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วิศวกรรมศาสตร์
Data Communication (chapter 1-8)
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Principle of data communications in summary
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Chapter Introduction packet envelope Data Communications 5 Components sender receiver messages transmission medium protocol data representation Lex. text, video data flow simplex wou half-duplex 2 way, not simul full-duplex: 2 way, simul → networks - interconnections of devices devices host ex. laptop connecting device ex. switch, router - criteria: performance, reliability, security physical structure L types of connection point-to-point - multipoint mesh topology network types. criteria: size, geo coverage, ownership Local Area Network (LAN) L privately owned Wide Area Network (WAN) switch - fox connecting devices. routes run by communication companies. Switching circuit-switched -voice communications reserve! establish data transfer disconnect packet - switched datagram NO reserve; bandwidth is shared BUT Ň a internet internetwork Lex. Internet full mesh -partial mesh ring topology ฟังแล้วกลับมาได้ไง (วนอีกทาง) #devices speed! / easy to form, self-healing Ľ star topology physical topology central equipment Hub hub fail all fails easy to implement bus topology cable end terminator LYNN signal bounce N I signal at once hard to add users link cut all cant communicate w/o terminator (ส่งพร้อมกันไม่ได้ - ต้องมี terminator prevent bounce) function of each layer purpose what filled in header topology bus: multipoint
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Chapter 2 Network Models TCP/IP < Protocol Layering oss intermediate protocol =rules of sender+ receiver + devices · must be layer-to-layer communication TCP/IP* Source host Header at transport layer Legend Healer at network layer 2 Header af date link lever ↑ Deplate Segment datagram/packet frame 04 no need cap Destination host Application Presentation Session Software Transport Network Datalink hardware Physical OSI 7 layers Less specific Names Port numbers Logical/IP addresses / Pretical address TCP/IP layers Link-layer/MAC addresses/Hardware/physical address More Specific 2 hes both header and trailer encapsule add info (capsule) @layer us, outest header = layer w multiplexing La protocol encapsulate a packet intralayer same layer comm. interlayer diff low laver mostly Hop to hop upper- end to end (dot hot physical present) end-to-end communication. Fi hop-to-hop communication Lourd
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Chapter 3 Introduction To Physical Layer Data and signals L Analog and digital data/signal analog discrete w digital continuous (Periodic) analog signals sine wave ·simple n Amp Composite ft (many sine wave รวมกัน) No Shannon/Nyquist Distinguish Analog/Digital its source Lu + nature ex. voice -How to convert (sampling, quantize, encode) finer quantization - more bit rate = more level basi band / ban paes = ? wavelength phase L c = fa deg 90 deg time and frequency domains. time domain signal amp / time. Amp $ -DC M • Time (6) 1 sec a cycles non-periodic Amp 1 ●time → digital signals Amp No phase Am 5V 1 •Freq (Hz) bandwidth = 3-0-3 f domain f He bandwidth max f - min f Amplitude 16 bis set in 1s 1 LA-1 bit Bit rase-16 bps 4-0-4 Level 4 Led) Level 2 b A dirital cimal with four levels LIAN 2²-4 levels Time 10' f bandwidth ! digital as composite analog บะ .. - In transmission of digital sianals analog disital frequency wave length # of bit sent in 1 sec bit rate bit length (in f domain) baseband med bandpass (shorter band) (bring to higher f→ can be sent to long distance) = analog/digital from min to max f of the medium input (o) + medium (f₁ + f₂) Bit Rate n = 1 kbps n = 10 kbps n=100 kbps 1/2 bit rate Harmonic 1 B=500 Hz B=5 kHz B=50 kHz + output (f₁+ f₂) ~ distort Harmonics 1.3 B=1.5 kHz B-15 kHz B=150 kHz base band in analog/digital Aras min-max f of medium wu. due to cutting off high frequency part, Harmonics 1. 3.5 B=2.5 kHz B=25 kHz B=250 kHz more #sine ft + ny 1,2,3,... harmonic If 1 cycle = 2 bits frequency N 2 (first harmonic)
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digi baseband digi/ena Converter wireless transmit (Misaaluf authority gain/loss of power ยลง Transmission impairment attenuation attenuation=lolog (P/P₁) (48) Amplifier -dbm-milli watt = loss Eheat bec. medium R amplifier solve the pb. Suppose a signal travels through a transmission medium and its power is reduced to one half. This means that P2 = 0.5 Pl. In this case, the attenuation (loss of power) can be calculated as 10 logo P₂/P-10 log10 (0.5 P₁)/P; – 10 log10 0.5 = 10 × (−0.3) = −3 dB. distortion m 1 frequency 1 phase shift form w pr A loss of 3 dB (-3 dB) is equivalent to losing one-half the power. noise M mw affect amplitude SNR = Prignal/Proise SNRB = 10 log (SNR) Data Rate Limits. = how fast we can send data higher bandwidth better signaling. Noiseless Channel Nyquist Rate bit rate 2 bandwidth = log. L bps. LL= # of levels Noisy Channel: Shannon use both got C cupper limit bit rate) choose a bit lower num as bit rate for Nyquist formula to find Level of signal Capacity-band width x 1092 (1+ SNR) - highest data rate possible -retreivable data ex. c-o-no data (Noise win) - Performance (ดูจากอะไรได้) bandwidth (bps, Hz) speed of data transmission Lex. B 4 kHz, can send 56,000 bps. Throughput (bps) LONLY useful data rate banwidth always throughput CIERA Net data arrived! dist. !) -packet length B Laio error, header, Latency/delay Latency propagation time + transmission time queueing time + processing delay & distance dtrans packet length d prop . link transmission rate length of physical link propagation speed bandwidth-delay product L= volume of pipe Jitter e L = variation of delay & bit rate (bs) & packet length & Congestion rate. d processing speed depends on medium Bx delay Delay: 5 Bandwidth delay-25 bits Bandwidth: 5 bps First 5 bits After 1 First 5 bits After 2 ▬▬▬▬▬▬▬ After 4 After 5 โอนข้อมูลต่างท่ง byte bits
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Chapter 4 Digital Transmission → Digital-to-digital conversion r. #data element #sional element N⭑ = = = = = = const. n = Nmax 1 1 x = 2 × 10 =26 1092 L digital u analog m ($) baud ratesisnal rate = signal/sec N = bit rate C Line Coding 10110 -Unipolar Nez filt Polar Polar NRZ -Polar NRZ-L Polar NRZ-I L next bit = 0 1 : 1011 DC problem #H - Polar RZ Frühl No inversion DC problem can set = pos ต 1-neg Inversion ( ปงบ - ว่าง - นาค) = บน ) or ( บน - ล่าง) NO DC problem in bit, r=1 Polar biphase in phase) manchester L↓ iso, I is 1 differential manchester 0100 Lnext bit No inversion Copp. Polar NRZ-I) 0: inversion -bipolar AMI L 1 can be both +/- but consecutive 1 cant be in the same sign Psuedo ternary L •1 is neutral, is +/- Block coding mB map to nb (multilevel, multiline Yazoon) Sender 48/58 encoding encoding (data) 48 to 58; 2 = 16 Digital signal NRZ4 Link decoding yellow-block coding blue line coding another 16 is used to signaling command control + error detection 2-32 Analog-to-digital conversion analos ↓ Sampling at rate 2x fmax quantizing the more the level, the less the error encoding ↓ PAM Pulse Amplitude Modulation)
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digital sampling methods SNRB = 6.02 (bits/sample) +1.76 1) ideal bits/sample ↑ SNRAS #level t 2) natural - find required SNRas when # # level w 3) flat-top Transmission Modes We want to digitize the human voice. What is the bit rate, assuming 8 bits per sample? parallel (one) 0=0 serie (ปจบ) asynchronous -start-stop bit not continuous -synchronous L Continuous stop 00001 start Solution The human voice normally contains frequencies from 0 to 4000 Hz. So the sampling rate and bit rate are calculated as follows: Sampling rate 4000x2=8000 samples/s Bit rate=8000x8 64,000 bps 64 kbps frame Synchronous receiver & sender มี clock tick ตรงกัน
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Chapter 5 Analog Transmission - Digital-to-analas transmission digital data modulator analog signal (wireless) demod digital data - Speed of signal transmission & bandwidth S-N NN-bit rate r = 1092 L baud rate = symbal rate rbit/baud ex. bits per signal element L= √26. level ex. signal elements ·Types of digital to analog conversion - Amplitude Shift Keying ASK Amplitude changes B = (1+0) 5 - symbol rate S-N, r= + Ld parameter btw. & 1 (depends on way) 6 - S - ; d = 0 ( 6 แถบสุด) If full-duplex in ASK L If band width -100 kHz ·Frequency Shift Keying FSK 1400 f changed A --1, S=N แบ่ง 2 ช่อง : ส่ง - รับ (full-duplex) 204 205 311 (+) (+) B- +2af (1)(1) B-=(1+d) S+2af na guard band Jaanu signal disturb - max bit rate 107 band anoin MFSK (multi) We need to send data 3 bits at a time at a bit rate of 3 Mbps. The carrier frequency is 10 MHz. Calculate the number of levels (different frequencies), the baud rate, and the bandwidth. 1 signal carry bits +1-20 - 8 baud rate SN3Mbps -1 Mbaud 3 r=3 bits/symbols B = (1+d) = 1 MHz + Level + 8×1 MHz 8MHz Phase Shift Keying PSK wirin r=1, SN, B (1) S ty pes กาหนดได้ Lapsk 4 levels + 2 bits/sis-r Quadrature Amplitude Modulation Constellation diagram 01 10 11 QAM change more than component at once. 45 4-QAM 3phase 3amp 4-QAM 1Amp phase 16-AM JAMP 12 phase
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analog-to-analos conversion modulation is needed if L medium is bandpass 3 ways of coversion AM (Amplitude Modulation) Modulating signal Carrier frequency Modulated signal -FM (frequency modulation) Amplitude Modulating signal (audio) Carrier frequency Time FM signal Time PM (Phase Modulation) Amplitude message in mod uns to (carrier frequency) Modulating signal (audio) Carrier frequency wwww PM signal j v v v v v v Time Time slope of audio F1 Time
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AAAA Chapter & Band width utilization +Multiplexing = techniques that allow simultaneous transmission of multiple signals across data link by t install higher 8 links add links each time a new channel is needed. categories frequency-division multiplexing (analog) bandwidths of a link bandwidths of signals concept: signals modulate diff carrier frequencies. The modulated signals are combined into a composite signal transported by the link. Five channels, each with a 100-kHz bandwidth, are to be multiplexed together. What is the minimum bandwidth of the link if there is a need for a guard band of 10 kHz between the channels to prevent interference? Solution For five channels, we need at least four guard bands. This means that the required bandwidth is at least 5x100 4x 10-540 kHz, as shown in Figure 6.7. 100100100 A: Shift and combine Modulator A 20 24 Filter and shift Bandpass filter 20 24 Modulator Modulator Higher-bandwidth link Bandpass filter Bandpass filter 04 28 Four data channels (digital), each transmitting at 1 Mbps, use a satellite channel of 1 MHz. Design an appropriate- configuration, using FDM. ch, 10 Hz 250 KHz ma channel BSN r 10250 K T r = 4 bits 1 Symbol 16QAM modulation Wavelength-Division Multiplexing optical fiber cable Time-Division Multiplexing Synchronous TDM / Fixed TDM demed AAAA 4 pass a block others 4 A bps = 11tz 1 Mbps Digital 250 kHz 16-QAM Analog 250kHz 16-QAM Digital Analog 1 MHz FDM 250 Ha 16-QAM Analog 1 Mbps 250kHz 16-QAM Analog bit dur. before MUX = Tros sMbps frame dur after MUX. D ... 00000 MUX Frames frame rate (#bits/frame) = bit rate A multiplexer combines four 100-kbps channels using a time slot of 2 bits. Show the output with four arbitrary inputs. What is the frame rate? What is the frame duration? What is the bit rate? What is the bit duration? 100 kbps Nek MUX 1 = 1.245 SM = 1 -0.2 MS SM bit dur. after MUX - 0.2 A 10 bits 2 0.02 MS bit rate = 400 kbps bits/frame. frame rate bit/frame = bit rate = 400k = 8xx + frame rate = 50,000 frame/s
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ow utilization I may contain empty time slot statistical TDM I have timerlot for identification, an bit rate + INww.n. label, higher utilization
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Chapter Transmission Media + Intro L trans mission media are controlled by physical layer L Connect physical layer of sender & receiver + 2 categories Guided Media transmit volt 3 categories twisted pair cable two conductore Carry signals ground twist to prevent magnetic field interference tighter twist better = 1100 indicator UTP & STP cables #twist/inch UTP (unshielded) performance L Gauge diameter attenuation↓ STP (shielded) L metal shield - Coaxial cable metal insulator preventing magnetic interference from outside heavy+ fat Vau. unnecessary due to Tech evolves high f categories AG-59 Cable TV -RG-58 + Thin Ethernet -Rb-11 Thick Ethernet performance diameter↓ L higher bandwidth, fiber-optic cable angle critical angle. - นัดเข้ามา Mode multimode Step index densitง คงเดิม graded index diameter disperse & interference t
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density varies; refract signals back to center · more than I signal = the fastest; the most straight singlemode small faster more straight. ·wavelength: 1310 nm, 1550 nm. unguided media (wireless) - propagation methods F ground <2MHE ( sky 2-30 MHZ 8 line-of-sight 30 MHZ EMW EMW antenne f radio waves omni 3kHz-16 He microwaves un 1 GHz - 300 GHz infared 300 GHz-400 THE Low interference.
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Chapter & Switching → Intro network a set of connected devices can happen at physical, dato - link, network layer methods of switching circuit switched networks physical links, dedicated path use FDM TDM 3 phases establish date transfer dis connect Fefficiency -voice (analos) √J worst guaranteed bit rate Circuit-switched network KHz kHz kHz delay ONLY once during construct circuit propagation processing delay. packet switching No reserving basic info -packet digital bit which has header & date. HD Y categories datagram D1 error efficiency t 17] ปจป. packet is treated independently interconnecting devices read header + process sent via proper route (flexible) Lhave buffer wait for the previous sequence to be sent to next layer first more delay from processing Hop + propagation delay + transmission delay. Virtual-circuit - If same start-dest → same route = same sequence ! bit rate depends VCI agreement must be done first. D Port VC Port VC setup request I Dont know VCI Ask for VCI from another interconnecting devices Switch 1 Switch 2 Incoming Outgoing Port VC Port C Switch 3 vel- Waiting time
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delay (set up VCI ) + delay (change VCI in header) same phases + structure of a switch Structure of circuit switches (disconnect ~ reset table) space division close - cross point physical contact = YN50 type Lmultistage box ex. N*N-100x100 Crossbars Crossbars Nin Crossbars nxk kxn N/nx Nin nxk kxn NinNin nxk Stage 1 Stage 2 (n-k) kxn Stage 3 + ken) 10 (105) + 5(1070) + 10(510) แบ่ง stage ต่างๆ reduce num of crosspoint cause bottleneck k can be anything Time-Division Switch Structure of Packet Switches 4 components i/p ports physical laver data-link layer เช้า Queue I wait for getting in to routing processor (layer 3) -o/p ports (same as i/p but app) ·routing processor type L (layer header process) read header of packet table. switching fabrics L Banyan switch Lbinars control = 14. cross point
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