Computer Networks - Alpen-Adria-Universität Klagenfurt. Physical... · László Böszörményi...

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László Böszörményi Computer Networks Physical Layer - 1 Computer Networks 7. Physical Layer

Transcript of Computer Networks - Alpen-Adria-Universität Klagenfurt. Physical... · László Böszörményi...

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László Böszörményi Computer Networks Physical Layer - 1

Computer Networks

7.Physical Layer

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Time

Superimposedsinus waves

Ampl

itude

• Information can be transmitted by varying some physical properties (e.g. current or frequency)

• Fourier Analysis (Fourier Series)– Any function of time with period T can be represented as

– f = 1/T (fundamental freq.)– The original signal (an,bn,c)

c can be reconstructed

László Böszörményi Computer Networks Physical Layer - 2

Theoretical Basis

s(t) = — c + Σ an sin(2πnft) + Σ bn cos(2πnft) 12 n=1 n=1

∞ ∞

an = — s(t) sin(2πnft) dt0

T2T

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László Böszörményi Computer Networks Physical Layer - 3

Non-periodic Signals

• Over long distance a periodic sinus signal (carrier) can be modulated (e.g. ampl.) for representing bits

• A bit can be represented e.g. by amplitude of current– E.g. the amplitude of “1” is +5mA that of “0” is -5mA

• Non-periodic signals (e.g. a bit pattern) can be handled as repeating the entire pattern endlessly– Example: Representation (and transmission) of character

“b“ as an 8-bit sequence “01100010“

0 1 1 10 0 0 0s(t)

t T

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László Böszörményi Computer Networks Physical Layer - 4

Reconstruction of Signals

• The Root-Mean-Square (RMS = √an

2 + bn2) of the

amplitudes is characteristic for the energy of a given frequency

• We need at least the first 4 harmonics for good signal reconstruction

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László Böszörményi Computer Networks Physical Layer - 5

Transmission Capacity

• Bandwidth (requirement) of a signal– Difference of maximum and minimum Fourier frequency

• Bandwidth of a channel (link, transmission media)– Frequency band within which transmitted signals are not

significantly distorted (up to link-specific cutoff frequency fc)• Signaling speed (baud rate) (Jean-Maurice Baudot)

– Number of times per second that a medium physicallychanges its value [baud]

• Data rate (bit rate)– Number of bits transmitted per second [bits/s, bps]– Bit rate b = (baud rate) * (number of bits per signal change)

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Maximum Data Rate on a Channel

Limited bandwidth of channel limits transmission rate• Assume voice grade line (cut-off frequency: fc ≈ 3000 Hz)• Bitrate is b [bit / sec] (Hz: After Heinrich Hertz)

b [bps] T = 8 / b [ms] f = b / 8 [Hz] nmax 300 26.67 37.5 80 600 13.33 75 40

1200 6.67 150 20 2400 3.33 300 10 4800 1.67 600 5 6000 1.33 750 4 9600 0.83 1200 2

19200 0.42 2400 1 38400 0.21 4800 0

• Time to transmit 8 bits: T = 8 / b sec

• Frequency of 1. harmonic: f = 1 / T = b / 8 Hz [1/sec]

• Frequency of n. harmonic: f(n) = n * b / 8 Hz

• Number of highest harmonic passed:nmax* b / 8 Hz ≈ 3000Hz ⇒nmax ≈ 3000 / (b/8) = 24.000 / b

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Harry Nyquist‘s Theorem (1924)• If the highest frequency of a signal is H [Hz] (filtered at

H) then it can be reconstructed by 2H samples/sec• If a noiseless channel can take V different values, the

maximum data rate bmaxNY of the channel is

bmaxNY = 2H ⋅ log2V [bps]

• Example– H = 3000 Hz (voice grade line)– V = 2 (binary signals, log2V =1)– bmax

NY = 6000 bps = 6 Kbps• Remark

– Noiseless channels do not exist

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Claude Shannon‘s Theorem (1948)• If the bandwidth of a noisy channel (subject to, e.g.

thermal noise) is H [Hz] and the signal-to-noise ratio (SNR) is S/N– The maximum data rate bmax

SH of the channel– bmax

SH = H ⋅ log2(1+S/N) [bps]• If S << N ⇒ S/N ≈ 0 ⇒ bmax

SH = 0 (log2(1) = 0)– This result is independent of signal levels / encoding!

• Example (cont‘d):– H = 3.000 Hz– S/N = 1.000 (i.e. SNR = 30 dB)

• SNR = 10•log10(S/N) [dB]) (log10(103) = 3)– bmax

SH ≈ 3.000 * log2(1+1000) ≈ 30.000 bps (log2(1024)=10)

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Cables• Twisted pair (UTP: Unshielded Twisted Pair)

– 2 insulated copper wires, twisted to avoid “antenna capabilities”– The most usual connection for telephones– Several kilometers without repeaters, several Mbit/sec– UTP3: 16 MHz, UTP5: 100 MHz, UTP6-7: 250 – 600 MHz

• Coaxial Cable– Up to 1 GHz

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László Böszörményi Computer Networks Physical Layer - 10

Fiber Optics

a) Three examples of a light ray from inside a silica fiber impinging on the air/silica boundary at different angles

b) Light trapped by total internal reflection• Several rays with different angles: multimode fiber

• Speed goes from 10 Gbps up to 50 Tbps …• Greatest problem: optical switching, optical ←→ electrical• Communication may become faster than computation!

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Fiber Cablesa) Side view of a single fiber.b) End view of a sheath with three fibers.

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Fiber Cables (2)

• A comparison of semiconductor laser and light emission diodes (LEDs) as light sources

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Fiber Optic Networks• A fiber optic ring with active repeaters

– Purely optical repeaters also available

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Wireless Transmission• In vacuum, electromagnetic waves travel with

– ~3*108m/sec (~ 1 foot/nsec) (= speed of light), ca. 2/3 of this elsewhere– λf = c (wavelength*frequency = constant)– λf ≈ 300 (if λ in meters, f in MHz)– Up to 8 bits at high frequencies (e.g. at 750MHz bandwidth: 6 Gb/s)

• Microwave Transmission (see also satellite)– Above 100 MHz waves travel nearly straight – well focused (TV dish)– With directed antennas excellent signal/noise can be reached– Widely used for long distance telephone, mobile phone, TV– Relatively inexpensive, microwave towers need few place– Do not pass buildings well and depends on weather (rain)

• Infrared and Millimeter Waves– For remote controls, do not pass walls

• Radio Transmission

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The Electromagnetic Spectrum• The electromagnetic spectrum used for communication

– Low, medium, high, very, ultra, super, extremely, tremendously high f.

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Radio Transmission

a) In VLF, LF, MF bands, radio waves follow the curvature of the earth (ca. 1000 km), power falls off sharply

b) In the HF and VHF band, they travel rather in straight lines and bounce off the ionosphere (at 100-500 km)

– Easy to generate, can long distances, penetrate buildings – Omnidirectional (can interfere with normal radio,e.g. police)

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Communication Satellites• Microwave repeaters in the sky

– Broadcast is for free, good for apps, bad for security• Geostationary Earth Orbit (GEO) Satellites

– Satellites at 35,800 km from earth remain “motionless”– Spaced by 2 degrees, the max. number is 180 (=360/2)– Slot/frequency allocation by ITU – highly political issue– Big latency (270 ms)

• Medium-Earth Orbit (MEO) Satellites– Move slowly (6 hours around the Earth), used for GPS

• Low-Earth Orbit (LEO) Satellites– Move fast: elements of a chain replace each other

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Communication Satellites

• Communication satellites and some of their properties, including altitude above the earth, round-trip delay time and number of satellites needed for global coverage.

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Low-Earth Orbit SatellitesIridium (77 LO satellites first, at 750 km)

a) The Iridium satellites form six necklaces around the earthb) 1628 moving cells (253,440 channels) cover the earth

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Globalstara) Iridium: Relaying in spaceb) Golbalstar: Relaying on the ground

– Complexity is managed in ground stations

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Public Switched Telephone System

• Structure of the Telephone System• The Local Loop

– Modems– ADSL– Wireless

• Trunks and Multiplexing• Switching

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Structure of the Telephone System

a) Fully-interconnected network (not practical!)b) Centralized switch (scales wrong)c) Two-level hierarchy ⇒ eventually 5 levels

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Structure of the Telephone Sys. (2)

1. Local loop– Usually analog, twisted pair, distance 1-10 km– Ca. 22.000 end offices in the USA

2. Trunks– Digital fiber optics, connecting switching offices

3. Switching offices

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Internet over the telephone system

• The fixed telephone system

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Local Loop: Modems, ADSL and Wireless• Direct current, digital signals in computer• Alternating current, analog signals for transmission• Conversion is done by the modems and codecs

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Modems

(a) A binary signal(b) Amplitude modulation

(c) Frequency modulation(d) Phase modulation

modulated sine wave

carrier

modulated sine wave

carrier

modulated sine wave

carrier

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Modems (2)

(a) QPSK (Quadrature Phase Shift Keying): 2 bits/baud(b) QAM-16 (Quadrature Amplitude Modulation): 4 bits/baud;

Can send 9,6 kbps over a (usual) 2400 baud line(c) QAM-64: 6 bits/baud (14,4 kbs)

45°135°

225° 315°

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Modems (3)

(a) (b)

(a) V.32 for 9, 6 kbps (4 data bits + 1 parity bit)(b) V.32bis (QAM-128): 14,4 kbps (6+1, in fax modems)• V.34bis: 33,6 Kbps – 35 Kbps is the limit by Shannon’s law• V90: 56Kbs – limit by Nyquist’s law, using 7-bit data bytes

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Digital Subscriber Lines (1)• The usual 4KHz (voice) filter is omitted (xDSL comm.)• Bandwidth versus distance over category 3 UTP for DSL• Dilemma of providers: distance vs. speed

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Digital Subscriber Lines (2)

• ADSL (Asymmetric DSL), discrete multitone modulation– 1. Channel for POTS (Plain Old Telephone Service)– 2-7. Channel: empty (separation) + control– Data: asymmetric partitioning for up/downstream (32/217)– Up/down speeds (Kbps): 768/8192, …1024/16384, 4096/30720

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Digital Subscriber Lines (3)• A typical ADSL equipment configuration

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Internet over Cable (Television)• Cable bandwidth is shared among the households

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Spectrum Allocation• Frequency allocation in a typical cable TV system used for Internet access• Analog modulation is used

– QAM-64 or QAM-256 downstream (≈30-40 Mbps), QPSK upstream

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Cable Modems• Typical details of the upstream and downstream channels in US• DOCSIS (Data Over Cable Service Interface Specification): standard• At switch on

– Modem looks for system-packet and announces itself. – In response, it gets its down/up channels from the head-end– Ranging: modem determines its distance from head for proper timing with minislots– For up-stream contention is possible: similar to slotted Aloha