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Antenna Arrays

Objectives

By the end of this lecture, we will be able to:

• Explain the motive for designing arrays.

• Design a uniform linear array using various constitute elements.

• Recognize various array typologies and their potential applications.

• Explain the impact of mutual coupling between array elements and how to combat

or avoid it.

Note: Boxed sections that start with the word ”Extra” are not required for tests. Neverthe-

less, they are very important to gain concise understanding of the underlying mathematical

and physical concepts.

1

Antenna Arrays

1. Why Arrays?

We use antennas to transmit signals between two systems mostly separated by free space.

Our goal is to ensure that the receiver can detect the transmitted signal. As we know, this

type of communication adheres to the well-known Friis transmission equation:

Preceived = PtransmittedGtransmitterGreceiver

(λ

4πR

)2

. (1.1)

Let us assume that we have no control over the transmitted power, there are two ways to

improve the detection of the received signal, control the gain of both transmitter and receiver

and operate at lower frequencies. If we set aside the frequency issue, the gain of an antenna

increases if we can focus its radiation pattern in a single direction. As we narrow the direction

of focus, the antenna gain in that direction increases (of course at the expense of reducing

the gain in other directions but we don’t care about these directions). Focusing the gain can

be done using two methods: Big antenna or an antenna array. Analogously, this problem is

similar to collecting rain using buckets as described by Haupt: ”A big antenna collects a lot

of electromagnetic waves just like a big bucket collects a lot of rain . . .Another approach to

collecting a lot of rain is to use many buckets rather than one large one. The advantage is

that the buckets can be easily carried one at a time. Collecting electromagnetic waves works

in a similar manner.” In this lecture, we will study antenna arrays since big antennas do

not differ from ”regular-sized” antennas. The advantage of arrays over big antennas is that

we can do a lot of ”cool” things with arrays that are not possible using big antennas as well

see in this class.

2. Two-Element λ/2-Dipole Array

At this point in the course, we know the electric field at an observation point (P ) generated

by a half-wave dipole placed at the origin (O) as shown in Figure 1.1, but for the sake of

completeness, let us rewrite it here

Ed = jηI exp (−jkr)

2πr

cos

(π2

cos θ

)sin θ

. (1.2)

c© Mohammad Alhassoun; [email protected] 2

https://goo.gl/DzFgQL

Antenna Arrays

Figure 1.1: A half-wave dipole located at the origin in a three-dimensional space.

Now, let us bring another antenna as in Figure 1.2 and study the the impact on the field

at P . To simplify the math, let us make the antennas symmetric with respect to the y-axis.

Since this problem is a far field problem, the amplitudes of the electric field at P do not

vary that much from one antenna to another, so we assume both antennas have the same

amplitude of the electric field at P . However, this assumption is invalid when we consider

the phase. Therefore, we can write the total electric field at P as

Etot = Ed1 exp jk(r • r1)+ Ed2 exp jk(r • r2)

= Ed

[exp jk(r • d

2y)+ exp jk(r • −d

2y)]

= Ed

[exp jk d

2sin θ sinφ+ exp −jk d

2sin θ sinφ

]= Ed

[2 cos(k d

2sin θ sinφ)

]= Ed︸︷︷︸

element factor

[2 cos(π

d

λsin θ sinφ)

]︸︷︷︸

array factor (AF)

(1.3)


Antenna Arrays

Figure 1.2: A two-element half-wave dipole array.

where

r1 = |R1 −R|y = d/2 y; r2 = |R2 −R| − y = −d/2 y

Equation (1.3) reveals a very important property of arrays, which is the total field is

the product of the array factor (AF) with the element factor. At this point in the course,

we know that the radiation pattern (or gain) is related to the power of the electromagnetic

wave, which is related to the square of the field. Thus, we can write the gain (G) of the

array as

Gain ∝ (|Element Factor|)2 × (|Array Factor|)2. (1.4)

The fact that we can decompose the gain of an array into two factors makes it easier to

generalize the concept of arrays to antenna structures that are not half-wave dipoles. The

array factor, as derived in Equation (1.3), does not depend on the elements of the array.

Now, let us study a specific case of Equation (1.3) with the assumption that d = λ/2. Let

us further investigate what happens on the xy-plane. By looking at Equation (1.2), we can

see that the electric field is independent on the variation along the azimuth (xy-plane), so it

is always constant. However, this not the case for the array factor of Equation (1.3), which


Antenna Arrays

depends on both θ and φ. Since we are only interested in the variation of the electric field on

the xy-plane1, θ = 90. In Figure 1.3, we superimposed the normalized radiation patterns2

of both a single half-wave dipole and a two-element array of dipoles. We notice that the

maximum radiation pattern of the array quadruples compared to that of the single antenna;

it is concentrated on the x direction at the expense of killing the transmitted power along

the y direction. In fact, this phenomenon follows from the law of conservation of energy.

1 2

3 4

y (?=90o)

-y (?=270o)

-x (?=180o) x (?=0o)

Figure 1.3: A comparison between the radiation pattern on the xy-plane of a single half-wavedipole (red) and a two-element array of dipoles (blue). Curves are in linear scale not in decibel.

3. N-Element Array

Now, we have seen how powerful bringing two antennas close to each other to form an array.

The next ”logical” step would be the generalization to an array of N antennas (of course N

is a positive integer; to the best of my knowledge, there is no such thing called one-quarter of

antenna!). Let us derive the array factor for a linear (means elements are distributed along

a straight line), equally-spaced, and identical arrays as depicted in Figure 1.4. The array

1The field at θ = 0 is zero for a half-wave dipole.2We are interested in the radiation patterns ratio rather than the absolute absolute pattern.


Antenna Arrays

factor is

AF = exp jk(r • r1)+ . . .+ exp jk(r • rN)

=[1 + exp jk(r • dy)+ exp jk(r • 2dy)+ . . .+ exp jk(r • (N − 1)dy)

]=

N∑n=1

exp jk[r • (n− 1)dy] =N∑n=1

exp j[(n− 1)kd sin θ sinφ].

(1.5)

Figure 1.4: An N elements array.

Equation (1.5) resembles a geometric series if we set m = n − 1 and ψ = kd sin θ sinφ.

Therefore, we can write Equation (1.5) as

AF =N−1∑m=0

exp (jmψ) =1− exp (jNψ)

1− exp (jψ)

= exp (jN−12ψ)

exp (−jN2ψ)− exp (jN

2ψ)

exp (−j 12ψ)− exp (j 1

2ψ)

= exp (jN−12ψ)

sin(N2ψ)

sin(ψ2

)(1.6)


Antenna Arrays

The phase factor (complex exponential) of Equation (1.6) does not affect the radiation

pattern3; thus, we can edit it out for radiation pattern calculations. Moreover—as in the

case of a two-element array—the gain of the array is related to the amplitude squared of

Equation (1.6). In other words,

G(θ, φ) ∝ |AF |2 =

[sin(N2ψ)

sin(ψ2

) ]2, (1.7)

which is a periodic squared sinc function with a period of ψ = 2π as indicated in Figure 1.5

for N = 4 and −3π ≤ ψ ≤ 3π. This property plays an important role for array synthesis

and orthogonal arrays. Unfortunately, such topics are beyond the scope of this class.

Ao

-2: -: 0 : 2:

|AF

|2

0

8

16

Figure 1.5: The amplitude squared of the array factor versus ψ. Curve is in linear scale not indecibel.

3In fact, the phase factor results from choosing the reference point, which is arbitrary. Interestingly, wecan see how to get rid of it when we discuss the relation between DTFT and array theory.


Antenna Arrays

4. Array Parameters

4.1. Type of constitute elements

The array elements can be any type of antennas. An important consideration is whether

the array factor is going to improve the radiation characteristics or not. For instance, the

two-element dipole array that we discussed in Section 1.2 is not going to radiate on the z

(θ = 0) direction because the dipole configuration in Figure 1.1 does not have any power

radiated in that direction. If we need to force the array to radiate in that direction, we have

to change the orientation of the dipoles (or use another type of antennas that does not have

a null in the z direction).

4.2. Element Spacing

The main beam of an antenna array is called the main lobe, which should point towards the

direction of interest. Choosing the separation distance d is not arbitrary. In fact, it affects

the radiation pattern. As an example, let us study three cases.

4.2.1 Closely-spaced elements (d << λ)

For simplicity, let us assume that we have two elements distributed along the y-axis, so the

array factor is exactly similar to that of Equation (1.3). Let us also assume that d = λ/M ,

where M is any positive number (it does not have to be an integer). Thus, we can rewrite

that array factor of Equation (1.3) as

AF = 2 cos(π

Msin θ sinφ). (1.8)

As M gets larger, the two elements get closer to each other and the argument of the cosine

in Equation (1.8) gets smaller, which allows us to use small angle approximation 4 (cosγ ≈1− γ2/2 for γ << 1). Therefore,

AF ≈ 2(1−( πM

sin θ sinφ)2

2) = 2− (

π

Msin θ sinφ)2. (1.9)

4The approximation is simply the Taylor series expansion of the cosine in which all terms but the firstand second are omitted because the remaining terms are very small.


Antenna Arrays

1

2

3

4y (?=90o)

-y (?=270o)

-x (?=180o) x (?=0o)

Figure 1.6: A comparison between the radiation pattern on the xy-plane of a closely-spaced two-element array; M = 5 (blue), M = 10 (red), M = 20 (orange), and M = 40 (dashed purple).Curves are in linear scale not in decibel.

The contribution of the second term in the right hand side of Equation (1.9) becomes negli-

gible as M increases. Thus, we can fairly assume that for closely-spaced elements, the array

factor is constant. Figure 1.6 plots four cases for M = 5, 10, 20, and 40 at θ = 90.

4.2.2 d = λ/2

When the elements are spaced half a wavelength apart, the array factor has a null as depicted

by the blue curve of Figure 1.3. The null results from the relationship between the space (d)

and wavenumber (k). Physically, let us think about a particular example, which is the two-

element array of Figure 1.2. Let us assume that the currents of the elements are identical and

in-phase with one another and d = λ/2. At a point (0, y1, 0), where (y1 > 0 and y1 >> d)5,

the waves generated by the two elements are 180 out of phase. Therefore, the waves cancel

5We need to ensure that (0, y1, 0) is a far-field point in order to assume that the amplitude of waves fromeach element is identical.


Antenna Arrays

each other at (0, y1, 0). We can use the same argument to explain why we have null at a

point (0, y2, 0), where (y2 < 0 and |y2| >> d).

4.2.3 d ≥ λ

This is the situation in which we have lobes that are of the same peak amplitude as the main

lobe. These lobes are called grating lobes.

Exercise

With your neighbor(s), explain physically and mathematically why for N = 2 and d = λ

we have grating lobes? Refer to Figure 1.2, determine the direction of the maximum

power on the yz-plane

Physical explanation:

Mathematical explanation: (Hint: use Equation (1.3) and assume φ = 90)

4.3. Phase of Elements

So far we have assumed that we are feeding the elements with waves of same amplitude

and phase. The question is what happens if we relax these assumptions? Let us start with

relaxing only the phase (feed the array with same amplitude but each element has different

phase 6). Relaxing the phase allows us to steer the main beam towards any direction.

Adding a progressive phase (β)7 to each element of the array requires reformulating the array

6Relaxing the amplitude results in something called ” amplitude tapering”. For example, binomial ta-pering, Dolph-Chebyshev tapering, . . . etc.

7Progressive phase means that the difference between the phase of two consecutive elements is fixed. Forinstance, if we have three elements on any of the principle axes, the phase difference between the element#1 and element #2 should be equal that between element #2 and element #3.


Antenna Arrays

factor of Equation (1.5) as

AF = exp jk(r • r1)+ exp j[k(r • r2) + β)]+ . . .+ exp j[k(r • rN) + (N − 1)β]

=[1 + exp j[k(r • dy) + β]+ . . .+ exp j[kr • (N − 1)dy + (N − 1)β]

]=

N∑n=1

exp j(n− 1)[kr • dy + β] =N∑n=1

exp j(n− 1)[kd sin θ sinφ+ β].

= exp (jN−12ψ)

[sin(N2ψ)

sin(ψ2

) ],(1.10)

where ψ = kd sin θ sinφ+β. Therefore, Equation (1.10) has two knobs (element spacing and

progressive phase) that control the array factor compared to one in Equation (1.5).

Exercise

So far we derived the array factor when the elements of the array are distributed along

the y-axis. With your neighbor(s), what will be the expression of ψ if the elements are

distributed along:

a. x-axis

b. z-axis

Hint: r • x = cosφ sin θ and r • z = cos θ

5. Polar Plot of Array Factor

An easy way to visualize the array factor is to plot it in polar coordinates. If we look back

at the array factor of Equation (1.10), we notice that it is periodic with respect to ψ, but it

does not give much information about the 3-dimensional pattern shape.

To extract information about the 3D pattern, let’s first assume that the array elements are


Antenna Arrays

distributed along the z-axis8. So we can write ψ as

ψ = kd cos θ + β, (1.11)

and the normalized array factor9 as

AF (ψ)n =

[sin(N2ψ)

N sin(ψ2

)]. (1.12)

Then we need to fellow these steps:

1. Plot the amplitude of |AF (ψ)n| versus ψ.

2. Below that plot, draw a circle of radius kd and centered at ψ = β. Make sure that

your plot in step 1 is aligned with circle.

3. Since we are using the normalized array factor, the circumference of the circle represents

an array factor of 1. (if we don’t normalize, the circumference would represent an array

factor of N).

4. Starting with the nulls of step 1, draw a vertical line from these nulls to the circum-

ference of the circle in step 2. Then, draw a radial line from the circumference of the

circle to its center. Note that if you cannot draw a vertical line between |AF (ψ)n| plot

and the circle, the point lies in what is so called the invisible region. Otherwise, the

point lies in the visible region10.

5. Now, repeat step 4 for peaks and smoothly connect these points.

These steps are better explained with an example. Let’s assume that we have N = 4,

d = λ/4, β = π/4 and let’s construct the polar plot of the radiation pattern using same

steps as above.

8In this distribution, we have only a variation on the θ direction. We can still plot with other distributionsbut we will have both θ and φ variations.

9Normalized array factor makes it easier to plot.10

• Visible region: The region that is electromagnetically visible, so the shape of the array factor at thatregion determines the radiation pattern of the array.

• Invisible region: Part of the array factor that does not affect the radiation pattern.


Antenna Arrays

1. The amplitude of the array factor is

AF (ψ)n =

[sin(2ψ)

4 sin(ψ2

)], (1.13)

which is plotted in Figure 1.7 part (a).

2. The next step is to draw a circle of radius r = kd =

(2π

λ

)(λ

4

)=π

2, and centered at

ψ =π

4as shown in Figure 1.7 part (b).

3. Finally, nulls and peaks location are drawn as shown in Figure 1.7 part (b).

4. To find the main beam angle, we use basic trigonometry. In this case, the main beam

is at θ = 180 − cos−1(π/4π/2

) = 120.

5. A final and yet unnecessary step is to mirror the results of the radiation pattern for

0 ≤ θ ≤ π to create the radiation pattern for −π ≤ θ ≤ 0.

It is apparent now why we have grating lobes for d > λ because we are expanding the

circle and hence expanding the visible region . In fact, it fellows from the 2π-periodicity of

the array fact that if d = λ/2, we are putting exactly a period in the visible region and if

d > λ we are putting at least two periods in the visible region.


Antenna Arrays

A-(3:)/2 -(5:)/4 -: -(3:)/4 -:/2 -:/4 0 :/4 :/2 (3:)/4 : (5:)/4 (3:)/2

|AF|

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0.2

0.4

0.6

0.8

1

:/6

-5:/6

:/3

-4:/6

:/2

-:/2

4:/6

-:/3

5:/6

-:/6

: 0:/2

0.2

0.4

0.6

0.8

1

:/6

-5:/6

:/3

-4:/6

:/2

-:/2

4:/6

-:/3

5:/6

-:/6

: 0:/2

(b)

(c)

(a)

Figure 1.7: Polar representation of the array factor for N = 4, d = λ/4, β = π/4; (a). |AF (ψ)|versus ψ. (b). Polar plot of the AF ; note that θ is defined over 0 ≤ θ ≤ π. However, to cover upthe whole space, we can mirror the results of 0 ≤ θ ≤ π to get the polar plot of −π ≤ θ ≤ 0. (c).A magnified polar plot of part (b)


Antenna Arrays

Exercise

With your neighbor(s), plot the polar plot of array factor for N = 4, d = λ/4, β = −π/4.

Determine angle of the main beam.


Antenna Arrays

Extra: Relation between DTFT and array factor

Recall that the discrete-time Fourier transform (DTFT) is

X(ω) =∞∑

n=−∞

x[n]e−jωn. (1.14)

We can write Equation (1.10) as

AF (ψ) =∞∑

n=−∞

w[n]e−jψn (1.15)

And letting wn = 0 ∀n ≤ 0 & n > N . Also, for a uniform linear array, w[n] = w ∀n ∈[1, N ].

The question now is how to use this resemblance to help us understand the array? Let

us start with some assumptions:

1. ψ = kd, we removed both phase and angular components.

2. We assume that x[n] is obtained by sampling a continuous time signal.

With that in mind, let us do the following mapping between the variables of Equations

(1.14) and (1.15)

t −→ u, ω −→ k

where u is the spacial coordinate; in other words u can be either x, y, or z. Let us

think about two scenarios under which this relation is helpful.

Choosing a reference point11

In the derivation of the array factor in Equation (1.5), we noticed that there is a

phase factor that results from choosing the reference point to be at the origin. If for

one reason we want to choose the reference point to be the center of the array and

let us assume N is odd, then we have two options: Rederive Equation (1.5) or use

the shift property of DTFT, which is x[n − k] −→ X(ω)e−jωk. In this case, we want


Antenna Arrays

w[n] −→ w[n+ N−12

] then the array factor would become

AF = exp (j(N − 1)ψ)sin(N2ψ)

sin(ψ2

)Element spacing

As we mentioned before, element spacing d should be at mostλ

2to avoid grating lobes.

In order to relate this condition to DTFT, we know that electromagnetic waves has

both spacial and temporal components. The spacial component defines the relation

between the spacial position u and wavenumber k, which—as we said before—a Fourier

transform pair. Therefore, from Nyquist sampling theorem, the spacing between ele-

ments should be

d ≤ π

Bk

=π2πλ

=λ

2. (1.16)

Reference point is the point that we call the origin

6. Special types of arrays:

6.1. Broadside Arrays

As we mentioned in the previous section, we have two knobs to control the radiation charac-

teristics of the array, phase and spacing. If we desire the array factor to peak on the direction

normal to the array axis, then we call this type of array a broadside array. The condition

under which an array is a broadside array is when ψ = 0 at the broadside. So, all what need

is to set ψ to zero at the broadside direction. For the array of Figure 1.4,

0 = ψ = kd sin θ sinφ+ β|φ=0,180 or θ=0,180 =⇒ β = 0 (1.17)

So, no matter what d is, as long as β = 0, the array will be broadside. However, if d ≥ λ,

there will be a maximum at the broadside along with other maximum or maxima in various

directions.


Antenna Arrays

Fact: Broadside array factor always has a fan beam.

Exercise

Work with your neighbor(s), what are the broadside radiation angles for an array of

half-wave dipoles if the array is distributed along the

a. x-axis?

b. z-axis?

Hint: Start with the definition of ψ for specific element distribution.

6.2. Ordinary End-Fire Arrays

In this case, we need the main beam to point towards the array axis. In other words, if the

array elements are distributed along the y-axis, we want the main beam to point towards

+y, −y, or both.

Let us start with making the beam points towards the +y. In this case, θ = 90 and

φ = 90; thus, we need to satisfy this condition

0 = ψ = kd sin θ sinφ+ β|φ=90,θ=90 =⇒ β = −kd. (1.18)

Of course, if we want to point the beam towards the −y, the new condition would be

0 = ψ = kd sin θ sinφ+ β|φ=−90,θ=90 =⇒ β = kd. (1.19)

Let us pause here and make the following observations:

(a). Although we think that we have two knobs to control the radiation pattern, we have

only one knob since the phase affects element spacing and vice versa.

(b). If d = λ/2, we will have two main beams, one points towards the +y, and the other

points towards the −y. (Prove it)


Antenna Arrays

(c). If d = nλ, n = 1, 2, . . ., we will have four maxima: two in the broadside and two along

the array axis.

(d). To get a single main beam, d ≤λ

2

(1−

1

2N

)with β = −kd for +y or β = kd for −y.

In Figure 1.8, we plot the an ordinary end-fire array factor for various spacing (or phasing)

for an array of N = 8 elements.

:/2

-:

: 0

(a) Blue curve is for β = −kd while red curve isβ = kd. In both cases, d = (0.47)λ.

:/2

-:/2

: 0

(b) Blue curve is for d = λ/2 while red curve isd = λ. In both cases, β = kd.

Figure 1.8: Radiation patterns for several ordinary end-fire arrays.

6.3. Hanson-Woodyard End-Fire Arrays

Let us ask the following question: can we make the end-fire array more directive? In other

words, can we make the main beam narrower? In 1938, Hanson and Woodyard derived the

conditions for directive end-fire array as


Antenna Arrays

β = −(kd+

2.94

N

)≈ −

(kd+

π

N

), For peak at φ = 90

β =

(kd+

2.94

N

)≈(kd+

π

N

), For peak at φ = −90

(1.20)

The spacing between elements should be

d =

(N − 1

N

)λ

4≈λ

4for large N (1.21)

To compare the directivity of Hanson-Woodyard (HW) end-fire array to that of ordinary

(ORD) end-fire array, the half-power beamwidth (HPBW) of these two designs are

ΘHW ≈ 2 cos−1(

1−0.1398λ

Nd

)(1.22)

ΘORD ≈ 2 cos−1(

1−1.391λ

πNd

). (1.23)

Note that, the increase in directivity observed by the Hanson-Woodyard array comes at the

expense of increasing the side-lobe level as shown in Figure 1.9.

Figure 1.9: Comparison between the radiation pattern of ORD and HW end-fire arrays. Thefigure is taken from Balanis


https://goo.gl/ED9EMN

Antenna Arrays

Exercise

For N = 8, compare the HPBW of both ordinary and Hanson-Woodyard end-fire arrays.

For fair comparison, fix d to the spacing required by the Hanson-Woodyard array. Work

with your neighbor(s)

6.4. Phased Arrays

In Section 5, we had two examples of two identical arrays. They both share same number of

elements and inter-element spacing; however, they have different phase factor (β). We saw

that the radiation patterns of the arrays are identical except that each one points the main

beam towards different direction. This observation gives us an idea on how to rotate the

main beam of an array. The question is what should be the value of β that corrsponds to

rotating (steering) the main beam towards a specifc direction. A simple answer would be to

find the steering angle and set ψ to zero at that angle. For example, let us assume that our

array is distributed along the z-axis and we want to steer the main beam towards θ = θ,

then β is given by

0 = ψ = kd cos θ + β|θ=θ =⇒ β = −kd cos θ. (1.24)


Antenna Arrays

For example, let’s assume we that have an array distributed along the z-axis with N = 4 and

d = 0.4λ. We want to scan broadside direction and 60 off the broadside from both directions.

Let’s do 30 increments, so the angles that we want to cover are θ = 30, 60, 90, 120, 150.The first step is to simplify the kd expression, so kd = 0.8π. Then, we need to find the phase

shifts associated with these angles, which are listed in Table 1.1.

Table 1.1: Phase factor β for each steering angle

θ β

30 −0.6928π

60 −0.4π

90 0

120 0.4π

150 0.6928π

The plots of the array factor for these cases are shown in Figure 1.10.

30

60

90

120

150

180

0

3=30o

3=60o

3=90o

3=120o

3=150o

Figure 1.10: Phased array radiation pattern.


Antenna Arrays

Exercise

With your neighbor(s), use the method discussed in Section to extract the polar plot of

the radiation pattern for θ = 60, 90, and 120. What is happening to the circle as the

phase changes?


Antenna Arrays

Extra: Element spacing of phased arrays

The rule of thumb that we have is the element spacing should adhere to the d < λ2.

However, let us assume that we have an array with its elements are distributed along

the z-axis, if we choose to steer the main beam towards the broadside θ = 90, we

can steer the main beam towards the broadside without any grating lobes even if the

space between elements, say 0.9λ. The general condition for minimum spacing—which

avoids grating lobes—is

d <λ

1 + | cos θ|, (1.25)

where θ is the maximum off broadside scanning angle.

7. Mutual Coupling

So far, we have assumed that the individual element of an array does not have any influence

on its neighbors. This assumption in reality is invalid, there must be an interaction between

array elements. As the spacing between the elements of an array gets smaller, the neighboring

elements become in the near-field region of each other. Thus, energy transfers from one

element to another. This interaction, consequently, impacts the following

1. The phase and magnitude of the element current.

2. As a result from the change in current, the antenna impedance is no longer equals to

the isolated antenna impedance12

3. Different currents alters the radiation pattern of the array; therefore, we need to ac-

count for it when we do the pattern multiplication.

Therefore, for closely-spaced elements, we need to modify the definition of Zin = V1/I1. For

N element array, the input impedance of the mth element is

Zm =VmIm

= Zm1I1Im

+ Zm2I2Im

+ . . .+ ZmmImIm

+ . . .+ ZmNINIm, (1.26)

12Isolated antenna is simply a stand alone antenna that does not belong to an array.


Antenna Arrays

where Zmn is called the mutual impedance if m 6= n and self impedance if m = n. It is

given by Zmn =Vm

In

∣∣∣∣∣Ii=0, ∀i 6=n

Equation (1.26) is derived from the general matrix that relates currents and voltages of

an array. Let us say that we have an array of N elements, the matrix that relates currents,

voltages, and mutual impedance in an array is an N ×N matrix given by

V1 = Z11I1 + Z12I2 + . . .+ Z1NIN

V2 = Z21I1 + Z22I2 + . . .+ Z2NIN

......

VN = ZN1I1 + ZN2I2 + . . .+ ZNNIN

(1.27)

From reciprocity, Znm = Zmn. For example, we know that λ/2-dipole has an input impedance

that is equal to 73 + j42.5Ω in which the radiation resistance is simply 73Ω. However, if

there are two dipoles separated by a distance of 0.2λ. The input impedance would be

Zin = Z11 − Z221/Z22 = (73 + j42.5)− (67.5)2/(73 + j42.5) = 26 + j70 Ω.13

References

Most of the material here are taken from:

(1). Prof. Gregory Durgin’s previous notes.

(2). Haupt, Randy L. Antenna Arrays: A Computational Approach: https://goo.gl/DzFgQL

(3). Balanis, Constantine A. Balanis Antenna Theory: Analysis and Design: https://goo.gl/ED9EMN

(4). Stutzman, Warren L. and Thiele, Gary A. Antenna Theory and Design: https://goo.gl/5Gi1xY

13Value of mutual impedance is taken from Kraus with the assumption that ratio between length of thedipole and its diameter is 73


https://goo.gl/DzFgQL

https://goo.gl/ED9EMN

https://goo.gl/5Gi1xY

https://goo.gl/FfMrBM

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