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CS 455/555: Spring 2007

CS 455/555: Spring 2007. Chapter 2: The Physical Layer. Topics. Theoretical Basis for Data Communication Transmission Media Wireless Transmission Telephone System Narrowband ISDN B-ISDN and ATM Cellular Radio Communication Satellites. Theoretical Basis for Data Communication.

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CS 455/555: Spring 2007

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  1. CS 455/555: Spring 2007 Chapter 2: The Physical Layer

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

  3. 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 sine and cosine functions.” g(t)=0.5c + ∑an sin(2∏nft) + ∑ bn cos(2 ∏nft) where f =1/T is the fundamental frequency and the a’s and b’s are amplitudes. (See page 86 for more details) RMS amplitude = Sqrt(an2 + bn2 )

  4. 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; (Frequency is measured in Hz, hertz, or cycles/sec) • Fundamental frequency f and harmonics (2 f, 3 f, …)

  5. 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.

  6. 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 rate 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 we determine that only the first 3 harmonics are of relevance, then we need a bandwidth of 3* f bandwidth.

  7. 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 (16 levels means 4 bits/signal) or 2.5 Mbaud. • What is the bandwidth needed? 5f = 2*5 = 10KHz

  8. Theoretical Basis for Data Communication (Contd.) • Nyquist’s 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.

  9. 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 3*20 or 60 kbps.

  10. 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. • Shannon’s result: Given a channel with a signal-to-noise ratio of S/N, maximum data rate = H log2 (1+S/N) Shannon’s result is independent of the number of levels in a signal and the rate of sampling of a signal.

  11. 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) =10*1.3=13 dB. • If this channel has a BW of 30 kHz, then maximum data rate is 30*log2(1+20)=30 log221 kbps

  12. 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)/0.3010=1.3222/0.3010= 4.39 kbps • So the above channel cannot deliver more than 4.39 kbps irrespective of how many levels there are per signal or how much ever the rate of sampling be.

  13. 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

  14. 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? 0.969/2=0.4845 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/x=100.04845=1.118; x=0.894 watt; • When the attenuation of a cable is specified, this is how you can compute the received power from the length of the cable.

  15. Wireless Transmission • Speed of light, c = 3*108 meters/sec • In copper or fiber it is about 2/3 of this: 2* 108 meters/sec

  16. The Telephone System • Hierarchy of switches (See Fig. 2-21) • 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

  17. 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 QAM-16 (Fig. 2-25b) each sample contains 4 bits. Hence, this will enable a 3 kHz line to send 12 kbps.QAM-64 sample contains 6 bits. • More complex coding results in more bits/sample, and hence higher data rate for a modem.

  18. 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.

  19. DSL: Digital Subscriber Lines • See Fig. 2-29 (page 133) • Goals of xDSL services (page 131) • See Fig. 2-28 to know how the frequency is divided among voice and data. • Telephone companies install a NID on customer’s premises • A splitter that is connected to NID separates the 0-4000 Hz signals used by telephones from the one used by the computer (via ADSL modem)

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

  21. Satellite Communication • Geo-synchronous satellites: • Signal travels at the speed of light: 3*108 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)

  22. Mobile Telephone System • Cellular network • Channels • 2nd Generation Mobile phones: D-AMPS, GSM, CDMA, and PDC • D-AMPS (Fig. 2-42) US and Japan • GSM-Global System for Mobile Communication---FDM and TDM are used • CDMA

  23. Cable Television

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