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Antenna and Wave Propagation

MirMuhammad Lodro, M.Res., M.E.

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[1] . 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

3.5 Radiation patterns of broadside array, intermediate array and end

ﬁrearray[2]............................... 72

4.1 Rhombic antenna geometrical structure and its radiation pattern[2] 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 EﬀectiveArea.......................... 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-ﬁreArray.......................... 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 Radio Wave Propagation 90

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

5.17FaradayRotation............................104

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

A−B= (Ax−Bx)ax+ (Ay−By)ay+ (Az−Bz)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[AyBz−AzBy] + ay[AzBx−AxBz] + az[AxBy−AyBx]

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

A plane φ=αradian

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 x−axis 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+y20≤ρ < ∞

φ= tan−1y

x,0≤φ < 2π

z=z0≤z < ∞

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

a cone θ=α(constant) radian

a plane φ=β(constant)radian

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+z20≤r≤ ∞

θ= cos−1z

px2+y2+z20≤θ≤π

φ= tan−1y

x0≤φ≤2π

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= tan−1r

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.

1.3.1 Gradient of Scalar V

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)

Faraday’s Law∇ × E=−∂B

∂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

is zero σ= 0. Therefore, above equations in diﬀerential form can adapted for free

space as follows:

Gauss’s Law∇ · B= 0 (2.4a)

Gauss’s Law∇ · D= 0 (2.4b)

Faraday’s Law∇ × E=−∂B

∂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=B0ejωt

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=−jωµH(2.7c)

∇ × H= (σ+jωε)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=−jωµ Zs

H·ds(2.8c)

IH·dl= (σ+jωε)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=−jωµH(2.9c)

∇ × H= (σ+jωε)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

∂t2∵D=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(jω)2E

(2.19)

∇2E=−ω2µ0ε0E(2.20)

Similarly, homogenous vector wave equation in complex time-harmonic form

for free space in H-ﬁeld is as follows:

∇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×

10−7A/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(z−ct)

| {z }

Term-I

+f2(z+ct)

| {z }

Term-II

(2.31)

In eq. 2.31 f1(z−ct) and f2(z+ct) are any two functions and may assume any

form like asin(βz −ct) and ae(βz−ct). 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(z−ct) (2.32)

22

Now putting general solution in trigonometric form and trying with particular

function

Ey= sin β(z−mt) + 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(t−βz

ω)

=E0sin ω(t−2π

λ·z

2πf )

(2.39)

Ey=E0sin ω(t−z

c) (2.40)

Ey=E0sin ω(t−tp) (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(t−tp) (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+ε∂E

∂t (2.44)

From eq. 2.43c

∇ × E=−µ∂H

∂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 jω. The wave equation

phasor form:

∇2E=µ0ε0

∂2E

∂t2(2.52)

∇2E=µ0ε0(jω)(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=A1e−jβz +A2ejβz (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(A1e−jβz +A2ej βz )ejωto

=<nA1ej(ωt−βz )+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 (jω)2

∇2E=µσ(jω)E+µε(jω)(jω)E

∇2E=jωµ(σ+jωε)E(2.62)

∇2E=γ2E

∇2H=γ2H(2.63)

27

Where γ2=jωµ(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

2πω∵ω= 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) = E0e−γz such that

˜

E(z, t) = <E0e−γ zejωt =<E0ejωte−γz

=<E0e(α+jβ)zejωt =<E0e−αzej(ωt−βz )(2.69)

˜

E(z, t) = e−αz <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+jωεE

(2.71)

Ratio of conduction current density to displacement current density is given by:

Jc

JD=|σE|

|jωεE|

=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= tan−1σ

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

εrad/m

•phase velocity:νp=ν01−σ2

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+jβ

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=E0e−γz

=E0e{(1+j)√ωµσ

2}z(2.76)

Ey=E0e√ωµσ

2z·e−j(√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.

e−αz =eαδe−1= 1/e (2.82)

Which implies that αδ = 1 and α= 1/δ. E-ﬁeld wave in y-plane when EM wave

is propagating in z-direction can be given as follows:

Ey=E0e−γz =E0e−(α+jβ)z

=E0e−αz ·e−jβz (2.83)

From our previous discussions we know that α=βin conductors. Therefore,

Ey=E0e−z/δ ·e−z/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=E0e−z/δ =E0e−z/z ∵z=δ

=E0e−1= (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 =µ∂Hy

∂t

∂

∂z (E0sin(ω−βz)) = µ∂Hy

∂t

E0cos(ωt −βz)(−β) = µ∂Hy

∂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[1]

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(108t−βz)ayA/m and E= 377 cos(108t−βz)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

λ= 0.3329 rad/m

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+jβ

=β∵For Free Space α= 0

We know that phase shift constant βcan be given as follows:

β=ω√µε for free space β=ω√µ0ε0

OR

β=2π

λ

γ=jβ

=jω√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

−40 rad/m

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

= 120πΩ for free space

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= 60πΩ

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:

For free space propagation constant is given by:

γ=jβ

γ=jω√µε =jωp9ε0µ0

γ=j2π×0.3×109

3×108

γ=j2π(1/m)

42

Intrinsic impedance η

η=rjωµ

σ+ωε =rjωµ

ωε =pµ09ε0

η=1

3×120π

η= 40πΩ

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= 10−2V/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= 10−2, 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 ×10−41

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 ×10−12 = 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

2= 117.2rad/m

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 ×106−12 = 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×10−7×5.8×107

δ= 8.53 ×10−3m

At f= 100 M Hz, the ratio

σ

ωε =5.8×107

2π×8.85 ×10−4= 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×10−7×5.8×107

δ= 6.608 ×10−6m

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 ×10−3×942 ×10−3

377

Pav = 2353 µW/m2

(b) Maximum energy density of the wave

WE=1

2εE2=ε0εrE2= 4 ×8.854 ×10−12 ×942 ×942 ×10−6

WE= 31.42 ×106×10−8

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 ×10−12 ×1×1002×π

WE= 13.9×10−8= 139 ×10−9

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 ×10−3V /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×10−7×2π×2×106

5.8×106

η=4π

√5.8×10−4=4π

2.40 ×10−4

η= 5.235 ×10−4Ω

Average Power

Pav =1

2

E2

|η|=0.5×4×10−6

5.235 ×10−4= 0.382 ×10−2W/m2

Pav = 3.82 mW/m2

48

Chapter 3

EM Radiation and Antennas

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 ωt−r

c(3.1)

For uniform plane wave travelling in +ve z direction sin(ωt −βz) but now

sin(ωt−βr) or sin ωt−r

cindicates travelling of spherical waves in radial direction.

Thus retarded current and retarded density in exponential form may be written

as:

[I] = Imejω(t−r

c)=Imej(ωt−βz )Amp (3.2)

[J] = Jmejω(t−r

c)=Imej(ωt−βr)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

Jmejω(t−r

c)dv (3.5)

[A] = µ

4πZv

J(t−r

c)

rdv (3.6)

50

For sinusoidal current element, retarded vector potential is

[A] = µ

4πZJ(t−r

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(t−r

c)

rdl

[A] = µ

4πZImsin ω(t−r

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

ρ0ejω(t−r

c)

rdv

(3.9)

Where [v] is retarded scalar potential and [ρ0] = ρ0ejω(t−r

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

receiver on receiver side.

•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

3.6.2 Radiation Resistance

Radiation resistance Rris a ﬁctitious or hypothetical resistance that would dissi-

pate an amount of power equal to radiate power. Mathematically

Rr=Power radiated

I2

rms

(3.12)

3.6.3 Directional Characteristics

These are called radiation characteristics or directional patterns. There are two

types of radiation patterns.

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

pattern. Antenna radiation pattern is three directional variation of the radiation

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

3.6.5 Radiation Intensity

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

radiation intensity. Mathematically,

Gd=RI

RIav

=RI

Wr/4π

Gd=4π(RI)

Wr

(3.16)

Where Wris radiated power.

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(XL−XC) WhereXL=ωL , XC=1

ωC (3.22)

The correspondent admittance

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

G=Power radiate by current element

Power radiated by isotropic antenna (3.25)

G=Prad

Wi/4πr2= 4πr2Prad

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

Pr(rad) = θImdl2

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

radiated by isotropic antenna. Mathematically

G=Power radiated by half-wavelength antenna

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= 900∵cos2θ= 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

2−cos βl

2

sin θ#(3.42)

The normalised radiation ﬁeld is

En="cos βl cos θ

2−cos 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 θ−cos β l

2

sin θ

En=

cos 2π

λλ/4 cos θ−cos 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 θ−cos 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

rA−rB

φArray Axis

Figure 3.4: Two-element antenna array diagram

rA−rB=dcos φ∵cos φ=rA−rB

d

rB=rA−dcos φ

(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. RA≈RB. 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 + ejψ !(3.50)

Magnitude of total ﬁeld strength when IA=IBis

|ER|=

EA 1 + ejψ !

=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 E1ejψ and the ﬁeld of antenna 3 is E1ej2ψ.

Similarly, the ﬁeld of antenna Nis

E1ej(N−1)ψ(3.56)

The total ﬁeld is vector sum of ﬁelds

ER=E1+E1ejψ +E1ej2ψ+·+E1ej(N−1)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 ejψ, we get

ERejψ =E1ejψ +ej2ψ+· ·· +ejN ψ (3.58)

Now subtracting eq. 5.16 from eq. 3.57

ERejψ =E1ejN ψ −1

ER

E1

=ejN ψ −1

ejψ −1

(3.59)

67

The normalised magnitude of ﬁeld strength

ER

E1=ejN ψ −1

ejψ −1

(3.60)

Now

ejψ −1=cos ψ+jsin ψ−1

=qcos ψ−12+ sin ψ

=p1 + cos ψ−2 cos ψ+ sin ψ

=q21−cos ψ

=√2p1−cos ψ

= 2 sin ψ

2sin ψ

2= 1 −cos ψ

(3.61)

Similarly, ejNψ −1=√2 sin N ψ

2(3.62)

From eq. 3.60

E=ejψ −1

ejψ −1

=

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 = ±kπ 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