CH- 3 Intro to Physical Layer of data communication.pdf

Physical Layer of Data Communication

1. Chapter 3
Introduction
To
Physical
Layer
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2. Chapter 3: Outline
3.1DATAANDSIGNALS
3.2PERIODICANALOGSIGNALS
3.3DIGITALSIGNALS
3.4TRANSMISSIONIMPAIRMENT
3.5DATARATELIMITS
3.6PERFORMANCE

3. Chapter 3: Objective
Thefirstsectionshowshowdataandsignalscanbeeither
analogordigital.Analogreferstoanentitythatiscontinuous;
digitalreferstoanentitythatisdiscrete.
Thesecondsectionshowsthatonlyperiodicanalogsignalscan
beusedindatacommunication.Thesectiondiscussessimple
andcompositesignals.Theattributesofanalogsignalssuchas
period,frequency,andphasearealsoexplained.
Thethirdsectionshowsthatonlynon-periodicdigitalsignals
canbeusedindatacommunication.Theattributesofadigital
signalsuchasbitrateandbitlengtharediscussed.Wealsoshow
howdigitaldatacanbesentusinganalogsignals.Basebandand
broadbandtransmissionarealsodiscussedinthissection.

4. Chapter 3: Objective (continued)
Thefourthsectionisdevotedtotransmissionimpairment.The
sectionshowshowattenuation,distortion,andnoisecanimpair
asignal.
Thefifthsectiondiscussesthedataratelimit:howmanybitsper
secondwecansendwiththeavailablechannel.Thedataratesof
noiselessandnoisychannelsareexaminedandcompared.
Thesixthsectiondiscussestheperformanceofdata
transmission.Severalchannelmeasurementsareexamined
includingbandwidth,throughput,latency,andjitter.
Performanceisanissuethatisrevisitedinseveralfuture
chapters.

5. 3.5
3-1 DATA AND SIGNALS
Figure3.1showsascenarioinwhicha
scientistworkinginaresearchcompany,
SkyResearch,needstoorderabook
relatedtoherresearchfromanonline
bookseller,ScientificBooks.

6. 3.6
Figure 3.1: Communication at the physical layer

7. 3.7
3.1.1 Analog and Digital Data
Datacanbeanalogordigital.Thetermanalogdata
referstoinformationthatiscontinuous;digitaldata
referstoinformationthathasdiscretestates.For
example,ananalogclockthathashour,minute,and
secondhandsgivesinformationinacontinuous
form;themovementsofthehandsarecontinuous.
Ontheotherhand,adigitalclockthatreportsthe
hoursandtheminuteswillchangesuddenlyfrom
8:05to8:06.

8. 3.8
3.1.2 Analog and Digital Signals
Likethedatatheyrepresent,signalscanbeeither
analogordigital.Ananalogsignalhasinfinitely
manylevelsofintensityoveraperiodoftime.Asthe
wavemovesfromvalueAtovalueB,itpasses
throughandincludesaninfinitenumberofvalues
alongitspath.Adigitalsignal,ontheotherhand,
canhaveonlyalimitednumberofdefinedvalues.
Althougheachvaluecanbeanynumber,itisoften
assimpleas1and0.

9. 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.
Note

10. 3.9
Figure 3.2: Comparison of analog and digital signals

11. 10
Analog vs. Digital Data
Analog data
Data take on continuous values
E.g., human voice, temperature reading
Digital data
Data take on discrete values
E.g., text, integers

12. 11
Analog vs. Digital Signals
Analog signals
have an infinite number of
values in a range
Digital signals
Have a limited number of
values
value
time
value
time
To be transmitted, data must be
transformed to electromagnetic signals.

13. 12
Data and Signals
Telephone
Analog Data Analog Signal
Modem
Digital Data Analog Signal
Codec
Analog Data Digital Signal
Digital
transmitter
Digital Data Digital Signal

14. 3.13
3.1.3 Periodic and Nonperiodic
Aperiodicsignalcompletesapatternwithina
measurabletimeframe,calledaperiod,andrepeats
thatpatternoversubsequentidenticalperiods.The
completionofonefullpatterniscalledacycle.A
nonperiodicsignalchangeswithoutexhibitinga
patternorcyclethatrepeatsovertime.Bothanalog
anddigitalsignals.

15. 3.14
3-2 PERIODIC ANALOG SIGNALS
Periodic analog signals can be classified
as simpleor composite.
A simple periodic analog signal, a sine
wave, cannot be decomposed into
simpler signals.
Acompositeperiodicanalogsignalis
composedofmultiplesinewaves.

16. 3.15
Sine Wave
A sine wave is the most fundamental
form of a periodic analog signal
It can be represented by three
parameters:
Peak Amplitude
Frequency
Phase

17. 3.16
Figure 3.3: A sine wave
Time
Value
• • •

18. 3.17
Figure 3.4: Two signals with two different amplitudes
Peak
amplitude
Peak
amplitude
Peak
amplitude
Peak
amplitude

19. Period and frequency
Period or Time Period:Time required
to complete one cycle. It is measured in
seconds.
Frequency:Number of cycles
completed in one second. It is measured
in Hz or cycles per second.
If “f”represents
frequency
and “T”
represents
period
then,
f= 1/Tand T= 1/f

21. Frequency and period are the inverse of
each other.

22. 3.18
Figure 3.5: Two signals with the same phase and frequency, but
different amplitudes

23. 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.

24. 3.19
Table 3.1: Units of period and frequency

25. 3.20
Expressaperiodof100msinmicroseconds.
Example 3.4
Solution
FromTable3.1wefindtheequivalentsof1ms(1msis10–
3s)and1s(1sis106μs).Wemakethefollowing
substitutions:

26. 3.21
Thepowerweuseathomehasafrequencyof60Hz(50Hz
inEurope).Theperiodofthissinewavecanbedetermined
asfollows:
Example 3.3
Thismeansthattheperiodofthepowerforourlightsat
homeis0.0116s,or16.6ms.Oureyesarenotsensitive
enoughtodistinguishtheserapidchangesinamplitude.

27. 3.22
Theperiodofasignalis100ms.Whatisitsfrequencyin
kilohertz?.
Example 3.5
Solution
Firstwechange100mstoseconds,andthenwecalculate
thefrequencyfromtheperiod(1Hz=10
–3
kHz).

28. 3.23
The term phase, or phase shift, describes
the position of the waveform relative to
time 0.
If we think of the wave as something that
can be shifted backward or forward along
the time axis, phase describes the amount
of that shift.
It indicates the status of the first cycle.
3.2.2 Phase

29. 3.24
Figure 3.6: Three sine waves with different phases

30. 3.25
Asinewaveisoffset1/6cyclewithrespecttotime0.What
isitsphaseindegreesandradians?
Example 3.6
Solution
Weknowthat1completecycleis360°.Therefore,1/6cycle
is

31. 3.26
Wavelength
Wavelength is the distance a signal can
travel in a period
Wavelength can be calculated if one is
given the propagation speed and the
period of the signal
In data communication, wavelength is
usually used to describe the transmission
of light in an optical fiber

32. Wavelength
Let λbe the wavelength, c be the speed
of light and f be the frequency, then:
(Recall, distance = speed x time)
wavelength=propagationspeed×period
=propagationspeed×
1
??????
�=
??????
??????

33. Wavelength
Wavelength is measured in micrometers
(i.e., microns)
Example:
We would like to calculate the wavelength
of red light in air when
f = 4 x 10
14
:
�=
??????
??????
=
3×10
8
4×10
14
=0.75×10
−6
m=0.75�m

34. 3.27
Figure 3.7: Wavelength and period
Direction of
propagation

35. 3.28
3.2.4 Time and Frequency Domains
Time domain plot is an amplitude vs. time
plot, i.e., it shows amplitude variations of
a signal with respect to time
Frequency domain plot is an amplitude vs.
frequency plot, i.e., it shows which
frequencies and of what amplitude are
present in the signal

36. 3.29
Figure 3.8: The time-domain and frequency-domain plots of a sine wave

37. 3.30
Thefrequencydomainismorecompactandusefulwhenwe
aredealingwithmorethanonesinewave.Forexample,
Figure3.9showsthreesinewaves,eachwithdifferent
amplitudeandfrequency.Allcanberepresentedbythree
spikesinthefrequencydomain.
Example 3.7

38. 3.31
Figure 3.9: The time domain and frequency domain of three sine waves

39. 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.
Fourier analysis is discussed in
Appendix E.

40. 3.32
3.2.5 Composite Signals
Sofar,wehavefocusedonsimplesinewaves.
Simplesinewaveshavemanyapplicationsindaily
life.Wecansendasinglesinewavetocarryelectric
energyfromoneplacetoanother.Forexample,the
powercompanysendsasinglesinewavewitha
frequencyof60Hztodistributeelectricenergyto
housesandbusinesses.Asanotherexample,wecan
useasinglesinewavetosendanalarmtoasecurity
centerwhenaburglaropensadoororwindowin
thehouse.Inthefirstcase,thesinewaveiscarrying
energy;inthesecond,thesinewaveisasignalof
danger.

41. 3.33
Figure3.10showsaperiodiccompositesignalwith
frequencyf.Thistypeofsignalisnottypicalofthosefound
indatacommunications.Wecanconsiderittobethree
alarmsystems,eachwithadifferentfrequency.Theanalysis
ofthissignalcangiveusagoodunderstandingofhowto
decomposesignals.Itisverydifficulttomanually
decomposethissignalintoaseriesofsimplesinewaves.
However,therearetools,bothhardwareandsoftware,that
canhelpusdothejob.Wearenotconcerned
abouthowitisdone;weareonlyinterestedintheresult.
Figure3.11showstheresultofdecomposingtheabove
signalinboththetimeandfrequencydomains.
Example 3.8

42. 3.34
Figure 3.10: A composite periodic signal

43. 3.35
Figure 3.11: Decomposition of a composite periodic signal
Time
Amplitude
• • •
Amplitude
f 3f 9f
b. Frequency-domain decomposition of the composite signal
Frequency

44. 3.36
Figure3.12showsanonperiodiccompositesignal.Itcanbe
thesignalcreatedbyamicrophoneoratelephonesetwhena
wordortwoispronounced.Inthiscase,thecomposite
signalcannotbeperiodic,becausethatimpliesthatweare
repeatingthesamewordorwordswithexactlythe
sametone.
Example 3.9

45. 3.37
Figure 3.12: Time and frequency domain of a non-periodic signal

46. 3.38
3.2.6 Bandwidth
Therangeoffrequenciescontainedinacomposite
signalisitsbandwidth.Thebandwidthisnormallya
differencebetweentwonumbers.Forexample,ifa
compositesignalcontainsfrequenciesbetween1000
and5000,itsbandwidthis5000−1000,or4000.

47. 3.39
Figure 3.13: The bandwidth of periodic and nonperiodic composite
signals

48. 3.40
Ifaperiodicsignalisdecomposedintofivesinewaveswith
frequenciesof100,300,500,700,and900Hz,whatisits
bandwidth?Drawthespectrum,assumingallcomponents
haveamaximumamplitudeof10V.
Example 3.10
Solution
Letf
hbethehighestfrequency,f
lthelowestfrequency,and
Bthebandwidth.Then

49. 3.41
Figure 3.14: The bandwidth for example 3.10

50. 3.42
Aperiodicsignalhasabandwidthof20Hz.Thehighest
frequencyis60Hz.Whatisthelowestfrequency?Drawthe
spectrumifthesignalcontainsallfrequenciesofthesame
amplitude.
Example 3.11
Solution
Letf
hbethehighestfrequency,f
lthelowestfrequency,and
Bthebandwidth.Then
Thespectrumcontainsallintegerfrequencies.Weshowthis
byaseriesofspikes(seeFigure3.15).

51. 3.43
Figure 3.15: The bandwidth for example 3.11

52. 3.44
Anonperiodiccompositesignalhasabandwidthof200
kHz,withamiddlefrequencyof140kHzandpeak
amplitudeof20V.Thetwoextremefrequencieshavean
amplitudeof0.Drawthefrequencydomainofthesignal.
Example 3.12
Solution
Thelowestfrequencymustbeat40kHzandthehighestat
240kHz.Figure3.16showsthefrequencydomainandthe
bandwidth.

53. 3.45
Figure 3.16: The bandwidth for example 3.12

54. 3.46
Anexampleofanonperiodiccompositesignalisthesignal
propagatedbyanAMradiostation.IntheUnitedStates,
eachAMradiostationisassigneda10-kHzbandwidth.The
totalbandwidthdedicatedtoAMradiorangesfrom530to
1700kHz.Wewillshowtherationalebehindthis10-kHz
bandwidthinChapter5..
Example 3.13

55. 3.47
Anotherexampleofanonperiodiccompositesignalisthe
signalpropagatedbyanFMradiostation.IntheUnited
States,eachFMradiostationisassigneda200-kHz
bandwidth.ThetotalbandwidthdedicatedtoFMradio
rangesfrom88to108MHz.Wewillshowtherationale
behindthis200-kHzbandwidthinChapter5.
Example 3.14

56. 3.48
Anotherexampleofanonperiodiccompositesignalisthe
signalreceivedbyanold-fashionedanalogblack-and-white
TV.ATVscreenismadeupofpixels(pictureelements)
witheachpixelbeingeitherwhiteorblack.Thescreenis
scanned30timespersecond.Ifweassumearesolutionof
525×700(525verticallinesand700horizontallines),
whichisaratioof3:4,wehave367,500pixelsperscreen.If
wescanthescreen30timespersecond,thisis367,500×30
=11,025,000pixelspersecond.Theworst-casescenariois
alternatingblackandwhitepixels.Inthiscase,weneedto
representonecolorbytheminimumamplitudeandtheother
colorbythemaximumamplitude.Wecansend2pixelsper
cycle.
Example 3.15

57. 3.49
Therefore,weneed11,025,000/2=5,512,500cyclesper
second,orHz.Thebandwidthneededis5.5124MHz.This
worst-casescenariohassuchalowprobabilityofoccurrence
thattheassumptionisthatweneedonly70percentofthis
bandwidth,whichis3.85MHz.Sinceaudioand
synchronizationsignalsarealsoneeded,a4-MHz
bandwidthhasbeensetasideforeachblackandwhiteTV
channel.AnanalogcolorTVchannelhasa6-MHz
bandwidth.
Example 3.15 (continued)

58. 3.23
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
voltage and a 0 as zero voltage.

-A digital signal can have more than two levels. In
this case, we can send more than 1 bit for each
level.

59. 3.51
Figure 3.17: Two digital signals: one with two signal levels and the
other with four signal levels

60. 3.52
Adigitalsignalhaseightlevels.Howmanybitsareneeded
perlevel?Wecalculatethenumberofbitsfromthe
followingformula.Eachsignallevelisrepresentedby3bits.
Example 3.16

61. 3.53
Adigitalsignalhasninelevels.Howmanybitsareneeded
perlevel?Wecalculatethenumberofbitsbyusingthe
formula.Eachsignallevelisrepresentedby3.17bits.
However,thisanswerisnotrealistic.Thenumberofbits
sentperlevelneedstobeanintegeraswellasapowerof2.
Forthisexample,4bitscanrepresentonelevel.
Example 3.17

62. 3.54
3.3.1 Bit Rate
Mostdigitalsignalsarenonperiodic,andthusperiod
andfrequencyarenotappropriatecharacteristics.
Anotherterm—bitrate(insteadoffrequency)—is
usedtodescribedigitalsignals.Thebitrateisthe
numberofbitssentin1s,expressedinbitsper
second(bps).Figure3.17showsthebitratefortwo
signals.
3.27
Bit Rate
Most digital signals are nonperiodic, and thus period
and frequency are not appropriate characteristics.
Another term— bit rate (instead of frequency)—is used
to describe digital signals.

The bit rate is the number of bits sent in 1s, expressed in
bits per second (bps).

63. 3.55
Assumeweneedtodownloadtextdocumentsattherateof
100pagespersecond.Whatistherequiredbitrateofthe
channel?
Example 3.18
Solution
Apageisanaverageof24lineswith80charactersin
line.Ifweassumethatonecharacterrequires8bits,the
rateis

64. 3.55
Assumeweneedtodownloadtextdocumentsattherateof
100pagespersecond.Whatistherequiredbitrateofthe
channel?
Example 3.18
Solution
Apageisanaverageof24lineswith80charactersin
line.Ifweassumethatonecharacterrequires8bits,the
rateis

65. 3.57
Whatisthebitrateforhigh-definitionTV(HDTV)?
Example 3.20
Solution
HDTVusesdigitalsignalstobroadcasthighqualityvideo
signals.TheHDTVscreenisnormallyaratioof16:9(in
contrastto4:3forregularTV),whichmeansthescreenis
wider.Thereare1920by1080pixelsperscreen,andthe
screenisrenewed30timespersecond.Twenty-fourbits
representsonecolorpixel.Wecancalculatethebitrateas
TheTVstationsreducethisrateto20to40Mbpsthrough
compression.

66. 3.58
3.3.2 Bit Length
Wediscussedtheconceptofthewavelengthforan
analogsignal:thedistanceonecycleoccupiesonthe
transmissionmedium.Wecandefinesomething
similarforadigitalsignal:thebitlength.Thebit
lengthisthedistanceonebitoccupiesonthe
transmissionmedium.

67. 3.59
3.3.3 Digital As Composite Analog
BasedonFourieranalysis(SeeAppendixE),a
digitalsignalisacompositeanalogsignal.The
bandwidthisinfinite,asyoumayhaveguessed.We
canintuitivelycomeupwiththisconceptwhenwe
consideradigitalsignal.Adigitalsignal,inthetime
domain,comprisesconnectedverticalandhorizontal
linesegments.Averticallineinthetimedomain
meansafrequencyofinfinity:ahorizontallinein
thetimedomainmeansafrequencyofzero.Going
fromafrequencyofzerotoafrequencyofinfinity
impliesallfrequenciesinbetweenarepartofthedomain.

68. 3.3.3 Digital As Composite Analog

69. 3.60
Figure 3.18: The time and frequency domains of periodic and
nonperiodic digital signals

70. 3.61
3.3.4 Transmission of Digital Signals

71. 3.62
Figure 3.19: Baseband transmission
Baseband transmission requires that we have a low-pass channel, a channel with
a bandwidth that starts from zero

72. 3.63
Figure 3.20: Bandwidth of two low-pass channels

73. 3.64
Figure 3.21: Baseband transmission using a dedicated medium

74. 3.65
Anexampleofadedicatedchannelwheretheentire
bandwidthofthemediumisusedasonesinglechannelisa
LAN.AlmosteverywiredLANtodayusesadedicated
channelfortwostationscommunicatingwitheachother.In
abustopologyLANwithmultipointconnections,onlytwo
stationscancommunicatewitheachotherateachmomentin
time(timesharing);theotherstationsneedtorefrainfrom
sendingdata.InastartopologyLAN,theentirechannel
betweeneachstationandthehubisusedforcommunication
betweenthesetwoentities.
Example 3.21

75. 3.66
Figure 3.22: Rough approximation of a digital signal (part 1)

76. 3.67
Figure 3.22: Rough approximation of a digital signal (part 2)

77. 3.68
Figure 3.23: Simulating a digital signal with first three harmonics
(part I)

78. 3.69
Figure 3.23: Simulating a digital signal with first three harmonics
(part II)

79. 3.70
Table 3.2: Bandwidth requirements

80. 3.71
Whatistherequiredbandwidthofalow-passchannelifwe
needtosend1Mbpsbyusingbasebandtransmission?
Example 3.22
Solution
Theanswerdependsontheaccuracydesired.
a.Theminimumbandwidth,aroughapproximation,isB
bitrate/2,or500kHz.Weneedalow-passchannelwith
frequenciesbetween0and500kHz.
b.Abetterresultcanbeachievedbyusingthefirstandthe
thirdharmonicswiththerequiredbandwidthB=3×500
kHz=1.5MHz.
c.Astillbetterresultcanbeachievedbyusingthefirst,
third,andfifthharmonicswithB=5×500kHz=2.5
MHz.

81. 3.72
Wehavealow-passchannelwithbandwidth100kHz.What
isthemaximumbitrateofthischannel?.
Example 3.23
Solution
Themaximumbitratecanbeachievedifweusethefirst
harmonic.Thebitrateis2timestheavailablebandwidth,or
200kbps.

82. 3.73
Figure 3.24: Bandwidth of a band-pass channel

83. 3.74
Figure 3.25: Modulation of a digital signal for transmission on
band-pass channel

84. 3.74
Figure 3.25: Modulation of a digital signal for transmission on
band-pass channel

85. 3.75
Anexampleofbroadbandtransmissionusingmodulationis
thesendingofcomputerdatathroughatelephonesubscriber
line,thelineconnectingaresidenttothecentraltelephone
office.Theselines,installedmanyyearsago,aredesignedto
carryvoice(analogsignal)withalimitedbandwidth.
Althoughthischannelcanbeusedasalow-passchannel,it
isnormallyconsideredabandpasschannel.Onereasonis
thatthebandwidthissonarrow(4kHz)thatifwetreatthe
channelaslow-passanduseitforbasebandtransmission,
themaximumbitratecanbeonly8kbps.Thesolutionisto
considerthechannelabandpasschannel,convertthedigital
signalfromthecomputertoananalogsignal,andsendthe
analogsignal.
Example 3.24

86. 3.76
Asecondexampleisthedigitalcellulartelephone.Forbetter
reception,digitalcellularphonesdigitizeanalogvoice.
Althoughthebandwidthallocatedtoacompanyproviding
digitalcellularphoneserviceisverywide,westillcannot
sendthedigitizedsignalwithoutconversion.Thereasonis
thatwehaveonlyaband-passchannelavailablebetween
callerandcallee.Forexample,iftheavailablebandwidthis
Wandweallow1000couplestotalksimultaneously,this
meanstheavailablechannelisW/1000,justpartofthe
entirebandwidth.Weneedtoconvertthedigitizedvoicetoa
compositeanalogsignalbeforetransmission.
Example 3.25

87. 3.28
TRANSMISSION IMPAIRMENT

-What is sent is not what is received

-Three causes of impairment are
attenuation, distortion, and noise

88. 3.78
Figure 3.26: Causes of impairment

89. 3.29
3.4.1 Attenuation
-Attenuation means a loss of energy.

-When a signal, simple or composite, travels
through a medium, it loses some of its energy in
overcoming the resistance of the medium.

-To compensate for this loss, amplifiers are used to
amplify the signal.

90. 3.80
Figure 3.27: Attenuation and amplification

91. 3.81
Supposeasignaltravelsthroughatransmissionmediumand
itspowerisreducedtoonehalf.ThismeansthatP2=0.5
P1.Inthiscase,theattenuation(lossofpower)canbe
calculatedas
Example 3.26
Alossof3dB(−3dB)isequivalenttolosingone-halfthe
power.

92. 3.82
Asignaltravelsthroughanamplifier,anditspoweris
increased10times.ThismeansthatP
2=10P
1.Inthiscase,
theamplification(gainofpower)canbecalculatedas
Example 3.27

93. 3.83
Figure 3.28: Decibels for Example 3.28

94. 3.84
Onereasonthatengineersusethedecibeltomeasurethe
changesinthestrengthofasignalisthatdecibelnumbers
canbeadded(orsubtracted)whenwearemeasuringseveral
points(cascading)insteadofjusttwo.InFigure3.28a
signaltravelsfrompoint1topoint4.Thesignalis
attenuatedbythetimeitreachespoint2.Betweenpoints2
and3,thesignalisamplified.Again,betweenpoints3and
4,thesignalisattenuated.Wecanfindtheresultantdecibel
valueforthesignaljustbyaddingthedecibelmeasurements
betweeneachsetofpoints.Inthiscase,thedecibelvalue
canbecalculatedas
Example 3.28

95. 3.85
Sometimesthedecibelisusedtomeasuresignalpowerin
milliwatts.Inthiscase,itisreferredtoasdB
mandis
calculatedasdB
m=10log10Pm,wherePmisthepowerin
milliwatts.CalculatethepowerofasignalifitsdB
m=−30.
Example 3.29
Solution
Wecancalculatethepowerinthesignalas

96. 3.86
Thelossinacableisusuallydefinedindecibelsper
kilometer(dB/km).Ifthesignalatthebeginningofacable
with−0.3dB/kmhasapowerof2mW,whatisthepowerof
thesignalat5km?
Example 3.30
Solution

97. 3.86
Thelossinacableisusuallydefinedindecibelsper
kilometer(dB/km).Ifthesignalatthebeginningofacable
with−0.3dB/kmhasapowerof2mW,whatisthepowerof
thesignalat5km?
Example 3.30
Solution
Thelossinthecableindecibelsis5×(−0.3)=−1.5dB.We
cancalculatethepoweras

98. 3.32
Distortion
-Signal changes its form or shape.

-Distortion can occur in a composite signal made
of different frequencies.

-Each signal component has its own propagation
speed through a medium and, therefore, its own
delay in arriving at the final destination.

-Differences in delay may create a difference in
phase if the delay is not exactly the same as the
period duration.

99. 3.88
Figure 3.29: Distortion

100. 3.89
3.4.3 Noise
Noiseisanothercauseofimpairment.Severaltypes
ofnoise,suchasthermalnoise,inducednoise,
crosstalk,andimpulsenoise,maycorruptthesignal.
Thermalnoiseistherandommotionofelectronsin
awire,whichcreatesanextrasignalnotoriginally
sentbythetransmitter.Inducednoisecomesfrom
sourcessuchasmotors.Crosstalkistheeffectofone
wireontheother.
3.34
Noise
-Noise is another cause of impairment.
-Several types of noise, such as thermal noise,
induced noise, crosstalk, and impulse noise, may
corrupt the signal.
-Thermal noise is the random motion of electrons in
a wire, which creates an extra signal not originally
sent by the transmitter.
-Induced noise comes from sources such as motors.
-Crosstalk is the effect of one wire on the other.

101. 3.90
Figure 3.30: Noise

102. 3.91
Figure 3.31: Two cases of SNR: a high SNR and a low SNR

103. 3.92
Thepowerofasignalis10mWandthepowerofthenoise
is1μW;whatarethevaluesofSNRandSNR
dB?
Example 3.31
Solution
ThevaluesofSNRandSNR
dBcanbecalculatedasfollows:

104. 3.93
ThevaluesofSNRandSNR
dBforanoiselesschannelare
Example 3.32
Solution
ThevaluesofSNRandSNRdBforanoiselesschannelare
Wecanneverachievethisratioinreallife;itisanideal.

105. 3.94
3-5 DATA RATE LIMITS
Averyimportantconsiderationindata
communicationsishowfastwecansend
data,inbitspersecond,overachannel.
Twotheoreticalformulasweredeveloped
tocalculatethedatarate:onebyNyquist
foranoiselesschannel,anotherby
Shannonforanoisychannel.
3.36
DATA RATE LIMITS

-How fast we can send data, in bits per second,
over a channel.

-Two theoretical formulas were developed to
calculate the data rate:
Nyquist for a noiseless channel
Shannon for a noisy channel.

106. 3.95
3.5.1 Noiseless Channel: Nyquist Rate
Foranoiselesschannel,theNyquistbitrateformula
definesthetheoreticalmaximumbitrate.

107. 3.96
DoestheNyquisttheorembitrateagreewiththeintuitivebit
ratedescribedinbasebandtransmission?
Example 3.33
Solution
Theymatchwhenwehaveonlytwolevels.Wesaid,in
basebandtransmission,thebitrateis2timesthebandwidth
ifweuseonlythefirstharmonicintheworstcase.However,
theNyquistformulaismoregeneralthanwhatwederived
intuitively;itcanbeappliedtobasebandtransmissionand
modulation.Also,itcanbeappliedwhenwehavetwoor
morelevelsofsignals.

108. 3.97
Consideranoiselesschannelwithabandwidthof3000Hz
transmittingasignalwithtwosignallevels.Themaximum
bitratecanbecalculatedas
Example 3.34

109. 3.98
Considerthesamenoiselesschanneltransmittingasignal
withfoursignallevels(foreachlevel,wesend2bits).The
maximumbitratecanbecalculatedas
Example 3.35

110. 3.99
Weneedtosend265kbpsoveranoiselesschannelwitha
bandwidthof20kHz.Howmanysignallevelsdoweneed?
Example 3.36
Solution
WecanusetheNyquistformulaasshown:
Sincethisresultisnotapowerof2,weneedtoeither
increasethenumberoflevelsorreducethebitrate.Ifwe
have128levels,thebitrateis280kbps.Ifwehave64
levels,thebitrateis240kbps.

111. 3.100
3.5.2 Noisy Channel: Shannon Capacity
Inreality,wecannothaveanoiselesschannel;the
channelisalwaysnoisy.In1944,ClaudeShannon
introducedaformula,calledtheShannoncapacity,
todeterminethetheoreticalhighestdataratefora
noisychannel:
3.41
Noisy Channel: Shannon Capacity
In reality, we cannot have a noiseless channel; the
channel is always noisy. In 1944, Claude Shannon
introduced a formula, called the Shannon capacity, to
determine the theoretical highest data rate for a noisy
channel:

112. 3.101
Consideranextremelynoisychannelinwhichthevalueof
thesignal-to-noiseratioisalmostzero.Inotherwords,the
noiseissostrongthatthesignalisfaint.Forthischannelthe
capacityCiscalculatedas
Example 3.37
Thismeansthatthecapacityofthischanneliszero
regardlessofthebandwidth.Inother
words,wecannotreceiveanydatathroughthischannel.

113. 3.102
Wecancalculatethetheoreticalhighestbitrateofaregular
telephoneline.Atelephonelinenormallyhasabandwidthof
3000Hz(300to3300Hz)assignedfordata
communications.Thesignal-to-noiseratioisusually3162.
Forthischannelthecapacityiscalculatedas
Example 3.38
Thismeansthatthehighestbitrateforatelephonelineis
34.860kbps.Ifwewanttosenddatafasterthanthis,wecan
eitherincreasethebandwidthofthelineorimprovethe
signal-to-noiseratio.

114. 3.103
Thesignal-to-noiseratioisoftengivenindecibels.Assume
thatSNR
dB=36andthechannelbandwidthis2MHz.The
theoreticalchannelcapacitycanbecalculatedas
Example 3.39

115. 3.104
WhentheSNRisveryhigh,wecanassumethatSNR+1is
almostthesameasSNR.Inthesecases,thetheoretical
channelcapacitycanbesimplifiedto For
example,wecancalculatethetheoreticalcapacityofthe
previousexampleas
Example 3.40

116. 3.105
3.5.3 Using Both Limits
Inpractice,weneedtousebothmethodstofindthe
limitsandsignallevels.Letusshowthiswithan
example.

117. 3.106
Wehaveachannelwitha1-MHzbandwidth.TheSNRfor
thischannelis63.Whataretheappropriatebitrateand
signallevel?
Example 3.41
Solution
First,weusetheShannonformulatofindtheupperlimit.
TheShannonformulagivesus6Mbps,theupperlimit.For
betterperformancewechoosesomethinglower,4Mbps.
ThenweusetheNyquistformulatofindthenumberof
signallevels.

118. 3.107
3-6 PERFORMANCE
Uptonow,wehavediscussedthetools
oftransmittingdata(signals)overa
networkandhowthedatabehave.
One important issue in networking is the
performance of the network— how good is it?

119. 3.108
3.6.1 Bandwidth
Onecharacteristicthatmeasuresnetwork
performanceisbandwidth.However,thetermcanbe
usedintwodifferentcontextswithtwodifferent
measuringvalues:bandwidthinhertzand
bandwidthinbitspersecond..
3.48
Bandwidth
- One characteristic that measures network performance

- can be used in two different contexts with two different
measuring values:
bandwidth in hertz
bandwidth in bits per second

121. 3.109
Thebandwidthofasubscriberlineis4kHzforvoiceor
data.Thebandwidthofthislinefordatatransmission
canbeupto56,000bpsusingasophisticatedmodemto
changethedigitalsignaltoanalog.
Example 3.42

122. 3.110
Ifthetelephonecompanyimprovesthequalityoftheline
andincreasesthebandwidthto8kHz,wecansend112,000
bpsbyusingthesametechnologyasmentionedinExample
3.42.
Example 3.43

123. 3.111
3.6.2 Throughput
Thethroughputisameasureofhowfastwecan
actuallysenddatathroughanetwork
.Although,at
firstglance,bandwidthinbitspersecondand
throughputseemthesame,theyaredifferent.Alink
mayhaveabandwidthofBbps,butwecanonly
sendTbpsthroughthislinkwithTalwayslessthan
B.

124. 3.112
3.6.3 Throughput
Thelatencyordelaydefineshowlongittakesforan
entiremessagetocompletelyarriveatthedestination
fromthetimethefirstbitissentoutfromthesource.
Wecansaythatlatencyismadeoffourcomponents:
propagationtime,transmissiontime,queuingtime
andprocessingdelay.
3.50
Throughput
-The latency or delay defines 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.

-We can say that latency is made of four
components: propagation time, transmission time,
queuing time and processing delay.

Propagation time measures the time required for a bi t to travel from the source to the
destination
Tranmission time between the first bit leaving the
sender and the last bit arriving at the receiver.

125. 3.113
Anetworkwithbandwidthof10Mbpscanpassonlyan
averageof12,000framesperminutewitheachframe
carryinganaverageof10,000bits.Whatisthethroughput
ofthisnetwork?
Example 3.44
Solution
Wecancalculatethethroughputas
Thethroughputisalmostone-fifthofthebandwidthinthis
case.

126. 3.114
Whatisthepropagationtimeifthedistancebetweenthetwo
pointsis12,000km?Assumethepropagationspeedtobe
2.4×108m/sincable.
Example 3.45
Solution
Wecancalculatethepropagationtimeas
TheexampleshowsthatabitcangoovertheAtlanticOcean
inonly50msifthereisadirectcablebetweenthesource
andthedestination.

127. 3.115
Whatarethepropagationtimeandthetransmissiontimefor
a2.5-KB(kilobyte)messageifthebandwidthofthe
networkis1Gbps?Assumethatthedistancebetweenthe
senderandthereceiveris12,000kmandthatlighttravelsat
2.4×108m/s.
Example 3.46
Solution

128. 3.115
Whatarethepropagationtimeandthetransmissiontimefor
a2.5-KB(kilobyte)messageifthebandwidthofthe
networkis1Gbps?Assumethatthedistancebetweenthe
senderandthereceiveris12,000kmandthatlighttravelsat
2.4×108m/s.
Example 3.46
Solution
Wecancalculatethepropagationandtransmissiontimeas
Notethatinthiscase,becausethemessageisshortandthe
bandwidthishigh,thedominantfactoristhepropagation
time,notthetransmissiontime.

129. 3.116
Whatarethepropagationtimeandthetransmissiontimefor
a5-MB(megabyte)message(animage)ifthebandwidthof
thenetworkis1Mbps?Assumethatthedistancebetween
thesenderandthereceiveris12,000kmandthatlight
travelsat2.4×108m/s.
Example 3.47
Solution
Wecancalculatethepropagationandtransmissiontimesas
Wecancalculatethepropagationandtransmissiontimesas

130. 3.117
3.6.4 Bandwidth-Delay Product
Bandwidthanddelayaretwoperformancemetrics
ofalink.However,aswewillseeinthischapterand
futurechapters,whatisveryimportantindata
communicationsistheproductofthetwo,the
bandwidth-delayproduct.Letuselaborateonthis
issue,usingtwohypotheticalcasesasexamples.
3.55
Bandwidth-Delay Product
-Bandwidth and delay are two performance metrics
of a link.

-What is very important in data communications is
the product of the two, the bandwidth-delay product.

131. 3.118
Figure 3.32: Filling the links with bits for Case 1
we can say that this product 1 × 5 is the maximum number of
bits that can fill the link. There can be no more than 5 bits at any time on the link.

132. 3.119
Figure 3.33: Filling the pipe with bits for Case 2

133. 3.120
Wecanthinkaboutthelinkbetweentwopointsasapipe.
Thecrosssectionofthepiperepresentsthebandwidth,and
thelengthofthepiperepresentsthedelay.Wecansaythe
volumeofthepipedefinesthebandwidth-delayproduct,as
showninFigure3.34.
Example 3.48

134. 3.121
Figure 3.34: Concept of bandwidth-delay product

135. 3.122
3.6.5 Jitter
Anotherperformanceissuethatisrelatedtodelayis
jitter.Wecanroughlysaythatjitterisaproblemif
differentpacketsofdataencounterdifferentdelays
andtheapplicationusingthedataatthereceiversite
istime-sensitive(audioandvideodata,forexample).
Ifthedelayforthefirstpacketis20ms,forthe
secondis45ms,andforthethirdis40ms,thenthe
real-timeapplicationthatusesthepacketsendures
jitter.
.