ANTENNAS AND

WAVE PROPAGATION

A.R. HARISH M. SACHIDANANDA

Assistant Professor Formerly Professor
Department of Electrical Engineering Department of Electrical Engineering
Indian Institute of Technology Kanpur Indian Institute of Technology Kanpur


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Contents

Preface
Symbols xv

CHAPTER 1 Electromagnetic Radiation 1

Introduction 1
1.1 Review of Electromagnetic Theory 4
1.1.1 Vector Potential Approach 9
1.1.2 Solution of the Wave Equation 11
1.1.3 Solution Procedure 18
1.2 Hertzian Dipole 19
Exercises 30

CHAPTER 2 Antenna Characteristics 31

Introduction 31
2.1 Radiation Pattern 32
2.2 Beam Solid Angle, Directivity, and Gain 44
2.3 Input Impedance 49
2.4 Polarization 53
2.4.1 Linear Polarization 54
2.4.2 Circular Polarization 56
2.4.3 Elliptical Polarization 58
2.5 Bandwidth 59

ix

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x Contents

2.6 Receiving Antenna 60

2.6.1 Reciprocity 60
2.6.2 Equivalence of Radiation and Receive Patterns 66
2.6.3 Equivalence of Impedances 67
2.6.4 Eective Aperture 68
2.6.5 Vector Eective Length 73
2.6.6 Antenna Temperature 80
2.7 Wireless Systems and Friis Transmission Formula 85
Exercises 90

CHAPTER 3 Wire Antennas 94

Introduction 94
3.1 Short Dipole 94
3.1.1 Radiation Resistance and Directivity 103
3.2 Half-wave Dipole 106
3.3 Monopole 115
3.4 Small Loop Antenna 117
Exercises 127

CHAPTER 4 Aperture Antennas 129

Introduction 129
4.1 Magnetic Current and its Fields 130
4.2 Some Theorems and Principles 133
4.2.1 Uniqueness Theorem 134
4.2.2 Field Equivalence Principle 134
4.2.3 Duality Principle 136
4.2.4 Method of Images 137
4.3 Sheet Current Distribution in Free Space 139
4.3.1 Pattern Properties 143
4.3.2 Radiation Pattern as a Fourier Transform of the Current
Distribution 149
4.4 Expressions for a General Current Distribution 154
4.5 Aperture in a Conducting Screen 155
4.6 Slot Antenna 158

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 Contents xi

4.7 Open-ended Waveguide Radiator 159

4.8 Horn Antenna 160
4.9 Pyramidal Horn Antenna 162
4.10 Reector Antenna 165
4.10.1 Flat-plate Reector 167
4.10.2 Corner Reector 172
4.10.3 Common Curved Reector Shapes 174
Exercises 193

CHAPTER 5 Antenna Arrays 195

Introduction 195
5.1 Linear Array and Pattern Multiplication 196
5.2 Two-element Array 199
5.3 Uniform Array 212
5.3.1 Polynomial Representation 220
5.4 Array with Non-uniform Excitation 227
5.4.1 Binomial Array 228
5.4.2 Chebyshev Array Synthesis 232
Exercises 240

CHAPTER 6 Special Antennas 242

Introduction 242
6.1 Monopole and Dipole Antennas 243
6.1.1 Monopole for MF and HF Applications 243
6.1.2 Monopole at VHF 247
6.1.3 Antenna for Wireless Local Area Network Application 247
6.2 Long Wire, V, and Rhombic Antennas 251
6.2.1 V Antenna 255
6.3 YagiUda array 262
6.4 Turnstile Antenna 270
6.4.1 Batwing and Super-turnstile Antennas 273
6.5 Helical Antenna 277
6.5.1 Axial Mode Helix 278
6.5.2 Normal Mode Helix 282

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xii Contents

6.6 Biconical Antenna 283

6.7 Log-periodic Dipole Array 285
6.7.1 Design Procedure 290
6.8 Spiral Antenna 295
6.9 Microstrip Patch Antenna 298
Exercises 302

CHAPTER 7 Antenna Measurements 303

Introduction 303
7.1 Antenna Measurement Range 304
7.2 Radiation Pattern Measurement 314
7.2.1 Antenna Positioner 315
7.2.2 Receiver Instrumentation 318
7.3 Gain and Directivity 319
7.3.1 Absolute Gain Measurement 320
7.3.2 Gain Transfer Method 323
7.3.3 Directivity 324
7.4 Polarization 324
7.5 Input Impedance and Input Reection Coecient 328
Exercises 329

CHAPTER 8 Radio Wave Propagation 330

Introduction 330
8.1 Ground Wave Propagation 332
8.1.1 Free Space Propagation 333
8.1.2 Ground Reection 334
8.1.3 Surface Waves 339
8.1.4 Diraction 341
8.1.5 Wave Propagation in Complex Environments 344
8.1.6 Tropospheric Propagation 348
8.1.7 Tropospheric Scatter 360
8.2 Ionospheric Propagation 364
8.2.1 Electrical Properties of the Ionosphere 367
8.2.2 Eect of Earths Magnetic Field 378
Exercises 380

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 Contents xiii

Appendix A Trigonometric Formulae 383

Appendix B Integration Formulae 385
Appendix C Series Expansions 387
Appendix D Vector Identities 389
Appendix E Coordinate Systems and Vector
Dierential Operators 390
Appendix F Coordinate Transformations 393
sin x
Appendix G The x Function 395

References 397
Index 399

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CHAPTER 1

Electromagnetic Radiation

Introduction
Most of us are familiar with cellular phones. In cellular communication
systems, there is a two-way wireless transmission between the cellular phone
handset and the base station tower. The cell phone converts the audio sig-
nals into electrical form using a microphone. This information is imposed on
a high frequency carrier signal by the process of modulation. The modulated
carrier is radiated into free space as an electromagnetic wave which is picked
up by the base station tower. Similarly, the signals transmitted by the tower
are received by the handset, thus establishing a two way communication.
This is one of the typical examples of a wireless communication system
which uses free space as a medium to transfer information from the trans-
mitter to the receiver. A key component of a wireless link is the antenna
which eciently couples electromagnetic energy from the transmitter to
free space and from free space to the receiver. An antenna is generally a
bidirectional device, i.e., the power through the antenna can ow in both the
directions, hence it works as a transmitting as well as a receiving antenna.
Transmission lines are used to transfer electromagnetic energy from one
point to another within a circuit and this mode of energy transfer is generally
known as guided wave propagation. An antenna acts as an interface between
the radiated electromagnetic waves and the guided waves. It can be thought
of as a mode transformer which transforms a guided-wave eld distribution
into a radiated-wave eld distribution. Since the wave impedances of the
guided and the radiated waves may be dierent, the antenna can also be
thought of as an impedance transformer. A proper design of this part is
necessary for the ecient coupling of the energy from the circuit to the free
space and vice versa.
One of the important properties of an antenna is its ability to transmit
power in a preferred direction. The angular distribution of the transmitted

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2 Chapter 1 Electromagnetic Radiation

Fig. 1.1 Parabolic dish antenna at the Department of Electrical Engineering,

Indian Institute of Technology, Kanpur, India (Courtesy: Dept of EE, IIT Kanpur)
power around the antenna is generally known as the radiation pattern (A
more precise denition is given in Chapter 2). For example, a cellular phone
needs to communicate with a tower which could be in any direction, hence
the cellular phone antenna needs to radiate equally in all directions. Sim-
ilarly, the tower antenna also needs to communicate with cellular phones
located all around it, hence its radiation also needs to be independent of the
direction.
There are large varieties of communication applications where the direc-
tional property is used to an advantage. For example, in point-to-point com-
munication between two towers it is sucient to radiate (or receive) only in
the direction of the other tower. In such cases a highly directional parabolic
dish antenna can be used. A 6.3 m diameter parabolic dish antenna used for
communication with a geo-stationary satellite is shown in Fig. 1.1. This an-
tenna radiates energy in a very narrow beam pointing towards the satellite.
Radio astronomy is another area where highly directional antennas are
used. In radio astronomy the antenna is used for receiving the electromag-
netic radiations from outer space. The power density of these signals from
outer space is very low, hence it is necessary to collect the energy over a very
large area for it to be useful for scientic studies. Therefore, radio astron-
omy antennas are large in size. In order to increase the collecting aperture,

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 Introduction 3

Fig. 1.2 A panoramic view of the Giant Metrewave Radio Tele-

scope (GMRT), Pune, India, consisting of 30 fully-steerable
parabolic dish antennas of 45 m diameter each spread over dis-
tances up to 25 km.1 (Photograph by Mr. Pravin Raybole, Cour-
tesy: GMRT, Pune, http://www.gmrt.ncra.tifr.res.in)
the Giant Metrewave Radio Telescope (GMRT) near Pune in India, has an
array of large dish antennas, as shown in Fig. 1.2.
The ability of an antenna to concentrate power in a narrow beam depends
on the size of the antenna in terms of wavelength. Electromagnetic waves of
wavelengths ranging from a few millimetres to several kilometres are used
in various applications requiring ecient antennas working at these wave-
lengths. These frequencies, ranging from hundreds of giga hertz to a few
kilo hertz, form the radio wave spectrum. Figure 1.3 depicts the radio wave
spectrum along with band designations and typical applications.
The radiation pattern of an antenna is usually computed assuming the
surroundings to be innite free space in which the power density (power
per unit area) decays as inverse square of the distance from the antenna.
In practical situations the environment is more complex and the decay is
not as simple. If the environment consists of well dened, nite number of
scatterers, we can use theories of reection, refraction, diraction, etc., to
predict the propagation of electromagnetic waves. However, in a complex
environment, such as a cell phone operating in an urban area, the eld
strength is obtained by empirical relations.
The atmosphere plays a signicant role in the propagation of electromag-
netic waves. The density of the air molecules and, hence, the refractive index
of the atmosphere changes with height. An electromagnetic wave passing
through media having dierent refractive indices undergoes refraction.
Thus, the path traced by an electromagnetic wave as it propagates through
1The GMRT was built and is operated by the National Centre for Radio Astrophysics (NCRA)
of the Tata Institute of Fundamental Research (TIFR).

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4 Chapter 1 Electromagnetic Radiation

Wavelength (m) 103 100 103 106 109 1012

Radio waves Infrared Ultraviolet X rays Gamma rays

Visible light

Radio wave bands

Band VLF LF MF HF VHF UHF SHF EHF

designation Very Low Medium High Very Ultra Super Extremely
low frequency frequency frequency frequency high frequency high frequency high frequency high frequency
Navigation Radio beacon AM broadcast Shortwave Television Television Radar Radar
Direction finding broadcast FM broadcast Satellite Microwave link Experimental studies
Maritime Amateur radio Air traffic control communication Satellite
Applications communication Aircraft Radar communication
Amateur radio communication Navigation Mobile
Radio astronomy Cellular telephone communication
Frequency: 3 kHz30 kHz 30 kHz300 kHz 300 kHz3 MHz 3 MHz30 MHz 30 MHz300 MHz 300 MHz3 GHz 3 GHz30 GHz 30 GHz300 GHz
Wavelength: 100 km to 10 km 10 km to 1 km 1 km to 100 m 100 m to 10 m 10 m to 1 m 1 m to 10 cm 10 cm to 1 cm 1 cm to 1 mm

Microwave bands

Millimeter wave
Band
designation
L S C X Ku K Ka
1 GHz2 GHz 2 GHz4 GHz 4 GHz8 GHz 8 GHz12.4 GHz 12.4 GHz18 GHz 18 GHz27 GHz 27 GHz40 GHz 40 GHz300 GHz

Fig. 1.3 Radio wave spectrum along with the band designations and typical ap-
plications.

the atmosphere is not a straight line. The air molecules also get ionized
by solar radiation and cosmic rays. The layer of ionized particles in the
atmosphere, known as the ionosphere, reects high frequency (3 MHz to
30 MHz) waves. A multi-hop communication link is established by repeated
reections of the electromagnetic waves between the ionosphere and the
surface of the earth. This is the mode of propagation of shortwave radio
signals over several thousand kilometres.
Both the radiation properties of the antennas and the propagation condi-
tions play a very important role in establishing a successful communication
link. This book addresses both these issues in some detail. It is assumed that
the students have some basic knowledge of electromagnetic theory. However,
in the following section some of the basic concepts of electromagnetic theory
used in the analysis of antennas are presented for easy reference as well as
for introducing the notation used in the book.

1.1 Review of Electromagnetic Theory

Electromagnetic elds are produced by time-varying charge distributions
which can be supported by time-varying current distributions. Consider sinu-
soidally varying electromagnetic sources. (Sources having arbitrary variation

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 1.1 Review of Electromagnetic Theory 5

with respect to time can be represented in terms of sinusoidally varying

functions using Fourier analysis.) A sinusoidally varying current i(t) can be
expressed as a function of time, t, as

i(t) = I0 cos(t + ) (1.1)

where I0 is the amplitude (unit: ampere, A), is the angular frequency

(unit: radian per second, rad/s), and is the phase (unit: radian, rad). The
angular frequency, , is related to the frequency, f (unit: cycle per second
or Hz), by the relation = 2f . One may also express the current i(t) as
a sine function

i(t) = I0 sin(t +  ) (1.2)

where  = + /2. Therefore, we need to identify whether the phase has

been dened taking the cosine function or the sine function as a reference.
In this text, we have chosen the cosine function as the reference to dene
the phase of the sinusoidal
 quantity.

Since cos(t + ) = Re ej(t+) where, Re{} represents the real part of
the quantity within the curly brackets, the current can now be written as
 
i(t) = I0 Re ej(t+) (1.3)
 
= Re I0 ej ejt (1.4)

The quantity I0 ej is known as a phasor and contains the amplitude and

phase information of i(t) but is independent of time, t.

EXAMPLE 1.1

Express i(t) = (cos t + 2 sin t) A in phasor form.

Solution: First we must express sin t in terms of the cosine function using
the relation cos(t /2) = sin t. Therefore
 

i(t) = cos t + 2 cos t
2
 
Using the relation cos(t + ) = Re ej(t+)
   
i(t) = Re ejt + Re 2ej(t/2)

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6 Chapter 1 Electromagnetic Radiation

For any two complex quantities Z1 and Z2 , Re{Z1 + Z2 } = Re{Z1 } +

Re{Z2 } and, hence, the current can be written as

i(t) = Re{(1 + 2ej/2 )ejt }

= Re{(1 j2)ejt }
= Re{2.24ej1.1071 ejt }

Therefore, in the phasor notation the current is given by

I = 2.24ej1.1071 A

EXAMPLE 1.2

Express the phasor current I = (I1 ej1 + I2 ej2 ) as a function of time.

Solution: The instantaneous current can be expressed as

i(t) = Re{Iejt }

Substituting the value of I

i(t) = Re{I1 ej1 ejt + I2 ej2 ejt }

= I1 cos(t + 1 ) + I2 cos(t + 2 )

The eld vectors that vary with space, and are sinusoidal functions of
time, can also be represented by phasors. For example, an electric eld vector
E(x, y, z, t), a function of space (x, y, z) having a sinusoidal variation with
time, can be written as
 
E(x, y, z, t) = Re E(x, y, z)ejt (1.5)

where E(x, y, z) is a phasor that contains the direction, magnitude, and

phase information of the electric eld, but is independent of time. In the text
that follows, ejt time variation is implied in all the eld and source quanti-
ties and is not written explicitly. In this text, bold face symbols (e.g., E) are
used for vectors, phasors, or matrices, italic characters for scalar quantities
(e.g., t), script characters (e.g., E) for instantaneous scalar quantities,
and script characters with an over-bar (e.g., E) for instantaneous vector
quantities.

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 1.1 Review of Electromagnetic Theory 7

Using phasor notation, Maxwells equations can be written for the elds
and sources that are sinusoidally varying with time as1

E = jH (1.6)
H = jE + J (1.7)
D= (1.8)
B=0 (1.9)

The symbols used in Eqns (1.6) to (1.9) are explained below:

E : Electric eld intensity (unit: volt per metre, V/m)

H : Magnetic eld intensity (unit: ampere per metre, A/m)
D : Electric ux density (unit: coulomb per metre, C/m)
B : Magnetic ux density (unit: weber per metre, Wb/m or tesla, T)
J : Current density (unit: ampere per square metre, A/m2 )
: Charge density (unit: coulomb per cubic metre, C/m3 )

The rst two curl equations are the mathematical representations of Fara-
days and Amperes laws, respectively. The divergence equation [Eqn (1.8)]
represents Gausss law. Since magnetic monopoles do not exist in nature,
we have zero divergence for B [Eqn (1.9)].
The current density, J, consists of two components. One is due to the
impressed or actual sources and the other is the current induced due to the
applied electric eld, which is equal to E, where is the conductivity of
the medium (unit: siemens per metre, S/m). In antenna problems, we are
mostly working with elds radiated into free space with = 0. Therefore, in
the analyses that follow, unless explicitly specied, J represents impressed-
source current density.
In an isotropic and homogeneous medium, the electric ux density, D,
and the electric eld intensity, E, are related by

D = E (1.10)

where  is the electric permittivity (unit: farad per metre, F/m) of the
medium. 0 is the permittivity of free space (0 = 8.854 1012 F/m) and
the ratio, /0 = r is known as the relative permittivity of the medium. It is

1See Cheng 2002, Hayt et al. 2001, Jordan et al. 2004, and Ramo et al. 2004.

2[IRUG8QLYHUVLW\3UHVV
8 Chapter 1 Electromagnetic Radiation

a dimensionless quantity. Similarly, magnetic ux density, B, and magnetic

eld intensity, H, are related by
B = H (1.11)
where = 0 r is the magnetic permeability (unit: henry per metre, H/m) of
the medium. 0 is the permeability of free space (0 = 4 107 H/m) and
the ratio, /0 = r , is known as the relative permeability of the medium.
For an isotropic medium  and are scalars and for a homogeneous medium
they are independent of position.
One of the problems in antenna analysis is that of nding the E and H
elds in the space surrounding the antenna. An antenna operating in the
transmit mode is normally excited at a particular input point in the an-
tenna structure. (The same point is connected to the receiver in the receive
mode). Given an antenna structure and an input excitation, the current
distribution on the antenna structure is established in such a manner that
Maxwells equations are satised everywhere and at all times (along with
the boundary conditions which, again, are derived from Maxwells equa-
tions using the behaviour of the elds at material boundaries). The
antenna analysis can be split into two parts(a) determination of the cur-
rent distribution on the structure due to the excitation and (b) evaluation
of the eld due to this current distribution in the space surrounding the
antenna. The rst part generally leads to an integral equation, the treat-
ment of which is beyond the scope of this book. We will be mainly concerned
with the second part, i.e., establishing the antenna elds, given the current
distribution.
Maxwells equations [Eqns (1.6)(1.9)] are time-independent, rst order
dierential equations to be solved simultaneously. It is a common practice
to reduce these equations to two second order dierential equations called
wave equations. For example, in a source-free region ( = 0 and J = 0) we
can take the curl of the rst equation [Eqn (1.6)], substitute it in the second
equation [Eqn (1.7)] to eliminate H, and get the wave equation, 2 E +
k 2 E = 0, satised by the E eld. Similarly, we can also derive the wave
equation satised by the H eld. (Start from the curl of Eqn (1.7) and
substitute in Eqn (1.6) to eliminate E). Thus, it is sucient to solve one
equation to nd both E and H elds, since they satisfy the same wave
equation.
To simplify the problem of nding the E and H elds due to a current
distribution, we can split it into two parts by dening intermediate potential
functions which are related to the E and H elds. This is known as the vector
potential approach and is discussed in the following subsection.

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 1.1 Review of Electromagnetic Theory 9

1.1.1 Vector Potential Approach

Given a current distribution on the antenna, the problem is one of deter-
mining the E and H elds due to this current distribution which satises all
four of Maxwells equations along with the boundary conditions, if any. In
the vector potential approach we carry out the solution to this problem in
two steps by dening intermediate potential functions. In the rst step, we
determine the potential function due to the current distribution and in the
second step, the E and H elds are computed from the potential function.
In the analysis that follows, the relationships between the vector potential
and the current distribution as well as the E and H elds are derived. All
four of Maxwells equations are embedded in these relationships.
Let us start with the last of the Maxwells equations, B = 0. Since
the curl of a vector eld is divergence-free (vector identity: A = 0),
B can be expressed as a curl of an arbitrary vector eld, A. We call this a
magnetic vector potential function because it is related to the magnetic ux
density, B, via the relationship

H = B = A (1.12)

or
1
H= A (1.13)

Substituting this into the equation E = jH, Maxwells rst equa-
tion is also incorporated

E = j(H) = j( A) (1.14)

or

(E + jA) = 0 (1.15)

Since the curl of a gradient function is zero (vector identity: V = 0),

the above equation suggests that the quantity in brackets can be replaced
by the gradient of a scalar function. Specically, a scalar potential function
V is dened such that

(E + jA) = V (1.16)

Using this we relate the E eld to the potential functions as

E = (V + jA) (1.17)

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10 Chapter 1 Electromagnetic Radiation

Equations (1.13) and (1.17) relate the H and E elds to the potential
functions A and V . Now, to satisfy Maxwells second equation, H =
jE + J, substitute the expression for the E and H elds in terms of the
potential functions [Eqns (1.13) and (1.17)]
1
( A) = j(V + jA) + J (1.18)

which is valid for a homogeneous medium. Expanding the left hand side
using the vector identity

A = ( A) 2 A (1.19)

we have

2 A + 2 A = J + ( A + jV ) (1.20)

So far we have satised three of Maxwells four equations. Note that only
the curl of A is dened so far. Since the curl and divergence are two
independent parts of any vector eld, we can now dene the divergence
of A. We dene A so as to relate A and V as well as simplify Eqn (1.20)
by eliminating the second term on the right hand side of the equation. We
relate A and V by the equation

A = jV (1.21)

This relationship is known as the Lorentz condition. With this the magnetic
vector potential, A, satises the vector wave equation

2 A + k 2 A = J (1.22)

where

k =  (1.23)

is the propagation constant (unit: radian per metre, rad/m) in the medium.
Now, to satisfy Maxwells fourth equation, D = , we substitute
E = (V + jA) in this equation to get

(V jA) = (1.24)

or

2 V + j( A) = (1.25)


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 1.1 Review of Electromagnetic Theory 11

Eliminating A from this equation using the Lorentz condition [Eqn (1.21)]

2 V + k 2 V = (1.26)

Thus, both A and V must satisfy the wave equation, the source function
being the current density for the magnetic vector potential, A, and the
charge density for the electric scalar potential function V .

1.1.2 Solution of the Wave Equation

Consider a spherically symmetric charge distribution of nite volume, V  ,
centred on the origin. Our goal is to compute the scalar potential V (x, y, z)
[or V (r, , )1 ] due to this source, which is the solution of the inhomogeneous
wave equation as given by Eqn (1.26). Since the charge is spherically sym-
metric, we will solve the wave equation in the spherical coordinate system.
The Laplacian 2 V in the spherical coordinate system2 is written as
   
1 V 1 V 1 2V
V = 2
2
r2 + 2 sin + (1.27)
r r r r sin r2 sin2 2

The scalar potential, V (r, , ), produced by a spherically symmetric charge

distribution is independent of and . Therefore, the wave equation,
Eqn (1.26), reduces to
 
1 V
2
r2 + k2 V = (1.28)
r r r 

The right hand side of this equation is zero everywhere except at the source
itself. Therefore, in the source-free region, V satises the homogeneous wave
equation
 
1 V
2
r2 + k2 V = 0 (1.29)
r r r

The solutions for V are the scalar spherical waves given by

ejkr
V (r) = V 0 (1.30)
r

where V 0 is a complex amplitude constant and ejkr/r is a spherical wave

+

travelling in the +r-direction. V 0 is the complex amplitude of the scalar
1(x, y, z):
rectangular co-ordinates; (r, , ): spherical co-ordinates.
2 SeeAppendix E for details on the coordinate systems and vector operations in dierent coordi-
nate systems.

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12 Chapter 1 Electromagnetic Radiation

spherical wave ejkr/r travelling in the r-direction. By substituting this in

the wave equation, it can be shown that it satises the homogeneous wave
equation [Eqn (1.29)].

EXAMPLE 1.3

Show that

ejkr
V (r) = V 0
r

are solutions of
 
1 V
2
r2 + k2 V = 0
r r r

Solution: Let us consider the wave travelling in the positive r-direction

+ ejkr
V (r) = V 0
r

Substituting into the left hand side (LHS) of the given equation
  
1 ejkr
+ + ejkr
LHS = 2 r2 V0 + k2 V 0
r r r r r
  
1
+ ejkr ejkr + ejkr
=V0 2 r2 2 jk + k2 V 0
r r r r r
+ 1
jkr jkr

2 +e
jkr
=V0 e jkre + k V 0
r2 r r
+ 1  jkr jkr 2 jkr

2 +e
jkr
=V0 jke jke k re + k V 0
r2 r
=0

Therefore, the positive wave is a solution of the given dierential equation.

Now, let us consider the wave that is travelling along the negative
r-direction

ejkr
V (r) = V 0
r

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 1.1 Review of Electromagnetic Theory 13

Substituting into the left hand side of the given dierential equation
  
1 ejkr ejkr
LHS = 2 r2 V0 + k2 V 0
r r r r r
 jkr 
1 1 e
r2 2 ejkr + jkrejkr + k 2 V 0

=V0 2
r r r r
1  jkr jkr 2 jkr

2 e
jkr
=V0 jke + jke rk e + k V 0
r2 r
=0

The wave travelling in the r-direction satises the dierential equation,

hence it is also a solution.
These are the two solutions of the wave equation in free space and rep-
resent spherical waves propagating away from the origin (+r-direction) and
converging on to the origin (r-direction). Taking physical considerations
into account, the wave converging towards the source is discarded.

The instantaneous value of the scalar potential V(r, t) for the wave prop-
agating in the +r-direction can be written as
 
ej(tkr)
+
V(r, t) = Re V 0 (1.31)
r

Since V 0 is a complex quantity, it can be expressed as, V 0 = |V 0 |ejv , where

+ + +

+
v is the phase angle of V 0 . The equation for the constant phase spherical
wave front is

v + t kr = const (1.32)

The velocity of the wave is the rate at which the constant phase front moves
with time. Dierentiating the expression for the constant phase front surface
with respect to time, we get

dr
j jk =0 (1.33)
dt
+
This follows from the fact that V 0 and, hence, the phase v , is independent
of time, i.e., dv /dt = 0. Therefore, the velocity (v, unit: metre per second,

2[IRUG8QLYHUVLW\3UHVV
14 Chapter 1 Electromagnetic Radiation

m/s) of the wave can be expressed as

dr
v= = (1.34)
dt k

Substituting the value of the propagation constant from Eqn (1.23), the
wave velocity is

1
v= = (1.35)
 

The velocity of the wave in free space is equal to 3 108 m/s. The distance
between two points that are separated in phase by 2 radians is known as
the wavelength (, unit: metre, m) of the wave. Consider two points r1 and
r2 on the wave with corresponding phases

1 = v + t kr1
2 = v + t kr2

such that

2 1 = k(r1 r2 ) = k = 2 (1.36)

Therefore, the wavelength and the propagation constant are related by

2
k= (1.37)

The velocity can be written in terms of the frequency and the wavelength
of the wave

2f
v= = = f (1.38)
k 2/

EXAMPLE 1.4

The electric eld of an electromagnetic wave propagating in a homogeneous

medium is given by

50
E(x, y, z, t) = a cos(4 106 t 0.063r) V/m
r

2[IRUG8QLYHUVLW\3UHVV
 1.1 Review of Electromagnetic Theory 15

Calculate the frequency, propagation constant, velocity, and the magnetic

eld intensity of the wave if the relative permeability of the medium is equal
to unity.
Solution: The -component of the electric eld can be expressed as
 
50 j(4106 t0.063r)
E = Re e
r

Comparing this with Eqn (1.31), = 4 106 rad/s, hence frequency

of the wave is f = /(2) = 2 MHz, and the propagation constant is
k = 0.063 rad/m. The velocity of the wave is given by v = /k = 4
106 /0.063 = 2 108 m/s.
Expressing the electric eld as a phasor

50 j0.063r
E = a e V/m
r

Substituting this in Maxwells equation, Eqn (1.6), and expressing the curl
in spherical coordinates
 
 a ra r sin a 
 r
1  
jH = E = 2  /r / / 

r sin  

 0 rE 0 

Expanding the determinant

 
1 (rE ) (rE )
E= 2 ar + a r sin
r sin r

Since r and E are not functions of

50
E = a (j0.063)ej0.063
r
Therefore, the magnetic eld is given by

1 0.063 50 0.063r
H= E = a e
j r

Substituting the values of = 4 106 rad/s and = 4 107 H/m

0.2 j0.063r
H = a e A/m
r

2[IRUG8QLYHUVLW\3UHVV
16 Chapter 1 Electromagnetic Radiation

The magnetic eld can also be expressed as a function of time.

0.2
H = a cos(4 106 t 0.063r) A/m
r

Consider a static point charge q kept at a point with position vector r as

shown in Fig. 1.4. The electric potential, V , at a point P (r, , ), with the
position vector r, is given by
q
V (r, , ) = (1.39)
4R
where R is the distance between the charge and the observation point,
R = |R| = |r r | (see Fig. 1.4). We are using two coordinate notations,
the primed coordinates (x , y  , z  ) for the source point and the unprimed
coordinates (x, y, z) or (r, , ) for the eld point.
If there are more than one point charges, the potential is obtained by
the superposition principle, i.e., summing the contributions of all the point
charges. If the source is specied as a charge density distribution over a
volume, the potential at any eld point is obtained by integration over the
source volume. To do this, we rst consider a small volume v  centered
on r . The charge contained in this volume is (r )v  , where (r ) is the

Source
q (x', y', z' )
R
P (x, y, z)

Field
r'

point
r

y
o

x
Fig. 1.4 Position vectors of source and eld points

2[IRUG8QLYHUVLW\3UHVV
 1.1 Review of Electromagnetic Theory 17

volume charge density distribution function. In the limit v  0 we can

consider the charge as a point charge and compute the potential at any eld
point r due to the charge contained in the volume v  using the expression
given in Eqn (1.39).

v 
V (r, , ) = (1.40)
4R

Let us now consider a time-varying charge v  with a sinusoidal time

variation represented by ejt . Heuristically, we can reason out that the eect
on the potential due to a change in the charge would travel to the eld
point with the propagation constant k. Hence for a point charge with an
exponential time variation of the form ejt , the phase fronts are spherical
with the point r as the origin. Therefore

(x , y  , z  )v  ejkR
V (r, , ) = (1.41)
4 R

The potential at point (r, , ) due to a charge distribution (x , y  , z  ) is

obtained by integrating Eqn (1.41) over the source distribution

1 ejkR 
V (r, , ) = (x , y  , z  ) dv (1.42)
4 R
V 

where V  is the volume over which (x , y  , z  ) exists, or the source volume.
The instantaneous value of the scalar potential V(r, , , t) is obtained by

  1  ejkR+jt
V(r, , , t) = Re V (r, , )ejt = Re (x , y  , z  ) dv 
4 R
V
(1.43)
Using the relation v = /k, this reduces to

1  ej(t R
v
)
V(r, , , t) = Re (x , y  , z  ) dv  (1.44)
4 R
V 

It is clear from this expression that the potential at time t is due to the
charge that existed at an earlier time R/v. Or the eect of any change
in the source has travelled with a velocity v to the observation point at a
distance R from the source. Therefore, V is also known as the retarded scalar
potential.

2[IRUG8QLYHUVLW\3UHVV
18 Chapter 1 Electromagnetic Radiation

In Section 1.1.1 it is shown that both, electric scalar potential, V and mag-
netic vector potential, A, satisfy the wave equation with the source terms
being / and J, respectively. Therefore, a similar heuristic argument can
be used to derive the relationship between the current density distribution
J(x , y  , z  ) and the vector potential A(r, , ). Starting from the expression
for the magnetic vector potential for a static current density we introduce
the delay time R/v to obtain the retarded vector potential expression
for the time-varying current density distribution J. The vector potential at
any time t is related to the current density distribution at time (t R/v).
Further, the vector A has the same direction as the current density J.
The relationship between the current density J(x , y  , z  ) and the vector
potential A(r, , ) is given by simply multiplying the static relationship
with the ejkR term. Thus, the retarded vector potential is given by

ejkR 
A(r, , ) = J(x , y  , z  ) dv (1.45)
4 R
V 

If the current density is conned to a surface with surface density Js (in

A/m), the volume integral in the vector potential expression reduces to a
surface integral


ejkR 
A(r, , ) = Js (x , y  , z  ) ds (1.46)
4 R
S

For a line current I (in A), the integral reduces to a line integral


ejkR 
A(r, , ) = I(x , y  , z  ) dl (1.47)
4 R
C

1.1.3 Solution Procedure

The procedure for computing the elds of an antenna requires us to rst
determine the current distribution on the antenna structure and then com-
pute the vector potential, A, using Eqn (1.45). In a source-free region, A is
related to the H eld via Eqn (1.13)

1
H= A (1.48)

2[IRUG8QLYHUVLW\3UHVV
 1.2 Hertzian Dipole 19

and H is related to the E eld by (Eqn (1.7) with J = 0 in a source-free

region)

1
E= H (1.49)
j

As mentioned in Section 1.1, the computation of the current distribution

on the antenna, starting from the excitation, involves solution of an integral
equation and is beyond the scope of this book. Here we assume an approx-
imate current distribution on the antenna structure and proceed with the
computation of the radiation characteristics of the antenna.

1.2 Hertzian Dipole

A Hertzian dipole is an elementary source consisting of a time-harmonic
electric current element of a specied direction and innitesimal length
(IEEE Trans. Antennas and Propagation 1983). Although a single current
element cannot be supported in free space, because of the linearity of
Maxwells equations, one can always represent any arbitrary current
distribution in terms of the current elements of the type that a Hertzian
dipole is made of. If the eld of a current element is known, the eld due to
any current distribution can be computed using a superposition integral or
summing the contributions due to all the current elements comprising the
current distribution. Thus, the Hertzian dipole is the most basic antenna
element and the starting point of antenna analysis.
Consider an innitesimal time-harmonic current element, I = az I0 dl, kept
at the origin with the current ow directed along the z-direction indicated
by the unit vector az (Fig. 1.5). I0 is the current and dl is the elemental
length of the current element. Time variation of the type ejt is implied
in saying the current element is time-harmonic. Consider the relationship
between the current distribution I and the vector potential A, as shown in
Eqn (1.47) and reproduced here for convenience

ejkR 
A(r, , ) = I(x , y  , z  ) dl (1.50)
4 R
C

Since we have an innitesimal current element kept at the origin

x = y  = z  = 0 (1.51)
 
R= (x x )2 + (y y  )2 + (z z  )2 = x2 + y 2 + z 2 = r (1.52)

2[IRUG8QLYHUVLW\3UHVV
20 Chapter 1 Electromagnetic Radiation

az Az
z

a r Ar

a A
P (x, y, z)

I0 dl
y

rs
in

X
Fig. 1.5 Components of the vector potential on the surface of a
sphere of radius r , due to a z -directed current element kept at the
origin

Now, the vector potential due to a current element can be written as

ejkr
A(r, , ) = az I0 dl = az Az (1.53)
4 r
Note that the vector potential has the same vector direction as the current
element. In this case, the az -directed current element produces only the
Az -component of the vector potential.
The H and E elds of a Hertzian dipole are computed using the rela-
tionships given by Eqns (1.48) and (1.49), respectively. The E and H elds
are generally computed in spherical coordinates for the following reasons
(a) the (ejkr/r) term indicates that the elds consist of outgoing spheri-
cal waves which are simple to represent mathematically in spherical coor-
dinates and (b) the spherical coordinate system allows easy visualization
of the behaviour of the elds as a function of direction and simplies the
mathematical representation of the radiated elds.

2[IRUG8QLYHUVLW\3UHVV
 1.2 Hertzian Dipole 21

From Fig. 1.5 we can relate the z-component of the vector potential, Az ,
to the components in spherical coordinates Ar , A , and A as

Ar = Az cos
A = Az sin (1.54)
A = 0

Taking the curl of A in spherical coordinates

 
 ar ra r sin a 

1  
A= 2  /r / /  (1.55)

r sin  

 Ar rA r sin A 

Substituting the components of A from Eqn (1.54) into Eqn (1.55), we get
 
 ar ra r sin a 
 
1  
A= 2  /r / /  (1.56)

r sin  
 Az cos rAz sin 0 

Since Az is a function of r alone, its derivatives with respect to and are

zero. Hence, the curl equation reduces to
 
1
A = a (rAz sin ) (Az cos ) (1.57)
r r

Substituting the expression for Az and performing the indicated dierenti-

ations in Eqn (1.57)

1
A = a Az sin (jkr + 1) (1.58)
r

Substituting the result in Eqn (1.48) and simplifying, we get the expres-
sions for the components of the H eld of a Hertzian dipole in spherical
coordinates as

Hr = 0 (1.59)
H = 0 (1.60)
 
I0 dl sin ejkr 1
H = jk 1+ (1.61)
4 r jkr

2[IRUG8QLYHUVLW\3UHVV
22 Chapter 1 Electromagnetic Radiation

The electric eld can be obtained from Maxwells curl equation. Substituting
the expression for H in Eqn (1.49) we get
 
 a ra r sin a 
 r 
1 1 1  
E= H=  /r / /  (1.62)
j 
j r2 sin  

 0 0 r sin H 

Expanding the determinant the equation reduces to

 
1 1
E= 2
ar (r sin H ) ra (r sin H ) (1.63)
j r sin r

After performing the indicated derivative operations, Eqn (1.63) can be sim-
plied to give the electric eld components of a Hertzian dipole in spherical
coordinates as
 
I0 dl cos ejkr 1
Er = 1+ (1.64)
2r r jkr
 
kI0 dl sin ejkr 1 1
E = j 1+ (1.65)
4 r jkr (kr)2
E = 0 (1.66)

where = k/() is the intrinsic impedance of the medium.

EXAMPLE 1.5

Show that = k/().

Solution: Substituting k = 2/ and = 2f and simplifying

k 2/ 1
= =
 (2f ) (f )

The velocity of the wave, v, is related to the frequency, f , and the wave-
length, , by v = f , and v is related to the permittivity, , and permeability,

, of the medium by v = 1/ . Substituting these in the above equation
and simplifying

k 1 
= = =
 v  

2[IRUG8QLYHUVLW\3UHVV
 1.2 Hertzian Dipole 23

The impedance of the medium is related to  and by = / and,
therefore
k
=


It is interesting to note that a z-directed current element kept at the origin

has only the H , Er , and E components and, further, the elds have com-
ponents that decay as 1/r, 1/r2 , and 1/r3 , away from the current element.
Thus, these expressions form a convenient basis for classifying the elds of
any antenna depending on the nature of decay away from the antenna.
To understand the nature of the eld behaviour as a function of r,
Eqns (1.64) and (1.65) can be re-written as
 
k 2 I0 dl cos jkr 1 1
Er = e 2
+ (1.67)
2 (kr) j(kr)3
 
k 2 I0 dl sin jkr 1 1 1
E = j e + 2
(1.68)
4 kr j(kr) (kr)3

A plot of 1/(kr), 1/(kr)2 , and 1/(kr)3 as functions of (kr) is shown in

Fig. 1.6. For large values of r, i.e., r  or kr  1, the terms contain-
ing 1/(kr)2 and 1/(kr)3 decay much faster than 1/(kr). Therefore, at large

6
10

4
10

2
10
Amplitude

0
10
1/(kr)

2 1/(kr)2
10

1/(kr)3
4
10

6
10 2 1 0 1 2
10 10 10 10 10
kr
Fig. 1.6 Dependence of 1/(kr), 1/(kr)2 , and 1/(kr)3 on (kr)

2[IRUG8QLYHUVLW\3UHVV
24 Chapter 1 Electromagnetic Radiation

distances from the Hertzian dipole, only the terms containing 1/r are re-
tained in the electric and magnetic eld expressions. The electric and the
magnetic elds of a z-directed Hertzian dipole for r  are given by

kI0 dl sin ejkr

E = j (1.69)
4 r
kI0 dl sin ejkr
H = j (1.70)
4 r

The ratio of E to H is equal to the impedance of the medium.

EXAMPLE 1.6

Calculate and compare the r and components of the electric eld inten-
sities at x = 100 m, y = 100 m, and z = 100 m produced by a Hertzian
dipole of length dl = 1 m kept at the origin, oriented along the z-axis,
excited by a current of i(t) = 1 cos(10 106 t) A, and radiating into free
space.
Solution: The frequency of excitation is = 10 106 rad/s, and therefore
f = 5 106 Hz. The dipole is radiating in free space with parameters =
4 107 H/m and  = 8.854 1012 F/m. Therefore, the impedance of
the medium is
 
4 107
= = = 376.73
 8.854 1012

and the propagation constant is



k =  = 10 10 6
4 107 8.854 1012 = 0.1047 rad/m

The distance r between the eld point and the dipole (which is at the
origin) is
 
r= x2 + y 2 + z 2 = 1002 + 1002 + 1002 = 173.2 m

Using the relation z = r cos , we can compute value of as

   
z 100
= cos1 = cos1 = 54.73
r 173.2

2[IRUG8QLYHUVLW\3UHVV
 1.2 Hertzian Dipole 25

Substituting these values in Eqn (1.64)

 
I0 dl cos ejkr 1
Er = 1+
2r r jkr
 
1 1 cos 54.73 ej0.1047173.2 1
= 376.73 1+
2 173.2 173.2 j0.1047 173.2
= 1.154 103 (1 j0.055)ej18.14
= 1.154 103 1.002 3.148 1 40.65
= 1.156 103  37.51 V/m

The -component of the electric eld is evaluated using Eqn (1.65)

 
kI0 dl sin ejkr 1 1
E = j 1+
4 r jkr (kr)2
 
0.1047 1 1 sin 54.73 ej18.14 1 1
= j376.73 1+
4 173.2 j18.14 18.142
= j0.0148ej18.14 (1 j0.055 3.04 103 )
= 1 90 0.0148 1 40.65 0.9985 3.158
= 0.0148 127.49 V/m

The wavelength of the EM wave is 60 m and, therefore, at a distance of

2.88 the -component of the electric eld is more than 10 times greater
than the r-component.

EXAMPLE 1.7

A vector A can be represented in rectangular coordinate system as A =

ax Ax + ay Ay + az Az and in spherical coordinates as A = ar Ar + a A +
a A . Express Ax , Ay , and Az in terms of Ar , A , and A and vice versa.
Solution: The position vector of any point r in the rectangular coordinate
system is given by
r = ax x + ay y + az z
From Fig. 1.5

x = r sin cos
y = r sin sin
z = r cos

2[IRUG8QLYHUVLW\3UHVV
26 Chapter 1 Electromagnetic Radiation

Therefore, the position vector can be written as

r = ax r sin cos + ay r sin sin + az r cos

At any point P (r, , ), ar , a , and a denote the unit vectors along the
r, , and directions, respectively. The unit vector along the r-direction is
given by
r
ar =
|r|

where |r| = (r sin cos )2 + (r sin sin )2 + (r cos )2 = r and hence

ar = ax sin cos + ay sin sin + az cos

The unit vector a is tangential to the -direction. The tangent to the

-direction is given by r/. Therefore, the unit vector along the -direction
is given by
r/
a = = ax cos cos + ay cos sin az sin
|r/|
Similarly, a can be written as

r/
a = = ax sin + ay cos
|r/|
This transformation from rectangular to spherical coordinates can be ex-
pressed in a matrix form as

ar sin cos sin sin cos ax

a = cos cos cos sin sin ay

a sin cos 0 az

Equating the rectangular and spherical coordinate representations of A

A = ax + ay + az = ar Ar + a A + a A

Substituting the expressions for ar , a , and a in terms of ax , ay , and az

ax + ay + az = (ax sin cos + ay sin sin + az cos )Ar

+ (ax cos cos + ay cos sin az sin )A
+ (ax sin + ay cos )A

2[IRUG8QLYHUVLW\3UHVV
 1.2 Hertzian Dipole 27

This can be rearranged as

ax Ax + ay Ay + az Az = ax (sin cos Ar + cos cos A sin A )

+ ay (sin sin Ar + cos sin A + cos A )
+ az (cos Ar sin A )

Equating the coecients of ax , ay , and az on both sides, we get a relationship

between (Ar , A , A ) and (Ax , Ay , Az ). This can be represented in matrix
form as

Ax sin cos cos cos sin Ar

Ay = sin sin cos sin cos
A
Az cos sin 0 A

The 3 3 matrix is known as the transformation matrix. Let us use the

symbol X to represent the transformation matrix.
The components of A in spherical coordinates can be written in terms of
its components in rectangular coordinates by pre-multiplying both the sides
of the above equation by the inverse of the transformation matrix.
1
Ar sin cos cos cos sin Ax

A = sin sin cos sin cos (1.7.1)
Ay
A cos sin 0 Az

Ax

1

=X
Ay
Az

The inverse of the transformation matrix is given by

      T
 cos sin cos   sin sin cos   sin sin cos sin 
     
     
 sin 0   cos 0   cos
sin 


     
   sin cos sin   sin cos cos cos 
1  cos cos sin     
X1 =      
 sin 0   cos 0   cos sin 


     
 cos cos sin   sin cos sin   sin cos cos cos 
     
     
 cos sin cos   sin sin cos   sin sin cos sin 

2[IRUG8QLYHUVLW\3UHVV
28 Chapter 1 Electromagnetic Radiation

where is the determinant of the transformation matrix and is equal to

unity. On simplifying
T
sin cos cos cos sin

X1 =
sin sin cos sin cos
cos sin 0

On taking the transpose of the matrix and substituting in Eqn (1.7.1)

Ar sin cos sin sin cos Ax

A = cos cos cos sin sin Ay

A sin cos 0 Az

The transformation matrix has the unitary property, i.e., X1 = XT . Using

this property we can transform the unit vectors in spherical coordinates into
unit vectors in rectangular coodrinates as

ax sin cos cos cos sin ar

ay = sin sin cos sin cos
a
az cos sin 0 a

EXAMPLE 1.8

Derive the expressions for the elds of a current element I0 dl kept at the
origin, oriented along the x-axis, and radiating into free space.
Solution: Since the current element is oriented along the x-direction, the
magnetic vector potential has only the x-component. Following the proce-
dure given in Section 1.2, we can write the magnetic vector potential as
0 ejkr
A = ax I0 dl
4 r
The unit vector ax can be written in terms of the unit vectors of the spherical
coordinates (see Example 1.7)

0 ejkr
A= I0 dl (ar sin cos + a cos cos a sin )
4 r
The magnetic eld is given by
 
 a ra r sin a 
 r
1 1  
H= A= 2  /r / / 

r sin  

 Ar rA r sin A 

2[IRUG8QLYHUVLW\3UHVV
 1.2 Hertzian Dipole 29

Expanding the determinant

     
1 I0 dl ejkr ejkr
H= 2 ar r sin sin r cos cos
r sin 4 r r
    
ejkr ejkr
a r r sin sin sin cos
r r r
     
ejkr ejkr
+ a r sin r cos cos sin cos
r r r

Performing the indicated dierentiations and simplifying

 
kI0 dl ejkr 1
H = j sin 1+
4 r jkr
 
kI0 dl ejkr 1
H = j cos cos 1+
4 r jkr

The electric eld can be calculated from Maxwells equation jE =

H
 
 a ra r sin a 
 r
1 1  
E=  /r / / 
2 
j r sin  

 0 rH r sin H 
On expanding the determinant, performing the indicated dierentiations,
and simplifying
 
I0 dl ejkr 1
Er = sin cos 1+
2r r jkr
 
kI0 dl ejkr 1 1
E = j cos cos 1+
4 r jkr (kr)2
 
kI0 dl ejkr 1 1
E = j sin 1+
4 r jkr (kr)2

So far we have learnt how to compute the elds due to a current distri-
bution using the vector potential approach. Every antenna can be looked
at as a current distribution producing electric and magnetic elds in the
surrounding space and, therefore, we have learnt the basics of computing the
elds of an antenna. In the following chapter we will learn about properties
of antennas and introduce the various terms associated with antennas.

2[IRUG8QLYHUVLW\3UHVV
 30 Chapter 1 Electromagnetic Radiation

Exercises
1.1 Prove that the spherical coordinate system 1.12 Show that at large distances from a radi-
is orthogonal. ating Hertzian dipole (r  ), the elec-
1.2 Show that for any twice dierentiable tric and magnetic elds satisfy Maxwells
scalar function, , = 0. equations.
1.3 Show that for any twice dierentiable vec- 1.13 A z -directed Hertzian dipole placed at
tor function A, A = 0. the origin has length dl = 1 m and is ex-
cited by a sinusoidal current of amplitude
1.4 Prove the vector identity A =
( A) 2 A. I0 = 10 A and frequency 1 MHz. If the
dipole is radiating into free space, calcu-
1.5 In a source-free region show that the E late the distance in the x-y plane from
and H elds satisfy 2 E + k 2 E = 0 and the antenna beyond which the magnitude
2 H + k 2 H = 0, respectively. of the electric eld strength is less than
1.6 Show that V (r) = V0 ejkr/r , where V0 is 1 103 V/m.
a complex constant and k is a real number, Answer: 6279 m
represents a wave travelling in the positive 1.14 Derive an expression for the elds of a
r-direction. Hertzian dipole of length dl carrying a
1.7 Plot the equiphase surfaces of the electric current of I0 which is located at the ori-
eld of an EM wave given by (a) E = gin of the coordinate system and oriented
a E0 ejkr/r and (b) E = ay E0 ejkx , along the y -axis.
where E0 is a complex constant. 1.15 Find the strength of the z -component of
1.8 Derive Eqns (1.59)(1.61) from Eqn (1.57). the electric eld at (0, 100 m, 0) produced
1.9 Derive Eqns (1.64)(1.66) from Eqn (1.63). by a z -directed Hertzian dipole of length
dl = 0.5 m, placed at the origin, carrying
1.10 For the Hertzian dipole considered in Sec-
a current of i(t) = 2 cos(6 106 t) A,
tion 1.2, compute the electric and magnetic
and radiating into free space. If the dipole
elds in the rectangular coordinate system
is oriented along the x-axis, what will be
directly by taking the curl of az Az . Now
strength of the x-component of the elec-
convert the elds into spherical coordinates
tric eld at the same point?
and compare them with the results given in
Answer: Ez = 0.01856 99.3 V/m;
Eqns (1.59)(1.61) and Eqns (1.64)(1.66).
Ex = 0.01856 99.3 V/m
1.11 Show that = k , where the symbols
have their usual meaning.

2[IRUG8QLYHUVLW\3UHVV