Computer Communication and Distributed Algorithms 202 2 1131

Computer Communication and Distributed Algorithms 202 -2 -1131 Second presentation – The physical layer Computer Communication and Distributed Algorithms (201213) © 1

The OSI Model (7 layers) Application Presentation Session Transport Network Link Physical The 7 -layer OSI Model Computer Communication and Distributed Algorithms (201213) © 2

Data transfer. How to we move bits? p Understanding n Physical the basics: aspect p Energy p Electromagnetic n Mathematical p Code Computer Communication and Distributed Algorithms (201213) waves aspect theory © 3

Signals l l In order to send data electro-magnetic signals are used. The communication channels transfer electromagnetic energy between the source and the destination. The channel may convert the electromagnetic energy to analog and digital signals. Analog signals may have infinite number of values. Digital signals include only a limited number of values. Computer Communication and Distributed Algorithms (201213) © 4

Analog vs. Digital • Vertical – Value or strength (e. g. Volts) • Horizontal - Time Computer Communication and Distributed Algorithms (201213) © 5

Analog Signals Simple – Cannot be decomposed to simpler signals p Complex – Composed of many sinus waves. p p Represented by: S(t) = A sin (2 ft + ) A – peak amplitude ( )גודל ההפרעה המקסימלי f – frequency ( )מספר המחזורים ליחידת זמן - phase ( )הפרש מופע הנמדד ברדיאנים Computer Communication and Distributed Algorithms (201213) © 6

Amplitude p Power – Electric signals (Volts / Watts) Computer Communication and Distributed Algorithms (201213) © 7

Period and frequency T = 1/f p f = 1/T p Period – The time it takes to do one cycle. Frequency – The number of periods done per second. The wave frequency is the change rate in time. A short Period means high frequency. A long period means low frequency. © 8 p p

Phase p The Phase describes the state of the waveform compared to a waveform starting at zero. Measured in degrees or radians. Computer Communication and Distributed Algorithms (201213) © 9

Example 1 Computer Communication and Distributed Algorithms (201213) © 10

Example 2 Computer Communication and Distributed Algorithms (201213) © 11

Example 3 Computer Communication and Distributed Algorithms (201213) © 12

Analog signal p Analog signal represented in the frequency domain. Computer Communication and Distributed Algorithms (201213) © 13

Signals p The frequency of a sinus signal by itself is unusable for data communication. n We need to control one or more of its properties in order to make it usable When we manipulate one or more of the signal properties it becomes a multiple frequency signal. p According to Fourier analysis, every complex signal may be represented by a composition of simple sinus waves with different frequency, amplitude and phase. p Computer Communication and Distributed Algorithms (201213) © 14

Square wave In order to use an electromagnetic signal for digital communication we need to change it to a square wave. Computer Communication and Distributed Algorithms (201213) © 15

Three harmonics Description of a harmonic wave -The waveform, Voltage or current A -Amplitude - Angular frequency - Phase Computer Communication and Distributed Algorithms (201213) © 16

Adding first three harmonics p By adding the three harmonics we create the square wave. Computer Communication and Distributed Algorithms (201213) © 17

Modulation of an analog signal p In order to transfer information or data over a communication line different modulation techniques were developed. Three analog techniques: . (amplitude modulation) אפנון משרעת . (frequency modulation) אפנון תדירות . (phase modulation) אפנון מופע p p p n n n In addition, a digital modulation was developed for digital channels. The analog techniques require a carrier signal and create a complex wave combining the carrier wave and the data. Every modulation technique requires a matching demodulation technique to be used by the receiver. Computer Communication and Distributed Algorithms (201213) 18

AM – Amplitude Modulation p Amplitude modulation is a transmission method over electromagnetic waves in radio frequency. It changes the amplitude of the wave according to the signal while the frequency of the wave remains unchanged. This method is usually used to transmit sound waves. It may also transmit codes and data. This method is common in the long, medium and short wave radio bands. An example of AM: The red wave Is the signal to be modulated, and the carrier wave (green). The lower plot shows the result of the modulation. Carrier wave Signal p Power Computer Communication and Distributed Algorithms (201213) © 19

FM – Frequency modulation p p Frequency modulation is a method used by controlling the frequency of the carrier wave. The frequency of the carrier changes according to the power of the data signal. Two major advantages over AM: n n p Bigger bandwidth (allows stereo broadcast) Durability of noise. Noise usually destroys the amplitude, and not the frequency. The amplitude of the data signal (red) is sent. The carrier (green) is a high frequency wave. The modulation adds The data signal to the frequency of the carrier. The modulated signal (blue) is not periodic. The momentary frequency changes according to the data signal amplitude. 20

PM – Phase modulation p Phase modulation requires two waves, a carrier and a modulating signal with a constant amplitude. האות המידע הגל הנושא תוצאת אפנון המשרעת בין שני הגלים Computer Communication and Distributed Algorithms (201213) © 21

Digital Signals Computer Communication and Distributed Algorithms (201213) © 22

Digital modulation p p Modulation techniques above do not apply on digital data. A digital wave holds a constant bit value 0 or 1 until an abrupt change occurs. The modulation should hold a high amplitude wave to allow weak digital signal. The modulation do not amplify noise and therefore efficiently enlarge the broadcast distance. Common modulation techniques: Computer Communication and Distributed Algorithms (201213) Amplitude shift key modulation (ASK) Frequency shift key modulation (FSK) Binary-phase shift key modulation (BPSK) Quadrature-phase shift key modulation (QPSK) Quadrature amplitude©modulation (QAM) n n n 23

Digital modulation methods ASK Amplitude-shift keying – just like AM, this method controls the amplitude of the signal. The data bit “ 1” is represented as the wave, while the data “ 0” is a break in the transmission. More complex implementations are represented by different combinations of different amplitudes. p FSK Frequency-shift keying – Bits are represented by controlling the frequency of the wave. Every frequency represents a bit or a set of bits. p FSK Computer Communication and Distributed Algorithms (201213) ASK 24 ©

Modulations Computer Communication and Distributed Algorithms A binary signal p Amplitude Modulation p Frequency Modulation p Phase Modulation p © 25

Digital channel capacity p p A digital communication channel has limits to the amount of state changes for a given time. These set the maximum data capacity. Every given time the channel has a capacity that allows it to send a constant number of bits for a given time. A digital system switch the signal between two energy or voltage states. The data has a sudden and sharp voltage changes. It is different than analog system where the voltage changes gradually and continously. Computer Communication and Distributed Algorithms (201213) © 26

Bit rate and bit interval p p Bit interval – The time needed to send one bit Bit rate – the number of bits sent in one second (bps – Bit per second). Computer Communication and Distributed Algorithms (201213) © 27

Bandwidth p The bandwidth is a property of the medium: it is the difference between the highest and the lowest frequency that the medium can transmit. We use the term bandwidth referring the medium. p Bandwidth: p n n n The highest rate that the hardware may change the signal. Measured in cycles per second (Hz) Bit rate and Bandwidth are proportional. Computer Communication and © Distributed Algorithms (201228 13)

Standard Transmission Rates p p Low Rates – 300 bps and multiplies – 1200, 2400, 9600, 19200, 54 K LANs – 2, 4, 10, 16, 1000 Mbps Wireless LANS – 11 Mbps, 54 Mbps Wireless MAN – 32 Mbps Computer Communication and Distributed Algorithms (201213) © 29

Propagation delay and transmission time p Propagation delay: n n p The time needed for a signal to move through the medium. For example: A speed of electric signal in vacuum C=3 x 108 m/s In copper wires the speed of the signal V=2 x 108 Transmission time n The time needed to transmit N bits in a given transmission rate. Transmission time = Message length / Transmission rate Computer Communication and Distributed Algorithms (201213) © 30

Calculating delay and transmission times p Example n n p p Message length: 500 Bytes Bit transmission Rate: 4 Mb/s Optical fiber length: 2 Km Speed of information in the fiber: 66% C ~ 2*108 m/s Transmission time=(Message length)/(Bit transmission Rate)= (500*8 bits) / (4*106 b/s) = 10 -3 s = 1 ms Propagation delay=d/(propagation speed)= (2000 m) / (2*108 m/s) = 0. 01 ms Computer Communication and Distributed Algorithms (201213) © 31

Throughput p The data transmission rate for a given channel n n n Units: bits/second. Considering the bandwidth Data transfer rate =1000 bits/second = 6 bits/6 ms Computer Communication and Distributed Algorithms (201213) © 32

Communication methods p A multilevel signal using one channel p p Data transfer over a channel must not be confined to a binary format- a number of voltage levels may be used. For example, 4 voltage levels allows to code 2 bits into the levels (level A=00, level B=01, level C=10, level D=11). A symbol state change sends two bits of data as opposed to one bit in binary systems. Twice throughput for a given bandwidth. A multilevel signal using parallel channels. p We may use multilevel signals with many channels. This enlarges the throughput. Computer Communication and Distributed Algorithms (201213) © 33

Multilevel representations We may use any number of symbols to transmit digital data. For example 1024 voltage levels sends: log 21024 = 10 bits. n The limit depends on our technical ability to identify the individual state accurately by the receiver. It depends on the noise and distortion levels. n Computer Communication and Distributed Algorithms (201213) © 34

Multilevel representation p The relation between bits and symbols: n The number of symbols need to singularly represent every pattern of n bits given by: p n M = 2 n symbol states 3 bits may be represented by: p M = 23 = 8 symbol states 4 bits by M = 24 = 16 symbol states § 5 bits by M = 25 = 32 symbol states § Computer Communication and Distributed Algorithms (201213) © 35

Maximum Data Rate of a Channel Simple Nyquist rule The relation between a bandwidth to the throughput: D = 2*B*log 2 K Where: D: Maximal data transfer rate (b/s). B: Bandwidth. K: Possible voltage values. Computer Communication and Distributed Algorithms (201213) © 36

Examples p For RS-232: n n p K = 2 because RS-232 uses only 2 values +15 V or -15 V to code data bits. log 22=1. D = 2*B For phase-shift encoding n n Assume K is 8. log 28=log 223 =3. D = 6*B Computer Communication and Distributed Algorithms (201213) © 37

Noise p An unwanted noise combined with the data transmission may fault the data. p Noise levels: Signal to Noise ratio (SNR) n n n S: A signal intensity level N: A noise intensity level Measured in decibels : 10 log 10 S/N. Computer Communication and Distributed Algorithms (201213) © 38

Noise effects Shannon rule: The capacity of a noisy channel C = B*log 2(1 + S/N) Where: C: a limit for the capacity B: The bandwidth of the channel S/N: Signal to noise ratio Computer Communication and Distributed Algorithms (201213) © 39

Summary p Shannon and Nyquist rules: n Engineering cannot overcome the basic and physical limits of a real transmission system. n We refer the data throughput as the bandwidth. n We encourage engineers to complex the signal coding. p n Enlarging K. Look for ways to encode more bits per cycle to enhance throughput Lowering noise p Detailing the limits of a real transmission systems. p Computer Communication and Distributed Algorithms (201213) © 40

Example 1 Question: We can calculate theoretic capacity of a standard phone line. A standard phone line has a bandwidth of (300 Hz to 3300 Hz) 3000 Hz and the signal to noise ratio is about 1000. p Solution: According to Shannon the capacity is, p C = B log 2 (1 + SNR) = 3000 log 2 (1 + 103) C = 29. 901 kbps Computer Communication and Distributed Algorithms (201213) © 41

Example 2 p p p Question: We have a communication channel with a bandwidth of 1 MHz, and signal to noise ratio of 63. What is the bit rate and how many levels are used. First we use Shannon to find the maximum capacity the transmission channel. C = B log 2 (1 + SNR) = 106 log 2 (1 + 63) = 106 log 2 (64) = 6 Mbps We use Nyquist to find the number of level used. D = 2*B*log 2 K 6 Mbps = 2 1 MHz log 2 L L = 8 Computer Communication and Distributed Algorithms (201213) © 42

Coding Schemes RZ & NRZ p RZ(Return to Zero) n n p A signal returns to zero after every coded bit 0 / 1: positive / negative pulse NRZ(Non-return to Zero) n The voltage level during a bit interval. p NRZ-L(NRZ Level) n positive voltage : 0 n negative voltage : 1 p NRZ-I(NRZ Inverted) n Differential coding scheme n 0: no transition n 1: transition p NRZ - simple and efficient. 43 © Computer Communication