Antennas and Wave Propagation

JNTUK III B.Tech- I sem Unit - II : Thin Linear Wire Antennas

1. ANTENNA AND WAVE PROPAGATION
B.TECH (III YEAR –I SEM)
Prepared by:
Mr. P.Venkata Ratnam.,M.Tech., (Ph.D)
Associate Professor
Department of Electronics and Communication Engineering
RAJAMAHENDRI INSTITUTE OF ENGINEERING & TECHNOLOGY
(Affiliated to JNTUK, Kakinada, Approved by AICTE -Accredited by NAAC )
Bhoopalapatnam, Rajamahendravaram, E.G.Dt, Andhra Pradesh

2. Unit-II
THIN LINEAR WIRE ANTENNAS
Introduction
Retarded Potentials
Radiation from Small Electric Dipole
Quarter wave Monopole and Half wave Dipole
Current Distributions.
Evaluation of Field Components.
Power Radiated.
Radiation Resistance.
Beam widths.
Directivity.
Effective Area and Effective Height.

3. Natural current distributions.
Fields and patterns of Thin Linear Center-fed
Antennas of different lengths.
Radiation Resistance at a point which is not current
maximum.
Antenna Theorems –Applicability and Proofs for
equivalence of directional characteristics.
Loop Antennas: Small Loops -Field Components.
Comparison of far fields of small loop and short
dipole.
Concept of short magnetic dipole.
D and Rrrelations for small loops.

4. Introduction :
Theelectricchargesarethesourcesofthe
electromagnetic(EM)fields.
Whenthesesourcesaretimevarying,theEMwaves
propagateawayfromthesourcesandradiationtakes
place.
Radiationcanbeconsideredasaprocessof
transmittingelectricenergy.
Theradiationofthewavesintospaceiseffectively
achievedwiththehelpofconductingordielectric
structurescalledantennasorradiators.

5. Anantennaisameansofradiatingorreceivingthe
EMwaves.
Anantennamaybeusedforeithertransmittingor
receivingEMenergy.
Inthedesignofantennasystems,wemustconsider
importantrequirementssuchastheantennapattern,
thetotalpowerradiated,theinputimpedanceofthe
radiator,theradiationefficiencyetc.
Thedirectsolutionfortheserequirementscanbe
obtainedbysolvingMaxwell'sequationswith
appropriateboundaryconditionsoftheradiatorandat
infinity.

6. Potential Functions and Electromagnetic Fields :
Incaseoftheelectrostaticfieldorthesteadymagnetic
field,theelectricfieldorthemagneticfieldcanbe
obtainedeasilybysettingthepotentialsintermsofthe
chargesorcurrents.
Thepotentialsareobtainedintermsofthechargesor
currentsandtheelectricormagneticfieldsare
obtainedfromthesepotentials.
Forobtainingthepotentialsfortheelectromagnetic
fieldtherearedifferentapproaches

7. Inthefirstapproach,usingtrialanderror,the
potentialsfortheelectricandmagneticfieldare
generalized.
Thenitisshownthatthesepotentialssatisfythe
Maxwell'sequations.
Thisapproachiscalledheuristicapproach.
ThesecondapproachistostartwiththeMaxwell's
equationsandthenderivethedifferentialequations
thatthepotentialssatisfy.
Thethirdapproachistoobtaindirectlythesolutions
ofthederivedequationsforthepotentials.

8. Retarded Potentials :
Heuristic Approach :
Considerauniformvolumecharge
densityρ
voverthegivenvolume
asshowninFig.
Considerthedifferentialvolumeatapointdistancer
fromorigin,wherethechargedensityisρ
v(r’)
ThentheScalarelectricpotentialVatpointPcanbe
expressedintermsofastaticchargedistributionas
……1

9. Then the fundamental electric field
can be obtained by finding the
gradient of a scalar potential V as
……. 2
Similarly for a steady magnetic field,
in a homogeneous medium, the vector
magnetic potential A can be expressed interms of
current distribution at constant with time as
…. 3

10. Then the fundamental magnetic field can be obtained
by finding the curl of the vector magnetic potential A
as
…… 4
ThepotentialsV(r)andA(r)representthepotentials
forthestaticelectricandMagneticfieldsrespectively
wherethechargeandcurrentdistributionsdonotvary
withtime.
Butthechargeandcurrentdistributionsproducingthe
electromagneticfieldvarywithtime.

11. Thus the time varying fields, the potentials are given
by
Where R = |r –r’ |
Ingeneral,thevelocityoftheelectromagnetic
fieldcanbedescribedintermsofμandεas
v=1/√με.

12. Thenthepotentialsatthesamepointwithatafinite
velocity.Henceitisnecessarytoaccomplishthisfinite
propagationtime,theequationscanbeexpressedby
introducingatimedelayR/vas
ThepotentialsaredelayedorretardedbytimeR/v,
Hencethepotentialsarecalledretardedpotentials

13. Maxwell’s Equation Approach :
Inthisapproach,startingfromMaxwell’sequations,
thedifferentialequationsarederived.
Fortimevaryingfields,Maxwell’sequationsaregiven
by

14. From Equation.12, It is clear that the divergence of H
is zero.
But from the vector identity the divergence of a Curl of
a vector is zero.
Thisclearlyindicatesthattosatisfy,Hmustbe
expressedasaCurlofvectorpotentialAas
Now, putting the value of μH in the Equation 9, We
get,

15. Interchanging the operators on RHS of the above
equation, We get,
FromvectoridentityCurlofagradientofscalaris
alwayszero,Soequation.14willbesatisfiedonlyif
theterm isdefinedasgradientofscalar.

16. Now, introduce a scalar potential V such that
Then the electric field strength is given by
So from equations 13 and 16, It is clear, the electric
and magnetic fields E and H can be expressed interms
of scalar potential V and vector potential A
Now substitute the values E and H from Eq 16 and 13
in Eq. 10, We get

17. Interchanging the operators, We have
From the vector identity
Rewrite the equation 17 using above identity

18. Now, Substitute E from Eq. 16 into Eq .11, We get
Theequations18and19arecalledCoupled
equations.
ThesearenotsufficientenoughtodefineAandE
completely.UsingHelmholtzTheoremwegetthe
uniquesolution.

19. TheHelmholtztheoremstatesthatAnyvectorfield
candefineduniquelyifthecurlanddivergenceofthe
fieldbothareknownatanypoint.
NowwemaychoosedivergenceofAfromequation18
as
ThusselectiondivergenceofAasgiveninequations
18,19,20becomeuncoupled.
TherelationshipbetweenthedivergenceofAandVis
knowasLorentzgaugeconditionorLorentzgaugefor
potentials.

20. Using this condition in equation 18 ,We get

21. SimilarlyusingLorentzgaugeconditioninequation
19,Weget,

22. Theequations21and22arethestandardwave
equationsincludingsourceterms.
The solutions of the equations 21 and 22 are given by

23. Potential Functions for Time Periodic Fields.
The potential functions obtained using Lorentz gauge
condition are given by,
In the sinusoidal steady state,we can rewrite these by
replacing ( ∂/∂t ) with ( jω) ,Thus the equation 1 can
be written as,

24. Similarly equation 2 can be written as
Similarly,forthewavetravellinginRdirectionwith
phasevariationrepresentedbye
-jβR.
Sothepotential
functionsforsinusoidaloscillationsaregivenby

25. Radiation from Alternating Current Element :
Tocalculatetheelectromagneticfieldradiationfrom
shortdipole,theretardedpotentialisused.
Ashortdipoleisanalternatingcurrentelement.Itis
alsocalledanoscillatingcurrentelement.
Analternatingcurrentelementisconsiderasthebasic
sourceofradiation.
Ingenerally,acurrentelementIdLisnothingbutan
elementlengthdLcarryingcurrentI.
Thislengthofwireisassumetobeveryshort.

26. Tocalculatetheelectromagneticfieldduetoalternating
currentelement,considerSphericalco-ordinatesystem
andthecurrentthroughisIdLcosωt,locateatcentre
isshowninfig.
The electromagnetic is calculate at a point P placed at a
distance R from origin.
The element is IdLcosωt is place along the z-axis.

27. Let, the vector potential A is given by
As the current element is placed along z-axis. Hence
the vector potential can be write as

28. Fromtheequation2,itisclearthatthevectorpotential
A
zcanbeobtainedbyintegratingthecurrentdensityJ
overthevolume.
•Now,thecurrentisassumedtobeconstantalongthe
lengthdL,theintegrationofJoverthelengthdLgives
valueIdL.
•Thus mathematically we can write,

29. Substitutetheintegrationofequation3inequation2,
thevectorpotentialinz-axisisgivenby
Now , the magnetic field is given by
Usingsphericalco-ordinatesystem,tofindcurlAinr,
θ,ɸdirection,wehave

30. Hence A is given by
Now, Aɸ= 0 and because of symmetry ∂/∂ɸ= 0. Thus
the first two terms of equation 7 can be neglected.

31. Now,PuttingvaluesofArandAθfromequation5,
Weget
SubstitutetheAzvalueinaboveequation,Wehave,

33. Therefore
Hence the magnetic field H is given by
Substitute the value of , We get,

34. From equation 10, It is clear that the magnetic field H
only in ɸdirection.
Now substitute , We get,

35. Let us now calculating electric field
Now, Integrating on both sides w.r.t variable, We get,

36. Let us calculate each term of separately
The component in a
rdirection is given by

37. NowsubstitutethevalueHɸfromequation12,
Wehave

38. Therefore
Letuscalculatethecomponentinaθ
direction

39. Now substitute Hɸfrom equation11, We have

40. Therefore

41. Now,Calculatetheelectricfieldcomponents,From
equation13,Einardirectionisgivenby

42. Therefore

45. Significance of Field components :

46. Power Radiated from Current Element :
Consider a current element placed at a centre of a
spherical co-ornate systems.
Then the power radiated per unit area at point P can
be calculated by using Poynting theorem .
The components of the Poynting vector are given by
P
r= E
θH
ɸ
P
θ= -E
rH
ɸ
P
ɸ= E
θH
r
But we know that, the current element is at orgin, then
the E
ɸ= 0

47. Let the field components of E
r , E
θ, H
ɸand replacing
vby c are given by

48. The Poynting vector inθdirection is given by

50. Therefore

51. Now calculate Radial power in r direction

55. Thereforethetotalpowerradiatedbycurrentelement
canbeobtainedbyintegratingtheradialpointing
vectoroverasphericalsurface.
The element area dsis given by
ds = 2πr
2
sinθdθ
The total power radiated is given by

56. In Spherical coordinate system θvaries from 0 to π.
Hence

58. Substitute this value, we get,
The power represented in terms of Maximum or Peak
current

61. Therefore the power radiated from the current
element is given by
Generally the power is expressed as
P = I
2
R
So, the Radiation resistance of current element is given
by

62. Short Linear Antenna:
Thecurrentelementthatwehaveconsideredpreviously
isnotapractical,butitishypothetical.
Itisusefulinthetheoreticalcalculationssuchasthe
componentsofthefields,radiationofpoweretc.
Thepracticalexampleofthecentre-fedantennaisan
elementarydipole.
Thelengthofsuchcentre-fedantennaisveryshortin
wavelength.

63. Thecurrentamplitudeonsuchantennaismaximum
atthecenteranditdecreasesuniformlytozeroatthe
ends.
IfweconsidersamecurrentIflowingthroughthe
hypotheticalcurrentelementandthepracticalshort
dipole,bothofsamelength.
Thenthepracticalshortdipoleradiateonlyone-
quarterofthepowerthatisradiatedbythecurrent
element.

64. The radiation resistance of short dipole is ¼ times of
that the current element
Hence the radiation resistance of short dipole is given
by

65. Monopole or Short vertical Antenna :
Consider the current I flows through a monopole of
length h and short dipole of length l =2h.
Then the field strength produced by both are same
above the reflecting plane.

66. The radiated power of monopole is half of that
radiated by short dipole.
Hence the radiation resistance of monopole is half of
the radiation resistance of short dipole

67. But h = l/2 for monopole

68. Half wave dipole and The Monopole :
Averycommonlyusedantennaisthehalfwavedipole
withalengthonehalfofthefreespacewavelengthof
theradiatedwave.
Itisfoundthelinearcurrentdistributionisnotsuitable
forthisantenna.
Butwhensuchantennaisfedatitscentrewiththe
helpofatransmissionline,
Itgivesacurrentdistributionwhichisapproximately
sinusoidal,withmaximumatthecentreandzeroatthe
ends.

69. The half wave clippie can be considered as a chain of
Hertzian dipoles.

70. Power Radiated by the Half Wave Dipole and Monopole
A dipole antenna is a vertical radiator fed in the centre.
It produces maximum is the overall length.
TheverticalantennaofheightH=L/2producesthe
radiationcharacteristicsabovetheplanewhichissimilar
tothatproducedbythedipoleantennaoflengthL=
2H.
The vertical antenna is referred as a monopole.
Thepracticallyusedantennasarehalfwavedipole
(λ/2)andquarterwavemonopole(λ/4).

71. The half wave dipole consists two legs each of length
L/2.
The physical length of the half wave dipole at the
frequency of operation is λ/2 in free space.

72. The quarter wave mono pole consists of single vertical
leg erected on the perfect ground.
The length of the leg of the quarter wave monopole is
λ/4.

73. Now consider the current element I dz is placed at
distance z from z=0.
Then the sinusoidal current distribution is given by
Now consider a point P located far distance R from
current element I dz. The vector potential Az is given
by

74. Now substitute I value in the above equation, we get
Wehavecertainassumptionstocalculateradiation
field,assumeR=r,replaceRindenominatoronly
withrand
R=r-zcosθinnumeratorterm.

75. Therefore, we have,


76. For quarter wave monopole

78. Therefore
.

80. Integrating w.r.t zand H =λ/4 , We get ,

81. Now finding the LCM, We have,

82. Substitute this property in above equation, We get,

83. Now find the magnetic field component Hɸusing
Maxwell equations.
The current element placed along the z-axis, then

84. Therefore
Now substitute Az value, We get,

85. ThemagnitudeofmagneticfieldofHalfwaveor
quarterwaveMonopoleisgivenby

86. Now, the relation between electric and magnetic fields
is given by
Therefore, the Electric field component is

87. The electric field component is
ThemagnitudeofElectricfieldofHalfwaveor
quarterwaveMonopoleisgivenby

88. The maximum power interms of effective current is
given by

89. Now, the average value of power is given by

90. The total radiated power from dipole antenna is

91. Therefore
The value of integral by the Simpsons rule is given by

92. Hence the radiated power is given by
Hence the radiation resistance of the quarter wave
Monopole is
Therefore the radiation resistance of the Half wave
dipole is given by

95. Antenna Theorems :
Anantennacanbeusedasbothtransmittingantenna
andreceivingantenna.
Whileusingso,wemaycomeacrossaquestion
whetherthepropertiesoftheantennamightchangeas
itsoperatingmodeischanged.
Thepropertiesofantennabeingunchangeableis
calledasthepropertyofreciprocity.

96. The properties of transmitting and receiving antenna
that exhibit the reciprocity are −
Equality of Directional patterns.
Equality of Directivities.
Equality of Effective lengths.
Equality of Antenna impedances.

97. Equality of Directional patterns :
Theradiationpatternoftransmittingantenna1,which
transmitstothereceivingantenna2isequaltothe
radiationpatternofantenna2,ifittransmitsand
antenna1receivesthesignal.
Equality of Directivities :
Directivityis same for both transmitting and receiving
antennas, if the value of directivity is same for both the
cases.
The directivities are same whether calculated from
transmitting antenna’s power or receiving antenna’s
power.

98. Equality of Effective lengths :
Thevalueofmaximumeffectiveapertureissamefor
bothtransmittingandreceivingantennas.
Equalityinthelengthsofbothtransmittingand
receivingantennasismaintainedaccordingtothevalue
ofthewavelength.
Equality in Antenna Impedances :
Theoutputimpedanceofatransmittingantennaand
theinputimpedanceofareceivingantennaareequal
inaneffectivecommunication.

99. Loop Antennas:
AnRFcurrentcarryingcoilisgivenasingleturnintoa
loop,canbeusedasanantennacalledasloop
antenna.
Thecurrentsthroughthisloopantennawillbein
phase.
The magnetic field will be perpendicular to the whole
loop carrying the current.
Itmaybeinanyshapesuchascircular,rectangular,
triangular,squareorhexagonalaccordingtothe
designer’sconvenience.

100. Loop antennas are of two types.
Large loop antennas
Small loop antennas
Large loop antennas are also called asresonant antennas.
They have high radiation efficiency. These antennas
have length nearly equal to the intended wavelength(L=λ).
Smallloopantennasarealsocalledasmagneticloop
antennas.Thesearemostlyusedasreceivers.
Theseantennasareofthesizeofone-tenthofthe
wavelength(L=λ/10)

101. Field Components :
Consider that the square loop is located at the centre
of the Spherical coordinate system.
Then the far field of the square loop will have only E
Φ
Component.

102. Now the far field radiation due to two point sources with
reference to centre O can be represented as
E
Φ= Field component + Field component
due to dipole 1,4 due to dipole 2, 3
The path difference between
from point L to M
Path difference = d cos( 90
0
–θ)
Then the path difference is
expressed interms of wavelength as
Path difference = d cos ( 90
0
–θ) / λ

103. Letψisthephaseangleanditisrelatewithpath
differenceisgivenby
Phase angle = ψ= 2π x path difference
The field component for any dipole is given by
Field component = Magnitude x e
j( phase angle )
Let the field component due to dipole 1,4 is given by

104. Similarlyfieldcomponentduetodipole2,3is
givenby
Hencethefarfieldradiationduetosquare
loopisgivenby

105. We have
Hence we can write
Now substitute phase angle, we get,

106. The other field components can be given by the
following relation

107. Comparison of far fields of small loop and short dipole :
Asmallloopcanbeconsiderasequivalenttoshort
magneticdipole.
ThusasmallloopofareaAcarryingcurrentIcanbe
replacedbyashortmagneticdipolelengthLand
carryingfictitiousmagneticcurrentI
m

108. ThemagneticmomentoftheloopisI.Aisthesmall
looparea,currentIistheuniformphasecurrent
throughoutloop
Henceequatingthismagneticmomentwiththe
magneticmomentofshortdipoleisgivenbyq
m.L
Hence we can write,
Theaboveequationrepresenttheequivalenceof
magneticdipoleoflengthLcarryingfictitiouscurrent
I
mwithsmallloopofareaAandcarryingcurrentI.

109. Radiation Resistance of Loop antenna :
Tofindtheradiationresistanceoftheloopantenna,the
poyntingvectorisintegratedoverlargesphere.
Thepowerradiatedisgivenby
P = I
2
rmsR
rad= ½ I
MR
rad
The average poynting power
is given by

110. The total power radiated over a large sphere is given
by
Now, equate this quation with power equation, we get,

111. But πa
2
is the areaA of loop. Hence the radiation
resistance is given by
The above expression can be written as

112. For circular loop antenna, the radiation resistance is
given by
Where C is the Circumference ( C= 2πa )

113. When C/λ>5, the loop consider large loop, then the
radiation resistance is given by

114. The directivity of circular Loop Antenna :
The directivity of an antenna is defined as the ratio of
maximum radiation intensity to the average radiation
intensity
Let us consider two cases one for small loop antenna
and other for large loop antenna

117. Application of loop antennas :
A small loop antenna is used as source for
paraboloid in many applications
Large loop antenna can be used as direction
finder.
Used in line of sight communication

118. Thank you