Wireless Communication and Cellular Communication

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1. Software Engineering Departments
COE (College of Engineering
3.1
Lecture 3
Dataand Signals
Computer Communication and Networks
Engr. Farhan Ghafoor

2. 3.2
To be transmitted, data must be
transformed to electromagnetic signals.
Note

3. 3.3
3-1 ANALOG AND DIGITAL
Datacanbeanalogordigital.Thetermanalogdatarefers
toinformationthatiscontinuous;digitaldatarefersto
informationthathasdiscretestates.Analogdatatakeon
continuousvalues.Digitaldatatakeondiscretevalues.
Analog and Digital Data
Analog and Digital Signals
Periodic and Nonperiodic Signals
Topics discussed in this section:

4. 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.

5. 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.

6. 3.6
Figure 3.1 Comparison of analog and digital signals

7. 3.7
3-2 PERIODIC ANALOG SIGNALS
Indatacommunications,wecommonlyuseperiodic
analogsignalsandnonperiodicdigitalsignals.
Periodicanalogsignalscanbeclassifiedassimpleor
composite.Asimpleperiodicanalogsignal,asinewave,
cannotbedecomposedintosimplersignals.Acomposite
periodicanalogsignaliscomposedofmultiplesine
waves.
Sine Wave
Wavelength
Time and Frequency Domain
Composite Signals
Bandwidth
Topics discussed in this section:

8. 3.8
Figure 3.2 A sine wave

9. 3.9
Figure 3.3 Two signals with the same phase and frequency,
but different amplitudes

10. 3.10
Frequency and period are the inverse of
each other.
Note

11. 3.11
Figure 3.4 Two signals with the same amplitude and phase,
but different frequencies

12. 3.12
Table 3.1 Units of period and frequency

13. 3.13
Thepowerweuseathomehasafrequencyof60Hz.
Theperiodofthissinewavecanbedeterminedas
follows:
Example 3.1

14. 3.14
Theperiodofasignalis100ms.Whatisitsfrequencyin
kilohertz?
Example 3.2
Solution
Firstwechange100mstoseconds,andthenwe
calculatethefrequencyfromtheperiod(1Hz=10
−3
kHz).

15. 3.15
Frequency
•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.

16. 3.16
If a signal does not change at all, its
frequency is zero.
If a signal changes instantaneously, its
frequency is infinite.
Note

17. 3.17
Phase describes the position of the
waveform relative to time 0.
Note

18. 3.18
Figure 3.5 Three sine waves with the same amplitude and frequency,
but different phases

19. 3.19
Asinewaveisoffset1/6cyclewithrespecttotime0.
Whatisitsphaseindegreesandradians?
Example 3.3
Solution
Weknowthat1completecycleis360°.Therefore,1/6
cycleis

20. 3.20
Figure 3.6 Wavelength and period

21. 3.21
Figure 3.7 The time-domain and frequency-domain plots of a sine wave

22. 3.22
A complete sine wave in the time
domain can be represented by one
single spike in the frequency domain.
Note

23. 3.23
Thefrequencydomainismorecompactand
usefulwhenwearedealingwithmorethanone
sinewave.Forexample,Figure3.8showsthree
sinewaves,eachwithdifferentamplitudeand
frequency.Allcanberepresentedbythree
spikesinthefrequencydomain.
Example 3.7

24. 3.24
Figure 3.8 The time domain and frequency domain of three sine waves

25. 3.25
Signals and Communication
A single-frequency sine wave is not
useful in data communications
We need to send a composite signal, a
signal made of many simple sine
waves.
According to Fourier analysis, any
composite signal is a combination of
simple sine waves with different
frequencies, amplitudes, and phases.

26. 3.26
Composite Signals and
Periodicity
If the composite signal is periodic, the
decomposition gives a series of signals
with discretefrequencies.
If the composite signal is nonperiodic, the
decomposition gives a combination of
sine waves with continuousfrequencies.

27. 3.27
Figure3.9showsaperiodiccompositesignalwith
frequencyf.Thistypeofsignalisnottypicalofthose
foundindatacommunications.Wecanconsiderittobe
threealarmsystems,eachwithadifferentfrequency.
Theanalysisofthissignalcangiveusagood
understandingofhowtodecomposesignals.
Example 3.4

28. 3.28
Figure 3.9 A composite periodic signal

29. 3.29
Figure 3.10 Decomposition of a composite periodic signal in the time and
frequency domains

30. 3.30
Figure3.11showsanonperiodiccompositesignal.It
canbethesignalcreatedbyamicrophoneoratelephone
setwhenawordortwoispronounced.Inthiscase,the
compositesignalcannotbeperiodic,becausethat
impliesthatwearerepeatingthesamewordorwords
withexactlythesametone.
Example 3.5

31. 3.31
Figure 3.11 The time and frequency domains of a nonperiodic signal

32. 3.32
Bandwidth and Signal
Frequency
The bandwidth of a composite signal is
the differencebetween the highest and the
lowest frequencies contained in that
signal.

33. 3.33
Figure 3.12 The bandwidth of periodic and nonperiodic composite signals

34. 3.34
Ifaperiodicsignalisdecomposedintofivesinewaves
withfrequenciesof100,300,500,700,and900Hz,what
isitsbandwidth?Drawthespectrum,assumingall
componentshaveamaximumamplitudeof10V.
Solution
Letfhbethehighestfrequency,flthelowestfrequency,
andBthebandwidth.Then
Example 3.6
Thespectrumhasonlyfivespikes,at100,300,500,700,
and900Hz(seeFigure3.13).

35. 3.35
Figure 3.13 The bandwidth for Example 3.6

36. 3.36
Aperiodicsignalhasabandwidthof20Hz.Thehighest
frequencyis60Hz.Whatisthelowestfrequency?Draw
thespectrumifthesignalcontainsallfrequenciesofthe
sameamplitude.
Solution
Letf
hbethehighestfrequency,f
lthelowestfrequency,
andBthebandwidth.Then
Example 3.7
Thespectrumcontainsallintegerfrequencies.Weshow
thisbyaseriesofspikes(seeFigure3.14).

37. 3.37
Figure 3.14 The bandwidth for Example 3.7

38. 3.38
Anonperiodiccompositesignalhasabandwidthof200
kHz,withamiddlefrequencyof140kHzandpeak
amplitudeof20V.Thetwoextremefrequencieshavean
amplitudeof0.Drawthefrequencydomainofthe
signal.
Solution
Thelowestfrequencymustbeat40kHzandthehighest
at240kHz.Figure3.15showsthefrequencydomain
andthebandwidth.
Example 3.8

39. 3.39
Figure 3.15 The bandwidth for Example 3.8

40. 3.40
Anexampleofanonperiodiccompositesignalisthe
signalpropagatedbyanAMradiostation.IntheUnited
States,eachAMradiostationisassigneda10-kHz
bandwidth.ThetotalbandwidthdedicatedtoAMradio
rangesfrom530to1700kHz.Wewillshowtherationale
behindthis10-kHzbandwidthinChapter5.
Example 3.9

41. 3.41
Anotherexampleofanonperiodiccompositesignalis
thesignalpropagatedbyanFMradiostation.Inthe
UnitedStates,eachFMradiostationisassigneda200-
kHzbandwidth.ThetotalbandwidthdedicatedtoFM
radiorangesfrom88to108MHz.Wewillshowthe
rationalebehindthis200-kHzbandwidthinChapter5.
Example 3.10

42. 3.42
Anotherexampleofanonperiodiccompositesignalis
thesignalreceivedbyanold-fashionedanalogblack-
and-whiteTV.ATVscreenismadeupofpixels.Ifwe
assumearesolutionof525×700,wehave367,500
pixelsperscreen.Ifwescanthescreen30timesper
second,thisis367,500×30=11,025,000pixelsper
second.Theworst-casescenarioisalternatingblackand
whitepixels.Wecansend2pixelspercycle.Therefore,
weneed11,025,000/2=5,512,500cyclespersecond,or
Hz.Thebandwidthneededis5.5125MHz.
Example 3.11

43. 3.43
Fourier analysis is a tool that changes a
time domain signal to a frequency
domain signal and vice versa.
Note
Fourier Analysis

44. 3.44
Fourier Series
Every composite periodicsignal can be
represented with a series of sine and cosine
functions.
The functions are integral harmonics of the
fundamental frequency “f” of the composite
signal.
Using the series we can decompose any
periodic signal into its harmonics.

45. 3.45
Fourier Series

46. 3.46
Examples of Signals and the
Fourier Series Representation

47. 3.47
Sawtooth Signal

48. 3.48
Fourier Transform
Fourier Transform gives the frequency
domain of a nonperiodictime domain
signal.

49. 3.49
Example of a Fourier
Transform

50. 3.50
Inverse Fourier Transform

51. 3.51
Time limited and Band limited
Signals
A time limited signal is a signal for which
the amplitude s(t) = 0 for t > T
1and t < T
2
A band limited signal is a signal for which
the amplitude S(f) = 0 for f > F
1and f < F
2

52. 3.52
3-3 DIGITAL SIGNALS
Inadditiontobeingrepresentedbyananalogsignal,
informationcanalsoberepresentedbyadigitalsignal.
Forexample,a1canbeencodedasapositivevoltage
anda0aszerovoltage.Adigitalsignalcanhavemore
thantwolevels.Inthiscase,wecansendmorethan1bit
foreachlevel.
Bit Rate
Bit Length
Digital Signal as a Composite Analog Signal
Application Layer
Topics discussed in this section:

53. 3.53
Figure 3.16 Two digital signals: one with two signal levels and the other
with four signal levels

54. 3.54
Adigitalsignalhaseightlevels.Howmanybitsare
neededperlevel?Wecalculatethenumberofbitsfrom
theformula
Example 3.16
Eachsignallevelisrepresentedby3bits.

55. 3.55
Adigitalsignalhasninelevels.Howmanybitsare
neededperlevel?Wecalculatethenumberofbitsby
usingtheformula.Eachsignallevelisrepresentedby
3.17bits.However,thisanswerisnotrealistic.The
numberofbitssentperlevelneedstobeanintegeras
wellasapowerof2.Forthisexample,4bitscan
representonelevel.
Example 3.17

56. 3.56
Assumeweneedtodownloadtextdocumentsattherate
of100pagespersec.Whatistherequiredbitrateofthe
channel?
Solution
Apageisanaverageof24lineswith80charactersin
eachline.Ifweassumethatonecharacterrequires8
bits(ascii),thebitrateis
Example 3.18

57. 3.57
Adigitizedvoicechannel,aswewillseeinChapter4,is
madebydigitizinga4-kHzbandwidthanalogvoice
signal.Weneedtosamplethesignalattwicethehighest
frequency(twosamplesperhertz).Weassumethateach
samplerequires8bits.Whatistherequiredbitrate?
Solution
Thebitratecanbecalculatedas
Example 3.19

58. 3.58
Whatisthebitrateforhigh-definitionTV(HDTV)?
Solution
HDTVusesdigitalsignalstobroadcasthighquality
videosignals.TheHDTVscreenisnormallyaratioof
16:9.Thereare1920by1080pixelsperscreen,andthe
screenisrenewed30timespersecond.Twenty-fourbits
representsonecolorpixel.
Example 3.20
TheTVstationsreducethisrateto20to40Mbps
throughcompression.

59. 3.59
Figure 3.17 The time and frequency domains of periodic and nonperiodic
digital signals

60. 3.60
Figure 3.18 Baseband transmission

61. 3.61
A digital signal is a composite analog
signal with an infinite bandwidth.
Note

62. 3.62
Figure 3.19 Bandwidths of two low-pass channels

63. 3.63
Figure 3.20 Baseband transmission using a dedicated medium

64. 3.64
Baseband transmission of a digital
signal that preserves the shape of the
digital signal is possible only if we have
a low-pass channel with an infinite or
very wide bandwidth.
Note

65. 3.65
Anexampleofadedicatedchannelwheretheentire
bandwidthofthemediumisusedasonesinglechannel
isaLAN.AlmosteverywiredLANtodayusesa
dedicatedchannelfortwostationscommunicatingwith
eachother.InabustopologyLANwithmultipoint
connections,onlytwostationscancommunicatewith
eachotherateachmomentintime(timesharing);the
otherstationsneedtorefrainfromsendingdata.Ina
startopologyLAN,theentirechannelbetweeneach
stationandthehubisusedforcommunicationbetween
thesetwoentities.
Example 3.21

66. 3.66
Figure 3.21 Rough approximation of a digital signal using the first harmonic
for worst case

67. 3.67
Figure 3.22 Simulating a digital signal with first three harmonics

68. 3.68
In baseband transmission, the required bandwidth is
proportional to the bit rate;
if we need to send bits faster, we need more bandwidth.
Note
In baseband transmission, the required
bandwidth is proportional to the bit rate;
if we need to send bits faster, we need
more bandwidth.

69. 3.69
Table 3.2 Bandwidth requirements

70. 3.70
Whatistherequiredbandwidthofalow-passchannelif
weneedtosend1Mbpsbyusingbasebandtransmission?
Solution
Theanswerdependsontheaccuracydesired.
a.The minimum bandwidth, is B = bit rate /2, or 500 kHz.
b.A better solution is to use the first and the third
harmonics with B = 3 ×500 kHz = 1.5 MHz.
c.Still a better solution is to use the first, third, and fifth
harmonics with B = 5 ×500 kHz = 2.5 MHz.
Example 3.22

71. 3.71
Wehavealow-passchannelwithbandwidth100kHz.
Whatisthemaximumbitrateofthis
channel?
Solution
Themaximumbitratecanbeachievedifweusethefirst
harmonic.Thebitrateis2timestheavailablebandwidth,
or200kbps.
Example 3.22

72. 3.72
Figure 3.23 Bandwidth of a bandpass channel

73. 3.73
If the available channel is a bandpass
channel, we cannot send the digital
signal directly to the channel;
we need to convert the digital signal to
an analog signal before transmission.
Note

74. 3.74
Figure 3.24 Modulation of a digital signal for transmission on a bandpass
channel

75. 3.75
Anexampleofbroadbandtransmissionusing
modulationisthesendingofcomputerdatathrougha
telephonesubscriberline,thelineconnectingaresident
tothecentraltelephoneoffice.Theselinesaredesigned
tocarryvoicewithalimitedbandwidth.Thechannelis
consideredabandpasschannel.Weconvertthedigital
signalfromthecomputertoananalogsignal,andsend
theanalogsignal.Wecaninstalltwoconvertersto
changethedigitalsignaltoanalogandviceversaatthe
receivingend.Theconverter,inthiscase,iscalleda
modemwhichwediscussindetailinChapter5.
Example 3.24

76. 3.76
Asecondexampleisthedigitalcellulartelephone.For
betterreception,digitalcellularphonesconvertthe
analogvoicesignaltoadigitalsignal(seeChapter16).
Althoughthebandwidthallocatedtoacompany
providingdigitalcellularphoneserviceisverywide,we
stillcannotsendthedigitalsignalwithoutconversion.
Thereasonisthatweonlyhaveabandpasschannel
availablebetweencallerandcallee.Weneedtoconvert
thedigitizedvoicetoacompositeanalogsignalbefore
sending.
Example 3.25

77. 3.77
3-4 TRANSMISSION IMPAIRMENT
Signalstravelthroughtransmissionmedia,whicharenot
perfect.Theimperfectioncausessignalimpairment.This
meansthatthesignalatthebeginningofthemediumis
notthesameasthesignalattheendofthemedium.
Whatissentisnotwhatisreceived.Threecausesof
impairmentareattenuation,distortion,andnoise.
Attenuation
Distortion
Noise
Topics discussed in this section:

78. 3.78
Figure 3.25 Causes of impairment

79. 3.79

80. 3.80
Attenuation
Means loss of energy -> weaker signal
When a signal travels through a medium it
loses energy overcoming the resistance of
the medium
Amplifiers are used to compensate for this
loss of energy by amplifying the signal.

81. 3.81
Measurement of Attenuation
To show the loss or gain of energy the unit
“decibel” is used.
dB = 10log
10P
2/P
1
P
1-input signal
P
2-output signal

82. 3.82
Figure 3.26 Attenuation

83. 3.83
Supposeasignaltravelsthroughatransmissionmedium
anditspowerisreducedtoone-half.ThismeansthatP
2
is(1/2)P
1.Inthiscase,theattenuation(lossofpower)
canbecalculatedas
Example 3.26
Alossof3dB(–3dB)isequivalenttolosingone-half
thepower.

84. 3.84
Asignaltravelsthroughanamplifier,anditspoweris
increased10times.ThismeansthatP
2=10P
1.Inthis
case,theamplification(gainofpower)canbecalculated
as
Example 3.27

85. 3.85
Onereasonthatengineersusethedecibeltomeasurethe
changesinthestrengthofasignalisthatdecibel
numberscanbeadded(orsubtracted)whenweare
measuringseveralpoints(cascading)insteadofjusttwo.
InFigure3.27asignaltravelsfrompoint1topoint4.In
thiscase,thedecibelvaluecanbecalculatedas
Example 3.28

86. 3.86
Figure 3.27 Decibels for Example 3.28

87. 3.87
Sometimesthedecibelisusedtomeasuresignalpower
inmilliwatts.Inthiscase,itisreferredtoasdB
mandis
calculatedasdB
m=10log10P
m,whereP
misthepower
inmilliwatts.CalculatethepowerofasignalwithdB
m=
−30.
Solution
Wecancalculatethepowerinthesignalas
Example 3.29

88. 3.88
Thelossinacableisusuallydefinedindecibelsper
kilometer(dB/km).Ifthesignalatthebeginningofa
cablewith−0.3dB/kmhasapowerof2mW,whatisthe
powerofthesignalat5km?
Solution
Thelossinthecableindecibelsis5×(−0.3)=−1.5dB.
Wecancalculatethepoweras
Example 3.30

89. 3.89

90. 3.90
Distortion
Means that the signal changes its form or shape
Distortion occurs in compositesignals
Each frequency component has its own
propagation speedtraveling through a medium.
The different components therefore arrive with
different delaysat the receiver.
That means that the signals have different phases
at the receiver than they did at the source.

91. 3.91
Figure 3.28 Distortion

92. 3.92

93. 3.93
Noise
There are different types of noise
Thermal-random noise of electrons in the wire
creates an extra signal
Induced-from motors and appliances, devices
act are transmitter antenna and medium as
receiving antenna.
Crosstalk-same as above but between two
wires.
Impulse-Spikes that result from power lines,
lighning, etc.

94. 3.94
Figure 3.29 Noise

95. 3.95
Signal to Noise Ratio (SNR)
To measure the quality of a system the SNR
is often used. It indicates the strength of the
signal wrt the noise power in the system.
It is the ratio between two powers.
It is usually given in dB and referred to as
SNR
dB.

96. 3.96
Thepowerofasignalis10mWandthepowerofthe
noiseis1μW;whatarethevaluesofSNRandSNR
dB?
Solution
ThevaluesofSNRandSNR
dBcanbecalculatedas
follows:
Example 3.31

97. 3.97
ThevaluesofSNRandSNR
dBforanoiselesschannel
are
Example 3.32
Wecanneverachievethisratioinreallife;itisanideal.

98. 3.98
Figure 3.30 Two cases of SNR: a high SNR and a low SNR

99. 3.99
3-6 PERFORMANCE
Oneimportantissueinnetworkingistheperformanceof
thenetwork—howgoodisit?Wediscussqualityof
service,anoverallmeasurementofnetworkperformance,
ingreaterdetailinChapter24.Inthissection,we
introducetermsthatweneedforfuturechapters.
Bandwidth -capacity of the system
Throughput -no. of bits that can be
pushed through
Latency (Delay) -delay incurred by a bit
from start to finish
Bandwidth-Delay Product
Topics discussed in this section:

100. 3.100
In networking, we use the term
bandwidth in two contexts.
Thefirst,bandwidthinhertz,referstothe
rangeoffrequenciesinacompositesignal
ortherangeoffrequenciesthatachannel
canpass.
Thesecond,bandwidthinbitspersecond,
referstothespeedofbittransmissionina
channelorlink.Oftenreferredtoas
Capacity.
Note

101. 3.101
Thebandwidthofasubscriberlineis4kHzforvoiceor
data.Thebandwidthofthislinefordatatransmission
canbeupto56,000bpsusingasophisticatedmodemto
changethedigitalsignaltoanalog.
Example 3.42

102. 3.102
Ifthetelephonecompanyimprovesthequalityoftheline
andincreasesthebandwidthto8kHz,wecansend
112,000bpsbyusingthesametechnologyasmentioned
inExample3.42.
Example 3.43

103. 3.103
Anetworkwithbandwidthof10Mbpscanpassonlyan
averageof12,000framesperminutewitheachframe
carryinganaverageof10,000bits.Whatisthe
throughputofthisnetwork?
Solution
Wecancalculatethethroughputas
Example 3.44
Thethroughputisalmostone-fifthofthebandwidthin
thiscase.

104. 3.104

105. 3.105
Propagation & Transmission delay
Propagation speed -speed at which a bit
travels though the medium from source to
destination.
Transmission speed -the speed at which all
the bits in a message arrive at the
destination. (difference in arrival time of
first and last bit)

106. 3.106
Propagation and Transmission Delay
Propagation Delay = Distance/Propagation speed
Transmission Delay = Message size/bandwidth bps
Latency = Propagation delay + Transmission delay +
Queueing time + Processing time

107. 3.107
Whatisthepropagationtimeifthedistancebetweenthe
twopointsis12,000km?Assumethepropagationspeed
tobe2.4×108m/sincable.
Solution
Wecancalculatethepropagationtimeas
Example 3.45
TheexampleshowsthatabitcangoovertheAtlantic
Oceaninonly50msifthereisadirectcablebetweenthe
sourceandthedestination.

108. 3.108
Whatarethepropagationtimeandthetransmission
timefora2.5-kbytemessage(ane-mail)ifthe
bandwidthofthenetworkis1Gbps?Assumethatthe
distancebetweenthesenderandthereceiveris12,000
kmandthatlighttravelsat2.4×108m/s.
Solution
Wecancalculatethepropagationandtransmissiontime
asshownonthenextslide:
Example 3.46

109. 3.109
Notethatinthiscase,becausethemessageisshortand
thebandwidthishigh,thedominantfactoristhe
propagationtime,notthetransmissiontime.The
transmissiontimecanbeignored.
Example 3.46 (continued)

110. 3.110
Whatarethepropagationtimeandthetransmission
timefora5-Mbytemessage(animage)ifthebandwidth
ofthenetworkis1Mbps?Assumethatthedistance
betweenthesenderandthereceiveris12,000kmand
thatlighttravelsat2.4×10
8
m/s.
Solution
Wecancalculatethepropagationandtransmission
timesasshownonthenextslide.
Example 3.47

111. 3.111
Notethatinthiscase,becausethemessageisverylong
andthebandwidthisnotveryhigh,thedominantfactor
isthetransmissiontime,notthepropagationtime.The
propagationtimecanbeignored.
Example 3.47 (continued)

112. 3.112
Figure 3.31 Filling the link with bits for case 1

113. 3.113
Wecanthinkaboutthelinkbetweentwopointsasa
pipe.Thecrosssectionofthepiperepresentsthe
bandwidth,andthelengthofthepiperepresentsthe
delay.Wecansaythevolumeofthepipedefinesthe
bandwidth-delayproduct,asshowninFigure3.33.
Example 3.48

114. 3.114
Figure 3.32 Filling the link with bits in case 2

115. 3.115
The bandwidth-delay product defines
the number of bits that can fill the link.
Note

116. 3.116
Figure 3.33 Concept of bandwidth-delay product