Antennas and Radiation

Course no. 0512.4861: Prof. Raphael Kastner.

School of Electrical Engineering, Tel Aviv University

Tel Aviv 69978, Israel. kast@eng.tau.ac.il

Main textbook:C. A. Balanis, Antenna Theory, Analysis and Design

3nd Edition, Wiley 2005.

Fall semester, 2009, Tav-Shin-Ayin.

These notes are for the sole use of students and staff of this course only.Unauthorized use of the notes for other purposes is strictly prohibited.

Contents

0 Introducing antennas 6

0.1 Definition of antenna . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

0.2 Antennas vs. sources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1 Radiation integrals and auxiliary potential functions 9

1.1 Maxwell Equations in the Frequency Domain . . . . . . . . . . . . . . . . 9

1.2 The wave equation for the electric field . . . . . . . . . . . . . . . . . . . 11

1.3 Magnetic sources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.4 Potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

1.5 Solution to A in free space . . . . . . . . . . . . . . . . . . . . . . . . . . 22

1.6 The far field approximation . . . . . . . . . . . . . . . . . . . . . . . . . 30

2 Fundamental parameters of antennas in TX mode 36

2.1 Radiation intensity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

2.2 Radiation patterns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

2.2.1 Power patterns . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

1

2

2.2.2 Field patterns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

2.2.3 Beamwidth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

2.2.4 Sidelobe level - SLL . . . . . . . . . . . . . . . . . . . . . . . . . . 40

2.2.5 Beam efficiency . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

2.3 Directivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

2.4 Radiation resistance and antenna impedance . . . . . . . . . . . . . . . . 42

2.5 Efficiency and Gain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

2.6 Effective radiated power (ERP) . . . . . . . . . . . . . . . . . . . . . . . 45

2.7 Polarization (TX mode) . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

2.7.1 Right handed and left handed elliptical polarizations: . . . . . . . 47

2.7.2 Linear polarizations: . . . . . . . . . . . . . . . . . . . . . . . . . 49

2.7.3 Circular polarizations: . . . . . . . . . . . . . . . . . . . . . . . . 49

2.7.4 Polarization states in the complex plane . . . . . . . . . . . . . . 49

2.7.5 Stokes’ parameters and the Poincare sphere . . . . . . . . . . . . 49

2.7.6 polarization characteristics in receive (RX) mode: . . . . . . . . . 52

3 The antenna in RX mode 53

3.1 The reciprocity theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

3.2 The open–circuit voltage at the antenna port . . . . . . . . . . . . . . . . 55

3.3 The effective length of the antenna . . . . . . . . . . . . . . . . . . . . . 57

3

3.4 Polarization loss factor (PLF) . . . . . . . . . . . . . . . . . . . . . . . . 60

3.5 Effective aperture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

3.5.1 Aperture efficiency . . . . . . . . . . . . . . . . . . . . . . . . . . 62

3.6 Frijs’s formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

3.7 Noise received at the antenna terminals . . . . . . . . . . . . . . . . . . . 64

4 Linear wire antennas 65

4.1 Line sources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

4.1.1 The infinitesimal electric dipole . . . . . . . . . . . . . . . . . . . 66

4.1.2 Uniform distribution . . . . . . . . . . . . . . . . . . . . . . . . . 67

4.1.3 Tapered distributions: The cosine distribution . . . . . . . . . . . 69

4.1.4 Tapered distributions: Triangular taper . . . . . . . . . . . . . . . 70

4.2 The equivalence theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

4.3 Perfectly conducting antennas and the equivalence theorem . . . . . . . . 74

4.4 Wire antennas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

4.5 Wire antennas over perfectly conducting ground planes . . . . . . . . . . 81

4.5.1 Half wave dipole over ground plane . . . . . . . . . . . . . . . . . 82

4.5.2 Quarter–wave Monopole antennas . . . . . . . . . . . . . . . . . . 82

4.6 Feeding wire antennas . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

4.7 The folded dipole . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

4

4.8 The Induced EMF method for the assessment of the input impedance of

a wire antenna . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

4.8.1 Example - the infinitesimal dipole . . . . . . . . . . . . . . . . . . 88

5 Aperture antennas 92

5.1 The equivalence theorem revisited . . . . . . . . . . . . . . . . . . . . . . 92

5.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

5.2.1 Uniform rectangular aperture . . . . . . . . . . . . . . . . . . . . 96

5.2.2 Other separable distributions . . . . . . . . . . . . . . . . . . . . 98

5.2.3 Non–separable distribution: a uniform circular aperture with radial

variation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

5.3 Approximate directivity and effective area calculations . . . . . . . . . . 100

5.4 Horn antennas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102

5.4.1 H–plane sectoral horns . . . . . . . . . . . . . . . . . . . . . . . . 103

5.4.2 E–plane sectoral horns . . . . . . . . . . . . . . . . . . . . . . . . 106

5.4.3 Pyramidal horn . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

5.5 Parabolic reflectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110

5.5.1 Constant phase . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

5.5.2 Tapered amplitude . . . . . . . . . . . . . . . . . . . . . . . . . . 113

5.5.3 Design procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

5.6 Microstrip antennas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

5

5.6.1 Slot antennas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

5.6.2 Microstrip patch antennas . . . . . . . . . . . . . . . . . . . . . . 118

6 Linear antenna arrays 121

6.1 The Array Factor (AF) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122

6.1.1 Scanning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

Chapter 0

Introducing antennas

Balanis: Chapter 1.

0.1 Definition of antenna

Definition 1 An antenna is a device that transmits and/or receives electromagnetic

waves.

More specifically, the antenna is a structure that coverts guided electromagnetic waves

in a transmission line into radiation in free space, and/or vice versa. it is represented

from the point of view of the feeding transmission line as a one–port network that can

be characterized by its input voltage Vin, current Iin, input impedance ZA, reflection

coefficient at the port Γ = S11 and VSWR (see Figure 0.1). This port is defined at a

reference plane in the feeding transmission line. If the antenna transmits well, then most

of the electromagnetic energy impinging on it from the transmission line is converted into

radiated energy. In this case, the Γ → 1 and the antenna is said to be matched to the

line.

6

7

Figure 0.1: The antenna from the point of view of the feeding transmission line intransmit (TX) mode: A one–port with an input impedance ZA. The feeding network isrepresented by its Thevenin equivalent Vs, Zs.

0.2 Antennas vs. sources

To make the antenna problem in transmit (TX) mode simpler, we replace it with an

equivalent (current) source distribution J(r, ω) (as well as equivalent magnetic source

distribution Jm(r, ω)) where r = xx + yy + zz that flows in in free space, instead of the

actual current that flows in the antenna structure, see Fig. 0.2. In general, J(r, ω) is not

known, but is rather a solution of the electromagnetic boundary problem of the antenna

8

Figure 0.2: Antenna structure needs to be transformed into an equivalent current sourcedistibution in free space in order to find the radiated fields in transmit (TX) mode.

excited by the feed. However, we will first solve a simplified problem, assuming that this

hard problem is solved and hence J(r, ω) is given, or is known via an appropriate ap-

proximations. This simplification is the basis for the upcoming chapters. Since J(r, ω) is

defined in the frequency domain, it has a sinusoidal time dependence that implies acceler-

ation. As noted in Balanis Chapter 1, accelerating charge causes radiation. Therefore, to

achieve powerful radiation, we need (1) rapid time dependence, and (2) a strong enough

source. The first condition is met by using ω in the range of radio frequencies (RF) that

includes microwaves and millimeter waves, and condition (2) will be met if the antenna

is well matched and is capable of directing the radiation into specified directions in space

in an efficient manner.

Chapter 1

Radiation integrals and auxiliarypotential functions

Balanis: Ch. 3.

1.1 Maxwell Equations in the Frequency Domain

We begin with the assumed source distribution J(r, ω) in TX mode. In magneto-

quasistatics, where ω → 0, the current distribution serves as the source for the magnetic

field according to Ampere’s law:

∇×H = J. (1.1)

A dual equation for the electo-quasitatic case for the curl of the electric field reads

∇× E = 0. (1.2)

This equation states that when time changes are slow, the electric field is irrotational,

hence the defintions of electostatic potential etc. Equations (1.1) and (1.2) are uncoupled,

as long as the time variations are slow. However, when the time changes are more rapid,

i.e., when the frequency is high, then Eq. (1.2) takes the more general form of Faraday’s

law:

∇× E = −ωµH. (1.3)

9

10

Similarly, the displacement current is added to Eq. (1.1) to form a generalized from of

Ampere’s law:

∇×H = J + ωεE. (1.4)

Equations (1.3) and (1.4), represent the electrodynamic case, as opposed to the qua-

sistatic case, where fields are coupled and none can exist without the other. Rapid time

variation is the key to this coupling, because the time derivative of the magnetic (or

electric) field serves as a source for the electric (or magnetic) filed on the right hand side

of both equations. These two equations are the Maxwell’s curl equations. We can also

add the two divergence equations, Gauss’s (Coulomb) law and the similar equation for

the magnetic field to form the compete system of Maxwell equations:

∇× E = −ωµH (1.5a)

∇×H = J + ωεE (1.5b)

∇ · (εE) = ρ (1.5c)

∇ · (µH) = 0. (1.5d)

The two Maxwell’s curl equations (1.5a) and (1.5b) contain most of the information

needed for developing the radiation theory. Eq. (1.5c) has the electric charge as another

source term. In electodymanics, though, this term is dependent on J. To see this, we

take the divergence of (1.5b):

∇ · ∇ ×H = ∇ · J + ω∇ · (εE) (1.6)

By using the identity ∇ · ∇ ×H = 0 and Eq. (1.5c), we have

∇ · J + ωρ = 0. (1.7)

11

Eq. (1.7) is the continuity equation, that also represents conservation of charge, and is

now seen to be a consequence of Maxwell’s equations.

In a source free space (J ≡ 0), Maxwell’s equations take the symmetrical form

∇× E = −ωµH (1.8a)

∇×H = ωεE (1.8b)

∇ · (εE) = 0 (1.8c)

∇ · (µH) = 0. (1.8d)

Then, they obey the principle of duality: If we use the following substitutions, we same

equations are recovered:

E −→ H (1.9a)

H −→ −E (1.9b)

ε −→ µ (1.9c)

µ −→ ε (1.9d)

1.2 The wave equation for the electric field

The electromagnetic field is known to be a wave traveling at the speed of light. This is

also a consequence of Maxwell’s equations, as follows. We use the two curl equations, as

we will ofter do, as the starting point. In order to get the wave equation for the electric

field, apply the curl operator to (1.5a):

∇×∇× E = −ω∇× (µH) = −ωµ∇×H− ω∇µ×H. (1.10)

Assume now that the medium is honogeneous in µ, i.e., µ 6= µ(r). Then, with (1.5b),

∇×∇× E = −ωµ(J + ωεE) (1.11)

12

from which we have the vector wave equation for the electric field:

∇×∇× E− ω2µεE = −ωµJ (1.12)

Denote the propagation constant k = ω√µε = ω

c, showing that the wavespeed c = 1√

µεis

equal to the speed of light.

We can further develop (1.12) using the identity

∇×∇× E = ∇(∇ · E)−∇2E. (1.13)

If the medium is also ε–homogeneous, (ε 6= ε(r), such that ∇ · (εE) = ε∇ ·E) we can use

(1.5c) in (1.13):

∇×∇× E =∇ρε−∇2E (1.14)

by which (1.12) takes the form

∇2E + k2E = ωµJ +∇ρε. (1.15)

Finally, bringing in the continuty equation (1.7) to substitute ρ = −∇·Jω

:

∇2E + k2E = ωµJ− ∇∇ · Jωε

(1.16)

written concisely as

(∇2 + k2)E = ωµ

(1 +

∇∇·k2

)J. (1.17)

1.3 Magnetic sources

In many application, it is mathematically convenient to replace a part of the electric

current sources by artificial magnetic currents Jm, that. Maxwell’s equations (1.5) then

13

become fullly symmetric:

∇× E = −ωµH− Jm (1.18a)

∇×H = ωεE + J (1.18b)

∇ · (εE) = ρ (1.18c)

∇ · (µH) = ρm (1.18d)

where ρm is the magnetic charge. These equations contain many physical principles,

including conservation of energy and charge and special relativity. Not included, though,

is the principle of causality, that we take for granted and use extensively below.

When re–deriving the wave equation for the electric field, instead of (1.17) we would

have

(∇2 + k2)E = ωµ

(1 +

∇∇·k2

)J +∇× Jm. (1.19)

The duality relationship (1.9) can now be extended to include sources:

E −→ H (1.20a)

H −→ −E (1.20b)

ε −→ µ (1.20c)

µ −→ ε (1.20d)

J −→ Jm (1.20e)

Jm −→ −J (1.20f)

ρ −→ ρm (1.20g)

ρm −→ −ρ (1.20h)

Using this duality relationship, we can write immediately the continuity equation for the

14

magnetic sources:

∇ · Jm + ωρm = 0 (1.21)

and also the dual equation to (1.19):

(∇2 + k2)H = ωε

(1 +

∇∇·k2

)Jm −∇× J. (1.22)

An example on the use of magnetic currents A plane wave propagating in the

+z direction and polarized in the x direction has the form

E(r) = xE0e−kz. (1.23)

Verify that this is a solution to the wave equation (1.17) for the case J = 0. The

accompaying magnetic field is then

H(r) = yE0

ηe−kz (1.24)

where η =√

µε

is the medium impedance. Equations (1.23) and (1.24) can be verified as

a solution to Maxwell’s source–free equations (1.8). This solution spans the entire three

dimensional space, and has no sources within this space.

Suppose now that we are only interested in the half–space z > 0. We wish to maintain

the solution (1.23) - (1.24) in this region, but for some reason may allow for a different

field to occupy the region z < 0. For example, suppose we wish to impose zero electric

field at z < 0. For example, create the following artificial field:

E(r) = xE0e−kzu(z) (1.25)

where u(z) is the unit step function, with the step at z = 0. If we only consider the

region z > 0, then u(z) = 1 and the total solution (1.23) - (1.24) is valid. However,

15

when looking at the entire space, the expression (1.24) needs updating and the space

may longer be source free. To see this, substitute (1.25) into Maxwell’s equation (1.18a):

∇× E = ∇× xE0e−kzu(z) = −x×∇E0e

−kzu(z) = −x× zE0e−kz (−ku(z) + δ(z))

= yE0e−kz (−ku(z) + δ(z)) = −ωµH− Jm (1.26)

For the artificial field H, assume the form

H = yE0

ηe−kzu(z) (1.27)

and substitue it back into (1.26). Then, we get an expression for the magnetic current:

Jm = −yE0δ(z). (1.28)

Before we look deeper into (1.28), use (1.27) with Maxwell’s equation (1.18b):

∇×H = ∇× yE0

ηe−kzu(z) = −y×∇E0

ηe−kzu(z) = −y× z

E0

ηe−kz (−ku(z) + δ(z))

= −xE0

ηe−kz (−ku(z) + δ(z)) = ωεE + J. (1.29)

Now E has been defined in (1.25), therefore (1.29) yields an expression for J:

J = −xE0

ηδ(z). (1.30)

We can define the surface electric and magnetic currents Js,Jms via

J = Jsδ(z) (1.31a)

J = Jmsδ(z). (1.31b)

Let’s sum up the expressions we have for the artificial fields and sources:

E = xE0e−kzu(z) (1.32a)

H = yE0

ηe−kzu(z) (1.32b)

Js = −xE0

η(1.32c)

Jms = −yE0 (1.32d)

16

Again, within the region z > 0 nothing has changed compared with the solution (1.23),

(1.24). However, in the region z < 0 we have imposed an artificial zero field. The artificial

discontinuEities created at z = 0 are compensated by the artificial sources Js,Jms that

clearly obey (see (1.32))

Js = z× (H|z=0+ − H|z=0−) (1.33a)

Jms = −z× (E|z=0+ − E|z=0−) (1.33b)

which are the familiar boundary conditions over a surface (with n = z).

In summary, we have been able to re-create the correct field over the half space region

z > 0, while anullling the field over z < 0 using artificial sources at the interface z = 0.

These sources will be termed equivalent sources, since they may not be physical, in

particular when we discuss magnetic sources.

It should be emphasized that if we only require the field to be correct at a part of

the space, there are many options for using equivalent sources, each option defining a

different field withing the complementary region. For example, for the region z < 0 we

could choose a plane wave propagating in the −z direction instead of a zero field:

E = xE0ekz, z < 0 (1.34a)

H = −yE0

ηekz, z < 0 (1.34b)

Then the equivalent currents would be

Js = −2xE0

η(1.35a)

Jms = 0 (1.35b)

reflecting the fact that H is discontinuous over z = 0 by the amount shown in (1.35)

17

while E is continuous. Yet another option is the back propagatin wave

E = −xE0ekz, z < 0 (1.36a)

H = yE0

ηekz, z < 0 (1.36b)

for which the equivalent sources would be

Js = 0 (1.37a)

Jms = −2yE0. (1.37b)

Many other options exist as well. The particular choice is a matter of simplicity and ease

of calculations.

1.4 Potentials

The vector wave equations (1.19) and (1.22) as developed above can be, and are used for

direct solution of the fields. In antenna theory, however, it is customary to use auxiliary

potential functions in the process. To define the potentials, we split the fields to ones

contributed by electric or magnetic sources:

E = Ee + Em (1.38a)

H = He + Hm. (1.38b)

Then, by virtue of (1.18c) and (1.18d),

∇ · εEm = 0 (1.39a)

∇ · µHe = 0. (1.39b)

18

Using again the identity ∇ · ∇ × A = 0, we can define the curl of the magnetic and

electric vector potentials F and A by

εEm = −∇× F (1.40a)

µHe = ∇×A. (1.40b)

With the definition of the two potentials, we can augment the duality relationships

(1.20) as follows:

E −→ H (1.41a)

H −→ −E (1.41b)

ε −→ µ (1.41c)

µ −→ ε (1.41d)

J −→ Jm (1.41e)

Jm −→ −J (1.41f)

ρ −→ ρm (1.41g)

ρm −→ −ρ (1.41h)

A −→ F (1.41i)

F −→ −A. (1.41j)

Consider first the electric potential A. Taking the curl of Ee,

∇× Ee = −ωµHe = −ω∇×A. (1.42)

Therefore,

∇× (Ee + ωA) = 0. (1.43)

19

In analogy with the quasistatic case, we can define a scalar potential φe by

Ee + ωA = −∇φe (1.44)

i.e.,

Ee = −∇φe − ωA (1.45)

Now take the curl on both sides of Eq. (1.40b):

∇× µHe = ∇×∇×A = ∇∇ ·A−∇2A (1.46)

For a µ–homogeneous medium,

∇×µHe = µ∇×He = ∇∇·A−∇2A = ωµεEe+µJ = ωµε (−∇φe − ωA)+µJ (1.47)

or

(∇2 + k2)A = −µJ +∇ (∇ ·A + ωµεφe) (1.48)

Now we choose the divergence of A as

∇ ·A = −ωµεφe (1.49)

Thereby arriving at a wave equation for A with a simple source term:

(∇2 + k2)A = −µJ. (1.50)

Equation (1.49) is called the Lorentz Gauge. This is an arbitrary choice, popular in

antenna theory but by no means the only one.

Once equation (1.50) has been solved for the potential, the electric field can be found

from (1.44):

Ee = −∇φe − ωA = ∇∇ ·Aωµε

− ωA = −ω(

1 +∇∇·k2

)A (1.51)

20

In summary, both fields are found from the potential via

Ee = −ω(

1 +∇∇·k2

)A (1.52a)

He =1

µ∇×A. (1.52b)

The magnetic potential and the corresponding fields can be found using the duality

relationships (1.41) with (1.50) and (1.52):

(∇2 + k2)F = −εJm (1.53)

and

Hm = −ω(

1 +∇∇·k2

)F (1.54a)

Em = −1

ε∇× F. (1.54b)

Example 1 Take the following (plane wave) solution to Equation (1.50):

A = xA0e−kz (1.55)

Then, the electric field is (1.52a)

E = −ω(

1 +∇∇·k2

)A = −ωA = −xωA0e

−kz (1.56)

and the magnetic field is

H =1

µ∇×A =

1

µ

∣∣∣∣∣∣

x y z∂∂x

∂∂y

∂∂z

Ax Ay Az

∣∣∣∣∣∣= −y

k

µA0e

−kz (1.57)

We can check that this is a valid plane wave:

|E||H| =

ωµ

k=

ωµ

ω√µε

=

õ

ε= η (1.58)

and E×H is z–directed.

21

Example 2 Given the potential A = xA0e−k|z|, (a) show that its source is a surface

electric current Js = xJs at z = 0 and find Js, (b) compute the E– and H–fields over

−∞ < z <∞.

(a) Let’s substitute A into the left hand side of (1.50):

(∇2 + k2)A = x

(d2

dz2+ k2

)A0e

−k|z|.

Now by the chain rule of differentiation,

d

dze−k|z| = −ke−k|z|d|z|

dz= −ke−k|z| sgn(z), where sgn(z) =

+1, z > 0

−1, z < 0.

Then,

d2

dz2e−k|z| = (−k)2e−k|z| sgn2(z)− ke−k|z|2δ(z)

since d[ sgn (z)]dz

= 2δ(z). Also, sgn2(z) = 1, therefore

d2

dz2e−k|z| = −k2e−k|z| − ke−k|z|2δ(z)

and

(∇2 + k2)A = x

(−k2 − 2kδ(z) + k2)A0e

−k|z| = −x2kA0e−k|z|δ(z) = −x2kA0δ(z).

By (1.50),

−x2kA0δ(z) = −µJ,

therefore

J = x2k

µA0δ(z) = xJsδ(z)

where

Js = 2k

µA0 (1.59)

which is the desired answer.

22

(b) To compute H, use (1.52b):

H =1

µ∇×A =

A0

µ∇× (

xe−k|z|)

= −A0

µx×∇e−k|z|

= −A0

µx× z(−k)e−k|z| sgn (z) = −y

kA0

µe−k|z| sgn (z).

Note that indeed the magnetic field is continuous across z = 0, the discontinuity being

offset by the surface current according to the boundary condition at z = 0:

z× (H|z=0+ − H|z=0−) = x

(kA0

µ−

(−kA0

µ

))= x2

k

µA0

Comparing with (1.59), we can verify the expected result

z× (H|z=0+ − H|z=0−) = xJs.

The electric field can be found from (1.52). Since ∇ ·A = 0,

E = −ω(

1 +∇∇·k2

)A = −ωA = −xωA0e

−k|z|. (1.60)

We can now verify that E H and J satisfy Maxwell’s equation (1.18b):

E =1

ωε(∇×H− J) =

1

ωε

(∇×

(−y

kA0

µe−k|z| sgn (z)

)− J

)

=1

ωε

(y × z

d

dz

(kA0

µe−k|z| sgn (z)

)− J

)= x

1

ωε

(kA0

µ

d

dz

(e−k|z| sgn (z)

)− 2k

µA0δ(z)

)

= x1

ωε

(kA0

µ

(−ke−k|z| sgn2(z) + e−k|z|2δ(z))− 2

k

µA0δ(z)

)

= xk2A0

ωµεe−k|z| = −xωA0e

−k|z|

which is the same result as in (1.60). You can also verify that |E||H| = η.

1.5 Solution to A in free space

We wish to solve Eq. (1.50) in free space. If we split the vector A into its Cartesian

components, then the unit vectors are constant and the equation can solved component

23

by component, e.g.,

(∇2 + k2)Ax = −µJx. (1.61)

We thus address the general scalar wave equation first. Denote Ax, Ay orAz as f , and

the corresponding scalar source as S. Then, the equation to be solved is

(∇2 + k2)f(r) = −S(r). (1.62)

The technique we use here is based on solving first for the Green’s function (sometimes

called the “fundamental solution” or, in system terms, the “impulse response”) which is

the solution to the equation

(∇2 + k2)G(r, r′) = −δ(r− r′). (1.63)

G(r, r′) is the solution for a point excitation located at r = r′. Once this problem is

solved, we can construct the solution to (1.62) as a superposition of Green’s functions

for a distribution of these point sources with varying r′.

The solution for (1.63) was obtained a long time ago. It is basicallly a perfect spherical

wave emanating from the point source:

G(r, r′) =e−k|r−r′|

4π|r− r′| (1.64)

where |r− r′| =√

(x− x′)2 + (y − y′)2 + (z − z′)2. We can verify this solution by direct

substitution in (1.63). However, to make this excercise simple, we take the special case

r′ = 0, understanding that the solution for a non–zero r′ can be obtained from this

solution by a simple translation as long as we deal with the free space problem. The

equation we are now looking at is

(∇2 + k2)G(r) = −δ(r) (1.65)

24

and the solution to be substituted into (1.65) is G(r, r′) = e−kr

4πr. This is a spherically–

symmetric expression, therefore we continue with the use of spherical coordinates.

(∇2 + k2)G(r) =

(∇2 + k2) e−kr

4πr(1.66)

Now,

∇2 e−kr

4πr= ∇ ·

(∇e

−kr

4πr

)= ∇ ·

(1

4πr∇e−kr + e−kr∇ 1

4πr

)

=1

4πr∇2e−kr + 2∇ 1

4πr· ∇e−kr + e−kr∇2 1

4πr(1.67)

Using the definition of the Laplacial in spherical coordinates,

∇2e−kr =1

r2

d

dr

(r2 d

dre−kr

)=−kr2

d

dr

(r2e−kr

)=

((−k)2 − 2

k

r

)e−kr

∇ 1

4πr= −r

1

4πr2; ∇e−kr = r(−k)e−kr −→ ∇ 1

4πr· ∇e−kr =

ke−kr

4πr2

therefore, after cancellation of terms,

(∇2 + k2) e−kr

4πr= e−kr∇2 1

4πr. (1.68)

However,

∇2 1

4πr= −δ(r) (1.69)

because

∇2 1

4πr= 0, r 6= 0 (1.70)

and

∆V

∇2 1

4πrdV =

∆V

∇ · ∇ 1

4πrdV =

S

∇ 1

4πr· rdS

= −π 2π

θ=0 φ=0

1

4πr2r2 sin θdθdφ = −1 (1.71)

25

where ∆V is a sphere around the origin whose surface is S. Thus,

(∇2 + k2) e−kr

4πr= −e−krδ(r) = −δ(r), (1.72)

and since the solution should be invariant under translation in free space,

(∇2 + k2)G(r, r′) = −δ(r− r′), G(r, r′) =

e−k|r−r′|

4π|r− r′| . (1.73)

The Green’s function is an outgoing spherical wave. Another solution to the second order

equation would include a + sign in the exponent: G(r, r′) = e+k|r−r′|4π|r−r′| . This would be an

incoming spherical wave that violates the priniple of causality, which we have to assume in

addition to Maxwell’s equations. The principle of causality is expressed mathematically

in the form

limr′→∞

r′[∂G(r, r′)∂r′

+ kG(r, r′)]

= 0. (1.74)

which is often referred to as the radiation condition.

In order to obtain to solution to (1.62), rewrite it with (1.73), add primes formally

and multiply as follows:

G(r, r′)[(∇′2 + k2

)f(r′) = − S(r′)] (1.75a)

f(r′)[(∇′2 + k2

)G(r, r′) = − δ(r− r′)] (1.75b)

having used ∇′2G = ∇2G. Subtracting (1.75b) from (1.75a), we have

G(r, r′)∇′2f(r′)− f(r′)∇′2G(r, r′) = −G(r, r′)S(r′) + f(r′)δ(r− r′) (1.76)

Eq. (1.76) is the same as

∇′ · [G(r, r′)∇′f(r′)− f(r′)∇′G(r, r′)] = −G(r, r′)S(r′) + f(r′)δ(r− r′). (1.77)

26

Integrating (1.77) over a volume V surrounded by a surface S, we get

f(r) =

V

G(r, r′)S(r′)dV ′ +S

[G(r, r′)

∂f(r′)∂n′

− f(r′)∂G(r, r′)∂n′

]dS ′ (1.78)

having used n′ · ∇′ = ∂∂n′ . Eq. (1.78) contain a volume integral, called the radiation

integral, and a surface integral. The contribution of the volume integral is readily inter-

preted as a superposition over source elements S(r′)dV ′, each one being weighted by the

Green’s function G(r, r′) see Fig. 1.1.

Figure 1.1: Volume integration in Eq. (1.78) is done over the shaded region where S(r′) 6=0. The surface integration is over S.

If all the sources are included in V , then Eq. (1.78) is greatly simplified according to

the following psedu-theorem.

Theorem 1 If all sources are included in V , then the surface integral in (1.78) vanishes.

27

Proof. This proof is based on the principle of causality. Consider two closed surfaces

S1 and S2, both enclosing all sources, as shown in Fig 1.2. The volume integral in (1.78)

is not affected by the choice of the closed surface, hence the surface integrals are the

same:

S1

[G(r, r′)

∂f(r′)∂n′

− f(r′)∂G(r, r′)∂n′

]dS ′ =

S2

[G(r, r′)

∂f(r′)∂n′

− f(r′)∂G(r, r′)∂n′

]dS ′.

Therefore, the choice of S is arbitrary as long as it encloses all sources. Choose a large

spherical surface with r′ À r, r′ ∈ S. Then

S

[G(r, r′)

∂f(r′)∂n′

− f(r′)∂G(r, r′)∂n′

]dS ′

=

π 2πθ=0 φ=0

[G(r, r′)

∂f(r′)∂r′

− f(r′)∂G(r, r′)∂r′

]r′2 sin θ′dθ′dφ

−→π 2π

θ=0 φ=0

e−k|r−r′|

4πr′

[∂f(r′)∂r′

+ kf(r′)]r′2 sin θ′dθ′dφ = 0.

provided that

limr′→∞

r′[∂f(r′)∂r′

+ kf(r′)]

= 0. (1.79)

Eq. (1.79) is an expression of the principle of causality, called the radiation condition.

It is similar in form (1.74), except it now applies to all causal functions, not just the

Green’s functions. ¥

For the vector fields, the radiation condition takes the form

limr′→∞

r′ (η r′ ×H + E) = 0 (1.80a)

limr′→∞

r′ (η r′ × E−H) = 0. (1.80b)

Summary: If the wave function f obeys the radiation condition, then the solution to

28

Figure 1.2: For Theorem 1: If all sources are contained within both S1 and S2 thenS1

[G(r, r′)∂f(r′)

∂n′ − f(r′)∂G(r,r′)∂n′

]dS ′ =

S2

[G(r, r′)∂f(r′)

∂n′ − f(r′)∂G(r,r′)∂n′

]dS ′ = 0.

(1.62) is simply the superposition integral

f(r) =

V

S(r′)e−k|r−r′|

4π|r− r′|dV′. (1.81)

Going back to (1.61), the solution for each one of the components of A is

Ax,y,z(r) = µ

V

Jx,y,z(r′)e−k|r−r′|

4π|r− r′|dV′ (1.82)

or, in vector form, we have the radiation integral:

A(r) = µ

V

J(r′)e−k|r−r′|

4π|r− r′|dV′ . (1.83)

Similary, for the magnetic potential, by duality

F(r) = µ

V

Jm(r′)e−k|r−r′|

4π|r− r′|dV′. (1.84)

29

Example: first visit with the infinitesimal electric dipole (see next visit in

Section 4). This is a source defined as

J(r′) = I0lδ(r′)z′. (1.85)

Apply (1.83):

A(r) = µ

V

I0lδ(r′)z′

e−k|r−r′|

4π|r− r′|dV′ = zµI0l

V

δ(r′)e−k|r−r′|

4π|r− r′|dV′ (1.86)

where the volume V contains the origin r′ = 0. Imporant note: The unit vector z could

be moved outside the integrand sign because it is a constant: z′ = z. Do not do this with

non–Cartesian unit vectors such as θ′, φ

′, because in general θ

′ 6= θ etc.

Continuing on with (1.83):

A(r) = zµI0l

V

δ(r′)e−k|r−r′|

4π|r− r′|dV′ = zµI0l

e−k|r−r′|

4π|r− r′|

∣∣∣∣r′=0

= zµI0le−kr

4πr. (1.87)

It seems that we have recovered the Green’s function, which suits the point source nature

of the dipole. However, note the addition of the unit vector z that destroys the sperical

symmetry of A. Transforming z to spherical unit vectors,

A(r) = (r cos θ − θ sin θ)µI0le−kr

4πr(1.88)

the expression for A is now writted fully in spherical coornitates, showing the θ–dependence

explicitely.

Proceed now to calculating the fields, using spherical coordinates formulas for the curl

30

operator:

H =1

µ∇×A =⇒

Hr = 0

Hθ = 0

Hφ = − I0lk2

4πe−kr

(1kr

+ 1(kr)2

)sin θ

(1.89a)

E =1

ωε∇×H =⇒

Er = − I0lk2η2π

e−kr(

1(kr)2

+ 1(kr)3

)cos θ

Eθ = − I0lk2η4π

e−kr(

1kr

+ 1(kr)2

+ 1(kr)3

)sin θ

Eφ = 0

(1.89b)

using, e.g.,

∇×A =

∣∣∣∣∣∣∣∣

brr2 sin θ

bθr sin θ

cφr

∂∂r

∂∂θ

∂∂φ

Ar rAθ r sin θAφ

∣∣∣∣∣∣∣∣=

φ

r

(∂

∂r(rAθ)− ∂

∂θAr

)

since ∂∂φ

= 0 and Aφ = 0. By duality, you can write down the expression for the potenial

and the fields of a magnetic elementary dipole defined by Jm(r′) = Im0lδ(r′)z′.

1.6 The far field approximation

When the distance from the source r is very large compared with both (a) the size of

the source and (b) the wavelength, we have the far field approximation for the Green’s

function.

The expression (1.73)i s clearly divided into amplitude and phase terms, as follows:

G(r, r′) =1

4π|r− r′|︸ ︷︷ ︸amplitude

e−k|r−r′|

︸ ︷︷ ︸phase

. (1.90)

The condition (a) tells us that r À r′. Use this to approximate the amplitude term:

1

4π|r− r′| '1

4πr. (1.91)

For the phase term, we cannot use the same approximation, since we cannot compare

absolute values of the phase. For example, 20 ≮ 7210. The phase is always compared to

2π. Therefore, we need a more subtle approximation.

31

Expand the phase (power) term as follows.

|r− r′| = [(r− r′) · (r− r′)]12 = [r2 − 2r · r′ + r′2]

12 = r

[1− 2

r · r′r2

+

(r′

r

)2] 1

2

= r

[1− 2r · r

′

r+

(r′

r

)2] 1

2

= r

1 +

1

2

(−2r · r

′

r+

(r′

r

)2)− 1

8

(−2r · r

′

r+

(r′

r

)2)2

+ · · ·

= r

[1− r · r

′

r+

1

2

(r′

r

)2

− 1

2

(r · r

′

r

)2

+1

2r · r′

(r′

r

)3

− 1

8

(r′

r

)4

+ · · ·]. (1.92)

Now let’s retain terms in the brackets up to the order of(r′r

)2:

|r−r′| ' r

[1− r · r

′

r+

1

2

(r′

r

)2

− 1

2

(r · r′ r

′

r

)2]

= r

[1− r · r

′

r+

1

2

(r′

r

)2

(1− r · r′)]

(1.93)

such that the total phase (power) term is

k|r− r′| ' kr − kr · r′︸ ︷︷ ︸keep

+kr

2

(r′

r

)2

(1− r · r′)︸ ︷︷ ︸

drop

. (1.94)

The term kr ·r′ cannot be neglected relative to ∼ π (we come back to this approximation

in (1.97)). This is the main difference between the approximations for the amplitude and

phase, pointed above. We will thus keep it, and neglect only the last term in addition the

the higher order terms already dropped. The Green’s function then takes the approsimate

form

G(r, r′) ' e−kr

4πrek·r

′(1.95)

where we define the vector k = kr = k(x sin θ cosφ + y sin θ sinφ + z cos θ). Insert this

expresseion into the radiation integral, and take outside the integral sign the terms that

are independent of r′:

A(r) ' µe−kr

4πr

V

J(r′)ek·r′dV ′ . (1.96)

32

Only the integral portion of this expression differes between different sources. It depends

on k, i.e. on (θ, φ) only. The e−kr

4πrdependence on r is universal.

Eq. (1.96) is interpreted as a superposition integral, where all contributions are added

up along parallel rays, see Fig. 1.3.

Figure 1.3: Superposition in the far field, showing the relative phase advance as kr · r′per current element.

What is the distance from the antenna for which this approximation (1.96) holds? We

look at the terms that has been neglected, and require for it the condition

kr

2

(r′

r

)2

(1− r · r′) ¿ π. (1.97)

The maximal value for the expression (1− r · r′) is 1. Taking this worst case, we require

kr

2

(r′

r

)2

¿ π (1.98)

33

or

r′2

λr¿ 1. (1.99)

Assume now that the sources are contained within a volume whose maximal linear di-

mension is L, i.e. r′ ≤ L ∀r′. Consider two cases:

1. L & λ/2. Then, (1.99) is satisfied for the upper bound on r′ if

r À L2

λ, L & λ/2. (1.100)

Practically, we use r > 2L2

λor so.

2. L . λ/2. Then, the upper bound on r′ is λ2, and (1.99) becomes

(λ/2)2

λr¿ 1 =⇒ r À λ

4, L . λ/2. (1.101)

A note on small antennas. Case 2 above shows that the far field boundary does

not depend on the size of the antenna once the antenna is smaller than ∼ λ/2. This

consclusion is typical of small antennas, where the dominant metric is λ rather than L.

The same applies to many other properties of small antenas, as we will see below.

Electric and magnetic fields in the far field zone The general formulas are given

in eqns. (1.52). The ∇ operator that appears there is approximated in the far field using

(1.95):

∇G(r, r′) ' ∇(e−kr

4πrek·r

′)

= ∇(e−kr

4πr

)ek·r

′+e−kr

4πr∇ek·r′

= −kr(e−kr

4πrek·r

′)− r

e−kr

4πr2ek·r

′+e−kr

4πr2

(θ∂

∂θ+ φ

1

sin θ

∂

∂φ

)ek·r

′. (1.102)

34

Only the first term drops off as 1r

as required by (1.91), therefore the ∇ operator is

approximated by

∇ ' −kr. (1.103)

Apply now (1.103) to (1.52) recalling that k = kr:

E(r) ' −ω (1− rr·)A(r) (1.104a)

H(r) ' η−1r×A(r) (1.104b)

What does eq. (1.104a) mean? Write it again as

E ' −ω (A(r)− r(r ·A)) .

Note that A(r)− r(r·A) = AT , where AT is the transversal component of A with respect

to r. We now have (1.104) in the final form

E(r) ' −ωAT (r) (1.105a)

H(r) ' η−1r× E(r) (1.105b)

Combining (1.105) with (1.96), we get the formulas for the radiation integral for the far

fields:

E(r) ' −ωµe−kr

4πr

V

JT (r′)ek·r′dV ′ (1.106a)

H(r) ' −ke−kr

4πrr×

V

J(r′)ek·r′dV ′ (1.106b)

Similar equations for magneric sources are readily obtained using the duality principle.

From here on, we replace the ' by =, keeping however in mind that we are still dealing

with the far field approximation.

35

Properties of the far field Inspection of equations (1.105) and (1.106) shows that

both E and H are transversal with respect to r. Also, they are perpendicular to each

other:

H = η−1r× E E = −ηr×H (1.107a)

E ·H = 0 r · E = 0, r ·H = 0 (1.107b)

There properties show that the far field behaves locally as a plane wave.

The Poyting vector, in the far field, is always real and is directed in the r–direction:

S(r) =1

2E×H∗ =

1

2ηE× r× E∗ =

1

2ηr (E · E∗)− 1

2ηE∗ (r · E)︸ ︷︷ ︸

=0

=1

2ηr (E · E∗)

= r1

2η

( ωµ4πr

)2

∣∣∣∣∣∣

V

JT (r′)ek·r′dV ′

∣∣∣∣∣∣

2

. (1.108)

The Poynting vector varies as 1r2

. This shows that energy is conserved as the distance

from the sources changes.

Example: second visit with the electric infinitesimal dipole. We can now com-

pute the far field of the infinitesimal electric dipole (1.85) from (1.106):

E(r) = −ωµI0l z|Te−kr

4πr

V

δ(r′)ek·r′dV ′ = −ωµI0l z|T

e−kr

4πrek·r

′∣∣∣r′=0

= θkηI0le−kr

4πrsin θ (1.109a)

H(r) = −kI0l e−kr

4πrr× z ek·r

′∣∣∣r′=0

= φkI0le−kr

4πrsin θ. (1.109b)

You can compare (1.109) with (1.89). Turns out that the 1r

terms there are indeed the

far field terms. The far field boundary for the infinitesimal dipole is r À λ/4, although

the length of the dipole is infinitesimal, as noted after Eq. (1.101).

Chapter 2

Fundamental parameters ofantennas in TX mode

Balanis: Chapter 2.

2.1 Radiation intensity

Assume that the current distribution J(r) is known and flows in free space. Then, the

far fields are expressed in (1.105). We have found the Poynting vector S(r) in (1.108).

In the far field, S(r) is r–directed, real and has a 1/r2 dependence:

S(r) = rS = r1

2ηE · E∗ = r

1

2η

( ωµ4πr

)2

∣∣∣∣∣∣

V

JT (r′)ek·r′dV ′

∣∣∣∣∣∣

2

(2.1)

where S = |S| is sometimes called the radiation density. Also, E ·E∗ = Eθ ·E∗θ +Eφ ·E∗

φ.

The radiation intensity U(θ, φ) is defined with the r–dependence removed. It thus

depends on (θ, φ) but also on the amplitude of the excitation:

U(θ, φ) = r2S =r2

2ηE · E∗ =

1

2η

(ωµ4π

)2

∣∣∣∣∣∣

V

JT (r′)ek·r′dV ′

∣∣∣∣∣∣

2

(2.2)

recalling that the (θ, φ)–dependence is obtained through k = kr = k(x sin θ cosφ +

y sin θ sinφ+ z cos θ).

36

37

The total power radiated by the antenna is

Prad =

π 2πθ=0 φ=0

S · r r2 sin θ dθ dφ =

π 2πθ=0 φ=0

U(θ, φ) sin θ dθ dφ =

π 2πθ=0 φ=0

U(θ, φ)dΩ (2.3)

where dΩ is a solid angle element in steradians.

Example: third visit with the electric infinitesimal dipole. In the second visit,

we found that the far field of the infinitesimal dipole is

E(r) = θkηI0le−kr

4πrsin θ

The radiation intensity is then

U(θ, φ)|inf dipole =r2

2ηE · E∗ =

r2

2ηEθE

∗θ =

(|I0|kl)2η

32π2sin2 θ =

(|I0|l)2η

8λ2sin2 θ. (2.4)

The total power radiated by the infinitesimal dipole is

Prad =

π, 2πθ=0, φ=0

U(θ, φ)|inf dipole sin θdθdφ =(I0l)

2η

8λ2

π, 2πθ=0, φ=0

sin2 θ sin θdθdφ =π

3

(l

λ

)2

|I0|2η.

(2.5)

Note that the radiated power is proportional to (l/λ)2, that can be very small. Therefore,

to achieve significant radiation, |I0| should be made very large.

2.2 Radiation patterns

2.2.1 Power patterns

The normalized value of the magnitude of U is the radiation power pattern:

Power RP =U(θ, φ)

U(θ, φ)|max

=

∣∣∣∣V

JT (r′)ek·r′dV ′

∣∣∣∣2

∣∣∣∣V

JT (r′)ek·r′dV ′∣∣∣∣2

max

(2.6)

One can draw a three dimensional surface plolt of the RP over the (θ, φ)–space. Two

dimensional cuts of the RP for fixed values of θ of φ are useful for more detailed inspection.

38

Example: fourth visit with the electric infinitesimal dipole. looking at equa-

tions (2.4) and (2.6), the radiation pattern of the infinitesimal dipole is

Power RP|inf dipole = sin2 θ.

2.2.2 Field patterns

The field pattern is defined as the normalized absolute value of the field (electric or

magnetic - same result) of a given polarization:

Field RP =|E(θ, φ)||E(θ, φ)|max

. (2.7)

Example: fifth visit with the electric infinitesimal dipole. The field pattern of

θ–polarized field is of the infinitesimal dipole is

Field RP|inf dipole = | sin θ|. (2.8)

If the chosen polarization is the actual polarization of the antenna, then (power pattern) =

(field pattern)2. It is then customary to measure the patterns in units of dB, defined as

pattern, dB = 10 log (power pattern) = 20 log (field pattern).

In many cases the pattern has several distinct main lobe, one of them being the “main

beam” (see Fig. 2.1). In these cases, we can define the following sub–parameters of the

radiation pattern:

2.2.3 Beamwidth

Beamwidth is usually defined for the main lobe. It can be defined in several ways, to suit

the system: First null beamwidth (FNBW), half–power (-3dB) beamwidth (HPBW), -

39

Figure 2.1: Balanis Fig. 2.4: (a) Radiation lobes and beamwidths of an antenna pattrern,shown in polar coordinates. (b) 2-D cut of the power pattern and its associated lobesand beamwidths, vs. Cartesian θ–coordinate.

2dB, -10db beamwidths and so on. If no specification is made, HPBW is usually assumed.

The standard IEEE definition for the HPBW is as follows:

Definition 2 (HPBW) In a plane containing the direction of the maximum of a beam,

the angle between the two directions in which the radiation intensity is one–half the max-

imum value of the beam.

40

Example: sixth visit with the infinetesimal dipole The HPBW for the infinetes-

imal dipole is calculated, say, from the power pattern sin2 θ. The −3dB points are at

θ1,2 = 450, 1350, therefore the beamwidth at any φ = const cut is

HPBW = θ2 − θ1 = 900 (2.9)

as seen in Fig, 2.2. For the cut θ = 900, the pattern is I(θ = 900, φ) = 1. Therefore, the

beamwidth is undefined, you can say that it is like 3600.

0 50 100 1500

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

θ (degrees)

Fiel

d pa

ttern

HPBW=900

Figure 2.2: Radiation pattern of the infinitesimal dipole Eq. (2.8) vs. Cartesian θ–coordinate (see Fig. 2.1(b)), showing the half power beamwidth HPBW= 900.

2.2.4 Sidelobe level - SLL

This is the level of the (first, RMS, highest ... you choose) sidelove relative to the main

lobe, usually measured in dB:

SLL = 10 log

(U(peak of sidelobe)

U(peak of main beam)

)(2.10)

41

where U is the intensity (2.2). Unfortunately, the infinitesimal dipole cannot provide an

example because it has no sidelobes to accompany its wide beam.

2.2.5 Beam efficiency

The beam efficiency is defind as

BE =

θ1 2πθ=0 φ=0

U(θ, φ)dΩ

π 2πθ=0 φ=0

U(θ, φ)dΩ

=power transmitted within cone angle

power transmitted by the antenna(2.11)

where the cone is defined by its half–angle θ1, assuming, for simplicity, a beam pointing

in the z–direction as in Fig. 2.1(a). Beam efficiency is a far field property, and is not a

measure of losses, in contrast to the “ordinary” efficiency η mentioned below.

2.3 Directivity

The directivity D has a standard IEEE definition:

Definition 3 Directivity is the ratio of the radiation intensity in agiven direction from

the antenna to the radiation intensity averaged over all directions. The average radiation

intensity is equal to the total power radiated by the antenna divided by 4π. If the direction

is not specified, the direction of maximum radiation intensity is implied.

In mathematical terms,

D(θ, φ) =U(θ, φ)

Uaverage

=U(θ, φ)Prad

4π

=U(θ, φ)

14π

π 2πθ=0 φ=0

U(θ, φ)dΩ

(2.12)

(see Eq. (2.3).) D is proportional to the power pattern, excpet it is normalized to the

average, not maximal power.

42

As defined in (2.12), if 10 log(·) is taken, then the units are called dBi. The reason:

a hypothetical isotropical antenna, whose pattern is constnant over all space, has the

directivity of 1, or 10 dBi. Thereofre, (2.12) can be seen as the “directivity over an

isotropical antnena”. Other normalizations are also possible.

Roughly speaking, the directivity is inversely proportional to the beamwidth. There

are some rules–of–thumb that relate the two. However:

A word of caution: The directivity is inversely proportional to the beamwidth only

in cases where the beam efficiency is high.

Example: seventh visit with the infinetesimal dipole The directivity can be

calculated via (check it out)

D(θ, φ) =sin2 θ

14π

π 2πθ=0 φ=0

sin2 θ sin θ dθ dφ

= 1.5 sin2 θ (2.13)

and the maximal value is

Dmax = D(θ = 900, φ) = 1.5 = 1.76 dBi. (2.14)

2.4 Radiation resistance and antenna impedance

The process of converting guided energy into radiation is viewed from the point of view

of the feed line as a loss mechanism, represented by the equivalent radiation resistnace:

Rrad =2Prad

|Iin|2 (2.15)

Note that we are using Prad. This is the power actually radiated into free space, as defined

in (2.3). it is not the input power Pin defined and measured at the port of the antenna.

43

Pin is the sum of the radiated power Prad and the loss power Ploss. dissipated in the loss

mechanism represented by the loss resistance Rloss:

Pin =1

2|Iin|2Rrad

︸ ︷︷ ︸Prad

+1

2|Iin|2Rloss

︸ ︷︷ ︸Ploss

. (2.16)

The total input resistance into the antenna is

RA = Rrad +RL (2.17)

In addition to the radiation and loss mechanisms, reactive energy is stored in the near

field region recall that the Poynting vector in the far field is purely real. This gives rise

to the antenna reactance XA. The total antenna impedance is then (see Fig. 0.1)

ZA = RA + XA. (2.18)

Example: eighth visit with the infinetesimal dipole We have seen that the total

radiated power for the infinetesimal dipole is (Eq. (2.5))

Prad =π

3

(l

λ

)2

|I0|2η.

The radiation resistance is, by (2.15)

Rrad, inf dipole =2π

3

(l

λ

)2

η Ω. (2.19)

Take, for example, l = 0.1λ. Then, Rrad, inf dipole ' 8 Ω. This is a very low value

compared, say with 50 Ω. Moreover, it varies rapidly with frequency, hence the difficulty

in matching of a small antenna over a finite bandwidth.

44

2.5 Efficiency and Gain

Based on the definitions of impedance, we define the radiation efficiency (also ohmic

efficiency) as

ηohmic =Prad

Pin

=Rrad

Rrad +RL

=Rrad

Re ZA . (2.20)

The radiation efficiency is a part of the overall efficiency. Another factor is the impedance

mismatch loss, that can be deduced from Fig. 0.1:

ηimpedance =4Re ZsRe ZA

|Zs + ZA|2 (2.21)

This factor becomes unity when ZL = Z?A. These two loss mechanisms, known so far, are

combined into the overall efficiency η = ηimpedanceηohmic.

The gain of the antenna is defined as

G(θ, φ) =U(θ, φ)

Pin

4π

. (2.22)

The difference with the directivity is in the term Pin that replaces Prad, where Pin = ηPrad.

Therefore,

G(θ, φ) = ηD(θ, φ)

Like the directivity, the gain is measured in units of dBi.

Note the difference in testing an antenna for gain and directivity: For the gain, we

need to measure Pin, which is easily done at the lab, while for the directivity we need to

integrate the intensity over the entire spherical space. This is why the efficiency of an

antenna is not always easy to calculate.

45

2.6 Effective radiated power (ERP)

This parameters is included in this chapter for completeness, although it is more of a

system, rather than strictly an antenna parameter.

Definition 4 The Effective radiated power is defined as

ERP = PTGmax (2.23)

where PT is the power fed into the antenna at the input port, and Gmax is the maximal

value of the gain over the angular space (θ, φ).

2.7 Polarization (TX mode)

The polarization of the radiated wave is defined in the far field per direction of prop-

agation k = kr. So far we have used the conventional spherical unit vectors θ and φ,

that indeed depend on k. In engineering practice, however, it more customary to refer

to “vertical” and “horizontal” polarization vectors v and h, respectively. If z points in

the vertical direction, then the transformation is quite simple, as can be seen in Fig. 2.3,

looking in a certain direction −k towards the antenna:

v = −θ, h = φ. (2.24)

Polarization is defined in terms of the electric field in the time domain. If the phasor

field is E = Ehh + Evv, and the complex amplitudes are related via

EvEh

= re−ψ. (2.25)

The temporal representation of E is

E(t) = <

(Ehh + Evv)eωt

= hEh(t) + vEv(t) (2.26)

46

Figure 2.3: Vertical and horizontal coordinates on a plane perpendicular to k. k pointsaway from the antenna to the viewer.

where the parametric relationship between Ev(t) and Eh(t) is

Eh(t) = eh cosωt (2.27a)

Ev(t) = ev cos (ωt− ψ). (2.27b)

where eh = |Eh|, ev = |Ev| = reh and we have taken ∠Eh = 0 as our reference.

With some trigonometric manipulations, we can extract the parameter t to obtain direct

relationship between Ev(t) and Eh(t) :

(Eh(t)

eh

)2

− 2Eh(t)Ev(t)

ehevcosψ +

(Ev(t)

ev

)2

= sin2 ψ (2.28)

47

which is an ellipse in the(Eh(t), Ev(t)

)–plane seen in Fig. 2.4. The ellipse has been fully

defined by the two parameters r and ψ. it can also be described by the angles describing

the tilt and (α) and eccentricity (β), to be mentioned in Section 2.7.5 below. In many

practical cases. α = 0, and the ellipse is defined solely by r or rather by the axial ratio

AR = 20 log |r|.

Figure 2.4: Polarization ellipse in the coordinates of Fig. 2.3.

Let’s consider a few special cases.

2.7.1 Right handed and left handed elliptical polarizations:

Let us distinguish between two cases for the value of ψ: (a) 0 ≤ ψ ≤ π, (b) π ≤ ψ ≤ 2π

The difference between these two cases is in the angular direction by which the tip of E(t)

traces the ellipse: it is clockwise (right–handed) and counter–clockwise (left–handed) for

48

cases (a) and (b), respectively, see Fig. 2.5. To see this, we compute the value of the

angle ϕ(t) between E(t) and the horizontal axis, as function of time:

tanϕ =ev cos (ωt− ψ)

eh cosωt= r

cos (ωt− ψ)

cosωt(2.29)

Then, the angular velocity is

Ω =∂ϕ

∂t= ω

r sinψ

cos2 ωt+ r2 cos2 (ωt− ψ)(2.30)

This angular velocity is positive if 0 ≤ ψ ≤ π and negative if π ≤ ψ ≤ 2π. The two case

are right and left handed elliptical polarizations, respectively. These are the “senses” of

the polarizations. Note that they are defined looking away from the antenna i.e., from

Fig. 2.5 in your direction. A complete revolution around the ellipse is done in T = 2πω

sec.

Figure 2.5: Right and left handed polarization ellipses. They are defined with respect tothe the direction of propagation, from the drawing to the viewer.

49

2.7.2 Linear polarizations:

If ψ = 0 or π, then the ellipse (2.28) becomes

Eh(t)

eh= ±Ev(t)

ev(2.31)

with the + and − signs pertain to ψ = 0 and ψ = π, respectively. In these cases Ω = 0

and the ellipse has degenerated into a line, that can be tilted in the first–third or the

second–fourth quadrants for the two cases, respectively. Specifically, if r = 0 or ∞, we

have linear horizontal or vertical polarization (Ev = 0 or Eh = 0), respectively.

2.7.3 Circular polarizations:

Right and left handed circular polarizations (RHCP and LHCP) are obtained when r = 1

and ψ = ±π2. In these cases, the ellipse becomes a circle with radius eh = ev and the tip

of the field traces this circle with a constant angular velocity Ω = ±ω (see (2.30)).

2.7.4 Polarization states in the complex plane

Let us look at the ratio reψ in its complex plane (the module is r and the argument is

ψ). Each point in this plane represents a polarization state, as shown in Fig. 2.6.

2.7.5 Stokes’ parameters and the Poincare sphere

We have seen that the polarization ellipse of Fig. 2.4 is defined by r = ev

ehand ψ. Alter-

natively, the ellipse can be defined by the angles α and β in the figure. The four Stokes’

50

Figure 2.6: Description of polarization states in the complex reψ–plane.

parameters are defined from both pairs of parametes as follows:

I =1

η(e2h + e2

v) (2.32a)

Q =1

η(e2h − e2v) = I cos 2α cos 2β (2.32b)

U =1

ηehev cosψ = I sin 2α cos 2β (2.32c)

V =1

ηehev sinψ = I sin 2β (2.32d)

with η =√

µε. Once can see immediately that

Q2 + U2 + V 2 = I2 (2.33)

51

such that (Q,U, V ) can serve as the Cartesian coordinates of the Poincare sphere, with

the angles (900 − 2β, 2α) playing the roles of the polar coordinates (θ, φ), see Fig. 2.7.

Each point on the sphere represents a polarization state that you can figure, e.g., by

translating its coordinates back to the (r, ψ) parameters with Eq. (2.32). The northern

and southern hemispheres are right– and left– handed polarization, respectively. In

particular, the north and south poles are β = +π4

and β = −π4, representing RHCP and

LHCP, respectively. The equator is β = 0 and contains the horizontal, vertical, +450

and −450 linear polarizations represented by α = 0, π2, π

4and − π

4, respectively.

Figure 2.7: First octant of the Poincare sphere, defined by Eqns (2.32)-(2.33).

52

2.7.6 polarization characteristics in receive (RX) mode:

Deferred to Section 3.4

Chapter 3

The antenna in RX mode

Balanis: Chapter 2.

3.1 The reciprocity theorem

As a receiving device, the antenna converts radiated electromagnetic waves back into

guided waves in the feed transmission line (recall Fig. 0.2). The properties of the receiving

antenna hang upon the following important theorem.

Figure 3.1: Definitions of sources within volume V enclosed by surface S for the reci-procity theorem.

53

54

Theorem 2 (The reciprocity theorem) Given a volume V enclosed by the closed

surface S (see Fig. 3.1). In this volume, two sets of sources are given: Set #1 is

J1(r), Jm1(r) that gives rise the fields E1(r), H1(r) and set#2, J2(r), Jm2(r) that gives

rise to E2(r), H2(r). Then the following applies:

−S

(E1 ×H2 − E2 ×H1) · dS =

V

(E1 · J2 − E2 · J1 −H1 · Jm2 + H2 · Jm1) dV .

(3.1)

Some notes on reciprocity theorem:

1. The set J1(r), Jm1(r) contributes its fields E1(r), H1(r) in the absence of J2(r), Jm2(r),

and vice versa. If the two sets are present together, then the total field is the su-

perposition E1(r) + E2(r), H1(r) + H2(r).

2. The medium within V can be an arbitrary composition of materials and boundary

conditions, not necessarily free space. However, the condition of reciprocal medium

has to be met. This condition will appear naturally in the course of the proof given

below.

Proof of Theorem 2. We start with the following expression, defined within V :

E1 ×H2 − E2 ×H1.

Taking its divergence and expanding it, we have

∇ · (E1 ×H2 − E2 ×H1) = E1 · ∇ ×H2 −H2 · ∇ × E1 − E2 · ∇ ×H1 + H1 · ∇ × E2

The curl terms can be replaced using Maxwell’s equations (1.18a) - (1.18b):

∇ · (E1 ×H2 − E2 ×H1)

= E1 · (ωεE2 + J2)−H2 · (−ωµH1 − Jm1)−E2 · (ωεE1 + J1)+H1 · (−ωµH2 − Jm2)

55

It seems now that some terms cancel out. But is E1 · εE2 = E2 · εE1? The answer is

yes, provided that the tensor ε is at least symmetric (can be diagonal). The same applies

for the terms H1 · µH2 and H2 · µH1. The condition that both ε and µ be at least

symmetric tensors is the condition for the medium to be reciprocal. Otherwise, the rest

of the development, and the reciprocity theorem, are not valid. Since we are dealing with

free space as the medium, it is inherently reciprocal. Therefore, the antennas themselves

need to be “reciprocal”.

Assuming now that the medium is reciprocal, we have

∇ · (E1 ×H2 − E2 ×H1) = E1 · J2 − E2 · J1 −H1 · Jm2 + H2 · Jm1

Now integrate the last result over the volume V . The divergence at the right hand side

becomes a surface integral using the divergence theorem, with dS pointing in the direction

of Fig. 3.1 We have thus obtained the reciprocity theorem of (3.1).¥

3.2 The open–circuit voltage at the antenna port

Going back to Fig. 0.2, suppose we remove the transmission line such that the input port

to the antenna is now open–circuited. When the antenna is subject to electromagnetic

radiation, an open circuit voltage V oc will develop over this port.

Armed with the reciprocity theorem, we can evaluate this voltage. Strangely enough,

the result will include the familiar current distribution J(r) in the TX mode.

We assume that the surface S has receded to infinity, and all sources are included

in V , such that the surface integral has vanished, and we also assume that the entire

problem can be described by electric current sources only. The reciprocity theorem is

56

then reduced to V

E1 · J2 dV =

V

E2 · J1 dV . (3.2)

Apply (3.2) twice, as shown in Fig. 3.2, using the actual model of the antenna and

the free space model from Fig. 0.2 in Figures 3.2(a) and 3.2(b), respectively. I1 is a

current source at the antenna terminal, used for transmission. In the equivalent case

(b), the J1 is the free space source representing the antenna while transmitting, such

that the fields E1,H1, transmitted by the antenna, are the same in both cases. Also, the

sources for the incident field, that the antennna receives, are J2 in both cases. For case

(a), the reciprocity theorem is applied not in free space. The antenna structure provides

boundary conditions of an arbitrary sort, and the sources are just J(a)1 = zI1lδ(r) '

zI1δ(x)δ(y), z ∈ [(1)−, (1)+] (see Section 4.1.1). Therefore, Eq. (3.2) becomes

V

E1 · J2 dV

=

∞x,y=−∞

z=(1)+z=(1)−

E2 · zI1δ(x)δ(y) dzdxdy = I1

(1)+(1)−

E2 · z dz = I1

(1)+(1)−

E2 · dl

= −I1V oc. (3.3)

Note that E2 is unknown. Now in Fig. 3.2(b), the medium is free space. This time the

sources J2(r) radiate the field Einc2 , which is the incident field that would have existed in

space if antenna #1 were not there. Eq. (3.2) is now

V

E1 · J2 dV =

∞x,y,z=−∞

Einc2 · J1dV . (3.4)

Equating the right–hand–sides of (3.3) and (3.4), we have formula for V oc:

V oc = − 1

I1

antenna 1

Einc2 · J1dV . (3.5)

The integration in (3.2) is done over the volume of antenna #1 where J1 is finite.

57

Figure 3.2: Two applications of the reciprocity theorem for the evaluation of V oc.(E1,H1) are the same in both cases, however they are generated by the actual antennastructure in (a) and by its free space source equivalent in (b) (see Fig. 0.2). (E2,H2) arenot the same.

3.3 The effective length of the antenna

As a special but important case, consider now plane wave incidence. The incoming plane

wave is

Einc2 = E0e

k·r

58

where the positive exponent is due that the incident field propagates in a direction op-

posite to k. Then, (3.2) becomes

V oc(k) = − 1

I1E0 ·

antenna 1

J1ek·rdV . (3.6)

Define now the effective length of the antenna:

Definition 5 The effective length of the antenna is defined by

`(k) =1

I1

antenna 1

J1T ek·rdV (3.7)

where the index T represents the transversal component of the integral, the only one that

interacgts with the transversal field E0. Eq. (3.6) is then written succinctly as

V oc(k) = −`(k) · E0(k). (3.8)

Note:

1. `(k) = `(θ, φ) depends on the direction from which the antenna receives the plane

wave.

2. Although being a parameter of the antenna in RX mode, `(k) is an integral of the

equivalent current in the TX mode. This is the outcome of reciprocity.

3. `(k) is proportional to the vector far field of the antenna in TX mode. Comparing

Equations (3.7) with (1.96) and (1.105a), the following relationship arises

E(r)|far field in TX = −ωµI1 e−kr

4πr`(k). (3.9)

4. E0(k) is the vector amplitude of a plane wave propagating in the −k–direction.

59

Example: ninth visit with the infinitesimal dipole Let’s substitute into (3.7) J1

of the infinitesimal dipole:

`(k) =1

I1I1lzT

antenna 1

δ(r)ek·rdV = −θl sin θ.

This effective length is very small, since l¿ λ, showing the inherently limited capability

of small antenna to convert radiated into guided energy.

Once you have computed V oc(k), you can use it in the receiving circuit behind the

antenna as a voltage source connected to the load, see Fig. 3.3

Figure 3.3: An equivalent circuit at the output of the antenna terminal in RX mode(compare with Fig. 0.1).

60

3.4 Polarization loss factor (PLF)

It is now evident from the scalar product in (3.8) and from (3.9) that the antenna should

be of the same polarization in TX mode as the polarization of the incident field. For

example, the infinitesimal dipole can receive signals that are linearly polarized. For best

performance, it should be aligned physically such that the vector θ in its coordinate

system is parallel to the polarization of the incident wave.

The power received at the antenna terminal is proportional to |V oc|2. The maximal

power will be received when the polarization of the antenna is matched to the incident

wave. Otherwise, the polarization mismatch us described by the polarization loss factor

according to the following definition.

Definition 6 The polarization loss factor (PLF) p is defined as

p =|` · E0|2|`|2|E0|2 = | · E0|2 (3.10)

Naturally, p ≤ 1. In Fig. 3.3, the power received at the load is

Pr =ReZL

2|ZL + ZA|2 |Voc|2 =

ReZL2|ZL + ZA|2 |` · E0|2 =

ReZL2|ZL + ZA|2p|`‖

2|E0|2 (3.11)

where the impedance mismatch and polarization mismatch losses are apparent.

Examples

1. The antenna has linear vertical polarization, the incident field is also linearly po-

larized, however at 450: ` ∼ v, E0 ∼ h + v. Then, p = 12

= −3dB.

2. The antenna is linearly polarized, the incident field is circularly polarized: ` ∼

v, E0 ∼ h± v. Then, p = 12

= −3dB.

61

3. The antenna is RHCP: ` ∼ h + v. What happens if also E0 ∼ h + v? Answer:

p = 0. Reason: the incident field propagates in the opposite direction to the TX

wave of the antenna, therefore its polarization as defined here is LHCP. Check also

that if E0 ∼ h− v (the incident field is RHCP) then p = 1.

3.5 Effective aperture

Assume that the incident field in the vicinity of the antenna has a Poynting vector whose

magnitude is Sinc = 12η

E0 · E∗0 = 1

2η|E0|2. When the antenna is subject to this field, the

received power at the load is Pr. Define the effective aperture as follows.

Definition 7 The effective aperture of the antenna is defined as

Ar =PrSinc

. (3.12)

This quantity of the antenna in RX mode can be linked to the gain parameter in the TX

mode.

Theorem 3 The antenna effective aperture in RX mode is proportional to the gain in

TX mode according to

Ar(θ, φ) =λ2

4πG(θ, φ). (3.13)

Proof we use (3.12) in conjunction of the definition of the gain defined in Equa-

tions (2.12) and (2.22). Combine (3.12) with (3.11):

Ar =PrSinc

=ηRe ZL|ZL + ZA|2p|h|

2. (3.14)

Now use the definition of directivity (2.2) with (3.9) and (2.15):

D(θ, φ) =U(θ, φ)Prad

4π

=

r2

2η|E|2Pin

4π

=

r2

2η| − ωµI1

e−kr

4πr`|2

Pin

4π

=ηπλ2 |`|2Rr

(3.15)

62

(recall ωµ = kη ), where I1 = Vs

Zs+ZA. Dividing (3.14) by (3.15), we have

Ar(θ, φ) =λ2

4π

Rr

Re ZA4Re ZLRe ZA

|ZL + ZA|2 p

︸ ︷︷ ︸total efficiency η

D(θ, φ) =λ2

4πG(θ, φ) (3.16)

with all loss mechanisms known so far are shown explicitly:

• Radiation (ohmic) loss: ηohmic = Rr

Re ZA (see Eq. (2.20));

• Impedance mismatch loss: ηimpedance = 4Re ZLRe ZA|ZL+ZA|2 = 1−|Γ|2, where Γ = ZL−ZA

ZL+ZA.

This factor becomes unity when ZL = Z?A (see Eq. (2.21));

• Polarization mismatch: p as defined in Section 3.4

or: η = ηimpedance ηohmic p. We thus have (3.13). ¥

Example: tenth visit with the infinitesimal dipole Using (3.13), the effective

aperture of the infinitesimal dipole, apart from losses, is

Ar(θ, φ) =λ2

4π1.5 sin2 θ = (0.345λ)2 sin2 θ.

Once again, we see that a property of a small antenna is related to the wavelength,

not the absolute size of the antenna. In this case, the effective aperture is finite even

though the antenna seems to be infinitely small. The electric size of an antenna is at

least(λ3

)× (λ3

).

3.5.1 Aperture efficiency

Definition 8 The aperture efficiency is defined as the ratio between the effective aperture

and the physical area of the antenna:

ηaperture =Ar

Aphysical

. (3.17)

63

Note that the aperture efficiency is due in part to the far field shape, not an actual loss

mechanism. In this sense it resembles the beam efficiency of Section 2.11. For large

antennas, ηaperture < 1. Small antennas can have ηaperture < 1, since the physical size does

not have much meaning, the metric being the wavelength, as we have seen above.

3.6 Frijs’s formula

Figure 3.4: TX - RX setup for the Frijs formula.

In Fig. 3.4, the magnitude of the Poynting vector at the receiving antenna is

ST (r, θT , φT )|at receiver =PTGT (θT , φT )

4πr2

where PT is the power at the port of theTX antenna and GT (θT , φT ) is the gain of the

TX antenna in the direction (θT , φT ) (see the figure).

The power received by the load of the RX antenna is

PR = ST (r, θT , φT )|at receiverAr(θR, φR) =

[PTGT (θT , φT )

4πr2

] [λ2

4πGR(θR, φR)

]

64

where Ar and GR(θR, φR) are the effective aperture and the gain in the direction (θR, φR)

of the RX antenna, respectively. This is the Frijs (communication) equation:

PR = PTGT (θT , φT )GR(θR, φR)(

4π rλ

)2 (3.18)

Some words of caution: As with many popular formulas, care should be taken not

to forget the assumptions that were made in the course of developing the formula:

1. The medium is free space. Ground, buildings and other objects are not present.

2. The antennas are at far field zone of each other.

3. Both TX and RX antennas are reciprocal.

4. All loss mechanisms have been incorporated into GT and GR. The polarization

mismatch factor p is applied once (either in GT or GR).

3.7 Noise received at the antenna terminals

If the brightness noise temperature (sky temperature) is TB, then the antenna tempera-

ture is

TA =

π,2πθ=0,φ=0

TB(θ, φ)G(θ, φ) sin θdθdφ

π,2πθ=0,φ=0

G(θ, φ) sin θdθdφ

. (3.19)

Then, the noise power received at the load is

Pnoise = kTABηohmic + kTpB(1− ηohmic) (3.20)

where k = 1.38×10−23 is Boltzmann’s constant and B is the bandwidth in Hz. The higher

the gain, the higher the signal–to–noise–ration (SNR) will be. The receiving antenna is

at the beginning of the RF chain, therefore a low antenna noise temperature is very

important when the signals are weak.

Chapter 4

Linear wire antennas

Balanis: Chapter 4.

4.1 Line sources

Line sources can be modeled via the following equivalent source in free space:

J(r′) = I(z′)δ(x′)δ(y′) (4.1)

with I(z′) given. Therefore, the dimension I(z′) is Ampere, and the radiation integral in

the far field becomes (see Eqns. (1.96) and (1.105a)) and Fig. 4.1:

Eθ(θ) = −ωµe−kr

4πrsin θ

L2

z′=−L2

I(z′)ekzz′dz′ (4.2)

with kz = k cos θ. Dropping the universal factor −ωµ e−kr

4πr, we will use the following

definitions:

Definition 9 The angular dependence of the line source is defined as

F (θ) = sin θ f(θ), f(θ) =

L2

z′=−L2

I(z′)ekzz′dz′. (4.3)

65

66

Note that F (θ) is a product of the slowly varying function sin θ, that is in fact the

contribution of an infinitesimal dipole, and the faster varying function f(θ) that describes

the superposition of such dipoles along the line source. Keep this note in mind for

Chapter ??.

Let’s focus now on f(θ) alone. You can notice that this in fact a Fourier transform of

the current from z′ to kz. This fact allows us to predict many important properties of

radiation patterns, sometimes without computations.

A word of caution: You may hear here and there that the radiation pattern is the

exact Fourier transform of the current. Not so! Remember the sin θ factor.

One apparent property is that the beamwidth (in units of kz, not θ) is inversely propor-

tional to L as long as the current distribution remains the same. This is approximately

true as long as the antenna is long enough, such that f(θ) dominates. If the antenna

is short, the beamwidth cannot become wider than the 900 of the infinitesimal dipole.

Once again, we see that the metric for small antennas is the wavelength, not the antenna

length.

Another word of caution: The beam will become narrower when you increase the

length of the antenna, but only if you the current distribution stays the same.

4.1.1 The infinitesimal electric dipole

This dipole was introduced in (1.85) as

J(r′) = I0lδ(r′)z. (4.4)

67

Let us assume now a slightly more realistic distribution in terms of a line source:

I(z′) =

I0, |z′| ≤ l/2, l ¿ λ

0, otherwise.(4.5)

Then, by (4.3),

f(θ) = I0

l2

z′=− l2

ekzz′dz′. (4.6)

Since l¿ λ, kzz′ ¿ π, approximate the exponent as ekzz′ ' 1.The integral (4.6) becomes

f(θ) = I0

l2

z′=− l2

dz′ = I0l (4.7)

and

F (θ) = I0l sin θ (4.8)

as expected.

4.1.2 Uniform distribution

Consider the distribution

I(z′) =

I0, |z′| ≤ L/2

0, otherwise.(4.9)

where L is not necessarily small. Then

f(θ) = I0

L2

z′=−L2

ekzz′dz′ = I0Lsin kzL

2kzL2

= I0Lsin kL cos θ

2kL cos θ

2

. (4.10)

and

F (θ) = I0Lsin kL cos θ

2kL cos θ

2︸ ︷︷ ︸sinc term

sin θ

︸ ︷︷ ︸inf. dipole term

. (4.11)

Let’s distinguish now between small and large antennas. If L¿ λ, we are back in the case

of section 4.1.1 and the infinitesimal dipole term dominates. This angular dependence

has no relation to the actual size of the antenna. For large antennas, the rapidly varying

sinc term is modulated by the slowly varying infinitesimal dipole term.

68

Beamwidth For large uniform distributions, whose beam is narrow compared with the

infinitesimal dipole beam of 900, we can assess the beamwidth as follows. Solve

sin kL cos θ2

kL cos θ2

=1√2⇒ kL cos θ

2= ±1.39. (4.12)

Then the half power angular points are,

cos θ±HP = ± 2

kL1.39 = ±0.443

λ

L⇒

θ+ = cos−1 0.443 λ

L

θ− = π − θ+(4.13)

Therefore,

HPBW = π − 2 cos−1 0.443λ

L= 2 sin−1 0.443

λ

L. (4.14)

If LÀ λ,

HPBW ' 0.886λ

L(rad) = 50.8

λ

L(degrees). (4.15)

It becomes clear that from a certain length L up, the beamwidth is inversely proportional

to L. A quantitative measure of L is given in Table 4.1.

Table 4.1: Beamwidths, in degrees, of uniform distributions for various values of L.n/a=not applicable.

Lλ

sin kL cos θ2

kL cos θ2

sin θsin kL cos θ

2kL cos θ

2

50.8 λL

L¿ λ 90 n/a n/a2 24.76 25.6 25.45 10.112 10.166 10.15310 5.071 5.080 5.075

Sidelobe level The level of the first sidelobe, adjacent to the main beam, is -13.3 dB

relative to the main lobe, for large Ls.

69

Directivity To compute the maximal directivity for large L, we use the definition in

(2.12):

D =1

14π

2π π0 0

∣∣∣ sin kL cos θ2

kL cos θ2

∣∣∣2

sin θdθdφ

. (4.16)

Change variables: u = kL cos θ2

. Then,

D = − 1

1kL

− kL2

kL2

∣∣ sinuu

∣∣2 du=

L

λ 12π

kL2

− kL2

∣∣ sinuu

∣∣2 du. (4.17)

Take kL→∞. Then, by Parseval’s theorem,

1

2π

∞−∞

∣∣∣∣sinu

u

∣∣∣∣2

du =1

2. (4.18)

Therefore,

D ' 2L

λ. (4.19)

Effective aperture Using (3.13), neglecting losses,

Ar(θ, φ) ' λ2

4π2L

λ=

λ

2πL. (4.20)

The antenna has an effective finite width of λ2π

, although it appears to be physically

infinitely thin.

4.1.3 Tapered distributions: The cosine distribution

Tapered distributions drop off (“taper off”) towards the edges of the line source. Com-

pared with a uniform distribution of the same size, we normally achieve lower sidelobes at

the expense of a wider beam and lower directivity. The cosine distribution is an example.

It is defined by

I(z′) =

I0 cos(πz

′L

), |z′| ≤ L2

0, otherwise(4.21)

70

The cosine distribution is considered a “sharp” taper, because it goes down to zero at

the edges. The far field is

f(θ) = I0

L2

z′=−L2

cos

(πz′

L

))ekz

′ cos θdz′ = I02L

π

cos(kL cos θ

2

)

1− (kL cos θ

π

)2 . (4.22)

Beamwidth: HPBW ' 68.2 λL.

Sidelobe level: SLL|first sidelobe = −23dB.

Directivity: D = 1.62Lλ. More generally, consider tapers of the type cosn

(πz′L

). Their

properties are summarized in Table 4.2.

Table 4.2: Main properties of cosn(πz′L

)tapers.

n Beamwidth(deg) SLL (dB) D/Duniform Type

0 50.8 λL

-13.3 1 Uniform

1 68.2 λL

-23.0 0.81 Cosine

2 83 λL

-31.7 0.667 Cosine squared

4.1.4 Tapered distributions: Triangular taper

Define the triangular taper as

I(z′) =

I0

(1− 2|z′|

L

), |z′| ≤ L

2

0, otherwise(4.23)

This is a self–covolution an L/2 long uniform distribution, therefore its normalized far

field is

f(θ) =

(sin

(kL cos θ

4

)kL cos θ

4

)2

(4.24)

Beamwidth: HPBW ' 73 λL.

71

Sidelobe level: SLL|first sidelobe = −26.6dB (explain).

Directivity: D = 1.5Lλ.

4.2 The equivalence theorem

Figure 4.1: Field equivalence theorem.

Let us return to Equations (1.77)-(1.78):

f(r) =

V

G(r, r′)S(r′)dV ′ +S

[G(r, r′)∇′f(r′)− f(r′)∇′G(r, r′)] · dS′. (4.25)

The solution (1.81) assumed that all sources are included in V . Conversely, let’s assume

now that all sources are outside S, as seen in Fig. 4.1. in that case, the following theorem

holds.

Theorem 4 Given a volume V enclosed by a closed surface S as in Fig. 4.1. In the

absence of sources within V , the electromagnetic field in V can be viewed as the outcome

of the following equivalent surface currents on S: Jeqs = n×H|S , Jeq

ms = − n× E|S.

72

Proof. Eq. (4.25) is now

f(r) =

S

[G(r, r′)∇′f(r′)− f(r′)∇′G(r, r′)] · dS′. (4.26)

If we take f(r) to be one of the components of A, and then combine back the three

components, then (4.27) becomes

A(r) =

S

[G(r, r′)∇′A(r′)−A(r′)∇′G(r, r′)] · dS′. (4.27)

The first term in the integrand, apart from the scalar multiplier G, is

∇′A(r′) · dS′ = dS′ × (∇′ ×A(r′)) +′ AdS′ · ∇′ (4.28)

Plugging (4.28) back into (4.27),

A(r) =

S

dS′ × (∇′ ×A(r′))G(r, r′) + AdS · ∇′G(r, r′)−A(r′)∇′G(r, r′) · dS︸ ︷︷ ︸=0

=

S

dS′ × (∇′ ×A(r′))G(r, r′) =

S

n× (∇′ ×A(r′))G(r, r′)dS ′

=

S

n× µH(r′)G(r, r′)dS ′ = µ

S

n×H(r′)︸ ︷︷ ︸Jeq

s

G(r, r′)dS ′. (4.29)

By duality, the magnetic vector potential is obtained from

F(r) = ε

S

−n× E(r′)︸ ︷︷ ︸Jeq

ms

G(r, r′)dS ′. (4.30)

This completes the proof to Theorem 4.¥

Using Theorem 4, Fig. 4.1 is now transformed to Fig. 4.2 for all points within V .

Outside V , the field in Fig. 4.2 becomes identically zero, since the fields just outside

S are zero, and the wave equation outside S gives a zero solution to a homogeneous

equation with zero boundary conditions. The equivalence theorem is restricted for usage

inside V .

73

Figure 4.2: Field equivalence theorem.

Example Suppose V : z > 0 and S : z = 0. Then, n = z. The field at z = 0+ is

recorded as

E = xE0 (4.31a)

H = yE0

η(4.31b)

Represent the actual sources of this field at z < 0 by the equivalent sources (1.32c)-(1.32d)

on S:

Js = −xE0

η(4.32a)

Jms = −yE0 (4.32b)

These sources will generate the solution (1.32a)-(1.32b):

E = xE0e−kzu(z) (4.33a)

H = yE0

ηe−kzu(z). (4.33b)

74

We have seen in the examples of Section 1.3 that equivalent sources can be chosen in

more than one way. More on that in Section 5.1 below.

4.3 Perfectly conducting antennas and the equiva-

lence theorem

An antenna made of a PEC, has a zero field inside its volume, and a surface current

Js is induced on its surface. On order to analyze these antennas by the tools provided

above, we need to perform the substtuition of the actual structure with an equivalent

source as in Fig. 0.2. The equivalence principle is applied by considering V as the

entire space excluding the antenna structure, and S as a closed surface just touching the

antenna from the outside, as in Fig. 4.3. Then, Jeqms = − n× E|S = 0 since, on the PEC

surface, n× E|S = 0. The field in V is thus determined solely by the electric current

Jeqs = n×H|S. This quantity is identical to the physical induced current Js.

A word of caution: The fact that the equivalent current in free space is identical to

the induced current on the structure is unique to the PEC.

4.4 Wire antennas

Wire antenna is shown in Fig. 4.4

The induced current distribution on PEC wire antennas is approximated as a standing

wave with zeros at the edges. By the reasoning in Section 4.3, we transform the wire

structure into the following line source distribution in free space:

I(z′) =

I0 sin

[k

(L2− |z′|)], |z′| ≤ L

2

0, otherwise(4.34)

75

Figure 4.3: A PEC antenna structure transformed into an equivalent current sourcedistibution in free space Js = n×H|S = Jind.

Figure 4.4: A wire antenna.

76

where I0 is the peak value of the current and Iin = I(z′ = 0) = I0 sin kL2

is the current at

the antenna input. Several cases are shown in Figs. 4.5–4.6.

(a) Small dipole antenna (b) Small dipole antennapattern

(c) λ2 dipole antenna (d) λ

2 dipole antenna pattern

Figure 4.5: Current distributions and patterns dipoles of infinitesimal and λ/2 lengths.

77

(a) λ dipole antenna. (b) λ dipole antenna pattern

(c) 3λ2 dipole antenna. (d) 3λ

2 dipole antenna pat-tern

Figure 4.6: Current distributions and patterns dipoles of λ and 3λ2

lengths.

The far field is

f(θ) =2I0k

cos(kL2

cos θ)− cos

(kL2

)

sin2 θ(4.35)

78

or

F (θ) =2I0k

cos(kL2

cos θ)− cos

(kL2

)

sin θ(4.36)

and the total far field is

E = −ηI0 e−kr

2πr

cos(kL2

cos θ)− cos

(kL2

)

sin θθ (4.37)

Special cases:

1. L¿ λ (Fig. 4.5(a)–4.5(b)):

F (θ) ' 2I0k

1− 12

(kL2

cos θ)2 − 1 + 1

2

(kL2

)2

sin θ∝ sin θ (4.38)

Bringing us back to the case of the infinitesimal dipole, except the current distri-

bution is now triangular.

2. L = λ2

(Fig. 4.5(c)–4.5(d). This is the most common case of wire antennas.

F (θ) ∝ cos(π2

cos θ)

sin θ. (4.39)

This distribution is in fact a cosine taper with L = λ2

(compare with (4.22) for

this case). The beamwidth is HPBW = 780 (compare with 900 for the infinitesimal

dipole).

3. L = λ (Fig. 4.6(a)–4.6(b)).

F (θ) ∝ cos (π cos θ) + 1

2 sin θ. (4.40)

Here, the beamwidth is 470.

4. L = 32λ (Fig. 4.6(c)–4.6(d)).

F (θ) ∝ 0.7148cos

(3π2

cos θ)

sin θ. (4.41)

The beamwidth is now split, as can be seen in the Figure.

79

The usefulness of dipoles l < λ is determined mainly by their impedance behavior. To

find the radiation resistance, evaluate first the radiated power as follows (see (4.37)):

Prad =1

2η

2π πφ=0 θ=0

|Eθ|2 r2 sin θdθdφ

=1

2η

2π πφ=0 θ=0

∣∣∣∣∣ηI02πr

cos(kL2

cos θ)− cos

(kL2

)

sin θ

∣∣∣∣∣

2

r2 sin θdθdφ

=η|I0|28π2

2πφ=0

dφ

πθ=0

∣∣∣∣∣cos

(kL2

cos θ)− cos

(kL2

)

sin θ

∣∣∣∣∣

2

sin θdθ

=η|I0|22π

π2

θ=0

∣∣cos(kL2

cos θ)− cos

(kL2

)∣∣2sin θ

dθ. (4.42)

Therefore, the radiation resistance is (see text after (4.34))

Rrad = 2Prad

|Iin|2 = 2

η2π|I0|2

π2

θ=0

|cos ( kL2

cos θ)−cos ( kL2 )|2

sin θdθ

|I0 sin kL2|2

=η

π sin2 kL2

π2

θ=0

∣∣cos(kL2

cos θ)− cos

(kL2

)∣∣2sin θ

dθ. (4.43)

The radiation resistance vs. the antenna length/frequecy is shown in Fig. 4.7(a). Note

that for integer muliples of λ, the antenna becomes an open circuit and does not radiate.

This can be understood by looking at Fig. 4.6(a) and seeing that we are trying to feed

the antenna at the point where I = 0. This won’t work. Good impedance matching can

be achieved at L ' λ2

or L ' 3λ2

. However, the directivity at L ' 3λ2

is too low, since

the power is split into two beams (see Fig. 4.6(d)). This leaves us with the almost only

choice of L ' λ2

(see Figs. 4.5(c)–4.5(d)). For this case,

Rrad|L=λ2

=η

π

π2

θ=0

cos2(π2

cos θ)

sin θdθ = 73 Ω. (4.44)

The imaginary part of the input impedance in the absence of losses is computed, say, by

the induced EMF method (see Section 4.16). Approximate data are shown in Fig. 4.7(b).

80

(a)

(b)

Figure 4.7: Radiation resistances (a) and input ractances (b) for wire antennas vs. L/λ.Radius of the wire: a ' 0.0005λ..

81

For the λ2

dipole (the “half wave dipole”), the final result is

ZA|L=λ2

= Rrad + XA = 73 + 43 Ω. (4.45)

The resonant frequency is defined where XA = 0. This happens at a length somewhat

smaller than λ2, as can be seen in Fig. 4.7(b). A very thin dipole will resonate at L =

0.48λ, where ZA = 70 + 0 Ω. If the radius is a = 0.00005λ, i.e., L2a

= 500, resonance is

achieved at L = 0.45λ. For L2a

= 10, resonance is at L = 0.41λ.

The 2:1 VSWR bandwidth is 8% for L2a

= 2500 and 16% for L2a

= 50. The thicker the

dipole, the wider the bandwidth and the lower the resonant frequency.

Directivity of the half–wave dipole:

D|L=λ2

= 1.64 = 2.15 dBi. (4.46)

Compare this number with the infinitesimal dipole for which D = 1.76 dBi. The big

difference between the two is the impedance value that allow for convenient matching of

the half–wave dipole, not the value of the directivity.

The folded dipole.

4.5 Wire antennas over perfectly conducting ground

planes

We consider a homogeneous half space bounded by a perfectly conducting plane, referred

to as a ground plane. Since any given wire antenna can be seen as a superposition

of infinitesimal dipoles, we look at the behavior of normal and tangential dipoles in the

presense of the ground plane. The case of the normal dipole is shown in Fig. . The ground

plane in this case can be replaced by an in–phase image of the ground plane, since the

82

tangnetial compoment of the E-field at the surface of the ground plane, superimposed

from the dipole and its image, is zero. Like wise, the tangential dipole has an out–of–

phase image, see Fig.. This is one way of seeing that a wire antenna tangential to the

ground plane and close to it produces zero field.

4.5.1 Half wave dipole over ground plane

A popular way of increasing the directivity of a half wave dipole is by placing it in parallel

to a ground plane, at about λ/4 away from it (see Fig. 4.8). Then, the field at the beam

peak is doubled compared with the isolated dipole. The total power at the antenna input

remains the same, therefore the directivity is now

D|λ2

diple over ground = 2.15 + 6 ' 8 dBi. (4.47)

The far field is the superposition of the fields of the dipole and its image. It is also

possible to increase the directivity further by adding “directors” in the direction of the

main beam, forming the so–called Yagi-Uda antenna (Fig. 4.9). Gain can become as high

as 11 dBi.

The impedance of the dipole backed by a ground plane or the Yagi-Uda antenna is

approximately that of the

4.5.2 Quarter–wave Monopole antennas

The monople antenna is shown in Fig. 4.10. The ground plane is an active part of the

antenna. The radiation pattern over the upper half space is the same as that of the

dipole antenna formed with the image, hence the total power radiated is half the power

of the corresponding dipole for the same input current Iin. This halves the value of the

83

Figure 4.8: Half wave dipole backed by a ground plane. The image (in dashed line) isout of phase.

Figure 4.9: Yagi antenna.

denominator in the directivity definition (2.12) therefore the directivity is

D|monopole = D|dipole + 3 dB = 5.15 dBi. (4.48)

For the same value of Iin, the voltage at the input is half that of the corresponding

84

Figure 4.10: Monopole antenna with its image (in dashed line). The image is in phasewith the monopole

dipole, or

Z|monopole =1

2Z|dipole ' 36 Ω. (4.49)

4.6 Feeding wire antennas

The monopole inoput being asymmetrical, is considered an “unbalanced” structure, that

can be fed by an unbalanced feed as in Fig. 4.11.

Figure 4.11: Monopole fed by a coaxial line.

Half wave dipoles, on the other hand, are “balanced” loads for an unbalanced feed

line such as a coax or microstrip (see Fig. 4.12). A direct connection between the two

can cause excitation of currents over the external skin of the coax and hence spurious

radiation.

To answer this problem, we design “BALance to UNbalance” transformers, or BALUNs.

85

Ideally, we would like to have I3 = 0 and I2 = I1. We add a “choke” to annul I3 = 0 top

Figure 4.12: Balance to unbalance feed.

create a “sleeve” balun as in Fig. 4.13. This balun further develops into the split coax

balun of Fig. 4.14.

Figure 4.13: Sleeve balun.

Figure 4.14: split-coax balun.

86

4.7 The folded dipole

The folded dipole is shown in Fig. 4.15. It is designed to provide wider bandwidth and

better matching to a 300 Ω line. The current distribution on it is approximated from a

shorted line that has been folded to produce this antenna. It is equivalent to a half–wave

dipole with a total current of I = 2Iin flowing in it. The input impedance is calculated

in comparison with the conventional half–wave dipole. Assuming the same input power

into both antenna,

Pin =1

2|Iin|2Zfolded dipole =

1

2|2Iin|2Zλ

2dipole = 4Zλ

2dipole ' 280 Ω (4.50)

The folded dipole thus enables straightforward matching to a balanced twin–line trans-

mission line with Zc = 300 Ω.

Figure 4.15: Folded dipole.

4.8 The Induced EMF method for the assessment of

the input impedance of a wire antenna

This is an approximate method for ontaining the complex impedance of a wire atnenna.

It is based both on reciporocity and equivalence. Consider Fig. 4.16, where we represent

the physical dipole with two equivalent sources: source #1 is the equivalent current Js

of Fig. 4.3, radiating in free space. Source #2 is a filamentary current I2(z) along the

87

z–axis, established in free space and radiating the same field as Js. Sources #1 and #2

give rise to the fields (E1, H1) and (E2, H2), resepectively. Both sources are contained

entirely within S, so the following form of the reciprocity theorem applies:

S

(E1 ×H2 − E2 ×H1

) · dS = 0 (4.51)

where the volume V is outside S. Assuming a slender dipole, Eq. (4.51) becomes

L2

−L2

2π0

(E1zH

2φ − E2

zH1φ

)adφdz = 0 (4.52)

Now, H1φ = Js, and

E1z =

−V δ(z), |z| ≤ L

2

0, otherwise.(4.53)

This balun further develops into the split coax balun of Fig. 4.14. Hence, since there is

no φ–dependence, Eq. (4.52) becomes

−2πaV H2φ(a, 0) =

L2

−L2

E2z (a, z)Js(z) 2πa dz. (4.54)

Using Ampere’s law,

V I2in = −

L2

−L2

E2z (a, z)I(z)dz. (4.55)

Set E2 = E1 (currents are the same):

V Iin = −L2

−L2

Ez(a, z)I(z)dz (4.56)

resulting in the induced EMF equation

ZA =V

Iin= − 1

I2in

L2

−L2

Ez(a, z)I(z)dz. (4.57)

The iduced EMF procedure is then:

88

Figure 4.16: Two equivalents of the dipole structure: (1) equivalent current Js, radiating(E1, H1) and (2) a slender wire with current I2, radiating (E2, H2).

1. Assess I(z).

2. Calculate Ez on the surface of the wire.

3. Compute ZA from (4.57).

4.8.1 Example - the infinitesimal dipole

Step 1:

Assume

J = Ilδ(r)z. (4.58)

89

Step 2:

A(r) = Azz =µIle−kr

4πrz (4.59)

E(r) = −ω(

1 +∇∇k2

)A(r)

= −ω(A +

∇k2

∂Az∂z

). (4.60)

Therefore,

Ez(r) = −ω(

1 +1

k2

∂2

∂z2

)Az(r) (4.61)

Take ρ = a (in cylindrical coordinates),

Ez(a, z) = −ω(

1 +1

k2

∂2

∂z2

)Az(a, z). (4.62)

Use

∂

∂z

e−kr

4πr=

(−k − 1

r

)e−kr

4πr

z

r(4.63)

and

∂2

∂z2

e−kr

4πr=

(−k − 1

r

)2e−kr

4πr

z

r+

1

r2

e−kr

4πr

z

r+

(−k − 1

r

)e−kr

4πr

r − z zr

r2

=

[(−k − 1

r

)2z

r+z

r3+

(−k − 1

r

)r2 − z2

r3

]e−kr

4πr. (4.64)

Substitute r = ρρ + zz, r =√ρ2 + z2, then, at ρ = a,

∂2

∂z2

e−kr

4πr=

[(−k − 1√

a2 + z2

)2z√

a2 + z2+

z

(a2 + z2)2/3

+

(−k − 1√

a2 + z2

)a2

(a2 + z2)2/3

]e−k

√a2+z2

4π√a2 + z2

. (4.65)

90

Step 3:

ZA = Rrad + XA = − 1

I2

l2

− l2

Ez(a, z)I(z)dz

=ωµIl

I2

l2

− l2

1 +

1

k2

[(−k − 1√

a2 + z2

)2z√

a2 + z2+

z

(a2 + z2)2/3

+

(−k − 1√

a2 + z2

)a2

(a2 + z2)2/3

]e−k

√a2+z2

4π√a2 + z2

Ilδ(z)dz

= ωµl2

1 +1

k2a2(−ka− 1)

e−ka

4πa= ωµl2

1− 1

(ka)2−

ka

e−ka

4πa. (4.66)

such that

Rrad = ωµl2

1

ka

cos ka

4πa+

(1− 1

(ka)2

)sin ka

4πa

' ωµl2

4πa

1− (ka)2

2

ka+

(1− 1

(ka)2

)(ka− 1

3!(ka)3

)

' ωµl2

4πa

−ka

2+ ka+

ka

3!

=

2π

3

(l

λ

)2

η Ω (4.67)

keeping terms of O(ka) and below. This result is independent of a and can be compared

with (2.19).

For the imaginary part,

XA =ωµl2

4πa

− sin ka

ka+

(1− 1

(ka)2

)cos ka

' ωµl2

4πa

−1 +

(1− 1

(ka)2

)(1− 1

2(ka)2

)' − ωµl2

4πa(ka)2= − 1

ωC(4.68)

where

C = ε16π3a3

l2(4.69)

showing that indeed the reactance behaves like a capacitor when the length of the dipole

91

approaces zero in terms of wavelengths. This reactance is strongly dependent on a; it

becomes larger as a becomes smaller.

Chapter 5

Aperture antennas

5.1 The equivalence theorem revisited

In Section 4.2, it was shown that a pair of equivalent sources Jeqs = n×H|S and Jeq

ms =

− n× E|S on a closed surface S can be used to represent the physical sources outside S

in order to find the electromagnetic field in the volume V enclosed by S. Outside this

volume, the field is identically zero. This choice is not unique, as we have already noticed

in the examples of Section 1.3. To see how electric and magnetic sources can be swapped,

we conduct the following “thought experiment”.

Assume, for simplicity, that the volume V is the half–space z > 0, and the surface S

is the plane z = 0 combined with the infinite hemisphere r →∞, z > 0, as can be seen

in Fig. 5.1. We also assume that the infinite hemisphere contributes nothing to the field.

in this case, n|S = n|z=0 = z.

Since the field at z < 0 is identically zero, we can place arbitraty objects in that region

without affecting the soultion in z > 0. Assume that fill the region z < 0 with a PEC

or PMC, as seen in Figures 5.2(a) and 5.2(b), respectively. In the case of the PEC, the

electric and magnetic sources are imaged out of phase and in phase, respectively, and the

92

93

Figure 5.1: The equivalent sources Js, Jms reconstruct the electromagnetic field withinV , that is caused by the physical sources outside V . The equivalent source produce zerofield outside V .

balance is 2Jeqms, with a dual result for the case of a PMC. It turns out that we only need

either one of the two sources. The result for all cases is the same within z > 0, however

in the cases of Fig. 5.2, the field is no longer zero in z < 0.

We thus have four options for computing the field in z > 0, resulting in the potentials

in the far field as follows:

94

Option 1: use both Jeqs and Jeq

ms:

A(r) =µe−kr

4πr

S

Js(x′, y′)ek·r

′(5.1a)

F(r) =εe−kr

4πr

S

Jms(x′, y′)ek·r

′(5.1b)

Option 2: use 2Jeqms only:

F(r) = εe−kr

4πr

S

2Jms(x′, y′)ek·r

′, A(r) = 0. (5.2)

Option 3: use 2Jeqs only:

A(r) = µe−kr

4πr

S

2Js(x′, y′)ek·r

′, F(r) = 0. (5.3)

A word of caution: This transfromation is valid only if S is an infinite planar surface.

(a) (b)

Figure 5.2: The “thought experiment” with a PEC (a) or PMC (b) filling the z < 0half–space of Fig. 5.1. The equivalent sources Jeq

s ,Jeqms are transformed into 2Jeq

ms and2Jeq

s , respectively, for computing the field in z > 0.

95

Example: a plane wave in z > 0 The plane wave is of the form

E =xE0e−kz (5.4a)

H =yE0

ηe−kz (5.4b)

The equivalent sources for the three options are:

Option 1: use both

Jeqs = z× y

E0

ηe−kz

∣∣∣∣z=0

= −xE0

ηand (5.5a)

Jeqms =− z× xE0e

−kz∣∣z=0

= −yE0. (5.5b)

Option 2: use only

2Jeqms = −2yE0. (5.6)

Option 3: use only

2Jeqs = −2x

E0

η. (5.7)

Compare these options to the examples in Section 1.3.¥

The far field is derived from the potential via Eq. (1.105) and its dual for F:

Eθ =− ωAθ − ωηFφ (5.8a)

Eφ =− ωAφ − ωηFθ (5.8b)

The computaion then requires the following Fourier transforms as building blocks:

fxy(kx, ky) =

S

Ea xy(x′, y′)e(kxx′+kyy′)dx′dy′ (5.9a)

gxy(kx, ky) =

S

Ha xy(x′, y′)e(kxx′+kyy′)dx′dy′ (5.9b)

96

where Ea x′y′

= Ex′y′

∣∣∣∣S

and Ha x′y′

= Hx′y′

∣∣∣∣S

. Combining (5.9) with (5.8) and the definitions

Jeqs (x′, y′) = n×H|a (x′, y′) and Jeq

ms(x′, y′) = − n× E|a (x′, y′) we have the following

working formulas:

Option 1:

Eθ =ke−kr

4πr[fx cosφ+ fy sinφ+ η cos θ (gy cosφ− gx sinφ)] (5.10a)

Eφ =ke−kr

4πr[cos θ (fy cosφ− fx sinφ)− η (gy sinφ+ gx cosφ)] (5.10b)

Option 2:

Eθ =kηe−kr

2πrcos θ (gy cosφ− gx sinφ) (5.11a)

Eφ =− kηe−kr

2πr(gy sinφ+ gx cosφ) (5.11b)

Option 3:

Eθ =ke−kr

2πr(fx cosφ+ fy sinφ) (5.12a)

Eφ =ke−kr

2πrcos θ (fy cosφ− fx sinφ) (5.12b)

5.2 Examples

5.2.1 Uniform rectangular aperture

Given a uniform distribution the aperture, and zero outside it:

Ea(x′, y′) =

E0x, |x′| ≤ a

2, |y′| ≤ b

2

0, otherwise(5.13)

We use Option 3:

fx(kx, ky) =E0absin

(kxa2

)kxa2

sin(kyb

2

)

kyb

2

(5.14a)

fy(kx, ky) =0. (5.14b)

97

Figure 5.3: A uniform aperture radiation into the half space z > 0. The aperturedistribution is given in (5.13).

An expression for the over the entire θ, φ range can be obtained by using (5.14) in (5.12)

and recalling kxy

= k sin θ

(cosφsinφ

). It is instructive, though, to take a look at specific

cuts, i.e.,

E–plane (also: φ = 0 ∪ φ = π, also x− z–plane):

Eθ|φ=0 =− E0absin

(ka sin θ

2

)ka sin θ

2

(5.15a)

Eθ|φ=π =E0absin

(ka sin θ

2

)ka sin θ

2

(5.15b)

H–plane (also: φ = π2∪ φ = −π

2, also y − z–plane):

Eφ|φ=π2

=E0ab cos θsin

(ka sin θ

2

)ka sin θ

2

(5.16a)

Eφ|φ=−π2

=− E0ab cos θsin

(ka sin θ

2

)ka sin θ

2

(5.16b)

98

The half power beamwidths in the two planes are approximated by

HPBW|E-plane ' 50.8λ

a(5.17a)

HPBW|H-plane ' 50.8λ

b(5.17b)

Effective aperure (not proven yet) is Ar = ab = physical area. This is the highest

practical effective area that can be obtained from a given physical area.

The directivity is thus

D ' 4π

λ2ab. (5.18)

5.2.2 Other separable distributions

The uniform distribution of Section 5.2.1 is an example of a separable rectangular distri-

bution, that looks generally like this:

Ea xy(x′, y′) = E1

xy(x′)E2

xy(y′) (5.19)

such that

fxy(kx, ky) =

S

Ea xy(x′, y′)e(kxx′+kyy′)dx′dy′

=

a2

−a2

E1xy(x′)ekxx′dx′

︸ ︷︷ ︸f1;xy

(kx)

b2

− b2

E2xy(y′)ekyy′dy′

︸ ︷︷ ︸f2;xy

(ky)

(5.20)

Example: an open–ended waveguide Assume the following aperture distribution

similar to the TE10–mode:

Ea(x′, y′) =

yE0 cos πx

′a

; |x′| ≤ a2, |y′| ≤ b

2

o, otherwise(5.21)

99

such that

fy = E0abπ2

cos(ka2

sin θ cosφ)

(π2

)2 − (ka2

sin θ cosφ)2

sin(kb2

sin θ sinφ)

kb2

sin θ sinφ(5.22)

The beamwidths are

E-pane: HPBW = 51λb

H-plane: HPBW = 68λa

5.2.3 Non–separable distribution: a uniform circular aperturewith radial variation

Consider the following aperture distribution, defined in polar coordinates (r, φ) in the

z = 0–plane:

Ea(r′, φ′) =

xE0r

′; |r′| ≤ a

o, otherwise(5.23)

leading to

fx =

a, 2πr′=0,φ′=0

E0(r′)ek·r

′d2r′ =

a, 2πr′=0,φ′=0

E0(r′)ekr

′ sin θ cos(φ−φ′)r′dr′dφ′

=

a0

r′dr′E0(r′)

2π0

dφ′ekr′ sin θ cos(φ−φ′) = 2π

a0

r′dr′E0(r′)J0(kr

′ sin θ) (5.24)

having taken φ = 0 and where J0(kr′ sin θ) is the zeroth–order Bessel function. For a

uniformly illuminated apertre, take E0 = 1:

fx = 2π

a0

r′dr′J0(kr′ sin θ). (5.25)

Using the identity xJ0(x)dx = xJ1(x), (5.26)

we have the normalized value

fx ∼ 2J1(ka sin θ)

ka sin θ(5.27)

100

and

E ∼ (θ cosφ− φ sinφ cos θ)ke−kr

2πrfx(θ, φ). (5.28)

For this pattern, the half power beamwidth is

HPBW = 58.44λ

2a

and the side lobe level is

SLL = −17.6 dBi.

The effective are is the same as the physical area and the directivity is, therefore,

D =4π

λ2πa2.

5.3 Approximate directivity and effective area cal-

culations

Assume that the aperture distribution varies slowly across the aperture in the form of a

TEM wave, i.e., the magnetic and electric fields are related via

Ha(x′, y′) ' 1

ηz× Ea(x

′, y′) (5.29)

The directivity is then

Dmax =|S|Prad

4ır2

'12η

(|Eθ|2 + |Eφ|2)Prad

4ır2

. (5.30)

We further approximate Prad as the power that passes through the aperture, rather than

the power radiated into the far field:

Prad ' <

aperture

S · z dx′dy′ ' 1

2η

aperture

E · E? dx′dy′ (5.31)

101

The numerator of (5.30) is approximated as follows:

|Eθ|2 =

(k

2πr

)2

|fx cosφ+ fy sinφ|2 (5.32a)

|Eφ|2 =

(k

2πr

)2

|fy cosφ− fx sinφ|2 (5.32b)

i.e.,

|Eθ|2 + |Eφ|2 =

(k

2πr

)2 (|fx|2 + |fy|2). (5.33)

Also, assuming a main beam direction at θ = 0,

fxy(θ = 0) =

aperture

Exyda (5.34)

Therefore

Dmax ' 4π

λ2

∣∣∣∣∣

aperture

Eadx′dy′

∣∣∣∣∣

2

aperture

|Ea|2 dx′dy′=

4π

λ2Ar. (5.35)

Note that for a uniform distribution, Ar = Aphysical. Therefore, it is said that practically

the uniform distribution provides the highest aperture efficiency.

Example: the open–ended waveguide Apply (5.35) to the aperture distribution of

(5.21):

Ar '

∣∣∣∣∣

aperture

Eadx′dy′

∣∣∣∣∣

2

aperture

|Ea|2 dx′dy′=E2

0

(2aπ

)2b2

E20a2b

=8

π2ab (5.36)

and the aperture efficiency is ηaperture ' 8π2 = 0.81, and the directivity is

Dmax ' 32

π

ab

λ2. (5.37)

102

5.4 Horn antennas

Figure 5.4: Common types of horn antennas: (a) E–plane sectoral horn, (b) H–planesectoral horn, (c) pyramidal horn, (d) conical horn.

Figure 5.5: E- and H-flare (pyramidal) horn antenna.

Horm antenna are useful for gains up to about 24 dBi, beyond which their length may

become an issue. Some typical horn antennas are shown in Fig. 5.4. They are basically

extensions of waveguides with dimensions a × b that are flared into dimensions a1 × b1

in order to increase directivity and effective aperture (see Fig. 5.5). We first address H–

and E– plane sectoral horns.

103

5.4.1 H–plane sectoral horns

The three dimensinal rendering and a top view of an H–plane sectoral horn is shown in

Fig. 5.6. The aperture is of size a1 × b. The following holds too:

ρ2h = ρ2

2 +(a1

2

)2

; ψh = tan−1 a1

2ρh(5.38)

The field in the feeding waveguide is considered the TE10 mode:

Ey = E0 cosπx

ae−βz, Hx = − Ey

ZTE(5.39)

where β =√k2 − (

πa

)2, ZTE = ηq

1−( λ2a)

2 .

Assuming an essentially free space prapagation within the horn region, we can add

the phase delay due to the distance δ(x′) (see Fig.5.6) at any point x′ at the aperture

and approximate is as quadratic phase delay:

δ(x′) =√ρ2

2 − x′2 − ρ2 ' ρ2

(1 +

1

2

(x′

ρ2

)2)− ρ2 =

x′2

2ρ2

. (5.40)

We thus assume the following separable aperture distribution:

Eay =

E0 cos πx

′a1e−k x′2

2ρ2 , |x′| ≤ a12, |y′| ≤ b

2

0 otherwise(5.41)

from which we can compute

fy = E0

a12

−a12

cosπx′

a1

e−k x′2

2ρ2 ekxx′dx′

b2

− b2

ekyy′dy′ (5.42)

Universal radiation patterns in the H– and E–planes are shown in Fig. 5.7. Note the

effect of the quadratic phase “error” on the pattern: Nulls are filled and the directivity

decreases accordingly. Define the maximum phase shift acroos the aperture in terms

of wavelengths, t = δmax12πk(a1

2 )2

2ρ2=

a218λρ2

, as a parameter. As ψh is increased with

104

ρ2 = const, the aperture becomes larger and the directivity increases accordingly, but

so does the quadratic phase error therefore the growth of the directivity is reversed at a

certain lare angle, where the directivity attains a maximal value. This point is topt = 38, or

a1opt =√

3λρ2 (see Fig. 5.8). This is considered the optimal design from the standpoint

of directivity only. The beamwidth also drops down to certain point, from which it

increases again, see Fig. 5.9. At the optimal point, HPBWopt ' 78 λa1

. The ideal SLL at

t = 0is −23 dB as fit for a cosine distribution. The aperture efficiency is of the order of

ηaperture ' 0.65.

Figure 5.6: H–plane sectoral horn.

105

Figure 5.7: H–plane sectoral horn universal pattern in the H–plane. A = a1.

Figure 5.8: Normalized directivity of H–plane sectoral horn as a function of aperture sizeand for different lengths.

106

Figure 5.9: HPBW of H–plane sectoral horn as a function of flare angle for differentlength.

5.4.2 E–plane sectoral horns

In a develpment analogous to the H–plane horn, the aperture distrbution for the E–plane

sectoral horn seen in Fig. 5.10 again separable and is equal to

Eay =

E0 cos πx

′a1e−k y′2

2ρ1 , |x′| ≤ a2, |y′| ≤ b1

2

0 otherwise(5.43)

In a similar way, we define the maximal phase shift across the aperture s =b21

8λρ2.

This time, the optimal size corresponds to sopt = 14, and b1opt =

√2λρ1. The universal

107

patterns, directivity and beamwidths are shown in Figs. 5.11, 5.12 and 5.13, respectively.

The E—plane beamwidth at the optimum is HPBWopt ' 54 λb1

. The aperture efficiency

is of the order of ηaperture ' 0.61.

Riddle: Why is sopt < topt?

Figure 5.10: E–plane sectoral horn.

108

Figure 5.11: E–plane sectoral horn universal pattern in the E–plane. B = b1.

Figure 5.12: Normalized directivity of E–plane sectoral horn as a function of aperturesize and for different lengths.

109

Figure 5.13: HPBW of E–plane sectoral horn as a function of flare angle for differentlength.

5.4.3 Pyramidal horn

We design a pyramidal horn (Fig. 5.5) approximately as a combination of the two sectoral

horns, with the following aperture distribution:

Eay =

E0 cos πx

′a1e− k

2

„x′2ρ2

+ y′2ρ1

«

, |x′| ≤ a12, |y′| ≤ b1

2

0 otherwise(5.44)

The E– and H–plane patterns are designed as if they corresponded to E and H sectoral

horns, respectively, each with its quadratic phase error s and t, that can be made optimal

separately. The overall aperture efficiency at optimum is roughly ηaperture ' 0.5. Hence

the optimal gain is

G =1

2

4π

λ2a1b1. (5.45)

110

Example Design a pyramidal horn, fed by a WR90 waveguide (a=0.9“=2.286cm,

b=0.4”=1.016cm), for G = 22.1 dBi at f = 9.3 GHz (λ = 3.226 cm).

Solution: We have aλ

= 0.7087, bλ

= 0.315, G = 102.21 = 162.18. The aperture area

should be

a1b1 = 24π

λ2162.18 = 268.62 cm2.

Assume ρ1 ' ρ2 (although not quite compatible mechanically). We have, for optimum

directivity,

a1 =√

3λρ1,2

b1 =√

2λρ1,2

or

a21

3=b212⇒

√2

3a2

1 = 268.62 (5.47)

and therefore

a1 =18.14 cm

b1 =14.8 cm

and ρ1,2 = 34 cm.

5.5 Parabolic reflectors

Reflector antennas fit within the gain regime of ∼ 22 dBi and up. They can produce

pencil beams fow long range communications, radar and radio astronomy, as well as

shaped beams such as cosec2. A reflector antenna comprises a large parabolic surface

(“dish”) and a much smaller feed antenna, that illuminates the dish. The main advantage:

large apertures with a short depth.

111

The aperture distributin is defined as the plane z = 0 in Fig. 5.14. It has a tapered

amplitude with a constant phase. To see how a constant phase is generated, we assume

free space proagation withing between the feed and the dish, i.e., the dish is considered

to reside in the far zone of the feed. We compute the optical lengths of rays emitted from

the feed and reflected of the dish, as follows.

Figure 5.14: Parabolic dish geometry.

112

5.5.1 Constant phase

We generate the equation for the dish surface by applying the constant phase condition

FP + PA = const:

z′ + ρ′ = z′ +√z′2 + r′2 = 2f (5.49)

from which we have the parabola equation

r′2 = 4f(f − z′), r′ ≤ a. (5.50)

This equation can also be expressed in spherical coordinates as

ρ′(1 + cos θ′) = 2f ⇒ ρ′ =2f

1 + cos θ′= fsec2

(θ′

2

)(5.51)

or, using a mixed coordinate system (r′, θ′),

r′ = ρ sin θ′ =2f

1 + cos θ′sin θ′ = 2f tan

θ′

2. (5.52)

Show also that the rays reflected off the dish are parallel to the z–axis. The normal to

the dish surface is

n =∇ (

f − ρ′ cos2(θ′2

))∣∣∇ (

f − ρ′ cos2(θ′2

))∣∣ = − cos2

(θ′

2

)ρ′ +

1

ρ′2ρ′

2cos

(θ′

2

)sin

(θ′

2

)θ′

= − cos

(θ′

2

)ρ′ + sin

(θ′

2

)θ′ (5.53)

therefore the condition

n · Si = n · ρ′ = − cos

(θ′

2

)= −n · Sr (5.54)

is satisfied by Sr = −z because

n · (−z) = − cos

(θ′

2

)ρ′ · (−z)︸ ︷︷ ︸

cos θ′

+ sin

(θ′

2

)θ′ · (−z)︸ ︷︷ ︸− sin θ′

= cos

(θ′

2

). (5.55)

113

5.5.2 Tapered amplitude

The amplitude along the line FP drops off like 1/ρ′, as the field in this region is the far

field of the feed that has a spherical phase front. (see Fig. 5.14). Over the remainder of

the optical path, PA, the phase front is planar is there is no drop attenuation. Therefoe,

the amplitude over the aperture is proporional to

|Ea|2 ∝ f

ρ′∝ cos2

(θ′

2

)=

1

1 + tan2(θ′2

) =1

1 +(r′2f

)2 . (5.56)

Eq. (5.56) is the radial dependence for an isortopic feed. If the feed is not isotropic, this

distribution is modulated in the feed pattern,:

|Ea|2 ∝ Gf (θ′, φ′)(

1 +(r′2f

)2)2 (5.57)

The feed pattern Gf (θ′, φ′) is usually approximated by cosq θ′ within the main beam, in

the dB scale.

The total taper (5.57) is charaterized by the extremal value, the Edge Taper (ET).

this parameter has direct bearing on the beamwidth, directivity, sidelobe level, aperture

efficiency and also “spillover efficiency”, that describes the losses due the power that is

not intercepted by the reflector. Effects of the ET on the patterns is seen in Fig. 5.15(a).

The SLL vs. ET is shown in Fig. 5.15(b). Deep ETs, of −20 dB and lower, are affected

more by the feed patterns than the geometrical attneuation.

5.5.3 Design procedure

For a given reflector of size D = 2a, the design procedure begins with the choice of the

f/D parameter, normally in the range of 0.25 to 0.5, and the specification of sidelobe

114

(a) (b)

Figure 5.15: Reflector far field patterns (a) and sidelobe levels (b) for different edgetapers..

level. The SLL determines the required ET, and this leads to the determination of the

size of the feed. Then, additional parameters such as blockage and spillover are evaluated,

and an iterative procedure ensues.

Design example (ex. 5.8) Suppose we set f/D = 0.45 and SLL= −27 dB. Then, from

Fig. 5.15(b), ET= −12 dB. The subtended angle (Fig. 5.16- 1 ) is 1200. Using Eq. (5.56)

or Fig. 5.16- 2 , the geometric ET is −2.5 dB. This leaves −12 + 2.5 = −9.5 dB for

the feed taper Gf (θ′, φ′). A universal pattern for a horn feed, based on Figures 5.7 and

5.11, is given in Fig. 5.16- 3 along with the cosq θ approximation. We find that for the

−9.5 dB drop, the angle is θ/θ10 dB = 0.95, hence θ10 dB = 600/0.95 = 630. From here,

we go to Figures 5.7 and 5.11 looking at the = 10 dB points for, say, the optimal t and

s. The results is, for the E-plane and H-plane flares, respectively, b1λ

sin 630 = 0.85 and

a1

λsin 630 = 1.3. We thus have for the feed horn a1 = 1.45λ and b1 = 0.953λ. The horn

length is then determined similarly to the example in Section 5.4.3.

115

Figure 5.16: Design procedure for a parabolic reflector feed.

116

5.6 Microstrip antennas

5.6.1 Slot antennas

Figure 5.17: Design procedure for a parabolic reflector feed.

Consider a rectangular resionant slot cut in in a PEC ground plane, as in Fig. 5.17.

The length of the slot is of L = λ/2 and its width is a¿ λ. The PEC boundary condition

preclude a uniform aperture distribution of the type described in Sec. 5.2.1. A reasonable

assumption for the aperture distribution would be of the type

Ea(x′, y′) =

xE0 cos ky′; |x′| ≤ a

2, |y′| ≤ λ

4

o, otherwise.(5.58)

This is the magnetic equivalent of the half wavelength dipole. The Fourier transform is

fx = E0aλ

π

cos(π2

sin θ sinφ)

1− (sin θ sinφ)2

sin(ka2

sin θ cosφ)

ka2

sin θ cosφ(5.59)

117

resulting in the far field (see (5.12))

E = ke−kr

2πrfx

(cosφ θ − cos θ sinφ φ

)(5.60)

or, using the condition a¿ λ,

E ∝cos

(π2

sin θ sinφ)

1− (sin θ sinφ)2

(cosφ θ − cos θ sinφ φ

). (5.61)

The patterns at the E– and H–planes, respectively, are slot-5

EE–plane ∝ θ (5.62a)

EH–plane ∝cos

(π2

sin θ)

cos θφ. (5.62b)

Note the following differences with respect to the far field of a half–wavelength dipole:

1. The polarization is perpendicular to the dipole

2. The sin θ term (that would new become cos θ) is not present in the E–plane cut.

3. A cosφ terms has appeared in the horizontal cut.

4. The slot radiates only into the half space z > 0.

The pattern of the slot and the half wavelength dipole are the same, except that the slot

only radiates into the half plane z > 0. Therefore, the directivity is roughly twice that

of the dipole:

Dmax = 2.15 + 3 = 5.15 dBi (5.63)

For the impedance, the following relationship applies to complementary screens:

Z1Z2 =η2

4. (5.64)

118

In our case, Z2 ' 70Ω, thus the slot impedance is

Zslot =1

4

3772

70' 500 Ω. (5.65)

This high impedance can become beneficial when a number of slots in ana array are to

be connected in parallel using a power divider.

5.6.2 Microstrip patch antennas

Figure 5.18: A microstip patch antenna with the E-field lines shown.

A typical microstrip patch antenna, pictured in Fig. 5.18, is printed on a grounded

substrate of thickness t. It operates in the resonance range, with the dimensions as shown.

Since the substrate environment is not free space, the currents distribution over the patch

should be integrated with the Green’s function of the substrate. This formulation leads to

integral equations that are outside the scope of this course. Alternateively, we can view

the patch from above as an aperture antenna made of an array of two slots, connected in

cascade by a transmission line of length λg

2, where λg is the wavelength in the micorstrip

structure. In view of the field line shown in Fig. 5.18, the aperture would look like

Fig. 5.19(a). The equivalent circuit, shown in Fig. 5.19(b), leads to an imput impedance

of the patch being equal to

Zpatch ≈ 250 Ω. (5.66)

119

Actual values are shown in Fig. 5.20. Directivity is 3 dB above that of a single slot, i.e.,

Dmax ≈ 8 dBi. Actual gain values are shown in Fig. 5.21(b). These values can be much

smaller than the directivity due to low efficiency (see Fig. 5.21(a)).

(a) (b)

Figure 5.19: A double slot model for the patch antenna (a) as viewed from above, (b)equivalent circuit.

Figure 5.20: Input resistances of a rectangular patch with εr = 2.2.

Basic design considerations One has to choose first the substrate over which the

antenna would be printed. Low loss is an obvious requirement. The dielectric constant

and thickness t have an important effect the resonant frequency, gain, efficiency and

bandwidth of the antenna. A lower ε provides higher efficiency, however it hard to realize.

120

The efficiency also increases with smaller thickness. The bandwidth, on the other hand,

increases with thickness. These interplay between these parameters is demonstrated in

Fig. 5.21.

(a) (b)

(c)

Figure 5.21: Effect of dielectric constant and substrate thickness on (a) the radiationefficiency, (b) gain and (c) bandwidth.

Chapter 6

Linear antenna arrays

The linear array in Fig. 6.1 is made of identical radiating elements aligned along the

z–axis at a fixed distance d between their centers. All elements have the same aperture

distribution a(z′) about their center. Suppose we excite the nth element in the array of

Fig. 6.1 by the voltage Vn, or the input current In. The aperture distribution can then

be written as a convolution between a(z′) and the infinite impulse comb as follows:

N∑n=−N

Inδ(z′ − nd) ? a(z′). (6.1)

Upon performing the Fourier transform,

Far field =N∑

n=−NIne

kznd

︸ ︷︷ ︸Arrary Factor

· a(kz)︸ ︷︷ ︸element pattern

. (6.2)

We thus have the “principle of pattern multiplication” between the Array Factor (AF)

and the element pattern.

A word of caution The element distribution a(z′) and the element pattern a(kz) are

defined for the element when embedded in the array, not for the isolated element.

We can treat the Array Factor and the element pattern seprately.

121

122

Figure 6.1: A linear array as a convolution between an infinite comb of impulses and5the element distribution a(z′).

6.1 The Array Factor (AF)

The array factor depends only on the number of identical elements, the distance between

then and the excitation. It can be interpreted as the far field pattern of an array made

123

of isotropic elements. It requires no electormagnetic computations (these are left for the

element pattern) and its analysis is purely “kinematic”.

Based on Eq. (6.2), we have the following one possible expressions for the AF:

AF =N∑

n=−NIne

kznd (6.3)

Another possible expression, based on Eq. (6.1), is

AF = F

N∑n=−N

Inδ(z′ − nd)

= F

∞∑n=−∞

I(z′)δ(z′ − nd)

=2π

dI(kz) ?

∞∑m=−∞

δ(kz −m2π

d) (6.4)

where I(z′) is a continuous distribution supported by the region |z′| ≤ −Nd, whose

samples at z′ = nd are In. Then, I(kz) is its far field and we can see that the pattern of

the sampled aperture is a periodic repetition of the continuous pattern, the period being

∆kz = 2πd

, seen in Fig. 6.2. This is the effect of “Grating Lobes”.

Example: A uniform distribution Take the continuous distribution

I(z′) =

I0, |z′| ≤ Nd

0, otherwise

The pattern of the continuous distru=ibution is

I(kz) = I0Pdsin kzPd

2kzPd

2

Once the aperture is sampled at the interval d, the AF can be expressed via either (6.3),

AF(kz) = I0

N∑n=−N

ekznd = I0sinP kzd

2

P sin kzd2

(6.5)

or via (6.4):

AF(kz) = I0Pd

∞∑m=−∞

sin (kz −m2πd

)Pd2

(kz −m2πd

)Pd2

124

Figure 6.2: A typical Array Factor (AF) with grating lobes spaced 2πd

apart, radiating inthe broadside (kz0 = 0) direction.

6.1.1 Scanning

A well known property of the Fourier tranform is this:

If F f(z′) = F (kz)

then Ff(z′)e−kz0z′

= F (kz − kz0).

Thus, adding a linear phase distribution kz0z over the other wise real aperture distribution

causes the beam to steer from broadside (θ0 = 900) to θ0 = cos−1(kz0

k

). The constant

phase difference between any two adjacent elements is ∆φ = kz0d. This phase is added to

the excitations, i.e., In ⇒ Ine−kz0nd. The Arrary Factor of Eq. (6.3) becomes expressions

for the AF:

AF =N∑

n=−NIne

(kz−kz0)nd = I(kz) ?∞∑

m=−∞δ(kz − kz0 −m

2π

d). (6.6)

The AF as a whole, main lobe and grating lobes, is scanned to θ0. Note that the element

pattern remains unchanged.

A criterion for choosing d for an array whose maximin scan range is kz0 can be seen

125

Figure 6.3: The PATRIOT phased array.

from Fig. 6.7. If we allow the peak of Grating Lobde #(m = −1) to appear at the edge

of the visible range or farther while the main beam (m = 0) is scanned to kz0, we have

required

2π

d≥ k + |kzo| = 2π

λ+ k| cos θ0| =⇒ d ≤ λ

1 + | cos θ0| . (6.7)

Broadside and endfire arrays For a boradside array (θ0 = 900), the criterion (6.7)

becomes d ≤ λ. For an endfire or backfire array (θ0 = 0 or θ0 = 1800), d ≤ λ2. The

126

Figure 6.4: The AN-SPY-1 (AEGIS) phased array.

conventional range of d is therefore λ2≤ d ≤ λ.

Uniform amplitude linear arrays

As seen in Eq. (6.5), the Array Factor of the normalized broadside uniform array is

AF(kz) =sin

(P kzd

2

)

P sin(kzd2

) =sin

(P kd

2cos θ

)

P sin(kd2

cos θ) (6.8)

Upon scanning to θ0, the AF becomes

AF(kz) =sin

(P (kz−kz0)d

2

)

P sin(

(kz−kz0)d2

) =sin

(P kd

2(cos θ − cos θ0)

)

P sin(kd2

(cos θ − cos θ0)) (6.9)

Examples

1. Find the AF of two elements with equal amplitudes and phases (∆φ = 0), spaced

127

Figure 6.5: Complete assembly of the AN-SPY-1 (AEGIS) phased array.

Figure 6.6: Elta EL/M-2080 phased array.

128

Figure 6.7: Derivation of the criterion for detemining d.

d = λ2

apart (see Fig. 6.8(a)). From (6.8),

AF(kz) =sin

(2π

2cos θ

)

2 sin(π2

cos θ) = cos

(π2

cos θ).

2. AF of two elements spaced d = λ2

apart, with equal amplitudes and ∆φ = kz0d =

1800 phase difference (Fig. 6.8(b)). The scan angle is then

θ0 = cos−1

(kz0k

)= cos−1

(kz0d

kd

)= cos−1

(ππ

)= 00

such that this is an endfire array. The AF is

AF(kz) =sin

(2π

2(cos θ − cos θ0)

)

2 sin(π2

(cos θ − cos θ0)) = cos

(π2

(cos θ − 1))

= sin(π

2cos θ

).

3. AF of two elements spaced d = λ2

apart, with equal amplitues and ∆φ = 900 phase

129

difference between them (Fig. 6.8(c)):

AF(kz) = cos(π

4(cos θ − 1)

).

4. AF of two elements spaced d = λ apart with equal amplitudes and phases (Fig. 6.8(d)):

AF(kz) = cos (π cos θ).

Properties of the AF of the uniform linear array The array factor in (6.8) - (6.9)

possesses the following properties:

1. As P increases with a fixed d, the beam narrows. For large P , HPBW' 51 λPd

(degrees).

2. As the array is scanned away from broadside, the beam widens as HPBW' 51 λPd

1cos θ0

(degrees).

3. In each period of the AF, there is one main lobe of grating lobe and P−1 sidelobes.

4. The width of the sidelobes between nulls in the kz–space is 2πP

. The width of the

main and grating lobes is twice this amount.

5. The sidelobe level decreases with increasing P . For P = 5 and P = 20, SLL=

−12 dB and SLL= −13 dB, respectively. SLL→ −13.3 dB as P →∞.

6. The AF is symmetric about kz = πd.

Effect of the element pattern

We now introduce a non–isotropic element to include the effect of the element pattern

as in Eq. (6.2). Take example 1 above, replacing the isotropic elements with elementary

130

(a) Example 1 (b) Example 2

(c) Example 3 (d) Example 4

Figure 6.8: Examples 1-4.

dipoles. The, if the dipoles are oriented in the z–direction, the total pattern is

Far field = cos(π

2cos θ

)

︸ ︷︷ ︸AF

· sin θ︸︷︷︸element pattern

(6.10)

In this case, the direction of the main beams of the AF and the element patterm, respec-

tively, are aligned (see Fig. 6.9(a)). If, on the other hand, the dipoles are oriented in a

131

(a) Example 1 with z–oriented dipole elements

(b) Example 1 with x–oriented dipole elements, pattern shownin the x− z plane

Figure 6.9: Examples 1-2 with dipole elements.

direction perpendicular to the array line, e.g., in the x–direction, then the main beams

are not aligned, and the pattern is

Far field = cos(π

2cos θ

)

︸ ︷︷ ︸AF

·√

1− sin2 θ cos2 φ︸ ︷︷ ︸

element pattern

. (6.11)

In the x− z (φ = 0⋃

φ = π) plane, this pattern is

Far field = cos(π

2cos θ

)· cos θ (6.12)

(compare with (6.10)), that causes the beam to split as shown in Fig. 6.9(b).

132

As another example, consider an uniform array with P elements spaced d apart along

the z–axis. Each element has a unifrom distribution of width d, such that the entire

aperture of width L = Pd is filled with a continuous uniform distribution (see Fig.6.10).

The combination of the AF and the element pattern looks like

Far field =sin

(P kzd

2

)

P sin(kzd2

)︸ ︷︷ ︸

AF

· sin(kxd2

)kxd2︸ ︷︷ ︸

element pattern

=sin

(kzPd

2

)kzPd

2

=sin

(kzL2

)kzL2

(6.13)

as expected.

Figure 6.10: A uniform array with P uniform elements of width d.

Array Feeds

Feeds are categorized as either constained or space feeds. Constrained feeds are seen in

Fig. 6.11. These are of three types:

More feeds are shown in Fig. 6.13

PPPPP

133

Figure 6.11: Constrained feeds for microstrip arrays.: (a) Parallel (“corporate”) feed, (b)series feed with impecance changes along the feed line, (c) Parallel networks connectedto series–fed lines and columns forming a dual–polarized array.

Figure 6.12: A circularly polarized antenna array.

134

Figure 6.13: Various array feeds.

135

Figure 6.14: Scheme of a corporate feed.