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CS 455/555: Spring 2001

CS 455/555: Spring 2001. 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 2001

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  1. CS 455/555: Spring 2001 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 sines and cosines.” 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 78 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; bandwidth=high.freq-low.freq (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 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.

  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 or 2.5 Mbaud. • What is the bandwidth needed? 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 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 the rate of sampling.

  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. 2-15) • Use of both analog and digital transmissions (see Fig. 2-17): 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 (Fig. 2-19b) 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.

  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. 2-26. • 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. Switching • Circuit Switching • Message switching • Packet switching • See Figure 2-35

  20. ATM • ATM --- virtual circuit network---when a circuit is to be setup between source and destination: 91) A route of switches is decided (2) All switches along the way make entries in their table about the VC; they may also reserve resources (Fig. 2-43) • Permanent VC: A VC that lasts for a long time; • Switched VC: Dynamic VC: set and break it. • In space-division switching, the line BW is actually reserved and no one else can use it.

  21. ATM (Cont.) • Synchronous vs. asynchronous transmission (See Fig. 2-44) • T1 carrier: Synchronous: Consists of 24 voice channels multiplexed together. In a single frame (125sec long), 24 channels are accommodated. 24 analog channels are fed through a Codec which samples and digitizes the data. Each sample is converted to 7-bits of data and 1 bit for control. There are 8000 frames/sec and hence a data rate of 56kbps and signaling data of 8 kbps. Channel 1 always occupies the same position in each frame. • ATM: Does not fix the channels with fixed positions. Cells from different channels may be sent in any arbitrary order. In case there are no cells to be sent between cells, idle cells are sent.

  22. ATM Switches • See Fig. 2-45 • Several input lines and several output lines (generally they are equal in #) • Each cell coming on each input line has a specific output line to be transmitted on. • Goal: (1) Switch all cells with as low discard rate as possible (2) Never reorder the cells on a virtual circuit. • Head-of-line blocking (Fig 2-46): If the cell ahead of the queue of an input line is blocked, so is everyone behind it even if their own output line is free. Solution: Keep output queues instead of input queues. • Output queueing at an ATM switch (Fig. 2-47)

  23. 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)

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