Contents

1 Phased Array Antennas 11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Array Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2.1 Discretization of Continuous Sources . . . . . . . . . . . . . . . . . . . 51.2.1.1 Discrete Uniformly Spaced Linear Array . . . . . . . . . . . . 51.2.1.2 Discrete Uniformly Spaced Planar Array . . . . . . . . . . . . 9

1.2.2 Beam Scanning using Progressive Phase Shift . . . . . . . . . . . . . . . 111.2.2.1 Grating-lobe Analysis – Discrete Uniformly Spaced Linear

Array . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.2.2.2 Grating-lobe Analysis – Discrete Uniformly Spaced Planar Ar-

ray . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.3 Array Synthesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.3.1 Binomial Array . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.3.2 Dolph-Chebyshev Array . . . . . . . . . . . . . . . . . . . . . . . . . . 16

Chapter 1

Phased Array Antennas

1.1 Introduction

According to IEEE standard definition, array antenna is an antenna comprised of a number ofidentical1 radiating elements in a regular arrangement2 and excited to obtain a prescribed radia-tion pattern.

Why Array Antennas ?

• Theory of array antennas is useful not only to understand discrete array antennas, but alsoto understand various commonly used antennas such as dipole antenna, horn antenna, etc(after all, all these antennas are nothing but continuous arrays of Hertzian dipoles).

• For applications where narrow beam radiation patterns are required, antenna size has to beelectrically large. That electrically large antenna could be a continuous source, for exam-ple, a parabolic reflector antenna. However, in the case of continuous sources, it is difficultto control amplitude and phase of the aperture distribution. So, such electrically large con-tinuous radiating sources can be approximated by discretising3 them. To understand theoperation of such discrete arrays, once again we need to understand the theory of arrayantennas.

• Another important feature of array antennas is, the possibility of scanning the beam elec-trically. Unlike the conventional mechanical beam scanning antennas, array antennas canuse the progressive phase shift concept to scan the beam within a fraction of second.

1The term array antenna is sometimes applied to cases where the elements are not identical or arranged in aregular fashion. For those cases qualifiers shall be added to distinguish from the usage implied in this definition.For example, if the elements are randomly located one may use the term random array antenna.

2The regular arrangements possible include ones in which the elements can be made congruent by simple trans-lation or rotation.

3In the case of discrete arrays, one can control amplitude and phase distributions with the help of power dividingnetworks and phase shifters.

1

Analysis Versus Synthesis

In the next section, definition of array factor will be provided. In the mean time, let us analyzethe equation given for array factor a little bit. Array factor of a continuous linear array orientedalong x axis is given as4

AF (kx) =

∫ +∞

−∞A (x′) exp (jkxx

′) dx′, (1.1)

where A (x′) is the aperture distribution and kx = k0 sin θ cosφ. From the above equation, it isclearly evident that the array factor and the aperture distribution form a Fourier transform pair.Here x′ and kx are analogous to t and ω, the parameters that we use in the signal processingand communication theory. Realizing this fact helps a lot in understanding the theory of arrayantennas, and the theory of antennas in general. So, ideally speaking, if array factor is knownin the entire kx domain, one can obtain the corresponding aperture distribution by taking inverseFourier transform as shown below:

A (x′) =1

2π

∫ +∞

−∞AF (kx) exp (−jkxx′) dkx (1.2)

Here, basically (1.1) indicates the case of analysis whereas (1.2) indicates the synthesis case. Inpractice, it is possible to provide the array factor specification in the visible space region only(i.e., −k0 ≤ kx ≤ k0). So, as a part of synthesis, antenna engineer has to decide the patternspecification in the invisible region (i.e., |kx| > k0).

1.2 Array Analysis

Consider a 3D array with all the array elements oriented along the same direction as shown inFig. 1.1. Even though the scenario depicted in Fig. 1.1 is a discrete one, the concept presented inthis section can be easily adapted for continuous arrays. Since all the elements are oriented alongthe same direction, one can apply the well known concept of pattern multiplication. So, fromnow onwards, let’s assume that the array is simply made up of isotropic radiators. Array factor isdefined as the cumulative contribution of all the isotropic radiating elements at the far-field pointP (r, θ, φ), and is given as

AF =∑n

An exp[jk0

(|~r| −

∣∣∣~r − ~r′n

∣∣∣)] . (1.3)

In the above equation, the term k0

(|~r| −

∣∣∣~r − ~r′n

∣∣∣) represents the phase difference5 between nth

element and the reference point (In the present case, origin is taken as the reference point, with

4In electromagnetism, it is customary to use (x, y, z) and (x′, y′, z′) notations for far-field and source positions,respectively. So, in this chapter the same notation will be followed.

5Remember the equation for electrical length βl! For free space, the propagation constant β = k0.

2

ReferencePoint

Figure 1.1: A typical 3D array antenna consisting of elements oriented along the same direction

out loss of generality). The termAn represents the amplitude as well as phase of the nth radiatingelement. Equation (1.3) can be generalized for continuous arrays6 as

AF =

∫∫∫A(~r′)

exp[jk0

(|~r| −

∣∣∣~r − ~r′∣∣∣)] dx′dy′dz′. (1.4)

In Cartesian co-ordinate system, source position is given as ~r′ = x′x + y′y + z′z. Similarly, thefar-field position P is given as ~r = r sin θ cosφx+r sin θ sinφy+r cos θz. From these equations,∣∣∣~r − ~r′∣∣∣ can be approximated as shown below:

∣∣∣~r − ~r′∣∣∣ = |r sin θ cosφx+ r sin θ sinφy + r cos θz − x′x− y′y − z′z|

=

√(r sin θ cosφ− x′)2 + (r sin θ sinφ− y′)2 + (r cos θ − z′)2

=√r2 − 2rx′ sin θ cosφ− 2ry′ sin θ sinφ− 2rz′ cos θ + x′2 + y′2 + z′2

≈√r2 − 2rx′ sin θ cosφ− 2ry′ sin θ sinφ− 2rz′ cos θ

= r

√1− 2x′ sin θ cosφ+ 2y′ sin θ sinφ+ 2z′ cos θ

r

≈ r

(1− 1

2

2x′ sin θ cosφ+ 2y′ sin θ sinφ+ 2z′ cos θ

r

)≈ [r − (x′ sin θ cosφ+ y′ sin θ sinφ+ z′ cos θ)] (1.5)

6Some times, array factor of a continuous array is also known as space factor.

3

Substituting (1.5) into (1.4) gives

AF =

∫∫∫A(~r′)

exp [jk0 (x′ sin θ cosφ+ y′ sin θ sinφ+ z′ cos θ)] dx′dy′dz′

=

∫∫∫A(~r′)

exp [jkxx′ + jkyy

′ + jkzz′] dx′dy′dz′, (1.6)

where kx = k0 sin θ cosφ, ky = k0 sin θ sinφ, and kz = k0 cos θ. From the above equation, one

can see that the aperture distribution A(~r′)

and the array factor AF form a Fourier transformpair.

Example 1. Calculate the array factor of a continuous linear array oriented along z′ axis andhaving an aperture distribution

A (z′) =

{1, |z′| ≤ L

2

0, elsewhere.

Also calculate (a) the directions in which array factor exhibits zero radiations (i.e., null direc-tions), and (b) null-to-null beamwidth.

Solution: From (1.6),

AF (kz) =

∫ +L/2

−L/2(1) ejkzz

′dz′

=ejkzz

′

jkz

∣∣∣∣+L/2−L/2

=ejkzL/2 − e−jkzL/2

jkz=

2j sin(kzL2

)jkz

=sin(kzL2

)kz2

.

Since kz = k0 cos θ, AF can be written in terms of θ and φ as

AF (θ, φ) =sin(k0 cos θL

2

)k0 cos θ

2

.

From the above equation, it can be seen that array factor is not a function of φ because array isalong z′ axis (can you visualize the cylindrical symmetry along z′ axis?!). Now, null positionsθnull can be obtained from the above equation as

k0 cos θnulln L

2= ±nπ, where n = 1, 2, 3, · · · .

⇒ θnulln = ± cos−1(

2nπ

k0L

)Note that n = 0 actually corresponds to the main beam positions. Finally null-to-null beamwidthis nothing but 2θnull1 , i.e., 2 cos−1

(2πk0L

).

4

Figure 1.2: A (x′) and AF (kx) corresponding to acontinuous source

1.2.1 Discretization of Continuous SourcesFrom the Fourier transform relation between A and AF , it can be concluded that the beam-widthof the radiation pattern is inversely proportional to the antenna’s electrical length. So, in order toachieve very narrow beam widths, continuous apertures should be electrically large. However,it is very difficult to implement such electrically large continuous aperture distributions. So,antenna engineer should opt for discrete antenna arrays instead. Discrete array concept is verysimilar to the concept of Nyquist-Shannon sampling technique in communication theory and isexplained in the following paragraph.

1.2.1.1 Discrete Uniformly Spaced Linear Array

Let us consider a linear continuous aperture distribution oriented along x axis and the corre-sponding array factor shown in Fig. 1.2. If this continuous source is discretized with an uniformspacing of a in between the elements, then the corresponding array factor can be obtained from(1.6) as

AF (kx) =∑m

Am exp (jkxx′m) =

∑Am exp (jkxma) . (1.7)

From the above equation, it can be seen that AF (kx) is a periodic7 function in the kx domain

7Array factor corresponding to a finite length continuous source oriented along x axis spreads from−∞ to +∞.

5

Figure 1.3: A (x′) and AF (kx) corresponding to a uniformly spaced discrete array

as shown in Fig. 1.3. Also, one can observe that the period of the array factor in the kx domainis 2π/a. Even though array factor is periodic and extends from −∞ to +∞, from applicationpoint of view kx is confined in the domain [−k0,+k0], because |k0 sin θ cosφ| ≤ 1 for the entirevisible domain 0 ≤ θ ≤ π and 0 ≤ φ ≤ 2π. This domain [−k0,+k0] is commonly known asvisible-space in antenna theory and is highlighted in Fig. 1.3. Also seen in the figure are infinitegrating-lobes placed on both sides of the main beam. If the spacing between array elements aincreases, then the separation between the main-lobe and the nearest grating lobes decreases.However, in order to avoid ambiguity in applications such as radar, grating-lobes should not en-ter into the visible space. So, spacing should be chosen as small as possible. However, if spacingbetween elements is too small, then the packing density as well as the mutual coupling betweenthe radiating elements increases. Increased mutual coupling increases the effective reflection co-efficient and thus reduces the efficiency of the array. So, spacing between elements a can neitherbe small nor be large. So, an optimum spacing has to be chosen by antenna engineer by takingmany things into consideration. This concept will be further explored in Sec. 1.2.2.1.

Example 2. Calculate the array factor of a discrete uniformly spaced linear array oriented alongx′ axis and having uniform aperture distribution. Assume that the array is having M number ofelements (M can be wither even or odd) and the spacing between any two consecutive elementsis a. Also calculate (a) the directions in which array factor exhibits zero radiations (i.e., nulldirections), and (b) null-to-null beamwidth.

So, for any discrete array, there always exists some aliasing effect as shown in Fig.1.3.

6

N - Odd N - Even

Figure 1.4: Even and odd linear arrays and the corresponding index values

Solution: First of all one should decide the indexes of array elements. If M is odd m =0,±1,±2, · · · , else m = ±1

2,±3

2, · · · as can be seen from Fig. 1.4. Now, using the definition of

array factor given by (1.7) gives

AF (kx) =

+(M−12 )∑

−(M−12 )

Amejkxma.

Remember that the above equation is valid for both even as well as odd cases. In either case,using the geometrical progression summation identity

∑N−1n=0 a0r

n = a01−rN1−r and substituting

Am = 1, one can prove that

AF (kx) =

+(M−12 )∑

−(M−12 )

Amejkxma

=

+(M−12 )∑

−(M−12 )

ejkxma

= e−jkx(M−1

2 )a + e−jkx(M−3

2 )a + · · ·+ ejkx(M−3

2 )a + ejkx(M−1

2 )a

= e−jkx(M−1

2 )a [1 + ejkxa + · · ·+ ejkx(M−2)a + ejkx(M−1)a]

= e−jkx(M−1

2 )a[

1− ejkxMa

1− ejkxa

]= e−jkx(

M−12 )a e

jkxMa/2

ejkxa/2

[ejkxMa/2 − e−jkxMa/2

ejkxa/2 − e−jkxa/2

]=

[ejkxMa/2 − e−jkxMa/2

ejkxa/2 − e−jkxa/2

]=

sin(kxMa

2

)sin(kxa2

) .7

Similar to Ex. 1, null positions can be obtained as

knullxn Ma

2= ±nπ, where n = ±1,±2, · · ·

⇒ knullxn = ±2nπ

Ma

⇒ k0 sin θnulln = ±2nπ

Ma⇒ θnulln = ± sin−1

(2nπ

Mak0

).

So, null-to-null beamwidth is given as 2 sin−1(

2πMak0

).

Example 3. Prove that array factor of any odd numbered array located symmetrically with re-spect to origin will be periodic in kx domain with a periodicity of 2π

a. Also prove that array factor

of any even numbered array located symmetrically with respect to origin will be periodic in kxdomain with a periodicity of 4π

a

Solution: From 1.7, array factor is defined as

AF (kx) =∑m

Am exp (jkxx′m) =

∑Am exp (jkxma) .

Since for odd numbered arrays, m = 0,±1,±2, · · · , array factor for odd numbered arrays isgiven as

AFOdd (kx) =

+M−12∑

−M−12

Am exp (jkxma)

= A−M−12

exp

[−jkx

(M − 1

2

)a

]+ · · ·+ A−1 exp (−jkxa) + A0 +

A+1 exp (+jkxa) + · · ·+ A+M−12

exp

[jkx

(M − 1

2

)a

].

All the terms in the above summation are of the form exp (jkxma), where m is an integer. So,periodicity of such terms in kx domain will be 2π

ma. So, greatest period is equal to 2π

a, which is

the periodicity of the array factor.In contrast to the odd numbered case, for even numbered arrays, m = 0,±1

2,±3

2· · · . So,

array factor for even numbered arrays is given as

AFOdd (kx) =

+M−12∑

−M−12

Am exp (jkxma)

= A−M−12

exp

[−jkx

(M − 1

2

)a

]+ · · ·+ A− 1

2exp

(−jkx

1

2a

)+

A+ 12

exp

(+jkx

1

2a

)+ · · ·+ A+M−1

2exp

[jkx

(M − 1

2

)a

].

8

(a) (b)

Figure 1.5: Array factors corresponding to uniform excitations with a = λ/2 : (a) M = 10 (b)M = 11.

All the terms in the above summation are of the form exp (jkxma), wherem is always a fraction,where m ∈

{±1

2,±3

2, · · ·

}. So, periodicity of such terms in kx domain will be 2π

ma. So, greatest

period is equal to 2π(1/2)a

= 4πa

, which is the periodicity of the array factor8.

1.2.1.2 Discrete Uniformly Spaced Planar Array

Discretization of continuous planar sources is similar to the case of linear arrays discussed inSec. 1.2.1.1. The only difference is that for planar arrays one should use 2D Fourier transforminstead of 1D. Let’s consider the discrete uniformly spaced planar array shown in Fig. 1.6. It canbe noticed that the planar array is having a very general lattice shape. If γ = 90◦, then it simplycorresponds to the planar array with rectangular lattice. For the planar array, it can be showedthat y′m = qb and x′m = pa+ qb

tan γ. Substituting these values into (1.6) gives

AF (kx, ky) =∑p

∑q

Apq exp(jkxx

′pq + jkyy

′pq

)=∑p

∑q

Apq exp

[jkx

(pa+

qb

tan γ

)+ jkyqb

]. (1.8)

For rectangular arrays, γ = 90◦. So, the above equation reduces to

AF (kx, ky) =∑p

∑q

Apq exp [jkxpa+ jkyqb] . (1.9)

Analogous to the linear case, planar discrete arrays also exhibit grating-lobe phenomena in kxkyspace. It can be understood by observing (1.8). Notice that AF (kx, ky) given by (1.8) will have

8 In Fig. 1.5, horizontal axis indicates normalized kx values, i.e., kx/k0 = sin θ cosφ = sin θ. So, in this domainperiods will be 2π

k0aand 4π

k0a, for odd numbered and even numbered arrays, respectively.

9

VISIBLE-SPACE DISK

ARBITRARY SCANSPECIFICATION

Figure 1.6: Discrete uniformly spaced planar array placed in the xy plane and the correspondingkxky domain

peaks when the following conditions are satisfied simultaneously:

kxa = 2µπ andkxb

tan γ+ kyb = 2νπ, where µ, ν = 0,±1,±2, · · · .

Rewriting the above set of equations gives{kx = 2µπ

a

ky = −cotγkx + 2νπb

= tan [− (90◦ − γ)] kx + 2νπb

. (1.10)

So, for discrete planar arrays, grating-lobes occur at the intersection points of the above set ofstraight lines as can be seen in Fig. 1.6.

One more thing that can be noticed in Fig. 1.6 is the visible space disk. From the definitionskx = k0 sin θ cosφ and ky = k0 sin θ sinφ, one can derive the following properties:{

k2x + k2y=k20 sin2 θ

tan−1(kykx

)= φ

. (1.11)

So, the visible space (i.e., 0 ≤ θ ≤ π and 0 ≤ φ ≤ 2π) corresponds to the disk k2x + k2y≤ k20as sin2 θ ≤ 1. This fact and the above properties are represented graphically in Fig. 1.7. Theconcept of visible-space will be explored further in the Sec. 1.2.2.2.Example 4. Calculate the array factor of a discrete uniformly spaced planar array (assume rect-angular lattice) placed in the x′y′ plane and having uniform aperture distribution along bothx′ and y′ axes. Assume that number of elements and spacing along x’ axis are M and a, re-spectively. Similarly, Assume that number of elements and spacing along y’ axis are N and b,respectively.

10

VISIBLE-SPACE DISK

Figure 1.7: Visible space representation in the kxky domain

Solution: Planar array having rectangular can be considered as a linear array of linear arrays.So, First one needs to calculate array factor along x-axis. And then using pattern multiplica-tion principle, this array factor should be multiplied with the array factor corresponding to thedistribution along y-axis. So, array factor for the given planar array is given as

AF (kx, ky) =sin(kxMa

2

)sin(kxa2

) × sin(kyNb

2

)sin(kyb

2

) .

1.2.2 Beam Scanning using Progressive Phase Shift

One main type of array antennas is phased array antenna, in which case each element of the arrayis excited with proper progressive phase shift so that the the main-beam steers in the desireddirection. In order to explain the beam scanning using progressive phase shift, the followingfrequency-shifting property of the Fourier transform9 can be used:

If f (t)⇐⇒ F (ω) ,

then f (t) exp (−ω0t)⇐⇒ F (ω − ω0) . (1.12)

In array antenna theory, (x′, y′, z′) and (kx, ky, kz) are analogous to time and angular frequencyparameters t and ω, respectively. So, analogous to the above frequency-shifting theorem, for

9Here Fourier transform is defined as F (ω) =∫∞−∞ f (t) exp (+ωt) dt.

11

Figure 1.8: Beam scanning using progressive phase shift

array antennas,

if A (x′, y′, z′) ⇐⇒ AF (kx, ky, kz) , then

A (x′, y′, z′) exp (−kx0x′ − ky0y′ − kz0z′) ⇐⇒ AF (kx − kx0, ky − ky0, kz − kz0) .(1.13)

From (1.13), it is evident that the array factor can be shifted/scanned in the (kx, ky, kz) domainby introducing proper linear progressive phase shifts along x′, y′, and z′- directions.

1.2.2.1 Grating-lobe Analysis – Discrete Uniformly Spaced Linear Array

Now that the general theory for beam scanning is explained, in this section grating-lobe analysiswill be presented for linear arrays. As mentioned earlier, for discrete arrays, spacing shouldneither be large nor be small. Antenna engineer has to decide an optimal spacing depending upon several factors, such as maximum scanning angle, packing density, cost of the antenna system,etc. In this section, theory is provided for deciding the optimum spacing for discrete uniformlyspaced linear arrays.

12

It is known that by providing proper progressive phase shift, array factor can be shifted inthe kxky domain. For a given spacing a, if the array factor is shifted as shown in Fig. 1.8,then the maximum shift that can be done without grating-lobes entering into the visible space is(2πa− k0

). So10,

kx0 ≤(

2π

a− k0

)⇒ k0 sin θ0 ≤

(2π

a− k0

)⇒ θ0 ≤ sin−1

(2π

ak0− 1

)⇒ θ0,max = sin−1

(2π

ak0− 1

). (1.14)

However, in many applications θ0,max is usually specified and one needs to decide the opti-mum spacing for the given maximum scan angle. In such cases,

kx0 ≤(

2π

a− k0

)⇒ k0 sin θ0,max ≤

(2π

a− k0

)⇒ a ≤ 1

k0

(2π

1 + sin θ0,max

)⇒ amax =

1

k0

(2π

1 + sin θ0,max

). (1.15)

Example 5. For a given linear array oriented along x axis and having an uniform spacing of15mm, calculate the progressive phase shift required to scan the beam to from broad-side direc-tion (i.e., θ = 0◦) to (a) θ0 = 30◦ (b) θ0 = 90◦ . Assume that the operating frequency of the arrayis 10 GHz.

Solution: From (1.13) as well as Fig. 1.8, progressive phase shift required to scan the beamto an arbitrary direction θ0 is kx0a = k0 sin θ0a. Also, wavelength at 10 GHz in free-space isλ = c

f= 30mm. So, a = λ

2and progressive phase shifts can be calculated as shown below:

(a) progressive phase required to scan to θ = 30◦ is

k0 sin θ0a =2π

λ× sin 30◦ × λ

2=π

2

(b) progressive phase required to scan to θ = 90◦ (i.e., end-fire direction) is

k0 sin θ0a =2π

λ× sin 90◦ × λ

2= π.

10Due to cylindrical symmetry, for linear arrays along x axis, it is sufficient to know the array factor for φ = 0◦.So, in this section kx = k0 sin θ cosφ = k0 sin θ.

13

Example 6. For a given linear array oriented along x axis and having an uniform spacing ofa = 0.75λ, calculate the maximum scanning that can be done so that grating-lobe doesn’t enterinto the visible-space.

Solution: From 1.14,

θ0,max = sin−1(

2π

ak0− 1

)= sin−1

(1

0.75− 1

)≈ 19.47◦.

Example 7. If the main beam corresponding to a given linear array has to be scanned to a max-imum angle of θ0 = 40◦, then what should be the maximum possible uniform spacing betweenconsecutive elements.

Solution: From 1.14,

amax =1

k0

(2π

1 + sin θ0,max

)=

λ

2π

(2π

1 + sin 40◦

)= λ

(1

1 + sin 40◦

)≈ 0.609λ.

1.2.2.2 Grating-lobe Analysis – Discrete Uniformly Spaced Planar Array

From (1.11), for a general two-dimensional lattice structure described in Fig. 1.6, grating lobesoccur in the kxky domain at

(kxg)µ = kx0 +2µπ

a

(kyg)µ,ν = ky0 +2νπ

b−

[(kxg)µ − kx0

tan γ

], (1.16)

where kx0 and ky0 are the shifts along kx and ky axes, respectively (see Sec. 1.2.2).Since by definition, kx = k0 sin θ cosφ and ky = k0 sin θ sinφ, radiating far fields are con-

fined to the circular disk(k2x + k2y

)1/2 ≤ k0, often known as visible space as mentioned earlier.The remaining kxky space, described as invisible space, is related to the stored energy in thenear field region, which is analogous to the phenomenon of evanescent modes in a waveguide.Usually, an antenna engineer intends to avoid the appearance of all the grating lobes (except forthe µ = ν = 0, main lobe case) within the visible space.

Fig. 1.6 shows the visible space and the grating lobe spaces placed according to (1.16) in thekxky domain. Assuming that V 1 represents the domain of the specified main lobe scan positions,the closed loopsG1–G6 represent contours of all the possible nearest grating lobe scan positions.From Fig. 1.6, it can be observed that all the grating lobe contours are just touching the visiblespace circle, except for G2 and G5. Thus, for the given scan specification V 1, the array latticearrangement shown in Fig. 1.6 is not optimal. In this chapter, optimal array arrangement isdefined as the configuration that maximizes the array’s unit cell area (ab). In other words, anoptimal configuration minimizes the number of elements needed in a given array aperture. Toachieve this optimal array lattice configuration, Fig. 1(b) should be modified according to thefollowing description.

14

1. All the left-hand side grating lobe contours should be moved upward and the right-handside contours downward.

2. After an optimal skew with respect to the ky axis, both left-hand side and right-hand sidegrating lobe contours should be moved horizontally toward each other, so that all wouldjust touch the visible space circle.

3. If V 1 is symmetric with respect to both the kx and the ky axes, then the contoursG2, G3, G5andG6 will be at the same distance from the kx axis for the optimal array lattice configura-tion. This optimal array element arrangement is a hexagonal array lattice

(γ = tan−1

(2ba

)).

For further information related to grating lobe analysis, please see [1].

1.3 Array Synthesis

In the previous section, analysis of arrays with uniform excitation has been provided. However,SLR achieved with uniform aperture distributions is approximately 13 dB only, which is notacceptable in many radar applications. In order to overcome this disadvantage, one has to optfor some other distributions, such as Binomial, Dolph-Chebyshev, Taylor n, etc. In this chaptertheory is provided to synthesize binomial and Dolph-Chebyshev arrays only. For Taylor n andsome advanced synthesis methods, please see [2].

1.3.1 Binomial Array

For array synthesis cases, array factor is provided (at least partially) and one has to calculate the(1D) aperture distribution using the formula

A (x′) =1

2π

∫ +∞

−∞AF (kx) exp (−jkxx′) dkx. (1.17)

For binomial arrays, array factor is given as

AF (kx) =(ejkxa − ejkx0a

)M−1. (1.18)

Reasons for choosing the above array factor are:• array factor should be a periodic function in kx domain with a period of 2π

a

• array factor should have an order of M − 1 in ejkxa domain• all the zeros should present at one single location, i.e., kx = kx0, so that array factor will

not have any side-lobesFor synthesizing such an array, one can apply either the Fourier transform given by (1.17) orthe definition of array factor AF (kx) =

∑Ame

jkxma itself. We will follow the second methodwhich is easier than calculating Fourier series coefficients of (1.18). The method to evaluate thearray coefficients corresponding to binomial array is explained by using the following example.

15

(a) (b)

Figure 1.9: A figure depicting the mapping process AF (kx) = TM−1(−c cos kxa

2

), where u =

kx/k0, a = λ/2, and M is an odd number.

Example 8. Design a 5-element Binomial array having all the zeros at θ0 = 30◦ and having anuniform spacing of a = λ/2.

Solution: From 1.18 array factor is given as,

F (kx) =(ejkxa − ejkx0a

)M−1=(ejkxa − ej

2πλ×sin 30◦×λ

2

)4=(ejkxa − ej

π2

)4=(ejkxa − j

)4= ejkx4a + 4ejkx3a (−j) + 6ejkx2a (−j)2 + 4ejkxa (−j)3 + (−j)4

= ejkx4a − 4jejkx3a − 6ejkx2a + 4jejkxa + 1.

Comparing the above equation with array factor definition AF (kx) =∑Ame

jkxma gives

A4 = 1, A3 = −4j, A2 = −6, A1 = 4j, and A0 = 1.

1.3.2 Dolph-Chebyshev ArrayEven though binomial array pattern doesn’t contain any side-lobes, it suffers from broad beamwidth,which in turn reduces the directivity. So, for many practical applications binomial distributionis not an optimal distribution. An optimal distribution which exhibits the maximum possibledirectivity for a given SLR is Dolph-Chebyshev distribution [3]. Array factor corresponding toDolph-Chebyshev array is given as

AF (kx) = TM−1

(−c cos

kxa

2

)(1.19)

16

(a) (b)

Figure 1.10: A figure depicting the mapping process AF (kx) = TM−1(−c cos kxa

2

), where

u = kx/k0, a = λ/2, and M is an even number.

where Chebyshev polynomial TN (x) is defined as

TN (x) =

{cos (N cos−1 x) , |x| ≤ 1

cosh(N cosh−1 x

), |x| ≥ 1

. (1.20)

The parameter c is decided by the given SLR. If the given SLR is R in linear scale, then c isdecided such that

TM−1 (c) = R⇒ cosh[(M − 1) cosh−1 c

]= R⇒ c = cosh

(cosh−1R

M − 1

). (1.21)

Both Chebyshev polynomials as well as the mapping given by (1.19) are given in Fig. 1.9 and1.10 for odd and even arrays, respectively.

Reasons for choosing array factor given by (1.19) are as given below:

• array factor should be a periodic function in kx domain with a period of 2πa

(4πa

)for odd

(even) numbered arrays• array factor should have an order of M − 1 in ejkxa domain• [−c,+c] region of Chebyshev polynomial should get mapped to

[0, 2π

a

]region of kx do-

main as shown in Fig. 1.9.

17

In order to synthesize Dolph-Chebyshev array for a given set of SLR and M values, all we needto know is M − 1 zeros of array factor (1.19). These zeros can be obtained as shown below:

TM−1

(−c cos

knullx a

2

)= 0

⇒ cos

[(M − 1) cos−1

(−c cos

knullx a

2

)]= 0

⇒ (M − 1) cos−1

(−c cos

knullx,n a

2

)= ± (2n− 1)

π

2, where n = ±1,±2,±3, · · ·

⇒ knullx,n =2

acos−1

{1

ccos

[(2n− 1

M − 1

)π

2

]}. (1.22)

Once array factor zeros are obtained, these zeros should be converted into zeros in Z transformdomain. Till now, it has been emphasized that AF is nothing but Fourier transform of aperturedistribution A (x′, y′, z′). This Fourier transform can be re-written in terms of Z transforms asshown below:

AF (kx) =∑

Amejkxma ⇒ AF (Z) =

∑AmZ

m, where Z = ejkxa. (1.23)

So, converting array factor zeros obtained using (1.22) into Z domain gives

Znulln = ejk

nullx,n a. (1.24)

Finally, one can obtain array coefficients by expanding array factor obtained by

AF (Z) =∏(

Z − Znulln

). (1.25)

An example describing the method given in this section is given below.e.

Example 9. Design a 5-element Dolph-Chebyshev array having SLR of 30 dB and spacinga = λ/2.

Solution: First of all, we need to calculate the parameter c using (1.21) and the given SLRin dB as shown below:

R = 10SLR/20 = 101.5 = 31.623.

So,

c = cosh

(cosh−1R

M − 1

)= 1.587.

We need to knowM − 1 zeros (i.e., 4 zeros) of the array factor to calculate Am values.However, due to symmetry, it is sufficient to calculate of 2 zeros. Because remaining two zeros

18

are nothing but mirror images of the first two with respect to kx = 0. So, from (1.22)

knullx,n a = 2 cos−1{

1

ccos

[(2n− 1

M − 1

)π

2

]}, n = 1, 2

= 2 cos−1[

1

1.587cos(π

8

)]and 2 cos−1

[1

1.587cos

(3π

8

)]= 1.899 and 2.654.

So, first two zeros in the Z domain are given as

Znull1,2 = ej1.899, and ej2.654.

Due to symmetry, other two zeros are given as

Znull−1,−2 = e−j1.899, and e−j2.654.

So, from (1.25), array factor is given as

AF (Z) =∏(

Z − Znulln

)=(Z − ej1.899

) (Z − e−j1.899

) (Z − ej2.654

) (Z − e−j2.654

)=(Z2 + 1− 2Z cos 1.899

) (Z2 + 1− 2Z cos 2.654

)= Z4 − 2Z3 (cos 2.654 + cos 1.899) + 2Z2 + 1.144Z2 − 2Z (cos 2.654 + cos 1.899) + 1

= Z4 + 2.412Z3 + 3.14Z2 + 2.412Z + 1.

Comparing the above equation with the definition of array factor AF (Z) =∑AmZ

m gives

A4 = 1, A3 = 2.412, A2 = 3.14, A1 = 2.412, A0 = 1.

19

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Bibliography

[1] S. R. Zinka, I. B. Jeong, J. H. Chun, and J. P. Kim, “A novel geometrical technique fordetermining optimal array antenna lattice configuration,” IEEE Transactions on Antennasand Propagation, vol. 58, no. 2, pp. 404–412, 2010. 1.2.2.2

[2] A. K. Bhattacharyya, Phased Array Antennas, Floquet analysis, Synthesis, BFNs, and ActiveArray Systems. Hoboken, NJ: John Willey, 2006. 1.3

[3] C. L. Dolph, “A current distribution for broadside arrays which optimizes the relationshipbetween beam width and side-lobe level,” Proceedings of the IRE, vol. 34, no. 6, pp. 335–348, 1946. 1.3.2

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