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Chapter 3

Data and Signals

To be transmitted, data must be transformed toelectromagnetic signals.

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3-1 ANALOG AND DIGITAL

Data can be analog or digital . The term analog data refers to

information that is continuous; digital data refers to information thathas discrete states. Analog data take on continuous values. Digitaldata take on discrete values.

Data can be analog or digital.

Analog data are continuous and take continuousvalues.

Digital data have discrete states and take discretevalues.

Signals can be analog or digital.Analog signals can have an infinite number of values

in a range; digital signals can have only a limitednumber of values.

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Figure 3.1 Comparison of analog and digital signals

In data communications, we commonly useperiodic analog signals and nonperiodic

digital signals.

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3-2 PERIODIC ANALOG SIGNALS

Periodic analog signals can be classif ied as simple or composite . A

simple periodic analog signal, a sine wave , cannot be decomposedinto simpler signals. A composite periodic analog signal iscomposed of multiple sine waves.

Figure 3.2 A sine wave

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Example 3.1

The power in your house can be represented by a sine wave with a peak ampl itude of 155 to 170 V.

H owever, i t i s common knowl edge that the voltage of the power in U .S. homes is 110 to120 V. This discrepancy is due to the fact that these are root mean square (rms) values.

The signal is squar ed and then the average ampli tude is calculated. The peak value isequal to 2 rms value.

Figure 3.3 Two signals wi th the same phase

and fr equency, but dif ferent amplitudes

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Figure 3.4 Two signal s wi th the sameampli tude and phase, but dif ferentfrequencies

Frequency and periodare the inverse of each

other.

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The power we use at home has a frequency of 60 H z . The per iodof this sine wave can be determined as follows:

Example 3.3

Example 3.4

Express a period of 100 ms in microseconds.

SolutionF rom Table 3.1 we f ind the equivalents of 1 ms (1 ms is 10 3 s)and 1 s (1 s is 10 6 s) . We make the following substi tutions: .

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The period of a signal is 100 ms. What is its frequency inkilohertz?

Example 3.5

SolutionF irst we change 100 ms to seconds, and then we calculate thefrequency from the period (1 H z = 10 3 kH z).

Frequency is the rate of change with respect to time.

Change in a short span of time means high frequency.

Change over a long span of time means low frequency.

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If a signal does not change at all, its frequency is zero.If a signal changes instantaneously, its frequency is infinite.

Phase describes the position of the waveform relative to time 0.

Figure 3.5 Three sine waves with thesame ampli tude and f requency, butdifferent phases

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A sine wave is offset 1/6 cycle with respect to time 0. What i s i tsphase in degrees and radians?

Example 3.6

SolutionWe know that 1 complete cycle is 360. Therefore, 1/6 cycle is

Figure 3.6 Wavelength and per iod

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Figure 3.7 The time-domain and frequency-domain plots of a sine wave

A complete sine wave in the time domain can berepresented by one single spike in the frequency

domain.

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The f requ ency d om ain i s m ore comp act and u sefu l wh enwe are deal ing with more than one s ine wave. For

exam ple, Figure 3 .8 sho w s th ree s ine waves, each w ithdifferent am pli tud e and frequency. A ll can be representedby th ree sp ikes in th e frequ ency dom ain .

Example 3.7

Figure 3.8 The time domain and f requency domain of three sine waves

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A single-frequency sine wave is not useful in data communications; weneed to send a composite signal, a signal made of many simple sine waves.

According to Fourier analysis, any composite signal is a combination ofsimple sine waves with different frequencies, amplitudes, and phases.

If the composite signalis periodic, the

decomposition gives aseries of signals with

discrete frequencies;if the composite signal

is nonperiodic, thedecomposition gives a

combination of sine

waves with continuousfrequencies.

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Figure 3.12 The bandwidth of per iodic and nonper iodic composite signals

The bandwidth ofa composite

signal is thedifference

between thehighest and the

lowest

frequenciescontained in thatsignal.

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I f a periodic signal is decomposed into five sine waves withfrequencies of 100, 300, 500, 700, and 900 H z, what is i ts

bandwidth? Draw the spectrum, assuming all components have amaximum ampli tude of 10 V.SolutionLet f h be the highest frequency, f l the lowest f requency, and B thebandwidth. Then

Example 3.10

The spectrum has only five spikes, at 100, 300, 500, 700, and 900H z (see F igur e 3.13).

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A periodic signal has a bandwidth of 20 Hz. The highestfrequency is 60 Hz. What is the lowest frequency? Draw the

spectrum if the signal contains all frequencies of the sameamplitude.SolutionLet f h be the highest frequency, f l the lowest f requency, and B thebandwidth. Then

Example 3.11

The spectrum contains all integer frequencies. We show this by aseries of spikes (see F igure 3.14).

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A nonperiodic composite signal has a bandwidth of 200 kH z, witha middle frequency of 140 kH z and peak amplitude of 20 V. Thetwo extreme frequencies have an amplitude of 0. Draw thefrequency domain of the signal.

Solution

The lowest f requency must be at 40 kH z and the highest at 240kH z. Figur e 3.15 shows the frequency domain and the bandwidth.

Example 3.12

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An example of a nonper iodic composite signal is the signal propagated by anAM radio station. I n the Uni ted States, each AM radio station i s assigned a 10- kH z bandwidth. The total bandwidth dedicated to AM radio r anges from 530 to1700 kH z. We will show the rationale behi nd this 10-kH z bandwidth in Chapter5.

Example 3.13

Example 3.14

Another example of a nonper iodic composite signal is the signal propagated by

an F M radio station. I n the United States, each F M radio station is assigned a200-kH z bandwidth. The total bandwidth dedicated to FM radio ranges from 88to 108 M H z. We will show the rationale behind thi s 200-kH z bandwidth inChapter 5.

Another example of a nonper iodic composite signal is the signal received by anold-f ashi oned analog black-and-white TV. A TV screen i s made up of pixels. I fwe assume a resolution of 525 700, we have 367,500 pixels per screen. I f wescan the screen 30 ti mes per second, th is is 367,500 30 = 11,025,000 pixels persecond. The worst-case scenar io is alternating black and white pixels. We can

send 2 pixels per cycle. Therefore, we need 11,025,000 / 2 = 5,512,500 cycles persecond, or H z. The bandwidth needed is 5.5125 M H z.

Example 3.15

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3-3 DIGITAL SIGNALS

I n addition to being represented by an analog signal, information can also berepresented by a digital signal . F or example, a 1 can be encoded as a positive

voltage and a 0 as zero voltage. A digi tal signal can have more than two levels.I n this case, we can send mor e than 1 bit for each level .

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A digital signal has eight levels. How many bits are needed perlevel? We calculate the number of bits from the formula

Example 3.16

Each signal level is represented by 3 bits.

Example 3.17

A digital signal has nine levels. How many bits are needed perlevel? We calculate the number of bits by using the formula. Each

signal level is represented by 3.17 bits. H owever, this answer is notrealisti c. The number of bits sent per level needs to be an i ntegeras well as a power of 2. For this example, 4 bits can r epresent onelevel.

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Assume we need to download text documents at the rate of 100pages per minute. What is the requi red bit rate of the channel?

SolutionA page is an average of 24 li nes with 80 characters in each l ine. I fwe assume that one character requi res 8 bits, the bit rate is

Example 3.18

A digiti zed voice channel, as we will see in Chapter 4, is made bydigitizing a 4-kH z bandwidth analog voice signal. We need tosample the signal at twice the highest f requency (two samples perhertz). We assume that each sample requires 8 bits. What i s therequired bit rate?

SolutionThe bit r ate can be calculated as

Example 3.19

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Figure 3.17 The time and frequency domains of per iodic and nonper iodicdigi tal signals

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Figure 3.18 Baseband tr ansmission

A digital signal is a compositeanalog signal with an infinitebandwidth.

Figure 3.19 Bandwidths of two low-pass channels

Fi 3 20 B b dt i i i d di t d di

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Figure 3.20 Baseband tr ansmission using a dedicated medium

Baseband transmission of a digital signal thatpreserves the shape of the digital signal is possible

only if we have a low-pass channel with an infinite orvery wide bandwidth.

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Figure 3.22 Simulating a digital signal wi th f ir st three harmonics

In baseband transmission, the required bandwidth is proportional to the bit

rate; if we need to send bits faster, we need more bandwidth.

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3-4 TRANSMISSION IMPAIRMENT

Signals travel through transmission media, which are not

perfect. The imperfection causes signal impair ment. Thismeans that the signal at the beginning of the medium isnot the same as the signal at the end of the medium.What i s sent i s not what i s received. Three causes ofimpair ment are attenuation , distortion , and noise .

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Figure 3.26 Attenuation

Example 3.26

Suppose a signal travels through a transmission medium and i tspower is reduced to one-half . This means that P 2 i s (1/2)P 1 . In thiscase, the attenuation (loss of power) can be calculated as

A loss of 3 dB ( 3 dB) i s equivalent to losing one-half the power.

A signal travels through an ampli f ier, and its power is increased10 times. This means that P 2 = 10P 1 . I n this case, theamplif ication (gain of power) can be calculated as

Example 3.27

E l 3 28

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One reason that engineers use the decibel to measure the changesin the str ength of a signal is that decibel numbers can be added

(or subtracted) when we are measuring several points (cascading)instead of j ust two. I n F igure 3.27 a signal travels from point 1 topoint 4. I n this case, the decibel value can be calculated as

Example 3.28

E l 3 29

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Sometimes the decibel is used to measure signal power inmilliwatts. I n this case, it i s referred to as dB m and is calculated asdB m = 10 log10 P m , where P m i s the power in mill iwatts. Calculatethe power of a signal with dB m = 30.SolutionWe can calculate the power in the signal as

Example 3.29

The loss in a cable is usual ly defined indecibels per ki lometer (dB/km). I f thesignal at the beginni ng of a cable with

0.3 dB/km has a power of 2 mW, what isthe power of the signal at 5 km?SolutionThe loss in the cable in decibels is 5 ( 0.3) = 1.5 dB. We can calculate thepower as

Example 3.30

Figure 3 28 Di t ti

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Figure 3.28 Distortion

Figure 3.29 Noise

E l 3 31

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The power of a signal is 10 mW and the power of the noise is 1 W ; what are the values of SNR and SNR dB ?

SolutionThe values of SNR and SNR dB can be calculated as follows:

Example 3.31

Example 3.32

The values of SNR and SNR dB for a noiseless channel are

We can never achieve thi s ratio in real l ife; i t i s an ideal.

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Figure 3.30 Two cases of SNR: a high SNR and a low SNR

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3-5 DATA RATE LIMITS

A very impor tant consideration in data communications

is how fast we can send data, in bits per second, over achannel. Data rate depends on three factors:

1. The bandwidth avail able2. The level of the signals we use3 . The quality of the channel (the level of noise)

Noiseless Channel: Nyquist Bit RateNoisy Channel: Shannon CapacityUsing Both Limits

Topics discussed in this section:

Example3 34

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Consider a noiseless channel with a bandwidth of 3000 H ztransmitting a signal wi th two signal levels. The maximum bit r ate

can be calculated as

Example 3.34

Example 3.35

Consider the same noiseless channel tr ansmitting a signal withfour signal levels (for each level, we send 2 bits). The maximumbit r ate can be calculated as

Example3 36

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We need to send 265 kbps over a noiseless channel with abandwidth of 20 kH z. H ow many signal levels do we need?SolutionWe can use the Nyquist f ormula as shown:

Example 3.36

Since this result i s not a power of 2, we need to either increase thenumber of levels or reduce the bit rate. I f we have 128 levels, thebit r ate is 280 kbps. I f we have 64 levels, the bit r ate is 240 kbps.

Increasing the levels of a signal may reducethe reliability of the system.

Example3 37

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Consider an extremely noisy channel in which the value of thesignal-to-noise ratio is almost zero. I n other words, the noise is sostrong that the signal is faint. For thi s channel the capacity C iscalculated as

Example 3.37

Example 3.38A telephone line normally has a bandwidth of 3000. The signal-to- noise ratio is usually 3162. For this channel the capacity iscalculated as

This means that the highest bi t r ate for a telephone li ne is 34.860kbps. I f we want to send data faster than this, we can eitherincrease the bandwidth of the l ine or improve the signal-to-noise

ratio.

Example3 39

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The signal-to-noise ratio is often given in decibels. Assume thatSNR dB = 36 and the channel bandwidth i s 2 M H z. The theoretical

channel capacity can be calculated as

Example 3.39

Example 3.40For practical purposes, when the SNR is very high, we canassume that SNR + 1 i s almost the same as SNR. I n these cases,the theoretical channel capacity can be simplif ied to

F or exampl e, we can calculate the theoreti cal capaci ty of the previous exampl e as

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3 6 PERFORMANCE

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3-6 PERFORMANCE

One important issue in networking is the performance of thenetwork how good is i t?

BandwidthThroughputLatency (Delay)

Bandwidth-Delay Product

Topics discussed in this section:

Example 3.44

A network with bandwidth of 10 Mbps can pass only an average of 12,000frames per minute with each frame carr ying an average of 10,000 bits. What i sthe throughput of thi s network?SolutionWe can calculate the throughput as

The throughput is almost one-

fifth of the bandwidth in thiscase.

In n etwor king , w e us e theterm b andw id th in two

contexts . hertz Bits per second,

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What is the propagation time if the distance between thetwo points is 12,000 km? Assume the propagation speedto be 2.4 108 m/s in cable.

SolutionWe can calculate the propagation time as

Example 3.45

The example shows that a bit can go over the AtlanticOcean in only 50 ms if there is a direct cable between thesour ce and the destination.

Example3 46

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What are the propagation time and the tr ansmission time for a2.5-kbyte message (an e-mail) i f the bandwidth of the network is 1

Gbps? Assume that the distance between the sender and thereceiver is 12,000 km and that l ight tr avels at 2.4 108 m/s.

SolutionWe can calculate the propagation and transmission time as shownon the next sl ide:

Example 3.46

Note that in this case, because the message is short and thebandwidth i s hi gh, the dominant factor is the propagation time,not the tr ansmission time. The transmission ti me can be ignored.

Example3 47

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What are the propagation time and the transmission time for a 5- M byte message (an image) i f the bandwidth of the network is 1

M bps? Assume that the distance between the sender and thereceiver is 12,000 km and that l ight tr avels at 2.4 10 8 m/s.

SolutionWe can calculate the propagation and transmission times asshown

Example 3.47

Note that in this case, because the message is very long and thebandwidth is not very high, the dominant factor is thetr ansmission time, not the propagation time. The propagation time

can be ignored.

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We can think about the l ink between two points as

a pipe. The cross section of the pipe represents thebandwidth , and the length of the pipe representsthe delay. We can say the volume of the pipedefines the bandwidth-delay product, as shown inF igure 3.33.

Example 3.48

The bandwidth-delay productdefines the numberof bits that can fill