Data Communication and Networking Chapter_3

3.1

Chapter 3Data and Signals

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

3.2

To be transmitted, data must betransformed to electromagnetic signals.

Note

3.3

3-1 ANALOG AND DIGITAL

Data can be analog or digital. The term analog data refersto information that is continuous; digital data refers toinformation that has discrete states. Analog data take oncontinuous values. Digital data take on discrete values.

Analog and Digital DataAnalog and Digital Signals Periodic and Nonperiodic Signals

Topics discussed in this section:

3.4

Analog and Digital Data

Data can be analog or digital. Analog data are continuous and take

continuous values. Digital data have discrete states and take

discrete values.

3.5

Analog and Digital Signals

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

of values in a range.• Digital signals can have only a limited

number of values.

3.6

Figure 3.1 Comparison of analog and digital signals

3.7

3-2 PERIODIC ANALOG SIGNALS

In data communications, we commonly use periodicanalog signals and nonperiodic digital signals.Periodic analog signals can be classified as simple orcomposite. A simple periodic analog signal, a sine wave,cannot be decomposed into simpler signals. A compositeperiodic analog signal is composed of multiple sinewaves.

Sine WaveWavelength Time and Frequency Domain Composite Signals Bandwidth


3.8

Figure 3.2 A sine wave

3.9

Figure 3.3 Two signals with the same phase and frequency,but different amplitudes

3.10

Frequency and period are the inverse ofeach other.

Note

3.11

Figure 3.4 Two signals with the same amplitude and phase,but different frequencies

3.12

Table 3.1 Units of period and frequency

3.13

The power we use at home has a frequency of 60 Hz.The period of this sine wave can be determined asfollows:

Example 3.1

3.14

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

Example 3.2

SolutionFirst we change 100 ms to seconds, and then wecalculate the frequency from the period (1 Hz = 10−3

kHz).

3.15

Frequency

• Frequency is the rate of change with respectto time.

• Change in a short span of time means highfrequency.

• Change over a long span oftime means low frequency.

3.16

If a signal does not change at all, itsfrequency is zero.

If a signal changes instantaneously, itsfrequency is infinite.

Note

3.17

Phase describes the position of thewaveform relative to time 0.

Note

3.18

Figure 3.5 Three sine waves with the same amplitude and frequency,but different phases

3.19

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

Example 3.3

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

3.20

Figure 3.6 Wavelength and period

3.21

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

3.22

A complete sine wave in the timedomain can be represented by one

single spike in the frequency domain.

Note

3.23

The frequency domain is more compact anduseful when we are dealing with more than onesine wave. For example, Figure 3.8 shows threesine waves, each with different amplitude andfrequency. All can be represented by threespikes in the frequency domain.

Example 3.7

3.24

Figure 3.8 The time domain and frequency domain of three sine waves

3.25

Signals and Communication A single-frequency sine wave is not

useful in data communications

We need to send a composite signal, asignal made of many simple sinewaves.

According to Fourier analysis, anycomposite signal is a combination ofsimple sine waves with differentfrequencies, amplitudes, and phases.

3.26

Composite Signals andPeriodicity

If the composite signal is periodic, thedecomposition gives a series of signalswith discrete frequencies.

If the composite signal is nonperiodic, thedecomposition gives a combination ofsine waves with continuous frequencies.

3.27

Figure 3.9 shows a periodic composite signal withfrequency f. This type of signal is not typical of thosefound in data communications. We can consider it to bethree alarm systems, each with a different frequency.The analysis of this signal can give us a goodunderstanding of how to decompose signals.

Example 3.4

3.28

Figure 3.9 A composite periodic signal

3.29

Figure 3.10 Decomposition of a composite periodic signal in the time andfrequency domains

3.30

Figure 3.11 shows a nonperiodic composite signal. Itcan be the signal created by a microphone or a telephoneset when a word or two is pronounced. In this case, thecomposite signal cannot be periodic, because thatimplies that we are repeating the same word or wordswith exactly the same tone.

Example 3.5

3.31

Figure 3.11 The time and frequency domains of a nonperiodic signal

3.32

Bandwidth and SignalFrequency

The bandwidth of a composite signal isthe difference between the highest and thelowest frequencies contained in thatsignal.

3.33

Figure 3.12 The bandwidth of periodic and nonperiodic composite signals

3.34

If a periodic signal is decomposed into five sine waveswith frequencies of 100, 300, 500, 700, and 900 Hz, whatis its bandwidth? Draw the spectrum, assuming allcomponents have a maximum amplitude of 10 V.SolutionLet fh be the highest frequency, fl the lowest frequency,and B the bandwidth. Then

Example 3.6

The spectrum has only five spikes, at 100, 300, 500, 700,and 900 Hz (see Figure 3.13).

3.35

Figure 3.13 The bandwidth for Example 3.6

3.36

A periodic signal has a bandwidth of 20 Hz. The highestfrequency is 60 Hz. What is the lowest frequency? Drawthe spectrum if the signal contains all frequencies of thesame amplitude.SolutionLet fh be the highest frequency, fl the lowest frequency,and B the bandwidth. Then

Example 3.7

The spectrum contains all integer frequencies. We showthis by a series of spikes (see Figure 3.14).

3.37


3.38

A nonperiodic composite signal has a bandwidth of 200kHz, with a middle frequency of 140 kHz and peakamplitude of 20 V. The two extreme frequencies have anamplitude of 0. Draw the frequency domain of thesignal.

SolutionThe lowest frequency must be at 40 kHz and the highestat 240 kHz. Figure 3.15 shows the frequency domainand the bandwidth.

Example 3.8

3.39


3.40

An example of a nonperiodic composite signal is thesignal propagated by an AM radio station. In the UnitedStates, each AM radio station is assigned a 10-kHzbandwidth. The total bandwidth dedicated to AM radioranges from 530 to 1700 kHz. We will show the rationalebehind this 10-kHz bandwidth in Chapter 5.

Example 3.9

3.41

Another example of a nonperiodic composite signal isthe signal propagated by an FM radio station. In theUnited States, each FM radio station is assigned a 200-kHz bandwidth. The total bandwidth dedicated to FMradio ranges from 88 to 108 MHz. We will show therationale behind this 200-kHz bandwidth in Chapter 5.

Example 3.10

3.42

Another example of a nonperiodic composite signal isthe signal received by an old-fashioned analog black-and-white TV. A TV screen is made up of pixels. If weassume a resolution of 525 × 700, we have 367,500pixels per screen. If we scan the screen 30 times persecond, this is 367,500 × 30 = 11,025,000 pixels persecond. The worst-case scenario is alternating black andwhite pixels. We can send 2 pixels per cycle. Therefore,we need 11,025,000 / 2 = 5,512,500 cycles per second, orHz. The bandwidth needed is 5.5125 MHz.

Example 3.11

3.43

Fourier analysis is a tool that changes atime domain signal to a frequency

domain signal and vice versa.

Note

Fourier Analysis

3.44

Fourier Series

Every composite periodic signal can berepresented with a series of sine and cosinefunctions.

The functions are integral harmonics of thefundamental frequency “f” of the compositesignal.

Using the series we can decompose anyperiodic signal into its harmonics.

3.45

Fourier Series

3.46

Examples of Signals and theFourier Series Representation

3.47

Sawtooth Signal

3.48

Fourier Transform

Fourier Transform gives the frequencydomain of a nonperiodic time domainsignal.

3.49

Example of a FourierTransform

3.50

Inverse Fourier Transform

3.51

Time limited and Band limitedSignals A time limited signal is a signal for which

the amplitude s(t) = 0 for t > T1 and t < T2

A band limited signal is a signal for whichthe amplitude S(f) = 0 for f > F1 and f < F2

3.1



3.2

3-3 DIGITAL SIGNALS

In addition to being represented by an analog signal,information can also be represented by a digital signal.For example, a 1 can be encoded as a positive voltageand a 0 as zero voltage. A digital signal can have morethan two levels. In this case, we can send more than 1 bitfor each level.

Bit Rate Bit Length Digital Signal as a Composite Analog SignalApplication Layer


3.3

Figure 3.16 Two digital signals: one with two signal levels and the otherwith four signal levels

3.4

A digital signal has eight levels. How many bits areneeded per level? We calculate the number of bits fromthe formula

Example 3.16

Each signal level is represented by 3 bits.

3.5

A digital signal has nine levels. How many bits areneeded per level? We calculate the number of bits byusing the formula. Each signal level is represented by3.17 bits. However, this answer is not realistic. Thenumber of bits sent per level needs to be an integer aswell as a power of 2. For this example, 4 bits canrepresent one level.

Example 3.17

3.6

Assume we need to download text documents at the rateof 100 pages per sec. What is the required bit rate of thechannel?SolutionA page is an average of 24 lines with 80 characters ineach line. If we assume that one character requires 8bits (ascii), the bit rate is

Example 3.18

3.7

A digitized voice channel, as we will see in Chapter 4, ismade by digitizing a 4-kHz bandwidth analog voicesignal. We need to sample the signal at twice the highestfrequency (two samples per hertz). We assume that eachsample requires 8 bits. What is the required bit rate?

SolutionThe bit rate can be calculated as

Example 3.19

3.8

What is the bit rate for high-definition TV (HDTV)?

SolutionHDTV uses digital signals to broadcast high qualityvideo signals. The HDTV screen is normally a ratio of16 : 9. There are 1920 by 1080 pixels per screen, and thescreen is renewed 30 times per second. Twenty-four bitsrepresents one color pixel.

Example 3.20

The TV stations reduce this rate to 20 to 40 Mbpsthrough compression.

3.9

Figure 3.17 The time and frequency domains of periodic and nonperiodicdigital signals

3.10

A digital signal is a composite analogsignal with an infinite bandwidth.

Note

3.11

Figure 3.18 Baseband transmission

The fundamental question is, How can we send a digital signal from pointA to point B?We can transmit a digital signal by using one of two different approaches :

Baseband transmissionBroadband transmission(using modulation).

3.12

Figure 3.19 Bandwidths of two low-pass channels

3.13

Figure 3.20 Baseband transmission using a dedicated medium

3.14

Baseband transmission of a digitalsignal that preserves the shape of the

digital signal is possible only if we havea low-pass channel with an infinite or

very wide bandwidth.

Note

3.15

Figure 3.21 Rough approximation of a digital signal using the first harmonicfor worst case

3.16

Figure 3.22 Simulating a digital signal with first three harmonics

3.17

In baseband transmission, the required bandwidth isproportional to the bit rate;

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

Note

In baseband transmission, the requiredbandwidth is proportional to the bit rate;if we need to send bits faster, we need

more bandwidth.

3.18

Table 3.2 Bandwidth requirements

3.19

What is the required bandwidth of a low-pass channel ifwe need to send 1 Mbps by using baseband transmission?

SolutionThe answer depends on the accuracy desired.a. The minimum bandwidth, is B = bit rate /2, or 500 kHz.

b. A better solution is to use the first and the thirdharmonics with B = 3 × 500 kHz = 1.5 MHz.

c. Still a better solution is to use the first, third, and fifthharmonics with B = 5 × 500 kHz = 2.5 MHz.

Example 3.22

3.20

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

channel?

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

Example 3.22

3.21

Figure 3.23 Bandwidth of a bandpass channel

3.22

If the available channel is a bandpasschannel, we cannot send the digital

signal directly to the channel;we need to convert the digital signal toan analog signal before transmission.

Note

3.23

Figure 3.24 Modulation of a digital signal for transmission on a bandpasschannel

3.24

An example of broadband transmission usingmodulation is the sending of computer data through atelephone subscriber line, the line connecting a residentto the central telephone office. These lines are designedto carry voice with a limited bandwidth. The channel isconsidered a bandpass channel. We convert the digitalsignal from the computer to an analog signal, and sendthe analog signal. We can install two converters tochange the digital signal to analog and vice versa at thereceiving end. The converter, in this case, is called amodem which we discuss in detail in Chapter 5.

Example 3.24

3.25

A second example is the digital cellular telephone. Forbetter reception, digital cellular phones convert theanalog voice signal to a digital signal (see Chapter 16).Although the bandwidth allocated to a companyproviding digital cellular phone service is very wide, westill cannot send the digital signal without conversion.The reason is that we only have a bandpass channelavailable between caller and callee. We need to convertthe digitized voice to a composite analog signal beforesending.

Example 3.25

3.1



3.2

3-4 TRANSMISSION IMPAIRMENT

Signals travel through transmission media, which are notperfect. The imperfection causes signal impairment. Thismeans that the signal at the beginning of the medium isnot the same as the signal at the end of the medium.What is sent is not what is received. Three causes ofimpairment are attenuation, distortion, and noise.

Attenuation Distortion Noise


3.3

Figure 3.25 Causes of impairment

3.4

3.5

Attenuation

Means loss of energy -> weaker signal When a signal travels through a

medium it loses energy overcoming theresistance of the medium

Amplifiers are used to compensate forthis loss of energy by amplifying thesignal.

3.6

Measurement of Attenuation

To show the loss or gain of energy theunit “decibel” is used.

dB = 10log10P2/P1P1 - input signal

P2 - output signal

3.7

Figure 3.26 Attenuation

3.8

Suppose a signal travels through a transmission mediumand its power is reduced to one-half. This means that P2

is (1/2)P1. In this case, the attenuation (loss of power)can be calculated as

Example 3.26

A loss of 3 dB (–3 dB) is equivalent to losing one-halfthe power.

3.9

A signal travels through an amplifier, and its power isincreased 10 times. This means that P2 = 10P1 . In thiscase, the amplification (gain of power) can be calculatedas

Example 3.27

3.10

One reason that engineers use the decibel to measure thechanges in the strength of a signal is that decibelnumbers can be added (or subtracted) when we aremeasuring several points (cascading) instead of just two.In Figure 3.27 a signal travels from point 1 to point 4. Inthis case, the decibel value can be calculated as

Example 3.28

3.11

Figure 3.27 Decibels for Example 3.28

3.12

Sometimes the decibel is used to measure signal powerin milliwatts. In this case, it is referred to as dBm and iscalculated as dBm = 10 log10 Pm , where Pm is the powerin milliwatts. Calculate the power of a signal with dBm =−30.

SolutionWe can calculate the power in the signal as

Example 3.29

3.13

The loss in a cable is usually defined in decibels perkilometer (dB/km). If the signal at the beginning of acable with −0.3 dB/km has a power of 2 mW, what is thepower of the signal at 5 km?SolutionThe loss in the cable in decibels is 5 × (−0.3) = −1.5 dB.We can calculate the power as

Example 3.30

3.14

3.15

Distortion Means that the signal changes its form or

shape Distortion occurs in composite signals Each frequency component has its own

propagation speed traveling through amedium.

The different components therefore arrivewith different delays at the receiver.

That means that the signals have differentphases at the receiver than they did at thesource.

3.16

Figure 3.28 Distortion

3.17

3.18

Noise There are different types of noise

Thermal - random noise of electrons in thewire creates an extra signal

Induced - from motors and appliances,devices act are transmitter antenna andmedium as receiving antenna.

Crosstalk - same as above but betweentwo wires.

Impulse - Spikes that result from powerlines, lighning, etc.

3.19

Figure 3.29 Noise

3.20

Signal to Noise Ratio (SNR)

To measure the quality of a system theSNR is often used. It indicates thestrength of the signal wrt the noisepower in the system.

It is the ratio between two powers. It is usually given in dB and referred to

as SNRdB.

3.21

The power of a signal is 10 mW and the power of thenoise is 1 μW; what are the values of SNR and SNRdB ?

SolutionThe values of SNR and SNRdB can be calculated asfollows:

Example 3.31

3.22

The values of SNR and SNRdB for a noiseless channelare

Example 3.32

We can never achieve this ratio in real life; it is an ideal.

3.23

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

3.1

3-5 DATA RATE LIMITS

A very important consideration in data communicationsis how fast we can send data, in bits per second, over achannel. 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)

Noiseless Channel: Nyquist Bit Rate Noisy Channel: Shannon Capacity Using Both Limits


3.2

Increasing the levels of a signalincreases the probability of an error

occurring, in other words it reduces thereliability of the system. Why??

Note

3.3

Capacity of a System The bit rate of a system increases with an

increase in the number of signal levels weuse to denote a symbol.

A symbol can consist of a single bit or “n”bits.

The number of signal levels = 2n. As the number of levels goes up, the spacing

between level decreases -> increasing theprobability of an error occurring in thepresence of transmission impairments.

3.4

Nyquist Theorem Nyquist gives the upper bound for the bit rate

of a transmission system by calculating thebit rate directly from the number of bits in asymbol (or signal levels) and the bandwidthof the system (assuming 2 symbols/per cycleand first harmonic).

Nyquist theorem states that for a noiselesschannel:

C = 2 B log22n

C= capacity in bpsB = bandwidth in Hz

3.5

Does the Nyquist theorem bit rate agree with theintuitive bit rate described in baseband transmission?

SolutionThey match when we have only two levels. We said, inbaseband transmission, the bit rate is 2 times thebandwidth if we use only the first harmonic in the worstcase. However, the Nyquist formula is more general thanwhat we derived intuitively; it can be applied to basebandtransmission and modulation. Also, it can be appliedwhen we have two or more levels of signals.

Example 3.33

3.6

Consider a noiseless channel with a bandwidth of 3000Hz transmitting a signal with two signal levels. Themaximum bit rate can be calculated as

Example 3.34

3.7

Consider the same noiseless channel transmitting asignal with four signal levels (for each level, we send 2bits). The maximum bit rate can be calculated as

Example 3.35

3.8

We need to send 265 kbps over a noiseless channel witha bandwidth of 20 kHz. How many signal levels do weneed?SolutionWe can use the Nyquist formula as shown:

Example 3.36

Since this result is not a power of 2, we need to eitherincrease the number of levels or reduce the bit rate. If wehave 128 levels, the bit rate is 280 kbps. If we have 64levels, the bit rate is 240 kbps.

3.9

Shannon’s Theorem

Shannon’s theorem gives the capacityof a system in the presence of noise.

C = B log2(1 + SNR)

3.10

Consider an extremely noisy channel in which the valueof the signal-to-noise ratio is almost zero. In otherwords, the noise is so strong that the signal is faint. Forthis channel the capacity C is calculated as

Example 3.37

This means that the capacity of this channel is zeroregardless of the bandwidth. In other words, we cannotreceive any data through this channel.

3.11

We can calculate the theoretical highest bit rate of aregular telephone line. A telephone line normally has abandwidth of 3000. The signal-to-noise ratio is usually3162. For this channel the capacity is calculated as

Example 3.38

This means that the highest bit rate for a telephone lineis 34.860 kbps. If we want to send data faster than this,we can either increase the bandwidth of the line orimprove the signal-to-noise ratio.

3.12

The signal-to-noise ratio is often given in decibels.Assume that SNRdB = 36 and the channel bandwidth is 2MHz. The theoretical channel capacity can be calculatedas

Example 3.39

3.13

For practical purposes, when the SNR is very high, wecan assume that SNR + 1 is almost the same as SNR. Inthese cases, the theoretical channel capacity can besimplified to

Example 3.40

For example, we can calculate the theoretical capacity ofthe previous example as

3.14

We have a channel with a 1-MHz bandwidth. The SNRfor this channel is 63. What are the appropriate bit rateand signal level?

SolutionFirst, we use the Shannon formula to find the upperlimit.

Example 3.41

3.15

The Shannon formula gives us 6 Mbps, the upper limit.For better performance we choose something lower, 4Mbps, for example. Then we use the Nyquist formula tofind the number of signal levels.

Example 3.41 (continued)

3.16

The Shannon capacity gives us theupper limit; the Nyquist formula tells us

how many signal levels we need.

Note

3.1

3-6 PERFORMANCE

One important issue in networking is the performance ofthe network—how good is it?.

Bandwidth - capacity of the system Throughput - no. of bits that can bepushed through Latency (Delay) - delay incurred by a bitfrom start to finish Bandwidth-Delay Product


3.2

In networking, we use the termbandwidth in two contexts.

The first, bandwidth in hertz, refers to therange of frequencies in a composite signalor the range of frequencies that a channelcan pass. The second, bandwidth in bits per second,refers to the speed of bit transmission in achannel or link. Often referred to asCapacity.

Note

3.3

The bandwidth of a subscriber line (telephone line) is 4kHz for voice or data.

The bandwidth of this line for data transmission can beup to 56,000 bps (5.6 Kbps) using a sophisticated modem(with SNR of 16000-16500) to change the digital signalto analog.i.e.C(b.w) = 4000*log2(1+16385)

= 56000 bps= 56 Kbps

Example 3.42

3.4

If the telephone company improves the quality of the lineand increases the bandwidth to 8 kHz, we can send112,000 bps by using the same technology as mentionedin Example 3.42.i.e.C(b.w) = 8000*log2(1+16385)

= 112001 bps= 112 Kbps

Example 3.43

3.5

Throughput Throughput is the measure of how fast we can

actually send data through a network?

Throughput and Bandwidth are not same. How & Why?

A link may have bandwidth of B bps, but we can only send T bps viathe link, where T < B.

In other words Bandwidth is the potential measurement of a Link,while the Throughput is actual measurement of how fast we can senddata.

For Example, we may have a link with a bandwidth of Mbps, but thedevices connected to end of the link may handle only 200 kbps. This200kps is the Throughput.

3.6

A network with bandwidth of 10 Mbps can pass only anaverage of 12,000 frames per minute with each framecarrying an average of 10,000 bits. What is thethroughput of this network?

SolutionWe can calculate the throughput as

Example 3.44

The throughput is almost one-fifth of the bandwidth inthis case.

3.7

Latency or Delay defines how long it takes for an entiremessage to completely arrive at the destination from thetime the first bit is sent out from the source.We can say that latency is made of four components i.e.• Propagation Time• Transmission Time• Queuing Time• Processing Delay

Latency (Delay)

Latency = PT+TT+QT+PD

3.8

Propagation Time

3.9

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 thesource and the destination.

3.10

Transmission Time

3.11

What are the propagation time and the transmissiontime for a 2.5-kbyte message (an e-mail) if thebandwidth of the network is 1 Gbps? Assume that thedistance between the sender and the receiver is 12,000km and that light travels at 2.4 × 108 m/s.

Solution

Example 3.46

3.12

Note that in this case, because the message is short andthe bandwidth is high, the dominant factor is thepropagation time, not the transmission time. Thetransmission time can be ignored.


3.13

What are the propagation time and the transmissiontime for a 5-Mbyte message (an image) if the bandwidthof the network is 1 Mbps? Assume that the distancebetween the sender and the receiver is 12,000 km andthat light travels at 2.4 × 108 m/s.

SolutionWe can calculate the propagation and transmissiontimes as shown on the next slide.

Example 3.47

3.14

Note that in this case, because the message is very longand the bandwidth is not very high, the dominant factoris the transmission time, not the propagation time. Thepropagation time can be ignored.


3.15

Figure 3.31 Filling the link with bits for case 1

Bandwidth-Delay Product

The bandwidth-delay product definesthe number of bits that can fill the link.

Note

3.16

Figure 3.32 Filling the link with bits in case 2

3.17

We can think about the link between two points as apipe. The cross section of the pipe represents thebandwidth, and the length of the pipe represents thedelay. We can say the volume of the pipe defines thebandwidth-delay product, as shown in Figure 3.33.

Example 3.48

3.18

Figure 3.33 Concept of bandwidth-delay product

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Data Communication and Networking Chapter_3