Antennas and Wave Propagation

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

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

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

Introduction :
 The electric charges are the sources of the
electromagnetic (EM) fields.
 When these sources are time varying, the EM waves
propagate away from the sources and radiation takes
place.
 Radiation can be considered as a process of
transmitting electric energy.
 The radiation of the waves into space is effectively
achieved with the help of conducting or dielectric
structures called antennas or radiators.

 An antenna is a means of radiating or receiving the
EM waves.
 An antenna may be used for either transmitting or
receiving EM energy.
 In the design of antenna systems, we must consider
important requirements such as the antenna pattern,
the total power radiated, the input impedance of the
radiator, the radiation efficiency etc.
 The direct solution for these requirements can be
obtained by solving Maxwell's equations with
appropriate boundary conditions of the radiator and at
infinity.

Potential Functions and Electromagnetic Fields :
 In case of the electrostatic field or the steady magnetic
field, the electric field or the magnetic field can be
obtained easily by setting the potentials in terms of the
charges or currents.
 The potentials are obtained in terms of the charges or
currents and the electric or magnetic fields are
obtained from these potentials.
 For obtaining the potentials for the electromagnetic
field there are different approaches

 In the first approach, using trial and error, the
potentials for the electric and magnetic field are
generalized.
 Then it is shown that these potentials satisfy the
Maxwell's equations.
 This approach is called heuristic approach.
 The second approach is to start with the Maxwell's
equations and then derive the differential equations
that the potentials satisfy.
 The third approach is to obtain directly the solutions
of the derived equations for the potentials.

Retarded Potentials :
Heuristic Approach :
 Consider a uniform volume charge
density ρv over the given volume
as shown in Fig.
 Consider the differential volume at a point distance r
from origin, where the charge density is ρv(r’)
 Then the Scalar electric potential V at point P can be
expressed in terms of a static charge distribution as
…… 1

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

 Then the fundamental magnetic field can be obtained
by finding the curl of the vector magnetic potential A
as
…… 4
 The potentials V(r) and A(r) represent the potentials
for the static electric and Magnetic fields respectively
where the charge and current distributions do not vary
with time.
 But the charge and current distributions producing the
electromagnetic field vary with time.

 Thus the time varying fields, the potentials are given
by
Where R = |r – r’ |
In general, the velocity of the electromagnetic
field can be described in terms of μ and ε as
v= 1/√ με .

 Then the potentials at the same point with at a finite
velocity. Hence it is necessary to accomplish this finite
propagation time, the equations can be expressed by
introducing a time delay R/v as
 The potentials are delayed or retarded by time R/v,
Hence the potentials are called retarded potentials

Maxwell’s Equation Approach :
 In this approach, starting from Maxwell’s equations,
the differential equations are derived.
 For time varying fields ,Maxwell’s equations are given
by

 From Equation.12, It is clear that the divergence of H
is zero.
 But from the vector identity the divergence of a Curl of
a vector is zero.
 This clearly indicates that to satisfy, H must be
expressed as a Curl of vector potential A as
 Now, putting the value of μH in the Equation 9, We
get,

 Interchanging the operators on RHS of the above
equation, We get,
 From vector identity Curl of a gradient of scalar is
always zero, So equation .14 will be satisfied only if
the term is defined as gradient of scalar.

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

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

 Now, Substitute E from Eq. 16 into Eq .11, We get
 The equations 18 and 19 are called Coupled
equations.
 These are not sufficient enough to define A and E
completely. Using Helmholtz Theorem we get the
unique solution.

 The Helmholtz theorem states that Any vector field
can defined uniquely if the curl and divergence of the
field both are known at any point.
 Now we may choose divergence of A from equation 18
as
 Thus selection divergence of A as given in equations
18,19,20 become un coupled.
 The relationship between the divergence of A and V is
know as Lorentz gauge condition or Lorentz gauge for
potentials.

Using this condition in equation 18 ,We get

 Similarly using Lorentz gauge condition in equation
19, We get,

 The equations 21 and 22 are the standard wave
equations including source terms.
 The solutions of the equations 21 and 22 are given by

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

 Similarly equation 2 can be written as
 Similarly, for the wave travelling in R direction with
phase variation represented by e-jβR. So the potential
functions for sinusoidal oscillations are given by

Radiation from Alternating Current Element :
 To calculate the electromagnetic field radiation from
short dipole, the retarded potential is used.
 A short dipole is an alternating current element. It is
also called an oscillating current element.
 An alternating current element is consider as the basic
source of radiation.
 In generally, a current element IdL is nothing but an
element length dL carrying current I.
 This length of wire is assume to be very short.

 To calculate the electromagnetic field due to alternating
current element, consider Spherical co-ordinate system
and the current through is IdL cosωt , locate at centre
is shown in fig.
 The electromagnetic is calculate at a point P placed at a
distance R from origin.
 The element is IdL cosωt is place along the z-axis.

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

 From the equation 2, it is clear that the vector potential
Az can be obtained by integrating the current density J
over the volume.
• Now , the current is assumed to be constant along the
length dL, the integration of J over the length dL gives
value IdL .
• Thus mathematically we can write,

 Substitute the integration of equation 3 in equation2,
the vector potential in z-axis is given by
 Now , the magnetic field is given by
 Using spherical co-ordinate system, to find curl A in r,
θ, ɸ direction, we have

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

 Now ,Putting values of Ar and Aθ from equation 5,
We get
 Substitute the Az value in above equation, We have,

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

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

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

 Let us calculate each term of separately
 The component in ar direction is given by

 Now substitute the value Hɸ from equation12,
We have

 Therefore
 Let us calculate the component in aθ
direction

 Now substitute Hɸ from equation11, We have

 Now , Calculate the electric field components, From
equation 13, E in ar direction is given by

 Significance of Field components :

Power Radiated from Current Element :
 Consider a current element placed at a centre of a
spherical co-ornate systems.
 Then the power radiated per unit area at point P can
be calculated by using Poynting theorem .
 The components of the Poynting vector are given by
Pr = Eθ Hɸ
Pθ = - Er Hɸ
Pɸ = Eθ Hr
 But we know that, the current element is at orgin, then
the Eɸ = 0

 Let the field components of Er , Eθ , Hɸ and replacing
v by c are given by

 The Poynting vector in θ direction is given by

 Now calculate Radial power in r direction

 Therefore the total power radiated by current element
can be obtained by integrating the radial pointing
vector over a spherical surface.
 The element area ds is given by
ds = 2πr2sinθ dθ
 The total power radiated is given by

 In Spherical coordinate system θ varies from 0 to π.
Hence

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

 Therefore the power radiated from the current
element is given by
 Generally the power is expressed as
P = I2 R
 So, the Radiation resistance of current element is given
by

Short Linear Antenna:
 The current element that we have considered previously
is not a practical, but it is hypothetical.
 It is useful in the theoretical calculations such as the
components of the fields, radiation of power etc.
 The practical example of the centre-fed antenna is an
elementary dipole.
 The length of such centre-fed antenna is very short in
wavelength.

 The current amplitude on such antenna is maximum
at the center and it decreases uniformly to zero at the
ends.
 If we consider same current I flowing through the
hypothetical current element and the practical short
dipole, both of same length.
 Then the practical short dipole radiate only one-
quarter of the power that is radiated by the current
element.

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

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

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

Half wave dipole and The Monopole :
 A very commonly used antenna is the half wave dipole
with a length one half of the free space wavelength of
the radiated wave.
 It is found the linear current distribution is not suitable
for this antenna.
 But when such antenna is fed at its centre with the
help of a transmission line,
 It gives a current distribution which is approximately
sinusoidal, with maximum at the centre and zero at the
ends.

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

Power Radiated by the Half Wave Dipole and Monopole
 A dipole antenna is a vertical radiator fed in the centre.
It produces maximum is the overall length.
 The vertical antenna of height H =L/2 produces the
radiation characteristics above the plane which is similar
to that produced by the dipole antenna of length L =
2H.
 The vertical antenna is referred as a monopole.
 The practically used antennas are half wave dipole
(λ / 2) and quarter wave monopole (λ / 4).

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

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

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

 Now substitute I value in the above equation, we get
 We have certain assumptions to calculate radiation
field, assume R = r, replace R in denominator only
with r and
 R = r - z cosθ in numerator term .

 Integrating w.r.t z and H = λ/4 , We get ,

 Now finding the LCM, We have,

 Substitute this property in above equation, We get,

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

 Therefore
 Now substitute Az value, We get,

 The magnitude of magnetic field of Half wave or
quarter wave Monopole is given by

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

 The electric field component is
 The magnitude of Electric field of Half wave or
quarter wave Monopole is given by

 The maximum power interms of effective current is
given by

 Now, the average value of power is given by

 The total radiated power from dipole antenna is

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

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

Antenna Theorems :
 An antenna can be used as both transmitting antenna
and receiving antenna.
 While using so, we may come across a question
whether the properties of the antenna might change as
its operating mode is changed.
 The properties of antenna being unchangeable is
called as the property of reciprocity.

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

Equality of Directional patterns :
 The radiation pattern of transmitting antenna 1, which
transmits to the receiving antenna 2 is equal to the
radiation pattern of antenna 2, if it transmits and
antenna1 receives the signal.
Equality of Directivities :
 Directivity is same for both transmitting and receiving
antennas, if the value of directivity is same for both the
cases.
 The directivities are same whether calculated from
transmitting antenna’s power or receiving antenna’s
power.

Equality of Effective lengths :
 The value of maximum effective aperture is same for
both transmitting and receiving antennas.
 Equality in the lengths of both transmitting and
receiving antennas is maintained according to the value
of the wavelength.
Equality in Antenna Impedances :
 The output impedance of a transmitting antenna and
the input impedance of a receiving antenna are equal
in an effective communication.

Loop Antennas:
 An RF current carrying coil is given a single turn into a
loop, can be used as an antenna called as loop
antenna.
 The currents through this loop antenna will be in
phase.
 The magnetic field will be perpendicular to the whole
loop carrying the current.
 It may be in any shape such as circular, rectangular,
triangular, square or hexagonal according to the
designer’s convenience.

 Loop antennas are of two types.
Large loop antennas
Small loop antennas
 Large loop antennas are also called as resonant antennas.
 They have high radiation efficiency. These antennas
have length nearly equal to the intended wavelength(L=λ).
 Small loop antennas are also called as magnetic loop
antennas. These are mostly used as receivers.
 These antennas are of the size of one-tenth of the
wavelength ( L = λ/10 )

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

 Now the far field radiation due to two point sources with
reference to centre O can be represented as
EΦ = Field component + Field component
due to dipole 1,4 due to dipole 2, 3
 The path difference between
from point L to M
Path difference = d cos( 900 – θ )
 Then the path difference is
expressed interms of wavelength as
Path difference = d cos ( 900 – θ ) / λ

 Let ψ is the phase angle and it is relate with path
difference is given by
Phase angle = ψ = 2π x path difference
 The field component for any dipole is given by
Field component = Magnitude x ej( phase angle )
 Let the field component due to dipole 1,4 is given by

 Similarly field component due to dipole 2,3 is
given by
 Hence the far field radiation due to square
loop is given by

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

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

Comparison of far fields of small loop and short dipole :
 A small loop can be consider as equivalent to short
magnetic dipole.
 Thus a small loop of area A carrying current I can be
replaced by a short magnetic dipole length L and
carrying fictitious magnetic current Im

 The magnetic moment of the loop is I . A is the small
loop area, current I is the uniform phase current
throughout loop
 Hence equating this magnetic moment with the
magnetic moment of short dipole is given by qm. L
 Hence we can write,
 The above equation represent the equivalence of
magnetic dipole of length L carrying fictitious current
Im with small loop of area A and carrying current I.

Radiation Resistance of Loop antenna :
 To find the radiation resistance of the loop antenna, the
poynting vector is integrated over large sphere.
 The power radiated is given by
P = I2
rms Rrad = ½ IM Rrad
 The average poynting power
is given by

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

 But πa2 is the area A of loop. Hence the radiation
resistance is given by
 The above expression can be written as

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

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

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

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

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