Antennas and Wave Propagation

JNTUK -Unit - III Antenna Arrays

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 -III
ANTENNA ARRAYS
Introduction
2 Element arrays –different cases.
Principle of Pattern Multiplication.
N element Uniform Linear Arrays –
Broadside,
End-fire Arrays.
EFA with Increased Directivity.
Derivation of their characteristics and
comparison

3. Concept of Scanning Arrays.
Directivity Relations (no derivations).
Related Problems.
Binomial Arrays,
Effects of Uniform and Non-uniform
Amplitude Distributions.
Design Relations.
Arrays with Parasitic Elements
Yagi-Uda Arrays
Folded Dipoles and their characteristics.

4. Introduction :
Antennaarrayistheradiatingsysteminwhichseveral
antennasarespacedproperlysoastogetgreaterfield
strengthatafardistancefromtheradiatingsystemby
combiningradiationsatpointfromalltheantennasin
thesystem.
Ingeneral,thetotalfieldproducedbytheantenna
arrayatafardistanceisthevectorsumofthefields
producedbytheindividualantennasofthearray.
Theindividualelementisgenerallycalledelementof
anantennaarray.

5. Themainfunctionofanarrayistoproducehighly
directionalradiation.
Theantennaarrayissaidtolineariftheelementsof
theantennaarrayareequallyspacedalongastraight
line.
Thefieldisavectorquantitywithbothmagnitudeand
phase.
Therelativephasesofindividualfieldcomponents
dependontherelativedistanceoftheindividual
clement.

6. Array Configurations :
Broadly,arrayantennascanbeclassifiedintofour
categories:
(a) Broadside array
(b) End-fire array
(c) Collinear array
(d) Parasitic array

7. Broadside Array :
Thisisatypeofarrayinwhichthenumberofidentical
elementsisplacedonasupportinglinedrawn
perpendiculartotheirrespectiveaxes.
Thespacingbetweenanytwoelementsisdenotedby-d.
Alltheelementsarefedwithcurrentswithequal
magnitudeandsamephase.
The direction of maximum radiation is perpendicular to
the array axis and to the plane containing the array
clement.

8. Now consider two isotropic point sources spaced
equally with respect to the origin of the co-ordinate
system as shown in the Fig.

9. Consider that point P is far away from the origin.
Let the distance of point P from
origin be r.
The wave radiated by radiator
A2 will reach point P as
Compared to that radiated
by radiator A1.
Thisisduetothepathdifferencethatthewave
radiatedbyradiatorA1hastotravelextradistance.

10. Hence the path difference is given by,
Path difference = d cos ϕ
Thispathdifferencecanbeexpressedintermsofwave
lengthas
Path difference = d/λcos ϕ
Fromtheopticsthephaseangleis2πtimesthepath
difference.
Hence the phase angle is given by
Phase angle = ψ=2π x Path difference

11. Therefore the phase angle is given by
End Fire Array :
Theendfirearrayisverymuchsimilartothe
broadsidearrayfromthepointofviewofarrangement.
Butthemaindifferenceisinthedirectionof
maximumradiation.

12. Inbroadsidearray,thedirectionofthemaximum
radiationisperpendiculartotheaxisofarray.
Whileintheendfirearray,thedirectionofthe
maximumradiationisalongtheaxisofarray.

13. Thusintheendfirearraynumberofidentical
antennasarespacedequallyalongaline.
Alltheantennasarefedindividuallywithcurrentsof
equalmagnitudesbuttheirphasesvaryprogressively
alongtheline.
CollinearArray:
Inthecollineararray,theantennasarearrangedco-
axially.
Theantennasarearrangedendtoendalongasingle
lineasshownintheFig

14. Theindividualelementsinthecollineararrayarefed
withcurrentsequalinmagnitudeandphase.

15. This condition is similar to the broadside array.
Incollineararraythedirectionofmaximumradiation
isperpendiculartotheaxisofarray.
Sotheradiationpatternofthecollineararrayandthe
broadsidearrayisverymuchsimilar
Buttheradiationpatternofthecollineararrayhas
circularsymmetrywithmainlobeperpendicular
everywheretotheprincipleaxis.
Thusthecollineararrayisalsocalledomnidirectional
arrayorbroadcastarray.

16. Parasitic Arrays :
Insomewayitissimilartobroadsidearray,butonly
oneelementisfeddirectlyfromsource.
Otherelementarcelectromagneticallycoupled
becauseofitsproximitytothefeedelement.
Feedelementiscalleddrivenelementwhileother
elementsarecalledparasiticelements.
Aparasiticelementlengthenedby5%todriven
elementactasreflectorandanotherelementshorted
by5%actsasdirector

17. 2 Element arrays –different cases :
Based on amplitude and phase conditions of isotropic
point sources, there are three types of arrays:
(a) Array with equal amplitude and phases
(b) Array with equal amplitude and opposite phases
(c) Array with unequal amplitude and opposite phases

18. Two Point Sources with Currents Equal in Magnitude
and Phase :
ConsidertwopointsourcesAlandA2separatedby
distancedasshownintheFigureoftwoelementarray.
Considerthatboththepointsourcesaresuppliedwith
currentsequalinmagnitudeandphase.
ConsiderpointPfarawayfromthearray.Letthe
distancebetweenpointPandpointsourcesAlandA2
ber1andr2respectively.

19. TheradiationfromthepointsourceA2willreach
earlieratpointPthanthatfrompointsourceAl
becauseofthepathdifference.
Theextradistanceistravelledbytheradiatedwave
frompointsourceAlthanthatbythewaveradiated
frompointsourceA2.

20. Hence the path difference is given by,
Path difference = d cos ϕ
Thispathdifferencecanbeexpressedintermsofwave
lengthas
Path difference = d/λcos ϕ
Fromtheopticsthephaseangleis2πtimesthepath
difference.
Hence the phase angle is given by
Phase angle = ψ=2π x Path difference

21. The phase angle is given by
Let E1be the far field at a distant point P due to point
source Al.
Similarly let E2be the far field at point P due to point
source A2.

22. ThenthetotalfieldatpointPbetheadditionofthe
twofieldcomponentsduetothepointsourcesA1and
A2.
RearrangingthetermonR.H.S,Weget

23. By the Trigonometric Identity
Therefore, The total field is
Now Substitute the value of ψ, We have,

24. The total amplitude of the field at point P is 2E0, while
the phase shift is βd cosϕ/2
Thearrayfactoristheratioofthemagnitudeofthe
resultantfieldtothemagnitudeofthemaximumfield.
But Maximum field is Emax=2E0

25. Maxima direction :
The total field is maximum when is maximum
As we know, the variation of cosine of a angle is ±1.
Hence the condition for maxima is given by
Let spacing between the two point sources be λ/2.
Then we can write

26. That is
If n = 0, then

27. Minima direction :
The total field is minimum when is minimum
That is 0 as cosine of angle has minimum value 0.
Hence the condition for minima is given by,
Again assuming d = λ/2 and β=2π/λ, we can write

28. If n=0 , Then,
Half power point direction:
When the power is half, the voltage or current is 1/√2
times the maximum value.
Hence the condition for half power point is given by,

29. Let d=λ/2 and β=2π/λ, then we can write,
If n=0 , Then,

30. ThefieldpatterndrawnwithETagainstϕford=λ/2,
thenthepatternisbidirectionalasShowninFig.
Thefieldpatternobtainedisbidirectionalanditisa
figureofeight.

31. Two Point Sources with Currents Equal in Magnitudes
but Opposite in Phase :
Considertwopointsourcesseparatedbydistanced
andsuppliedwithcurrentsequalinmagnitudebut
oppositeinphase.
Thephaseofthecurrentsisoppositei.e.180°.With
thiscondition,thetotalfieldatfarpointPisgivenby,

32. •Assuming equal magnitudes of currents, the fields at
point P due to the point sources A1 and A2 can be
written as,
•Therefore total field is given by
•Rearranging the above equation, we have

33. The above equation can be written as
Now Substitute phase angle, we get,
Maxima direction :
Thetotalfieldismaximumwhen is
maximumi.e.±1asthemaximumvalueofsineof
angleis±1.Henceconditionformaximaisgivenby,

34. Let the spacing between two isotropic point sources be
equal to d=λ/2
Substituting d=λ/2 and β=2π/λ, in above equation, we
get,
If n=0, Then

35. Minima direction :
Total field strength is minimum when is
minimum i.e. 0.
Assuming d=λ/2 and β=2π/λ, we get,
If n= 0 ,Then

36. Half Power Point Direction (HPPD) :
Whenthepowerishalfofmaximumvalue.Hencethe
conditionforthehalfpowerpointcanbeobtained.
Letd=λ/2andβ=2π/λ,wecanwrite,
Ifn=0,Then

37. Thusfromtheconditionsofmaxima,minimaandhalf
powerpoints,thefieldpatterncanbedrawnasshown
intheFig.
Ascomparedwiththefieldpatternfortwopoint
sourceswithinphasecurrents,themaximahaveshifted
by90°alongX-axisincaseofout-phasecurrentsintwo
pointsourcearray.

38. Two point sources with currents unequal in magnitude and
with any phase :
LetusconsiderFig.shownbelow.
Assumethatthetwopointsourcesareseparatedby
distancedandsuppliedwithcurrentswhichare
differentinmagnitudesandwithanyphasedifference
sayα.

39. Considerthatsource1isassumedtobereferencefor
phaseandamplitudeofthefieldsE1andE2,whichare
duetosource1andsource2respectivelyatthedistant
pointP.
LetusassumethatE1isgreaterthanE2inmagnitude
asshowninthevectordiagraminFig.

40. Nowthetotalphasedifferencebetweentheradiationsby
thetwopointsourcesatanyfarpointPisgivenby,
Assumevalueofphasedifferenceas0<α<180
0
.
ThentheresultantfieldatpointPisgivenby,

41. Note that E1> E2, the value of k is less than unity.
Moreover the value of kis given by, 0 ≤ k ≤ 1
ThemagnitudeoftheresultantfieldatpointPisgiven
by,
ThephaseanglebetweentwofieldsatthefarpointP
isgivenby,

42. n Element Uniform Linear Arrays :
Highlydirectivesinglebeampatterncanbeobtained
byincreasingthepointsourcesinthearrowfrom2to
nsay.
Anarrayofnelementsissaidtobelineararrayifall
theindividualelementsarespacedequallyalongaline.
Anarrayissaidtobeuniformarrayiftheelementsin
thearrayarefedwithcurrentswithequalmagnitudes
andwithuniformprogressivephaseshiftalongthe
line.

43. Considerageneralnelementlinearanduniformarray
withalltheindividualelementsspacedequallyat
distancedfromeachother.
Allelementsarefedwithcurrentsequalinmagnitude
anduniformprogressivephaseshiftalongline

44. ThetotalresultantfieldatthedistantpointPis
obtainedbyaddingthefieldsduetonindividual
sourcesvectorically.
Hence we can write,
Ifα=0
0
wegetnelementuniformlinearbroadside
array.
Ifα=180
0
wegetnelementuniformlinearEndfire
array.

45. Multiplying above equation by e
jψ
, we get
Now Subtracting equations, we get,

46. Simplify mathematically, we get
The resultant field is given by,

47. Themagnitudeoftheresultantfieldisgivenby,
ThephaseangleθoftheresultantfieldatpointPis
givenby,

48. Array of n elements with Equal Spacing and Currents
Equal in Magnitude and Phase -Broadside Array :
Consider'n'numberofidenticalradiatorscarries
currentswhichareequalinmagnitudeandinphase.
Hencethemaximumradiationoccursinthedirections
normaltothelineofarray.
HencesuchanarrayisknownasUniformbroadside
array.

49. Considerabroadsidearraywithnidenticalradiatorsas
shownintheFig.
TheelectricfieldproducedatpointPduetoan
elementA0isgivenby

50. NowtheelectricfieldproducedatpointPduetoan
elementA1willdifferinphaseasr0andr1arenot
actuallysame.
HencetheelectricfieldduetoA1isgivenby,

51. Thesimilarlineswecanwritetheelectricfield
producedatpointPduetoanelementA2as,
ButtheterminsidethebracketrepresentE1

52. Now, Substituting the value of E1, we get
TheelectricfieldproducedatpointPduetoelement
An-1isgivenby
ThetotalelectricfieldatpointPisgivenby

53. Let ,rewriting above equation with Phase angle (ψ)
Therefore ETis given by

54. Using the trigonometric identities, We can write the
above equation as
Now considering magnitudes of the electric fields, we
can write,

55. Properties of Broadside Array
1.Major lobe
In case of broadside array, the field is maximum in the
direction normal to the axis of the array.
Thus the condition for the maximum field at point P is
given by,

56. 2. Magnitude of major lobe
The maximum radiation occurs when ϕ=0. Hence we
can write,

57. 3.Nulls
Theratiooftotalelectricfieldtoanindividualelectric
fieldisgivenby
NowEquatingratioofmagnitudesofthefieldstozero
Theconditionofminimaisgivenby

58. Therefore

59. 4. Side Lobes Maxima
The directions of the subsidary maxima or side lobes
maxima can be obtained by
.
.
.

60. Now equation for ϕcan be written as
The equation (15) represents directions of subsidary
maxima or side lobes maxima

61. 5. Beamwidth of Major Lobe
Beamwidth is defined as the angle between first nulls.
Alternatively beamwidth is the angle equal to twice the
angle between first null and the major lobe maximum
direction.
Hence beamwidth between first nulls is given by,

62. Now , Taking cosine of angle on both sides, we get
.

63. But nd≈ (n-1)d if n is very large. This L= (nd) indicates
total length of the array.
.
.
.

64. 6. Directivity
The directivity in case of broadside array is defined as
Where
From the expression of ratio of magnitudes we can write


65. Hence normalized field pattern is given by
Hence we can write electric field due to n array as

66. Therefore radiation intensity is given by
.
‘

67. Hence
For large array, nis large hence nβdis also very large
.

68. Therefore
The directivity is given by
.
.

69. ArrayofnElementswithEqualSpacingandCurrents
EqualinMagnitudebutwithProgressivePhaseShift-
EndFireArray:
Considernnumberofidenticalradiatorssuppliedwith
equalcurrentwhicharenotinphase

70. Consider that the current supplied to first element A0
be I0.
Then the current supplied to A1is given by
Similarly the current supplied to A2is given by
Thus the current supplied to last element is

71. The electric field produced at point P, due to A0is
given by
The electric field produced at point P, due to A1 is
given by
.

72. .
TheelectricfieldproducedatpointP,duetoA2is
givenby
Similarly electric field produced at point P, due to An-1
is given by
The resultant field at point p is given by

73. Therefore , We get
Considering only magnitude we get

74. Properties of End Fire Array
1.Major lobe
Incaseoftheendfirearray,theconditionof
principlemaximaisgivenby
.
2.Magnitudeofthemajorlobe
Themaximumradiationoccurswhenψ=0.

75. 3. Nulls
Theratiooftotalelectricfieldtoanindividualelectric
fieldisgivenby
NowEquatingratioofmagnitudesofthefieldstozero
Theconditionofminimaisgivenby

76. Hence we can write
Substituting value ofψwe get,

77. Thus equation gives direction of nulls

78. 4. Side Lobes Maxima
Thedirectionsofthesubsidarymaximaorsidelobes
maximacanbeobtained
Nowputtingthevalueofψ,weget

79. Therefore

80. 5. Beamwidth of Major Lobe
Beamwidth is defined as the angle between first nulls
.

81. TheL=(nd)indicatestotallengthofthearraySo
equationbecomes
.
.

82. 6. Directivity
Thedirectivityincaseofendfirearrayisdefinedas
Where,U0isaverageradiationintensitywhichisgiven
by

83. End Fire Array with Increased Directivity :
Themaximumradiationcanbeobtainedalongthe
axisoftheuniformendfirearray
Iftheprogressivephaseshiftabetweentheelementsis
givenby,
Itisfoundthatthefieldproducedinthedirection
θ=0°ismaximum;butthedirectivityisnotmaximum.

84. HansenandWoodyardproposedcertainconditions
fortheendfireforenhancingthedirectivitywithout
alteringothercharacteristicsoftheendfirearray.
AccordingToHansen-Woodyardconditions,the
phase-shiftbetweencloselyspacedradiatorsofavery
longarrayshouldbe

85. TheenhanceddirectivityduetoHansen-Woodyard
conditionscanberealizedbyusingaboveequation.

86. Itcanbesatisfiedbyusingequationoffirstsetforθ=
0°andθ=180°byselectingthespacingbetweentwo
elementsas,
.
Henceforlargeuniformarray,theHansen-Woodyard
conditionsillustrateenhanceddirectivityifthespacing
betweenthetwoadjacentelementsisλ/4

87. Consider n element array. The array factor of the n-
element array is given by
.
.

88. Let the progressive phase shift be α= -pd, where p is
constant. Then above equation becomes
.
.

89. .
.
.

90. By simplifying we can get
.

91. The variation g(v) as a function of v is as figure below

92. Directivity of end fire array with increased directivity
For end fire array with increased directivity and
maximum radiation in ϕ= 0°direction.
the radiation on intensity for small spacing between
elements (d<<λ) is given
.

93. Pattern Multiplication Method :
Thesimplemethodofobtainingthepatternsofthe
arraysisknownaspatternmultiplicationmethod.
Consider4elementarrayofequispacedidentical
antennaswiththespacingbetweentwounitsbe
d=λ/2.

94. Theradiationpatternoftheantennas(1)and(2)
treatedtobeoperatingasasingleunitisasshownin
theFigure
Similarlytheradiationpatternoftheantennas(3)and
(4),spacedd=λ/2distanceapartandfedwithequal
currentinphase,treatedtobeoperatedassingleunit

95. Nowtheresultantradiationpatternoffourelement
arraycanbeobtainedasthemultiplicationofpattern
asshownintheFigure.
Notethatthismultiplicationispolargraphical
multiplicationfordifferentvaluesofϕ.

96. Binomial Array :
Inordertoincreasethedirectivityofanarrayitstotal
lengthneedtobeincreased.
Inthisapproach,numberofminorlobesappears
whichareundesiredfornarrowbeamapplications.
Insomeofthespecialapplications,itisdesiredto
havesinglemainlobewithnominorlobes.
Thatmeanstheminorlobesshouldbeeliminated
completely.

97. Toachievesuchpattern,thearrayisarrangedinsuch
awaythatthebroadsidearrayradiatemorestronglyat
thecentrethanthatfromedges.
Incaseofuniform4-elementarray,theresultant
patternshowsfoursidelobes.
Thesecondarylobesappearintheresultantpattern,
becausetheelementsproducingthegrouppattern
haveaspacinggreaterthanone-halfwavelength.

98. Pattern for 2-element array and 4-element array :
The two element arrays are spaced λ/ 2 distance
apart from each other.
Such array produces increased radiation pattern with
no secondary lobes.
Hereantenna2and3coincideatthecentreasshown
intheFigure.

99. Henceitcanbereplacedbyasingleelementcarrying
doublecurrentcomparedwithotherelements.
ThusasshownintheFigure,theresultantarray
consiststhreeelementswithcurrentratio1:2:1.
Thesameconceptcanbeextendedfurtherby
consideringthreeelementarrayasaunitandwitha
secondsimilarthreeelementarrayspacedhalf-wave
lengthfromit.
Thisresultsin4-elementarrayasshownintheFig.
Inthisarray,thecurrentratioisgivenby1:3:3:1.

100. Ifwecontinuethisprocess,wecanobtainthepattern
witharbitrarilylargedirectivitywithoutminorlobes.
Butitisnecessarytoadjusttheamplitudesofthe
currents.
Inthearraycorrespondingtothecoefficientofthe
binomialseries.
Ingeneralthepatternforthebinomialarrayisgiven
by

101. Phased Arrays or Scanning Arrays :
Incaseofthebroadsidearrayandtheendfirearray,
themaximumradiationcanbeobtainedbyadjusting
thephaseexcitationbetweenelements.
Sowecanobtainanarraywhichgivesmaximum
radiationinanydirectionbycontrollingphase
excitationineachelement.
Suchanarrayiscommonlycalledphasedarray.
Thearrayinwhichthephaseandtheamplitudeof
mostoftheelementsisvariable,providedthatthe
directionofmaximumradiationiscalledasphased
array.

102. Suppose the array gives maximum radiation in the
direction ϕ=ϕ0then the phase shift that must be
controlled can be obtained as follows.
.
Thusfromequation,itisclearthatthemaximum
radiationcanbeachievedinanydirectionifthe
progressivephasedifferencebetweentheelementsis
controlled.
The electronic phased array operates on the same
principle.

103. Consider a three element array, the element of array is
considered as λ/2 dipole.
All the cables used are of same length. All the three
cables are brought together at common feed point.

104. Inmanyapplicationsphaseshifterisusedinsteadof
controllingphasebyswitchingcables.
It can be achieved by using ferrite device.
Theconductingwiresarewrappedaroundthephase
shifter.
Thearraywhichautomaticallyreflectsanincoming
signalbacktothesourceiscalledretroarray.
Itactsasaretroreflectorsimilartothepassivesquare
cornerreflector.

105. Attaarraywith8identicalλ/2dipoleelementsare
used,withpairsformedbetweenelements1and8,2
and7,3and6,4and5usingcablesofequallength.

106. Dipoles with Parasitic Elements :
LetI1becurrentinthedrivenelementD.SimilarlyI2
bethecurrentinducedintheparasiticelementP.
Therelationbetweenvoltagesandcurrentscanbe
writtenonthebasisofcircuittheoryas,

107. The impedances Z11and are the self-impedances of
the driven element D and the parasitic element P.
The impedances Z12and Z21is the mutual impedance
between the two elements such that, Z12= Z21=ZM

108. Therefore
The current
The resistance is given by

109. Yagi-Uda Arrays :
Yagi-UdaarraysorYagi-Udaantennasarehighgain
antennas.
TheantennawasfirstinventedbyaJapaneseProf.S.
UdaandProf.H.YagiwasdescribedinEnglish.
A basic Yagi-Uda antenna consists a driven element,
one reflector and one or more directors.
Basically it is an array of onedriven element and one
of moreparasitic elements.

110. Generally the spacing between the driven and the
parasitic elements is kept nearly 0.1λto 0.15 λ.
A Yagi-Uda antenna uses both the reflector (R) and
the director (D) elements in same antenna.

112. Thelengthsofthedifferentelementscanbeobtained
byusingfollowingformula
ThelengthofthedipoleisL=150/f(MHZ)meter
FordipolethelengthL=143/f(MHZ)meter
ForreflectorlengthL=152/f(MHZ)meter
ForfirstdirectorD1=L=137/f(MHZ)meter
SpacingbetweenRandDR=0.25λ=40/f(MHz)meter
SpacingbetweenDandDR=0.25λ=40/f(MHz)meter
SpacingbetweenD1andD2=0.25λ=40/f(MHz)meter

113. Advantages of Yagi-Uda Antenna :
It has excellent sensitivity.
Itsfronttobackratioisexcellent.
Itisusefulastransmittingantennaathighfrequency
forTVreception.
It has almost unidirectional radiation pattern.
Duetouseoffoldeddipole,theYagi-Udaantennais
broadband.

114. Disadvantages of Yagi-Uda Antenna :
Gain is limited.
Bandwidth is limited.
The gain of antenna increases with reflector and
director.

115. Folded dipole :
Folded dipole and radiation pattern

116. Three wire folded dipole
Input impedance o folded dipole antenna is 292ohm

117. Different types of folded dipole antenna :
In practice, the folded dipoles of several different types
are possible.
Some of the folded dipoles consists all dipoles of length
λ/2 but number of dipoles may varies.
While in some other cases the folded dipoles consists
dipoles with different lengths.

119. Thank you