Unit-II:

LINEAR WIRE ANTENNAS AND ARRAYS

• Wire Antennas: Short Dipole, Radiation Resistance and Directivity,

Half Wave Dipole, Monopole, Small Loop Antennas.

• Antenna Arrays: Linear Array and Pattern Multiplication, Two-Element

Array, Uniform Array, BSA and EFA, EFA With increased Directivity.
BSA with Non- uniform Amplitude Distributions and Binomial Arrays.

Mr. P Pradeep
Assistant Professor
ECE Dept.
SNIST
Half wave dipole antenna Loop antenna Monopole antenna
Current Distribution on Wires

Wire Antennas. (2021, January 30). Retrieved May 14, 2021, from https://phys.libretexts.org/@go/page/25034
Electric scalar potential
A potential is a function whose derivative gives a field. Fields are associated with forces;
potentials are associated with energy

The electric scalar potential 𝑉 is defined so that the electric field 𝐸ത is given by:
𝑑𝐴ҧ
ത
𝐸 = −𝛻𝑉 − 𝑑𝑡

Magnetic vector potential

The magnetic vector potential 𝐴ҧ is defined so that the magnetic field 𝐵ത is given
by:
𝐵ത = 𝛻𝑋𝐴ҧ

A vector function whose curl gives rise to a magnetic vector field is called vector potential

𝜇 𝐼ⅆ𝑙 𝜇 𝐾ⅆ𝑠 𝜇 𝐽
𝐴ҧ = න 𝐴ҧ = ඵ 𝐴ҧ = ම ⅆ𝑣
4𝜋 𝑟 4𝜋 𝑟 4𝜋 𝑟
Retarded time

• The retarded time is the time when the field began to propagate from the
point where it was emitted to an observer.

• Let the instantaneous current in the 𝑃 (𝑜𝑏𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛 𝑝𝑜𝑖𝑛𝑡)

𝜃
element be 𝐼 = 𝐼𝑚 cos 𝜔𝑡 then the retarded
current , wherein retardation time is taken ⅆ𝑙
𝑟
𝑟
account is given as 𝐼 = 𝐼𝑚 cos 𝜔(𝑡 − )
𝑐
𝑟
Where term (𝑡 − ) is retarded time 𝐈
𝑐
Infinitesimal Dipole

• An infinitesimal linear wire (l≪λ) is positioned symmetrically at

the origin of the coordinate system and oriented along the z
axis, as shown in Figure. Although infinitesimal dipoles are not
very practical, they are used to represent capacitor-plate (also
referred to as top-hat-loaded) antennas.

Δ𝑙 = 𝜆/50
Hertzian dipole antenna

• an informal derivation of the electromagnetic field radiated by a

Hertzian dipole represented by a zero-length current moment.

• A Hertzian dipole is commonly defined as an electrically-short and

infinitesimally-thin straight filament of current, in which the density
of the current is uniform over its length.

• The strategy is to first identify a solution for a distribution of current

that exists at a single point. This distribution is the current moment.
Radiation From A Small Current Element And Hertizan Dipole

• Let us consider a current element of length

ⅆ𝑙 placed at the origin of a spherical
coordinate system and let the current
flowing through it as 𝐼 = 𝐼𝑚 cos 𝜔𝑡

• In general, a current element Iⅆ𝑙 is nothing

but an element of the ⅆ𝑙 carrying
filamentary current I.

• The current element possess

electromagnetic field. The components of
the electric and magnetic fields can be
obtained as follows:
Magnetic field components

• To find the fields radiated by the current element, It will be required to determine first A
then find the E and H.

• Since the source only carries an electric current Ie,

• To find A we write

𝑟
𝜇 𝐼(𝑡−𝑐)
• A(r)= ‫׬‬ ⅆ𝐿
4𝜋 𝑟

• The current flows along Z-direction so A(r) has only Z component

• 𝐴𝑧 ≠ 0, 𝐴𝑥 = 0, 𝐴𝑦 = 0
• Let the current element excited by the current 𝐼 = 𝐼𝑚 cos 𝜔𝑡
• The vector potential A along Z direction for given current,
𝑟
𝜇 𝐼𝑚 cos 𝜔(𝑡−𝑐)
𝐴𝑧 = ‫׬‬ ⅆ𝐿
4𝜋 𝑟

𝑟 ‫ ׬‬ⅆ𝐿 = 𝐿 where L= ⅆ𝑙
𝜇 𝐼𝑚 cos 𝜔(𝑡− )
𝐴𝑧 = 𝑐
‫ ׬‬ⅆ𝐿
4𝜋 𝑟

𝑟
𝜇 𝐼𝑚 cos 𝜔(𝑡−𝑐)
𝐴𝑧 = ⅆ𝑙
4𝜋 𝑟
• 𝐴𝑟 = 𝐴𝑧 𝑐𝑜𝑠 𝜃
• 𝐴𝜃 = −𝐴𝑧 𝑠𝑖𝑛 𝜃
• 𝐴∅ = 0
𝜕
• Where =0
𝜕∅
• 𝐵 = 𝛻Χ𝐴
• 𝛻Χ𝐴ҧ = μ𝐻
ഥ
 𝑎𝑟 𝑟𝑎𝜃 𝑟 sin 𝜃 𝑎𝜙
1 𝜕 𝜕 𝜕
= 𝜇𝐻
𝑟 2 𝑠𝑖𝑛𝜃 𝜕𝑟 𝜕𝜃 𝜕𝜙
𝐴𝑟 𝑟𝐴𝜃 𝑟 sin 𝜃 𝐴𝜙

𝜕
𝐴𝜙 = 0 =0
𝜕𝜙
𝑎𝑟 𝑟𝑎𝜃 𝑎𝜙
1 𝜕 𝜕
0 = 𝜇 𝐻𝑟 𝑎𝑟 + 𝐻𝜃 𝑎𝜃 + 𝐻𝜙 𝑎𝜙
𝑟 2 𝑠𝑖𝑛𝜃 𝜕𝑟 𝜕𝜃
𝐴𝑟 𝑟𝐴𝜃 0

1 𝜕(𝑟𝐴𝜃 ) 𝜕𝐴𝑟
𝐻𝑟 = 0; 𝐻𝜃 = 0; 𝐻𝜙 = [ − ]
𝜇𝑟 𝜕𝑟 𝜕𝜃
 1 𝜕(𝑟𝐴𝜃 ) 𝜕𝐴𝑟
𝐻𝜙 = [ − ]
𝜇𝑟 𝜕𝑟 𝜕𝜃

Where 𝐴𝑟 = 𝐴𝑧 𝑐𝑜𝑠 𝜃 𝑎𝑛ⅆ 𝐴𝜃 = −𝐴𝑧 𝑠𝑖𝑛 𝜃

1 𝜕(−𝑟𝐴𝑧 𝑠𝑖𝑛 𝜃) 𝜕𝐴𝑧 𝑐𝑜𝑠 𝜃

𝐻𝜙 = [ − ]
𝜇𝑟 𝜕𝑟 𝜕𝜃
𝑟
𝜇 𝐼𝑚 cos 𝜔(𝑡−𝑐)
𝑊ℎ𝑒𝑟𝑒 𝐴𝑧 =4𝜋 ⅆ𝑙
𝑟

𝐼𝑚 ⅆ𝑙𝑠𝑖𝑛𝜃 −𝜔 𝑠𝑖𝑛(𝜔𝑡1 ) 𝑐𝑜𝑠 𝜔𝑡1

𝐻𝜙 = [ + ]
4𝜋 𝑐𝑟 𝑟2

The only magnetic field component which exists in the current elements(Hertzian dipole) is 𝐻𝜙
Electric field components
𝜕𝐷
𝛻Χ𝐻 =
𝜕𝑡

𝜕𝐸
𝛻Χ𝐻 = 𝜀
𝜕𝑡
𝐻𝑟 = 0
𝑎𝑟 𝑟𝑎𝜃 𝑟 sin 𝜃 𝑎𝜙 𝜕
1 𝜕 𝜕 𝜕 𝜕𝐸 =0
𝜕𝜙 𝐻𝜃 = 0
=𝜀
𝑟 2 𝑠𝑖𝑛𝜃 𝜕𝑟 𝜕𝜃 𝜕𝜙 𝜕𝑡
0 0 𝑟 sin 𝜃 𝐻𝜙

𝑎𝑟 𝑟𝑎𝜃 𝑟 sin 𝜃 𝑎𝜙
1 𝜕 𝜕 𝜕 𝜕(𝐸𝑟 𝑎𝑟 + 𝐸𝜃 𝑎𝜃 + 𝐸𝜙 𝑎𝜙 )
=𝜀
𝑟 2 𝑠𝑖𝑛𝜃 𝜕𝑟 𝜕𝜃 𝜕𝜙 𝜕𝑡
0 0 𝑟 sin 𝜃 𝐻𝜙
 1 𝜕(𝑟 sin 𝜃𝐻𝜙 ) 𝜕𝐸 1 𝜕 𝑟 sin 𝜃𝐻𝜙 𝜕𝐸𝜃
= 𝜀 𝜕𝑡𝑟 ; − =𝜀 ; 𝐸𝜙 = 0
𝑟 2 𝑠𝑖𝑛𝜃 𝜕𝜃 𝑟𝑠𝑖𝑛𝜃 𝜕𝑟 𝜕𝑡

1 𝜕(𝑟𝐴𝜃 ) 𝜕𝐴𝑟
Where 𝐻𝜙 = 𝜇𝑟 [ − ]
𝜕𝑟 𝜕𝜃

2𝐼𝑚 ⅆ𝑙 𝑐𝑜𝑠𝜃 𝑐𝑜𝑠𝜔𝑡1 𝑠𝑖𝑛 𝜔𝑡1

𝐸𝑟 = [ + ]
4𝜋𝜀 𝑐𝑟 2 𝜔𝑟 3

𝐼𝑚 ⅆ𝑙 𝑠𝑖𝑛𝜃 −𝜔 𝑠𝑖𝑛 𝜔𝑡1 𝑐𝑜𝑠 𝜔𝑡1 𝑠𝑖𝑛 𝜔𝑡1

𝐸𝜃 = [ + + ]
4𝜋𝜀 𝑐2𝑟 𝑐𝑟 2 𝜔𝑟 3

The electric field components which exists in the current elements (Hertzian
dipole) are 𝐸𝑟 𝑎𝑛ⅆ 𝐸𝜃 𝑎𝑛ⅆ 𝐻𝜙 .

Only three field components of current elements (Hertzian dipole) are

𝐸𝑟 , 𝐸𝜃 𝑎𝑛ⅆ 𝐻𝜙 exist out of 𝐸𝑟 , 𝐸𝜃 , 𝐸𝜙 , 𝐻𝑟 , 𝐻𝜃 𝑎𝑛ⅆ 𝐻𝜙 .
Power radiated by the current element

• The pointing power flow can be obtained by multiplying 𝐸𝜃 𝑎𝑛ⅆ𝐻∅

• i.e. 𝑊𝑟 = 𝐸𝜃 𝐻∅
• 𝑊𝑟 =𝐼𝑚𝑑𝑙4𝜋𝜀𝑠𝑖𝑛𝜃 [−𝜔 𝑠𝑖𝑛
𝑐 2𝑟
𝜔𝑡1
+
𝑐𝑜𝑠 𝜔𝑡1
𝑐𝑟 2
+
𝑠𝑖𝑛 𝜔𝑡1 𝐼𝑚 𝑑𝑙𝑠𝑖𝑛𝜃 −𝜔 𝑠𝑖𝑛 𝜔𝑡1 )
𝜔𝑟 3
]
4𝜋
[
𝑐𝑟
+
𝑐𝑜𝑠 𝜔𝑡1
𝑟2
]

(𝐼𝑚 𝑑𝑙 𝑠𝑖𝑛𝜃)2 𝜔2
• 𝑊𝑟 (𝑎𝑣𝑔)= +0
(4𝜋)2 𝜀 2𝐶 3 𝑟 2
𝜂(𝐼𝑚 𝑑𝑙 𝑠𝑖𝑛𝜃)2 1
• 𝑊𝑟 (𝑎𝑣𝑔)= where 𝜂 =
8𝜆2 𝑟 2 𝜀𝐶
 𝐸𝜃
•𝜂=
𝐻𝜙
• For field region
𝐼𝑚 𝑑𝑙 𝑠𝑖𝑛𝜃 −𝜔 𝑠𝑖𝑛 𝜔𝑡1
[ ]
𝑐2 𝑟
•𝜂= 4𝜋𝜀
𝐼𝑚 𝑑𝑙𝑠𝑖𝑛𝜃 −𝜔 𝑠𝑖𝑛(𝜔𝑡1 )
[ ]
4𝜋 𝑐𝑟
1
•𝜂= = 120𝜋 = 377Ω
𝜀𝐶
• The total power radiated by the current element is given by
• Poynting vector over the surface of the sphere is

𝜋 2𝜋
• 𝑃𝑟𝑎𝑑 = ‫=𝜃׬‬0 ‫=𝜙׬‬0 𝑊𝑟 . ⅆ𝑠
𝜋 2𝜋 𝜂(𝐼𝑚 𝑑𝑙 𝑠𝑖𝑛𝜃)2 2
• 𝑃𝑟𝑎𝑑 = ‫=𝜃׬‬0 ‫=𝜙׬‬0 𝑟 sin 𝜃 ⅆ𝜃 ⅆ𝜙
8𝜆2 𝑟 2
2 𝑑𝑙 2 2
• 𝑃𝑟𝑎𝑑 = 80𝜋 𝐼𝑟𝑚𝑠
𝜆
2
• 𝑃𝑟𝑎𝑑 = 𝑅𝑟𝑎𝑑 𝐼𝑟𝑚𝑠
2 𝑑𝑙 2
• 𝑅𝑟𝑎𝑑 =80𝜋
𝜆
Directivity of current element

Directivity” of an antenna: the ratio of the maximum radiated intensity to the average radiated
intensity. Directivity gives a measure of how strongly directional is the radiation pattern.

𝑅𝑎ⅆ𝑖𝑎𝑡𝑖𝑜𝑛 𝐼𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦 𝑚𝑎𝑥

𝐷=
𝑅𝑎ⅆ𝑖𝑎𝑡𝑖𝑜𝑛 𝐼𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦 𝑎𝑣𝑔
4𝜋𝑃𝑚𝑎𝑥
𝐷=
𝑃𝑟𝑎𝑑
2
2
ⅆ𝑙 2
𝑃𝑟𝑎𝑑 = 80𝜋 𝐼𝑟𝑚𝑠
𝜆
𝑃𝑚𝑎𝑥 = 𝑟 2 𝑊𝑟𝑎𝑑
2
2
𝐸𝑚𝑎𝑥
𝑃𝑚𝑎𝑥 = 𝑟
𝜂
 The electric field in the current element 𝐼ⅆ𝑙 𝑟 ≫ 𝑙 𝑖. 𝑒 𝑓𝑜𝑟 𝑓𝑖𝑒𝑙ⅆ 𝑟𝑒𝑔𝑖𝑜𝑛
𝐼𝑚 𝑑𝑙 𝑠𝑖𝑛𝜃 −𝜔 𝑠𝑖𝑛 𝜔𝑡1
𝐸𝜃 = 4𝜋𝜀
[ 𝑐 2𝑟 ]
𝐼𝑚 ⅆ𝑙 𝑠𝑖𝑛𝜃 −𝜔 𝑠𝑖𝑛 𝜔𝑡1
𝐸𝑚𝑎𝑥 = | |
4𝜋𝜀 𝑐2𝑟
𝐼𝑚 ⅆ𝑙𝜔
𝐸𝑚𝑎𝑥 =| |
4𝜋𝜀𝑐 2 𝑟

1
𝜔 = 2π𝑓, 𝑐 = 𝜆𝑓, 𝜂 =
𝜀𝑐
𝐼𝑚 ⅆ𝑙
𝐸𝑚𝑎𝑥 = .𝜂
2𝜆𝑟
2
𝐸𝑚𝑎𝑥
𝑃𝑚𝑎𝑥 = 𝑟2
𝜂
2
ⅆ𝑙
𝑃𝑚𝑎𝑥 = 30𝜋 2 2
𝐼𝑟𝑚𝑠
𝜆
 4𝜋𝑃𝑚𝑎𝑥
𝐷=
𝑃𝑟𝑎𝑑

ⅆ𝑙 2 2
4𝜋. 30𝜋 2 𝐼𝑟𝑚𝑠
𝜆
𝐷=
ⅆ𝑙 2 2
80𝜋 2 𝐼𝑟𝑚𝑠
𝜆

3
𝐷=
2
𝐷 = 1.5 = 1.76 ⅆ𝐵
Short dipole
• The Short dipole is the dipole antenna having the length of its wire
shorter than the wavelength
λ 𝜆
• A current element whose length < 𝑙≤ is
is called small dipole
50 10
antenna
L = 𝜆/10
• The wire that leads to the antenna must be less than one-tenth of the
wavelength.
Where
•L is the length of the wire of the short dipole.
•λ is the wavelength.

2 𝑑𝑙 2
𝑅𝑟𝑎𝑑 =20𝜋 𝜆
 Half-Wave Dipole Antenna
• The half wave dipole is formed from a conducting element which is wire or metal
tube which is an electrical half wavelength long. The half wave dipole is normally
fed in the middle where the impedance falls to its lowest. In this way, the antenna
consists of the feeder connected to two quarter wavelength elements in line with
each other.
• It is considered as a very popular form of a dipole antenna and is sometimes known
as Hertz antenna.
• Generally, it is known to be the one with the simplest resonance structure in
antennas for the transmission and receptions applications. It is present as the
fundamental element of all the antenna shapes, as these are used for constructing
various complex antennas.
• The operating frequency range of the half wave dipole antennas lies between 3 kHz
to 300 GHz.
𝑅𝑟 = 73 𝑜ℎ𝑚 ; D = 1.643 = 2.156 dB
 Monopole
• A monopole antenna is one half of a dipole antenna,
almost always mounted above some sort of ground
plane. The case of a monopole antenna of
length L mounted above an infinite ground plane is
shown in Figure 1(a).
• The radiation pattern of monopole antennas above a
ground plane are also known from the dipole result.
The only change that needs to be noted is that the
impedance of a monopole antenna is one half of
that of a full dipole antenna.
• 𝑅𝑟 = 36.5 ohm ; D = 3.286 = 5.167 dB

https://www.antenna-theory.com/antennas/monopole.php
 Loop antenna
• 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.
• The frequency range of operation of loop antenna is around 300MHz to 3GHz. This
antenna works in UHF range.
• A loop antenna is a coil carrying radio frequency 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

• 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
Where, L is the length of the antenna, λ is the wavelength
• The main parameter of this antenna is its perimeter length, which is
about a wavelength and should be an enclosed loop. It is not a good
idea to meander the loop so as to reduce the size, as that increases
capacitive effects and results in low efficiency.
Small loop antennas

• Small loop antennas are also called as magnetic loop antennas. These
are less resonant. These are mostly used as receivers.
• These antennas are of the size of one-tenth of the wavelength.
• 𝐿 = 𝜆/10
Where, L is the length of the antenna, λ is the wavelength
The features of small loop antennas are −
• A small loop antenna has low radiation resistance. If multi-turn ferrite
core constructions are used, then high radiation resistance can be
achieved.
• It has low radiation efficiency due to high losses.
• Its construction is simple with small size and weight.
• Due to its high reactance, its impedance is difficult to match with the
transmitter. If loop antenna have to act as transmitting antenna, then
this impedance mis-match would definitely be a problem. Hence,
these loop antennas are better operated as receiver antennas.

https://www.tutorialspoint.com/antenna_theory/antenna_theory_loop.htm
Antenna Arrays

An antenna array can be designed to give a particular

shape of radiating pattern. Control of the phase and current
driving each array element along with spacing of array
elements can provide beam steering capability.

For simplification:
 All antenna elements are identical
 The current amplitude is the same feeding each element.
 The radiation pattern lies only in xy plane, θ=π/2
The radiation pattern then can be controlled by:
 controlling the spacing between elements or
 controlling the phase of current driving for each element
Antenna Arrays (Cont’d..)

For simple example, consider a pair of dipole antennas driven in

phase current source and separated by λ/2 on the x axis.
Assume each antenna radiates
independently, at far field point P, the
fields from 2 antennas will be 180 out-
of-phase, owing to extra λ/2 distance
travel by the wave from the farthest
antenna  fields cancel in this
direction. At point Q, the fields in phase
and adds. The E field is then twice from
single dipole, fourfold increase in power
 broadside array  max radiation is
directed broadside to axis of elements.
Antenna Arrays (Cont’d..)

Modify with driving the pair of dipoles

with current sources 180 out of phase.
Then along x axis will be in phase and
along y axis will be out of phase, as
shown by the resulting beam pattern
 endfire array  max radiation is
directed at the ends of axis containing
array elements.
• Broadside – Propagates on Z-
Axis, perpendicular to antenna
array. Signal from individual
elements transmitted in phase.

• End Fire – Radiates only on X-

Axis, in-line with antenna array.
Element signals transmitted 180
degs out of phase

• Individual phase shifts between 0

– 180 degs will result in antenna
pattern shifting
Linear Array Electronic Steering
• By inducing phase shift between elements, can electronically focus
beam and steer it
• No mechanical breakdown, physical restrictions, and can use same
system for multiple roles (Search, Tracking, Fire Control)
simultaneously
 TWO ELEMENT ARRAY ANTENNA

• Consider two point sources A1 and A2 separated by ‘d’

distance as shown in figure.
• The path difference can be expressed in terms of wavelength
𝑑𝑐𝑜𝑠𝜙
Path difference =
𝜆
• Hence the phase angle due to path difference is given by
𝑑𝑐𝑜𝑠𝜙
Phase angle(𝜓) = (2𝜋) 2𝜋
𝜆 𝛽=
𝜆
𝜓 = 𝛽ⅆ𝑐𝑜𝑠𝜙
• the phase angle due to path difference and in currents is given by

𝜓 = α + 𝛽ⅆ𝑐𝑜𝑠𝜙
 ORIGIN AT ELEMENT 1
𝑬𝑻

𝜓
𝑬𝑻 = 2𝐸0 𝑐𝑜𝑠
2

A.F

Source:Prof. Girish Kumar Electrical Engineering Department, IIT Bombay

 𝜓 𝜓
−𝑗 2 𝑗2
𝐸𝑇 = 𝐸0 𝑒 +𝑒
𝜓 𝜓
−𝑗 2 𝑗2
𝑒 +𝑒
𝐸𝑇 = 2𝐸0
2
𝜓
𝐸𝑇 = 2𝐸0 𝑐𝑜𝑠
2

𝑤ℎ𝑒𝑟𝑒, 𝜓 = α + 𝛽ⅆ𝑐𝑜𝑠𝜙

Source:Prof. Girish Kumar Electrical Engineering Department, IIT Bombay

Array Factor = It is the ratio of the magnitude of the resultant total field to the
magnitude of the maximum field.
𝐸𝑇
𝐴𝐹 =
𝐸𝑚𝑎𝑥
Two element sources with Equal Amplitude and In Phase (EAIP)

• two point sources A1 and A2 are fed with equal amplitude and
same phase.
i.e α=0

Total phase angle at observation is 𝜓 = α + 𝛽ⅆ𝑐𝑜𝑠𝜙

Finally, the phase difference is 𝜓 = 𝛽ⅆ𝑐𝑜𝑠𝜙 For two element sources with EAIP
 𝜓
𝑬𝑻 = 2𝐸0 𝑐𝑜𝑠
2
𝛽ⅆ𝑐𝑜𝑠𝜙 2𝜋
𝑬𝑻 = 2𝐸0 𝑐𝑜𝑠 𝑤𝑒 𝑘𝑛𝑜𝑤 𝛽 =
2 𝜆
2𝜋 ⅆ𝑐𝑜𝑠𝜙 𝑡ℎ𝑒 𝑠𝑒𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡𝑤𝑜 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑠ℎ𝑜𝑢𝑙ⅆ 𝑏𝑒
𝑬𝑻 = 2𝐸0 𝑐𝑜𝑠
𝜆 2 𝜆
ⅆ=
2𝜋 𝜆 𝑐𝑜𝑠𝜙 2
𝑬𝑻 = 2𝐸0 𝑐𝑜𝑠
𝜆 2 2
𝜋
𝑬𝑻 = 2𝐸0 𝑐𝑜𝑠 𝑐𝑜𝑠𝜙
2 In order to draw radiation pattern the
𝜋 directions of Maxima, minima and half
𝑬𝒏𝒐𝒓𝒎 = 𝑐𝑜𝑠 𝑐𝑜𝑠𝜙
2 power points should be calculated
Maxima Direction
𝜋
• The total field is maximum when 𝑐𝑜𝑠 𝑐𝑜𝑠𝜙 is maximum.
2
• As we know, the variation of cosine of a angle is ±1.
• Hence the condition of maxima is given by
𝜋
𝑐𝑜𝑠 𝑐𝑜𝑠𝜙 =±1
2
𝜋
𝑐𝑜𝑠𝜙 =𝑐𝑜𝑠 −1 ±1
2
𝜋
𝑐𝑜𝑠𝜙 = ±n 𝜋 where n=0,1,2…….
2

If n=0 then

𝜋
𝑐𝑜𝑠𝜙 =0
2
i,.e 𝑐𝑜𝑠𝜙𝑚𝑎𝑥 = 0

𝜙𝑚𝑎𝑥 = 90𝑜 𝑜𝑟 270𝑜

Minima(null) Direction
𝜋
• The total field is minimum when 𝑐𝑜𝑠 𝑐𝑜𝑠𝜙 is minimum. i.e. 0
2
• Hence the condition of minima is given by
𝜋
𝑐𝑜𝑠 𝑐𝑜𝑠𝜙 =0
2
𝜋
𝑐𝑜𝑠𝜙 =𝑐𝑜𝑠 −1 0
2
𝜋 𝜋
𝑐𝑜𝑠𝜙 = ±(2n+1) where n=0,1,2…….
2 2

If n=0 then

𝜋 𝜋
𝑐𝑜𝑠𝜙 =±
2 2
𝑐𝑜𝑠𝜙𝑚𝑖𝑛 = ±1
𝜙𝑚𝑖𝑛 = 𝑐𝑜𝑠 −1 (±1)

𝜙𝑚𝑖𝑛 = 0𝑜 𝑜𝑟 180𝑜
Half Power Point Direction
1
• When the power is half ,the voltage or current is ± times the maximum
2
value.
• Hence the condition of Half Power point is given by
𝜋 1
𝑐𝑜𝑠 𝑐𝑜𝑠𝜙 =±
2 2
𝜋 1
𝑐𝑜𝑠𝜙 =𝑐𝑜𝑠 −1 ±
2 2
𝜋 𝜋
𝑐𝑜𝑠𝜙 = ±(2n+1) where n=0,1,2…….
2 4
If n=0 then
𝜋 𝜋
𝑐𝑜𝑠𝜙 =±
2 4
1
𝑐𝑜𝑠𝜙𝐻𝑃𝑃𝐷 = (± )
2
1
𝜙𝐻𝑃𝑃𝐷 = 𝑐𝑜𝑠 −1 (± )
2

𝜙𝐻𝑃𝑃𝐷 = 60𝑜 𝑜𝑟 120𝑜

Two Isotropic Point Sources of Same Amplitude and Phase
𝜙𝑚𝑎𝑥 =
𝜋
𝑬𝒏𝒐𝒓𝒎 = 𝑐𝑜𝑠 𝑐𝑜𝑠𝜙 𝝓𝑯𝑷𝑷𝑫 =
2
𝜙 0° 90° 60°
𝑬𝒏𝒐𝒓𝒎 0 1

HPBW = 120°

𝜙𝑚𝑖𝑛 =
 The field pattern drawn with 𝑬𝒏𝒐𝒓𝒎 against 𝜙
𝜆
for ⅆ = , EAIP is shown in figure
2

Source:Prof. Girish Kumar Electrical Engineering Department, IIT Bombay

Two element sources with Equal Amplitude and Out of Phase (EAOP)

• two point sources A1 and A2 are fed with equal amplitude and
out of phase.
𝜆
For this case, currents are out of phase α= 𝜋 and spacing between antennas ⅆ = 2
We know 𝜓 = α + 𝛽ⅆ𝑐𝑜𝑠𝜙

𝜓 = 𝜋 +𝛽ⅆ𝑐𝑜𝑠𝜙

Finally, the phase difference is 𝜓 =𝜋 +𝛽ⅆ𝑐𝑜𝑠𝜙For two element sources with

EAOP at any observation point
 𝜓
𝑬𝑻 = 2𝐸0 cos
2
𝜋+𝛽𝑑𝑐𝑜𝑠𝜙
𝑬𝑻 = 2𝐸0 cos( ) 2𝜋
2 𝑤𝑒 𝑘𝑛𝑜𝑤 𝛽 =
𝜆
𝜋 2𝜋 𝑑𝑐𝑜𝑠𝜙 𝑡ℎ𝑒 𝑠𝑒𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡𝑤𝑜 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑠ℎ𝑜𝑢𝑙ⅆ 𝑏𝑒
𝑬𝑻 = 2𝐸0 cos( + )
2 𝜆 2
𝜆
ⅆ=
𝜋 2𝜋 𝜆 𝑐𝑜𝑠𝜙 2
𝑬𝑻 = 2𝐸0 cos( + )
2 𝜆 2 2

𝜋 𝜋
𝑬𝑻 = 2𝐸0 𝑐𝑜𝑠 + 𝑐𝑜𝑠𝜙
2 2 In order to draw radiation pattern the
𝜋 directions of Maxima, minima and half
𝑬𝒏𝒐𝒓𝒎 = 𝑠𝑖𝑛 𝑐𝑜𝑠𝜙
2 power points should be calculated
Maxima Direction
𝜋
• The total field is maximum when 𝑠𝑖𝑛 𝑐𝑜𝑠𝜙 is maximum.
2
• As we know, the variation of sin of a angle is ±1.
• Hence the condition of maxima is given by
𝜋
sin 𝑐𝑜𝑠𝜙 =±1
2
𝜋
𝑐𝑜𝑠𝜙 =𝑠𝑖𝑛−1 ±1
2
𝜋 𝜋
𝑐𝑜𝑠𝜙 =±(2n+1) where n=0,1,2…….
2 2

If n=0 then
𝜋 𝜋
𝑐𝑜𝑠𝜙 =±
2 2
i.e. 𝑐𝑜𝑠𝜙𝑚𝑎𝑥 = ±1

𝜙𝑚𝑎𝑥 = 0𝑜 𝑜𝑟 180𝑜
Minima Direction
𝜋
• The total field is maximum when 𝑐𝑜𝑠 𝑐𝑜𝑠𝜙 is minimum. i.e. 0
2
• Hence the condition of minima is given by
𝜋
𝑠𝑖𝑛 𝑐𝑜𝑠𝜙 =0
2
𝜋
𝑐𝑜𝑠𝜙 =𝑠𝑖𝑛−1 0
2
𝜋
𝑐𝑜𝑠𝜙 = ±n 𝜋 where n=0,1,2…….
2

If n=0 then

𝜋
𝑐𝑜𝑠𝜙 =0
2
𝑐𝑜𝑠𝜙𝑚𝑖𝑛 =0
𝜙𝑚𝑖𝑛 = 𝑐𝑜𝑠 −1 (0)

𝜙𝑚𝑖𝑛 = 90𝑜 𝑜𝑟 270𝑜

Half Power Point Direction
1
• When the power is half ,the voltage or current is ± times the maximum
2
value.
• Hence the condition of Half Power point is given by
𝜋 1
sin 𝑐𝑜𝑠𝜙 =±
2 2
𝜋 1
𝑐𝑜𝑠𝜙 =𝑠𝑖𝑛−1 ±
2 2
𝜋 𝜋
𝑐𝑜𝑠𝜙 = ±(2n+1) where n=0,1,2…….
2 4
If n=0 then
𝜋 𝜋
𝑐𝑜𝑠𝜙 =±
2 4
1
𝑐𝑜𝑠𝜙𝐻𝑃𝑃𝐷 = (± )
2
1
𝜙𝐻𝑃𝑃𝐷 = 𝑐𝑜𝑠 −1 (± )
2

𝜙𝐻𝑃𝑃𝐷 = 60𝑜 𝑜𝑟 120𝑜

Two Isotropic Point Sources of Same Amplitude and Out of Phase
𝜋
𝑬𝒏𝒐𝒓𝒎 = 𝑠𝑖𝑛 𝑐𝑜𝑠𝜙
2
𝜙 0° 90° 60°
𝜙𝑚𝑖𝑛 =90° 𝝓𝑯𝑷𝑷𝑫 = 60°
𝑬𝒏𝒐𝒓𝒎 1 0

HPBW = 120°

 The field pattern drawn with 𝑬𝒏𝒐𝒓𝒎 𝜙𝑚𝑎𝑥 = 0°

𝜆
against 𝜙 for ⅆ = , EAOP is shown
2
in figure

Source:Prof. Girish Kumar Electrical Engineering Department, IIT Bombay

 𝜓 𝜓
−𝑗 2 𝑗2
𝐸𝑇 = 𝐸0 𝑒 +𝑒
𝜓 𝜓
−𝑗 2 𝑗2
𝑒 +𝑒
𝐸𝑇 = 2𝐸0
2
𝜓
𝐸𝑇 = 2𝐸0 𝑐𝑜𝑠
2

𝑤ℎ𝑒𝑟𝑒, 𝜓 = α + 𝛽ⅆ𝑐𝑜𝑠𝜙

Source:Prof. Girish Kumar Electrical Engineering Department, IIT Bombay

Two Isotropic Point Sources of Same Amplitude with 900
Phase Difference at λ/2

ф 0° 60° 90° 120° 180°

E 0 1

Source:Prof. Girish Kumar Electrical Engineering Department, IIT Bombay

Two Isotropic Point Sources of Same Amplitude with 900
Phase Difference at λ/4

Spacing between the sources is reduced to λ/4

ф 0° 90° 120° 150° 180°

E 0 0.924 0.994 1

HPBW = 180°
Source:Prof. Girish Kumar Electrical Engineering Department, IIT Bombay
Two element sources with Equal Amplitude and in Phase with spacing d = 𝜆

• two point sources A1 and A2 are fed with equal amplitude and
in phase with spacing d = 𝜆.
ⅆ =𝜆
For this case, currents are out of phase α= 0 and spacing between antennas

We know 𝜓 = α + 𝛽ⅆ𝑐𝑜𝑠𝜙

Finally, 𝜓 = 𝛽𝜆𝑐𝑜𝑠𝜙
 𝜓
𝑬𝑻 = 2𝐸0 cos
2
𝛽𝜆𝑐𝑜𝑠𝜙
𝑬𝑻 = 2𝐸0 cos( ) 2𝜋
2 𝑤𝑒 𝑘𝑛𝑜𝑤 𝛽 =
𝜆
2𝜋 𝜆𝑐𝑜𝑠𝜙 𝑡ℎ𝑒 𝑠𝑒𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡𝑤𝑜 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑠ℎ𝑜𝑢𝑙ⅆ 𝑏𝑒
𝑬𝑻 = 2𝐸0 cos( )
𝜆 2
ⅆ =𝜆
2𝜋 𝜆 𝑐𝑜𝑠𝜙
𝑬𝑻 = 2𝐸0 cos( )
𝜆 2 2

𝑬𝑻 = 2𝐸0 𝑐𝑜𝑠 𝜋𝑐𝑜𝑠𝜙

In order to draw radiation pattern the
𝑬𝒏𝒐𝒓𝒎 = 𝑐𝑜𝑠 𝜋𝑐𝑜𝑠𝜙 directions of Maxima, minima and half
power points should be calculated
Maxima, minima and half power directions of two element sources with Equal Amplitude and in Phase with
spacing d = 𝜆

Maxima direction 𝜙𝑚𝑎𝑥 = 0°/90°/180°/270°

minima direction 𝜙𝑚𝑖𝑛 = 60°/120°/240°/300°

and half power points direction 𝜙𝐻𝑃𝑃𝐷 = 𝟒𝟓°/135°/-45°/-135°

Fig: Pattern of two isotropic elements separated by 𝜆

Source: https://buzztech.in/principle-of-pattern-multiplication/
 Two Same Dipoles and Pattern Multiplication
Dipole Pattern:

For δ = 0, Array Factor (AF)

will give max. radiation in
Broadside Direction

Dipole AF Final Pattern

Source:Prof. Girish Kumar Electrical Engineering Department, IIT Bombay
 PATTERN MULTIPLICATION
Dipole E-Field for Vertical Orientation:

Combined E-Field

Array of two dipole

antennas

Dipole Pattern AF Product of Patterns

Source:Prof. Girish Kumar Electrical Engineering Department, IIT Bombay
Source: https://buzztech.in/principle-of-pattern-multiplication/
Source: https://buzztech.in/principle-of-pattern-multiplication/
Source: https://buzztech.in/principle-of-pattern-multiplication/
 N Isotropic Point Sources of Equal Amplitude and Spacing

𝐸𝑇 = 𝐸0 (1 + 𝑒 𝑗𝜓 + 𝑒 𝑗2𝜓 + 𝑒 𝑗3𝜓 +…………. 𝑒 𝑗(𝑛−1)𝜓 )

𝐸𝑇 𝑒 𝑗𝜓 = 𝐸0 (𝑒 𝑗𝜓 + 𝑒 𝑗2𝜓 + 𝑒 𝑗3𝜓 +…………. 𝑒 𝑗𝑛𝜓 ) We know 𝜓 = α + 𝛽ⅆ𝑐𝑜𝑠𝜙

𝐸𝑇 − 𝐸𝑇 𝑒 𝑗𝜓 =𝐸0 {(1 + 𝑒 𝑗𝜓 + 𝑒 𝑗2𝜓 + 𝑒 𝑗3𝜓 +……. 𝑒 𝑗(𝑛−1)𝜓 )-(𝑒 𝑗𝜓 + 𝑒 𝑗2𝜓 + 𝑒 𝑗3𝜓 +…. 𝑒 𝑗𝑛𝜓 )

𝐸𝑇 − 𝐸𝑇 𝑒 𝑗𝜓 =𝐸0 (1-𝑒 𝑗𝑛𝜓 )

(1−𝑒 𝑗𝑛𝜓 )
𝐸𝑇 = 𝐸0
(1−𝑒 𝑗𝜓 )

Source:Prof. Girish Kumar Electrical Engineering Department, IIT Bombay

 (1−𝑒 𝑗𝑛𝜓 )
𝐸𝑇 = 𝐸0
(1−𝑒 𝑗𝜓 )
𝜓 𝜓 𝜓
𝑗𝑛 2 −𝑗𝑛 2 𝑗𝑛 2
𝑒 (𝑒 −𝑒 )
𝐸𝑇 = 𝐸0 𝜓 𝜓 𝜓
𝑗 −𝑗 𝑗
𝑒 2 (𝑒 2 −𝑒 2 )

𝜓 𝜓 𝜓
𝑗𝑛 𝑗𝑛 −𝑗𝑛
𝑒 2 (𝑒 2 −𝑒 2) 2𝑗
𝐸𝑇 = 𝐸0 𝜓 𝜓 𝜓
𝑋
𝑗 𝑗 −𝑗 2𝑗
𝑒 2 (𝑒 2 −𝑒 2 )

𝜓
𝜓 si𝑛(𝑛
𝐸𝑇 = 𝐸0 𝑒 𝑗(𝑛−1) 2 2)
𝜓
si𝑛( 2 )

𝜓
𝑗(𝑛−1) 2
If the reference point is shifted to the centre of the array then 𝑒 term will be eliminated

𝜓
si𝑛(𝑛 2 )
𝐸𝑇 = 𝐸0
𝜓
si𝑛( 2 )
𝐸𝑇 is maximum when 𝜓=0, if 𝑠𝑖𝑛0 ≈ 0 we can write

𝜓
𝑛2
(𝐸𝑇 )𝑚𝑎𝑥 = 𝐸0
𝜓
2
(𝐸𝑇 )𝑚𝑎𝑥 = 𝑛𝐸0

Array Factor= 𝐸𝑛𝑜𝑟𝑚

𝐸𝑇
𝐸𝑛𝑜𝑟𝑚 =
(𝐸𝑇 )𝑚𝑎𝑥
𝜓
si𝑛(𝑛 2 )
𝐸0 𝜓
si𝑛( 2 )
𝐸𝑛𝑜𝑟𝑚 =
𝑛𝐸0

sin(𝑛𝜓2)
𝐸𝑛𝑜𝑟𝑚 =
nsin(𝜓2)
Array of N elements with Equal Spacing, Amplitude and In Phase (Broadside array)

• In broadside array, the field is maximum in the direction

normal to the axis of the array 𝜓 = α + 𝛽ⅆ𝑐𝑜𝑠𝜙

The condition for maximum field at point P is given by

i.e. 𝜓=0

𝛽ⅆ𝑐𝑜𝑠𝜙=0
i.e. 𝑐𝑜𝑠𝜙=0
𝜙 = 90𝑜 𝑜𝑟 270𝑜

Thus 𝜙 = 90𝑜 𝑜𝑟 270𝑜 are called direction of principle maxima(Major lobe)

Maxima Direction(minor lobe maxima)
• For array of N elements with equal amplitude, spacing and in phase, we have
𝑛𝜓
sin
2
𝐸𝑇 = 𝐸0
𝜓
sin
2
𝑛𝜓 𝜓
• The minor lobe maxima can be obtained when sin is max but not sin ≠0
2 2
𝑛𝜓
si𝑛 =±1
2
𝑛𝜓
= sin−1 (±1)
2
𝑛𝜓 3𝜋 5𝜋 7𝜋
=± , ± , ± ,…
2 2 2 2
1𝜋
Here ± givers direction of principle maxima.
2
We know 𝜓 = 𝛽ⅆ𝑐𝑜𝑠𝜙
Hence 𝜓 can be written as ,
2𝜋 3𝜋 5𝜋 7𝜋
ⅆ𝑐𝑜𝑠𝜙=± ,± , ± ,…
𝜆 𝑛 𝑛 𝑛
𝜆 (2𝑚+1)𝜋
𝑐𝑜𝑠𝜙= ± Where m=1,2,3….
2𝜋𝑑 𝑛
 𝜆 (2𝑚+1)𝜋
𝑐𝑜𝑠𝜙=2𝜋𝑑 ± 𝑛

𝜆 (2𝑚+1)𝜋
𝜙=𝑐𝑜𝑠 −1 ±
2𝜋𝑑 𝑛

−1 (2𝑚+1)𝜆
𝜙𝑚𝑎𝑥 𝑚𝑖𝑛𝑜𝑟 =𝑐𝑜𝑠 ± 2𝑛𝑑

The above equation represents the directions of where certain radiation which is not maxima
i.e. the direction of minor lobe maxima's

For example
𝜆
Let n=4, d=2 and α=0
−1 ± (2𝑚+1)𝜆
Then 𝜙𝑚𝑎𝑥 𝑚𝑖𝑛𝑜𝑟 =𝑐𝑜𝑠 for m=1
2𝑛𝑑
−1 ± (2+1)𝜆
Then 𝜙𝑚𝑎𝑥 𝑚𝑖𝑛𝑜𝑟 =𝑐𝑜𝑠 𝜆
2(4)2
−1 ± 3
𝜙𝑚𝑎𝑥 𝑚𝑖𝑛𝑜𝑟 =𝑐𝑜𝑠 4
𝜆
Thus +41.4°, +138.6°, -41.4°, -138.6° are the four minor lobe maxima of 4-isotropic sources with in phase and spaced 2
Minima Direction(Nulls)
• For array of N elements with equal amplitude, spacing and in phase, we have
𝑛𝜓
sin
2
𝐸𝑇 = 𝐸0
𝜓
sin
2
𝑛𝜓 𝜓
• To find the direction of minima sin = 0 but not sin ≠0
2 2
𝑛𝜓
si𝑛 =0
2
𝑛𝜓
= sin−1 (0)
2
𝑛𝜓
=±𝑚𝜋
2
We know 𝜓 = 𝛽ⅆ𝑐𝑜𝑠𝜙
Hence 𝜓 can be written as,
𝑛 2𝜋
ⅆ𝑐𝑜𝑠𝜙=±𝑚𝜋
2 𝜆
𝑛𝑑
𝑐𝑜𝑠𝜙=±𝑚
𝜆
𝜆
𝑐𝑜𝑠𝜙= ±𝑚 Where m=1,2,3….
𝑛𝑑
 𝜆
𝑐𝑜𝑠𝜙=𝑛𝑑 ±𝑚

𝑚𝜆
𝜙=𝑐𝑜𝑠 −1 ± 𝑛𝑑

−1 𝑚𝜆
𝜙𝑚𝑖𝑛 𝑛𝑢𝑙𝑙𝑠 =𝑐𝑜𝑠 ± 𝑛𝑑

The above equation represents the directions of nulls where certain radiation which is minima

For example
𝜆
Let n=4, d=2 and α=0
−1 𝑚𝜆
Then 𝜙𝑚𝑖𝑛 𝑛𝑢𝑙𝑙𝑠 =𝑐𝑜𝑠 ± 𝑛𝑑
−1 ± (1)𝜆
Then 𝜙𝑚𝑖𝑛 𝑛𝑢𝑙𝑙𝑠 =𝑐𝑜𝑠 𝜆 if m=1
(4)
2

−1 1
𝜙𝑚𝑖𝑛 𝑛𝑢𝑙𝑙𝑠 =𝑐𝑜𝑠 ± 2 if m=1

Thus ± 60°, ± 120° if m=1

𝜆
± 0°, ± 180° if m=2 here 4 linear isotropic sources are fed with in phase and spaced 2 distance
Beamwidth of major lobe
Half Power BeamWidth is the angle equal to twice the angle between first null

FNBW=2ɣ where ɣ=90- 𝜙

−1 ± 𝑚𝜆
𝜙𝑚𝑖𝑛 𝑛𝑢𝑙𝑙𝑠 =𝑐𝑜𝑠 𝑛𝑑
𝜙𝑚𝑖𝑛 =90- ɣ
𝑚𝜆 𝑚𝜆
−1 ⟹ cos 90− ɣ = ±
90- ɣ=𝑐𝑜𝑠 ±
𝑛𝑑 𝑛ⅆ
𝑚𝜆
sin(ɣ)=±
𝑛𝑑
if ɣ is very small, sin(ɣ) ≈ ɣ
𝜆
For first null i.e. 𝑚=1 ɣ=
𝑛𝑑
2𝜆
FNBW=
𝑛𝑑
The total length of array 𝐿 =(𝑛 − ⅆ) ≈ 𝑛ⅆ if n is very large
2𝜆
FNBW=
𝐿
2
FNBW= 𝑟𝑎ⅆ
(𝐿/𝜆)
𝑭𝑵𝑩𝑾 𝒊𝒏 𝒅𝒆𝒈𝒓𝒆𝒆𝒔
2 1 𝑟𝑎ⅆ=57.3 degree
FNBW=(𝐿/𝜆) 𝑟𝑎ⅆ

2x57.3
FNBW= degree
(𝐿/𝜆)

114.6
FNBW= degree
(𝐿/𝜆)

𝑯𝑷𝑩𝑾 is given by,

FNBW 1
HPBW= = 𝑟𝑎ⅆ
2 (𝐿/𝜆)

57.3
HPBW= (𝐿/𝜆) degree
 Broadside Array (Sources In Phase)

ф Ψ E
0° π 0
90° π/2 0
120° 0 1
Field pattern of 4 isotropic point sources with the
same amplitude and phase and spacing of /2.
Source:Prof. Girish Kumar Electrical Engineering Department, IIT Bombay
Array of N elements with Equal Spacing, Amplitude and Progressive Phase
(End fire array)
• In end fire array, the field is maximum in the direction normal
to the axis of the array 𝜓 = α + 𝛽ⅆ𝑐𝑜𝑠𝜙
The condition for maximum field at point P is given by
i.e. 𝜓= 𝛽ⅆ(𝑐𝑜𝑠𝜙-1)
The condition of principle maxima is given by 𝜓=0

𝛽ⅆ(𝑐𝑜𝑠𝜙 − 1)=0
i.e. 𝑐𝑜𝑠𝜙=1
𝜙 = 0𝑜 𝑜𝑟 180𝑜
𝜓 = α + 𝛽ⅆ𝑐𝑜𝑠𝜙
0= α + 𝛽ⅆ𝑐𝑜𝑠𝜙
α= −𝛽ⅆ𝑐𝑜𝑠0𝑜 α= −𝛽ⅆ𝑐𝑜𝑠180𝑜
α= −𝛽ⅆ α= 𝛽ⅆ
𝑜 𝑜
Thus 𝜙 = 0 𝑜𝑟 180 are called direction of principle maxima(Major lobe)
Maxima Direction(minor lobe maxima)
• For array of N elements with equal amplitude, spacing and in phase, we have
𝑛𝜓
sin
2
𝐸𝑇 = 𝐸0
𝜓
sin
2
𝑛𝜓 𝜓
• The minor lobe maxima can be obtained when sin is max but not sin ≠0
2 2
𝑛𝜓
si𝑛 =±1
2
𝑛𝜓
= sin−1 (±1)
2
𝑛𝜓 3𝜋 5𝜋 7𝜋
=± , ± , ± ,…
2 2 2 2
𝑛𝛽𝑑(𝑐𝑜𝑠𝜙−1) (2𝑚+1)𝜋
=±
2 2
(2𝑚+1)𝜋
𝑐𝑜𝑠𝜙 − 1=± Where m=1,2,3….
𝑛𝛽𝑑
 (2𝑚+1)𝜋
𝑐𝑜𝑠𝜙 − 1=± 𝑛𝛽𝑑

(2𝑚+1)𝜋
𝑐𝑜𝑠𝜙=± +1
𝑛𝛽𝑑

−1 (2𝑚+1)𝜋
𝜙𝑚𝑎𝑥 𝑚𝑖𝑛𝑜𝑟 =𝑐𝑜𝑠 ± +1
𝑛𝛽𝑑

The above equation represents the directions of where certain radiation which is not maxima
i.e. the direction of minor lobe maxima's

For example
𝜆
Let n=4, d=2 and α=- 𝜋
−1
𝜙𝑚𝑎𝑥 𝑚𝑖𝑛𝑜𝑟1 =𝑐𝑜𝑠 0.25 = 75.5°, for m=1

−1
𝜙𝑚𝑎𝑥 𝑚𝑖𝑛𝑜𝑟1 =𝑐𝑜𝑠 −0.25 = −75.5°, for m=2
Minima Direction(Nulls)
• For array of N elements with equal amplitude, spacing and in phase, we have
𝑛𝜓
sin
2
𝐸𝑇 = 𝐸0
𝜓
sin
2
𝑛𝜓 𝜓
• To find the direction of minima sin = 0 but not sin ≠0
2 2
𝑛𝜓
si𝑛 =0
2
𝑛𝜓
= sin−1 (0)
2
𝑛𝜓
=±𝑚𝜋
2
We know 𝜓 = 𝛽ⅆ𝑐𝑜𝑠𝜙
Hence 𝜓 can be written as,
𝑛 2𝜋
ⅆ𝑐𝑜𝑠𝜙=±𝑚𝜋
2 𝜆
𝑛𝑑
(𝑐𝑜𝑠𝜙 − 1)=±𝑚
𝜆
𝜆
𝑐𝑜𝑠𝜙-1= ±𝑚 Where m=1,2,3….
𝑛𝑑
 𝑚𝜆
𝑐𝑜𝑠𝜙= ± 𝑛𝑑 + 1

𝑚𝜆
𝜙=𝑐𝑜𝑠 −1 ± 𝑛𝑑 + 1 (1)

−1 𝑚𝜆
𝜙𝑚𝑖𝑛 𝑛𝑢𝑙𝑙𝑠 =𝑐𝑜𝑠 1 ± 𝑛𝑑

From above equation1 𝜙𝑚𝑖𝑛 𝑚𝜆

2𝑠𝑖𝑛2 =±
2 𝑛ⅆ

𝜙𝑚𝑖𝑛 𝑚𝜆
𝑠𝑖𝑛 = ±
2 2𝑛ⅆ
−1 𝑚𝜆
𝜙𝑚𝑖𝑛 𝑛𝑢𝑙𝑙𝑠 =2𝑠𝑖𝑛 ± 2𝑛𝑑

The above equation represents the directions of nulls where certain radiation which is minima
For example
𝜆
Let n=4, d= and α=𝜋
2
−1 1 ± 𝑚𝜆
Then 𝜙𝑚𝑖𝑛 𝑛𝑢𝑙𝑙𝑠 =𝑐𝑜𝑠 𝑛𝑑
𝜆
𝜙𝑚𝑖𝑛 𝑛𝑢𝑙𝑙𝑠 = ± 60°, ±90°, ±120°, ± 180° here 4 linear isotropic sources are fed with in phase and spaced 2 distance
Beamwidth of major lobe
Half Power BeamWidth is the angle equal to twice the angle between first null
FNBW=2𝜙𝑚𝑖𝑛
−1 ± 𝑚𝜆
𝜙𝑚𝑖𝑛 𝑛𝑢𝑙𝑙𝑠 =2𝑠𝑖𝑛 2𝑛𝑑

𝜙𝑚𝑖𝑛 𝑚𝜆 𝜙𝑚𝑖𝑛 =90- ɣ

𝑠𝑖𝑛 = ±
2 2𝑛ⅆ
𝜙𝑚𝑖𝑛 𝜙𝑚𝑖𝑛
if𝜙𝑚𝑖𝑛 is very small, sin( )≈
2 2
𝜙𝑚𝑖𝑛 𝑚𝜆
= ±
2 2𝑛ⅆ

2𝑚𝜆
𝜙𝑚𝑖𝑛 =±
𝑛ⅆ
The total length of array 𝐿 =(𝑛 − ⅆ) ≈ 𝑛ⅆ if n is very large
2𝑚𝜆
𝜙𝑚𝑖𝑛 = ±
𝐿

2𝑚𝜆
FNBW= ± 2
𝐿
𝑭𝑵𝑩𝑾 𝒊𝒏 𝒅𝒆𝒈𝒓𝒆𝒆𝒔
2𝑚𝜆 1 𝑟𝑎ⅆ=57.3 degree
FNBW= ± 2 𝐿
𝑟𝑎ⅆ

2𝑚𝜆
If m=1, FNBW=± 2 𝐿
x57.3 degree

2𝑚𝜆
FNBW=± 144 degree
𝐿

𝑯𝑷𝑩𝑾 is given by,

FNBW 2𝑚𝜆
HPBW= = 𝑟𝑎ⅆ
2 𝐿

2𝑚𝜆
HPBW=57.3 degree
𝐿
 Ordinary Endfire Array

BWFN=120°

Field pattern of ordinary end-fire array of 4

isotropic point sources of same amplitude.
Spacing is /2 and the phase angle  = -.
Source:Prof. Girish Kumar Electrical Engineering Department, IIT Bombay
𝛼 𝛽 𝛽

𝛼 𝛽 𝛽
𝛽 𝛼 𝛽 𝛼

𝛽 𝛼 𝛽 𝛼
Non uniform Amplitude Distribution:
Binomial array
• The binomial BSA was investigated and proposed by J.S.Stones to synthesize patterns
without side lobes.
• Various cases of uniform(equal) amplitude distributions were discussed in arrays. The
study of arrays with non uniform amplitudes is also essential and Binomial array is one.
• The excitation amplitude distribution can be obtained easily by the expansion of the
binome in Making use of Pascal’s triangle:
• Binomial arrays are mainly used to eliminate the minor lobes.
𝜆
• The spacing between two consecutive radiating sources should not exceed
2
Advantages of Binomial arrays
• In binomial arrays, minor lobes are eliminated so radiation can be achieved in the desired direction

Disadvantages of Binomial arrays

• HPBW increases and hence directivity decrease
• For designing larger arrays, larger amplitude ratio of sources is required.