CS 455555: Spring 2004

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CS 455/555 Spring 2004

• Chapter 2 The Physical Layer

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Topics

• Theoretical Basis for Data Communication
• Transmission Media
• Wireless Transmission
• Telephone System
• Narrowband ISDN
• B-ISDN and ATM
• Cellular Radio
• Communication Satellites

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Theoretical Basis for Data Communication

• Fourier Analysis
• Any reasonably behaved periodic function, g(t),
  with period T can be constructed by summing a
  (possibly infinite) number of sines and cosines.
• g(t)0.5c ?an sin(2? nft) ? bncos(2 ? nft)
  where f 1/T is the fundamental frequency and the
  as and bs are amplitudes. (See page 78 for more
  details)
• Rms amplitude Sqrt(an2 bn2 )

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Theoretical Basis for Data Communication (Contd.)

• Attenuation The power of a signal diminishes as
  it travels along the medium. Higher frequencies
  may be subjected to higher attenuation than lower
  frequencies.
• Bandwidth-limited signals The bandwidth of a
  signal is generally limited by filters which
  cut-off frequencies above certain limit. If the
  cut-off is high, then more harmonics are
  transmitted otherwise less are transmitted
  bandwidthhigh.freq-low.freq (Frequency is
  measured in Hz, hertz, or cycles/sec)
• Fundamental frequency f and harmonics (2 f, 3 f,
  )

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Theoretical Basis for Data Communication (Contd.)

• Signal vs. data Signal is the actual voltage
  pattern sent on a transmission medium data is
  what the signal conveys
• Example Suppose two groups standing apart on two
  mountain tops of a valley use colored flags to
  send information to each other. Suppose they
  choose 4 types of flags (e.g., Red, Blue, Green,
  and Yellow) for this purpose. Suppose the
  flaggers can change the flags at the rate of
  3/minute, then the signal rate is 3/minute. What
  is the data rate?
• Since each color can convey 2 bits of
  information, the data rate is 6 bits/minute.

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Theoretical Basis for Data Communication (Contd.)

• The rate at which signal changes is referred to
  as Baud rate measured in bauds.It represents
  signal changes/sec.
• The data is measured in bits/second or bps.
• If only voltage levels 0 and 1 are used by
  signals, then baud rate bit rate. This is not
  always the case.
• Suppose a periodic signal with a period of T sec
  or a frequency of f Hz (1/T) is to be
  transmitted over a channel, then we first should
  determine how much bandwidth is needed for this
  signal. After a Fourier analysis, if determine
  that only the first 3 harmonics are of relevance,
  then we need a bandwidth of 3 f bandwidth.

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Theoretical Basis for Data Communication (Contd.)

• Example You wish to send data at a rate of 10
  Mbps using a signaling method that uses 16 levels
  (e.g., voltage).
• A Fourier analysis of the signal revealed that
  the fundamental frequency is 2 kHz and up to 5
  Harmonics are significant.
• What is the signaling rate (baud rate) we need
  for the signal? A minimum of 10/4 or 2.5 Mbaud.
• What is the bandwidth needed? 25 10KHz

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Theoretical Basis for Data Communication (Contd.)

• Nyquists sampling theorem If an arbitrary
  signal has been passed through a low-pass filter
  of bandwidth H Hz, then the filtered signal can
  be completely reconstructed by sampling it at the
  source at the rate of 2H samples/sec. Sampling at
  a rate higher than this is not any more
  beneficial as other higher harmonics have already
  been removed from the filtered signal.

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Theoretical Basis for Data Communication (Contd.)

• Suppose we have a channel with a bandwidth of 10
  kHz (channels behave like low-pass filters), and
  we use 8-level signals to pass through the
  channel, what is the maximum data rate we can
  obtain using the channel/signal combination? A
  signal with the highest harmonic of 10 kHz needs
  only a sampling rate of 20 K samples/sec. Each
  sample of 8-level signal can represent 3 bits. So
  maximum data rate is 320 or 60 kbps.

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Theoretical Basis for Data Communication (Contd.)

• In general, maximum data rate of a noiseless
  channel 2H log2V bit/sec
• Where H is the channel bandwidth and V is
  levels/signal.
• Shannons result Given a channel with a
  signal-to-noise ratio of S/N,
• maximum data rate H log2 (1S/N)
• Shannons result is independent of number of
  levels in a signal and the rate of sampling of a
  signal.

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Theoretical Basis for Data Communication (Contd.)

• Signal-to-noise ratio (S/N) This is a ratio of
  signal power to noise power present in a signal.
  This noise is referred to as thermal noise,
  random noise, white noise, Johnson noise, etc.
• In practice, this is measured in decibels (dB) or
  10 log10 (S/N).
• For example, if a channel has a signal power of
  10 watts and noise power of 0.5 watts, then S/N
  is 10/0.5 20. In decibels, the same is
  expressed as 10 log10 (20) 101.313 dB.
• If this channel has a BW of 30 kHz, then maximum
  data rate is 30log2(120)30 log221 kbps

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Theoretical Basis for Data Communication (Contd.)

• How to find log2(21) since calculators only have
  log to the base of 10 or e?
• log2(21) log10(21)/ log10(2)
• log10(2) 0.3010
• So, maximum data rate in the previous example
  log2(21) log10(21)/0.30101.3222/0.3010 4.39
  kbps
• So the above channel cannot deliver more than
  4.39 kbps irrespective of how many levels there
  are per signal or the rate of sampling.

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Transmission Media

• Magnetic media (disks, floppies, tapes, etc)
• Twisted pair (e.g., telephones)
• Baseband coaxial cable For digital
  transmission---1-2Gbps
• Broadband coaxial cable For analog
  transmission--- up to 300-450 MHz (bandwidth)
• Fiber-optics Almost infinite bandwidth
  (certainly 50,000 Gbps and more)---No more
  limitation of Nyquist and Shannon

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Transmission Media (Contd.)

• Attenuation introduced by a transmission medium
  is measured in decibels (dB)
• Attenuation in decibels
• 10log10(transmitted power/received power)
• If over a 1 km cable, the transmitted power was 1
  Watt and received power was 0.8 watt, then
  attenuation of the wire
• 10log10(1/0.8)0.969 dB/1 km
• What is the attenuation over 0.5 km cable?
  0.969/20.4845 dB. So if the transmitted power at
  one end of a 0.5km is 1 watt, what is the power
  at the other end?
• 0.4845 10log10(1/x) 1/x100.048451.118
  x0.894 watt
• When the attenuation of a cable is specified,
  this is how you can compute the received power
  from the length of the cable.

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Wireless Transmission

• Speed of light, c 3108 meters/sec
• In copper or fiber it is about 2/3 of this
• 2 108 meters/sec

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The Telephone System

• Use of both analog and digital transmissions (see
  Fig. 2-23) Codec Code/decode For digital
  transmission Modem Modulator/demodulator for
  analog transmission
• Transmission impairments Attenuation, delay
  distortion, and noise

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Modems

• Digital data is converted to analog signals using
  modems.
• At the sending end, the stream of bits are used
  to modulate a sine wave carrier.
• At the receiving end, the analog signal is
  sampled to derive the bit stream.
• Amplitude modulation, frequency modulation, phase
  modulation
• A 3000-Hz telephone line allows a frequency of at
  most 3 kHz. Hence, to reconstruct the original
  signal we need at most 6000 samples/sec. The bps
  now depend on the coding of more bits/sample.
• In quadrature amplitude modulation (Fig. 2-25b)
  each sample contains 4 bits. Hence, this will
  enable a 3 kHz line to send 12 kbps.
• More complex coding results in more bits/sample,
  and hence higher data rate for a modem.

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Trunks and Multiplexing

• Trunks have large bandwidth, so they can carry
  multiple channels simultaneously
• Multiplexing Frequency division multiplexing,
  time division multiplexing, and wavelength
  division multiplexing (fiber-optics)
• TDM Pulse-code modulation (PCM) to convert
  analog signals to digital signals (codec) one
  sample of the signal is converted to a string of
  bits. A 7-bit PCM can digitize a sample into one
  of 27 or 128-levels to produce a 7-bit stream.
  This is used in TDM as shown in Fig. 2-33.
• DPCM is an alternate to PCM where the difference
  in levels of the present sample from the previous
  is measured. Delta modulation is a special case
  of DPCM where only higher or lower are recorded.
• DM needs least bits, DPCM needs some more, and
  PCM needs the most.

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Switching

• Circuit Switching
• Message switching
• Packet switching
• See Figure 2-39

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Satellite Communication

• Geo-synchronous satellites
• Signal travels at the speed of light 3108 m/sec
• The time for a signal to traverse from source to
  the satellite, reflected back, and reach the
  destination is about 270 milliseconds. This is
  referred to as end-to-end delay
  (source-destination) or as a round-trip delay
  (i.e., ground-satellite-ground)

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DSL Digital Subscriber Line

• A means by which telephone companies are offering
  high-speed access over telephone lines
• In a normal telephone line, the filter at the end
  office cuts off frequencies below 300 HZ and
  above 3400 Hz. Thus, data were are restricted to
  this limited BW.
• The trick used by DSL (or ADSL) is to remove this
  restriction for data. Thus, filter BW is no
  longer the limitation.
• They have a monthly charge and not per-minute
  connection charge.
• See Fig. 2-29 for a typical DSL configuration.
  NIDSplitter are installed by the tel. Company
  and ADSL modem is also needed.
• ADSL modem acts as a 250 QAM (I.e., 250 points in
  the configuration map)

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Wireless Local Loops

• Local loop The connection between a home and
  the end-office. Typically, it is a twisted-pair.
  However, this limits the BW and data rates.
• The idea to improve the bandwidth is to make the
  local loop wireless via an antenna at home. The
  antenna transmits and receives from the telephone
  companys close by tower.

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Mobile Telephone Systems

• First generation mobile phones Analog voice
• Cell phones---area is divided into cells
• Frequencies are not reused in adjacent cells
• Each cell has a base station.
• Handoff A base station handing off a phone to
  its neighboring cells base station.
• MTSO Mobile Telephone Switching Offices they
  communicate with the base stations, each other,
  and the rest of the packet-switching network.

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Second-Generation Mobile Phones Digital Voice
(PCS or personal communication services

• D-AMPS, GSM, CDMA, PDC (different standards)
• D-AMPS The Digital Advanced Mobile Phone System
  (USA) ATT
• Uses 30 kHz channels Frequency division
  multiplexing
• On the mobile phone, voice is digitized, and
  compressed, resulting in 8kbps or less
• Each frequency pair supports 25 frames/sec of 40n
  msec each each frame holds three users.
• GSM The Global System for Mobile Communication
  (Rest of the world)
• Similar to D-AMPS uses FDM frequency pairs
  using TDM to hold multiple users
• GSM channel 200kHz (versus 30 kHz) supporting 8
  separate connections with TDM
• CDMA Code Division Multiple Access (Qualcomm,
  Inc.) Sprint PCS
• Does not use FDM channels are separated using
  coding theory

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Third-Generation Mobile Phones Digital Voice and
Data

• New applications data, video conf., group game
  playing, M-commerce, etc.
• W-CDMA (Wideband CDMA) by Ericsson
• CDMA 2000 by Qualcomm, Inc.