BookPDF Available
Antenna and Wave Propagation
Assistant Professor,
Department of Electrical Engineering, Sukkur IBA
Jan 01, 2016
List of Figures
2.1 Loss tangent (conduction current density and displacement current
density vectors are orthogonal to eachother) . . . . . . . . . . . . . 30
2.2 SkinDepth ............................... 34
2.3 Types of EM wave polarization . . . . . . . . . . . . . . . . . . . . 37
2.4 Circularly polarized EM wave with right hand sense of rotation . 38
3.1 Half-wave dipole antenna . . . . . . . . . . . . . . . . . . . . . . . . 61
3.2 Radiation Pattern of Vertical Dipole (a)normalised E-plane or ver-
tical pattern (φ= 0) (b) normalised H-plane or horizontal pattern
(φ=π/2) (c) three-dimensional plane . . . . . . . . . . . . . . . . . 63
3.3 The monopole antenna . . . . . . . . . . . . . . . . . . . . . . . . . 64
3.4 Two-element antenna array diagram . . . . . . . . . . . . . . . . . 65
rearray............................... 72
4.1 Rhombic antenna geometrical structure and its radiation pattern 86
4.2 Helical antenna operating modes . . . . . . . . . . . . . . . . . . . 88
4.3 WhipAntenna ............................. 89
1
Contents
1 Introduction and Mathematical Preliminaries 6
1.1 Fundamentals of Scalars and Vector . . . . . . . . . . . . . . . . . . 6
1.1.1 DotProduct........................... 7
1.1.2 CrossProduct.......................... 7
1.2 CoordinateSystem ........................... 7
1.2.1 Cartesian Coordinate System . . . . . . . . . . . . . . . . . 8
1.2.2 Properties of Unit Vectors . . . . . . . . . . . . . . . . . . . 8
1.2.3 Cylindrical Coordinate System . . . . . . . . . . . . . . . . . 8
1.2.4 Spherical Coordinate System . . . . . . . . . . . . . . . . . . 10
1.3 DEL Operator ............................ 12
1.3.1 Gradient of Scalar V...................... 12
2 Propagation of EM Waves 13
2.1 Introduction............................... 13
2.2 Maxwell’s Field Equations . . . . . . . . . . . . . . . . . . . . . . . 14
2.3 Maxwell’s Field Equations in Free Space . . . . . . . . . . . . . . . 15
2.4 Maxwell’s Equations for Harmonically Varying Fields . . . . . . . . 16
2.5 EM Wave in Homogeneous Medium . . . . . . . . . . . . . . . . . . 17
2.6 Wave Equations for Lossless Medium . . . . . . . . . . . . . . . . . 17
2.7 Uniform Plane Wave in Free Space . . . . . . . . . . . . . . . . . . 20
2.8 Solution of Maxwell’s Equation for Uniform Plane Wave . . . . . . 21
2.9 EM Wave Equation for Conducting Media . . . . . . . . . . . . . . 24
2.10 Propagation of EM Wave in Perfect Dielectrics . . . . . . . . . . . . 26
2.11 Propagation of Uniform Plane EM Wave in Conducting Medium . . 27
2
2.12 Conductor and Dielectrics . . . . . . . . . . . . . . . . . . . . . . . 29
2.13 Propagation of Plane EM Waves in Good Dielectrics . . . . . . . . 30
2.14 Propagation of Plane EM Waves in Good Conductors . . . . . . . . 31
2.14.1 SkinDepth ........................... 32
2.15 Impedance of Homogenous Isotropic Perfect Dielectric Medium . . . 34
2.16 Electromagnetic Wave Polarization . . . . . . . . . . . . . . . . . . 36
2.16.1 Linear Polarization . . . . . . . . . . . . . . . . . . . . . . . 36
2.16.2 Elliptical Polarization . . . . . . . . . . . . . . . . . . . . . . 37
2.16.3 Circular Polarization . . . . . . . . . . . . . . . . . . . . . . 37
2.17Examples ................................ 38
3 EM Radiation and Antennas 49
3.1 Introduction............................... 49
3.2 Short Electric Dipole or Hertzian Antenna . . . . . . . . . . . . . . 49
3.3 Retarded Vector Potential . . . . . . . . . . . . . . . . . . . . . . . 50
3.4 AntennaFunctions ........................... 51
3.5 AntennaProperties........................... 52
3.6 Antenna Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . 52
3.6.1 Antenna Impedance . . . . . . . . . . . . . . . . . . . . . . 52
3.6.2 Radiation Resistance . . . . . . . . . . . . . . . . . . . . . . 53
3.6.3 Directional Characteristics . . . . . . . . . . . . . . . . . . . 53
3.6.4 Eﬀective Length of Antenna . . . . . . . . . . . . . . . . . . 53
3.6.5 Radiation Intensity . . . . . . . . . . . . . . . . . . . . . . . 54
3.6.6 DirectiveGain.......................... 54
3.6.7 Directivity............................ 55
3.6.8 PowerGain ........................... 55
3.6.9 Antenna Eﬃciency . . . . . . . . . . . . . . . . . . . . . . . 55
3.6.10 EectiveArea.......................... 56
3.6.11 Antenna Equivalent Circuit . . . . . . . . . . . . . . . . . . 56
3.6.12 Antenna Bandwidth . . . . . . . . . . . . . . . . . . . . . . 56
3.6.13 Front-to-Back Ratio (FBR) . . . . . . . . . . . . . . . . . . 57
3.6.14 Polarization ........................... 57
3.7 Basic Antenna Elements . . . . . . . . . . . . . . . . . . . . . . . . 57
3
3.8 Directivity of Electric Current Element . . . . . . . . . . . . . . . . 58
3.9 Gain of Half-wavelength Antenna . . . . . . . . . . . . . . . . . . . 59
3.10 Radiation Pattern of Alternating Current Element . . . . . . . . . . 60
3.11 Radiation Pattern Expression of Center-fed Vertical Dipole of Finite
Length.................................. 61
3.12 Radiation Pattern of Center-fed Vertical Dipole . . . . . . . . . . . 63
3.13 Radiation Pattern of Center-fed Horizontal Dipole . . . . . . . . . . 64
3.14 Radiation Pattern of Vertical Monopole . . . . . . . . . . . . . . . . 64
3.15 Two-element Uniform Array . . . . . . . . . . . . . . . . . . . . . . 64
3.16 Field Strength of Uniform Linear Array . . . . . . . . . . . . . . . . 67
3.16.1 First Side-lobe Ratio (FSR) . . . . . . . . . . . . . . . . . . 69
3.17 Broadside Array and End-ﬁre Array . . . . . . . . . . . . . . . . . . 70
3.17.1 Broadside Array . . . . . . . . . . . . . . . . . . . . . . . . . 70
3.17.2 End-reArray.......................... 72
3.18Examples ................................ 74
4 Antennas for HF, VHF and UHF 82
4.1 Introduction............................... 82
4.2 Yagi-UdaAntenna ........................... 82
4.3 FoldedDipole.............................. 83
4.4 V-antenna................................ 84
4.5 InvertedV-antenna........................... 85
4.6 RhombicAntenna............................ 85
4.7 HelicalAntenna............................. 87
4.8 WhipAntenna ............................. 88
5.1 Factors Involved in Propagation of Radio Waves . . . . . . . . . . . 90
5.2 Factors that Inﬂuence the Propagation . . . . . . . . . . . . . . . . 91
5.3 Ground Wave Field Strength . . . . . . . . . . . . . . . . . . . . . . 91
5.4 Reﬂection of Radio Waves by the Surface of Earth . . . . . . . . . . 93
5.4.1 Roughness of Earth . . . . . . . . . . . . . . . . . . . . . . . 93
5.4.2 Reﬂection Factors of Earth . . . . . . . . . . . . . . . . . . . 94
4
5.5 Space Wave or Tropospheric Wave Propagation . . . . . . . . . . . 94
5.6 Field Strength Due to Space Wave . . . . . . . . . . . . . . . . . . 95
5.7 DuctPropagation............................ 97
5.8 DuctPropagation............................ 98
5.9 Troposcatter............................... 99
5.10 Fading of EM Waves in Troposphere . . . . . . . . . . . . . . . . . 99
5.11LineofSight(LOS)........................... 99
5.12 Ionospheric Wave Propagation . . . . . . . . . . . . . . . . . . . . . 100
5.13 Characteristics of Ionosphere . . . . . . . . . . . . . . . . . . . . . . 100
5.13.1 Characteristics of D-Layer . . . . . . . . . . . . . . . . . . . 100
5.13.2 Characteristics of E-Layer . . . . . . . . . . . . . . . . . . . 101
5.13.3 Characteristics of Es-Layer...................101
5.13.4 Characteristics of F1-Layer...................101
5.13.5 Characteristics of F2Layer...................101
5.14 Refractive Index of Ionosphere . . . . . . . . . . . . . . . . . . . . . 102
5.14.1 Critical Frequency . . . . . . . . . . . . . . . . . . . . . . . 102
5.15 Mechanism of Ionospheric Propagation—Reﬂection and Refraction . 102
5.16 Characteristics Parameters of Ionospheric Propagation . . . . . . . 103
5.18 Ionospheric Abnormalities . . . . . . . . . . . . . . . . . . . . . . . 104
5.18.1 Normal .............................104
5.18.2 Abnormal ............................104
5.19 Ionospheric Storms . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
5.20 Sudden Ionospheric Disturbance (SID) . . . . . . . . . . . . . . . . 105
5
Chapter 1
Introduction and Mathematical
Preliminaries
1.1 Fundamentals of Scalars and Vector
A scalar has magnitude and an algebraic sign. For example temperature, mass,
charge, work, and so on. Whereas a vector has both magnitude and direction.
For example velocity, force, electric ﬁeld, magnetic ﬁeld and so on. A vector A
is expressed in two forms A=Ax, Ay, Azand A=Axax+Ayay+Azaz, where
Ax, Ay, Azand ax,ay+azare known as components of vector Aand unit vectors
along the coordinate axes. The magnitude of Ais written as A=|A|. The unit
vector of Ais aand is given by
a=A
A
The sum and diﬀerence of two vectors are given by
A+B= (Ax+Bx)ax+ (Ay+By)ay+ (Az+Bz)az
AB= (AxBx)ax+ (AyBy)ay+ (AzBz)az
6
1.1.1 Dot Product
The dot product of two vectors is given by
A·BorB·A
A·B=B·A=AB cos θ
A·B=AxBx+AyBy+AzBz
Where θis the angle between vectors Aand B. The dot product of two vectors is
scalar.
1.1.2 Cross Product
The cross product of two vectors is denoted by A×Bas follows:
A×B=AB sin θan
Where anis unit vector perpendicular to Aand B. Moreover
A×B=
axayaz
AxAyAz
BxByBz
=ax[AyBzAzBy] + ay[AzBxAxBz] + az[AxByAyBx]
Cross product of two vectors is a vector.
1.2 Coordinate System
Coordinate system is deﬁned as a system used to represent a point in space. Ba-
sically coordinate systems are of three types namely cartesian coordinate system,
cylindrical coordinate system and spherical coordinate system.
7
1.2.1 Cartesian Coordinate System
In this system a point P is represented by P(x, y, z). The variables are x, y, z. A
point is obtained by intersection of three planes given by x=k1, y =k2, z =k3.
The unit of x,yand zis meter. The three axes x,y,zare mutually perpendicular.
These are said to be orthogonal to each other.
1.2.2 Properties of Unit Vectors
Following are the possible combinations of the dot and cross multiplication of the
unit vectors.
ax·ax= 1 ay·az= 0
ay·ay= 1 az×ay=az
az·az= 1 ay×az=ax
ax×ax= 0 az×ax=ay
ay×ay= 0 ay×ax=az
az×az= 0 az×ay=ax
ax·ay= 0 ax×az=ay
ax·az= 0
1.2.3 Cylindrical Coordinate System
In this system a point Pis represented by P(ρ, φ, z) where ρrepresents radius
of cylinder, φis called azimuthal angle and zis same as in Cartesian coordinate
system. The unit of ρis meter,φis measured in degree or radian and zis given in
meters. In Cylindrical coordinate system, appoint is obtained by intersection of
three surface namely:
A cylindrical surface ρ=k1meter
Another plane z=k2meter
8
All the three surface are mutually perpendicular. These are said to be mutually
orthogonal. A point in Cylindrical coordinate system is shown as P(ρ, φ, z). The
coordinate ρis radius of the cylinder, φis measured from xaxis and zis measured
the same way as for cartesian system. Here aρ,aφ,azrepresents unit vectors along
the coordinates ρ,φand z. Their magnitude is unity and they are in the increasing
directions of ρ,φ,zrespectively. It is obvious that increase in ρresults in cylinders
of greater radius, φincreases in anti-clockwise direction, z is same as in cartesian
system. The relations between x,y,zand ρ,φ,z:
x=ρcos φ
y=ρsin φ
z=z
and
ρ=px2+y20ρ <
φ= tan1y
x,0φ < 2π
z=z0z <
Dot products of ax,ayand azwith aρ,aφand azare given by
ax·aρ= cos φ
ax·aφ=sin φ
ay·aρ= sin φ
ay·aφ= cos φ
az·aρ= 0
az·aφ= 0
The unit vectors of cylindrical coordinates in terms of cartesian coordinates
are given by
9
aρ= cos φax+ sin φay
aφ=sin φax+ cos φay
az=az
A vector A= (Axax+Ayay+Azaz) is expressed in cylindrical coordinates as
A= (Aρ, Aφ, Az) = [(Axcos φ+Aysin φ),(Axsin φ+Aycos φ), Az]
That’s is
Aρ=Axcos φ+Aysin φ
Aφ=Axsin φ+Aycos φ
Az=Az
Ais also written as:
A= (Axcos φ+Aysin φ)aρ+ (Axsin φ+Aycos φ)aφ+Azaz
1.2.4 Spherical Coordinate System
In this system P(r, θ, φ) represents a point, rrepresents the radius of the sphere, θ
is the angle of deviation measured from z-axis and φis azimuthal angle measured
from x-axis. A point is obtained by intersection of three surfaces as follows:
Aspherical surface r=k(constant) meter
Three surfaces are mutually perpendicular. Three vector ar,aθ,aφrepresents
10
unit vectors along the coordinate axes. Their magnitude is unity and they are
in the increasing directions of r,θand φaxes. Increase of rresults in sphere of
large radius, θincreases in clockwise direction and φincreases in in anti-clockwise
direction. The variables of cartesian and spherical coordinates system are related
by:
x=rsin θcos φ, −∞ <x<
y=rsin θsin φ, −∞ < y <
z=rcos θ, −∞ < z <
and
r=px2+y2+z20r≤ ∞
θ= cos1z
px2+y2+z20θπ
φ= tan1y
x0φ2π
The relations between the variables of cylindrical and spherical coordinates are
given by:
ρ=rsin θ
φ=φ
z=rcos θ
r=pρ2+z2
θ= tan1r
z
φ=φ
11
1.3 DEL Operator
The del or nabla is known as diﬀerential vector operator and is deﬁned as
=ax
∂x +ay
∂y +az
∂z
Del has unit of 1/m.
Gradient of scalar is a vector and is deﬁned as:
V=∂V
∂x +Ay
∂y +Az
∂z
12
Chapter 2
Propagation of EM Waves
2.1 Introduction
This chapter starts with Maxwell’s equations in point form and diﬀerential form
from which we derive the vector wave equation whose solution is a wave propa-
gating with the speed of light. We give detailed treatment of the behaviour of
uniform plane wave in lossless dielectric media or free space and the behaviour of
uniform plane wave inside the conductor. The behaviour of uniform plane wave
in dielectric and conductors has been substantiated with the help of some mathe-
matical formulas for attenuation constant, phase shift constant, phase velocity and
intrinsic impedance. In the end, depth of penetration or skin depth has been taken
into consideration when the EM wave encounters conductor which attenuates the
E and H ﬁelds of EM wave with 1/e of it maximum value.
13
2.2 Maxwell’s Field Equations
We have following Maxwell’s equations in diﬀerential form:
Gauss’s Law∇ · B= 0 (2.1a)
Gauss’s Law∇ · D=ρv(2.1b)
∂t (2.1c)
Ampere’s Law∇ × H=J+D
∂t (2.1d)
Where Bis magnetic ﬂux density W eber/m2,Dis electric ﬂux density in
C/m2,Eand Hare electric ﬁeld strength in V /m and magnetic ﬁeld strength
in A/m.ρvstands for volume charge density in C/m3,Jis conduction current
density, D
tis displacement current density. The constituent parameters are the
following relations between E, and H:
D=εE(2.2a)
B=µH(2.2b)
J=σE(2.2c)
We can also write Maxwell’s equations in integral forms as follows:
Is
B·ds= 0 Net ﬂux is zero (2.3a)
Is
=D·ds=Zv
ρvdv(2.3b)
IE·dl=Is
B
∂t (2.3c)
IH·dl=Zs
(J+D
∂t )·ds(2.3d)
Maxwell’s ﬁrst equation says that net ﬂux passing through closed surface is
14
zero i.e. there is no magnetic monopole. Maxwell’s second equation can be in-
terpreted as total electric ﬂux density through surface enclosing volume is equal
to total charge within that volume. From Maxwell’s third equation we see that
electromagnetic force around closed path is equal to the time derivative of mag-
netic ﬂux density which is followed by corrected version of Ampere’s law which
is magnetic force around closed path is equal to conduction current density plus
time derivative of electric ﬂux density or displacement current.
2.3 Maxwell’s Field Equations in Free Space
We can rewrite above set of Maxwell’s ﬁeld equations for free space in both diﬀer-
ential or point form and integral form. We know that in free space conductivity
space as follows:
Gauss’s Law∇ · B= 0 (2.4a)
Gauss’s Law∇ · D= 0 (2.4b)
∂t (2.4c)
Ampere’s Law∇ × H=D
∂t (2.4d)
Similarly, integral form of Maxwell’s ﬁeld equations can be written as:
Is
B·ds= 0 Net ﬂux is zero (2.5a)
Is
=D·ds= 0 (2.5b)
IE·dl=Is
B
∂t (2.5c)
IH·dl=Zs
D
∂t ·ds(2.5d)
15
2.4 Maxwell’s Equations for Harmonically Vary-
ing Fields
We shall now rewrite Maxwell’s ﬁeld equations in diﬀerential form and integral
forms as harmonically varying ﬁelds. We know that D=D0ejωt and B=B0et
where D0and H0are the magnitudes of electric ﬂux density and magnetic ﬂux
density respectively.
D=D0ejωt Diﬀ. w.r.t. time
D
∂t =D0ω ej ωt
D
∂t =jωD
(2.6)
By comparing LHS and RHS we can see that
∂t =jω. Keeping this in view
we can rewrite Maxwell’s equations in diﬀerential form and integral form as har-
monically varying ﬁelds as follows:
∇ · B= 0 (2.7a)
∇ · D=ρ(2.7b)
∇ × E=µH(2.7c)
∇ × H= (σ+ε)E(2.7d)
Similarly, the integral form of harmonically varying ﬁelds:
Is
B·ds= 0 (2.8a)
Is
D·ds=Zv
ρvdv(2.8b)
IE·dl=µ Zs
H·ds(2.8c)
IH·dl= (σ+ε)Zs
E·ds(2.8d)
16
2.5 EM Wave in Homogeneous Medium
Let us obtain electromagnetic wave equations from Maxwell’s ﬁeld equations con-
sidering a homogenous and isotropic medium.
Homogeneous Medium: In homogenous medium the medium parameters i.e. per-
mittivity ε, permeability µand conductivity σare constant throughout medium.
Isotropic Medium: In isotropic medium εis scalar constant so that Dand Epos-
sess same direction such that D=εE
In any electromagnetic phenomenon, the following Maxwell’s ﬁeld equations
must be satisﬁed:
∇ · B= 0 (2.9a)
∇ · D=ρ(2.9b)
∇ × E=µH(2.9c)
∇ × H= (σ+ε)E(2.9d)
2.6 Wave Equations for Lossless Medium
For free space or lossless or non-conducting medium or more general perfect di-
electric medium there is no conduction current i.e. J= 0 and no charge density
ρ= 0. Thus, Maxwell’s equations can be reduced to:
∇ · B= 0 (2.10a)
∇ · D= 0 (2.10b)
∇ × E=B
∂t (2.10c)
∇ × H=D
∂t (2.10d)
Now, let us diﬀerentiate eq. 2.10d with respect to time:
17
∂t (∇ × H) =
∂t (D
∂t ) (2.11)
Curl is diﬀerentiable with respect to space hence order can be changed
∇ × H
∂t =2D
∂t2
∇ × H
∂t =ε0
2E
∂t2D=ε0E
(2.12)
Now, let us take curl of eq. 2.10c
∇ × ∇ × E=−∇ × B
∂t
=µ0(∇ × H
∂t )B=µ0Hin free space
(2.13)
From eq. 2.12 and using vector identity ∇×∇×A=(∇ · A)− ∇2Aabove
equation can be rewritten as:
∇ × ∇ × E=µ0ε0
2E
∂t2
(∇ · E)− ∇2E=µ0ε0
2E
∂t2
−∇2E=µ0ε0
2E
∂t2∇ · E=0
(2.14)
2E=µ0ε0
2E
∂t2(2.15)
This is wave equation for E-ﬁeld in free space. Similarly, the wave equation can
also be derived in terms of magnetic ﬁeld as follows:
2H=µ0ε0
2H
∂t2(2.16)
Equations 2.15 and 2.16 are homogenous vector wave equations for free space
(vacuum)
In problem of rectangular coordinate system, the wave equation assume the
18
form of scalar wave equations in terms of its components. Therefore, the scalar
wave equations:
2Ex=µ0ε0
2Ex
∂t2(2.17a)
2Ey=µ0ε0
2Ey
∂t2(2.17b)
2Ez=µ0ε0
2Ez
∂t2(2.17c)
2Hx=µ0ε0
2Hx
∂t2(2.18a)
2Hy=µ0ε0
2Hy
∂t2(2.18b)
2Hz=µ0ε0
2Hz
∂t2(2.18c)
In terms of complex time harmonic wave equations i.e. replace
∂t with jω
2E=µ0ε0
2E
∂t2
2E=µ0ε0()2E
(2.19)
2E=ω2µ0ε0E(2.20)
Similarly, homogenous vector wave equation in complex time-harmonic form
2H=ω2µ0ε0H(2.21)
Velocity of Wave
19
We know that homogenous vector wave equations in free space are as follows:
2E=µ0ε0
2E
∂t2
2H=µ0ε0
2H
∂t2
(2.22)
Comparing eq. 2.22 with standard partial diﬀerential wave equation
2X=1
ν2
2X
∂t2(2.23)
Where νis velocity of wave.
1
ν2=µ0ε0
=ν2=1
µ0ε0
ν=1
µ0ε0
(2.24)
Substituting the permittivity ε0=1
36π×109F/m and permeability µ0= 4π×
107A/m values for free space we get velocity of wave ν= 3 ×108m/sec =c
which is equal to speed of light. EM waves propagate with speed of light!
2.7 Uniform Plane Wave in Free Space
An EM wave originates from a point and spread out uniformly in all directions
and forms spherical wave. An observer at a greater distance observes only small
part of wave in his immediate vicinity.
Properties of Plane Wave in Free Space
At every point in space, the electric ﬁeld vector Eand magnetic ﬁeld vector
Hare perpendicular to each other and to the direction of propagation.
The direction of Eand His perpendicular to everywhere in plane. A uniform
plane wave is one in which Eand Hlie in a plane and have the same value
everywhere in that plane at any ﬁxed instant.
20
Velocity of propagation of a wave in free space is given by 3 ×108m/s.
Eand Hoscillate in phase and ratio of their amplitudes is constant and is
equal to:
η=E
H=rµ
ε=η0=rµ0
ε0
= 120 πor 377 Ω (2.25)
Whatever may be the frequency, the EM wave travels in space with velocity
of light.
2.8 Solution of Maxwell’s Equation for Uniform
Plane Wave
If the phase of the wave is same for all points on a plane surface then such wave
is called plane wave. Additionally, if amplitude of the wave is also constant over
the plane surface then such wave is known as uniform plane wave.
Let us consider the EM wave propagates in z-direction such that E-ﬁeld lies in
x-plane or y-plane and amplitude of the ﬁeld is constant i.e. Ex=kand Hy=k
such that ∂Ex
∂x = 0 and ∂Ey
∂y = 0. We also know that there is no ﬁeld component in
the direction of the propagation of EM wave i.e.Ez= 0. The vector wave equation
in free space is:
2E=µ0ε0
2E
∂t2(2.26)
We know that ∇·∇=2= ( 2
∂x2+2
∂y2+2
∂z2) and E=axEx+ayEy+azEz:
ax(2
∂x2+2
∂y2+2
∂z2)Ex+ay(2
∂x2+2
∂y2+2
∂z2)Ey
az(2
∂x2+2
∂y2+2
∂z2)Ez=µ0ε0
2
∂t2(axEx+ayEy+azEz)
(2.27)
By deﬁnition of uniform plane wave progressing in z-direction we know that
Ez= 0,∂Ex
∂Ex= 0, Ey
∂Ey= 0, 2Ex
∂E2
x= 0, 2Ey
∂E2
y= 0. Therefore, eq.2.27 can be modiﬁed
as:
21
ax
2Ex
∂E2
z
+ay
2Ey
∂z2=µ0ε0
2
∂t2(axEx+ayEy) (2.28)
Now by comparing LHS and RHS of eq. 2.28 we can rewrite E-ﬁeld wave
equations as follows:
2Ex
∂z2=µ0ε0
2Ex
∂t2(2.29a)
2Ey
∂z2=µ0ε0
2Ey
∂t2(2.29b)
Equation 2.29 relates space and time variation of scalar magnitude Exof the
electric ﬁeld intensity and is known as a wave equation. Exand Eyrepresent
scalar magnitudes of electric ﬁeld intensity of a plane EM wave travelling in z-
direction. Equation 2.29 is most general view of describing the motion of this ﬁeld
as function of time and space. Similarly, the magnetic ﬁeld is identical:
2Hx
∂z2=µ0ε0
2Hx
∂t2(2.30a)
2Hy
∂z2=µ0ε0
2Hy
∂t2(2.30b)
Equation 2.29 and eq. 2.30 are second-order partial diﬀerential equation and
the general solution of such diﬀerential can be written as:
E=f1(zct)
| {z }
Term-I
+f2(z+ct)
| {z }
Term-II
(2.31)
In eq. 2.31 f1(zct) and f2(z+ct) are any two functions and may assume any
form like asin(βz ct) and ae(βzct). Term-I represents wave travelling in +ve z
direction and Term-II represents wave traveling -ve z direction. General solution
of wave equation in this case has two waves one travelling in positive direction of
z away from source and the other travelling in negative z direction . If there’s no
reﬂection surface then term-II is zero. Therefore
E=f1(zct) (2.32)
22
Now putting general solution in trigonometric form and trying with particular
function
Ey= sin β(zmt) + sin β(z+mt) (2.33)
We can rewrite eq. 2.33 by using βc =2π
λ·fλ = 2πf =ω:
Ey= sin(βz ωt) + sin(βz +ωt) (2.34)
If only positive direction wave is taken, then there may be number of choice of
solution in trigonometric form:
Ey= sin(βz ωt) (2.35a)
Ey= sin(ωt βz) (2.35b)
Ey= cos(βz ωt) (2.35c)
Ey= cos(ωt βz) (2.35d)
The diﬀerence between 2.35b and 2.35c is diﬀerence of πradians.
Ey= sin(βz ωt) = sin(ωt βz)
Ey= sin(ωt βz +π)(2.36)
If we disregard the phase
Ey= sin(ωt βz) (2.37)
In above equation maximum amplitude is unity. We can rewrite with amplitude
E0:
Ey=E0sin(ωt βz) (2.38)
This is preferable solution and has got advantage because tis positive when the
wave is travelling in +ve and -ve direction
23
The solution could be more attractive in which time of propagation is visible
Ey=E0sin ω(tβz
ω)
=E0sin ω(t2π
λ·z
2πf )
(2.39)
Ey=E0sin ω(tz
c) (2.40)
Ey=E0sin ω(ttp) (2.41)
Where tp=z
cis time of propagation of wave from origin of the wave to a point P
of observation. Similarly, magnetic ﬁeld at point P is:
Hy=H0sin ω(ttp) (2.42)
2.9 EM Wave Equation for Conducting Media
We know the maxwell’s ﬁeld equations in point form:
∇ · B= 0 (2.43a)
∇ · D=ρv(2.43b)
∇ × E=B
∂t (2.43c)
∇ × H=J+D
∂t (2.43d)
From eq. 2.43d
∇ × H=σE+εE
∂t (2.44)
From eq. 2.43c
∇ × E=µH
∂t (2.45)
24
Now by taking curl of eq. 2.45 and using vector identity ∇×∇×E=(∇·E)−∇2E
∇ × ∇ × E=µ(∇ × H
∂t )
(∇ · E)− ∇2E=µ
∂t (∇ × H)
− ∇2E=−∇(∇ · E)µσ E
∂t µε 2E
∂t2
2E=(∇ · E) + µσ ∂E
∂t +µε 2E
∂t2
(2.46)
Now using eq. 2.43b we can rewrite eq. 2.46 as follows:
2E=(ρ
E) + µσ E
∂t +µε 2E
∂t2(2.47)
There is no net charge within conductor although it may exists on surface.
Therefore, ﬁrst term becomes zero i.e. ρ= 0
2E=µσ E
∂t +µε 2E
∂t2(2.48)
This is wave equation in conducting medium in E. Similarly, wave equation in
terms of His obtained by taking curl of eq. 2.43d
∇ × ∇ × H=σ(∇ × E) + ε(∇ × E
∂t )
(∇ · H)− ∇2H=σ(H
∂t ) + ε(
∂t (∇ × E))
(2.49)
Using ∇×E=B
∂t and ·B= 0 and vector identity ∇×∇×H=(·H)2H
we can write
− ∇2H=σµ(H
∂t ) + ε(
∂t (B
∂t ))
− ∇2H=σµ H
∂t µε 2H
∂t2
(2.50)
Hence
2H=σµ H
∂t +µε 2H
∂t2(2.51)
25
This is wave equation for conducting medium in H.
2.10 Propagation of EM Wave in Perfect Dielectrics
For sinusoidal time-variation wave equation for perfect dielectric or lossless medium
may be obtained by replacing time-derivative with a factor . The wave equation
phasor form:
2E=µ0ε0
2E
∂t2(2.52)
2E=µ0ε0()(jω)E
2E=µ0ε0ω2E
2E+ω2µ0ε0E= 0
(2.53)
Since β2=ω2µ0ε0. Therefore,
2E+β2E(2.54)
For uniform plane wave propagating in +ve z-direction there are no x and y com-
ponents and also Ezcomponent. Thus wave equation in Eand Hreduces to
2Ex
∂z2=β2Ex(2.55a)
2Ey
∂z2=β2Ey(2.55b)
Similarly for H
2Hx
∂z2=β2Hx(2.56a)
2Hy
∂z2=β2Hy(2.56b)
Taking y component of E ﬁeld only , the solution of wave equation 2.55 can be
written as:
Ey=A1ez +A2ez (2.57)
Where A1and A2are arbitrary complex constants. This is solution in phasor
26
form. The solution in terms of time-varying ﬁelds may be written as:
˜
Ey(z, t) = <nEy(z)ej ωto=<n(A1ejβz +A2ej βz )ejωto
=<nA1ej(ωtβz )+A2ej(ωt+βz )o(2.58)
˜
Ey(z, t) = <"A1ncos(ωt βz) + jsin(ωt βz)o+
A2ncos(ωt +βz) + jsin(ωt +βz)o#(2.59)
˜
Ey(z, t) = A1cos(ωt βz) + A2cos(ωt +βz) (2.60)
We can see that a time-varying solution consists of forward moving sinusoid
and backward moving sinusoid in z direction.
2.11 Propagation of Uniform Plane EM Wave in
Conducting Medium
For conducting medium the solution of wave equation is easily obtained by em-
ploying phasor:
2E=µσ E
∂t +µε 2E
∂t2(2.61)
Converting above equation in phasor form by replacing ∂/∂t with jω and 2/∂t2
with ()2
2E=µσ(jω)E+µε(jω)(jω)E
2E=µ(σ+jωε)E(2.62)
2E=γ2E
2H=γ2H(2.63)
27
Where γ2=µ(σ+jωε) and γ=α+jβ is propagation constant. According
to phasor notation, sinusoidally time-varying electric and magnetic ﬁelds ˜
Ey(z, t)
and ˜
Hx(z, t) are
˜
Ey(z, t) = <nEy(z)ej ωto(2.64a)
˜
Hx(z, t) = <nHx(z)ejωto(2.64b)
Ey(z) and Hx(z) are complex space vectors called phasors of time-dependent
real ﬁelds E(z, t) and H(z, t) respectively.
Phase Shift Factor: We can write phase shift factor for plane EM wave in con-
ducting medium as following:
β=2π
λ(2.65)
Phase Velocity: We can write phase velocity of uniform plane EM wave in con-
ducting medium as:
νp=fλ =ω
2πλ
=λ
2πωω= 2πf =f=ω
2π
(2.66)
Refractive Index: Refractive index of the medium is ratio of the speed of light cto
phase velocity of the wave νp. Mathematically
η=c
νp
=3×108m/s
νp
(2.67)
For the solution of plane wave progressing in +ve z-direction, the wave equation
becomes as usual
2Ex
∂z2=γ2Ex(2.68a)
2Ey
∂z2=γ2Ey(2.68b)
28
We have in phasor notation Ey(z) = E0eγz such that
˜
E(z, t) = <E0eγ zejωt =<E0eteγz
=<E0e(α+)zejωt =<E0eαzej(ωtβz )(2.69)
˜
E(z, t) = eαz <hE0ncos(ωt βz) + jsin(ωt βz)oi (2.70)
2.12 Conductor and Dielectrics
In electromagnetics, materials are roughly classiﬁed into two categories, the con-
ductors and dielectrics or insulators. However, dividing line is not sharp between
conductors and insulators. While some media (for example earth) are considered
as conductors in one part of the radio frequency range and as a dielectric in another
part of radio frequency range. Maxwell’s ﬁrst equation for time-varying ﬁelds in
phasor form can be expressed as:
∇ × H=J+D
∂t =Jc+JD
=σE+εE
(2.71)
Ratio of conduction current density to displacement current density is given by:
Jc
JD=|σE|
|εE|
=σ
ωε = Loss Tangent
(2.72)
Moreover, the dividing line between conductors and dielectrics is when
Jc
JD=σ
ωε = 1 (2.73)
If conductivity σ6= 0 then we may arbitrarily deﬁne three conditions as follows:
σ
ωε 1 which leads to ωε σi.e. good dielectrics
29
~
JD=jω ε ~
E
~
Jc=σ~
E
θ= tan1σ
ωε
Figure 2.1: Loss tangent (conduction current density and displacement current
density vectors are orthogonal to eachother)
σ
ωε 1 which leads to ωε σi.e. good conductors
σ
ωε = 1 which leads to ωε
=σi.e. quasi-conductors
When σ= 0 the medium is perfect or lossless dielectric and when σ6= 0 the
medium is lossy or imperfect dielectric. The two vectors in our loss tangent un-
derstanding are 90 deg or π
2out of phase as shown in ﬁg. 2.1.
If loss tangent is very small then some useful expressions can be obtained for
α,βand η.
2.13 Propagation of Plane EM Waves in Good
Dielectrics
We know the condition for good dielectric is when ωε σthen we can write
expressions for attenuation constant α, phase constant βand phase velocity νpas
follows:
attenuation constant:α=σ
2pµ
εNp/m
30
phase constant:β=σ
2pµ
phase velocity:νp=ν01σ2
8ω2ε2+···m/s
2.14 Propagation of Plane EM Waves in Good
Conductors
We know that the condition for good conductors σ
ωε 1i.e. σωε then we can
develop expressions for attenuation constant, phase shift constant, phase velocity
and skin depth. We can write propagation constant is complex number γ=α+
and for the case of conductors it can mathematically expressed as:
γ= (1 + j)rωµσ
2(2.74)
We can deduce expressions for αand β
α=β=rωµσ
2(2.75)
The E-ﬁeld wave equation in y-plane for an EM wave propagating in z-direction
can be given as follow:
Ey=E0eγz
=E0e{(1+j)ωµσ
2}z(2.76)
Ey=E0eωµσ
2z·ej(ωµσ
2)z(2.77)
We deﬁne the notion of skin depth δin order to foster the understanding of EM
wave experiencing attenuation when it propagates through conductor. Mathemat-
ically,
31
δ=1
α=r2
ωµσ =r1
πf µσ
=1
πf µσ
(2.78)
Similarly, velocity of propagation can also be expressed in terms of media
parameters i.e. µ,σand angular frequency ωas
ν=ω
β=ω
pωµσ
2
=sω2·2
ωµσ
=r2ω
µσ
(2.79)
We can also represent velocity of propagation in terms of skin depth δand
angular frequency ωas follows:
ν=s2·ω2
ωµσ =r2
ωµσ ·ω=r2
2πf µσ ·ω
=1
πf µσ ·ω
=ν=δω
(2.80)
Wavelength in conducting medium can be given in terms of skin depth as
λ=2π
β=2π
pωµσ
2
=2π
q2πf µσ
2
= 2π·1
πf µσ
=λ= 2πδ
(2.81)
2.14.1 Skin Depth
Depth of penetration may be deﬁned as that depth in which the wave has been
attenuated by amount 1/e or 37% of its initial value. Amplitude decreases ex-
32
ponentially as eαz at distance zwhich makes αz = 1, the amplitude is only 1/e
times its original value at z= 0. According to deﬁnition this distance is depth of
penetration or skin depth i.e. z=δ.
eαz =eαδe1= 1/e (2.82)
Which implies that αδ = 1 and α= 1. E-ﬁeld wave in y-plane when EM wave
is propagating in z-direction can be given as follows:
Ey=E0eγz =E0e(α+jβ)z
=E0eαz ·ez (2.83)
From our previous discussions we know that α=βin conductors. Therefore,
Ey=E0ez/δ ·ez/δ (2.84)
Now, applying boundary condition i.e. z= 0. The above equation reduces to
Ey=E0(2.85)
Which is maximum amplitude of the wave at the surface of conducting medium.
Now, at certain distance inside the conductor the magnitude of E-ﬁeld wave equa-
tion is:
Ey=E0ez/δ =E0ez/z z=δ
=E0e1= (1/e)E0= 37%E0
(2.86)
When wave penetrates distance δ, the ﬁeld Eydecreases to 37% of its initial value.
33
Distance along propagation vector (Skin Depths)
0.5 1 1.5 2 2.5 3 3.5 4 4.5
E/e field
1
2
3
4
5
6
Figure 2.2: Skin Depth
2.15 Impedance of Homogenous Isotropic Per-
fect Dielectric Medium
Ratio of E-ﬁeld and H-ﬁeld is constant and is mathematically deﬁned as:
η=rµ
ε=η0=rµ0
ε0
for perfect dielectrics (2.87)
This result can be obtained using solution to Maxwell’s equation
Hy=H0sin(ωt βz) (2.88)
34
This is the solution of plane wave progressing in +ve z-direction and the solution
for Exrepresenting the a wave in +ve z-direction is given by:
Ex=E0sin(ωt βy) (2.89)
∂Ex
∂z =µHy
∂t
∂z (E0sin(ωβz)) = µ∂Hy
∂t
E0cos(ωt βz)(β) = µHy
∂t =∂Hy
∂t =E0β
µcos(ωt βz)
(2.90)
Integrating both sides of eq. 2.90
Hy=βE0
µZcos(ωt βz)dt
Hy=βE0
µ
sin(ωt βz)
ωconsidering k= 0
(2.91)
Now the ratio of Exto Hycan be given by
Ex
Hy
=E0sin(ωt βz)
βE0
µ0sin(ωt βz)
Ex
Hy
=µω
β=µ2πf
2π/λ =µc =µ1
µε
(2.92)
Hence
Ex
Hy
=rµ
ε(2.93)
Similarly, it can be shown that
Ey
Ex
=rµ
ε(2.94)
Now the wave equation can be written as
Ex=E0sin(ωt βz) (2.95)
35
From eq. 2.93
rµ
εHy=E0sin(ωt βz)
Hy=rε
µE0sin(ωt βz)
(2.96)
Exand Hyare identical functions of xand tbut their magnitudes diﬀer by a
factor qε
µand its reciprocal pµ
ε. The dimension of ηis Ω in SI units. Thus free
space impedance is
η0=rµ0
ε0
= 120π= 377Ω (2.97)
2.16 Electromagnetic Wave Polarization
The polarisation of uniform wave refers to time-varying behaviour of the electric
ﬁeld intensity vector.
2.16.1 Linear Polarization
In linear polarisation, the electrical vector is under all times remain only in one
direction. In terrestrial communication linear polarization can further be divided
into horizontal polarization where E-ﬁeld vector is parallel to surface of earth and
vertical polarization where E-ﬁeld vector is perpendicular to surface of earth. E-
ﬁeld of linearly polarised wave progressing in z-direction:
Ey=E2sin(ωt βz) (2.98)
E-ﬁeld varies between +ve and -ve value of E and the direction conﬁned to y-
direction only. Figure 2.3 shows E-ﬁeld vector which trace line, circle and elliptical
polarization
36
2.16.2 Elliptical Polarization
E-ﬁeld rotates as a function of time. The tip of E-ﬁeld vector describes an ellipse
which is known as ”polarisation ellipse ”. Ellipse is deﬁned by axial ratio which is
the ratio of major axis to the minor axis of polarisation.
AR =E2
E1
(2.99)
Where where E2is magnitude of E-ﬁeld along major axis and E1is magnitude of
E-ﬁeld along minor axis.
Figure 2.3: Types of EM wave polarization
2.16.3 Circular Polarization
Circular polarization and linear polarization are special case of elliptical polariza-
tion. In circular polarization E2=E1hence we get E2
E1= 1. In case of linear
polarization
AR =E2
E1
=E2
0=(2.100)
Figure 2.4 shows circularly polarized wave with right-hand sense of rotation.
37
Figure 2.4: Circularly polarized EM wave with right hand sense of rotation
2.17 Examples
Example#01
If the electric ﬁeld strength of a radio broadcast signal of a TV is given by
E= 5 cos(ωt βy)ayV /m. Determine displacement current density, If the same
ﬁeld exists in a medium whose conductivity is given by 2 ×103mho/cm ﬁnd the
conduction current density.
Solution:
We know that
E= 5 cos(ωt βy)azV /m
38
Electric ﬂux density
D=ε0E= 5ε0cos(ωt βy)azC/m2
Displacement current density
JD=D
∂t =
5ε0cos(ωt βy)az
=5ε0sin(ωt βy)A/m2
The conduction current density
Jc=σE
= 2 ×105×5 cos(ωt βy)az
Example#02
If H= cos(108tβz)ayA/m and E= 377 cos(108tβz)axV /m in free space.
Find the frequency, wavelength, phase shift constant and intrinsic impedance of
the medium.
Solution:
We know the angular frequency
ω= 2πf = 108=f=108
2π= 15.9MHz
Now, the wavelength λ
λ=c
f=3×108
15.9×106=300
15.9= 18.86 m
Phase shift constant β
β=2π
39
Impedance η
η0=
E
H=rµ0
0
= 377Ω
Example#03
Find the propagation constant for a wave with 100 MHz of frequency that prop-
agates in free space.
Solution:
We know that propagation constant
γ=α+
We know that phase shift constant βcan be given as follows:
OR
β=2π
λ
γ=
=µ0ε0=jω
ν0
=2πf
ν0
=2π×100 ×106
3×108
=j2π×108
3×108=j2.094
γ=j2.094(1/m)
40
Example#04
If H ﬁeld is given by H(z, t) = 48 cos(108t+ 40z)ayA/m. Identify the amplitude,
frequency and phase constant. Find the wavelength.
Solution:
Amplitude of the magnetic ﬁeld is 48 A/m
Angular frequency ωis
ω= 108=2πf = 108=f=108
2π= 15.9MHz
Phase shift constant β
Wavelength λ
λ=2π
β=2π
40 = 0.157 m
Example#05
When the amplitude of the magnetic ﬁeld in a plane wave is 2A/m (a) determine
the magnitude of the electric ﬁeld for plane wave in free space (b) determine magni-
tude of electric ﬁeld when the wave propagates in a medium which is characterised
by σ= 0, µ =µ0and ε= 4ε0.
Solution:
We have
41
(a) σ= 0, µr= 1, εr= 1
E
H=η0=rµ0
ε0
E=ηH
E= 120π×2 = 240π V/m
E= 240π V/m
(b) σ= 0, ε = 4ε0, µr= 1
η=rµ
ε=rµ0
ε0
=rµ0
4ε0
η=1
2rµ0
ε0
=1
2×120π= 60π
E=ηH = 60π×2 = 120π V /m
Example#06
If εr= 9, µ =µ0for the medium in which a wave with a frequency f= 0.3GHz
is propagating. Determine Propagation constant γ, intrinsic impedance ηof the
medium when σ= 0.
Solution:
γ=
γ=µε =jωp9ε0µ0
γ=j2π×0.3×109
3×108
γ=j2π(1/m)
42
Intrinsic impedance η
η=rµ
σ+ωε =rjωµ
ωε =pµ09ε0
η=1
3×120π
η= 40π
Example#07
The wavelength of x-directed plane wave in a lossless medium is 0.25 mand the
velocity of propagation is 1.5×1010 cm/s. The wave has z-directed electric ﬁeld
with an amplitude equal to 10 V/m. Find the frequency and permittivity of the
medium. The medium has µ=µ0.
Solution: Frequency of wave
λ=c
f=f=c
λ
f= 600 MHz
and we have
ν=1
µε =1
µrµ0εrε0
ν=1
µ0εrε0
=ν0
εr
=3×108
εr
=ε=3×108
1.5×108
εr= 1.141
43
Example#08
Earth has conductivity of σ= 102V/m, εr= 10, µr= 2. What are the conduct-
ing characteristics of the earth at:
(a) f= 50 Hz (b) f= 1 KHz (c) f= 1 MHz (d) f= 100 M H z
(e) f= 1 MHz
Solution:
The parameters of Earth are σ= 102, εr= 10, µr= 2. Let us write the ratio
σ
ωε in a more convenient way:
(a) f= 50 Hz
σ
ωε =18 ×106
50 = 3 ×1051
Hence at f= 50 Hz Earth behaves like good conductor.
(b) f= 1 KH z
σ
ωε =18 ×106
103= 18 ×1031
Hence at f= 1 KH z Earth behaves like good conductor.
(c) f= 1 MHz
σ
ωε =18 ×106
106= 18 >1
Hence at f= 1 MHz Earth behaves like moderate conductor.
(d) f= 100 MHz
σ
ωε =18 ×106
100 ×106= 0.18 1
Hence at f= 100 MHz Earth behaves like quasi-dielectric.
44
(c) f= 10 GHz
σ
ωε =18 ×106
10 ×109= 18 ×1041
Hence at f= 10 GHz Earth behaves like good dielectric.
Example#09
A medium like copper conductor which is characterised by the parameters σ=
5.8×107mho/m, εr= 1, µr= 1 supports uniform plane wave of frequency 60 Hz.
Find the attenuation constant, wavelength and phase velocity of wave.
Solution:
Let us decide based on ratio:
σ
ωε =5.8×107
2π×60 ×8.854 ×1012 = 173 ×1014 1
It’s very good conductor. Therefore, for good conductors:
Attenuation constant α=rωµσ
2= 117.2Np/m
Phase shift constant β=rωµσ
Propagation constant γ=α+jβ = 117.2 + j117.2 (1/m)
Wavelength λ=2π
β=2π
117.2= 0.053 m
Example#10
Find the depth of penetration δof an EM wave in copper at f= 60 Hz and
100 MHz. For copper σ= 5.8×107mho/m, µr= 1, εr= 1.
Solution: For frequency f= 60 Hz, the ratio
45
σ
ωε =5.8×107
2π60 ×8.85 ×10612 = 175 ×1014 1
Therefore, at f= 60 Hz copper is very good conductor. The depth of pene-
tration:
δ=1
α=r2
ωµσ =r2
2π×60 ×4π×107×5.8×107
δ= 8.53 ×103m
At f= 100 M Hz, the ratio
σ
ωε =5.8×107
2π×8.85 ×104= 0.104 ×1011 1
Hence, copper is very good conductor at f= 100 M Hz. Now, the depth of
penetration
δ=1
α=r2
ωµσ =r2
2π×100 ×106×4π×107×5.8×107
δ= 6.608 ×106m
Example #11
The magnetic ﬁeld Hof a plane wave has magnitude of 5 mA in a medium deﬁned
by εr= 4, µr= 1. Determine (a) average power ﬂow (b) maximum energy density
in the plane wave.
Solution:
(a) We know that
46
E
H=rµ
ε=rµ0µr
ε0εr
E
H=rµ0
ε0εr
=120π
εr
= 60π= 188.4Ω =E= 188.4H
Now the average power
Pav =E2
2×18.5=942 ×103×942 ×103
377
Pav = 2353 µW/m2
(b) Maximum energy density of the wave
WE=1
2εE2=ε0εrE2= 4 ×8.854 ×1012 ×942 ×942 ×106
WE= 31.42 ×106×108
WE= 41.42 pJ/m3
Example#12
A plane wave transmitting in a medium of εr= 1, µr= 1 has an electric ﬁeld
intensity of 100 ×π V /m. Determine the energy density in magnetic ﬁeld and
also the total energy density.
Solution: The electric energy density is given by
WE=1
2εE2=1
2ε0εrE2
WE=1
2×8.85 ×1012 ×1×1002×π
WE= 13.9×108= 139 ×109
WE= 139 nJ/m3
47
As the energy of electric density is equal to that of magnetic ﬁeld for a plane wave:
WH= 139 nJ/m3
So total energy density is
WT= 278 nJ/m3
Example#13
A plane wave of frequency 2 MHz is incidental normally upon a copper conduc-
tor. The wave has electric ﬁeld amplitude of E= 2 ×103V /m. Copper has
µr= 1, εr= 1 and σ= 5.8×107mho/m. Find the average power density absorbed
by the copper.
Solution:
Copper is good conductor
η=rµω
σ=r4π×107×2π×2×106
5.8×106
η=4π
5.8×104=4π
2.40 ×104
η= 5.235 ×104
Average Power
Pav =1
2
E2
|η|=0.5×4×106
5.235 ×104= 0.382 ×102W/m2
Pav = 3.82 mW/m2
48
Chapter 3
3.1 Introduction
This section shall explain the basic mechanism of radiation and the fundamental
types of antennas.
3.2 Short Electric Dipole or Hertzian Antenna
Any linear antenna may be considered as large number of very short conductors
and hence it’s important to consider radiation properties of short conductors. A
short linear conductor is so short that current may be assumed to be constant
throughout its length.
Hertzian dipole is hypothetical antenna and is deﬁned as short isolated con-
ductor carrying alternating current. When length of short dipole is vanishingly
small the term inﬁnitesimal dipole is used. If Idl be inﬁnitesimal small length and
Ibe the current then Idl is current element required for mathematical analysis.
Since I=Imsin ωt or I0cos ωt so the current element be referred to as I0dl cos ωt
or Imdl sin ωt.
””An oscillating current will result in a an oscillating voltage as well or vice-
versa.If the current oscillation is sinusoidal the voltage oscillation will be also
sinusoidal and approximately 900lagging the current in phase angle or short dipole
is capacitive in nature from current/voltage point of view
49
3.3 Retarded Vector Potential
If the expression for vector potential is integrated it follows that potential due to
various current elements are added up. Let the instantaneous current Iin element
be sinusoidal function of time as I=Imsin ωt, where Imus maxim current and
Iis instantaneous current. The eﬀect reaching a distance point Pfrom a given
element at an instant tis due to current value which followed at an earlier time or
current eﬀective in producing for ﬁeld. ﬁnite amount of time must be taken into
consideration. Mathematically,
I=Imsin ωtr
c(3.1)
For uniform plane wave travelling in +ve z direction sin(ωt βz) but now
sin(ωtβr) or sin ωtr
cindicates travelling of spherical waves in radial direction.
Thus retarded current and retarded density in exponential form may be written
as:
[I] = Ime(tr
c)=Imej(ωtβz )Amp (3.2)
[J] = Jme(tr
c)=Imej(ωtβr)Amp/m2(3.3)
According to expression for vector magnetic potential Awhich is applicable in
time-varying conditions where distance travelled are signiﬁcant in terms of wave-
length.
A=µ
4πZV
J
rdv (3.4)
Vector magnetic potential can also be written in terms of exponential form and
general form respectively as:
[A] = µ
4πZv
Jme(tr
c)dv (3.5)
[A] = µ
4πZv
J(tr
c)
rdv (3.6)
50
For sinusoidal current element, retarded vector potential is
[A] = µ
4πZJ(tr
c)
rds·dl(3.7)
Since dv =ds·dl, where dsis cross sectional area and dlis the diﬀerential length
and I=RJds
[A] = µ
4πZI(tr
c)
rdl
[A] = µ
4πZImsin ω(tr
c)
rdl
(3.8)
Similarly, scalar potential into the form of retarded scalar potential is written as:
[V] = 1
4πε Z[ρ]
rdv
[V] = 1
4πε Zv
ρ0e(tr
c)
rdv
(3.9)
Where [v] is retarded scalar potential and [ρ0] = ρ0e(tr
c)is retarded charge
density C/m3
3.4 Antenna Functions
Antenna performs the following functions:
It is used as a transducer i.e. it converts electrical energy into EM energy at
the transmitting end and it converts EM energy back into electrical energy
at the receiving end.
It’s used as an impedance matching device i.e. it matches the transmitter
and free space on the transmitting side and it matches free space and the
It’s used to direct radiated energy into desired directions and suppress it in
unwanted directions.
51
It’s used to sense the presence of electromagnetic waves.
3.5 Antenna Properties
The properties of antenna are as follows:
It has identical impedance when used for transmitting and receiving pur-
poses. This is called equality of impedances.
It has identical directional patterns when it’s used for transmitting and re-
ceiving purposes. This property is called equality of directional patterns.
It has same eﬀective length when it’s used for transmitting and receiving
purposes. This property is called equality of eﬀective lengths.
3.6 Antenna Parameters
3.6.1 Antenna Impedance
It’s deﬁned as ratio of input voltage to input current. Mathematically
Za=Vi
Ii
(3.10)
Where Zais complex quantity and it’s written as
Za=Ra+jXa(3.11)
Here the reactive part Xaresults from ﬁelds surrounding the antenna. The resistive
part Rarepresents losses in the antenna. Rris called radiation resistance.
52
Radiation resistance Rris a ﬁctitious or hypothetical resistance that would dissi-
pate an amount of power equal to radiate power. Mathematically
I2
rms
(3.12)
3.6.3 Directional Characteristics
These are called radiation characteristics or directional patterns. There are two
Field Strength Pattern
It’s variation of the absolute value of the ﬁeld strength as a function of θ. Therefore.
Eversus θis called ﬁeld strength pattern.
Power Pattern
It’s variation of radiated power with θ. Therefore, Pversus θis called power
ﬁeld. It’s pattern drawn as a function of θand φ. The pattern consists of one
main lobe and a number of side lobes.
3.6.4 Eﬀective Length of Antenna
It’s used to indicate eﬀectiveness of the antenna as a radiator or receiver for elec-
tromagnetic energy.
Eﬀective length of transmitting antenna
It’s equal to length of an equivalent linear antenna which radiates the same ﬁeld
strength as the actual antenna and the current is constant throughout the length
of the linear antenna. If leof transmitting antenna is deﬁned as
le(Tx) = 1
IZl/2
l/2
I(z)dz m (3.13)
53
Eﬀective length of receiving antenna
It’s deﬁned as ratio of the open circuit voltage developed at the terminals of the
antenna under the received ﬁeld strength E i.e.
le(Rx) = VOC
Em(3.14)
Where VOC is an open-circuit voltage. Eﬀective length of an antenna is always less
than the actual length that is le< L
It’s deﬁned as power radiated in a given direction per unit solid angle. Mathemat-
ically
RI =r2P=r2E2
η0
Watts/unit solid angle (3.15)
Where η0is intrinsic impedance of medium, ris the radius of the sphere, Pis power
radiated instantaneously, Eelectric ﬁeld strength, RI =RI(θ, φ) is a function of
θand φ.
3.6.6 Directive Gain
It’s deﬁned as ratio of radiation intensity in a speciﬁed direction to the average
Gd=RI
RIav
=RI
Wr/4π
Gd=4π(RI)
Wr
(3.16)
54
3.6.7 Directivity
It’s deﬁned as ratio of maximum radiation intensity to the average radiation in-
tensity. Mathematically
D= (Gd)max and in dB
D[dB] = 10 log10(Gd)max
(3.17)
Directivity is also deﬁned as maximum directive gain.
3.6.8 Power Gain
It’s deﬁned as ratio of 4πtimes radiation intensity to the total input power. There-
fore, power gain can be given as follows:
Gp=4π(RI)
Wt
(3.18)
Where Wr+Wtand Wlis ohmic loss in the antenna.
3.6.9 Antenna Eﬃciency
It’s deﬁned as ratio of radiate power to the input power. Therefore, antenna
eﬃciency ξcan be written as:
ξ=I2
rmsRr
I2
rms(Rr+Rloss)
ξ=Rr
(Rr+Rloss)
(3.19)
Or equivalently
ξ=Wr
Wt
=Wr
Wr+Wl
=Gp
Gd
(3.20)
Therefore, antenna eﬃciency can be deﬁned as ratio of power gain to directive
gain.
55
3.6.10 Eﬀective Area
Eﬀective area of an antenna depends on wavelength and directive gain. Hence
Ae=λ2
4πGdm2
Ae=WR
Pm2
(3.21)
Where WRis received power in watts and Pis power ﬂow per square meter
(Watts/m2) for incident wave.
3.6.11 Antenna Equivalent Circuit
Antenna be equivalently represented in circuit domain and is a series connection
of Ra,Laand Cain a circuit. The main diﬀerence between the antenna equivalent
circuit and an RLC circuit is that Ra,Laand Cavary with frequency. As a result
antenna conductance peak appears not at resonant frequency but at a frequency
slightly away from fr. The antenna impedance is combination of resistive real
component and imaginary reactive component as follows:
Za=Rj(XLXC) WhereXL=ωL , XC=1
ωC (3.22)
Ya=1
Za
=g+jb (3.23)
Where gis conductance and bis susceptance.
3.6.12 Antenna Bandwidth
It’s deﬁned as range of frequencies over which the antenna maintains its charac-
teristics and parameters like gain, front-to-back lobe ratio, standing wave ratio,
radiation pattern, polarisation, impedance and so on without considerable change.
56
3.6.13 Front-to-Back Ratio (FBR)
FBR is deﬁned as ratio of radiated power in the desired direction to the radiated
power in the opposite direction.
F BR =Radiated power in desired direction
Radiated power in opposite direction (3.24)
3.6.14 Polarization
Polarization of an antenna is deﬁned as the direction of the electric vector of the
EM wave produced by an antenna. It’s of the following three types:
Linear Polarization
Circular Polarization
Elliptical Polarization
Linear polarization is of three types i.e. horizontal polarization, vertical polar-
ization and theta polarization.Circular and elliptical polarization are described by
their sense of rotation.The sense of rotation can be right-handed and left-handed.
Accordingly, they are right-handed and left-handed circular/elliptical polariza-
tions.
3.7 Basic Antenna Elements
Basic antenna elements are
Current element or Hertzian dipole: It’s short linear antenna in which the
current along its length can be considered constant.
Short dipole: It’s linear antenna whose length is less than λ/4 and the current
distribution is assumed to be triangular.
Short monopole: It’s linear antenna whose length is λ/8 and the current
distribution is assumed to be triangular.
57
Half-wave dipole: It’s linear antenna whose length is λ/2 and the current
distribution is assumed to be sinusoidal. It’s usually center-fed.
Quarter-wave monopole: It’s a linear antenna whose length is λ/4 and the
current distribution is assumed to be sinusoidal. It’s fed at one end with
respect to earth.
3.8 Directivity of Electric Current Element
Directivity is deﬁned as maximum directive gain of electric current element ob-
tained by comparing it with isotropic radiator
Power radiated by isotropic antenna (3.25)
Wi
(3.26)
Where 4πr2is area of the sphere since isotropic radiates uniformly in all direc-
tions and Wiis input power. It’s known that power radiated by current element
Pr(av) = ηImdl sin θ2
8λ2r2W att/m2(3.27)
Power will be maximum when θ=π/2
8λ2r2W att/m2(3.28)
Also input power is given by
Wi= 80π2dl
λ2, I2
rms = 40π2dl
λ2I2
m(3.29)
Putting equations 3.28 and 3.29 in eq.3.26, we have
G= 4πr2η(Imdl)2
8λ2r2×1
40π2(dl
λ)2I2
m
G=η
90π=120π
80π= 3/2=1.5
(3.30)
58
The linear value of gain can also be converted in dB value as follows:
GdB = 10 log G= 10 log(3/2) = 1.761 dB (3.31)
3.9 Gain of Half-wavelength Antenna
Gain of λ/2 antenna is ratio of power radiated by λ/2 antenna to the power
Power radiated by isotropic antenna (3.32)
Equivalently
G=Pr
Wi/4πr2
G= 4πr2Pr
Wi
(3.33)
Poynting vector for λ/2 antenna can be given as:
Pr(av) = 30I2
rms
πr2 cos2(π/2 cos θ)
sin2θ!(3.34)
The above expression for Pr(av) becomes maximum when θ= 900cos2θ= 1.
Therefore, eq. 3.34 can be rewritten as:
Pr(av) = 30I2
rms
πr2(3.35)
Input power to λ/2 antenna is given by
Wi= 73.14I2
rms (3.36)
59
Hence, the gain of λ/2 antenna is given by
Gλ/2= 4πr2Pr(av)
Wi
=4πr2
73.14I2
rms ·30I2
rms
πr2
Gλ/2= 1.641
(3.37)
In terms of dB, the gain of λ/2 antenna is
Gλ/2= 10 log G= 10 log 1.641 = 10 ×0.2151
Gλ/2= 2.141 dB (3.38)
3.10 Radiation Pattern of Alternating Current
Element
The alternating current element is also called dipole or oscillating dipole or current
element. The radiation ﬁeld of z-directed current element is
E=60πIdl
λr sin θ V/m (3.39)
Where Idl is current element, dl is diﬀerential length,ris far-ﬁeld distance, λ
is operating wavelength, and θangle between dipole axis and the line of far-ﬁeld
point.
Let the eq. 3.39 is be represented by
E=Emsin θ(3.40)
Where Em=60πIdl
λr . The normalised ﬁeld is
En=E
Em
= sin θ(3.41)
The horizontal pattern of elementary dipole is a circle. This could be obtained
for θ= 900. It is evident from above expression that the ﬁeld is independent of φ.
60
3.11 Radiation Pattern Expression of Center-fed
Vertical Dipole of Finite Length
Half-wave dipole antenna is shown in the following ﬁgure 3.1 which shows the
current passing through half-wave dipole fed through transmission line and also
the current distribution along the arms of half-wave dipole.
Figure 3.1: Half-wave dipole antenna
The magnitude of the radiation ﬁeld of a vertical dipole lis given by:
E=60Im
r"cos βl cos θ
2cos βl
2
sin θ#(3.42)
En="cos βl cos θ
2cos βl
2
sin θ#(3.43)
We can substitute the required of length and can get the corresponding nor-
malised ﬁeld strength value. For example, the normalised ﬁeld strength of half-
wave dipole l=λ/2 is
61
En=
cos βλ/4 cos θcos β l
2
sin θ
En=
cos 2π
λλ/4 cos θcos 2π
λλ/4
sin θ
En=π
2cos θ
sin θ
(3.44)
Moreover, the horizontal pattern of any length is represented by
En="cos βl
2cos θcos βl
2#
sin θθ=900
En=constant
(3.45)
Therefore, horizontal pattern of dipole is circle.
62
3.12 Radiation Pattern of Center-fed Vertical Dipole
Figure 3.2: Radiation Pattern of Vertical Dipole (a)normalised E-plane or vertical
pattern (φ= 0) (b) normalised H-plane or horizontal pattern (φ=π/2) (c) three-
dimensional plane
63
3.13 Radiation Pattern of Center-fed Horizontal
Dipole
3.14 Radiation Pattern of Vertical Monopole
Figure 3.3: The monopole antenna
3.15 Two-element Uniform Array
The array of radiators is deﬁned as a system of antennas which are similar or
non-similar and either similarly oriented or diﬀerently oriented. Arrays are used
to increase the directivity and gain. We shall discuss arrays of similar antennas
and similar orientation.
Expression for Resultant Radiation Pattern of Two-element Array
The expression for resultant radiation pattern of two-element array can be given
as follows:
ER= 2EAcos πd cos φ
λ+αe
2!(3.46)
Where dis spacing between antennas, φis angle between the axis of the array
and line of the observer, EAﬁeld strength due to antenna A alone, λis operating
wavelength, αeis excitation phase. If point Pas shown in Fig. 3.4 is far away
from the array, Ray A and Ray B can be assumed to be parallel. Hence the path
diﬀerence between two rays is
64
P(Observer)
rA
RayA
rB
RayB
d
rArB
φArray Axis
Figure 3.4: Two-element antenna array diagram
rArB=dcos φcos φ=rArB
d
(3.47)
This exact expression must be used in the phase term of the ﬁeld. But in the
magnitude term of the ﬁeld, we can use the approximation i.e. RARB. Now
the resultant phase diﬀerence due to spacing of antennas is given by
αd=β×dcos φ(3.48)
If the excitation phase diﬀerence is αe, the total phase diﬀerence is
ψ=βd cos φ+αe(3.49)
Here, αeis the phase angle by which current IBin antenna Bleads current IA
in antenna A. The resultant ﬁeld in phasor form when two antennas are uniformly
excited is given by:
65
ER=EA 1 + e !(3.50)
Magnitude of total ﬁeld strength when IA=IBis
|ER|=
EA 1 + e !
=EA1 + cos ψ+jsin ψ
=EAq(1 + cos ψ)2+ sin2ψ
=EAq1 + cosψ+2 cos ψ+ sinψ
=EAp2 + 2 cosψ=EAp2(1 + cos ψ)
(3.51)
1 + cos ψ= 2 cos2ψ
2
E=EAq2×2 cosψ
2
= 2EAcos ψ
2
(3.52)
Now, substituting eq. 3.49 into eq. 5.20 we can write ﬁeld strength expression
as
E= 2EAcos βd cos φ+αe
2!
= 2EAcos βd cos φ
2+αe
2!(3.53)
Hence
EA= 2EAcos πd cos φ
λ+αe
2!(3.54)
In practical applications, two-element array is rarely used. Mostly arrays with
66
more number of elements are used to get high directivity and gain and have con-
trol over more parameters like spacing, current amplitude, phase and antenna
conﬁguration.
3.16 Field Strength of Uniform Linear Array
The normalised ﬁeld strength of uniform linear array is
E=
sin Nψ
2
sin ψ
2
(3.55)
Here Nis number of elements in the array, ψ=βd cos φ+αe,βis wave number, d
is spacing between elements, ψis angle between axis of array and line of observer
and αeis excitation phase (progressive phase shift). r1,r2,r3and rNare ray
paths of 1,2, 3, and Nth antenna to point P. Let Pbe point far-away. If ﬁeld of
antenna 1 is E1the ﬁeld of antenna 2 is E1eand the ﬁeld of antenna 3 is E1ej2ψ.
Similarly, the ﬁeld of antenna Nis
E1ej(N1)ψ(3.56)
The total ﬁeld is vector sum of ﬁelds
ER=E1+E1e+E1ej2ψ+·+E1ej(N1)ψ(3.57)
Where ψ=βd cos φ+αeand αeis progressive phase shift between antennas. Eq.
3.57 is geometric progression and now multiplying both sides by e, we get
ERe=E1ejψ +ej2ψ+· ·· +ejN ψ (3.58)
Now subtracting eq. 5.16 from eq. 3.57
ERe =E1ejN ψ 1
ER
E1
=ejN ψ 1
e1
(3.59)
67
The normalised magnitude of ﬁeld strength
ER
E1=ejN ψ 1
e1
(3.60)
Now
e1=cos ψ+jsin ψ1
=qcos ψ12+ sin ψ
=p1 + cos ψ2 cos ψ+ sin ψ
=q21cos ψ
=2p1cos ψ
= 2 sin ψ
2sin ψ
2= 1 cos ψ
(3.61)
Similarly, ej1=2 sin N ψ
2(3.62)
From eq. 3.60
E=e1
e1
=
sin Nψ
2
sin ψ
2
(3.63)
The normalised ﬁeld strength is plotted as a function of ψand typical variation
lis shown below.
Salient Features of Uniform Linear Array
1. Maximum value of normalized ﬁeld strength
E=
ER
E1(3.64)
2. Maximum value N at ψ= 0 is called principle maximum of the array
3. Minimum value of Ecalled nulls occur at Nψ/2 = ± k = 1,2,3,···
4. Secondary maximums occurs approximately between nulls. The secondary
68
maximum occur when numerator of
E=
sin Nψ
2
sin ψ
2
(3.65)
becomes maximum i.e. secondary maximums occur at N ψ/2 = ±(2m+
1)π/2m= 1,2,3,···
5. Ratio of ﬁrst secondary maximum to principle maximum is called side lobe
ratio (SLR) and the ﬁrst SRL of ULA is 13.47 dB
3.16.1 First Side-lobe Ratio (FSR)
First side lobe ratio is deﬁned as the ratio of the ﬁrst side lobe level to the main
side lobe level i.e.
SLR = First Side Lobe Level
Main Side Lobe Level (3.66)
SLR of uniform array is 13.5dB. We know that
E=
sin Nψ
2
sin ψ2
(3.67)
Secondary maximum occur approximately at the centre between nulls i.e. they
occur at Nψ
2=±22m+ 1π/2m= 1,2,3,··· (3.68)
Hence the ﬁrst SRL maxim occurs at
Nψ
2=±2+1π/2 = 3π
2(3.69)
Putting eq. 3.69 in eq. 3.67
E=
sin 3π
2
sin sin 3π
2N
=
1
sin 3π
2N
(3.70)
For large values of N, 3π/2Nis very small so
69