Array of Point Source

7/28/2019 Array of Point Source

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EEL 338 2008EEL 338 2008--0909 Antennas & PropagationAntennas & Propagation 55

The far-field approximation of the two-element array problem:

Assumptions: The array elements are

Identical, i.e.,

Oriented in the same way in space(they have identical polarization),i.e.,

excitation is of the same amplitude,i.e.,


Then, the total field is:

The total field of the array is equal to the product of the fieldcreated by a single element located at the origin and the Ar r a y

f a c t o r ,AF.


The normalizedAF,

Using the normalized field pattern of a single element, En( ,), thenormalized field pattern of the array is expressed as:

Since, the array factor does not depend on the directionalcharacteristics of the individual elements, it can beformulated by replacing the actual elements with isotropic(point) sources assuming that each point source hasamplitude, phase and location of the corresponding elementit is replacing.


The field pattern of an array of nonThe field pattern of an array of non--isotropic but similar pointisotropic but similar pointsources is thesources is the product ofproduct of thethe pattern of thepattern of the individual sourcesourceand theand the pattern of an array of isotropic point sourcespattern of an array of isotropic point sources havinghavingthe same locations, relative amplitudes and phase as thethe same locations, relative amplitudes and phase as the

nonnon--isotropic sources.isotropic sources.

PATTERNMULTIPLICATION

The total field pattern of an array of non-isotropic but similarpoint sources is the product of the pattern of the individualsource and the pattern of the array of isotropic point sourceshaving the same locations, relative amplitudes and phase,while the total phase pattern is the sum of the phasepatterns of the individual source and the array of isotropicsources.


3/28


Example 1: An array consists of two horizontal infinitesimaldipoles located at a distance d= / 4 from each other. Findthe nulls of the total field, if the excitation magnitudes are thesame and the phase difference is:

a) = 0;

b) =/2;c) = /2


The element factor En(,) does not depend on , and it produces thesame null in all three cases. Since En(,) =|cos|, the null is at1 =/ 2.

TheAFdepends on and produces different results in the 3 cases:

a) = 0

A solution with a real-valued angle does not exist.

In this case, the total field pattern has only 1 null at

=90.

EEL 338 2008EEL 338 2008--0909 Antennas & PropagationAntennas & Propagation 1111 EEL 338 2008EEL 338 2008--0909 Antennas & PropagationAntennas & Propagation 1212

0.0

0.2

0.4

0.6

0.8

1.0

0

30

60

90

120

150

180

210

240

270

300

330

0.0

0.2

0.4

0.6

0.8

1.0

d=/2d=/4d=/8

Array factor for various values of d (=0):


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b) = /2

The equation

does not have a solution.

The total field pattern has 2 nulls: 1 = 90 and 2 = 0



0.0

0.2

0.40.6

0.8

1.0

0

30

60

90

120

150

180

210

240

270

300

330

0.0

0.2

0.4

0.6

0.8

1.0

d=/2d=/4d=/8

Array factor for various values of d (=/2):


c) = /2

The total field pattern has 2

nulls: 1 = 90 and at 2 =180.


5/28


Example 2: Consider a 2-element array of identical(infinitesimal) dipoles oriented along the y-axis. Find theangles of observation, where the nulls of the pattern occur,as a function of the distance between the dipoles, d, andthe phase difference, .

In order to find the nulls:

The element factor, produces one null at 1 =/2


The array factor leads to the following solution:


NN--Element Linear Array: Uniform AmplitudeElement Linear Array: Uniform Amplitude

and Spacingand Spacing

d

d

r1

r2

rN

1

2

3

N

r3

All elements have identical amplitudesbut each succeeding element has aprogressive phase-lead current

excitation equal to relative to thepreceding one.

An array of identical elements all ofidentical magnitude and each with aprogressive phase is referred to as aUniform Array


TheThe Array Factor (AF)Array Factor (AF) is given by:is given by:

( ) ( ) ( )( )

( )( ) ( )

cos 2 cos 1 cos

1 cos 1

1 1

1

cos

j kd j kd j N kd

N Nj n kd j n

n n

AF e e e

AF e e kd

+ + + + + +

+

= =

= + + + +

= = = +

KK

Since the total array factor for a uniform array is a sum ofexponentials, it can be represented by the vector sum of N phasorseach of unit amplitude and progressive phase = relative to the

previous one.

..(1)

Phase terms ofthe partialfields:


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#1

1 0 o1

1 2

1 3

( )1 1N

AF

#2

#3

#4

2

3

It is apparent from thephasor diagram that theamplitude and phase of theAF can be controlled in

uniform arrays by properlyselecting the relative phasebetween the elements.

In non-uniform arrays, theamplitude as well as thephase can be used to controlthe formation anddistribution of the total array

factor


( )1

1

cos

Nj n

n

AF e

kd

=

=

= +

The array factor can be expressed in more recognizable form:

..(1)

Multiply both sides of (1) by , subtract the original equationfrom the resulting equation and rearranging,

je

( )[ ] [ ]

[ ] [ ]

( )

2 21 2

1 2 1 2

1 2

1

1

sin

21sin

2

j N j NjNj N

j j j

j N

e e eAF e

e e e

N

AF e

= =

=

..(2)


sin2

1sin

2

N

AF

=

For small values of , :sin

2

2

N

AF

=

..(3)

..(4)

Here, Nshows the location of the last element with respect tothe reference point in steps with length d.

The phase factor exp[j(N1) / 2] represents the phaseshift of the arrays phase centre relative to the origin, and itwould be one if the origin coincides with the array centre.

Neglecting the phase factor gives (taking physical center ofthe array as phase reference:


To normalize equation (3) or(4), we need the maximum oftheAF. Re-write equation (3)as:

( )

sin

21

sin2

N

AF N

N

=

The function f(x) has itsmaximum atx= 0, ,, and

the value of this maximum isfmax =1.

( )maxAF N=

( )( )

( )

sin

sin

Nxf x

N x=


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The maximum value of equation 3 or 4 is equal to N. Tonormalize the array factors so that the maximum value isequal to unity, equations 3 and 4 are written in normalizedform as:

and

( )sin

1 21

sin2

n

N

AFN

=

..(5)

( )sin

1 2

2

n

N

AF

N

=

..(6)


Nulls of the ArrayNulls of the Array: Equations 5 and 6 are set equal to zero.: Equations 5 and 6 are set equal to zero.

That is,That is,

1

sin 02 2

2cos2

n

N Nn

nd N

= =

= 1,2,3,....n =

,2 ,3 , ....n N N N with equation 5, because for these values of n,equation 5 attains its maximum value as it reduces toform.

( )sin 0 0

The values of determine the order of the nulls (first, second,etc.). For a zero to exist, the argument of arccosine must be

between 1 and+1. Thus the number of nulls that can existwill be a functions of element separation and phase excitationdifference.

n

..(7)


TheThe maximum of equation 5maximum of equation 5 occurs when,occurs when,

( )

( )1

1cos |

2 2

cos 2

2

m

m

kd m

m

d

=

= + =

=

0,1,2,m = K

1cos2

md

=

..(9)

Equation 6Equation 6 has only one maximum and occurs when,has only one maximum and occurs when,

That is, the observation angle that makesThat is, the observation angle that makes

0m =0 =

..(8)


TheThe 33--dB pointdB point for the array factor offor the array factor ofequation 6equation 6::

( )

1

cos | 1.3912 2

2.782cos

2

h

h

N Nkd

d N

=

= + =

=

which can also be written aswhich can also be written as

For large values of it reduces to:For large values of it reduces to:

1 2.782sin2 2

hd N

= ( )d d >>

2.782

2 2h d N

..(10)

The half power beamwidth for a symmetrical pattern is:

2h m h = ..(11)


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Maximum radiation of an array directed normal to the axis of thearray. ( =90 in the present case)

Broadside Array

Maximum of the array factor occurs when (equations 5 and 6):

cos 0kd = + = For broadside array,

90cos | 0

0

kd

== + =

=

o

For broadside pattern, all elements should have same phase andamplitude excitation.

To ensure that there are no maxima in other directions, which are

referred to as grating lobes, the separation between the elementsshould not be equal to multiples of a wavelength when 0 =


Example: N=10

d=/4 d=


Parameters for Broadside ArrayParameters for Broadside Array

Nulls

1cos

1,2,3,...

, 2 , 3 ,...

nn

N d

n

n N N N

=

=

Maxima

1cos

0,1,2,...

m

m

d

m

=

=

1 1.391cos

1

hNd

d


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AFpattern of an End Fire Array (EFA): N= 10, d= /4


Parameters for Ordinary EndParameters for Ordinary End--Fire ArrayFire Array

Nulls

1cos 1

1,2,3,...

, 2 , 3 , ...

nn

N d

n

n N N N

=

=

Maxima

1cos 1

0,1,2,...

m

m

d

m

=

=

1 1.391cos 1

1

hNd

d


10/28


1 10 0

2

2.782 2.782cos cos cos cos

k

HPBWNkd Nkd

=

= +

If L is length of the array:

L dN

d

+=

1 10 0cos cos 0.443 cos cos 0.443HPBW

L d L d

= + + +

These equations can be used to calculate the HPBWof a broadsidearray, too (

0=90=const ). However, they are not valid for end-

fire arrays.



Hansen-Woodyard End-Fire Array

End-fire arrays (EFA) have relatively broad HPBW as comparedto broadside arrays Directivity is low.

To enhance the directivity of an end-fire array, Hansen andWoodyard proposed that the phase shift of an ordinary EFA beincreased:

2.94

2.94

kd kd N

kd kd N

= = +

= + = + +

For maximum in =0

For maximum in =180

Hansen-Woodyard conditions for end-

fire radiation (do not necessarily yieldthe maximum possible directivity)


Hansen-Woodyard conditions, ensure minimumbeamwidth (maximum directivity) in the end-firedirection.

There is, however, a trade-off in the side-lobe level,which is higher than that of the ordinary EFA.

These conditions have to be complemented by additionalrequirements, to ensure low level of the radiation in thedirection opposite to the main lobe:


11/28


Maximum at = 0

In order to ensure a maximum in the =180 direction, we musthave:

0

180 N

=

=

o

o

Hence, Hansen-Woodyard conditions for maximum directivityin the = 0 direction are:

0

180

N

=

=

o

o


N-Element Linear Array: Directivity

1. Broadside Array:

( )

sin cos

1 21sin cos

2

n

Nkd

AFN

kd

=

0 =

( )sin cos

2

cos2

n

d

Nkd

AFN

kd


12/28


22

0 2

1 sinNkd

Nkd

ZU dZ

Nkd Z

=

for a large array the above equation can beapproximated by extending the limits to infinity.

( )2 LargeNkd

2

01 sinZ

U dZNkd Z

+

2

sinZdZ

Z

+

=

0U Nkd

since


The directivity is then,The directivity is then,

max0

0

2U Nkd d

D NU

= =

( )1L N d=

using

overall length of the array

0 2 2 1d L d

D Nd

+

For a large array, L d>>

0 2L

D


2. Ordinary End-Fire Array:


0 4L

D

3. Hansen-Woodyard End-Fire Array:

0 1.789 4L

D



NN--Element Linear Array: ThreeElement Linear Array: Three--Dimensional CharacteristicsDimensional Characteristics

Generally, it is assumed that the linear-array elements are locatedalong thez-axis, which is symmetrical around thez-axis.

If the array axis has an arbitrary orientation, the array factor canbe expressed as

( )( ) ( )1 cos 1

1 1

cos

N Nj n kd j n

n n

n n

AF a e a e

kd

+

= =

= =

= +

is the angle subtended between the array axis and theposition vector to the observation point.


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Thus, if the array axis is along the unit vector a

sin cos sin sin cosa a a a aa x y z = + +

and the position vector to the observation point is

sin cos sin sin cosr x y z = + +

( )

cos sin cos sin cos

sin sin sin sin cos cos

cos sin sin cos cos cos

a a

a a a

a a a

a r x

y z

= = +

+

= +

To Do:Express the array factor for elements along x-axis and y-axis.


NN--Element Linear Array: Uniform Spacing, NonElement Linear Array: Uniform Spacing, Non--uniformuniform

Ampl itudeAmpl itude

The most often used Broad-Side Arrays (BSAs), are classifiedaccording to the type of their excitation amplitude:

1. The uniform BSA relatively high directivity, but the side-lobe levels are high;

2. DolphTschebyscheff BSA for a given number of elementsmaximum directivity is next after that of the uniform BSA;side-lobe levels are the lowest in comparison with the othertwo types of arrays for a given directivity;

3. Binomial BSA does not have good directivity but has verylow side-lobe levels (when d= /2, there are no side lobes at

all).


( )( )

( )

( )( )

( )( )

( ) ( )

2 11 3cos cos cos

2 2 21 2

2 11 3cos cos cos

2 2 21 2

1

1

1

.....

.....

2 12 cos cos

2

2 1cos cos

2

1cos 2 1 c2

Mj kd j kd j kde

M

Mj kd j kd j kd

M

Me

n

n

Me

nnn

M

e nnn

AF a e a e a e

a e a e a e

nAF a kd

nAF a kd

AF a n u u kd

+ + +

=

=

=

= + + + +

+ + +

=

=

= =

os cosd =

Even number (2M) of elements, located symmetrically along thez-axis, with excitation symmetrical with respect toz= 0.

For a broadside array( =0),


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( )

( ) ( )

( ) ( )

( ) ( )

cos 2 cos cos1 2 3 1

cos 2 cos cos2 3 1

1

1

1

1

1

1

2 .....

.....

2 cos 1 cos

cos 1 cos

1cos 2 1 cos c

2

o jkd j kd jMkd M

jkd j kd jMkdM

Mo

n

n

Mo

nnn

Mo

nnn

AF a a e a e a e

a e a e a e

AF a n kd

AF a n kd

dAF a n u u kd

+ + ++

+

+

=+

=+

=

= + + + + +

+ + +

=

=

= = =

os

Odd number (2M+1) of elements, located symmetrically alongthez-axis,

For a broadside array( =0),


Binomial Array

( ) ( )( )( ) ( )( )( )1 2 31 2 1 2 31 1 1 ...

2! 3!

m m m m m mx m x x x

+ = + + + +

The positive coefficients of the series expansion:

m=6

m=5m=5

m=4m=4

m=3m=3

m=2m=2

m=1m=1

15101051

14641

1331

121

11

1


An approximate closed-form expression for the HPBW with d= /2:

1.06 1.06 1.75

1 2HPBW

N L L = = =

The directivity with spacing d= /2 is

( )0 1.77 1.77 2 1D N L = = +



15/28


DolphDolph--Tschebyscheff ArrayTschebyscheff Array

A compromise between uniform and binomial arrays.A compromise between uniform and binomial arrays.

Excitations coefficients are related to TschebyscheffExcitations coefficients are related to Tschebyscheff

polynomials.polynomials.

A DolphA Dolph--Tschebyscheff array with no sidelobes (or sideTschebyscheff array with no sidelobes (or side

lobes of dB) reduces to the binomial design. Thelobes of dB) reduces to the binomial design. The

excitation coefficients for this case, as obtained by bothexcitation coefficients for this case, as obtained by both

methods would be identical.methods would be identical.


Array Factor for symmetric amplitude excitation:Array Factor for symmetric amplitude excitation:

( ) ( ) ( )

( ) ( ) ( )

21

1

2 1 1

cos 2 1

cos 2 1

cos

M

nMn

M

nM n

AF even a n u

AF odd a n u

du

=+

+ =

=

=

=

Summation of M or (M+1) cosine terms.

Largest harmonic of the cosine terms is one less than thetotal no. of elements of the array.

Each cosine term, whose argument is an integer times a

fundamental frequency, can be rewritten as a series ofcosine functions with the fundamental frequency as theargument..


( )cos sin cos sinmjmu

e mu j mu u j u= + = +On taking real parts:

Expanding as binomial series:

( )cos Re cos sinm

mu u j u= +

( ) ( )( )

( ) ( )

( )( )( )( ) ( )

2 2

4 4

1cos cos cos sin

2!

1 2 3cos sin ...

4!

m m

m

m mmu u u u

m m m mu u

=

+

Putting and substituting particular values ofm, we have ( ) ( )

2 2sin 1 cosu u=

By de Moivres theorem:


( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )( ) ( )

( ) ( )

( ) ( )

2

3

4 2

5 3

6 4 2

7

0 cos 1

1 cos cos

2 cos cos 2 2cos 1

3 cos cos 3 4cos 3cos

4 cos cos 4 8cos 8cos 15 cos cos 5 16cos 20cos 5cos

6 cos cos 6 32cos 48cos 18cos 1

7 cos cos 7 64cos

m mu

m mu u

m mu u u

m mu u u u

m mu u u um mu u u u u

m mu u u u u

m mu u

= =

= =

= = =

= = =

= = = += = = +

= = = +

= = =

( ) ( )

( ) ( )

5 3

8 6 4 2

9 7 5 3

112 cos 56 cos 7 cos

8 cos cos 8 128cos 256cos 160cos 32cos 1

9 cos cos 9 256cos 576cos 432cos 120cos 9cos

u u u u

m mu u u u u u

m mu u u u u u u

+

= = = + +

= = = + +


16/28


If we let above equations can be rewritten as:cosz u=

( ) ( )

( ) ( )

( ) ( )

( ) ( )( ) ( )

( ) ( )

( ) ( )

( ) ( )

( )

0

1

22

33

4 24

5 35

6 4 26

7 5 37

8 6 4

0 cos 1

1 cos

2 cos 2 1

3 cos 4 34 cos 8 8 1

5 cos 16 20 5

6 cos 32 48 18 1

7 cos 64 112 56 7

8 cos 128 256 160 3

m mu T z

m mu z T z

m mu z T z

m mu z z T z

m mu z z T z

m mu z z z T z

m mu z z z T z

m mu z z z z T z

m mu z z z

= = =

= = =

= = =

= = == = + =

= = + =

= = + =

= = + =

= = + ( )

( ) ( )

28

9 7 5 39

2 1

9 cos 256 576 432 120 9

z T z

m mu z z z z z T z

+ =

= = + + =

And each is related to a Tschebyscheff (Chebyshev) polynomial ( )mT z


These relations between cosine functions and Tschebyscheffpolynomials are valid only in the range: 1 1z +

( )

( )

cos 1

1m

mu

T z

Q

for 1 1z +

The recursion formula for Tschebyscheff polynomials is:

( ) ( ) ( )1 22m m mT z zT z T z =

The polynomials can also be computed using:

( ) ( )

( ) ( )

1

1

cos cos 1 1

cosh cosh 1, 1

m

m

T z m z z

T z m z z z

= +

= < > +


-1.5 -1.0 -0.5 0.0 0.5 1.0 1.5

-10

-8

-6

-4

-2

0

2

4

6

8

10

Tm(z)

z

T0 T1 T2 T3

T4 T5 T6 T7

The first seven TschebyscheffThe first seven Tschebyscheff

polynomials have been plottedpolynomials have been plotted

below:below:

For1 1z +

( )1 1mT z +

All polynomials pass through (1,1)

All roots occur within 1 1z +

Properties

All maxima and minima withinthis range have values 1


Since, the array factor of an even or odd number ofelements (summation of cosine terms) has the same formas the Tschebyscheff polynomials: The unknowncoefficients of the array factor can be determined byequating the cosine series of the array factor to theappropriate Tschebyscheff polynomial.

The order of the polynomial should be one less than thetotal number of elements of the array.


17/28


18/28


The AF of an N-element array is identical with a Tschebyscheffpolynomial if

( )

( )

( )

11

1

cos 2 1 , 2 ,

cos 2 1 , 2 1,

1cos cos

2

M

n

nN M

n

n

a n u N M even

T z

a n u N M odd

du kd

=

=

= =

= +

= =

Let the side-lobe level (voltage ratio) be

max0

max

1 1sl sl sl

ERE E E AF

= = =


Then, the maximum ofTN1 is fixed at an argumentz0, where

( )max1 0 0NT z R =

This corresponds to

( ) ( )max 0AF u AF u=

Obviously,z0 must satisfy the condition

0 1z >

Then, the portion ofAF(u), which corresponds to TN1

(z) for |z|


Example of DExample of D--T distribution for an array for Ten sourcesT distribution for an array for Ten sources

An array of2M=10 in-phase isotropic sources, with spacingbetween the elements d is to have a side-lobe level of 26 dBbelow the main lobe maximum. Find: (a) the amplitude distributionfulfilling this requirement that produces the minimum beamwidthbetween the first nulls.

SolutionSolution:: The array factor is given by (even case):The array factor is given by (even case):

( ) ( )

( ) ( ) ( ) ( )

( ) ( )

5

21

1 2 310

4 5

cos 2 1

cos cos 3 cos 5

cos 7 cos 9

M

nMn

AF a n u

AF a u a u a u

a u a u

=

=

=

= + +

+ +

cosd

u

=


19/28


Replace cosine terms with their expansions:

( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

3 5 31 2 310

7 5 34

9 7 5 3

5

cos 4cos 3cos 16cos 20cos 5cos

64cos 112cos 56cos 7cos

256cos 576cos 432cos 120cos 9cos

AF a u a u u a u u u

a u u u u

a u u u u u

= + + +

+ +

+ + +

( ) ( )[ ] ( )[ ]

( )[ ] ( )[ ] ( )[ ]

31 2 3 4 5 2 3 4 510

5 7 93 4 5 4 5 5

cos 3 5 7 9 cos 4 20 56 120

cos 16 112 432 cos 64 576 cos 265

AF u a a a a a u a a a a

u a a a u a a u a

= + + + +

+ + + +

After rearranging we get,

( )( )

0 10 0

0

R (dB) = 26 = 20logvoltage ratio 20

RR =

Determine Zo by equating R0 to T9(z):


( ) ( )

( ) ( )

1

1

cos cos 1 1

cosh cosh 1, 1

m

m

T z m z z

T z m z z z

= +

= < > +

Using

( ) ( )

( )

10 9 0 0

10

20 cosh 9cosh

1cosh cosh 20 1.0851

9

R T z z

z

= = =

= =

Substitute:

( ) 0cos 1.0851z z

u z= =


( ) [ ]

[ ] [ ]

[ ] [ ]

1 2 3 4 510

3 5

2 3 4 5 3 4 5

7 9

4 5 5

3 5 7 91.0851

4 20 56 120 16 112 4321.0851 1.0851

64 576 2651.0851 1.0851

zAF a a a a a

z z

a a a a a a a

z za a a

= + + +

+ + + +

+


( ) [ ]

[ ]

[ ]

[ ]

[ ]

( )

1 2 3 4 510

3

2 3 4 5

5

3 4 5

7

9 7

4 5

9

5

5 39

3 5 7 91.0851

4 20 56 1201.0851

16 11

256 57

2 4321.0851

64 5761.0851

2651.08

6 432 120

1

9

5

zAF a a a a a

za a a a

za a a

za

z z z z z

za

z T

a

= + +

+ +

+ +

+

+

= + +

=

Equate this array factor to T9(z):


20/28


Match similar terms to get the coefficients ans:

5

4

3

2

1

2.0860

2.8308

4.1184

5.20735.8377

a

a

a

a

a

=

=

=

==

Normalization 1: Withrespect to amplitude of theelements at the edge

Normalization 2: Withrespect to amplitude of theelement at the centre

5

4

3

2

1

1

1.357

1.974

2.496

2.798

a

a

a

a

a

=

=

=

=

=

5

4

3

2

1

0.357

0.485

0.706

0.890

1

a

a

a

a

a

=

=

=

==


The array factor can be written as:

( ) ( ) ( ) ( ) ( ) ( )10 2.798cos 2.496cos 3 1.974cos 5 1.357cos 7 cos 9

cos

AF u u u u u

du

= + + + +

=


0 0cos cos cos 1.0851cos cosd d

z z u z

= = =


21/28


Examples of AFs of arrays of non-uniform amplitude distribution(N=5, d=/2, 0 = 90)

Uniform amplitude distribution


Triangular (1:2:3:2:1) amplitude distribution


Binomial (1:4:6:4:1) amplitude distribution


Dolph-Tschebyschev (1:1.61:1.94:1.61:1) amplitude distribution


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Dolph-Tschebyschev (1:2.41:3.14:2.41:1) amplitude distribution


Calculate beamwidth of a uniform array with same no. of elementsand element spacing:

multiply it by beam-broadening factor f

Beamwidth of D-T Array

( )

221 2

00

2

1 0.636 cosh coshf RR

= +

For large D-T arrays, scanned near-broadside and side lobes in the -20to -60 dB range:

1 10 0cos cos 0.443 cos cos 0.443HPBW

L d L d

= + + +


The directivity of D-T array, with a given side-lobe level,

increases as the array size or no. of elements increases.

For a given array length (or no. of elements in the array), thedirectivity does not necessarily increase as side lobe leveldecreases.

( )( )

20

02

0

2

1 1

RD

R fL d

=

+ +

Directivity of D-T Array


Example: For a D-T array of 10 elements with spacing d=/2and side lobe level of 26 dB, calculate HPBW and thedirectivity.

R0 = 26 dB R0 = 20 (voltage ratio)

The beam broadening factor f=1.079

Beamwidth for uniform BSA with L+d=5 = 10.17Beamwidth for DT array = 10.17f = 10.97

Directivity = 9.18 (dimensionless) = 9.63 dB


23/28


PLANAR ARRAYS

Planar arrays provide directional beams, symmetricalpatterns with low side lobes, much higher directivity (narrowmain beam) than that of their individual element.

In principle, they can point the main beam toward anydirection.

Applications tracking radars, remote sensing,communications, etc.



The AF of a linear array ofMelements along thex-axis is

( )( )1 sin cos1

1

sin cos cos

x x

Mj m kd

m

m

x

AF I e

+

=

=

=

directional cosine with respect tox-axis.

All elements are equispaced with an interval ofdxand aprogressive shift x.

Im denotes the excitation amplitude of the element at thepoint with coordinates: x=(m-1)dx, y=0.

This is the element of the m-th row and the 1st column ofthe array matrix.


IfNsuch arrays are placed next to each other in the ydirection, a rectangular array is formed.

We assume again that they are equispaced at a distancedyand there is a progressive phase shift along each row ofy.

We also assume that the normalized current distributionalong each of thex-directed arrays is the same but theabsolute values correspond to a factor of I1n (n=1,...,N).

Th tt f t l i th d t f th


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( )( ) ( )( )

( )( )

( )( )

1 sin sin1 sin cos1 1

1 1

1 sin cos1 1

1

1 sin sin1 1

1

sin cos cos

sin sin cos

y yx x

M N

x x

M

y y

MN

N Mj n kdj m kd

n m

n m

x y

Mj m kd

x x m

m

Nj n kd

y y n

n

x

y

AF I I e e

AF S S

S AF I e

S AF I e

x r

y r

+ +

= =

+

=

+

=

=

=

= =

= =

= =

= =

Then, the AF of the entire MNarray is


The pattern of a rectangular array is the product of thearray factors of the linear arrays in thexand ydirections.

In the case of a uniform planar (rectangular) array, allelements have the same excitation amplitudes: m1 1 0I nI I= =

( )( ) ( )( )1 sin sin1 sin cos01 1

y yx xM N j n kdj m kd

m n

AF I e e + +

= =

=

The normalized array factor is obtained as:

( )sin sin

1 12 2,

1 1sin sin

2 2

sin cos

sin sin

x y

n

x y

x x x

y y y

M N

AFM N

kd

kd

=

= +

= +


The major lobe (principal maximum) and grating lobes of are

located at angles such that

sin sin1 12 2

1 1sin sin

2 2

M N

x y

x y

x y

M N

S SM N

= =

sin cos 2 0,1,...

sin sin 2 0,1,...

x m m x

y n n y

kd m m

kd n n

+ = =

+ = =

The principal maxima correspond to m=0, n=0.

In general, x and y are independent from each other.


If it is required that the main beams ofSxMand SyN intersect(which is usually the case), then the common main beam isin the direction:

0 0 0m n = = = =

If the principal maximum is specified by (0,0) , then theprogressive phases x and y must satisfy

0 0

0 0

sin cos

sin sin

x x

y y

kd

kd

=

=

h d f d h b d

3 D PATTERN OF A 5 ELEMENT SQUARE PLANAR UNIFORM ARRAY


25/28


When x and y are specified, the main beam directioncan be found by:

22

0 0tan siny x yx

x y x y

d

d kd kd

= = +

The grating lobes can be located by

0 0

0 0

0 0

0 0

sin sin

tansin cos

sin cossin

cos

sin sin

sin

ymn

x

xmn

mn

y

mn

nd

md

md

nd

=

=

=

In order a true grating lobeto occur, these equationsmust have a real solution(mn,mn).

To avoid grating lobes, thespacing between theelements must be less than

(dx


26/28


For a square array (M=N, x0=y0) with amplitudedistributions of the same type along thexand yaxes,

0 0 0 0

0 0

sec sech x y

h x y

= =

= =

The beam solid angle of the planar array can be approximated by

0 0 0

1 2 1 22 202 2 2 20

0 0 0 02 20 0

sec

sin cos sin cos

A h h

x yA

y x

x y

=

=

+ +


Directivity of Planar Array

The general expression for the calculation of the directivity ofan array is:

( )

( )

20 0

0 2 2

0 00 0

,4

, sin

AFD

AF d d

=

For large planar arrays, which are nearly broadside:

0 0cos x yD D D =

Dx is the directivity of the respective linear BSA,x-axis;Dy is the directivity of the respective linear BSA, y-axis.

Approximate directivity of a nearly broadside planar array:

( ) ( )2

0 2 232,400

degA A

Drads

=


ARRAYS

9The main beam direction is controlled throughthe phase shifts, xand y.

9The beamwidth and side-lobe levels are

controlled through the amplitude distribution.


Example: Compute the HPBWs, beam solid angle anddirectivity of a planar square array of 100 isotropic elements(10 10). Assume a D-T distribution, /2 spacing betweenelements, -26 dB side lobe level, and the maximum orientedalong 0 = 30, 0 = 45.

S l i 6 10


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1. number of elements,2. overall length of the array (in wavelengths),3. approximate half-power beamwidth (in

degrees),4. amplitude level (compared to the maximum of

the major lobe) of the first minor lobe (in dB)5. progressive phase between the elements (in

degrees).

Problem 6.10: Design an ordinary end-fire uniform lineararray with only one maximum so that itsdirectivity is 20 dBi (above isotropic). Thespacing between the elements is d= /4, andits length is much greater than the spacing.

Determine the:


Solution 6.10:

0 4d

D N

=

0 20 100 100D dB N= = =

( )1 24.75L N d L = =

1 1.391cos 1 10.8

2 21.6

h h

h

N d

HPBW

= =

= =

o

o


Problem 6.15: Show that in order for a uniform array ofNelements not to have any minor lobes, thespacing and the progressive phase shiftbetween the elements must be:

1. d=/N, =0 for a broadside array,2. d=/(2N), =kdfor an ordinary end-firearray.


( )sin

1 2cos

1sin

2

n

N

AF kdN

= = +

Solution 6.15:

For =0:

( )sin cos1 2

1sin cos

2

n

N

kdAF

Nkd

=

In order for the array not to have any minor lobes, we canassume that its first null occurs at = 0 or 180. Thus,

( )sin

1 20

12sin 2

n

Nkd

NAF kd d

N Nkd

= = = =

Fo kd ma ima occ s at 180


28/28


For =kd, maxima occurs at =180:

( )( )

( )

sin cos 11 2

1sin cos 1

2

n

Nkd

AFN

kd

+ = +

In order for the array not to have any minor lobes, we canassume that its first null occurs at = 0. Thus,

( )( )

( )

sin10

sin 2nNkd

AF Nkd dN kd N

= = = =

Array of Point Source

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