. Sending parity, Odd or even parity. Two dimensional parity.

42. Univ. of TehranIntroduction to Computer Network42 Error Detecting (goals) Goals: Reduce overhead, i.e., reduce the number of redundancy bits Increase the number and the type of bit error patterns that can be detected

43. Univ. of TehranIntroduction to Computer Network43 Internet Checksum Algorithm View message as a sequence of 16-bit integers; sum using 16-bit ones-complement arithmetic; take ones- complement of the result. u_short cksum(u_short *buf, int count) { register u_long sum = 0; while (count--) { sum += *buf++; if (sum & 0xFFFF0000) { /* carry occurred, so wrap around */ sum &= 0xFFFF; sum++; } return ~(sum & 0xFFFF); }

44. Univ. of TehranIntroduction to Computer Network44 Even Parity Check EDC field has 1 bit Sender: If number of 1’s in payload is odd, the check bit is 1, else it is 0 Receiver: Accept the packet if payload number of 1’s match the value of the check bit. Can detect an odd number of bit errors

45. Univ. of TehranIntroduction to Computer Network45 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 (for even parity, and odd for odd parity) Add a parity byte for the packet Example: five 7-bit character packet, even parity 0110100 1011010 0010110 1110101 1001011 1 0 1 1 0 10001101

46. Univ. of TehranIntroduction to Computer Network46 How Many Errors Can you Detect? All 1-bit errors Example: 0110100 1011010 0000110 1110101 1001011 1 0 1 1 0 10001101 error bit odd number of 1’s Can detect AND correct 1-bit errors

47. Univ. of TehranIntroduction to Computer Network47 Cyclic Redundancy Code (CRC) T M m R MSB i.e. T = M.2 r + R Modulo-2 addition (XOR) Add r bits of redundant data to an n-bit message Where r << n e.g., r= 32 and n = 12,000 (1500 bytes) r

48. Univ. of TehranIntroduction to Computer Network48 CRC(cont) Represent n-bit message as n-1 degree polynomial e.g., MSG=10011010 as M(x) = x 7 + x 4 + x 3 + x 1 Let k be the degree of some divisor polynomial e.g., C(x) = x 3 + x 2 + 1 Transmit polynomial T (x) that is evenly divisible by C(x) shift left k bits, i.e., M(x)x k subtract remainder of M(x)x k / C(x) from M(x)x k

49. Univ. of TehranIntroduction to Computer Network49 CRC (cont) Receiver polynomial T(x) + E(x) E(x) = 0 implies no errors Divide (T(x) + E(x)) by C(x); remainder zero if: E(x) was zero (no error), or E(x) is exactly divisible by C(x) All operation is done in modulo 2 in which there is no carry. Then, the operation can be done by XOR only.

50. Univ. of TehranIntroduction to Computer Network50 CRC(cont) Example: M(x)= 110101, C(x) = 1001 1 0 0 11 1 0 1 0 1 0 0 0 1 0 0 1 1 0 0 0 1 0 0 1 0 0 1 1 0 0 0 0 0 1 1 0 0 0 0 0 1 1 0 0 1 0 0 1 1 0 1 0 0 1 0 1 1 R The final transmitted message is: T(x) = 1 1 0 1 0 1 0 1 1

51. Univ. of TehranIntroduction to Computer Network51 Selecting C(x) All single-bit errors, as long as the x k and x 0 terms have non-zero coefficients. All double-bit errors, as long as C(x) contains a factor with at least three terms Any odd number of errors, as long as C(x) contains the factor (x + 1) Any ‘burst’ error (i.e., sequence of consecutive error bits) for which the length of the burst is less than k bits. Most burst errors of larger than k bits can also be detected See Table 2.6 on page 102 for common C(x)

52. Univ. of TehranIntroduction to Computer Network52 Hamming Distance Given codewords 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 Example H(011101,000000)=4 If all codewords are at least d Hamming distance apart, then up to d-1 bit errors can be detected and (d-1)/2 error can be corrected.

53. Univ. of TehranIntroduction to Computer Network53 Reliable Transmission Error detection code Very costly Acknowledgment and timeout Send Acks in opposite direction. Retransmit if sender did not receive Ack.

54. Univ. of TehranIntroduction to Computer Network54 Acknowledgements & Timeouts Sender Receiver Frame ACK T Timeout Time Sender Receiver Frame ACK T Timeout Frame ACK T Timeout Sender Receiver Frame ACK T imeout Frame ACK T imeout Sender Receiver Frame T imeout Frame ACK T imeout (a) (c) (b)(d)

55. Univ. of TehranIntroduction to Computer Network55 Stop-and-Wait Problem 1: keeping the pipe full Example 1.5Mbps link x 45ms RTT = 67.5Kb (8KB) 1KB frames imples 1/8th link utilization Problem 2: How about repeated frames? Add a sequence number to frames. Sender Receiver

56. Univ. of TehranIntroduction to Computer Network56 Sliding Window Allow multiple outstanding (un-ACKed) frames Upper bound on un-ACKed frames, called window SenderReceiver T ime … …

57. Univ. of TehranIntroduction to Computer Network57 SW: Sender Assign sequence number to each frame ( SeqNum ) Maintain three state variables: send window size ( SWS ) last acknowledgment received ( LAR ) last frame sent ( LFS ) Maintain invariant: LFS - LAR <= SWS Advance LAR when ACK arrives Buffer up to SWS frames  SWS LARLFS ……

58. Univ. of TehranIntroduction to Computer Network58 SW: Receiver Maintain three state variables receive window size ( RWS ) largest frame acceptable ( LFA ) last frame received ( LFR ) Maintain invariant: LFA - LFR <= RWS Frame SeqNum arrives: if LFR < SeqNum < = LFA accept if SeqNum LFA discarded Send cumulative ACKs  RWS LFR LFA ……

59. Univ. of TehranIntroduction to Computer Network59 Sequence Number Space SeqNum field is finite; sequence numbers wrap around Sequence number space must be larger then number of outstanding frames SWS <= MaxSeqNum-1 is not sufficient suppose 3-bit SeqNum field (0..7) SWS=RWS=7 sender transmit frames 0..6 arrive successfully, but ACKs lost sender retransmits 0..6 receiver expecting 7, 0..5, but receives second incarnation of 0..5 SWS < (MaxSeqNum+1)/2 is correct rule Intuitively, SeqNum “slides” between two halves of sequence number space

60. Univ. of TehranIntroduction to Computer Network60 Concurrent Logical Channels Multiplex 8 logical channels over a single link Run stop-and-wait on each logical channel Maintain three state bits per channel channel busy current sequence number out next sequence number in Header: 3-bit channel num, 1-bit sequence num 4-bits total same as sliding window protocol Separates reliability from order