Data Communication Data and Signals Behrouz A. Forouzan 1Data Communication - Data and Signals.

Data Communication - Data and Signals 1

Data CommunicationData and Signals

Behrouz A. Forouzan


Index

• ANALOG AND DIGITAL• PERIODIC ANALOG SIGNALS• DIGITAL SIGNALS• TRANSMISSION IMPAIRMENT• DATA RATE LIMITS• PERFORMANCE


Data and Signals

• major functions of physical layer is to move data in the form of electromagnetic signals across a transmission medium

• data must be transformed to electromagnetic signals for transmission over medium


ANALOG AND DIGITAL

• Analog and Digital Data– computer file is digital

• Can be transmitted through analog or digital signal

– sound is analog• Can be transmitted through analog or digital signal

• Analog and Digital Signal– Analog signals have infinite number of values in a

range; – digital signals have only a limited number of values


ANALOG AND DIGITAL


ANALOG AND DIGITALPeriodic and Nonperiodic Signals

• Both analog and digital signals can be: – periodic– non-periodic

• periodic– repeats a pattern over identical periods – The completion of one full pattern is called a cycle

• non-periodic– changes without exhibiting a pattern or cycle


ANALOG AND DIGITALPeriodic and Nonperiodic Signals

• In data communications, we commonly use periodic analog signals (because they need less bandwidth, Chapter 5) and nonperiodic digital signals (because they can represent variation in data, (Chapter 6)


PERIODIC ANALOG SIGNALS

• Simple– cannot be decomposed into simpler signals

• Composite– composed of multiple sine waves


PERIODIC ANALOG SIGNALSSine Wave

• fundamental form of a periodic analog signal• sine wave can be represented by three

parameters: – peak amplitude– frequency– phase



• Peak Amplitude– absolute value of its highest intensity, proportional to

the energy it carries, measured in volts.• The power in your house can be 110 to 120V• peak value of an AA battery is normally 1.5 V

• Period and Frequency– Period : amount of time, in seconds, a signal needs to

complete 1 cycle. (Second)– Frequency : number of periods in 1 second (HZ)

• Phase



• Power at home has a frequency of 60 Hz



• Peak Amplitude• Period and Frequency• Phase– position of the waveform relative to time 0– measured in degrees or radians– Phase of 180° corresponds to a shift of one-half of

a period



• Example:– A sine wave is offset 1/6 cycle with respect to time

0. What is its phase in degrees and radians?• 1/6 * 360 degree = 60 degree =


PERIODIC ANALOG SIGNALS Wavelength

• Distance a signal can travel in one period• depends on both frequency and medium• often used wavelength to describe

transmission of light in an optical fiber• in a vacuum, light speed is . speed is

lower in air and even lower in cable


PERIODIC ANALOG SIGNALS Wavelength

• Example:

• In coaxial or fiber-optic cable, wavelength is shorter (0.5 micron) because propagation speed is decreased


PERIODIC ANALOG SIGNALS Time and Frequency Domains

• Time Domain:– amplitude versus time plot– Phase is not explicitly shown

• Frequency Domain:– relationship between amplitude and frequency– concerned with only peak value of the frequency– Amplitude changes during one period are not shown– frequency domain conveys the information in time domain

plot– The advantage of the frequency domain is that we can

immediately see values of frequency and peak amplitude



• Example:– frequency domain is more compact and useful

when dealing with more than one sine wave


PERIODIC ANALOG SIGNALS Composite Signals

• Simple sine– 60 Hz as energy– A signal of danger

• Composite Sine– Send composite signal to communicate data



• periodic composite signal– can be decomposed into a series of simple sine

waves with discrete frequencies– not typical of those found in data communications

• nonperiodic composite signal– can be decomposed into infinite number of simple

sine waves with continuous frequencies having real values


PERIODIC ANALOG SIGNALS periodic Composite Signals


PERIODIC ANALOG SIGNALS periodic Composite Signals

• Fundamental frequency: (first harmonic)– frequency of the composite signal– In Previous example , composite signal

fundamental frequency is f and it includes 1st, 3rd and 9th harmonic of frequency of f


PERIODIC ANALOG SIGNALS non-periodic Composite Signals

• Example: – signal created by a microphone or a telephone


PERIODIC ANALOG SIGNALS Bandwidth

• difference between the highest and the lowest frequencies contained in a composite signal


PERIODIC ANALOG SIGNALS Bandwidth

• AM Radio– 530 to 1700 kHz– each AM radio station is 10kHz bandwidth

• FM Radio– ranges from 88 to 108 MHz– each FM radio station is 200-kHz bandwidth

• black-and-white TV has 3.85 MHz bandwidth• analog color TV channel has 6-MHz bandwidth


DIGITAL SIGNALS

• information can be represented by an analog or digital signal

if a signal has L levels, each level needs (log L based2 ) bits


DIGITAL SIGNALSBit Rate

• number of bits sent in 1 s, expressed in bps• digital signals are nonperiodic, thus period

and frequency are not appropriate Characteristics

• bit rate is used to describe digital signals.• A digitized voice channel bit rate id 64kbps• high-definition TV (HDTV) bit rate is 20 to

40Mbps


DIGITAL SIGNALS Bit Length

• wavelength for an analog signal = bit length in digital signal

• distance one bit occupies on the transmission medium


Digital Signal as aComposite Analog Signal

usually rare

• Frequency components od the square wave is as follows


Transmission of nonperiodic Digital Signals

• Baseband Transmission• sending a digital signal over a channel without changing

the digital signal to an analog signal

– Case1: Low-Pass Channel with Wide Bandwidth (dedicated medium)

– Case 2: Low-Pass Channel with Limited Bandwidth • Broadband Transmission (Using Modulation)– changing the digital signal to an analog signal for

transmission (band-pass channel)


Digital Signal Baseband Transmition

• A digital signal is a composite analog signal with an infinite bandwidth

• requires a low-pass channel, with a bandwidth that starts from zero to infinity

• Design based on medium (channel) property:– Case 1: Low-Pass Channel with Wide Bandwidth

(dedicated medium)– Case 2: Low-Pass Channel with Limited Bandwidth


Digital Signal Baseband Transmition dedicated medium

• Example: LAN (medium channel is time-division between users)

• entire spectrum of a medium is required (however limited in spectrum)

• amplitudes of higher frequencies in digital signal is so small that they can be ignored

• So over a dedicated medium, such as a coaxial cable or fiber optic sending digital signals with very good accuracy is possible


Digital Signal Baseband Transmition dedicated medium


Digital Signal Baseband Transmition limited bandwidth medium

• we approximate the digital signal with an analog signal

• The level of approximation depends on the bandwidth available



• Let us assume that we have a digital signal of bit rate N

• roughly simulate this signal, we need to consider the worst case, a maximum number of changes in the digital signal 01010101 ... or the sequence 10101010· ...

• So analog signal of frequency f = N/2• however, just this one frequency cannot make

all patterns; we need more components



• Example: a digital signal with a 3-bit pattern is simulated by using analog signals.

• 000: frequency f =0 and a phase of 180° • 111: frequency f =0 and a phase of 0°• The two worst cases – 010: frequenc of =NI2 and phases of 180°– 101: frequenc of =NI2 and phases of 0°.

• other four cases: can only be simulated with an analog signal with f = NI4 and phases of 180°, 270°, 90°, and 0°.



• rough approximation is referred to as using the first harmonic (NI2) frequency

• Bandwidth required is N/2 – 0 = N/2 Hz



• Better Approximation• To make the shape of the analog signal look

more like that of a digital signal, we need to add more harmonics of the frequencies



• Example: required bandwidth of a low-pass channel to send 1 Mbps by using baseband transmission?

• a. The minimum bandwidth, is B =: bit rate /2, or 500 kHz.

• b. A better result can be achieved by using the first and the third harmonics with the required bandwidth B =: 3 x 500 kHz =: 1.5 MHz.

• c. Still a better result can be achieved by using the first, third, and fifth harmonics with B =: 5 x 500 kHz =2.5 MHz



• Example: We have a low-pass channel with bandwidth 100 kHz. What is the maximum bit rate of this channel?

• The maximum bit rate can be achieved if we use the first harmonic. The bit rate is 2 times the available bandwidth, or 200 kbps


Broadband Transmission

• Modulation allows us to use a bandpass channel (a channel with a bandwidth that does not start from zero)

• This type of channel is more available than a low-pass channel



• We have used a single-frequency analog signal (called a carrier);

• The result, is not a single-frequency signal; it is a composite signal, (Chapter 5)

• At the receiver, the received analog signal is converted to digital


Broadband

• Example: sending of computer data through a telephone subscriber line

• designed to carry voice (analog signal) with a limited bandwidth (frequencies between 0 and 4 kHz).

• Although this channel can be used as a low-pass channel, it is normally USED AS a bandpass channel.

• Because the bandwidth is so narrow (4 kHz), the maximum bit rate can be only 8 kbps

• Solution: is to consider the channel a bandpass channel.



• Example: digital cellular telephone• bandwidth allocated to a company providing

digital cellular phone service is very wide, but bandpass


TRANSMISSION IMPAIRMENT

• What is sent is not what is received• causes of impairment are attenuation,

distortion, and noise


TRANSMISSION IMPAIRMENTAttenuation

• loss of energy in overcoming resistance of the medium.

• To compensate for this loss, amplifiers are used



• use the unit of the decibel to show lost or gained strength of a signal

• decibel numbers can be added (or subtracted) when we are measuring several points (cascading)

• decibel is negative if a signal is attenuated and positive if a signal is amplified

• A loss of 3 dB (-3 dB) is equivalent to losing one-half the power

• Gain of 10 dB (+10 dB) is equivalent to increasing 10 times the power



• dB=-3+7-3=+1

• dBm = 10 log Pm – where Pm is the power in milliwatts



• Example : If the signal at the beginning of a cable with -0.3 dBm/km has a power of 2 mW, what is the power of the signal at 5 km?– Power in source = 2mW = +3 dBm– Loss = 5 * (-0.3dBm) = -1.5dBm– Power in dst = 3 – 1.5 = 1.5dBm = 1.4mW


TRANSMISSION IMPAIRMENTDistortion

• signal changes its form or shape• Differences in delay in a compound signal at

different frequencies may create a difference in phase


TRANSMISSION IMPAIRMENT Noise

• thermal noise, • induced noise, – comes from sources such as motors and appliances

• crosstalk, – effect of one wire on the other

• impulse noise• spike (a signal with high energy in a very short

time) that comes from power lines, lightning, and so on


TRANSMISSION IMPAIRMENT Noise

• Signal-to-Noise Ratio (SNR)• SNR= (average signal power / average noise

power)• SNR dB(m) = lOlog SNR (m)W

• The values of SNR and SNRdB for a noiseless channel are which never achieve this ratio in real life


DATA RATE LIMITS

• Data rate depends on three factors:1. The bandwidth available2. The level of the signals we use3. The quality of the channel (the level of noise)• calculate the data rate: – Nyquist for a noiseless channel. – Shannon for a noisy channel


DATA RATE LIMITSNoiseless Channel: Nyquist Bit Rate

• Maximum BitRate = 2 x bandwidth x 10g2 L– bandwidth is the bandwidth of the channel, – L is number of signal levels used to represent data,– BitRate is the bit rate in bits per second

• Increasing levels of a signal may reduce the reliability of the system

• it can be applied to baseband transmission and modulation not just baseband transmit ion previously described


DATA RATE LIMITS Noisy Channel: Shannon Capacity

• Highest Capacity =bandwidth X log2 (1 +SNR)– Bandwidth is the bandwidth of the channel, Hz

– SNR is the signal-to-noise ratio, in unit of W not dB

– capacity is the capacity of the channel in bits per second bps

• no matter how many levels we have, we cannot achieve a data rate higher than the capacity of the channel



• Example: Consider an extremely noisy channel in which the value of the signal-to-noise ratio is almost zero. C=B log2 (1 + SNR) =B log2 (l + 0) =0

• Example: regular telephone line. Normally has a bandwidth of 3000 Hz (300 to 3300 Hz). The signal- to-noise ratio is usually 3162. C =B log2 (1 + SNR) =3000 log2 (l + 3162) = 34,860 bps– highest bit rate for a telephone line is 34.860 kbps



• when the SNR is very high– C = B * (SNR dB/3)

• Example: SNRdB = 36 and channel bandwidth is 2 MHz. – C= 2 MHz X 36/3 =24 Mbps


DATA RATE LIMITS Using Both Limits

• Example: Bandwidth = 1MHz, SNR=63– Shanon for upper limit: C= B log2 (l + SNR) =6Mbps– For better performance we choose something

lower, 4 Mbps– use the Nyquist formula to find the number of

signal levels. 4Mbps=2x 1 MHz x log2 L L=4


PERFORMANCE

• Measure performance of the network-how good is it?– Bandwidth– Throughput– Latency (Delay)– Bandwidth-Delay Product– Jitter


PERFORMANCEBandwidth

• Bandwidth in Hertz– range of frequencies contained in a signal– range of frequencies a channel can pass

• Bandwidth in Bits per Seconds– number of bits per second that a channel can transmit

• increase in bandwidth in hertz means an increase in bandwidth in bps– depends on whether we have baseband transmission

or transmission with modulation (Chapter 4 & 5)


PERFORMANCE Throughput

• bandwidth is a potential measurement of a link; the throughput is an actual measurement of how fast we can send data both in bps

• Because of congestion or sender limitation


PERFORMANCE Latency (Delay)

• how long it takes for an entire message to completely arrive at the destination from the time the first bit is sent out from the source

• Latency =propagation time +transmission time +queuing time + processing delay



• Propagation Time– Propagation time = (Distance/Propagation-speed)

• propagation speed of electromagnetic signals depends on the medium and frequency of the signaL– in a vacuum, light is propagated with a speed of 3

x 108 m/s



• Transmition Time– time between the first bit and the last bit leaving

the sender – Transmission time = (Message size /Bandwidth)– Message size in bits , Bandwidth in bps



• Queuing Time– time needed for each intermediate /end device to

hold the message before it can be processed• The queuing time is not a fixed factor• When there is heavy traffic on the network,

the queuing time increases


PERFORMANCEBandwidth-Delay Product

• in data communications product of the two (Bandwidth and Delay) is important

• maximum number of bits that can fill the link



• Example: bandwidth of 1 bps, link delay is 5 s no more than 5 bits at any time on the link



• This measurement is important in sending data in bursts and wait for the acknowledgment of each burst before sending the next one.

• To use the maximum capability of the link, we need to make the size of our burst 2 times the product of bandwidth and delay;

• we need to fill up the full-duplex channel• The sender should send a burst of data of (2 x

bandwidth x delay) bits then waits for receiver acknowledgment


PERFORMANCEJitter

• Jitter is a problem if different packets of data encounter different delays

• It is important for time-sensitive applications (audio and video)

• Discussed in multimedia section

Data Communication Data and Signals Behrouz A. Forouzan 1Data Communication - Data and Signals.

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