8. Antennas and Radiating Systems · Antennas and Radiating Systems 1 Electromagnetic Field Theory...

Prof. Rakhesh Singh Kshetrimayum

8. Antennas and Radiating Systems

Prof. Rakhesh Singh Kshetrimayum

4/24/20181 Electromagnetic Field Theory by R. S. Kshetrimayum

8.1 Introduction We use mobile phones everyday

Mobile phone converts our voice into electrical signal using microphone

This signal is modulated and radiated to free space

by antennas as EM waves

4/24/2018Electromagnetic Field Theory by R. S. Kshetrimayum2

by antennas as EM waves

which is picked up by the base station antennas

We generally use transmission line like tv cables

for transferring EM energy from one point to another within a circuit

8.1 Introduction This mode of energy transfer is called guided wave propagation

It basically means that wave inside transmission line like coaxial cable is guided inside it and

will not come out from it into free space


will not come out from it into free space

Hence antenna is also called as mode transformer which

transforms guided-wave field into a radiated wave field for transmitting antenna and vice versa

8.1 Introduction


Fig. Antenna as mode transformer

8.1 Introduction An important property of antenna is its ability to transmit power in a preferred direction like in microwave towers

where we align the transmitting antenna and receiving antenna


antenna

for line of sight (LOS) communication

8.1 Introduction


Fig. Microwave tower: LOS communication

8.1 Introduction Radiation pattern shows how power is radiated from the antenna in 3-dimension

Unlike the previous case, ideally base station (BS) and mobile station (MS) antennas should radiate equally in all directions


directions

as well as they can pick up signals from all directions

Such isotropic antennas do not exist in practice

Omnidirectional directional antennas are used for such cases

8.1 Introduction


Fig. Mobile communication

8.1 Introduction

Antennas

Radiation fundamentals

Antenna pattern and parameters

Types of antennas

Antenna arrays

When does a charge radiate?


Fig. 8.1 Antennas (cover antenna pattern and parameters after types of antennas)

charge radiate?

Wave equation for potential functions

Solution of wave equation for potential functions

Hertz dipole

Dipole antenna

Loop antenna

8.2 Radiation fundamentals8.2.1 When does a charge radiates?

accelerating/decelerating charges or time-varying currents in a conductor radiate EM waves

r


Fig. 8.2 A giant sphere of radius r with a source of EM wave at its origin

Source

8.2 Radiation fundamentals Consider a giant sphere of radius r which encloses the source of EM waves at the origin (Fig. 8.2)

The total power passing out of the spherical surface is given by Poynting theorem,


( ) ( )∫∫ •×=•= ∗sdHEsdSrP avgtotal

rrrrrRe

( )lim

totalP P r

r=

→∝

8.2 Radiation fundamentals This is the energy per unit time that is radiated into infinity and it never comes back to the source

The signature of radiation is irreversible flow of energy away from the source


away from the source Let us analyze the following three cases:CASE 1: A stationary charge will not radiate

no flow of charge =>no current=>no magnetic field=>no radiation (for EM waves we need both E and H)

8.2 Radiation fundamentalsCASE 2: A charge moving with constant velocity will not radiate

The area of the giant sphere of Fig. 8.2 is 4= r2

So for the radiation to occur Poynting vector must decrease no faster than 1/r2

power remains constant in that case


power remains constant in that case

irrespective of the distance from the source

8.2 Radiation fundamentals From Coloumb’s law, electrostatic fields decrease as 1/r2,

whereas Biot Savart’s law also states that magnetic fields decrease as 1/r2

So the total decrease in the Poynting vector is


So the total decrease in the Poynting vector is proportional to 1/r4

Hence power decreases as 1/r2

It dies out after some distance from the source

implies no radiation

8.2 Radiation fundamentalsCASE 3: A time varying current or acceleration (or deceleration) of charge will radiate

To create radiation

there must be a time varying current or


there must be a time varying current or

acceleration (or deceleration) of charge

8.2 Radiation fundamentals

Basic radiation equation:

where L=length of current carrying element, m

dt

dvQ

dt

diL =

dt

diL

dt

dvQ+

Fig. Fundamental law of radiation


L=length of current carrying element, m =time changing current, As-1(units)

Q=charge, C

=acceleration of charge, ms-2

dt

di

dt

dv

8.2 Radiation fundamentals For this case (we will show this later),

a time varying field (both E and H) is produced

which varies as 1/r

whose field direction is along and θ φ


Hence the direction of Poynting vector is radiallyoutward

Since Poynting vector varies as 1/r2, total power is always constant

It can go to infinite distance

8.2 Radiation fundamentals Two conditions for EM waves:

1) Fields produced by the EM source should have components varying as 1/r

2) Field direction should not be radial but transversal so that the power flow or Poynting vector should be radial


the power flow or Poynting vector should be radial

It can be shown that for an infinitesimally small current carrying element (Hertz dipole) which is the building block for antenna, it indeed produces such fields when supplied with time varying currents

8.2 Radiation fundamentals But it is a difficult process to find such fields directly from current density and calculations are highly complex

A major simplification is possible when we find the magnetic vector potential first and


we find the magnetic vector potential first and find the fields from it

It is similar to find electric field from electric potential than directly finding electric field

This way it is easier


Jr

HErr

,

Ar


Fig. Two ways to find fields

In order to find magnetic vector potential,

we need to write down the wave equation for magnetic vector potential first

8.2 Radiation fundamentals8.2.2 Wave equation for potential functions

One of the Maxwell’s divergence equation

Hence, we can write0=•∇ B

r


It means that we can find magnetic flux density

from the curl of magnetic vector potential

ABrr

×∇=


Putting this in the following Maxwell’s curl equation

t

BE

∂

∂−=×∇

rr ( )

t

AE

∂

×∇∂−=×∇⇒

rr


which can be rewritten as

For time varying fields,

t∂ t∂

0A

Et

∂⇒∇× + =

∂

rr

t

AVE

∂

∂−−∇=

rr

8.2 Radiation fundamentals Putting this in the following Maxwell’s divergence equation

ε

ρ=•∇ E

r


ε

ρ−=

∇+

∂

∂•∇ V

t

Ar

2AV

t

ρ

ε

∂⇒∇ • + ∇ = −

∂

r

8.2 Radiation fundamentals Applying Lorentz Gauge condition

Applying above condition

0=∂

∂+•∇

t

VA µεr

∂ ∂ ∂r

r r


22

2

V AA J V

t t tµε µ µε

∂ ∂ ∂ ⇒∇ − − ∇ = − + ∇

∂ ∂ ∂

rr r

22

2

AA J

tµε µ

∂⇒∇ − = −

∂

rr r


Another Maxwell’s curl equation

t

EJB

∂

∂+=×∇

rrr

µεµ


Simplifies to

( )t

Vt

A

JAAA∂

∇+

∂

∂∂

−=∇−•∇∇=×∇×∇

r

rrrrµεµ2

8.2 Radiation fundamentals Applying Lorentz Gauge condition

Applying above condition

0=∂

∂+•∇

t

VA µεr


( )ε

ρ−=∇+

∂

•∇∂V

t

A 2

r

22

2

VV

t

ρµε

ε

∂⇒∇ − = −

∂


From Maxwell’s equations for time varying fields,

we have derived the two wave equations for potential functions

magnetic vector and


magnetic vector and

electric potentials

Why find potential functions instead of fields?

Jt

AA

rr

rµµε −=

∂

∂−∇

2

22

22

2;

VV

t

ρµε

ε

∂∇ − = −

∂

8.2 Radiation fundamentals8.2.3 Solution of wave equation for potential functions

For time harmonic functions of potentials,

JAArrr

µβ −=+∇ 22


where

To solve the above equation, we can apply Green’s function technique

JAA µβ −=+∇

µεωβ =

8.2 Radiation fundamentals8.2.3 Solution of wave equation for potential functions

Green’s function G is the solution of the above equation with the R.H.S equal to a delta function

( )spaceGG δβ =+∇ 22


Once we obtain the Green’s function,

we can obtain the solution for any arbitrary current source by applying the convolution theorem

( )spaceGG δβ =+∇ 22

( ) ( ) ( ) ( )rGrrArJrrrrrr

→→− δµ ;

8.2 Radiation fundamentals Since the medium surrounding the source is linear,

we can obtain the potential for any arbitrary current input

by the convolution of the impulse function (Green’s function) with the input current


function) with the input current

( ) ( ) ( )( ) ( ) '

V

'

rrjβ

''dv

rr

erJ

4π

µrJµrGrA

'

∫−

=−∗=

−−

rrrrrrrr

rr

8.2 Radiation fundamentals Notation:

The prime coordinates denote the source variables

unprimed coordinates denote the observation points

The modulus sign in is to make sure that rr


is positive

since the distance in spherical coordinates is always positive

'rrrr

−

8.2 Radiation fundamentalsDigression:

LTI system

For such system with an impulse response h(t) and input signal x(t),

the output signal is given by y(t)=h(t)*x(t)


the output signal is given by y(t)=h(t)*x(t)

Note that is LTI system,

we consider x(t), h(t) and y(t) are functions of time

In magnetic vector calculation,

are functions of space ( ) ( ) ( )rA,r,GrJµ

rrrrr−


Sl. No. System LTI Magnetic vector potential calculation

Table: Analogy of LTI and Magnetic vector potential calculation


calculation

1 x(t)

2 h(t)

3 y(t)

( )rJµrr

−

( )rGr

( )rArr

8.2 Radiation fundamentals For radiation problems,

the most appropriate coordinate system is spherical since the wave moves out radially in all directions


Fig. An antenna radiating equally in all directions


It has also symmetry along θ and φ directions

0G G

θ φ

∂ ∂= =

∂ ∂


Hence, the above equation reduces to

( )rGr

Gr

rrδβ =+

∂

∂

∂

∂ 22

2

1


Putting Ψ= G r,

For r not equal to 0 (field should not be obtained at the

( )rrr

δβ =Ψ+Ψ∂

∂ 2

2

2


source itself),

Therefore,

02

2

2

=Ψ+Ψ∂

∂β

r

rjrjBeAe

ββ +− +=Ψ

8.2 Radiation fundamentals Since the radiation moves radially in positive r direction negative r direction is not physically feasible for a source of a field, we get,

rjAe

β−=Ψ AeG

rjβ−

=


we can find the constant A and hence

which is magnetic vector potential produced by a delta source

Ae=Ψr

G =

r

eGA

rj

ππ

β

4;

4

1 −

−=−=

8.4 Kinds of antennas8.4.1 Hertz dipole Electrically small antennas are small relative to wavelength

whereas electrically large antennas are large relative to wavelength


wavelength Hertz dipole is not of much practical use

but it is the basic building block of any kind of antennas

An infinitesimally small current element is called a Hertz dipole

8.4 Kinds of antennas8.4.1 Hertz dipole

Let us find the fields of a small current carrying element of length dl

The procedure involves

Determining the current on the antenna


Determining the current on the antenna

Then compute

EHAJrrrr

→→→

8.4 Kinds of antennas The infinitesimal time-varying current in the Hertz dipole is assumed as

where ω is the angular frequency of the current

Since the current is assumed along the z-direction,

( ) zeItItj ˆ0

ω=


Since the current is assumed along the z-direction,

the magnetic vector potential at the observation point P is along z-direction

zr

edleIzAA

tjrj

z ˆ4

ˆ 00

π

µ ωβ−

==r

8.4 Kinds of antennas Note that for this case

For infinitesimally small current element at the origin

( )' '

0

j tJ r dv I dle

ω=r r

'r r r− =r r r


r r r− =

( ) ( ) ( )( ) ( ) '

V

'

rrjβ

''dv

rr

erJ

4π

µrJµrGrA

'

∫−

=−∗=

−−

rrrrrrrr

rr

zr

edleIzAA

tjrj

zˆ

4ˆ 00

π

µ ωβ−

==⇒r

8.4 Kinds of antennas Fig. 8.5 Hertz dipole

located at the origin and

oriented along z-axis

( ), ,P r θ φ

r

θ

φ


0

j tI e

ω

8.4 Kinds of antennas

sin cos sin sin cos sin cos sin sin cos 0

cos cos cos sin sin cos cos cos sin sin 0

sin cos 0 sin cos 0

xr

y

zz

AA

A A

AA A

θ

φ

θ φ θ φ θ θ φ θ φ θ

θ φ θ φ θ θ φ θ φ θ

φ φ φ φ

= − = − − −

r


Coordinate transformation

cos ; sin ; 0r z z

A A A A Aθ φθ θ⇒ = = − =

r

θ


Using the symmetry of the problem (no variation in ),

0

AH

µ

∇×=

rr

Q2

0

ˆ ˆˆ sin

1

sin

sinr

r r r

r r

A rA r Aθ φ

θ θφ

µ θ θ φ

θ

∂ ∂ ∂ = ∂ ∂ ∂


Using the symmetry of the problem (no variation in ), we have,

2

0

ˆ ˆˆ sin

10

sin

cos sin 0z z

r r r

Hr r

A rA

θ θφ

µ θ θ φ

θ θ

∂ ∂ ∂ = ≡ ∂ ∂ ∂

−

r

φ


( ) ( )2

0

sin0; 0; sin cos

sinr z z

rH H H rA A

r rθ φ

θθ θ

µ θ θ

∂ ∂ ⇒ = = = − −

∂ ∂

( )0 0ˆ ˆcos sin

sin sin4 4

j t j tj r j rj r j rI dle I dlee e

H e j er r r r r

ω ωβ ββ β

φ

φ φθ θθ β θ

π θ π

+ +− −− −

∂ ∂ ⇒ = − − = +

∂ ∂


The Hertz dipole has only component of the magnetic field,

i.e., the magnetic field circulates the dipole

0

2

ˆ sin 1

4

j t j rI dle e j

r r

ω βφ θ β

π

+ − = +

φ

8.4 Kinds of antennas The electric field for ( in free space, we don’t have any conduction current flowing) can be obtained as

0J =r

ωεj

HE

rr ×∇

=


Using the symmetry of the problem (no variation in ) like before, we have,

φ

( ) ( )2 2

ˆ ˆˆ sin

1 1 ˆˆ0 sin sinsin sin

0 0 sin

r r r

E r H r r H rj r r j r r

r H

φ φ

φ

θ θφ

θ θ θωε θ θ φ ωε θ θ

θ

∂ ∂ ∂ ∂ ∂ = ≡ = − ∂ ∂ ∂ ∂ ∂

r


Note that Er has only 1/r2 and 1/r3 variation with r

( )20 0

2 2 2 2

1 1sin 2sin cos

sin 4 sin 4

j t j r j t j r

r

I dle e I dle ej jE r r

j r r r j r r r

ω β ω ββ βθ θ θ

ωε θ π θ ωε θ π

− −∂ = + = +

∂

( )

+∂

∂−=

∂

∂−= − rj

tj

er

jrrj

dleIHr

rrj

rE

βω

φθ βπωε

θθ

θωε

1

4

sinsin

sin

0

2


We see that electric field is in the (r, θ) plane

whereas the magnetic field has component only

( )

+

∂−=

∂−= e

rj

rrjHr

rrjE φθ β

πωεθ

θωε 4sin

sin2

2

0

2 3

sin

4

j tj rI dle j j

er r r

ωβθ β β

πε ω ω ω−

= + −

φ

8.4 Kinds of antennas Therefore, the electric field and magnetic field are

perpendicular to each other

Which wave is this?

Points to be noted:


Fields can be classified into three types

Radiation fields (spatial variation 1/r)

Induction fields (spatial variation 1/r2) and

Electrostatic fields (spatial variation 1/r3)

8.4 Kinds of antennas1) Field variation with frequency since

Electrostatic fields (1/r3) are also inversely proportional to the frequency ( )

µεωβ =

β

ω

1


Induction field (1/r2) is independent of frequency ( )

Radiation field (1/r) is proportional to frequency ( )ω

β 2

ω

β

8.4 Kinds of antennas2) Field variation with r

For small values of r,

electrostatic field is the dominant term and

Induction field is the


transition from electrostatic field from radiation fields

For large values of r,

radiation field is the dominant term

8.4 Kinds of antennas Near fields:

We can show that Hertz dipole has reactive near field

2

0

4

sin

r

edleIH

rjtj

nf π

θβω

φ

−

=

− sin jedleIrjtj βθβω


−=−

32

0

4

sin

r

j

r

edleIE

rjtj

nf

β

πεω

θβω

θ

+=−

32

0 1

4

cos2

rr

j

j

edleIE

rjtj

rnf

β

πεω

θβω

8.4 Kinds of antennas It looks like the static magnetic field produced by a current carrying element using Biot Savart’s law

2

0

4

sin

r

edleIH

rjtj

nf π

θβω

φ

−

=


It resembles the electric field produced by an electric dipole

−=−

3

0

4

sin

r

jedleIE

rjtj

nf πεω

θβω

θ

=−

3

0 1

4

cos2

rj

edleIE

rjtj

rnf πεω

θβω

8.4 Kinds of antennas Note that 1/r5 term in Poynting vector is purely reactive

Note that the 1/r4 term in the Poynting vector is

due to induction fields

which will die out after some distance from the source


Far fields:

r

edleIjH

rjtj

ff π

θβ βω

φ4

sin0

−

=

r

edleIjE

rjtj

ff πεω

θβ βω

θ4

sin0

2 −

=

8.4 Kinds of antennas Note that the 1/r2 term in the Poynting vector

gives total power over a giant sphere always constant

Hence such waves from the radiation fields

will go to infinite distance in free space


They also satisfy two conditions for EM waves:

Transversal fields (Eθ and Hφ gives Poynting vector along radial direction)

Fields vary as 1/r (Poynting vector varies as 1/r2)

8.4 Kinds of antennas We can also observe that the three types of fields are equal in magnitude when

β2/r= β/r2=1/r3

=> r=1/β= λ/2=

λ − 2sin jjedleIrjtj ββθβω


For r< λ/2=,

1/r3 term dominates

For r>> λ/2=,

the 1/r term dominates

−+=−

32

2

0

4

sin

r

j

rr

jedleIE

rjtj

r

ββ

πεω

θβω

8.4 Kinds of antennas Near field region:

For r<< λ/2=

in fact the near field region distance r= λ/2= is for D<<λ

for an ideal infinitesimally small Hertz dipole


for an ideal infinitesimally small Hertz dipole

electrostatic fields dominate


as r<< λ/2=

1→− rje

βQ

0

3

cos2 ;

4

j t

r

I dl eE j

r

ωθ

πεω≈ − 0

3

sin

4

j tI dl e

E jr

ω

θ

θ

πεω≈ −


The magnitude of the near field is

34r

rπεω 34E j

rθ

πεω≈ −

2 2 2 20

34cos sin

4r

I dlE E E

rθ θ θ

πεω= + = +

8.4 Kinds of antennas A polar plot of the near field can be generated by writing a MATLAB program for plotting

( ) θθθ 22 sincos4 +=F


Maximum field is along

θ=00, θ=1800 and

minimum is along

θ=900, θ=2700 (see Fig. 8.6(a))


Fig. 8.6 (a) Near field pattern plot of a Hertz dipole located at the origin and oriented along z-axis (maximum radiation along z-axis)

8.4 Kinds of antennas Far field region:

For r>> λ/2= (in between reactive near field and Fraunhofer far field region, there exists the Fresnel near field region that’s why we have chosen an r>> λ/2=), radiation field is the dominant term


radiation field is the dominant term

In other words kr>>1, we have,

2

0 0sin sin;

4 4

j t j r j t j rI dl e e I dl e e

E j H jr r

ω β ω β

θ φ

θ β θ β

πεω π

− −

= =

8.4 Kinds of antennas The electric fields and magnetic fields are in phase with each other

They are 90˚ out of phase with the current

due to the (j) term in the expressions of Eθ and HφIt is interesting to note that the ratio of electric field and


It is interesting to note that the ratio of electric field and magnetic field is constant

E

H

θ

φ

ω µεβ µη

ωε ωε ε= = = =


Hence, the fields have sinusoidal variations with θ

They are zero along θ=0

No radiation along z-axis unlike near field case

They are maximum along θ=π /2



Fig. 8.6 (b) E-plane radiation pattern of a Hertz dipole in far field (H-plane radiation will look like a circle)

8.4 Kinds of antennas Power flow:

( ) * *1 1 ˆ ˆˆRe Re2 2

avg rS E H E r E Hθ ϕθ φ= × = + ×

r r

*1ReS E H r= $

2 3

0 sin1 I dlr

θ β =

$


Antenna power flows radially outward

Power density is not same in all directions

The net real power is only due to

the radiations fields (i.e. jβ2/r and jβ/r) of electric and magnetic fields

*Re2

avgS E H rθ φ= $ 0

2 4r

rπ ωε=

$

8.4 Kinds of antennas Total radiated power:

The total radiated power from a Hertz dipole

2 sinavg

W S r d dθ θ φ= ∫∫


∫∫2

2 2

040dl

W Iπλ

∴ =

8.4 Kinds of antennas Power radiated by the Hertz dipole is proportional to

the square of the dipole length and

inversely proportional to the dipole wavelength

It implies more and more power is radiated as


the frequency and

the length

of the Hertz dipole increases

8.4 Kinds of antennas Radiation resistance of a Hertz Dipole:

Hertz dipole can be equivalently modeled as a radiation resistance

Since W=1/2 I02 Rrad


implies that Rrad = 80π2

2

2 2

040dl

W Iπλ

∴ =

2dl

λ

Fig. Equivalence of Hertz dipole and radiation resistance


Radiation pattern of a Hertz Dipole:

F(θ )=sin θ for a Hertz dipole

The 3D plot of sin θ looks like an apple (see Figure 8.6


(c))


Fig. 8.6 (c) A typical 3-D radiation pattern of a Hertz dipole in the far field

8.4 Kinds of antennas Two principal planes radiation patterns (2-D) are normally plotted

E-plane (vertical cut)

H-plane (horizontal cut) radiations patterns

are sufficient to describe the radiation pattern of a Hertz


are sufficient to describe the radiation pattern of a Hertz dipole

H-plane (xy-plane) radiation pattern is in the form of circle of radius 1 since F(θ, ) is independent of

E-plane (xz-lane) radiation pattern looks like 8 shape

φ φ


00=θ00=φ

090=φ0270=φ


Fig. H-plane and E-plane radiation patterns of Hertz dipole

0270=θ0180=θ

090=θ

0180=φ

8.4 Kinds of antennas To get the 3-D plot from the 2-D plot

you need to rotate the E-plane pattern along the H-plane pattern

For this case it will give the shape of an apple

Note that θ is also known as elevation angle and as φ


Note that θ is also known as elevation angle and as azimuth angle

E-plane pattern for a dipole is also known as elevation pattern

H-plane pattern as azimuthal pattern

φ

8.3 Antenna pattern and parameters

Which parameters are used for specifying an antenna?

What parameters differentiate one antenna from others?

How do one characterize an antenna?

Antenna has two types of characteristics from


Field concepts

Circuit concepts


An antenna can launch free space wave in a desired direction

This directional characteristics of an antenna can be interpreted from its radiation characteristics

We also know that an antenna is connected to a


We also know that an antenna is connected to a transmission line

and it converts guided wave to free space wave

Hence it can be thought as a load to the transmission line

One can find its equivalent circuit (input characteristics)



Fig. Antenna characteristics: (a) Radiation characteristics (b) input characteristics


Radiation characteristics

Radiation pattern

Directivity

Gain

Polarization, etc.


Polarization, etc.


Input characteristics

Input impedance

Bandwidth

Reflection coefficient

Voltage standing wave ratio, etc.


Voltage standing wave ratio, etc.


Radiation characteristics

Radiation intensity:

The most fundamental parameter is its radiation intensity

Based on this parameter,


Based on this parameter,

gain and

directivity

of the antenna will be defined


rdA

dΩ 2

r

dAd =Ω


Fig. Solid angle (Ratio of the area subtended by the solid angle to the squared radius)

2r

d =Ω


Radiation intensity is defined as power crossing per unit solid angle

Power crossing over the area dA is ( )dAS φθ ,

( ) ( ) ( ) ( )SrWrSdAS

U /,,

, 2φθφθ

φθ ==


Radiation intensity can be calculated by

multiplying the Poynting vector by r2

( ) ( ) ( ) ( )SrWrSd

dASU /,

,, 2φθ

φθφθ =

Ω=


For example

Hertz dipole

Note that U(θ,φ) is independent of r

( ) ( )0

2

222

0

22

32

sin,,,

ηπ

θβφθφθ

dlIrrSU ==


Normalized radiation intensity is a dimensionless quantity

For Hertz dipole

( ) ( )( )φθ

φθφθ

,

,,

maxU

UU n =

( ) θφθ 2sin, =nU


It is usually expressed in dB

For Hertz dipole

( ) ( )θφθ 2

10 sinlog10, =dBnU


The power or radiation intensity pattern is

the angular distribution of antenna’s radiated power

per unit solid angle


One could also find the total radiated power of an antenna as

2

4 0 0

( , ) ( , ) sinradP U d U d d

π π

π θ φ

θ φ θ φ θ θ φΩ= = =

= Ω =∫ ∫ ∫


Average radiation intensity is defined as

4 0 0π θ φΩ= = =

2

0 0

1( , )sin

4 4

radavg

PU U d d

π π

θ φ

θ φ θ θ φπ π

= =

= = ∫ ∫


8.3.1 What is antenna radiation pattern?

The radiation pattern of an antenna is a 3-D graphical representation of the radiation properties of the antenna as a function of position (usually in spherical coordinates)

If we imagine an antenna is placed at the origin of a spherical


If we imagine an antenna is placed at the origin of a spherical coordinate system, its radiation pattern is given by measurement of the magnitude of the electric field over a surface of a sphere of radius r


Dipole AntennaOmni-directional radiation pattern


Horn Antenna Directional radiation pattern


For a fixed r, electric field is only a function of θ and

Two types of patterns are generally used: (a) field pattern (normalized or versus spherical coordinate position) and

φ

( ),E θ φr

Er

Hr


coordinate position) and

(b) power pattern (normalized power versus spherical coordinate position).


3-D radiation patterns are difficult to draw and visualize in a 2-D plane like pages of this book

Usually they are drawn in two principal 2-D planes which are orthogonal to each other

Generally, xz- and xy- plane are the two orthogonal


Generally, xz- and xy- plane are the two orthogonal principal planes

E-plane (H-plane) is the plane in which there are maximum electric (magnetic) fields for a linearly polarized antenna


For example,

Hertz dipole

How to decide E- and H- planes?

Current is flowing in z-direction

( ), ,P r θ φ

0

j tI e

ω

r

θ

φ


Current is flowing in z-direction

Magnetic vector potential follows the current direction

Magnetic field will be along φ-direction (H-plane is in x-y plane)

Electric field will be along θ-direction (E-plane is in x-z plane)


00=θ00=φ

090=φ0270=φ


Fig. H-plane and E-plane radiation patterns of Hertz dipole

0270=θ0180=θ

090=θ

0180=φ


Besides 2-D and 3-D polar plots,

radiation pattern may be plotted as rectangular plots

In this case,

horizontal axis is in degrees


vertical axis is normalized radiated power in dB


0 dB

-3 dB


Fig. 8.3 (c) Typical radiation pattern of an antenna (rectangular plots)

maxθ

θ00 500-500 1000-1000

-15 dB


A typical antenna radiation pattern looks like as in Fig. 8.3 (c)

It could be a polar plot as well

An antenna usually has either one of the following patterns: (a) isotropic (uniform radiation in all directions, it is not possible to realize this practically)


possible to realize this practically)

(b) directional (more efficient radiation in one direction than another)

(c) omnidirectional (uniform radiation in one plane)


8.3.2 Direction of the main beam (θmax)

A radiation lobe is a clear peak in the radiation intensity surrounded by regions of weaker radiation intensity

Main beam is the biggest lobe in the radiation pattern of the antenna


of the antenna

It is the radiation lobe in the direction of maximum radiation


θmax is the direction in which maximum radiation occurs

Any lobe other than the main lobe is called as minor lobe

The radiation lobe opposite to the main lobe is also termed as back lobe


termed as back lobe

This will be more appropriate for polar plot of radiation pattern


8.3.3 Half power beam width (HPBW)

It is the angular separation between the half of the maximum power radiation in the main beam

At these points, the radiation electric field reduces by of the maximum electric field

1

2


the maximum electric field

Its shows how sharp is the beam

Half power is also equal to -3-dB

We also call HPBW as -3-dB beamwidth

They are measured in the E-plane and H-plane radiation patterns of the antenna


8.3.4 Beam width between first nulls (BWFN)

It is the angular separation between the first two nulls on either side of the main beam

For same values of BWFN, we can have different values of HPBW for narrow beams and broad beams


of HPBW for narrow beams and broad beams

HPBW is a better parameter for specifying the effective beam width

It gives an idea of the main beam shape


8.3.5 Side lobe level (SLL)

The side lobes are the lobes other than the main beam and

it shows the direction of the unwanted radiation in the antenna radiation pattern


antenna radiation pattern

The amplitude of the maximum side lobe in comparison to the main beam maximum amplitude of the electric field is called as side lobe level (SLL)

It is normally expressed in dB and a SLL of -30 dB or less is considered to be good for a communication system



Fig. Microwave tower: LOS communication


Back lobe Sometimes antenna may have back lobe radiation

It is specified by front-to-back ratio

It is basically the ratio of the peak of the main lobe


over the peak of the back lobe

It gives an idea abut the directivity of an antenna

Like side lobe, back lobe is also an unwanted radiation


Horn Antenna Directional radiation pattern


At what distance from the antenna, we may assume far field region

Region surrounding an antenna may be divided into three regions:

Reactive near field


Reactive near field

Radiating near field

Far field

3

max0 0.62D

r< ≤


Fig. 8.3 (a) Antenna field regions

max10 0.62

nf

Dr

λ< ≤

3 2

max max2

20.62

nf

D Dr

λ λ< ≤

2

max2ff

Dr

λ<


The antenna field regions could be divided broadly into three regions (see Fig. 8.3 (a)):

Reactive near field region:

This is the region immediately surrounding the antenna where the reactive field (stored energy-standing waves)


where the reactive field (stored energy-standing waves) dominates

Reactive near field region is for a radius of

where Dmax is the maximum antenna dimension

3

max10 0.62

nf

Dr

λ< ≤


Radiating near field (Fresnel) region:

The region in between the reactive near field and the far-field (the radiation fields are dominant)

the field distribution is dependent on the distance from the antenna


antenna

Radiating near field (Fresnel) region is usually for a radius of

3 2

max max2

20.62 nf

D Dr

λ λ< ≤


Far field (Fraunhofer) region:

This is the region farthest from the antenna where the field distribution is essentially independent of the distance from the antenna (propagating waves)

Fraunhofer far field region is usually for a radius of2

max2ff

Dr

<


Fraunhofer far field region is usually for a radius of

In the far field region, the spherical wavefront radiated from a source antenna can be approximated as plane wavefront

The phase error in approximating this is π/8

maxffr

λ<


Fig. 8.3 (b) Illustration of

far field region

(antenna under test: AUT)



We can calculate the distance rff by equating the maximum error (which is at the edges of the AUT of maximum dimension Dmax) in the distance r by approximating spherical wavefront to plane wavefront to λ/16 (Exercise 8.1)



8.3.7 Directivity

The directivity of an antenna is defined as the ratio of the radiation intensity in a given direction from the antenna

to the radiation intensity averaged over all directions


to the radiation intensity averaged over all directions which equivalent to the radiation intensity of an isotropic antenna

2

0 0

( , ) 4 ( , )( , )

( , )sinavg

U UD

UU d d

π π

θ φ

θ φ π θ φθ φ

θ φ θ θ φ= =

= =

∫ ∫


D (θ, ) is maximum at θmax and minimum along θnull

is also known as beam solid angle

φ

( ) max maxmax 2

0 0

4 ( , ) 4,

( , )sin A

U UD

U d d

π π

θ φ

π θ φ πθ φ

θ φ θ θ φ= =

= = =Ω

∫ ∫avgU

Ω


is also known as beam solid angle

It is also defined as the solid angle through which all the antenna power would flow if the radiation intensity was for all angles in

AΩ

max ( , )U θ φ AΩ


Given an antenna with one narrow major beam, negligible radiation in its minor lobes

where and are the half-power beam widths in

HPBW HPBW

rad rad

A θ φΩ ≈ ×

θ φ


where and are the half-power beam widths in radians which are perpendicular to each other

For narrow beam width antennas

It can be shown that the maximum directivity is given by

HPBWθHPBWφ

( ), 1HPBW HPBW

θ φ <<

max

4

HPBW HPBW

rad radD

π

θ φ≅

×


If the beam widths are in degrees, we have

8.3.8 Gain

2

max deg deg deg deg

1804

41,253

HPBW HPBW HPBW HPBW

D

ππ

θ φ θ φ

≅ =

× ×


8.3.8 Gain

In defining directivity, we have assumed that the antenna is lossless

But, antennas are made of conductors and dielectrics


It has same in-built losses accompanied with the conductors and dielectrics

Thereby, the power input to the antenna is partly radiated and remaining part is lost in the imperfect conductors as well as in


remaining part is lost in the imperfect conductors as well as in dielectrics

The gain of an antenna in a given direction is defined as the ratio of the intensity in a given direction

to the radiation intensity that would be obtained if the power accepted by the antenna were radiated isotropically


Note that definitions of the antenna directivity and gain are essentially the same

( ) radrad

4 ( , )e4 ( , ), e ( , )

input rad

UUG D

P P

π θ φπ θ φθ φ θ φ= = =


essentially the same except for the power terms used in the definitions

Directivity is the ratio of the antenna radiated power density at a distant point to the total antenna radiated power radiated isotropically

Gain is the ratio of the antenna radiated power density at a distant point to the total antenna input power radiated isotropically


The antenna gain is usually measured based on Friistransmission formula and it requires two identical antennas

One of the identical antennas is the radiating antenna, and the other one is the receiving antenna

Assuming that the antennas are well matched in terms of


Assuming that the antennas are well matched in terms of impedance and polarization, the Friis transmission equation is

2

10 10

1 420log 10log

4 2

r rt r t r

t t

P PRG G G G G G

P R P

λ π

π λ

= = = ∴ = +

Q


Friis transmission equation states that the ratio of the received power at the receiving antenna and transmitted power at the transmitting antenna is: directly proportional to both gains of the transmitting (Gt) and receiving (Gr) antennas


receiving (Gr) antennas

inversely proportional to square of the distance between the transmitting and receiving antennas (1/R2) and

directly proportional to the square of the wavelength of the signal transmitted (λ2)


Assumptions made are:

(a) antennas are placed in the far-field regions

(b) there is free space direct line of sight propagation between the two antennas


(c) there are no interferences from other sources and

no multipaths between the transmitting and receiving antennas due to

reflection,

refraction and

diffraction


8.3.9 Polarization

Let us consider antenna is placed at the origin of a spherical coordinate system

and wave is propagating radially outward in all directions


directions

In the far field region of an antenna,

( ) ( ) ( )ˆ ˆ, , ,E E Eθ φθ φ θ φ θ θ φ φ= +r


Putting the time dependence, we have,

where δ is the phase difference between the elevation and azimuthal components of the electric field

( ) ( ) ( ) ( ) ( ) ( ) ( )ˆ ˆ ˆ ˆ, , , cos , cos , , , ,E t E t E t E t E tθ φ θ φθ φ θ φ ω θ θ φ ω δ φ θ φ θ θ φ φ= + + = +r


and azimuthal components of the electric field

The figure traced out by the tip of the radiated electric field vector

as a function of time for a fixed position of space can be defined as antenna polarization


a) LP

When δ=0, the two transversal electric field components are in time phase

The total electric field vector


makes an angle with the -axis

( ) ( ) ( ) ( ) ( )( )ˆ ˆ, , , cos , cosE t E t E tθ φθ φ θ φ ω θ θ φ ω φ= +r

LPθ θ

( )( )

( )( )

1 1, , , ,

tan tan, , , ,

LP

E t E t

E t E t

φ φ

θ θ

θ φ θ φθ

θ φ θ φ− −

= =


The tip of the total radiated electric field vector traces out a line

Therefore, the antenna’s polarization is LP

b) CPπ

δ = ±


When ,

the two transversal electric field components are out of phase in time

and if the two transversal electric field components are of equal amplitude

2

πδ = ±

( ) ( ) ( )0, , ,E E Eθ φθ φ θ φ θ φ= =


The total electric field vector

makes an angle with the -axis

( ) ( ) ( ) ( ) ( )( )ˆ ˆ, , , cos , sinE t E t E tθ φθ φ θ φ ω θ θ φ ω φ= +r

CPθ θ


This implies that the total radiated electric field vector of the antenna traces out a circle as time progresses from 0 to 2 and so on

( )( )

( )1 1sin

tan tan tancos

CP

tt t

t

ωφ ω ω

ω− −

= = =

π

ω


If the vector rotates counterclockwise (clockwise), then the antenna polarization is RHCP (LHCP)

For

the total electric field vector traces out an ellipse and

( ) ( ), , 0,E E andθ φθ φ θ φ α π≠ ≠ πδ ,0≠


the total electric field vector traces out an ellipse and hence it is elliptically polarized (EP)

The ratio of the major and minor axes of the ellipse is called axial ratio (AR)

For instance, AR=0 dB for CP and AR= ∞ dB for LP


Antenna polarization:

To sum up,

The polarization of a radiated wave in the far field region of an antenna defines antenna polarization

Assume a transmitting antenna transmitting electric field of


Assume a transmitting antenna transmitting electric field of maximum magnitude Em in the far field region of an antenna

where is the unit vector which gives polarization for the transmitting antenna

tm

teEE ˆ=

r

te


It is the complex vector representation of field normalized to unity

Assume that there is a receiving antenna in the far field region of this antenna

The field of some magnitude Em that would give maximum


The field of some magnitude Em that would give maximum received power at the received antenna terminal is

where is the unit vector which gives polarization for the receiving antenna

r

mreEE ˆ=

r

re


Polarization efficiency is defined as

Receiving LP wave with an LP receiving antenna: Let us consider a case where the transmitting antenna is

2*ˆˆ.. rt eeEP •=


Let us consider a case where the transmitting antenna is placed at an angle θt w.r.t z-axis

and the receiving antenna is placed at an angle θr w.r.t z-axis

Both transmitting and receiving antennas are LP antennas


tθrθ



Therefore, polarization unit vectors for the transmitting and receiving antennas are

Polarization efficiency can be calculated as

yzeyze rrrtttˆsinˆcosˆ;ˆsinˆcosˆ θθθθ +=+=


( )rtrt eeEP θθ −=•= 22

* cosˆˆ..rt θθ −


θt-θ

rP.E. Illustration

0 1


π/2 0



Fig. Receiving a CP wave with a CP receiving antenna


Receiving a CP wave with a CP receiving antenna

Consider a RHCP receiving antenna is kept in the far field region of RHCP transmitting antenna

Assume wave propagation along positive y-axis

Hence the polarization unit vectors for transmitting and


Hence the polarization unit vectors for transmitting and receiving antennas are

( ) ( )zjxezjxe rtˆˆ

2

1ˆ;ˆˆ

2

1ˆ −=−=


Polarization efficiency

For instance, the receiving antenna is LHCP

1ˆˆ..2

* =•= rt eeEP


Polarization efficiency

( ) ( )zjxezjxe rtˆˆ

2

1ˆ;ˆˆ

2

1ˆ +=−=

0ˆˆ..2

* =•= rt eeEP


It is like two persons shaking hands

Right to right hand shake is matched but right to left hand shake is not matched

Receiving an LP wave by a CP receiving antenna:

Consider an LP wave transmitting antenna which is oriented


Consider an LP wave transmitting antenna which is oriented at an angle θt with the z-axis

Receiving antenna is a RHCP antenna

Hence

( )zjxezxe rtttˆˆ

2

1ˆ;ˆsinˆcosˆ −=+= θθ


tθ


Fig. Receiving an LP wave by a CP receiving antenna


Therefore

Polarization efficiency is

2

1sincos

2

1ˆˆ..

22* =+=•= ttrt jeeEP θθ


Hence there is a 3-dB signal loss

The same case when one use a RHCP/LHCP transmitting antenna and a LP receiving antenna

2sincos

2ˆˆ.. =+=•= ttrt jeeEP θθ


Let us try to understand two terms (co- and cross-polarization) which are important for the antenna radiation pattern

Co-polarization means you measure the antenna with another antenna oriented


another antenna oriented

in the same polarization with the antenna under test (AUT)

Cross-polarization means that you measure the antenna with antenna oriented

perpendicular w.r.t. the main polarization


Cross-polarization is the polarization orthogonal

to the polarization under consideration

For example,

if the field of an antenna is horizontally polarized,

the cross-polarization for this case is vertical polarization


the cross-polarization for this case is vertical polarization

If the polarization is RHCP,

the cross-polarization is LHCP

Let us put this into mathematical expressions:

We may write the total electric field propagating along z-axis as

( )ˆ ˆ j z

co co cr crE E u E u eβ−= +

r


where the co- and cross-polarization unit vectors satisfy the orthonormality condition

Therefore,

* * * *ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ1, 1, 0, 0co co cr cr co cr cr co

u u u u u u u u• = • = • = • =


Therefore,

the co- and cross-polarization components of the electric fields can be obtained as

* *ˆ ˆ,co co cr crE E u E E u= • = •r r


a) LP

For a general LP wave, we can write,

For a x-directed LP wave, =0, hence,

ˆ ˆ ˆ ˆ ˆ ˆcos sin , sin cosco LP LP cr LP LP

u x y u x yφ φ φ φ= + = −

LPφ


For a y-directed LP wave, =900, hence,

* *ˆ ˆ ˆ ˆ ˆ ˆ, ; ,co cr co co x cr cr y

u x u y E E u E E E u E= = − = • = = • = −r r

LPφ

* *ˆ ˆ ˆ ˆ ˆ ˆ, ; ,co cr co co y cr cr xu y u x E E u E E E u E= = = • = = • =r r


b) CP

For a RHCP wave, we can write,

ˆ ˆ ˆ ˆˆ ˆ,

2 2co cr

x jy x jyu u

− += =


For a LHCP wave, co- and cross-polarization unit vectors and components of the electric field will interchange

* *ˆ ˆ,2 2

x y x y

co co cr cr

E jE E jEE E u E E u

+ −= • = = • =r r


c) EP

For a EP wave, we can write,

2 2

ˆ ˆ ˆ ˆˆ ˆ,

1 1

EP EPj j

co cr

x Ae y Ae x yu u

A A

φ φ−+ − += =

+ +


In order to determine the far-field radiation pattern of an AUT, two antennas are required

The one being tested (AUT) is normally free to rotate and it is connected in receiving mode



Fig. Antenna measurement set up


Note that AUT as a receiving antenna measurement will generate the same radiation pattern

to that of AUT used as a transmitting antenna (from reciprocity theorem)

Another antenna is usually fixed and it is connected in


Another antenna is usually fixed and it is connected in transmitting mode

The AUT is rotated by a positioner and

it can rotate 1-, 2- and 3-degrees of freedom of rotation


The AUT is rotated in usually two principal planes (elevation and azimuthal)

The received field strength is measured by a spectrum analyzer or power meter which will be used to generate the antenna radiation pattern in


which will be used to generate the antenna radiation pattern in two principal planes also known as E- and H- planes

The antenna radiation patterns in these two principal planes can be used to generate the 3-D radiation pattern of an antenna


Let us consider Hertz dipole which is a vertically polarized antenna

For E-plane co-polar measurements,

source and test antenna are oriented vertically

so that the main beam of the test antenna (receiving


so that the main beam of the test antenna (receiving antenna) is directed towards y-axis

Source antenna is kept stationary and

we rotate the test antenna in the x-z plane

and record the electric fields variation over θ


Fig. E-plane co-polar measurements

Test antenna

z

Rotate around

x‐z plane


x

ySource

antenna


For E-plane cross-polar measurements,

source is oriented horizontally and test antenna is oriented vertically





and record the electric fields variation over θ


Fig. E-plane cross-polar measurements

Test antenna

z

Rotate in the

x‐z plane


x

ySource

antenna


For H-plane co-polar measurements,

source and test antenna are oriented vertically


we rotate the test antenna in the x-y plane


and record the electric fields variation over φ


Fig. H-plane co-polar measurements

z

Rotate around

x‐y plane


Test antenna

x

ySource

antenna


For H-plane cross-polar measurements,

source is oriented horizontally and test antenna is oriented vertically





and record the electric fields variation over φ


Fig. H-plane cross-polar measurements

z

Rotate around

x‐y plane


Test antenna

x

y

Source

antenna

Cross-polar

Co-polar

Fig. Typical E-plane pattern4/24/2018Electromagnetic Field Theory by R. S. Kshetrimayum153

Cross-polar

Co-polar


polar

Fig. Typical H-plane pattern


Let us look at the antenna input characteristics

Antenna is considered as a load to the transmission line

Hence one can find out the equivalent circuit of an antenna

Antenna is a resonator

It can be equivalently modelled as a series RLC circuit


It can be equivalently modelled as a series RLC circuit



Fig. Equivalent circuit of an antenna


8.3.10 Quality factor and bandwidth

The equivalent circuit of a resonant antenna can be

approximated by a series RLC resonant circuit

where R=Rr+RL are the radiation and loss resistances,


r L

L is the inductance and

C is the capacitance of the antenna


For a resonant antenna like dipoles,

the FBW is related to the radiation efficiency and quality factor Q (FBW=1/Q)

The quality factor of an antenna is defined as 2πf0 (f0 is the resonant frequency) times the energy stored over the


the resonant frequency) times the energy stored over the power radiated and Ohmic losses

( )

( ) ( )

( )

2 2

2

0 00

20

0

1 1 1

4 4 2 2 12

1 2

2

1

2

r L r Lr L

rlossless rad

r r L

I L If L f L

Q fR R f R R C

I R R

RQ e

f R C R R

π ππ

π

π

+

= = =+ ++

= = •+


where Qlossless is the quality factor when the antenna is lossless (RL=0) and

erad is the antenna radiation efficiency.

Note that the radiation efficiency of an antenna is defined

as the ratio of the power delivered


as the ratio of the power delivered

to the radiation resistance Rr to the power delivered to Rr and RL

( ) ( )

2

2

1

21

2

rr

rad

r Lr L

I RR

eR R

I R R

= =++


Small antennas are a necessity for portable devices For example, We need small antennas for mobile devices Smaller antenna is an attractive feature But the fundamental question remains


But the fundamental question remains Can we keep on reducing antenna size?

Is there some limit beyond which we can reduce the antenna size?

Consider a smallest sphere which enclose a small antenna as shown in next slide


Fig. Smallest sphere which can enclose a small antenna


There is a very important concept on designing electrically small antennas

When kr<1 (electrically small antennas), the quality factor Q of a small antenna can found from the J. L. Chu’s relation


where k is the wave number and r is the radius of the smallest sphere enclosing the antenna

( )

( ) ( )

2

3 2

1 2

1rad

krQ e

kr kr

+= •

+


The above relation gives the relationship between the antenna size, efficiency and quality factor.

This expression can be reduced further for smallest Q for a LP very small antenna (kr<<1) as follows:


Harrington gave also a practical upper limit to the gain of a small antenna for a reasonable BW as

( )min 3

1 1Q

krkr= +

( ) ( )2

max 2G kr kr= +


CASE STUDY of a small antenna

for a Hertz dipole of

very small length 0.01λ,

Qmin =32283


It has very high Q and

hence a very narrow FBW (0.000031) (very narrow)

Gmax=0.0638 or -12dB (very small no practical use)

Note that in the above calculations r=0.005λ has been used


Very small antenna like Hertz dipole has no practical use For practical use, we need to have a sufficiently large antennaAntenna size, quality factor,


quality factor, bandwidth and radiation efficiency is interlinked

There is no complete freedom to optimize each one of them independently


RECAP:

Antenna acts as an interface between a guided wave and a

free-space wave

One of the most important characteristics of an antenna is its

directional property

Ability to concentrate radiated power in a certain directionAbility to concentrate radiated power in a certain direction

Or receive power from a preferred direction

This directional property is characterized in Radiation pattern

From reciprocity theorem, we can show that

the pattern characteristics of an antenna are the same in the

transmit and

the receive modes


Antenna equivalent circuit can be thought of as

Radiation resistance

Loss resistance

Reactive parts


So the feed line also has a characteristic impedance Z0

usually of 50 Ohm

At the input of the antenna, the impedance seen by the feed line can be assumed as ZL


Then the reflection coefficient and

VSWR may be calculated as

1

1;

0

0

Γ−

Γ+=

+

−=Γ VSWR

ZZ

ZZ

L

L


2:

1,10

10

≤

∞≤≤≤Γ≤

Γ−+

VSWRBW

VSWR

ZZ L

%100×−

=c

lh

f

ffBW


fl fh

fc

( )dBΓ

8.4 Kinds of antennas 8.4.2 Dipole antenna

The next extension of a Hertz dipole is

a linear antenna or a dipole of finite length as depicted in Fig. 8.7 (a)

It consists of a conductor of length 2L


It consists of a conductor of length 2L

fed by a voltage or current source at its center

We will first find the current distribution on such antennas

Fig. 8.7 (a) Dipole of length 2L

(Dipole can be θ


(Dipole can be assumed

to composed of many Hertz dipoles)

8.4 Kinds of antennas Let us assume a transmission line loaded with a load ZL

Since the line impedance and load impedance may be different

Any wave incident from the line to the load will be reflected


reflected

Hence at any location along the line,

there will be both reflected and incident wave which may be expressed as

( ) zjzj eVeVzV ββ +−−+ += 00



Fig. A transmission line loaded with an impedance of ZL

8.4 Kinds of antennas Corresponding electric current wave along the line may be expressed as

( ) ( )0

0

0

0

0

0

0

0

0

0 ;ZZ

ZZ

V

Vee

Z

Ve

Z

Ve

Z

VzI

L

Lzjzjzjzj

+

−==ΓΓ−=−=

+

−+−

++

−−

+ββββ


The characteristic impedance of the line is defined as

For o.c. transmission line

( )0 0

0 0

( ) 2 sinj z j zV VI z e e j z

Z Z

β β β+ +

− += − = −

−

−

+

+

−==0

0

0

00

I

V

I

VZ

8.4 Kinds of antennas The current is zero at z= L

since at the ends, there is no path for the current to flow

so, we can write,

( )0 00 0( ) 2 sin( ( )) sin ; 2

V VI z j L z I L z I j

Z Zβ β

+ +

= − − = − = −


How to plot the current distribution of dipoles of various length?

( )0 0

0 0Z Z

2

λ

2

λ

o.c.

Cut at any point before 8

λ


Fig. Current distribution of a 2L<<λ dipole

8

L

L

2

λ

2

λ

4

λ


Fig. Current distribution of a λ/2 dipole

4

4

λ

4

λ

2

λ

2

λ

o.c.

Cut at 2

λ


Fig. Current distribution of a λ dipole

2

λ

2

λ

2

λ

2

λ

o.c.

Cut at

4

3λ

2

λ

2

λ


Fig. Current distribution of a 3λ/2 dipole

2

2

λ2

λ

8.4 Kinds of antennasHow to find the electric field?

The electric field due to the current element dz

it has the same expression of the Hertz dipole of the previous section

except that now we have a length of dz and current of I(z)


except that now we have a length of dz and current of I(z)

at far away observation point or in the far field can be written as

12

1 0

sin ( );

4

j R dEj I z dzedE dH

R

βθ

θ φ

β θ

πεω η

−

= =

Fig. 8.7 (a) Dipole of length 2L

(Dipole can be θ


(Dipole can be assumed

to composed of many Hertz dipoles)

zcosθ

8.4 Kinds of antennas Since the observation point P is at a very far distance, the lines OP and QP are parallel and therefore

Note that for the amplitude, we can approximate

since the dipole size is quite small in comparison to the

1 cosR R z θ∴ ≅ −

1

1 1

R R≅


since the dipole size is quite small in comparison to the distance of the observation point P from the origin

Hence

2 cossin ( )

4

j R j Zj I z dze e

dER

β β θ

θ

β θ

πεω

−

≅

8.4 Kinds of antennas Since we have assumed that the dipole of length 2L is composed of many Hertz dipoles as depicted in Fig. 8.7 (a)

(this is one of the reasons why we say that Hertz dipole or infinitesimal dipole is the building block for many


or infinitesimal dipole is the building block for many antennas),

we can write the total radiated electric field as

2 cos2 cos0sin sin( ( ))sin ( )

4 4

j R j ZL L Lj R j Z

z L z L z L

j I L z e ej I z e eE dE dz dz

R R

β β θβ β θ

θ θ

β θ ββ θ

πεω πεω

−−

=− =− =−

−= = =∫ ∫ ∫

8.4 Kinds of antennas It can be shown that (see textbook for derivations)

( )( )0 0

cos cos cos60 60

sin

j R j RL Le eE j I j I F

R R

β β

θ

β θ βθ

θ

− − −⇒ ≅ =


In the previous equation, the term under bracket is F(θ) and it is the variation of electric field as a function of θand

it is the E-plane radiation pattern

8.4 Kinds of antennas In the H-plane like that of Hertz dipole,

Eθ is a constant and it is not a function of

hence it is a circle

The E–plane radiation pattern of the dipole

φ


varies with the length of the dipole as depicted in Fig. 8.8

8.4 Kinds of antennas Note that according to Fig. 8.7 (a),

we have considered the total dipole length is 2L

Fig. 8.8 E-plane radiation pattern for dipole of length

(a) 2L=2×λ/4= λ/2

λ λ


(b) 2L=2×λ=2λ

(c) 2L=2×2λ=4λ

( )( )0 0

cos cos cos60 60

sin

j R j RL Le eE j I j I F

R R

β β

θ

β θ βθ

θ

− − −⇒ ≅ =


(a) 2L=2×λ/4= λ/2


(b) 2L=2×λ=2λ


(c) 2L=2×2λ=4λ

8.4 Kinds of antennasPoints to be noted:

1. Input impedance of the dipole (z=0, center of dipole)

For dipole of length 2L, where L =odd multiples of

0 sin

in inin

in

V VZ

I I Lβ= =


For dipole of length 2L, where L =odd multiples of

=1,

For dipole of length 2L, where L =even multiples of .

, , sin4 2

mL L

λ πβ β=

0

inin

VZ

I=

, , sin 04

L m Lλ

β π β= = inZ⇒ = ∞

8.4 Kinds of antennas That’s why

it is preferable to have dipoles of length odd multiples of λ/2,

otherwise

it is difficult to have a source with infinite impedance


it is difficult to have a source with infinite impedance

2. Since increasing the dipole length more and more current is available for radiation,

• the total power radiated increases monotonically

8.4 Kinds of antennas3. The electric field has only component

and hence it is linearly polarized

4. The radiation pattern have nulls

and it can be calculated by equating F(θ)=0

$θ



cos( cos ) cos0

sin

null

null

L Lβ θ β

θ

−=

nullθcos⇒ 1mλ

π= ± ±


o For m=0,

But, in denominator is also zero

So let us take the limit of F(θ) as θ→0, and see

πθ θ ,0cos ,1 =±= nullnull

sinnull

θ

8.4 Kinds of antennas( ) ( ) ( ) ( ) ( ) ( )

0, 0,

2 4 2 4cos cos cos cos cos1

1 1sin sin 2! 4! 2! 4!

L L L L L L

Lim Limθ π θ π

β θ β β θ β θ β β

θ θ→ →

− ≅ − + − − +

( ) ( ) ( ) ( ) ( ) ( )0, 0,

4 42 24 22 1 cos sin 1 cossin sin10

2! 4! sin 2! 4!

L LL L

Lim Limθ π θ π

β θ β θ θβ θ β θ

θ→ →

− + = − × = − =


5. To find θ for maximum radiation, we have to find the solution of

=0

We can also take the mean of the first two nulls to approximate

( )dF

d

θ

θ

maxθ

8.4 Kinds of antennas What is the radiation resistance of a dipole antenna?

Radiated power can be obtained from the radiation intensity as

2

0

2

I

PR rad

r =


as

( ) φθθφθπ

φ

π

θ

ddUPrad sin,

2

0 0

∫ ∫= =

=

8.4 Kinds of antennas Radiation intensity can be obtained from the Poynting vector as

Therefore,

( ) ( ) ( )θ

θπ

πθ

πφθφθ

2

2

2

0

22

0

2

sin

cos2

cos3030

,,

=== IFISrU


Therefore,

∫=

=π

θ

θθ

θπ

π0

2

2

2

0sin

cos2

cos30

dIPrad

8.4 Kinds of antennasMonopole antennas:

A monopole is a dipole that has been divided in half at its center feed point and fed against a ground plane

Monopole is usually fed from a coaxial cable (see Fig. 8.7 (b))


Monopole is usually fed from a coaxial cable (see Fig. 8.7 (b))

A monopole of length L placed above a perfectly conducting and infinite ground plane will have the same field distribution to that of a dipole of length 2L without the ground plane


Monopole

Ground plane


(b) Monopole of length L over a ground plane

Ground plane

Coaxial cable

Dipole in free space

Image of monopole

8.4 Kinds of antennas This is because an image of the monopole will be formed inside the ground plane (similar to the method of images in chapter 2)

The monopole looks like a dipole in free space (see Fig. 8.7 (b))


(b))

Since this monopole is of length L only, it will radiate only half of the total radiated power of a dipole of length 2L

Hence, the radiation resistance of a monopole is half that of a dipole

8.4 Kinds of antennas Similarly, directivity of the monopole is twice that of a dipole

Since the field distributions are the same for a monopole and dipole, the maximum radiation intensity will be also same for both cases


cases

But for monopole, the total radiated power is half that of a dipole

Hence, the directivity of a monopole above a conducting ground plane is twice that of dipole in free space

8.4 Kinds of antennas8.4.3 Loop antenna

Loop antennas could be of various shapes: circular,

triangular,

square,


square,

elliptical, etc.

They are widely used in applications up to 3GHz

Loop antennas can be classified into two: electrically small (circumference < 0.1 λ) and

electrically large (circumference approximately equals to λ)

8.4 Kinds of antennas Electrically small loop antennas have

very small radiation resistance

They have very low radiation and

are practically useless


are practically useless

Electrically small loop antennas could be analyzed assuming that it is equivalently represented as a Hertz dipole

8.5 Antenna Arrays

One of the disadvantages of single antenna is that it has fixed radiation pattern

That means once we have designed and constructed an antenna, the beam or radiation pattern is fixed


the beam or radiation pattern is fixed

If we want to tune the radiation pattern, we need to apply the technique of antenna arrays

Antenna array is a configuration of multiple antennas (elements) arranged

to achieve a given radiation pattern

8.5 Antenna Arrays

There are several array design variables which can be changed to achieve the overall array pattern design

Some of the array design variables are:

(a) array shape linear, circular,


circular, planar, etc.

(b) element spacing

(c) element excitation amplitude

(d) element excitation phase

(e) patterns of array elements

φ

8.5 Antenna Arrays

Given an antenna array of identical elements, the radiation pattern of the antenna array may be found according to the pattern multiplication principle

It basically means that array pattern is equal to φ


It basically means that array pattern is equal to the product of the

pattern of the individual array element into

array factor , a function dependent only on

the geometry of the array

the excitation amplitude and phase of the elements

φ

8.5 Antenna Arrays

8.5.1 Two element array

Let us investigate an array of two infinitesimal dipoles positioned along the z axis as shown in Fig. 8.10 (a)

The field radiated by the two elements, assuming no coupling between the elements


assuming no coupling between the elements

is equal to the sum of the two fields

where the two antennas are excited with current

1 1 2 22 2

1 21 2

1 2

sin sinˆ ˆ4 4

j r j j r j

total

jI dl e e jI dl e eE E E

r r

β δ β δθ β θ βθ θ

πεω πεω

− −

= + = +r r r

1 1 2 2I and Iδ δ< <

8.5 Antenna Arrays

1rr

rr

1θ

θ


Fig. 8.10 (a) Two Hertz dipoles

2rr2θ

8.5 Antenna Arrays

For 1 2 oI I I= =

1 2,2 2

α αδ δ= = −

2 cos cossinˆd dj r j jjI dl e

β α β αβ θ θθβθ

− + − + = +

r


2 cos cos2 2 2 20 sinˆ

4

j r j j

total

jI dl eE e e

r

β θ θθβθ

πεω

− + − +

= +

r

2

0 sinˆ 2cos cos4 2 2

j rjI dl e d

r

βθβ β αθ θ

πεω

− = +

8.5 Antenna Arrays

Hence the total field of the array is equal to

the field of single element positioned at the origin

multiplied by a factor which is called as the array factor

Array factor is given


Normalized array factor is

( )1

2cos cos2

AF dβ θ α

= +

( )2

1cos cos

2AF dβ θ α

= +

8.5 Antenna Arrays

8.5.2 N element uniform linear array (ULA) This idea of two element array can be extended

to N element array of uniform amplitude and spacing

Let us assume that N Hertz dipoles are placed along a straight line along z-axis at positions


straight line along z-axis at positions 0, d, 2d, …, (N-2) d and (N-1) d respectively

8.5 Antenna Arrays Current of equal amplitudes

but with phase difference of 0, α, 2α, … , (N-2) α and (N-1) α


are excited to the corresponding dipoles at 0, d, 2d, …, (N-2) d and (N-1) d respectively

8.5 Antenna Arraysz

2d

(N-1)d

.

.

.

2I α⟨

( 1)I N α⟨ −

0I ⟨

I α⟨

( 1)

2

NI α

−⟨


Fig. 8.10 (b) ULA 1 (c) ULA 2 (assume N is an odd number)

d

2d

00I ⟨

I α⟨

2I α⟨0I ⟨

I α⟨−

( 1)

2

NI α

−⟨−

8.5 Antenna Arrays

Then the array factor for the N element ULA of Fig. 8.10 (b) will become

( ) ( ) ( )cos 2 cos ( 1) cos1 .....

j d j d j N dAF e e e

β θ α β θ α β θ α+ + − +∴ = + + + +

( )( )ψ−1 e

jNN


( )( ) αθβψψ

ψαθβ +=

−

−=⇒=⇒ ∑

=

+− cos;1

1

1

cos1d

e

eAFeAF

j

jN

N

N

n

dnj

1

1

jN

j

e

e

ψ

ψ−

=−

( ) ( )1 2 2( )2

1 1( ) ( )2 2

N Nj jN

j

j j

e ee

e e

ψ ψψ

ψ ψ

−−

−

−=

−

1( )

2

sin( )2

sin( )2

Nj

N

eψ

ψ

ψ

−

=

8.5 Antenna Arrays

If the reference point is at the physical

center of the array as depicted in

Fig. 8.10 (c), the array factor is

( )sin( )

2

Nψ

0I ⟨

I α⟨

( 1)

2

NI α

−⟨


For small values of

( )sin( )

2

sin( )2

NAF

ψ

ψ=

( )sin( )

2

2

N

N

AF

ψ

ψ=

0I ⟨

I α⟨−

( 1)

2

NI α

−⟨−

Ψ

8.5 Antenna Arrays

The maximum value of AF is for and its value is N

Apply L’ Hospital rule since it is of the form

To normalize the array factor so that the maximum value is equal to unity, we get,

0ψ =

0

0sin


( )sin( )

1 21

sin2

N

N

AFN

ψ

ψ

=

sin( )2

2

N

N

ψ

ψ≅

8.5 Antenna Arrays

This is the normalized array factor for ULA As N increases, the main lobe narrows The number of lobes is equal to N

one main lobe and other N-1 side lobes


other N-1 side lobes

in one period of the AF The side lobes widths are of 2=/N and main lobes are two times wider than the side lobes The SLL decreases with increasing N This can be verified from Fig. 8.11 (see textbook)

8.5 Antenna Arrays

Null of the array

To find the null of the array,

( )sin( ) 0 cos2

Ndψ ψ β θ α= = +Q

( )sin( )

1 21

sin2

N

N

AFN

ψ

ψ

=


( )

1

2cos cos

2 2

1 2cos 1,2,3,......

n

N N nn d n d

N

nn

d N

πψ π β θ α π β θ α

πθ α

β−

⇒ = ± ⇒ + = ± ⇒ = − ±

⇒ = − ± =

8.5 Antenna Arrays

Maximum values

It attains the maximum values for 0ψ =

( )1

cos2 2

m

dθ θ

ψβ θ α

=

= + 0=

( )sin( )

1 21

sin2

N

N

AFN

ψ

ψ

=


8.5.3 Broadside array

We know that when

mθ θ=

1cosmd

αθ

β−

−

⇒ =

8.5 Antenna Arrays

the maximum radiation occurs

It is desired that maximum occurs at θ=90˚

c o s 0dψ β θ α= + =

0

0

90cos 0 0d

θψ β θ α α

== + = ⇒ =


8.5.4 Endfire array

We know that when


090cos 0 0d

θψ β θ α α

== + = ⇒ =

cos 0dψ β θ α= + =

8.5 Antenna Arrays

It is desired that maximum occurs at θ=0˚,

8.5.5 Phase scanning array

We know that when

00cos 0d d

θψ β θ α α β

== + = ⇒ = −


We know that when


It is desired that maximum occurs at θ=θ0

cos 0dψ β θ α= + =

00cos 0 cosd d

θ θψ β θ α α β θ

== + = ⇒ = −

8.5 Antenna Arrays

Draw the polar plot of radiation pattern for the following

uniform linear array (ULA)

of N isotropic radiating antennas spaced λ/2

apart for the following cases: θ

I α⟨

( 1)

2

NI α

−⟨


(a) Broadside array (Maximum field is at θ=90˚)

(b) End fire array (Maximum field is at θ=0˚)

(c) Maximum field is at θ=60˚ and

(d) Null at θ=60˚

( )sin( )

1 21

sin2

N

N

AFN

ψ

ψ

=

0I ⟨

I α⟨−

( 1)

2

NI α

−⟨−

8.5 Antenna Arrays 0

0

90cos 0 0d

θψ β θ α α

== + = ⇒ =


(a) Broadside array (Maximum field is at θ=90˚)

8.5 Antenna Arrays 00cos 0d d

θψ β θ α α β

== + = ⇒ = −


(b) End fire array (Maximum field is at θ=0˚)

8.5 Antenna Arrays


(c) Maximum field is at θ=60˚

00cos 0 cosd d

θ θψ β θ α α β θ

== + = ⇒ = −

8.5 Antenna Arrays


(d) Null at θ=60˚L,3,2,1,cos

2=−= nd

N

nnullθβ

πα

8. Antennas and Radiating Systems · Antennas and Radiating Systems 1 Electromagnetic Field Theory...

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Transcript of 8. Antennas and Radiating Systems · Antennas and Radiating Systems 1 Electromagnetic Field Theory...