Computer Networks - Physical Layerpcosta.azurewebsites.net/cn_slides/cn_02-4.pdf · Fourier Analysis Example The transmission of the ASCII characterb: 01100010 (a) A binary signal

Computer NetworksPhysical Layer

Paolo [email protected]

http://www.cs.vu.nl/∼costa

Vrije Universiteit Amsterdam

(Version April 23, 2008)

Paolo Costa 02 - Physical Layer 1 / 94

Physical Layer

Provide the means to transmit bits from sender to receiverinvolves a lot on how to use (analog) signals for digital information

Overview

Theoretical background: signal transmission and Fourier analysisTransmission media wires and no wiresModulation techniques: the actual encoding (multiplexing, andswitching)

Paolo Costa 02 - Physical Layer 2 / 94

Analog vs. Digital

The continuous signal might represent speechThe discrete signal might represent binary 1s and 0s

Paolo Costa 02 - Physical Layer Theoretical Background 3 / 94

Transmitting Signals (1/2)

We’re living in a digital world, meaning that we’d preferably want tosend digital (i.e. two-valued) signals through wires.Wires are pretty much physical, meaning that Mother Nature willprobably impose a few constraints here and there.

Observation: Signals are not entirely transmitted through a wire as youwould expect:


Transmitting Signals (2/2)

Effect of frequency-dependent transmission delays:

Effect of frequency-dependent attenuation:

Overall effect including noise:


Sin WaveDefinition

s(t) = Asin(2πft + φ)

The sine wave is the fundamental periodic signal. A general sine wavecan be represented by three parameters:

peak amplitude A - the maximum value or strength of the signalover time; typically measured in volts.frequency f - the rate [in cycles per second, or Hertz (Hz)] at whichthe signal repeats. An equivalent parameter is the period (T) of asignal, so T = 1/f.phase φ - measure of relative position in time within a singleperiod of a signal, illustrated subsequently


Sin WaveExample

(a) f = 1Hz,A = 1, φ = 0(b) f = 1Hz,A = 0.5, φ = 0

(c) f = 2Hz,A = 1, φ = 0(d) f = 1Hz,A = 1, φ = π/4


Fourier AnalysisDefinition

A periodic function with period T (and frequency f = 1/T ) g(t) can bewritten as:

g(t) =12

c +∞∑

n=1

an sin(2πnft) +∞∑

n=1

bn cos(2πnft)

Example: g(t) =∑n

k=11

2k−1 sin[(2k − 1)t ] (n is the number ofharmonics we take into account)


Fourier AnalysisExample

The transmission of theASCII character b:01100010

(a) A binary signal andits root-mean-squareFourier amplitudes√

a2n + b2

n

(b)-(e) Successiveapproximations to theoriginal signal


Fourier AnalysisDemo

Let’s play:

http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=17

http://www.falstad.com/fourier/


Bandwidth

What does this all mean?

Digital signal transmission can be thought of as being constructedas an infinite number of periodic analog signals.The quality of transmission is frequency dependent

not all parts of the digital signal get through the wire as you wouldexpect.

Digital signal transmission is subject to attenuation, distortion, etc.This is partly caused by disallowing high-frequency components topass through.The range of frequencies transmitted without being stronglyattenuated is called bandwidth


BandwidthDemo

Let’s play:

http://www2.egr.uh.edu/~glover/applets/Sampling/Sampling.html




http://www.falstad.com/fourier/



BandwidthExample

The transmission of theASCII character b:01100010

(a) the original signal(b) the signal thatresults from a channelthat allows only the firstharmonic f1 to passthrough(c)-(e) reconstructedsignals fro higherbandwidth channel


BandwidthExample

We are trying to transmit a single byte:With a bit rate of b bits/sec, it takes 8/b seconds to send a byte.

i.e., 8/b seconds is the period T of the sin waveThe frequency f1 of the first harmonic is b/8 Hz;

Assume maximum supported frequency is 3000 Hz.since all harmonics are multiple of the first harmonic, we have(roughly) 3,000/(b/8) or 24,000/b harmonicse.g., b = 300bps⇒ f1 = 37.5Hz with 80 harmonics

bps T (ms) f1 # har.300 26.67 37.5 80600 13.33 75.0 40

1200 6.67 150.0 202400 3.33 300.0 104800 1.67 600.0 59600 0.83 1200.0 2

19200 0.42 2400.0 138400 0.21 4800.0 0


BandwidthExample (contd)

bps T (ms) f1 # har.300 26.67 37.5 80600 13.33 75.0 40

1200 6.67 150.0 202400 3.33 300.0 104800 1.67 600.0 59600 0.83 1200.0 2

19200 0.42 2400.0 138400 0.21 4800.0 0

Most telephone carriers cut off the highest frequency at 3000 Hzwe can never transmit at 9600 bps (or at a higher speed) on thetelephone line

QuestionYou gotta be kidding. How about my modem at 56 Kbps ? We areassuming a simple encoding technique based on the fact that theline supports only two signal values.


BandwidthEncoding

Improvement: If there are four signal values available, we couldencode 2 bits at a time:

00→ 0 volt 01→ 2 volt10→ 4 volt 11→ 6 volt

The number changes in a signal per second is called the baud.Example 1: A 2400 bauds line (modem) can make a bit rate of 9600 bps provided ituses 16 (24) signal values, i.e., 4 bits per signal:

S bits S bits S bits S bits0 0000 4 0100 8 1000 12 11001 0001 5 0101 9 1001 13 11012 0010 6 0110 10 1010 14 11103 0011 7 0111 11 1011 15 1111

Example 2: A 2400 bauds line (modem) can make a bit rate of 33,600 bps provided ituses 16,384 (214) signal values, i.e., 14 bits per signal.

56 kbps modems use a 8,000 baud line (4000 Hz) with 8 bits per sample (1 bit isreserved for control purpose)


Nyquist & Shannon

Why cannot we increase the number of samples per seconds ?Nyquist showed that if the cut-off frequency is H Hz, the filteredsignal can be reconstructed by making 2H samples. No more, noless.

maximum transmission rate = 2H log2 V bps(where V is the number of signal values)http://www2.egr.uh.edu/~glover/applets/Sampling/Sampling.htmle.g., H = 3,000Hz and V = 2 (binary transmission over atelephone line) data rate cannot exceed 6000 bps

Group Quorum System Properties

Shannon showed that a noisy channel with a signal-to-noise rationS/N, has a limit with respect to the bit rate:

maximum transmission rate = H log2(1 + S/N) bpse.g., A telephone line with H = 3000 and10 log10(S/N) = 30 dB, can do no better than 30 kbps, no matterhow you do your encoding (excluding compression).56 kbps is still possible because with H = 4000 and little noise butno more than 70 kbps


Nyquist & Shannon

Shannon showed that a noisy channel with a signal-to-noise rationS/N, has a limit with respect to the bit rate:

maximum transmission rate = H log2(1 + S/N) bpse.g., A telephone line with H = 3000 and10 log10(S/N) = 30 dB, can do no better than 30 kbps, no matterhow you do your encoding (excluding compression).56 kbps is still possible because with H = 4000 and little noise butno more than 70 kbps

QuestionNo way. My brand new ADSL goes at 4 Mbps! ADSL uses up to 2244-kHz channels. With 15 bits/baud and 4000 baud, the downstreambandwidth would be 13.4 Mbps (more details later on)


Transmission MediaMagnetic Tape

Never underestimate the bandwidth of a station wagon full oftapes hurtling down the highway

Take a standard videotape that can carry about 7 gigabytes ofdata.A box of 50 × 50 × 50 cm can hold about 1000 tapes, whichcorresponds to 7000 gigabytes.Sending such a box can be done within 24 hours, worldwide.

We’ve got a transmission rate of 648 Mbps!

QuestionWhat is overlooked in this reasoning? We’re overlooking latency.

Paolo Costa 02 - Physical Layer Transmission Media 19 / 94

Throughput vs. Latency

Throughput is the rate at which information can be moved

Latency is the amount of time between request and response




Twisted Pair (TP)

Twisted pair: Two insulated copper wires, twisted like a DNA string(reduces electrical inference). Often, twisted pairs go by the bundle.Comparable to telephone wiring at home.

(a) category 3:traditional phone wires,10 Mbps Ethernet(b) category 5:100Mbps Ethernet


Twisted PairCharacteristics

limited distanceneeds a repeater every 2-3km

limited bandwidth1MHz

limited data rate

For long-distance, data rates of up to a few Mbps are possible; forvery short distances, data rates of up to 10 Gbps have beenachieved in commercially available products.

susceptible to interference and noise


Coaxial Cable

Coaxial cable, like twisted pair, consists of two conductors, but isconstructed differently to permit it to operate over a wider range offrequencies

Coax is better than twisted pair when you need more bandwidth, but isnow rapidly being replaced with fiber.

Like the one you usefor your TV Set


Coaxial CableCharacteristics

superior frequency characteristics to TP

performance limited by attenuation and noise

amplifiers every few km

bandwidth: up to 500MHz


Fiber OpticsPrinciple

Rather than using electrical signals, we use optical ones that arepassed through optical fiber. Principal working is based on therefraction property of light:

When a light ray passes from one medium to another (e.g., silicato air) the ray is refracted at the boundary.

e.g., the ray incident at an angle α1 emerges at an angle β

For angle of incidence above a certain critical value, the light isrefracted back into the silica


Fiber OpticsMedium

An optical fiber is a thin (2 to 125micron), flexible medium capable ofguiding an optical rayVarious glasses and plastics can beused to make optical fibers.


Fiber OpticsAttenuation

As it turns out, attenuation is extremely well in optical fiber. Thismeans that they can be used for long distances. In addition, thebandwidth is enormous.


Fiber OpticsBandwidth

c = λ · f ⇒ f = cλ ⇒

dfdλ = − c

λ2 ⇒ ∆f = − c·∆λλ2

The wider the range, and the shorter the wavelength, the higherthe bandwidth.Fiber optics often work at λ = 1.3 × 10−6 with ∆λ = 0.17 × 10−6

leading to 30 THz bandwidth!Paolo Costa 02 - Physical Layer Transmission Media 28 / 94

Fiber Connections

An interface consists of a receiver (photodiode) which transforms lightinto electrical signals, and/or a transmitter (LED or laserdiode)

Passive interface: A computer is directly connected to the opticalfiberActive interface: There’s an ordinary electrical repeater connectedto two fiber segments and the computer:


Optical Fiber vs Copper Wire (1/2)

The optical fiber cable in the foreground has the equivalentinformation-carrying capacity of the copper cable in the background.


Optical Fiber vs Copper Wire (2/2)

Bandwidth:Fiber can support enormous bandwidths, exactly what we needwith upcoming image-based applications (video-on-demand).Attenuation:Because the attenuation in fiber is less than in copper (can youimagine why?), we don’t need to boost the signal as often. Inpractice, fiber requires an active repeater every 30 km, copperevery 5 km.External influences:That’s right, no more interference from other cables, radios, powerfailures, etc. Crosstalk (you hearing another conversation) is outof the question.Weight:Fiber simply doesn’t weigh as much. Good for backs, bones, andthe use of heavy maintenance equipment.


Guided Transmission: Recap


Guided Transmission: RecapAttenuation


Wireless Transmission (1/5)

Wireless transmission is really great for all of us who can’t sit still,or feel they have to be on-line all the time.It’s also convenient when wiring is needed where it can’t be done,or isn’t really worth the trouble (jungles, islands, mountains), orbecause it’s just user-unfriendly (homes).

signal carried in electromagnetic spectrumno physical “wire”bidirectionalpropagation environment effects:

reflectionobstruction by objectsinterference

Paolo Costa 02 - Physical Layer Wireless Transmission 34 / 94


Wireless transmissions travel at the speed of light (c), uses afrequency (f ) which has a wavelength (λ):

c = λ · fThe larger the wavelength is, the longer the distance it can travelwithout attenuation. Also, the dispersion of higher frequencies is muchlower.

QuestionWhy don’t we go further ? X-rays and gamma rays are hard toproduce and modulate, do not propagate through buildings andare dangerous to living things !!!


Wireless Transmission (2/5)Naming

The terms LF, MF, and HF refer to low, medium, and highfrequency.Nobody expected to go above 10 MHzThe higher bands were later named the Very, Ultra, Super,Extremely, and Tremendously High Frequency bands.



The amount of information that an electromagnetic wave can carryis related to its bandwidth

∆f = −c ·∆λλ2

The wider the range, and the shorter the wavelength, the higherthe bandwidth.

wireless transmission will generally have a much lower bandwidth(in practice: 50-100 Mbps maximum).

Fiber optics operate in the high frequency range, which explainsthe transmission rates of gigabits per second.



Frequency hopping:Use a wide band, but let the transmitter hop from frequency tofrequency (hundreds of times per second). Good for avoidingcontinuous interference and reducing the effect of reflectedsignals (you won’t be listening to them).

Direct sequence:Simply spread the signal over a wide frequency band (and allowseveral signals with different encoding/modulation techniques tobe transmitted simultaneously).



Observation: Radio transmission (VLF–VHF) is extremely popular forits cheapness and range. Also, waves just go all over the place.

(a) In the VLF, LF and MF (AM radio) bands, radio waves followthe curvature of the earth(b) In the HF band, they bounce off the ionosphere


Microwave transmissions (100 MHz - 10 GHz)

Microwave transmission is also popular for telecommunication systems

It is good for long distances (repeaterevery 10-100 km) but requiresline-of-sight transmission:

unlike radio waves, they don’t passthrough buildingsthey don’t tolerate the curvature of theearth (e.g., San Francisco - Amsterdam)

they propagate in straight lineshigh signal-to-noise ration

high frequencyhigh bandwidth

Problemthe density in the spectrum requires higher frequency (up to 10GHz)hard for unguided transmissionswaves are few centimeters long and are absorbed by rain


Multipath Fading

Some waves may be refracted and may take slightly longer to arriveThe delayed waves may arrive out of phase with the direct wave andthus cancel the signal

it is weather and frequency dependentPaolo Costa 02 - Physical Layer Wireless Transmission 41 / 94

Communication Satellites

Observation: Satellites are attractive because they provide a relativelysimple model of communication: one signal up can be broadcast tomany receivers downwards.Taking Mother Nature into account (i.e., avoiding belts around theearth consisting of highly-charged particles that would destroy asatellite), there are three types of satellites:

Paolo Costa 02 - Physical Layer Communication Satellites 42 / 94

Geostationary Orbit Satellites

Feature: GEO satellites are placed at 35,800 km above the earthwhere their rotational speed is the same as that of the earth. Theeffect is that they appear to remain motionless in the sky.VSATs: Very Small Aperture Terminals – simple systems that output 1Watt at 19.2 kbps but can download as much as 512 kbps. To allow theVSATs to communicate with each other, hubs are used:


Medium-Earth Orbit Satellites

Example: The Global Positioning System (GPS) orbit at 18,000 km. Ittakes about 6 hours for a satellite to circle the earth. They are not usedfor telecommunications.


Low-Earth Orbit Satellites (1/2)

Essence: We throw in a relatively large number of low-orbit satelliteswhich jointly cover the surface of the earth; when you are out of yourcurrent satellite’s spot beam, you should be in that of the next satellite(Iridium):

Note: Iridium uses 66 satellites, each having a maximum of 48 cells(i.e., spot beams), totaling 1628 cells.Observation: This approach is virtually the same as that of cellularradio, except that the cells are moving instead of the subjects.


Low-Earth Orbit Satellites (2/2)

Alternative: In Globalstar, much of the complexity is handled by groundstations that pick up a connection from a satellite, and pass it on to theone closest to the receiver:

Observation: This scheme avoids much of the complexity for(managing) inter-satellite communication.


Where (not) to use Satellites

Bandwidth:Fiber wins, but not everyone has access to all the availablebandwidth. Satellites may make it easier to transfer data anywayMobility and remote locations:Satellites win, although it isn’t clear whether simple cellulartechniques may do just fineBroadcasting:Satellites win easily: broadcasting essentially comes for freeFast and reliable:Give credits to fiber: satellites are pretty bad due to inherent highlatency (230 ms round-trip for geostationary satellites), and toomuch Mother Nature (rain!)


The Local Loop

When it comes the telephone system, from a networking perspectivethe local loop (a.k.a. the last mile) is the most interesting to look at.The general structure is as follows:

Paolo Costa 02 - Physical Layer Telephone System 48 / 94

Modulation Techniques (1/3)

Problem: How can we encode our signals when we can effectivelyuse only a single frequency (or better: small frequency range)?Answer: Apply modulation techniques

Change the amplitude (strength) of the signal: changing amplitudemeans a binary 1, constant amplitude a binary 0.Use different frequencies to encode your bits (these frequenciescan be put “on top” of your base frequency).Change the phase of the wave (cf. sine and cosine) to do signalencoding.


Modulation techniques (2/3)

(a) A binary signal(b) Amplitude modulation

(c) Frequency modulation(d) Phase modulation


Modulation techniques (3/3)

Observation: Modulation is strongly related to not being able to set a(wide-frequency-ranges) DC signal value on the wire as directencoding of binary signals:

becomes


Modem

Modem derives frommodulator-demodulator)

modulates an analog carriersignal to encode digitalinformationdemodulates such a carriersignal to decode the transmittedinformation

Used for digital datatransmission through thepublic telephone network

replacing the analoginfrastructure would be toocostly

Two different frequencies areused to enable traffic in bothdirection (full duplex)


Increasing Transmission Rates

To get to higher and higher speeds, it is not possible to justincrease the sampling rate

Nyquist says that even with a perfect 3000 Hz line, there is no pointin sampling more than 6000 Hz

=⇒ most modems sample 2400 times / sec and focus on getting morebits per sample

The number of samples per second is measured in baud

an n-baud line transmits n symbols / sec

Baud is NOT the modem’s speed (but they are related)

e.g., if only two symbols (e.g., 0 volt and 1 volt) are sent on a 2400baud line, the bit rate is 2400 bpsif the voltages 0, 1, 2, and 3 volts are used, each symbol consists of2 bits and the bit rate becomes 4800 bpsinstead of using 4 different voltages, it is possible to use 4 phaseshifts [Quadrature Phase Shift Keying (QPSK)]


Let’s Recap...

BandwidthThe range of frequency that pass through a given medium withminimum attenuation (measured in Hz)Baud RateThe number of samples (symbols) / sec made and is constrainedby the bandwidth (Nyquist’s theorem). The modulation technique(e.g., QPSK) determines the number of bits per symbol.Bit (Data) RateThe amount of information sent over the channel during a second(= number of symbols / sec × number of bits / symbol)


Modulation Techniques

All modern modems use a combination of modulation techniques totransmit multiple bits per baud.

(a) Quadrature Phase Shift Keying (QPSK)

The phase of a dot is indicated by the angleThe amplitude is the distance from the origin4 phases (45, 135, 225, and 315 degrees) with constant amplitude2 bits / symbol =⇒ 4800 bps (assuming 2400 baud line)


Modulation Techniques

All modern modems use a combination of modulation techniques totransmit multiple bits per baud.

(b) Quadrature Amplitude Modulation (QAM-16)4 amplitudes and 4 phases16 (24) possible combinations leading to 4 bits per symbolused to transmit 9600 bps over a 2400 baud line

(c) QAM-648 amplitudes and 8 phases64 (26) possible combinations leading to 6 bits per symbolused to transmit 14400 bps over a 2400 baud line


Higher Speeds

With many dots, even a small amount of noise can result in an erroneousdecode of phase / amplitudeAdd extra bits to perform error correction (Trellis Code Modulation)

(a) V.32

V.32 modem uses 32 (25) dots to transmit 4 data bits and 1 parity bit9,600 bps with error correction at 2400 baudrotation was done only for engineering reason

(b) V.32 bis

V.32 bis modem uses 128 (27) dots to transmit 6 data bits and 1 parity bit14,400 bps with error correction at 2400 baud (fax modems)

Higher speeds (28,800 and 33,600 bps) are achieved using 12 and 14 bits /symbol (V.34 and V.34 bis)Further improvements are made by also using compression techniques


Higher Speeds56K modem

Standard modems stop at 33,600 because the Shannon limit for a telephonesystem is about 35 Kbps

A call from “Computer” to “ISP 1” goes through two local loops

each of them adds noise

“ISP 2”, instead, has a digital trunk

less noise =⇒ maximum data rate now becomes 70 kbps

V.90 modems use 8 bits per symbol (7 data bit and 1 control bit) over a 8,000baud line (assuming 4,000 Hz band for phone cable)

56,000 bps in downstream33,000 bps in upstream (using V.34 to reduce noise)


Beyond 56 K

Cable TV industry was offering speeds up to 10 Mbps and satellitecompanies were planning to offer 50 MbpsTelephone companies realized that they need to offer higherspeeds to their customers to remain competitive

56 kbps were not appealing any longerThe new service went under the name of xDSL (Digital SubscriberLine)

the most popular is ADSL (Asymmetric DSL)


Digital Subscriber Lines

Traditional telephone cables are artificially limited to a 3000 Hzbandwidth

good enough to carry human voicea filter in the end office attenuates all frequencies below 300 Hzand above 3400 Hz

With xDSL, the incoming line of a customer is connected to adifferent kind of switch, that does not have any filter

the limiting factor then becomes the physics of the local loop

The service can be offered only within near proximityPaolo Costa 02 - Physical Layer Telephone System 60 / 94

Asymmetric DSLBasics

The spectrum of the local loop is about 1.1 Mhz spectrumWe can divide it into 256 channels of 4,000 Hz each (like intraditional phone systems)

each one plays a different role

channel 1-5 are not used to avoid interferencesince most users download more data than they upload 80 % ofchannels are allocated to downstream

this explains the “A” of Asymmetricdifferent combinations are possible.


Asymmetric DSLData Rate

Within each channel a modulation scheme similar to V.34 is used15 bits / baud on 4000 baud line

=⇒ with 224 downstream channels, the downstream bandwidth is13.44 Mbps !

In practice, the signal-to-noise ratio is never good enoughbut 8 Mbps is possible on short-runs over high quality loops


Asymmetric DSLArchitecture

The NID (Network Interface Device) is a small plastic box, marking the borderbetween company’s and customer’s propertyThe splitter is a filter that separates the voice band (0-4000 Hz) from the restThe ADSL modem can be seen as 250 QAM modems operating in parallel atdifferent frequenciesThe DSLAM (DSL Access Multiplexer) recovers the signal into a bit stream andsends off to the ISP


Asymmetric DSLHardware

ADSL Model

4: CPU 5: JTAG interface6: 8 Mb RAM 7: Flash memory13: Ethernet port 16: USB port17: Telephone port

DSLAM


Asymmetric DSLADSL2(+)

The bandwidth of the local loop has been expanded to 2.2 MHzADSL2 allows to a maximum of 12 Mbps download speedADSL2+ allows to a maximum of 25 Mbps, with a 3–7 km distance to becrossed by copper.Maximum upload speed is 1.2–3.5 Mbps

Highly dependent on the cable length


Wireless Local Loop

Problem: Suppose you want to start an ISP but using the localloop is out of the question because it is owned by your competitor.Solution: Set up a wireless direct connection between one of yourantennas and your subscribers (in a so-called sector):

e.g., WiMax (IEEE 802.16 standard)

A sector can operate at 36 Gbps downstream bandwidth and 1Mbps upstream, to be shared by subscribers.

The range is about 2–5 km.Practice: single users get 10 Mbps over 10 km.


Multiplexing: FDM

Problem: Considering that the bandwidth of a channel can behuge, wouldn’t it be possible to divide the channel intosub-channels?Frequency Division Multiplexing: Divide the available bandwidthinto channels through frequency filtering, and apply modulationtechniques per channel:


Multiplexing: FDMExample

Frequency Division Multiplexing is used by Radio and TVbroadcasting

Station Channel FrequencyNederland 1 22 479,25Nederland 2 25 503,25Nederland 3 26 511,25RTL 4 28 527,25RTL 5 36 591,25SBS 6 37 599,25RTL 7 38 607,25Veronica / Jetix 39 615,25Net 5 40 623,25Tien 41 631,25Casema Service Kanaal 29 535,25TV West 44 655,25Eén 42 639,25Ketnet / Canvas 43 647,25Discovery Channel 46 671,25National Geographic / CNBC 47 679,25Animal Planet 48 687,25Eurosport 54 735,25MTV 50 703,25Nickelodeon / The BOX 51 711,25TMF 53 727,25BBC 1 55 743,25BBC 2 56 751,25


Multiplexing: WDM

Wavelength Division Multiplexing: Actually the same as FDM, butused for fiber optics.

Light waves have their own frequency range; they are simplycombined and separated using standard (de)fraction properties


Multiplexing: TDMBasics

Time Division Multiplexing: Simply merge/split streams of digitaldata into a new stream. Data is handled in frames – a fixed seriesof consecutive bits:

This is a full-digital solution in contrast to FDM and WDM


Multiplexing: TDMPCM

Analog signals need to be digitalized before being eligible for TDMPCM (Pulse Code Modulation)

The magnitude of the signal is sampled regularly at uniformintervals, then quantized to a series of symbols in binary codeBut because the quantized values are only approximations, it isimpossible to recover the original signal exactly

8 bits / sample are used for digitalized voice on the phone16 bits / sample are used for Audio CD


Multiplexing: TDMT1 system

Example: The T1 system samples at 8000 Hz, and encodes eachsample as a 7-bit number (i.e. 128 different values). With some extracontrol bits, we merge samples into 193-bit frames, every 125 µsec:

Observation: T1 supports a total of 193 × 1125× 106 = 1.544 Mbps


Multiplexing: TDMQoS

TDM also makes it easy to offer individual senders higherbandwidth, by simply putting more data into a frame:

or to combine several trunks into higher-bandwidth trunks:


SwitchingPacket vs. Circuit

Circuit switching: Make a true physical connection from sender toreceiver. This is what happens in traditional telephone systems.Packet switching: (1) Split any data (i.e. message) into smallpackets, (2) route those packets separately from sender toreceiver, and (3) assemble them again.


SwitchingMessage Switching

Message (Store-and-forward) switching: a message is completelyreceived at a router, stored, and then put into an outgoing queuefor further routing

analogous to packet switching but here a message must be fullyreceived before being forwarded

(a) circuit-switching; (b) store-and-forward; (c) packet-switchingPaolo Costa 02 - Physical Layer Telephone System 75 / 94

SwitchingComparison

NoteCheck also what discussed in the Introduction.


The Mobile Telephone SystemOverview

We will discuss the following technologies:

AMPS (Advanced Mobile Phone System)

first generation of mobile phonesanalog voicea.k.a. (E)TACS

GSM (Global System for Mobile communication)digital voicedigital data (EDGE and GPSR)

UMTS (Universal Mobile Telecommunication System)

third generation (3G)high speeddigital voice and data

Paolo Costa 02 - Physical Layer Mobile Telephone System 77 / 94

Advanced Mobile Phone System (AMPS)Overview

The whole idea is to break up an area into small regional cells,each has their own frequency rangeno two adjacent cells have the same frequency.

Pretty good for handling different densitiescells may be split in sub-cells

The problem is frequency allocation and energy emission


Advanced Mobile Phone System (AMPS)Channels

The AMPS uses 832 full-duplex channels, each consisting of apair of simplex channels

832 simplex channels from 824 to 849 Mhz832 simplex channels from 869 to 894 Mhzeach channel is 30kHz wide

Thus, FDM is used to separate channelsChannels are used for different tasks

control (base to mobile) to manage the systemspaging (base to mobile) to notify callsaccess (bidirectional) for call setup and channel assignmentdata (bidirectional) for voice

Since the same frequencies cannot be reused in nearby cells, theactual number of voice channels per cell is about 45


GSMOverview

GSM: Global System for Mobile communications, is a full-blowndigital cellular radio transmission systemFDM is used with each mobile transmitting on one frequency andreceiving on a higher one

also, a single frequency pair is split by time-division multiplexinginto time-slots shared by multiple mobiles

GSM uses 124 downlink channels, and 124 uplink channels percell, each channel multiplexed by TDM (GSM-900):

this gives 8× 124 = 992 full duplex channels.a lot of them are not used to avoid interference with neighboring cells


GSMChannels

There are also separate channels for:

broadcasting cell info (so that a mobile station can see whether ithas changed cells).cell maintenance (the base station has to know who’s in its cell).call setup (incoming, and outgoing).


GSMData Transmission

EDGE (Enhanced Data rates for GSM Evolution) is GSM withmore bits per baud.

more bits per baud also mean more errors per baudEDGE introduces nine different schemes for modulation and errordetection

GPSR (General Packet Radio Service) is an overlay packetnetwork on top of GSM

it allows mobile stations to send and receive IP packetswhen GPSR is operating, some time slots on some frequencies arereserved for packet transmissionthe number of these slots are dynamically managed by thebase-station according to the traffic conditions


CDMAOverview

Code Division Multiple Access (CDMA) allows transmissions to beinterleaved, but avoids interference.

no message collisions

Principle: assign a chip sequence to a station, which is just anm-bit code (each bit in this code is called chip)

to transmit a 1 bit, a station sends its chip sequenceto transmit a 0 bit, it sends the one’s complement of the chipsequencee.g., assuming m = 8, if A is assigned the chip sequence00011011, it sends 1 by sending 00011011 and 0 by sending11100100

Let’s introduce bipolar notation:

rewrite a binary 0 as -1, and a binary 1 as +1so, 1 bit for station A becomes (−1− 1− 1 + 1 + 1− 1 + 1 + 1)


CDMAMath

Let’s indicate with S the m−chip vector for station S and S for itsnegationKey: chip sequences S and T belonging to two stations S and Tmust be orthogonal

i.e., the normalized inner product must be 0

S • T =1m

m∑i=1

SiTi = 0

note that if S • T = 0, then also S • T = 0the normalized product of any chip sequence with itself is 1:

S • S =1m

m∑i=1

SiSi =1m

m∑i=1

(Si )2 =

1m

m∑i=1

(±1)2 = 1

also note that S • S = −1


CDMASimultaneous Transmissions

When two stations transmit at the same time, their signals addlinearly

remember that no FDM or TDM is usede.g., if three stations output +1 and one station outputs -1, the resultis +2

To recover the bit stream of an individual station, say C, thereceiver computes the inner product of the received chipsequence and the chip sequence of C

e.g., assume two stations A and C transmitting a 1 bit at the sametime that B transmits a 0 bitthe receiver sees the sum S = A + B + C and computes:

S • C = (A + B + C) • C = A • C + B • C + C • C = 0 + 0 + 1 = 1

the first two terms vanish thanks to the orthogonality


CDMAExample


CDMAExample

Hidden Assumptionsthe receiver must known the chip sequencesender and receiver are synchronizedpower levels of all stations are the same as perceived by thereceiver


CDMAAnalogy

Let’s make an analogy:TDM is comparable to all people being in the middle of the roombut talking at different times

FDM is comparable to people being in widely separated groups,each group holding separate conversations but at the same time

CDMA is comparable to everybody being in the middle of theroom, talking at the same time but in different languages


CDMABandwidth

The amount of information grows from b bits / sec to mb bits / secThis can be done only if the available bandwidth is increased by afactor mCDMA uses a form of spread spectrum communication

e.g., if we have 1-MHz band available for 100 stationswith FDM, each one would have 10 kHz and could send at 10 kbps(assuming 1 bit per Hz)with CDMA, each station uses the full 1 MHz, so the chip rate is 1megachips / second (chip is a bit in the chip sequence=if we use less than 100 bit per chip sequence, the effectivebandwidth for CDMA is higher than FDM and the channel allocationproblem is solved


UMTS

UMTS adopts a CDMA scheme:it runs a 5 MHz bandwidth and is compatible with GSMit offers 2 Mbps for indoor stationary basestation, 384 kbps forpeople walking, and 144 kbps for connections in cars

It is supposed to deliver the following services:

High-quality voice transmissionsMessaging (e-mail, fax, SMS, chat, ...)Multimedia (playing music, viewing videos, films, televisions, ...)Internet access

on mobile phones ?


Wi-Fi: IEEE 802.11gOFDM

Frequency division multiplexing (FDM) transmits multiple signalssimultaneously over a single transmission medium

each signal travels within its own unique frequency range (carrier)Orthogonal FDM’s (OFDM) distributes the data over a largenumber of carriers that are spaced apart at precise frequencies

the sub-carrier frequencies are chosen so that the sub-carriers areorthogonal to each other

At the center frequency of any particular subcarrier all othersubcarriers attain a null point

no guardbands requiredno special filter to demodulate each frequency carriers


Cable Television

TV cable companies areactively working to increasetheir marketThey use a mix of fiber (for thelong-haul runs) and coaxialcable (to the house)=⇒ HFC (Hybrid Fiber Coax)Issue: every cable is sharedamong multiple users (a) whilein the telephone systemsevery house has its ownprivate cable (b)

Paolo Costa 02 - Physical Layer Cable Television 92 / 94

Cable TelevisionPrinciple

There’s a lot of unused bandwidth that can be allocated to sendingbits over the wire:

Because downstream (television) starts at 54 MHz, there is limitedbandwidth that can be used for upstream data

most traffic, however, is likely to be downstream (like in ADSL)QAM-64 (or QAM-256 if the cable quality is really high) is used oneach downstream 6-Mhz channel

36 Mbps (of which 27 Mbps for payload) data rateFor upstream QAM-64 does not work well (too noisy)

QPSK is used2 bits / baud instead of 6 or 8 bits of QAM

Paolo Costa 02 - Physical Layer Cable Television 93 / 94

Cable TelevisionModem

Internet access require a cable modemThe downstream data always comes from one source, whichmakes it easier to handle.Upstream data requires that subscribers contend for availableslots:

Each modem is assigned a minislot to use to request upstreambandwidth

this may be subject to contention⇒ retry mechanism

Security is also a concernPaolo Costa 02 - Physical Layer Cable Television 94 / 94

Documents

Computer Networks - Physical Layerpcosta.azurewebsites.net/cn_slides/cn_02-4.pdf · Fourier Analysis Example The transmission of the ASCII characterb: 01100010 (a) A binary signal