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# CS 455/555: Spring 2007 - PowerPoint PPT Presentation

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

Chapter 2: The Physical Layer

• Theoretical Basis for Data Communication

• Transmission Media

• Wireless Transmission

• Telephone System

• Narrowband ISDN

• B-ISDN and ATM

• Communication Satellites

• 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 )

• 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, …)

• 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.

• 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.

• 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

• 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.

• 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.

• 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.

• 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

• 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.

• 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

• Attenuation introduced by a transmission medium is measured in decibels (dB)

• Attenuation in decibels =

• 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.

• Speed of light, c = 3*108 meters/sec

• In copper or fiber it is about 2/3 of this:

2* 108 meters/sec

• 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

• 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.

• 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.

• 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)

• Circuit Switching

• Message switching

• Packet switching

• See Figure 2-39

• 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)

• 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