course notes
TRANSCRIPT
Antennas and Radiation
Course no. 0512.4861: Prof. Raphael Kastner.
School of Electrical Engineering, Tel Aviv University
Tel Aviv 69978, Israel. [email protected]
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.