Antenna Theory & DesignECED 4360 Lecture Notesc©2019 Sergey A. Ponomarenko

February 25, 2019

Contents

1 Vector algebra & complex numbers: A brief review 31.1 Scalars and vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Vector addition and subtraction . . . . . . . . . . . . . . . . . . . . . 41.3 Vector multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . 51.4 Complex numbers and phasors . . . . . . . . . . . . . . . . . . . . . 9

2 Electromagnetic fields and Maxwell’s equations 122.1 Charges, currents and electromagnetic fields . . . . . . . . . . . . . . 122.2 Electromagnetic fields in materials . . . . . . . . . . . . . . . . . . . 152.3 Global or integral form of Maxwell’s equations . . . . . . . . . . . . 172.4 Local or differential form of Maxwell’s equations . . . . . . . . . . . 212.5 Electromagnetic boundary conditions . . . . . . . . . . . . . . . . . 252.6 Conservation laws in electromagnetic theory . . . . . . . . . . . . . . 27

3 Introduction to antenna theory. Vector and scalar potentials. 323.1 Antenna: the concept and implementation . . . . . . . . . . . . . . . 323.2 Plane electromagnetic waves in free space . . . . . . . . . . . . . . . 343.3 Scalar and vector potentials for time-harmonic fields . . . . . . . . . 413.4 Electric dipole moment of any charge distribution . . . . . . . . . . . 453.5 Elemental electric dipole antenna . . . . . . . . . . . . . . . . . . . . 493.6 Magnetic dipole antenna . . . . . . . . . . . . . . . . . . . . . . . . 533.7 Antenna parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

4 Linear antenna and linear antenna arrays 604.1 Linear antenna . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

4.1.1 Standing wave antenna . . . . . . . . . . . . . . . . . . . . . 604.1.2 Traveling wave antenna . . . . . . . . . . . . . . . . . . . . . 63

4.2 Linear antenna as a boundary-value problem . . . . . . . . . . . . . . 644.3 Linear antenna arrays . . . . . . . . . . . . . . . . . . . . . . . . . . 67

4.3.1 Two-element arrays . . . . . . . . . . . . . . . . . . . . . . . 674.3.2 Dipole radiator near perfect conductor . . . . . . . . . . . . . 694.3.3 Multi-element arrays . . . . . . . . . . . . . . . . . . . . . . 71

4.4 Non-uniform antenna array synthesis . . . . . . . . . . . . . . . . . . 744.4.1 Binomial arrays . . . . . . . . . . . . . . . . . . . . . . . . . 75

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4.4.2 Schelkunoff polynomial method . . . . . . . . . . . . . . . . 754.4.3 Fourier series method . . . . . . . . . . . . . . . . . . . . . . 774.4.4 Woodward-Lawson method . . . . . . . . . . . . . . . . . . 77

5 Beyond linear antannas 795.1 Aperture antenna and angular spectrum . . . . . . . . . . . . . . . . 795.2 Spherical nanoparticle as a simple nano-antenna . . . . . . . . . . . . 835.3 Nano-antenna applications . . . . . . . . . . . . . . . . . . . . . . . 90

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Chapter 1

Vector algebra & complexnumbers: A brief review

1.1 Scalars and vectorsIn general, there are two kinds of objects one deals with in vector algebra: scalarsand vectors. While the former have only magnitude, the latter are characterized bytheir magnitudes and directions. Physical quantities such as mass, density, temper-ature, and charge, say, are scalars, whereas a velocity, or a force is a vector. A unitvector–which has a unit magnitude–can always be formed by dividing a vector by itsmagnitude. For instance,

eA =A

|A|,

is a unit vector directed along A. A vector eA can be geometrically represented as anarrow; the length of the arrow equals the magnitude of A, and the arrow points in thedirection of A as is seen in Fig. 1.1.

Another way to represent a vector is through its three–in a three-dimensional space,of course–components in a suitably chosen coordinate system. For instance, in theCartesian coordinates, any vector A can be represented in terms of its three coordinates(Ax, Ay, Az), or alternatively,

A = Axex +Ayey +Azez .

Here ex, ey and ez are three mutually orthogonal unit vectors, (see Fig.1.2). The vectormagnitude can be determined using the Pythagoras’s theorem,

|A| =√A2x +A2

y +A2z .

3

A

Figure 1.1: Geometric representation of a vector A.

A

za

yaxa o

x

y

z

Figure 1.2: Decomposition of a vector A in the Cartesian coordinate system.

1.2 Vector addition and subtractionTwo vectors A and B can be added and/or subtracted component by component,

A + B = (Ax +Bx)ex + (Ay +By)ey + (Az +Bz)ez.

Geometrically, the vector addition can be represented using either a parallelogram ruleor a head-to-tail rule as depicted in Fig. 1.3. The subtraction is inverse to addition. Asfollows from the definition, the vector addition/ subtraction obeys commutativity andassociativity properties, implying that

A + B = B + A, (commutativity),

4

C

B

A

(a)

C

B

A

(b)

Figure 1.3: Illustrating vector addition: (a) parallelogram rule and (b) head-to-tail rule.

andA + (B + C) = (A + B) + C, (associativity).

Also a vector can be multiplied by a scalar, implying each vector component ismultiplied by a scalar,

kA = kAxex + kAyey + kAzez.

A product of a scalar and a vector sum/difference obeys the distributive law,

k(A + B) = kA + kB.

1.3 Vector multiplicationThere are two kinds of vector products: dot or scalar product and cross or vector prod-uct.Definition. The dot product of two vectors A and B, written as A ·B is defined asa product of the vector magnitudes times the cosine of the smaller angle between themwhenever the two are drawn tail,

A ·B = |A||B| cos θAB . (1.1)

Implication. As follows from the definition, the two vectors are orthogonal if theirscalar product is equal to zero.

In terms of the vector coordinates,

A ·B = AxBx +AyBy +AzBz. (1.2)

Note that the dot product always results in a scalar quantity. The dot product obeys thecommutative and distributive rules

• A ·B = B ·A, commutativity;

• A · (B + C) = A ·B + A ·C, distributivity.

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As a corollary of the definition,

A ·A = |A|2,

implying an alternative way of determining the vector magnitude without resorting tovector components,

|A| =√A ·A .

Further, the mutual orthogonality of the Cartesian unit vectors implies that

ex · ey = ey · ez = ex · ez = 0 , (1.3)

andex · ex = ey · ey = ez · ez = 1 . (1.4)

Exercise 1.1. Show that (A + B) · (A−B) = |A|2 − |B|2.Solution. Using the properties of the dot product: (A + B) · (A−B) = A ·A + B ·A−A ·B−B ·B = |A|2 − |B|2.

Definition. The cross product of two vectors A and B, written as A × B isa vector whose magnitude is the area of the parallelogram formed by A and B–seeFig.1.4–and is in the direction determined by the right-handed cork screw rule illus-trated in Fig.1.5.It follows that

A×B = AB sin θABen ,

where en is a unit normal to the plane containing A and B.

A

B

BA

Figure 1.4: Illustrating the cross-product.

In the coordinate representation,

A×B =

∣∣∣∣∣∣ex ey ezAx Ay AzBx By Bz.

∣∣∣∣∣∣The cross product obeys the following rules

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Figure 1.5: The right-handed cork-screw rule.

• A×B = −B×A;

• A× (B + C) = A×B + A×C;

• A×A = 0.

Also, the mutual cross products of the Cartesian unit vectors obey the rule

ex × ey = ez , (1.5)

with cyclic permutations for the right-handed Cartesian system as is shown in Fig. 1.6.

za

yaxa

ya

xa

za

Figure 1.6: Illustrating unit vector cross products under cyclic permutations.

Exercise. 1.2. Show that (A + B)× (A−B) = 2B×A.Solution. Using the properties of the cross product: (A + B)× (A−B) = A×A +B×A−A×B−B×B = 2B×A.

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Exercise. 1.3. Given, ex ×A = −ey + 2ez and ey ×A = ex − 2ez , Find A.Solution. Assume that A = aex + bey + cez . It follows that ex ×A = b(ex × ey) +c(ex × ez) = bez − cey = −ey + 2ez . Hence, c = 1 and b = 2. Similarly, ey ×A =−aez + cex = ex − 2ez , implying that c = 1 and a = 2. Thus A = 2ex + 2ey + ez .

As a consequence of the scalar and cross product definitions, we can infer that thescalar triple product can be represented as

A · (B×C) = (A×B) ·C = B · (C×A) . (1.6)

In the Cartesian coordinates, the scalar triple product can be written as

A · (B×C) =

∣∣∣∣∣∣Ax Ay AzBx By BzCx Cy Cz

∣∣∣∣∣∣ (1.7)

Finally, the vector triple product can be expressed as using “bac-cab” mnemonicrule in the form

A× (B×C) = B(A ·C)−C(A ·B). (1.8)

Exercise 1. 4. Show that A · B × C is a volume of a parallelepiped having A, B,and C as three contiguous edges.Solution. A ·B×C = |A| cos θ︸ ︷︷ ︸

height

|B×C|︸ ︷︷ ︸area

, see the sketch below.

C

B

A

|| CB

Figure 1.7: Geometric illustration of the scalar triple product.

Exercise 1.5. Given A ·B = A ·C and A×B = A×C, and A is not a null vector,show that B = C.Solution. Choose the x-axis along the direction of A. It follows that A = Aex whereA 6= 0. Assume further that B = Bxex+Byey+Bzez and C = Cxex+Cyey+Czez .A ·B = A ·C then implies thatBx = Cx, and A×B = A×C implies thatBy = Cyas well as Bz = Cz . As the components are the same, the vectors are equal.

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1.4 Complex numbers and phasorsDefinition. A complex number z can be expressed in the so-called rectangular formas

z = u+ jv = Re(z) + jIm(z) , (1.9)

where j =√−1 and u and v are real and imaginary parts of a complex number,

respectively, both being real numbers. Alternatively, it can be expressed in the polarform as

z = rejφ = r(cosφ+ j sinφ) , (1.10)

where the magnitude r and phase φ can be written as

r =√u2 + v2, φ = tan−1 v/u . (1.11)

Geometrically, z can be represented as a ray in the uv plane making the angle φ withthe u-axis, see Fig. 1.8.

Z

u

v

Figure 1.8: Polar form of a complex number.

Given two complex numbers, z1 = u1+jv1 = r1ejφ1 and z2 = u2+jv2 = r2e

jφ2 ,the result of their addition or subtraction can be most easily expressed in the rectangularform:

z1 ± z2 = u1 ± u2 + j(v1 ± v2). (1.12)

On the other hand, their multiplication and division are more naturally expressed in thepolar form as

z1z2 = r1r2ej(φ1+φ2),

z1

z2=r1

r2ej(φ1−φ2). (1.13)

One can also introduce complex conjugation by the definition

z∗ = u− jv = re−jφ . (1.14)

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It follows at once from Eqs. (1.9) and (1.14) that

u = Re(z) = 12 (z + z∗) . (1.15)

Eq. (1.15) gives a convenient representation of a real number in terms of a complexnumber and its conjugate.

In the polar form, a complex number is not uniquely defined such that

z = rejφej2πk, k = 0,±1,±2,±3 . . . . (1.16)

This is because ej2πk = 1 for any integer k. The latter form comes in handy wheneverwe want to find roots of a complex number. In general all nth roots can be representedas

z1/n = r1/nejφ/nej2πk/n. (1.17)

For example, if n = 2, there are only two distinct roots corresponding to k = 0 andk = 1; since ej0 = 1 and ejπ = cosπ + j sinπ = −1, we obtain

√z = ±

√rejφ/2. (1.18)

Definition. A time-harmonic signal varies sinusoidally with time.Definition. A phasor represents a complex signal with a time-harmonic phase.Thus any physical time-harmonic signal ψ(t) = a cos(ωt + θ), where ω and θ areconstant frequency and initial phase, respectively, can be represented in terms of acomplex phasor ψ0e

jωt as

ψ(t) = Re(ψ0e−jωt) = 1

2 (ψ0e−jωt + ψ∗0e

jωt) , (1.19)

where the last line is obtained with the aid of Eq. (1.15). Here Re denotes the real partof the complex signal and the complex amplitude ψ0 can be represented as

ψ0 = a ejθ , (1.20)

where a is a real amplitude. The generalization to the phasor form of a vector time-harmonic signal is straightforward:

E(t) = Re(E0e−jωt), E0 = |E0|ejθ. (1.21)

Exercise 1. 6. The complex impedance of a monochromatic electromagnetic waveof frequency ω, propagating in a lossy medium is defined as

η =

√µ/ε

1 + jσεω

.

Here µ, η and σ are constitutive parameters of the medium. Express η in the polarform.

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Solution. Multiplying the numerator and denominator inside the square root by (1 −jσ/εω), we obtain

η =

√µ/ε

(1− jσ

εω

)1/2[1 +

(σεω

)2]1/2 =

√µ/ε ejθη[

1 +(σεω

)2]1/4 = |η|ejθη ,

where

|η| =√µ/ε[

1 +(σεω

)2]1/4 , tan 2θη =σ

εω.

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Chapter 2

Electromagnetic fields andMaxwell’s equations

2.1 Charges, currents and electromagnetic fieldsDefinition. An electric charge Q quantifies the capacity of an object for electro-magnetic interactions—the greater the charge the stronger the interaction. The chargescould be positive or negative; the charges of the opposite signs attract to each otherwhile those of the same sign repel from one another.The interaction force between the two point charges Q1 and Q2, separated a distanceR12 is determined by the Coulomb law

F =1

4πε0

Q1Q2R12

R312

, (2.1)

where R12 is a radius vector from charge Q1 to Q2 and ε0 is the so-called free spacepermittivity, given in the SI units by the expression

ε0 =10−9

36π, F/m (2.2)

Definition. An electric current is a flow of electric charges past a point or within aconductor. The current I is a time rate of change of the charge Q,

I =dQ

dt. (2.3)

The charges are measured in Coulombs, C and the currents are measured in Amperes,abbreviated A. The smallest charge encountered in nature is the electron charge e,which is equal to −1.60219× 10−19 C.

The electric charges and currents (moving charges) are the sources of electric Eand magnetic B fields, respectively. The vector field E is known as the electric field

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intensity or electric field strength and it is measured in volts per meter, V/m. Thefield B is more precisely referred to as the magnetic flux density for the reasons thatbecome clear shortly and it is measured in Webers per square meter, Wb/m2 To helpvisualize the behavior of electric and magnetic fields in space, we introduce the conceptof electric and magnetic field lines.Definition. The electric field lines are, in general, curves in space such that at anygiven point on the line, the electric field is tangential to the line. As electric chargesare sources/sinks of the field, the electric field lines start at the positive (source) andend at the negative (sink) charges. Alternatively, if the electric field is generated by atime-dependent magnetic field, its lines are closed. These possibilities are illustrated inFig. 2. 1.

E

E

+ Q - Q

(a) (b)

E

B

Figure 2.1: Lines of the electric field generated by (a) static electric charges and (b) atime-dependent magnetic field.

Definition. The magnetic flux density lines are, in general, continuous curves inspace such that at any given point on the line, the magnetic flux density is tangentialto the line. As no static magnetic charges have so far been found in nature, thereare no static sources of magnetic fields—the latter are generated by moving electriccharges. Therefore, the magnetic flux density lines are either closed or go to infinity.A natural question then arises regarding the quantitative description of electric andmagnetic fields: How can one quantify and measure E and B at a given point in space?

To answer this question, let us consider a small point test charge q at rest. It isknown from the experiment that the charge placed in an electric field E generated bysome other charges experiences the force

Fe = qE . (2.4)

It follows at once from Eq. (2.4) that the electric field at a position of the test charge issimply the force per unit charge and it can be determined as

E =Feq. (2.5)

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Note that the definition (2.5) is unambiguous, provided the test charge is so small thatit does not alter the field at its location. Thus, the strength of the electric field at theposition of the charge can be determined by measuring the force acting on a small testcharge at rest.

Next, if a small test charge moves with the velocity v in a magnetic field B, it isknown to experience the magnetic force

Fm = q(v ×B). (2.6)

Exercise 2.1. Determine the components of the charge velocity v‖ and v⊥, paralleland perpendicular to the magnetic field B, respectively.Solution. Introduce a unit vector along B, b = B/B. It follows from the definition ofthe dot product that the projection of v onto b is v · b. Hence the vector projectionalong b is v‖ = (b · v)b. Consequently, v⊥ = v − v‖ = v − (v · b)b. Note thatv⊥ · b ≡ 0.

On taking a cross product of both sides of Eq. (2.6) with v⊥ we obtain

Fm × v⊥ = −qBv⊥ × (v × b) = −qv(v⊥ · b) + qB(v · v⊥) = qB(v · v⊥).

Note also that v⊥ ·v = v2−v2‖. Thus we arrive at the expression for the magnetic field

B =(Fm × v⊥)

q(v2 − v2‖)

. (2.7)

Thus, we can determine B by measuring the force on a moving charge and the chargevelocity. Note that Eq. (2.7) is indeterminate whenever v = v‖, because in this casethe force equals to zero according to Eq. (2.6). So the charge velocity should havea component at an angle to the magnetic field to unambiguously determine the latter.We note that Eqs. (2.5) and (2.7) serve as the operational definitions of E and B,respectively. The E and B fields characterize the strength of electric and magneticinteractions at a given point in space—described by the position radius vector r— andhence are local measures of the electromagnetic interactions in a given system. Thus,E and B are functions of the space coordinate; in general, they can also vary with time,

E = E(r, t) and B = B(r, t).

If a test charge is moving in both electric and magnetic fields, which are, in general,functions of time, it experiences the combined Lorentz force,

FL = qE + qv ×B ,

and the electric and magnetic fields can be thought of as components of a commonentity called the electromagnetic field.Exercise 2.2. A point charge Q with a velocity v = v0ex enters a region of spacewith a uniform magnetic field. The magnetic flux density in the region is B =Bxex +Byey +Bzez . What E should exist in the region for the charge to proceedwithout change of its velocity.

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Solution. Assuming E = Exex + Eyey + Ezez and writing down the second law ofNewton in components, we arrive at the equations,

mv′x = QEx +Q(vyBz −Byvz),

mv′y = QEy +Q(vzBx −Bzvx),

andmv′z = QEz +Q(vxBy −Bxvy).

Here the prime stands for a time derivative. The charge will proceed with the samevelocity if all components of the acceleration vanish at all times, i. e, v′x = v′y = v′z =0. Since at t = 0 vy = vz = 0, it follows that Ex = 0, Ey = Bzv0 and Ez = −v0By .Thus, E = v0(Bzey −Byez).

2.2 Electromagnetic fields in materialsThe response of a material to an applied electric field depends on whether the materialhas free electrons and therefore can conduct currents or not. Materials of the first kindare called conductors whereas the rest are known as dielectrics.

In conductors, the electrons are free to move and their motion past heavy ions ofa crystal lattice constitutes a conduction current. One can introduce a local quantitycharacterizing the current, the current density J measured in Amperes per square meter,which is just a current per unit cross-section of a conductor. The total current is then

I =

∫dS · J , (2.8)

where dS = endS is an oriented elementary surface, en being a unit normal to thesurface as is indicated in Fig. 2.2.

s

dS

J

Figure 2.2: Illustrating the current density definition.

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The integral on the r.h.s of Eq. (2.8) is an example of a flux of the vector field—inthis instance J—through an open surface S.Exercise 2. 3. Show that J = ρvv where ρv is the volume charge density and v isthe drift velocity of charge carriers.Solution. Consider a small volume element dv = dS · vdt. The amount of chargeinside a cylindrical volume of height v · endt with a finite cross-section S is dQ =∫S

(dS · v)dtρv . By definition, the current through the cross-section S is then I =dQ/dt =

∫SdS · vρv =

∫SdS · J. It follows that J = ρvv.

The current density is related to the electric field via the local form of Ohm’s law,

J = σE , (2.9)

where σ is called the electric conductivity, measured in Siemens per meter, S/m.In the dielectrics, the electrons are bound to nuclei, forming neutral atoms. The

application of an external electric field, however, causes spatial displacement of neg-atively charged electron clouds away from positively charged nuclei; the latter beingso heavy that they remain immobile. The medium is then said to be polarized. Thisprocess is illustrated in Fig. 2.3.

E

(a) (b) (c)

Q

E

d

Figure 2.3: Illustrating the polarization of a nonpolar dielectric.

The polarization can be quantitatively described in terms of individual atom dipole mo-ments.Definition. An individual dipole moment vector p is defined as the product of anelectron cloud charge and a position vector from the the nucleus to the electron cloudcenter. For instance, for an atom having just one bound electron, p = −er.

The dielectrics with the atoms that have no dipole moments in the absence of theapplied field are called nonpolar. Alternatively, the medium atoms of polar dielectricscan have nonzero dipole moments even in the absence of E, but they are randomlyoriented. As the external electric field is applied, though, the dipoles align along thefield resulting in the medium polarization.

Regardless of a specific polarization origin, we can define a macroscopic polariza-tion field.Definition. The polarization field P(r, t) is a dipole moment per unit volume at theposition r within a polarized medium.The polarized medium alters (reduces) the external electric field E such that the effec-tive field inside the medium is described in terms of the electric flux density D,

D = ε0E + P , (2.10)

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where ε0 = 8.854×10−12 farad per meter (F/m) is the so-called dielectric permittivityof free space. Eq. (2.10) works for any dielectric; throughout this course we will bedealing with linear, homogeneous, isotropic dielectrics for which P is linearly relatedto E viz.,

P = ε0χeE , (2.11)

where χe is the electric susceptibility. It follows from Eqs. (2.10) and (2.11) that

D = ε0(1 + χe)E = ε0εrE = εE . (2.12)

Here ε is the dielectric permittivity of the medium, and εr is a dielectric constant(relative permittivity). Note that while ε has the same units as ε0, i.e, farads per meter,εr is dimensionless.

The phenomenological treatment of macroscopic medium response to the magneticfield parallels that we just presented. An external magnetic field causes the mediummagnetization: the atomic magnetic moments align along the applied field causinga finite macroscopic average dipole moment density. The latter called magnetizationM, and is a magnetic analog of P. By analogy, the magnetic field intensity H withinthe magnetized medium can be introduced as

H = B/µ0 −M . (2.13)

Eq. (2.13) holds true for any medium. Here,

µ0 = 4π × 10−7 , H/m, (2.14)

is known as the free space permeability. In the case of a linear, homogeneous,isotropic magnetic, we postulate the following relation between the magnetization andmagnetic field intensity

M = χmH , (2.15)

where χm is the magnetic susceptibility, implying that

B = µ0(1 + χm)H = µ0µrH = µH . (2.16)

Here µ and µr are the magnetic permeability and relative magnetic permeability ofthe medium. Note that while E and B are directly related to measurable quantities, theforces on charges, D and H are auxiliary fields.

2.3 Global or integral form of Maxwell’s equationsThe first two Maxwell’s equations are mathematical expressions of the fact that staticcharges are sources of the electric field and there are no static magnetic charges. Inparticular, the first Maxwell equation—also known as the electric Gauss law—statesthat the total flux of D through any closed surface S is equal to the total enclosedcharge, ∮

S

dS ·D = Qenc =

∫v

dvρv . (2.17)

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Here the circle around the integral implies that the surface for the surface integrationmust be closed. The choice of the oriented elementary surface dS = endS used onthe l.h.s of Eq. (2.17) is ambiguous as the unit normal can be directed either inside oroutside the volume enclosed by S. By convention, we choose en to be the outwardunit normal as is indicated in Fig. 2.5. Also ρv is the volume density, in C/m3, of thecharge inside S.

D

na

S

dS

Figure 2.4: Outward unit normal to a closed surface S.

Since there are no static magnetic charges, the second Maxwell equation (magneticGauss’s law) states that ∮

S

dS ·B = 0 . (2.18)

It is now clear from Eqs. (2.17) and (2.18) why D and B are referred to as the electricand magnetic flux densities, respectively.

In the Cartesian coordinate system, for example, the infinitesimally small surfaceand volume elements required in Eqs. (2.17) and (2.18) can be expressed as

dS =

dydzexdxdzeydxdyez

anddv = dxdydz.

The surface element calculation is illustrated in Fig. 2.6.The third Maxwell equation, or the Faraday’s law, relates the electric field circu-

lation around any closed path C with the time rate of change of the magnetic fluxthrough an open surface S bounded by the path,∮

C

dl ·E︸ ︷︷ ︸emf

= − d

dt

∫S

dS ·B . (2.19)

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(a) (b) (c)

za

ya

xa

x

y

z

dx

dzdx

dz

dy

dy

Figure 2.5: Illustrating the elementary surfaces in the Cartesian coordinates.

In the circulation integral on the l.h.s. of Eq. (2.19), dl is an oriented infinitesimallysmall path element which can be expressed, for instance, in the Cartesian coordinatesas

dl = dxex + dyey + dzez.

The fourth Maxwell equation, sometimes referred to as Ampere’s law, links thecirculation of the magnetic field along a closed path with the flux of an overall enclosedcurrent—conduction current plus the so-called displacement current—through an opensurface S rimmed by the path,∮

C

dl ·H︸ ︷︷ ︸mmf

= Ienc =

∫S

dS · J︸ ︷︷ ︸conduction

+d

dt

∫S

dS ·D︸ ︷︷ ︸displacement

. (2.20)

Here Ienc denotes the enclosed current. To illustrate the displacement current role, weconsider the following exercise.Exercise 2.4. Consider a parallel-plate capacitor filled with an ideal dielectric andconnected to an ac voltage source and show that the displacement current insidethe capacitor ensures Ampere’s law holds there.Solution.—According to Ampere’s law, the circulation of H through any closed pathenclosing a given current must be the same. If we consider an open surface S1 outsidethe capacitor bounded by the path L that encloses the current I , the Ampere law gives∮Ldl · H = I , where I is a conduction current flowing through the wire connecting

the capacitor to the voltage source. On the the hand, if we apply Ampere’s law to L,rimming any open surface S2 lying partially within the capacitor and we ignore thedisplacement current, we obtain

∮Ldl · H = Ienc = 0. This is because no conduc-

tion current can flow between the plates of a capacitor filled with an ideal dielectric(with zero conductivity). The apparent contradiction is resolved at once, if we take thedisplacement current into consideration as follows∮

L

dl ·H =d

dt

∮S2

dS ·D =d

dt

∮S

dS ·D =dQ

dt= I, (2.21)

19

where we replaced the open surface S2, displayed in the figure below, with a closedone, S = S2 + Scomp, which includes a complementary surface Scomp. We then ap-plied Gauss’s law, Eq. (2.17) to S. Thus the displacement current ensures Ampere’slaw works in conducting as well as dielectric media.

Figure 2.6: Illustrating the displacement current concept. Reproduced from M. Sadiku,Elements of Electromagnetics (Oxford U. Press, 2010).

To summarize, the first and third Maxwell equations quantitatively describe howelectric fields can be generated by two types of sources:

• static or time-dependent charges;

• time-dependent magnetic fields.

The electric field generated by a time-varying field gives rise to an electromotiveforce (emf),

Eemf ≡∮C

dl ·E , (2.22)

in a given closed loop determined by the time rate of change of the magnetic fluxthrough the loop, that is

Eemf = − d

dt

∫S

dS ·B . (2.23)

The second Maxwell equation states the absence of magnetic charges. The fourthMaxwell equation quantifies magnetic field generation by two types of sources:

• conduction currents;

• time-varying electric fields via displacement currents.

20

The displacement currents generate a magnetic field with the magnetomotive force(mmf),

Emmf ≡∮C

dl ·H, (2.24)

in a closed loop determined by the overall conduction current and the time rate ofchange of the electric flux through the loop. In particular, in the absence of conductioncurrents J = 0, the time-varying electric fields can generate magnetic fields. Thus thepropagation of electromagnetic waves in source-free space, (ρv = 0, J = 0) is a directconsequence of Eqs. (2.19) and (2.20). Note also that the displacement current is afictitious current that has to do with time-varying electric fields.Exercise 2.5. A magnetic flux density is given by B = eyB0/xWb/m2, whereB0 isa constant. A rigid rectangular loop is situated in the xz-plane with the corners atthe points (x0, z0), (x0, z0 + b), (x0 + a, z0 + b), (x0 + a, z0). If the loop is movingwith the velocity v = v0ex, determine the induced emf.Solution. At the time t the corners of the loop will be at the points (x0 + vt, z0), (x0 +vt, z0 + b), (x0 + a+ vt, z0 + b), (x0 + a+ vt, z0). Using the Faraday’s law, Eemf =∮Cdl ·E = − d

dt

∫SdS ·B. In our case, dS = dxdzey implying that∫

S

dS ·B = B0

∫ z0+b

z0

dz

∫ x0+a+vt

x0+vt

dx

x= B0b ln

x0 + a+ vt

x0 + vt.

It then follows that

Eefm = B0bv

(1

x0 + vt− 1

x0 + a+ vt

).

Exercise 2.6. Solve the previous problem for a stationary loop in the time-varyingmagnetic field B = ey(B0/x) cosωt Wb/m2.Solution. If the loop is at rest, by analogy with the previous example,

Eemf = − d

dt

∫S

dS ·B = ωB0 sinωt

∫ z0+b

z0

dz

∫ x0+a

x0

dx

x

= ωbB0 sinωt lnx0 + a

x0. (2.25)

Thus,Eemf = ωbB0 sinωt ln

x0 + a

x0.

2.4 Local or differential form of Maxwell’s equationsThe Maxwell equations can also be cast into a differential (local) form in which theypertain to any spatial point within a given region of space. Although local Maxwell’sequations are less physically intuitive, they are more suitable to mathematically de-scribe versatile electromagnetic problems. We begin by introducing local measures of

21

the vector field flux and circulation, the flux and circulation densities, or the divergenceand curl of the vector field.

Definition. The divergence of a vector field A at a given point is the net outwardflux of A per unit volume at the point. Mathematically,

divA ≡ lim∆v→0

∮SdS ·A∆v

. (2.26)

It is known from the vector calculus that the divergence can also be written in terms ofthe Del operator, denoted ∇, as

divA = ∇ ·A. (2.27)

In the Cartesian coordinates, for example, the latter is defined as

∇ = ex∂x + ey∂y + ez∂z.

And since A = Axex +Ayey +Azez , we conclude that

divA = ∇ ·A = ∂xAx + ∂yAy + ∂zAz. (2.28)

In practice, the following divergence theorem is often handy in working out fluxes ofvector fields through closed surfaces.Divergence Theorem. The flux of a vector field through a closed surface equals theintegral of the vector field divergence over the volume enclosed by the surface,∮

S

dS ·A =

∫v

dv∇ ·A. (2.29)

We are now in a position to express the first two Maxwell’s equations in the localform. Applying the divergence theorem to the l.h.s of Eq. (2.17), we obtain∮

S

dS ·D =

∫v

dv∇ ·D =

∫v

dvρv. (2.30)

It can be inferred from Eq. (2.30) that∫v

dv(∇ ·D− ρv) = 0. (2.31)

Since the integral equation (2.31) holds for any volume, we conclude that the integrandmust be equal to zero at any point within the volume,

∇ ·D = ρv . (2.32)

By the same token,∇ ·B = 0 . (2.33)

Next, we introduce the curl of a vector field asDefinition. The curl of a vector field A at a given point is a vector with a magnitude

22

dS

na

dl

Figure 2.7: Illustrating the choice of unit normal in curl evaluation.

equal to the maximum net circulation of A per unit area at the point. The curl isdirected along a unit normal to the infinitesimal area around the point which is orientedto maximize the curl. The unit normal is chosen to conform to the right-hand rule:whenever the fingers of your right hand follow the direction of dl along the area border,your thumb points in the direction of the unit normal.

Mathematically, curl can be defined as

curlA = ∇×A = lim∆S→0

∮Cdl ·A∆S

en. (2.34)

In the Cartesian coordinates, for example, the curl can be expressed as

∇×A =

∣∣∣∣∣∣ex ey ez∂x ∂y ∂zAx Ay Az

∣∣∣∣∣∣ . (2.35)

The following curl theorem is often useful in determining the circulation of a vectorfield around a closed loop.Curl Theorem. The circulation of a vector field along a closed path is equal to theflux of the vector field curl through an open surface bounded by the path,∮

C

dl ·A =

∫S

dS · (∇×A). (2.36)

With the aid of Eq. (2.36) and assuming that the loop C is stationary, one cantransform the third Maxwell equation as∮

C

dl ·E =

∫S

dS · (∇×E) = − d

dt

∫S

dS ·B = −∫S

dS · ∂tB,

implying that locally∇×E = −∂tB . (2.37)

23

Similarly, ∮C

dl ·H =

∫S

dS · (∇×H) =

∫S

dS (J + ∂tD) .

It can then be inferred at once that

∇×H = J + ∂tD . (2.38)

Here the second term on the r.h.s. is the displacement current (density) defined as

Jd = ∂tD . (2.39)

Finally, in the antenna theory, we often encounter time-harmonic sources such that

ρv(r, t) = 12 [ρv(r)e−jωt + c. c.]; J(r, t) = 1

2 [J(r)e−jωt + c. c.], (2.40)

and the corresponding time-harmonic fields of the form

E(r, t) = 12 [E(r)e−jωt + c. c.]; D(r, t) = 1

2 [D(r)e−jωt + c. c.], (2.41)

as well as

H(r, t) = 12 [H(r)e−jωt + c. c.]; B(r, t) = 1

2 [B(r)e−jωt + c. c.]. (2.42)

Hereafter the abbreaviation “c.c” stands for a complex conjugate. While the Gausselectric and magnetic laws—Eqs. (2.32) and (2.33)—remain unchanged,

∇ ·D(r) = ρv(r) , ∇ ·B(r) = 0 ; (2.43)

the other two Maxwell equations transform to

∇×E(r) = jωB(r) , (2.44)

and∇×H(r) = J(r)− jωD(r) . (2.45)

Exercise 2.7. Given B = eyB0z cosωt and its is known that E has only an x-component, determine the electric field generated by this magnetic field.Solution. The Faraday law implies

∇×E = −∂tB = eyωB0z sinωt. (2.46)

As there is only an x-component of E, E = exE(x, y, z, t), say, we have

∇×E = ey∂zE − ez∂yE. (2.47)

On comparing Eqs. (2.46) and (2.47), we conclude that

∂yE = 0, ∂zE = ωB0z sinωt. (2.48)

It follows thatE(x, z, t) = 1

2ωB0z2 sinωt+ f(x, t).

Here f(x, t) is an arbitrary function of time. In the limit ω = 0, the magnetic fieldis static and hence it cannot generate any electric field. Therefore, we conclude thatf(x, t) = 0. Thus, E = exωB0(z2/2) sinωt.

24

2.5 Electromagnetic boundary conditionsWe consider an interface separating two media. The boundary conditions linking theelectromagnetic fields on both sides of the interface can be derived from the Maxwellequations in the integral form. To this end, introduce a set of three mutually orthogonalunit vectors: the outward unit normal pointing into medium 2, en12, the unit tangentialvector eτ and unit bi-normal vector eb such that (see Fig. 2. 7.),

eb = en12 × eτ . (2.49)

Let us decompose all fields into normal and tangential components to the interface suchthat

E = En + Eτ , B = Bn + Bτ , (2.50)

with the similar expressions for D and H. It can be inferred from geometry that

En = en(E · en) (2.51)

andEτ = E− en(E · en) = en × (E× en), (2.52)

where “bac-cab” rule was used on the r.h.s of Eq. (2.52).Applying electric Gauss’s law, Eqs. (2.17), to the cylindrical Gaussian pillbox S

shown in Fig. 2.7 and taking the limit of a very shallow pillbox, we obtain∮S

dS en12 ·D = en12 · (D2 −D1)∆S =

∫dvρv = ρs∆S, (2.53)

where ρs is the surface charge density on the interface. It follows at once from Eq. (2.53)that

en12 · (D2 −D1)|s = ρs ; (2.54)

hereafter the subscript “s” implies that the corresponding fields are evaluated at theinterface. By the same token,

en12 · (B2 −B1)|s = 0 , (2.55)

because there are no magnetic charges. Eqs. (2.54) and (2.55) relate the normal com-ponents of the electric and magnetic flux densities on both sides of the interface.

Applying now the Faraday law (2.19) to the Stockesian loop C, we obtain in thelimit of a very small loop the expression∮

C

dl ·E = eτ · (E2 −E1)∆l = 0, (2.56)

since ∂tB is finite on the surface of C and the surface area vanishes as we shrink theloop sides. Thus,

E2τ −E1τ = 0, (2.57)

25

11 , BE

11 , HD

C

a

ba

ss J

,

22 , BE

22 , HDna

S1

2

Figure 2.8: Electromagnetic fields at the interface between the two homogeneous me-dia.

or, alternatively, with the help of Eq. (2.52),

en12 × [en12 × (E2 −E1)] = 0, (2.58)

implying for an arbitrary point of the surface that

en12 × (E2 −E1)|s = 0 . (2.59)

At the same time, the Ampere equation tells us that∮C

dl ·E = eτ · (H2 −H1)∆l = (eb × en12) · (H2 −H1)∆l =

= eb · [en12 × (H2 −H1)]∆l =

∫SC

dSeb · (J + ∂tD)

= eb · Js∆l (2.60)

where Js is the surface current at the interface. It can be inferred from (2.62) that

en12 × (H2 −H1)|s = Js . (2.61)

Eqs. (2.54), (2.55), (2.59) and (2.61) constitute the general boundary conditions in theelectromagnetic theory.Exercise 2.8. The upper (lower) half-space z > 0 (z < 0) contains a mediumwith the permeability µ1 (µ2). The magnetic field in the media is given by theexpression

H =

exH0e

−z/a, z > 0;eyH0e

z/a, z < 0,

26

where H0 and a are known constants. What is a surface current density, if any, atz = 0?Solution.—In this case, the interface corresponds to z = 0. We may then label z < 0as region “1”, and z > 0 as region “2”. It follows that en12 = ez and

H1|s = H0ey, H2|s = H0ex. (2.62)

We can the infer from Eqs. (2.61) and (2.62) that

Js = ez × (exH0 − eyH0) = [(ez × ex)︸ ︷︷ ︸ey

− (ez × ey)︸ ︷︷ ︸−ex

]H0. (2.63)

and finally,Js = H0(ex + ey). (2.64)

Exercise 2.9. Establish the electromagnetic boundary conditions at the interfaceof a dielectric and a perfect conductor.Solution.—In the perfect conductor, σ → ∞. The Ohm law, J = σE implies thenthat E = 0 inside the conductor to prevent unphysical infinite current magnitudes.Labelling the conductor as medium 2 and dielectric as 1, we can infer at once fromEqs. (2.54) and (2.59) that

en21 ×E1 = 0; ρs = en12 ·D1 , (2.65)

i.e., there can be no tangential component of the electric field on a perfect conduc-tor surface and the surface charge density on a perfect conductor is determined bythe normal component of the electric flux density just outside the surface.

2.6 Conservation laws in electromagnetic theoryWe will now examine two important conservation laws encountered in the electromag-netic theory: the charge and electromagnetic energy conservation. While the formeris a fundamental law of nature, independent of Maxwell’s equations, the latter is theirdirect consequence.

The charge conservation law states that charges cannot be created, nor can theybe annihilated. In a global sense, this statement implies that an overall charge withinany finite volume must be conserved. Therefore the time rate of change of the chargewithin the volume is equal to the current flux through the surface enclosing the volume,

d

dt

∫dvρv = −

∮dS · J . (2.66)

The minus sign in Eq. (2.66) indicates the fact that the charge within the volume de-creases (increases) if the current flows outside (inside) the volume, see Fig. 2.9.

27

)(a )(b

J

0dt

dQ 0dt

dQ

J

Figure 2.9: Illustrating the charge conservation law.

Assuming the volume in Eq. (2.66) is at rest, the local form of the charge conser-vation law follows from Eq. (2.66) on the application of the divergence theorem to ther.h.s,

d

dt

∫dvρv =

∫v

dv∂tρv = −∮dS · J = −

∫dv∇ · J. (2.67)

It follows at once from Eq. (2.67) that∫v

dv (∂tρv +∇ · J) = 0. (2.68)

As Eq. (2.68) is satisfied for any volume v, the local continuity equation follows

∂tρv +∇ · J = 0 . (2.69)

It follows at once that for a time-harmonic charge and current densities defined byEq. (2.40), the continuity equation reads

∇ · J(r) = jωρv(r) . (2.70)

Exercise 2.10. Assume a charged lump of density ρv(r, 0) is created inside a homo-geneous medium with the permittivity ε and conductivity σ. Determine the timeevolution of the lump, that is work out ρv(r, t).Solution.—It follows from the Ohm law and the constitutive relation for a homoge-neous medium that J = σE and D = εE. Combining this with the electric Gauss’slaw, Eq. (2.32), we obtain

∇ · J = σρv/ε. (2.71)

On substituting from Eq. (2.71) into the continuity equation (2.69), we arrive at

∂tρv + σε ρv = 0. (2.72)

28

The latter equation can be straightforwardly integrated, yielding the result

ρv(r, t) = ρv(r, 0)e−t/τ , (2.73)

where we used the initial condition ρv(r, t)|t=0 = ρv(r, 0) and introduced a charac-teristic charge relaxation time τ = ε/σ inside the medium. It follows at once fromEq. (2.73) that in the case of a poor conductor (or a good dielectric) σ → 0, imply-ing that τ → ∞, that is it takes forever for a charged lump to disappear in a gooddielctric. This is the reason we can assume good dielectrics to hold static (nearly time-independent) volume charges. In the opposite limit of a good conductor, σ → ∞,implying that τ → 0. In other words, a charged lump, created inside a very good con-ductor, momentarily disappears. This is the reason we assume that in the static case,all charges in a perfect conductor reside on its surface.

As a corollary, let us estimate relative strengths of the conduction and displacementcurrents induced in a homogeneous dielectric with permittivity ε and conductivity σ bya time-harmonic electric field, E = E0e

−jωt. By definition, J = σE and JD =ε∂tE = jεωE. It follows at once that

|J||JD|

=σ

ωε=

1

ωτ. (2.74)

Thus, the displacement current is dominant at high frequencies, ω 1/τ , while theconduction one dominates at low frequencies, ω 1/τ . Alternatively at a givenfrequency, displacement currents are negligible in good conductors, whereas one canneglect conduction currents in good dielectrics. In reality, the situation is more subtleas both σ and ε are, in general, frequency dependent with potentially far-reaching andinteresting consequences as we will learn when we discuss nano-atennae.

Let us now explore the electromagnetic energy propagation in a material medium.We assume that the medium is linear, homogeneous and isotropic as far as its electro-magnetic properties are concerned,

D = εE, B = µH, (2.75)

and the current obeys Ohm’s lawJ = σE. (2.76)

In view of Eqs. (2.75) and (2.76), Maxwell’s equations can be cast into the form

∇ ·E = ρv/ε, ∇ ·H = 0; (2.77)

∇×E = −µ∂tH (2.78)

and∇×H = σE + ε∂tE (2.79)

Next, taking a dot product of Eq. (2.78) with H and Eq. (2.79) with E, we obtain

H · (∇×E) = −µH · ∂tH = −µ2 ∂tH2 (2.80)

29

andE · (∇×H) = σE2 + εE · ∂tE = σE2 + ε

2∂tE2 (2.81)

Recalling the vector calculus identity,

∇ · (R×Q) = Q · (∇×R)−R · (∇×Q), (2.82)

for any vector fields R and Q, and choosing Q = H and R = E, and subtractingEq. (2.81) from Eq. (2.80), we arrive at

∇ · (E×H) = −∂t(

12εE

2 + 12µH

2)− σE2. (2.83)

We can now re-write Eq. (2.83) as

∂twem +∇ ·P = −σE2 , (2.84)

where we introduced the elecromagnetic energy density in J/m3 as

wem = 12εE

2 + 12µH

2 , (2.85)

and the so-called Poynting vector representing the instantaneous electromagneticpower flow, i.e., the electromagnetic power flowing per unit cross section in the medium,W/m2 by the expression

P = E×H . (2.86)

Eq. (2.84) represents the eleectromagnetic energy conservation in the local form.The right-hand side describes local ohmic losses per unit volume.

Integrating Eq. (2.83) over the volume and using the divergence theorem on thel.h.s, we obtain∮

S

dS · (E×H) = − d

dt

∫v

dv

(1

2εE2 +

1

2µH2

)−∫v

dv σE2. (2.87)

Finally, rearranging terms we obtain the electromagnetic energy conservation law inthe global form as

d

dt

∫v

dv wem = −∮S

dS · P −∫v

dvσE2 . (2.88)

The second term on the r.h.s. of Eq. (2.88) describes global ohmic losses within agiven volume. Thus, the electromagnetic energy conservation law asserts that the elec-tromagnetic energy inside a finite volume can only change if the energy flows in or outof the volume through its surface and is lost inside to ohmic losses. Note the conserva-tion law (2.88) is a direct consequence of Maxwell’s equations.

In the case of time harmonic fields defined by Eqs. (2.41) and (2.42), it makes senseto introduce a time-averaged Poynting vector which defines a power flow averaged overa time period via

〈P(r)〉 =1

T

∫ T

0

dtP(r, t), (2.89)

30

where T = 2π/ω is a period associated with the frequency ω. Using the fact that forany complex number, Re(z) = (z + z∗)/2, we can rewriting Eq. (3.5) as

E(r, t) = 12

[E(r)e−jωt + E∗(r)ejωt

], (2.90)

andH(r, t) = 1

2

[H(r)e−jωt + H∗(r)e−jωt

]. (2.91)

We can then obtain for an instantaneous Poynting vector the expression

P = 14 (E×H∗ + E∗ ×H)

+ 12

[(E×H)e−2jωt + (E∗ ×H∗)e2jωt

], (2.92)

where we dropped the spatial argument for brevity. It follows from Eqs. (2.89) and (2.92)that the last two terms on the r.h.s. of Eq. (2.92) average to zero as complex exponen-tials oscillate at twice the frequency and are averaged over the full oscillation period.As a result, an average Poynting vector of time-harmonic electromagnetic fields takesthe form

〈P〉 =1

2Re(E×H∗) . (2.93)

31

Chapter 3

Introduction to antenna theory.Vector and scalar potentials.

3.1 Antenna: the concept and implementation

Figure 3.1: Antenna as a transducer between a transmission line and free space. Re-produced from C. A. Blanis Antenna Theory: Analysis & Design (Wiley, 2016).

As time-dependent charges or currents generate electromagnetic fields, the fieldsradiate away from the sources in the form of electromagnetic waves. The processof efficient electromagnetic energy transmission away from the source is in practiceaccomplished with the aid of antannas. In many cases, the sources—that is, time-dependent charges or currents—are themselves excited by external electromagneticfields transported to the region by waveguides or transmission lines. Thus, an antennacan be viewed as a transducer facilitating the match between electromagnetic waves ofthe waveguide or transmission line and the electromagnetic waves radiated away into

32

free space, see Fig. 3.1. The antannas serve two major purposes:

1. Efficient radiation of electromagnetic waves into free space;

2. Impedance matching between energy supplying waveguides/transmission linesand free space to maximize energy throughput.

The antannas can operate in a transmitting or receiving mode, as is illustrated in thesketch below. A specific antenna design heavily depends on a particular application; weare listing below several popular configurations, including elemental electric dipoles,magnetic dipole loops, helical antannas, horn antannas and satellite dishes, see Fig.3.3.

Figure 3.2: Antenna as (a) transmitter and (b) receiver. Reproduced from C. A. BlanisAntenna Theory: Analysis & Design (Wiley, 2016).

Figure 3.3: Some basic antenna configurations. Reproduced from C. A. Blanis AntennaTheory: Analysis & Design (Wiley, 2016).

For nano-scale applications, nano-antannas are commonly used. A simplest nano-antenna example is served by a small spherical nano-particle attached to the tip of a

33

tapered waveguide or a fiber. In the transmitting mode, the electromagnetic waves inthe taper excite the nano-antenna. If such a radiating nano-antenna is brought into thevicinity of a probe atom or molecule, the nano-antenna excites the probe causing itto radiate. By detecting the probe radiation, we can learn the atom/molecule opticalproperties. Alternatively, whenever one brings a taper with the nano-antenna into thevicinity of an excited atom or molecule, the probe field excites the nano-antenna, caus-ing it to radiate electromagnetic waves. The radiated waves, captured within the taper,can propagate toward a detection device and the information on the excited probe isgathered. This is a receiving mode of the nano-antenna. Both modes are schematically

Figure 3.4: Receiver/transmitter mode of a nano-antenna. Reproduced from P. Bharad-waj et. al, Adv. in Opt. and Photon., 1, 438-483, (2009).

illustrated in the figure above.Hereafter, we will focus on radiating nano-antenna properties such as the radiated

field strength, angular distribution, polarization etc. The receiving properties of theantannas can be deduced from their transmitting ones by the virtue of the so-calledreciprocity theorem that states: The angular distribution of an antenna for reception isidentical to that for transmission.

In the following, we proceed to study both the nature of electromagnetic wavespropagating in free space and the process of their generation by oscillating antennacurrents/charges. We start by treating plane electromagnetic wave propagation in freespace

3.2 Plane electromagnetic waves in free spaceIn the absence of charges and currents, Maxwell’s equations in free space take the form

∇ ·E = 0, (3.1)

∇ ·H = 0, (3.2)

∇×E = −µ0∂tH (3.3)

and∇×H = ε0∂tE. (3.4)

34

We look for plane-wave solutions to the Maxwell equations in the form

E(r, t) = ReE0ej(k·r−ωt), H(r, t) = ReH0e

j(k·r−ωt). (3.5)

These solutions are called plane waves because in any plane transverse to the propa-gation direction of the wave, specified by the wave vector k, the electric and magneticfields magnitudes are constants. Indeed, in any transverse plane, by definition, k ·r = 0and |E| = |E0| = const, and by the same token, |H| = |H0| = const. By linearityof Maxwell’s equations in free space, we can drop the real part and deal with com-plex phasors describing the waves directly. The real part can be taken at the end of allcalculations to yield physical (real) electric and magnetic fields of a plane wave.

To proceed, we require the following relations.Exercise 3.1. Show that for a plane wave given by Eq. (3.5), ∇ · E = jk · E and∇×E = jk×E.Solutions.—In the Cartesian coordinates,

∇ ·E = (E0x∂x + E0y∂y + E0z∂z) ej(kxx+kyy+kzz)e−jωt

= j(kxE0x + kyE0y + kzE0z)ej(k·r−ωt) = jk ·E. (3.6)

The second relation is proven by analogy using the Cartesian coordinate representationof the curl.The Maxwell equations in the plane-wave form can then be rewritten as

k ·E0 = 0 , (3.7)

k ·H0 = 0 , (3.8)

k×E0 = ωµ0H0 , (3.9)

andk×H0 = −ωε0E0 . (3.10)

In Eqs. (3.7) – (3.10) we dropped plane-wave phasors on both sides.Next, we can exclude the magnetic field from the fourth Maxwell equation leading

tok× (k×E0) = −ε0µ0ω

2E0. (3.11)

Using the “bac-cab” rule on the l.h.s of Eq. (3.11), we arrive at

k(k ·E0)− k2E0 = −ε0µ0ω2E0. (3.12)

With the aid of Eq. (3.7), we obtain

(k2 − µ0ε0ω2)E0 = 0, (3.13)

implying thatk = ω

√ε0µ0 = ω/c (3.14)

35

where we introduced the speed of light in vacuum

c =1

√ε0µ0

= 3× 108 m/s. (3.15)

Equation (3.14) is a dispersion relation for plane electromagnetic waves in freespace; it relates the wave number to the wave frequency. The complex amplitudes E0

and H0—which determine the directions of E and H—are not independent, but arerelated by the Maxwell equations (3.9) or (3.10). For instance, from the knowledge ofE0 one can determine H0 using Eq. (3.9),

H0 =(ek ×E0)

η0, (3.16)

where ek = k/k and η0 is the free space impedance defined as

η0 =

õ0

ε0= 120π ' 377 Ω. (3.17)

By the same token, E0 can be inferred from H0 with the help of Eq. (3.10):

E0 = −η0(ek ×H0) . (3.18)

Exercise 3.2. Show that E0, k and H0 are mutually orthogonal for a plane wavein free space.Solution.— It follows at once from the Maxwell equations, Eq. (3.7) and (3.8) thatE0⊥k and H0⊥k. Taking a dot product of Eq. (3.9), say, with E0 we obtain, E0 ·(k× E0) = (E0 × E0) · k = 0 = ωµ0(E0 ·H0). It follows that E0 ·H0 = 0. Thus,E0⊥H0⊥k. See Fig. 3.5.

K

E

Η

Figure 3.5: Mutual orientation of E, H and k of a plane wave propagating in freespace.

Definition. The time evolution of the electric field vector is called polarization.Let us consider a plane wave propagating along the z-axis in free space. As, k = kez ,

36

and E⊥k, the electric field in the phasor form reads

E(z, t) = Re(ex|E0x|ejφ0x + ey|E0y|ejφ0y )ej(kz−ωt), (3.19)

We will now show that, in general, the tip of the electric field vector moves aroundan ellipse as the time evolves. This general polarization is called elliptic. To proceed,we rewrite the complex amplitude in the rectangular form as

E0xex + E0yey = (ex|E0x| cosφ0x + ey|E0y| cosφ0y)︸ ︷︷ ︸U

+ j (ex|E0x| sinφ0x + ey|E0y| sinφ0y)︸ ︷︷ ︸V

. (3.20)

Note that U and V are not orthogonal which makes the situation tricky. We can how-ever introduce a transformation from U and V to u, v involving an auxiliary parameterθ such that

U + jV = (u + jv)ejθ, (3.21)

It follows at once from Eq. (3.21) that

U = u cos θ − v sin θ, V = u sin θ + v cos θ. (3.22)

Inverting Eqs. (3.22), we obtain

u = U cos θ + V sin θ, v = U sin θ −V cos θ. (3.23)

We can now use our freedom to choose θ wisely. In particular, choosing it such thatu · v = 0 (orthogonal axes), we obtain by taking the dot product of u and v,

tan 2θ =2U ·VU2 − V 2

=⇒ θ =1

2tan−1

(2U ·VU2 − V 2

). (3.24)

Here we made use of the trigonometric identities, sin 2θ = 2 sin θ cos θ and cos 2θ =cos2 θ − sin2 θ. By combining Eqs. (3.20) and (3.21), we can rewrite our field as

E(z, t) = Re(u + jv)ej(kz−ωt+θ). (3.25)

Using the orthogonality of u and v, we can write the two orthogonal components ofthe field, Eu and Ev as

Eu = u cos(ωt− kz − θ), Ev = v sin(ωt− kz − θ). (3.26)

It follows from Eq. (3.26) thatE2u

u2+E2v

v2= 1, (3.27)

where u and v are given by Eq. (3.23) and θ by Eq. (3.24). Eq. (3.27) manifestlyrepresents an ellipse with the semi-major axis making the angle θ with the x-axis as isshown in Fig. 3.6. The tip of E can move either clockwise or counterclockwise alongthe ellipse; depending on the direction of motion of E, the polarization is left-hand

37

E

vE

uE

Figure 3.6: Illustrating elliptic polarization.

E

xE

yE

xa

ya

Figure 3.7: Illustrating plane polarization.

or right-hand elliptical. In the left-hand (right-hand) elliptical polarization, the fingersof your left (right) hand follow the direction of rotation and the thumb points to thewave propagation direction. Thus, for a general elliptic polarization, the electric fieldamplitude takes the form

E(z, t) = ex|E0x| cos(kz − ωt+ φ0x) + ey|E0y| cos(kz − ωt+ φ0y) . (3.28)

Although, in general, the electric field is elliptically polarized, there are two impor-tant particular cases.Definition. The electric field is said to be linearly polarized if the phases of two or-thogonal components of the field in Eq. (3.19) are the same, φ0x = φ0y .In this case,

E(z, t) = (ex|E0x|+ ey|E0y|) cos(kz − ωt+ φ0) , (3.29)

and the electric field is always directed along the line making the angle

θ = tan−1(E0y/E0x) (3.30)

38

with the x-axis as is shown in Fig. 3.7.Definition. If the phases of the two orthogonal components in Eq. (3.20) differ byπ/2, and |E0x| = |E0y|, the wave is said to be circularly polarized.In this case

E(z, t) = |E0|[ex cos(kz − ωt+ φ0)∓ ey sin(kz − ωt+ φ0)] . (3.31)

In a circularly polarized wave, the E has the same magnitude but is moving along

E

xE

yE

o

Figure 3.8: Illustrating circular polarization.

the circle. In the case of “-” sign in Eq. (3.31), E moves counterclockwise around thecircle and the wave is left circularly polarized; for the “+” sign it is right circularlypolarized.Exercise 3.3. Determine the polarization of the electromagnetic wave E(x, t) =(eyA− ezB) sin(kx− ωt).Solution.—The wave is linearly polarized at the angle θ = tan−1(A/B) with the z-axis; it propagates in the positive x-direction.

Consider the power flow associated with a plane wave, specified by the Poyntingvector, P = E × H. In general, both fields oscillate with rather high frequenciessuch that a more sensible—and actually detectable—measure of the power flow is thetime-averaged Poynting vector, defined as

〈P(r)〉 =1

T

∫ T

0

dtP(r, t), (3.32)

where T is the wave period. Using the fact that for any complex number, Re(z) =(z + z∗)/2, we can re-write Eq. (3.5) as

E =1

2

[E0e

j(k·r−ωt) + E∗0e−j(k·r−ωt)

]; H =

1

2

[H0e

j(k·r−ωt) + H∗0e−j(k·r−ωt)

].

(3.33)

39

We can then obtain for the instantaneous Poynting vector the expression

P =1

4(E0 ×H∗0 + E∗0 ×H0)

+1

2

[(E0 ×H0)e2j(k·r−ωt) + (E∗0 ×H∗0)e−2j(k·r−ωt)

]. (3.34)

It follows from Eqs. (3.32) and (3.34) that the last two terms on the r.h.s. of Eq. (3.34)average to zero as complex exponentials oscillate at twice the frequency and are aver-aged over a full period of the wave. As a result, the average Poynting vector of anyplane wave, regardless of its polarization, takes the form

〈P〉 =1

2Re(E0 ×H∗0) . (3.35)

Exercise 3.4. Show that the instantaneous Poynting vector of a circularly polar-ized plane wave in free space is independent of either time or the propagationdistance.Solution.—For a circularly polarized plane wave, propagating in the positive z-direction,say,

E = |E0|[ex cos(kz−ωt+φ0)±ey sin(kz−ωt+φ0)] = Re[E0(ex ∓ jey)ej(kz−ωt)

],

(3.36)implying that

E0 = E0(ex ∓ jey). (3.37)

Applying Eq. (3.16), with Eq. (3.37), we obtain

H0 =(ez ×E0)

η0=E0

η0(ey ± jex).

It follows that

H = Re[E0

η0(ey ± jex)ej(kz−ωt)

]=|E0|η0

[ey cos(kz − ωt+ φ0)∓ ex sin(kz − ωt+ φ0)]. (3.38)

Using Eqs. (3.36) and (3.38), we obtain

P = E×H =|E0|2

η0[(ex × ey) cos2(kz − ωt+ φ0)

− (ey × ex) sin2(kz − ωt+ φ0)] = ez|E0|2

η0. (3.39)

40

3.3 Scalar and vector potentials for time-harmonic fieldsConsider time-harmonic fields defined by Eqs. (2.41) and (2.42). We assume thatwe have time-harmonic charge and current sources in free space and recall the time-harmonic Maxwell equations

∇ ·E = ρv/ε0, ∇ ·H = 0; (3.40)

and∇×E = jωµ0H, (3.41)

as well as∇×H = J− jωε0E. (3.42)

Here we eliminated D and B in favor of E and H using the free space definitionsD = ε0E and B = µ0H; for the brevity sake, we will hereafter drop the explicitdependence of the time-harmonic field amplitudes on the space coordinates wheneverit does not cause confusion.

To tackle radiation problems, it proves convenient to introduce a vector potential

A(r, t) = 12 [A(r)e−jωt + c.c.], (3.43)

such thatB(r) = ∇×A(r),=⇒ H(r) = µ−1

0 ∇×A(r) . (3.44)

Exercise 3.5. Show that the magnetic field of Eq. (3.44) automatically satisfies themagnetic Gauss’s equation, .i.e.,∇ ·H = 0.Solution—Introducing an auxiliary vector field F = ∇×A; its Cartesian componentsread

Fx = ∂yFz − ∂zFy, (3.45)

Fy = ∂zFx − ∂xFz, (3.46)

andFz = ∂xFy − ∂yFx. (3.47)

It then follows from the definition,

∇ · F = ∂xFx + ∂yFy + ∂zFz = ∂x(∂yFz − ∂zFy) +

+∂y(∂zFx − ∂xFz) + ∂z(∂xFy − ∂yFx) =

∂2xyFz − ∂2

xzFy + ∂2yzFx − ∂2

yxFz + ∂2zxFy − ∂2

zyFx = 0. (3.48)

This follows by noticing that mixed second-order partials are the same. Thus,∇ ·H =µ−1

0 ∇·F = 0, and the result follows from the divergence independence of a coordinatesystem.Exercise 3.6. Show that for any scalar field F , ∇ × (∇F ) = 0, thereby showingthat the vector potential A′ = A±∇F gives the same magnetic field as A.Solution—Let us apply Stokes’s theorem to the vector field ∇F and an arbitrary opensurface S, yielding,∫

dS · ∇ × (∇F ) =

∮dl · ∇F =

∮dF = F (f)− F (i) = 0, (3.49)

41

where the final “f” and initial “i” points coincide for a closed path and we used arelation between the gradient and infinitesimal change in a scalar field. Alternativelyin the Cartesian coordinates,

∇× (∇F ) =

∣∣∣∣∣∣ex ey ez∂x ∂y ∂z∂xF ∂yF ∂zF

∣∣∣∣∣∣= ex(∂2

yzF − ∂2zyF )− ey(∂2

xzF − ∂2zxF ) + ez(∂

2xyF − ∂2

yxF ) = 0,(3.50)

and the result follows from the symmetry of cross partials—∂2xy ≡ ∂2

yx etc—and thecurl independence of a coordinate system.On substituting from Eq. (3.44) into Eqs. (3.41) and (3.42) we arrive at

∇× (∇×A) = µ0J−jω

c2E, (3.51)

and∇×E = jω∇×A. (3.52)

It follows at once from Eq. (3.52) that we can introduce a scalar potential

V (r, t) = 12 [V (r)e−jωt + c.c.], (3.53)

such thatE = jωA−∇V . (3.54)

On substituting from Eq. (3.54) into Eqs. (3.40) and (3.51), we can eliminate the fieldscompletely in favor of the potentials, yielding the equations

∇× (∇×A)− ω2

c2 A = µ0J− jωc2∇V, (3.55)

as well as−∇2V + jω∇ ·A = ρv/ε0. (3.56)

Recall now the definition of a Laplacian of a vector field,

∇2A ≡ ∇(∇ ·A)−∇× (∇×A). (3.57)

With the aid of Eq. (3.57), Eqs. (3.55) and (3.56) can be cast into the form

∇2A + ω2

c2 A = −µ0J +∇(∇ ·A− jω

c2 V), (3.58)

and∇2V − jω∇ ·A = −ρv/ε0. (3.59)

Exercise 3.7. Show that in the Cartesian coordinates,∇2A = ex∇2Ax+ey∇2Ay+ez∇2Az .Solution.—Because of the form symmetry in the Cartesian coordinates, it will suffice toprove the assertion for a particular coordinate, x, say. To this end,

∇(∇ ·A) = ex∂x(∂xAx + ∂yAy + ∂zAz) + . . . (3.60)

42

The Cartesian components of∇×A read

(∇×A)x = ∂yAz − ∂zAy, (3.61)

(∇×A)y = ∂zAx − ∂xAz, (3.62)

and(∇×A)z = ∂xAy − ∂yAx. (3.63)

It follows that

∇× (∇×A) =

∣∣∣∣∣∣ex ey ez∂x ∂y ∂z

∂yAz − ∂zAy ∂zAx − ∂xAz, ∂xAy − ∂yAx

∣∣∣∣∣∣= ex[∂y(∂xAy − ∂yAx)− ∂z(∂zAx − ∂xAz)] + . . . (3.64)

Hence,

∇2A ≡ ∇(∇ ·A)−∇× (∇×A) = ex[∂2xxAx + ∂2

xyAy + ∂2xzAz

−(∂2yxAy − ∂2

yyAx − ∂2zzAx + ∂2

zxAz)] + . . .

= ex(∂2xxAx + ∂2

xyAy + ∂2xzAz − ∂2

yxAy + ∂2yyAx + ∂2

zzAx − ∂2zxAz) + . . .

= ex(∂2xx + ∂2

yy + ∂2zz)Ax + . . . = ex∇2Ax + . . . . (3.65)

Thus, we conclude that

∇2A = ex∇2Ax + ey∇2Ay + ez∇2Az. (3.66)

Further, as we have just shown the vector potential is not uniquely defined: anyvector potential up to a gradient of a scalar field gives the same magnetic field. We canuse this so-called gauge freedom, to simplify the form of the wave equations in termsof the potentials. One popular choice is the Lorentz gauge defined by

∇ ·A− jωc2 V = 0. (3.67)

If one utilizes the Lorentz gauge, the wave equations, Eqs. (3.58) and (3.59), take on aparticularly simple symmetric form, effectively decoupling the two potentials,

∇2A + k2A = −µ0J, (3.68)

and∇2V + k2V = −ρv/ε0, (3.69)

where k = ω/c. Eqs. (3.68) and (3.69) are known as vector and scalar inhomogeneousHelmholtz equations, respectively. A general solution to the scalar Helmholtz equa-tions, describing a scalar potential generated by a time-harmonic charge source has theform

V (r) =

(1

4πε0

)∫dv′ ρv(r

′)ejk|r−r

′|

|r− r′|. (3.70)

43

Note that in the static case, k = 0, Eq. (3.70) reduces to a familiar result from the fieldscourse,

V (r) =

(1

4πε0

)∫dv′

ρv(r′)

|r− r′|. (3.71)

To find the vector Helmholtz equation solution, we notice that in the Cartesiancoordinates, Eq. (3.68) can be written separately for each component with the help ofEq. (3.66), i.e., for the x-component, we obtain

∇2Ax + k2Ax = −µ0Jx, (3.72)

On comparing Eq. (3.69) and (3.72) and using Eq. (3.70), we infer that the solution forthe x component of A reads,

Ax(r) =µ0

4π

∫dv′ Jx(r′)

ejk|r−r′|

|r− r′|. (3.73)

By symmetry, the other Cartesian components of A take the same form, implying that

A(r) =µ0

4π

∫dv′ J(r′)

ejk|r−r′|

|r− r′|. (3.74)

This general expression for the vector potential can now be evaluated using any conve-nient coordinate system, Cartesian or curvilinear. We also mention in passing that insome cases, for instance with perfect conductors involved, the currents can flow onlyon the conductor surface. We can then introduce a surface current density Js, measuredin Amps per meter. In these cases, Eq, (3.74) is replaced with

A(r) =µ0

4π

∫dS′ Js(r

′)ejk|r−r

′|

|r− r′|, (3.75)

where the integration is extended over the conductor surface.In many cases, one deals with the currents flowing in fine cross-section conductors

where the detailed knowledge of a current distribution across the conductor is irrele-vant, and a filamentary current approximation is valid. Under this approximation, weassume a characteristic transverse conductor size δ to be so small, δ |r′| that thefactor ejk|r−r

′|/|r − r′| can be assumed constant across the conductor. Writing theelementary volume dv = dSdl, where dl is an infinitesimal length of the conductingfilament, it follows that

A(r) =µ0

4π

∫dl′

ejk|r−r′|

|r− r′|

∫dS′J(r′). (3.76)

Next, writing J = el(J · el), where el is a local unit tangential vector to the filament,we observe that∫

dS′J(r′) = el

∫dS′(J · el) = el

∫dS′ · J = Iel, (3.77)

44

where dS′ = eldS′ is an oriented infinitesimal element of the surface and we introduce

the total current by definition, cf. Eq. (2.8). On substituting from Eq. (3.77) intoEq. (3.76) and introducing an oriented infinitesimal line element, dl = eldl, we arriveat the expression for the vector potential induced by a filamentary current distributionas

A(r) =µ0

4π

∫dl′ I(l′)

ejk|r−r′|

|r− r′|. (3.78)

Here I(l′) is, in general, an inhomogeneous current along the filamentary conductor.Finally, once the vector potential has been determined using either Eq. (3.74) or (3.78)

depending on the type of the current source, the magnetic field can be readily workedout from Eq. (3.44)

H =1

µ0(∇×A) , (3.79)

It then follows that outside the source, J = 0, the electric field can be obtained fromEq. (3.42) viz,

E =j

ωε0(∇×H) . (3.80)

In principle, Eqs. (3.74), (3.78), (3.79), and (3.80) solve the problem of determiningthe electromagnetic fields radiated by any time-harmonic source. Alternatively, we canfind the potentials A and V independently from Eqs. (3.74) and (3.70), respectively,and the fields from Eqs. (3.79) and (3.54). This is usually a longer route, though.

3.4 Electric dipole moment of any charge distributionWe start by recalling the concept of an elemental dipole consisting of two equal andopposite point charges, separated by a finite distance. As the spatial distributions oftime-harmonic electromagnetic fields do not depend on their frequency, we consider,for simplicity, the static limit, ω = 0. Assume further a negative charge is located atthe origin and the positive is at a position r′. The scalar potential due to the elementaldipole at any point r follows from Coulomb’s law and the superposition principle,

V (r) = − Q

4πε0r+

Q

4πε0|r− r′|. (3.81)

Far away from the dipole, r′ r, we can approximate

|r− r′| =√

(r− r′) · (r− r′) '√r2 − 2(r · r′) ' r(1− r · r′/r2). (3.82)

Here we used a Taylor series expansion to the lowest order in the small parameter,r′/r 1; in particular, (1 + x)n ' 1 + nx for x 1. It then follows at once fromEq. (3.82) that

|r− r′|−1 ' r−1(1 + r · r′/r2). (3.83)

45

Figure 3.9: Illustrating the concept of an elemental dipole.

On substituting from Eq. (3.83) into (3.81) we arrive at

V (r) =Q

4πε0

[−1

r+

1

r

(1 +

r · r′

r2

)]=

p · r4πε0r3

, (3.84)

wherep = Qr′, (3.85)

is a dipole moment of such an elemental dipole.Consider now a generic localized neutral charge distribution such that

Qtot =

∫dvρv(r) = 0. (3.86)

In the static limit, it follows from Eqs. (3.71), (3.83) and (3.86) that the scalar potentialfar away from the charge, |r′| |r|, is given by the expression

V (r) '(

1

4πε0r

)∫dv′ρv(r

′)

(1 +

r · r′

r2

)=

p · r4πε0r3

, (3.87)

where

p =

∫dv′r′ρv(r

′) , (3.88)

is a dipole moment of the neutral charge distribution ρv . Eq. (3.88) should be viewedas a natural generalization of Eq. (3.85) to a localized continuous charge distributionpolarized by an external electric field. This definition will be useful in our discussionof nanoantannas and their interaction with single atoms and/or molecules.

46

Figure 3.10: Dipole moment of a localized neutral charge distribution.

Let us now show that an elemental filamentary current can be viewed as a dipoleas well and let us calculate the corresponding dipole moment. To this end, consider anelemental filamentary current of the length ∆l . We assume the filament so short thatwe can neglect the change in the unit tangential vector direction along the filament. Letus choose the z-axis of our coordinate system to lie along the unit tangential filamentvector . Hence,

J = ez(J · ez) = ezJz. (3.89)

Consider now an integral,∫dv′J(r′) =

∫dl

∫dSJ(r) =

∫dzIez = I∆lez. (3.90)

Here we used Eqs. (3.77) and (3.90) on the right-hand side of Eq. (3.90). On the otherhand, ∫

dv′ezJz = ez

∫dS′

∫ ∞−∞

dz′Jz, (3.91)

Integrating by parts on the right-hand side, we can transform the inner integral as∫ ∞−∞

dzJz = z′∂z′Jz|∞−∞︸ ︷︷ ︸=0

−∫ ∞−∞

dz′z′∂z′Jz. (3.92)

The first term is zero because the current and charge distributions are localized so that

47

Figure 3.11: a) Small nanoparticle and b) elemental filamentary current as effectivedipole sources.

Jz → 0 as z′ → ±∞. By charge conservation (continuity equation), specified to ourcase, we have

∂zJz = jωρv, (3.93)

On combining Eqs. (3.90) through (3.93), we finally arrive at

I∆lez =

∫dv′ezJz = −jωez

∫dv′z′ρv. (3.94)

Therefore,I∆lez = −jωp, (3.95)

wherep ≡ ez

∫dv′z′ρv(r

′), (3.96)

is recognized as a dipole moment in the z-direction. Thus, as far as radiation problemsare concerned, an elemental filamentary current can be viewed as a dipole with aneffective dipole moment given by

p = jI∆lω ez . (3.97)

Exercise 3.8 Determine the line charge distribution on a straight filamentary con-ductor aligned with the z-axis which carries a time-harmonic current I(z, t) =I0 cosβz e−jωt.Solution.—The continuity equation for time-harmonic charges and currents impliesthat ∇ · J(r) = jωρv(r). As the wire is straight, the current density has only a z-component, so that J(r) = Jz(z) ez , implying that ∇ · J = ∂zJz . Introducing the linecharge density, ρl =

∫dS ρv and integrating the continuity equation with respect to

the wire cross-section, we obtain at once

jωρl =

∫dS∂zJz = ∂z

∫dSJz︸ ︷︷ ︸

=I(z)

= ∂zI(z). (3.98)

48

It follows at once from Eq. (3.98) that

ρl = jβI0ω sinβz (3.99)

3.5 Elemental electric dipole antenna

Figure 3.12: Radiating elemental electric dipole.

We now discuss the electromagnetic fields radiated by an elemental filamentarycurrent. We take our z-axis directed along the current such that the elemental currentreads I∆lez . As we just showed, the elemental current can be viewed as an elementaldipole, schematically sketched in the figure. Considering the localized nature of thesource, we can determine the vector potential far away from the source, |r′| |r|using Eq. (3.78). In this approximation, Eq. (3.78) in the leading order in the smallparameter r′/r 1 readily reduces to

A ' µ0I∆l

4πrejkrez. (3.100)

Eq. (3.100) can be re-writtten in the spherical coordinates as

A =µ0I∆l

4π

(ejkr

r

)(er cos θ − eθ sin θ). (3.101)

The radiating dipole magnetic field follows from Eqs. (3.79) and (3.101),

H =I∆l

4πr2 sin θ

∣∣∣∣∣∣∣er r eθ r sin θ eφ∂r ∂θ ∂φ(

ejkr

r

)cos θ −ejkr sin θ 0

∣∣∣∣∣∣∣ . (3.102)

Thus,

H =I∆l

4πr2 sin θr sin θ

(−jkejkr sin θ + sin θ e

jkr

r

)eφ, (3.103)

49

After elementary algebra, Eq. (3.103) simplifies to

H =kI∆l

4π

(ejkr

r

)sin θ

(−j +

1

kr

)eφ. (3.104)

It then follows from Eqs. (3.80) and (3.104) that

E =jkI∆l

4πε0ωr2 sin θ

∣∣∣∣∣∣er r eθ r sin θ eφ∂r ∂θ ∂φ0 0

(−j + 1

kr

)ejkr sin2 θ

∣∣∣∣∣∣ . (3.105)

It follows that

E =jI∆l

4πε0c

1

r2 sin θ

−eθr∂r

[ejkr

(−j +

1

kr

)]+ ere

jkr

(−j +

1

kr

)2 sin θ cos θ

.

(3.106)Doing an elementary derivative on the r. h. s. of Eq. (3.106), we obtain, after simplealgebra, the general expression for the electric field radiated by the dipole as

E =jI∆l/c

4πε0r2ejkr

[2 cos θ

(−j +

1

kr

)er − kr sin θ

(1 +

j

kr− 1

k2r2

)eθ

].

(3.107)Let us now discuss the radiated field behavior in two important limiting cases: near-

zone, where kr 1 and far zone, defined by kr 1. In the near field, we can obtainfrom Eq. (3.104) and (3.107)

H =I∆l sin θ

4πr2eφ, (3.108)

which is reminiscent of the magnetostatic field a distance r away from an elementalfilamentary current I∆lez located at the origin, c.f., the Bio-Savart law of magneto-statics. By the same token,

E =jI∆l/ω

4πε0r3(2er cos θ + eθ sin θ). (3.109)

It then follows from Eqs. (3.97) and (3.109) that

E =p

4πε0r3(2er cos θ + eθ sin θ), (3.110)

which has the same spatial distribution as that of an electrostatic field due to a dipolewith the moment p = jI∆l/ω located at the origin. We can infer from Eqs. (3.108)and (3.110) that

|E| ∝ η0

kr|H|, (3.111)

implying that in the near-zone behavior at kr 1 is quasi-static, dominated by anelectrostatic field. These results will be very useful in our discussion of nanoantannas.

In the far zone, kr 1, on the other hand, Eqs. (3.104) and (3.107) simplify to

H = −ω2p sin θ

4πc

(ejkr

r

)eφ , (3.112)

50

and

E = −ω2p sin θ

4πcη0

(ejkr

r

)eθ , (3.113)

respectively, implying that the far-fields satisfy the relation,

E = −η0(er ×H), (3.114)

where we employed the relation eφ = er × eθ. We notice that while the nanonanettaeapplications are mainly concerned with near-field properties of dipole radiators, themacroscopic applications—radio, TV, microwave, etc,—require the knowledge of far-field radiation patterns. We also observe that the far-zone angular distribution of dipoleradiation is dominated by the factor sin θ, implying that the radiation maximum takesplace at the angle θ = π/2 to the (charge) current oscillation direction and there is noradiation in the current oscillation direction. This pattern is sketched in the followingfigure.

Figure 3.13: Angular pattern of an elemental electric dipole antenna.

Further, the radial distribution of the far-zone electromagnetic fields of a radiatingdipole is described by an outgoing spherical wave

E(r, t) ∝ eθej(kr−ωt)

r; H(r, t) ∝ eφ

ej(kr−ωt)

r, (3.115)

which has concentric spheres as wavefronts, kr − ωt = const, with the wavefrontradius r growing with time t ≥ 0. Note that the electromagnetic field magnitudes areconstant on the surface of any sphere of radius r, making it a spherical wave. At thesame time, the amplitudes of the electric and magnetic fields in the far-zone in anyradiation direction θ are related through the free-space impedance, η0, just like the cor-responding quantities of a plane electromagnetic wave, c.f., Eqs. (3.18) and (3.114).

51

Hence, the radiation of an elemental dipole travels as a plane wave in any given propa-gation direction θ.

Let us now generalize our results to the case of an arbitrarily oriented elementaldipole. We assume that an elemental current direction is specified by the unit vectorep—implying that the correspnoding effective dipole moment can be written as p =pep—which immediately gives us a generalization of Eq. (3.100) as

A =µ0I∆l

4π

(ejkr

r

)ep. (3.116)

It then follows from Eqs. (3.79) and (3.116) that

H =I∆l

4π∇×

(epejkr

r

). (3.117)

We will work out the curl in a general vector form using the following vector calculusidentity, valid for any scalar ψ and vector Q fields,

∇× (ψQ) = ψ∇×Q +∇ψ ×Q. (3.118)

In our case, we identify, Q = ep and ψ = ejkr/r. Since ep is a constant unit vector, itfollows from Eqs. (3.117) and (3.118) that

H =I∆l

4π∇(ejkr

r

)× ep. (3.119)

Further, in the far zone,

∇(ejkr

r

)' jk e

jkr

rer. (3.120)

Combining Eqs. (3.119) and (3.120) and using the definition of p, Eq. (3.97), we arriveat the final answer

H =ω2p

4πc

(ejkr

r

)(er × ep) , (3.121)

Since the far-zone electric and magnetic fields are locally those of a plane wave, Eqs. (3.114)and (3.121) immediately give us the electric field in the far zone as

E =ω2p

4πcη0

(ejkr

r

)(er × ep)× er . (3.122)

Thus, we derived the expressions for the radiating dipole electromagnetic fields in thefar-zone in a form independent of a particular coordinate system.Exercise 3.9 Derive Eq. (3.122) directly from Eqs. (3.121) and (3.80).Solution—To take the curl in a general form, we use the identity (3.118) with ψ =ejkr/r and Q = er × ep. It then follows at once that

E = jωp

4πε0c∇(ejkr

r

)× (er × ep) = − ω2p

4πε0c2

(ejkr

r

)er × (er × ep)

=ω2p

4πε0c2

(ejkr

r

)(er × ep)× er. (3.123)

52

Using the fact that c = (µ0ε0)−1/2, η0 =√µ0/ε0, we arrive at Eq. (3.122) after

elementary re-arrangement.Exercise 3.10. Determine the scalar potential of a radiating dipole and explore itsbehavior in the near and far zones.Solution—We look for the magnitude of V far away from the dipole, |r′| |r|. Em-ploying Eq. (3.82), we obtain the following approximation

ejk|r−r′| ' exp

[jkr

(1− r · r′

r2

)]= ejkr exp

(−jk r · r

′

r

). (3.124)

Expanding the second exponential on the right-hand side of Eq. (3.116) to the firstpower in the small parameter r · r′/r2 1 , we arrive at

ejk|r−r′| ' ejkr

(1− jk r · r

′

r

). (3.125)

Combining Eqs. (3.83) and (3.117), we obtain to the first order in r · r′/r2 1,

ejk|r−r′|

|r− r′|' ejkr

r

(1 +

r · r′

r2

)(1− jkr r · r

′

r2

)' ejkr

r

[1 + (1− jkr)r · r

′

r2

].

(3.126)It then follows at once from Eq. (3.70) and (3.118) that at r r′,

V =ejkr

4πε0r

∫dv′ρv(r

′)︸ ︷︷ ︸=0

+r

r2(1− jkr)

∫dv′ r′ρv(r

′)︸ ︷︷ ︸=p

, (3.127)

implying that

V =p · r

4πε0r2(1− jkr)

(ejkr

r

). (3.128)

In the near zone, kr 1 and it follows at once from Eq. (3.120) that

V =p · r

4πε0r3, (3.129)

which is the electrostatic limit, c.f. Eq. (3.87). In the far zone, on the other hand,kr 1 and we arrive at

V = −jk(p · r)

4πε0r

(ejkr

r

), (3.130)

which amounts to an outgoing spherical wave.

3.6 Magnetic dipole antennaWe now examine the radiation pattern of a small loop carrying time-harmonic filamen-tary current I and show that this source radiates as a magnetic dipole. For simplicity,

53

Figure 3.14: Radiating elemental magnetic dipole (small current loop).

we assume the loop has a circular shape of radius a and we study the electromagneticfields far away from the loop, a r. We take our z-axis directed normal to the planeof the loop. The antenna geometry is illustrated in the figure on the next page. It canbe inferred from the figure that

r = rer; r′ = ae′ρ, and dl′ = e′φadφ′. (3.131)

It then follows at once from Eq. (3.118) and (3.131) that to the first order in a/r 1,

ejk|r−r′|

|r− r′|' ejkr

r

[1 +

a

r(er · e′ρ)(1− jkr)

]. (3.132)

We can then infer at once from Eqs. (3.78), (3.131) and (3.132) that

A =µ0I

4π

∫ 2π

0

dφ′ e′φ

(ejkr

r

)[1 +

a

r(er · e′ρ)(1− jkr)

]. (3.133)

Using the formula sheet relations,

e′φ = −ex sinφ′ + ey cosφ′, (3.134)

er = sin θ (ex cosφ+ ey sinφ) + ez cos θ, (3.135)

54

ande′ρ = ex cosφ′ + ey sinφ′, (3.136)

we observe thater · e′ρ = sin θ (cosφ cosφ′ + sinφ sinφ′). (3.137)

On substituting from Eqs. (3.134) and (3.137) into Eq. (3.133), we obtain after minimalalgebra the expression

A =µ0Ia

2

4πr2sin θ (1− jkr)ejkr

∫ 2π

0

dφ′ (−ex sinφ sin2 φ′ + ey cosφ cos2 φ′).

(3.138)In deriving Eq. (3.138) we used the fact that

∫ 2π

0dφ cosφ =

∫ 2π

0dφ sinφ = 0 as the

integrals of periodic functions over the period. On the other hand,∫ 2π

0

dφ cos2 φ =

∫ 2π

0

dφ sin2 φ = π, (3.139)

which follows at once from the trigonometric identities cos2 φ = (1 + cos 2φ)/2 andsin2 φ = (1 − cos 2φ)/2 and the fact that the sine and cosine of the double argumentyield zero when integrated over the period. Hence,

A =µ0Ia

2

4r2sin θ (1− jkr)ejkr (−ex sinφ+ ey cosφ)︸ ︷︷ ︸

=eφ

. (3.140)

Finally, by introducing the magnetic moment of the loop, defined as

m ≡ ISen, (3.141)

where S is the loop area and en is a unit normal to the loop; in our case S = πa2 anden = ez , we can cast Eq. (3.140) into the form

A = eφµ0m

4πr2sin θ (1− jkr)ejkr. (3.142)

The result, expressed in terms of the loop magnetic moment, is quite general and doesnot depend on a particular shape of the loop as long as its characteristic size is dwarfedby the radiated wavelength. As part of your homework, you will show, in particular thata small rectangular loop carrying a time-harmonic current gives rise to electromagneticfields with the same vector potential as that in Eq. (3.142).

Let us now discuss the electromagnetic field behavior in the far and near zones. Inthe far zone, kr 1 and we obtain for the vector potential the expression

A = −eφjµ0km

4πsin θ

(ejkr

r

). (3.143)

It then follows from Eqs. (3.79) and (3.143) that

H = − jµ0km

4πr2 sin θ

∣∣∣∣∣∣er r eθ r sin θ eφ∂r ∂θ ∂φ0 0 ejkr sin2 θ

∣∣∣∣∣∣ , (3.144)

55

implying that

H = − jµ0km

4πr2 sin θ

(er 2 sin θ cos θ − eθ jkr sin2 θ

)ejkr. (3.145)

In the far zone, the second term on the r.h.s. of Eq. (3.145) dominates, yielding thefollowing far-field behavior

H = −eθω2m

4πc2sin θ

(ejkr

r

). (3.146)

The electric field distribution in the far zone follows directly from Eqs. (3.80) and (3.146)as

E = − j

ε0ωr2 sin θ

∣∣∣∣∣∣er r eθ r sin θ eφ∂r ∂θ ∂φ0 ω2m

4πc2 ejkr sin θ 0

∣∣∣∣∣∣ = −eφjωm

4πε0c2jkr sin2 θejkr

r2 sin θ.

(3.147)Simplifying, and recalling the definitions c−1 =

√µ0ε0 and η0 =

√µ0/ε0, we arrive

at

E = eφω2m

4πc2η0 sin θ

(ejkr

r

). (3.148)

On comparing Eqs. (3.146) and (3.148, we notice that

E = −η0(er ×H), (3.149)

implying the plane wave behavior of the magnetic dipole radiation—see the far-fieldelectric dipole radiation pattern discussion of the previous section.Exercise 3.11. Discuss the electromagnetic field behavior of the magnetic dipolein the near zone, kr 1.Solution.— It follows at once from Eq. (3.142) that the vector potential in the near zonesimplifies to

A = eφµ0m

4πr2sin θ. (3.150)

Eqs. (3.79) and (3.150) then yield

H =m

4πr2 sin θ

∣∣∣∣∣∣er r eθ r sin θ eφ∂r ∂θ ∂φ0 0 r−1 sin2 θ

∣∣∣∣∣∣ . (3.151)

Working out the curl in Eq. (3.150) results in the magnetic field expression

H =m

4πr3(er 2 cos θ + eθ sin θ), (3.152)

which is a magnetic analog of the (quasi) static electric dipole field of Eq. (3.110 ).The electric field expression follows from Eqs. (3.80) and (3.152)

E =j

ε0ωr2 sin θ

∣∣∣∣∣∣er r eθ r sin θ eφ∂r ∂θ ∂φ

2 cos θr3

sin θr2 0

∣∣∣∣∣∣ . (3.153)

56

It follows that

E =j

ε0ωr2 sin θ

[eφ r sin θ

(−2 sin θ

r3+

2 sin θ

r3

)]= 0. (3.154)

Thus, to the first order in the parameter kr 1, the electric field of a magnetic dipoleis zero in the near zone.

3.7 Antenna parametersDefinition.—The region between the far-zone electric field maximum and its first nullis called the main antenna beam. Its width is known as an antenna beamwidth.We now introduce a solid angle concept. The concept is illustrated in the figure below.

Figure 3.15: Illustrating the solid angle concept. Reproduced from C. A. Blanis An-tenna Theory: Analysis & Design (Wiley, 2016).

The solid angle, measured in Steradians, is, basically, a measure of how much of thefield of view a given patch on a sphere of radius r covers, as viewed from the center ofthe sphere. In other words, it is a measure of how large the patch looks to an observerlocated at the center of the sphere. More precisely, if a Radian is an angle such that theassociated arc on a circle of radius r has the length of r, the solid angle of 1 Steradiancorresponds to a spherical patch of the area r2 on a sphere of radius r. An infinitesimalsolid angle dΩ can be expressed in the spherical coordinates as

dΩ ≡ (er · dS)/r2 = sin θ dφ dθ . (3.155)

57

Definition.—The region around secondary maxima are called sidelobes. We can definethe radiant intensity J as the amount of time-averaged power radiated per unit solidangle, i.e.,

J dΩ = r2〈P〉dΩ, =⇒ J = r2〈P〉. (3.156)

We can then define the directive gain Gd(Θ, φ) of an antenna as its radiant intensitynormalized to the average radiant intensity, so that

Gd(θ, φ) ≡ J1

4π

∫dΩJ

=4π〈P〉∫dΩ〈P〉

, (3.157)

where we used Eq. (3.156) to obtain the r.h.s., of Eq. (3.157). The directed gain maxi-mum is known as the antenna directivity,

D = maxGd(θ, φ) . (3.158)

Recall that a time-averaged Poynting vector—energy flow—of an electromagneticfield is given by the expression

〈P〉 = 12 ReE ·H∗. (3.159)

It then follows from Eqs. (3.114) and (3.149) that for any radiation field behaving asan outgoing spherical wave far away from the source, the magnituges of electric andmagnetic fields in free space are related as

E = η0|H|, =⇒ |H| = 1η0|E|. (3.160)

We can then readily infer from Eqs. (3.159) and (3.163) the following useful expression

〈P〉 = 12η0|E|2. (3.161)

It then follows from Eq. (3.157) and (3.161) that

Gd(θ, φ) =4π|E|2∫dΩ|E|2

, (3.162)

which furnishes a useful formula for the antenna directive gain.Finally, we define the radiation resistance of an antenna as the resistance of a

hypothetical resistor that would dissipate the amount of power equal to the antennaradiated power when the current through the resistor is equal to the maximum currentalong the antenna. In other words,

12I

2maxR =

∫dΩ r2〈P〉. (3.163)

We can infer at once from Eq. (3.161) and (3.163) that

R =

(r2

η0I2max

)∫dΩ |E|2 . (3.164)

58

Let us now illustrate these concepts by calculating the directive gain, directivityand radiation resistance of an elemental dipole antenna. It follows from Eq. (3.113)that

|E|2 =ω4p|2 sin2 θ

16π2c2r2η2

0 . (3.165)

On substituting from Eq. (3.165) into Eq. (3.162), we obtain after elementary algebra,

Gde(θ, φ) =4π sin2 θ∫dΩ sin2 θ

. (3.166)

Consider now∫dΩ sin2 θ =

∫ 2π

0

dφ

∫ π

0

dθ sin θ sin2 θ = 2π

∫ 1

−1

d(cos θ)(1− cos2 θ)

= 2π

∫ 1

−1

dx(1− x2) = 4π

∫ 1

0

dx(1− x2) = 8π/3. (3.167)

On substituting from Eq. (3.167) into Eq. (3.166), we finally obtain

Gde(θ, φ) = 32 sin2 θ. (3.168)

implying at once thatDe = maxGde(θ, φ) = 3/2. (3.169)

The electric dipole radiation resistance follows from Eqs. (3.164) and (3.165)

Re =r2

η0I2

ω4|p|2

16π2c2r2η2

0

∫dΩ sin2 θ. (3.170)

Using Eq. (3.167) and (3.97) on the r.h.s. of Eq. (3.170), we obtain, after minor algebra,for the radiation resistance the expression

Re = 80π2

(∆l

λ

)2

, (3.171)

where we used the fact that η0 = 120π and λ = 2π/k is the radiation wavelength atfrequency ω in free space.Exercise 3.12. Determine the radiation resistance of a magnetic dipole loop withthe current I and the radius a.Solution.— It follows from Eq. (3.148) that

|E|2 =ω4I2π2a4 sin2 θ

16π2c2r2η2

0 . (3.172)

On substituting from Eq. (3.172) into Eq. (3.164) we arrive at

Rm =r2

η0I2

ω4I2π2a4

16π2c4r2η2

0

∫dΩ sin2 θ. (3.173)

Hence, using Eq. (3.167), we obtain after simple algebra

Rm = 320π6(aλ

)4

. (3.174)

59

Chapter 4

Linear antenna and linearantenna arrays

4.1 Linear antenna

4.1.1 Standing wave antenna

Figure 4.1: Illustrating center-fed linear antenna geometry.

We now consider a short—in the precise sense to be specified below—linear an-tenna and determine its radiation pattern first approximately by assuming a currentdistribution along the antenna. We assume the antenna is center-fed so that the currentterminates at the antenna ends. This corresponds to a standing wave current patterngiving rise to the antenna name, standing wave linear antenna. We can then assume asinusoidal current distribution of the form

I(z′) = Im sin k(l − |z′|), (4.1)

60

where Im is a maximum current in the antenna, 2l is the antenna length and k = ω/c.We can then assume the antenna is aligned along the z-axis and break up the antennainto small segments of length dl′ = ezdz

′. The geometry is illustrated in the sketch onthe previous page. Each such segment radiates as an elemental dipole and contributes

dE = −e′θjkI(z′)dz′

4πη0 sin θ

(ejk|r−r

′|

|r− r′|

), (4.2)

to the electric field of the antenna in the far zone, |r′| |r|; where in this case,

r = rer, r′ = z′ez. (4.3)

Here z′ is a position of an elemental dipole along the antenna.To proceed, we make the following approximations

1. The antenna is short enough, l r so that in the far zone, we can neglect thepolarization difference among all elemental dipole contributions,

e′θ ' eθ. (4.4)

2. In the denominator on the r.h.s. of Eq. (4.2), we can assume that

|r− r′| ' r. (4.5)

3. Using the approximation (3.82) in the exponent of the exponential function onthe r.h.s. of Eq. (4.2), we obtain

ejk|r−r′| ' exp

[jkr

(1− r · r′

r2

)]' ejkr exp (−jkz′ cos θ) , (4.6)

where we used Eq. (4.3) to arrive at the last expression on the r.h.s of Eq. (4.6).

On substituting from Eqs. (4.1), (4.4), (4.5) and (4.6) into Eq. (4.2), we obtain afterelementary rearrangement,

E = −eθjkImη0 sin θ

4πrejkr

∫ l

−ldz′ sin k(l − |z′|)e−jkz

′ cos θ. (4.7)

Using the Euler formula (1.10),

e−jkz′ cos θ = cos(kz′ cos θ)− j sin(kz′ cos θ), (4.8)

On substituting Eq. (4.8) onto the r.h.s. of Eq. (4.7), and observing that an integral ofan odd function over a symmetric interval equals to zero, we arrive at

E = −eθjkImη0 sin θ

2πrejkr

∫ l

0

dz′ cos(kz′ cos θ) sin k(l − z′). (4.9)

61

Employing the trigonometric identity,

sinβ cosα ≡ 12 [sin(β + α) + sin(β − α)], (4.10)

on the r.h.s. of Eq. (4.10), we transform Eq. (4.9) into

E = −eθjkImη0 sin θ

4πrejkr

∫ l

0

dz′ sin k[l−z′(1− cos θ]+sin k[l−z′(1+cos θ)].(4.11)

Finally, performing elementary integrations on the r.h.s. of Eq. (4.11), we obtain forthe short linear antenna electric field the expression

E = −eθjImη0

2πrejkr F (kl, θ) , (4.12)

where we introduced an angular pattern function F (kl, θ) by the expression

F (kl, θ) =cos(kl cos θ)− cos kl

sin θ. (4.13)

Exercise 4.1. Determine the magnetic field generated by a short linear antenna.Solution.—As each dipole contribution has the same direction, we can conclude thatthe far-zone electric and magnetic fields of a short linear antenna are related the sameway E and H of an elemental dipole, i,e, E = −η0(er ×H). Taking a cross productof er and E, we obtain, er × E = −η0er × (er ×H). Using the bac-cab rule on ther.h.s., we obtain,

er ×E = −η0[er (er ·H)︸ ︷︷ ︸=0

−H(er · er)] = η0H,=⇒ H = 1η0

(er ×E). (4.14)

It then follows from er × eθ = eφ and Eqs. (4.12) and (4.14) that

H = −eφjImη0

2πrejkr F (kl, θ). (4.15)

Exercise 4.2. Determine the angular distribution pattern of the half-wave an-tenna.Solution.—In this case, 2l = λ/2, implying that kl = (2π/λ)× λ/4 = π/2. It followsat once that

F (π/2, θ) =cos(π/2 cos θ)

sin θ. (4.16)

To elucidate the angular dependence, we look into two limits: θ → 0 and θ → π/2.As θ → 0, we can use the L’Hopital rule to resolve the uncertainty of the ratio on ther.h.s. of Eq. (4.16). Taking derivatives in the numerator and denominator, we obtainF (π/2, θ) → (π/2) sin θ → 0 as θ → 0. On the other hand, as θ → π/2, F → 1.Thus, the angular radiation patterns resembles that of an elemental dipole: there is noradiation in the direction of the antenna current and there is a maximum in the planeorthogonal to the antenna axis.

62

4.1.2 Traveling wave antennaThe sinusoidal current distribution along a linear antenna considered in the previoussection implies a standing current with the nodes at the antenna ends. In case of verylong linear antannas, l λ, a more physically sound excitation method is via a travel-ing current of the form

I(z′) = Imejkz′ , (4.17)

where we assumed that the wave number of the wave propagating along the antennawire coincides with that of the wave radiated into free space. Breaking up the antennainto infinitesimal dipole elements, we can re-write Eq. (4.2) in the far zone as

dE = −eθjkI(z′)dz′

4πη0 sin θ

(ejkr

r

)e−jkz

′ cos θ, (4.18)

where we made use of Eq. (4.7).The total radiated field follows from Eqs. (4.17) and (4.18) as

E = −eθjkImη0 sin θ

4π,

(ejkr

r

) ∫ l

−ldz′ e−jkz

′(cos θ−1). (4.19)

An elementary integration yields

E = −eθjkImη0 sin θ

4π

(ejkr

r

)ejkl(cos θ−1) − e−jkl(cos θ−1)

jk(cos θ − 1), (4.20)

which can be cast into the form

E = −eθjkImη0 sin θ

2πsinc[kl(cos θ − 1)]

(ejkr

r

). (4.21)

The magnetic field reads then

H =(er ×E)

η0. (4.22)

We can now work out the radiated power flow as

P =1

2η0|E|2 =

k2l2I2mη0

8π2r2sin2 θ sinc2[kl(cos θ − 1)]. (4.23)

As long as kl 1, the radiation maxima are determined by the sinc function maximawhich are specified by the condition

sin2[kl(cos θm − 1)] = 1, (4.24)

implying that

kl(cos θm − 1) = ± 12 (2m+ 1)π, m = 0, 1, 2, 3 . . . (4.25)

The major lobe takes place at m = 0, corresponding to

kl(1 + θ20/2− 1) = π/2,=⇒ θ0 '

√λ2l 1. (4.26)

63

Eq. (4.26) implies that the traveling wave antenna maxima, in particular, its major lobepeak, are all clustered around the forward direction, θmax 1 which justifies ourexpanding a cosine in a Taylor series in (4.26). The traveling wave antenna radiationpattern is thus very different from the standing wave antenna one which has symmetriclobes in the forward and backward directions. Finally, the radiation pattern nulls aregiven by the condition

sin2[kl(cos θn − 1)] = 0, (4.27)

orkl(cos θn − 1) = ±nπ, n = 1, 2, 3, 4 . . . (4.28)

Again, in the limit, l λ, the first null is approximately at

θ1 '√λ/l 1, (4.29)

which is also nearly in the forward direction.Direct numerical modelling indicates that the current distribution along the antenna

can be chosen according to the antenna length as follows

• constant current for infinitesimal dipoles, l ≤ λ/50;

• linear (triangular) current, illustrated in your homework assignment for shortantannas, λ/50 ≤ l ≤ λ/10;

• sinusoidal current with the nodes at the ends for fairly long antannas, l ≥ λ/10;

• traveling wave current for very long antannas, l λ.

4.2 Linear antenna as a boundary-value problemIn the previous section, we studied the linear antenna problem assuming we know thecurrent along the antenna. This is strictly true only for infinitely thin antannas. If theantenna has a finite thickness, we can assume no apriori knowledge of the current dis-tribution in the antenna. Rather, we have to determine this distribution self-consistentlywhile solving for the antenna radiation field. This problem can be rigorously stated forsufficiently thin antannas under the following assumptions.

1. We deal with a straight linear antenna of cylindrical shape with a circular cross-section;

2. The antenna radius a is assumed to be sufficiently small compared to its lengthl, a l;

3. The antenna is assumed to be made of a perfectly conducting material;

4. The antenna current distribution is assumed to be azimuthally symmetric.

64

Under assumptions 1 through 4, we can derive an integral equation for the antennacurrent distribution. To this end, let us recall that the vector potential outside of aperfect conductor, carrying a surface current of density Js is given by Eq. (3.75). Inour case, we choose the z-axis of our coordinate system along the antenna axis. In thiscase,

Js = ezJs(z′), (4.30)

where we used assumptions 1,2& 4. It follows from Eqs. (3.75) and (4.30) that thevector potential of the radiated field has only a z-component. On the antenna surface,ρ = a, the z-component of A is then given by the expression

Az(rs) =µ0

4π

∫dS′ Js(z

′)ejk|rs−r

′|

|rs − r′|, (4.31)

where we can write in the cylindrical coordinates on the antenna surface

rs = aeρ + zez; r′ = ae′ρ + z′ez. (4.32)

Next, for a cylindrical surface of radius a, we have

dS′ = adφ′dz′. (4.33)

On combining Eqs. (4.31) and (4.33), we arrive at

Az(rs) =µ0

4π

∫ 2π

0

dφ′∫ l/2

−l/2dz′ aJs(z

′)ejk|rs−r

′|

|rs − r′|. (4.34)

Further, it follows from Eq. (4.32) that

|rs − r′| =√

(rs − r′) · (rs − r′) =√

2a2[1− (eρ · eρ′)] + (z − z′)2. (4.35)

Consider now the radial unit vectors displayed in the figure below. To evaluate thedot product on the r.h.s. of Eq. (4.35), we re-express the radial unit vectors in thecylindrical coordinates in terms of the corresponding Cartesian unit vectors viz.,

eρ = ex cosφ+ ey sinφ; e′ρ = ex cosφ′ + ey sinφ′. (4.36)

We can then infer at once from Eq. (4.36) that

eρ · e′ρ = cosφ cosφ′ + sinφ sinφ′ = cos(φ− φ′). (4.37)

We can now introduce an effective filamentary current through the antenna by theexpression

I(z′) = 2πaJs(z′). (4.38)

On substituting from Eqs. (4.35), (4.37) and (4.38) into Eq. (4.34), we obtain for thevector potential on the antenna surface the expression

Az(ρ = a) =µ0

4π

∫ l/2

−l/2dz′ I(z′)K(z, z′). (4.39)

65

Figure 4.2: Illustrating the mutual orientation of two radial unit vectors in the xy-plane.

Here we introduced the angle averaged kernel as

K(z, z′) =

∫ 2π

0

dφ′

2π

ejkR

R, (4.40)

whereR =

√4a2 sin2 φ′/2 + (z − z′)2. (4.41)

In writing Eq. (4.41) we took into account the azimuthal symmetry of the problemwhich implies that the kernel cannot depend on φ. The φ-dependence can be eliminatedby choosing the x-axis along our observation direction eρ.

We shall now make full use of assumption 3. As the antenna is a prefect conductor,the tangential component of the electric field on the surface must vanish,

Ez(ρ = 0) = 0. (4.42)

We can re-write Eq. (4.42) in terms of the vector potential. With this purpose, weexpress E in terms of A. First we eliminate the scalar potential using the Lorentzgauge condition (3.67),

V = −jck∇ ·A, (4.43)

where k = ω/c. On substituting from Eq. (4.43) into Eq. (3.54) we readily obtain

E =jc

k

[∇(∇ ·A) + k2A

]. (4.44)

Since, only Az is nonzero, we can infer at once from Eq. (4.44) that

Ez(ρ = a) =jc

k(k2 + ∂2

z )Az(ρ = a). (4.45)

On combining Eqs. (4.39), (4.42) and (4.45), we obtain an integro-differential equationfor the unknown current I(z) in the Pocklington form(

d2

dz2+ k2

)∫ l/2

−l/2dz′ I(z′)K(z, z′) = 0 . (4.46)

66

Alternatively, Eq. (4.46) can be viewed as a linear ODE for the variable F (z) =∫ l/2−l/2 dz

′ I(z′)K(z, z′): (d2

dz2+ k2

)F (z) = 0, (4.47)

which can be integrated at once yielding an integral equation for the antenna currentdistribution in the Hallen form as

a1 cos kz + a2 sin kz =

∫ l/2

−l/2dz′ I(z′)K(z, z′) , (4.48)

where a1 and a2 are constants determined by the boundary conditions at the antennaends and the current excitation method.

A solution to either Eq. (4.46) or (4.48) is not straightforward. Although in the limitof a → 0, we can show the current to be sinusoidal, careful inspection of Eq. (4.40)reveals that at z ' z′, the finite size of the antenna has to be taken into account toavoid kernel singularities. Detailed numerical simulations reveal that the solution isvery sensitive to the magnitude of the antenna size. In particular, even for the antannasas thin as a/l ∼ 10−3, the corrections of the order of 10-15% are expected to theideal infinitely thin antenna results. Moreover, when the current distribution has a nodewithin the excitation region, the sinusoidal current approximation leads to significanterrors in evaluating the antenna parameters.

4.3 Linear antenna arrays

4.3.1 Two-element arraysConsider the radiation pattern of two dipole radiators located sufficiently close to eachother. We assume that there is a phase difference ξ between the radiating sources, re-sulting in the same phase difference between the radiated fields. Ignoring polarizationeffects—that is assuming the dipoles are identically polarized—and assuming, for sim-plicity that one dipole is located at the origin while the other is placed along the x-axisand is located a distance d away from the origin, i.e., r′ = dex, we can write down thefar-field as

E = E(r, θ)ejkr

r+ E(r, θ)e−jξ

ejkr1

r1= E(r, θ)

(ejkr

r+ e−jξ

ejkr1

r1

), (4.49)

where E(r, θ) is a far field of an individual dipole. The geometry is illustrated in thefigure below. As we are interested in the far-field pattern, we can assume that |r′| rand make use of Eq. (3.82) to write

r1 = |r− r′| ' r(

1− r · r′

r2

)= r − d sin θ cosφ. (4.50)

Further, in the denominator of Eq. (4.49), we can let approximately, r1 ' r.

67

Figure 4.3: Illustrating the geometry of a linear two-antenna array.

On substituting from Eq. (4.50) into (4.49), we obtain at once that

E = E(r, θ)

(ejkr

r

)(1 + e−jψ), (4.51)

where we introduced the phase difference

ψ = kd sin θ cosφ+ ξ. (4.52)

It follows that Eq. (4.51) can be cast into the form,

E = E(r, θ)

(ejkr

r

)e−jψ/2(ejψ/2 + e−jψ/2), (4.53)

implying that

|E| = 2|E(r, θ)|r

| cosψ/2|. (4.54)

Hence, the far-zone field pattern is determined by the individual field of each dipole, aso-called space factor as well as by an array factor. The latter, in this two-antennaarray case, is given by cosψ/2. This is a general rule for arbitrary arrays, i. e.,

Array Field = Space Factor︸ ︷︷ ︸SF

×Array Factor︸ ︷︷ ︸AF

. (4.55)

As an example, let us consider a two half-wave dipole array for which

|E| ∝∣∣∣∣cos(π/2 cos θ)

sin θcosψ/2

∣∣∣∣ . (4.56)

We can infer at once from Eqs. (4.52) and (4.56) that in the xy-plane, θ = π/2, thefar-field angular distribution of the array is completely determined by the array fac-tor. Notice that the array factor can be the dominant factor shaping the array angulardistribution in the far zone. Hereafter, we will be mainly focusing on the array factor.

68

Figure 4.4: Illustrating the geometry of an elemental dipole located above a perfectconductor.

4.3.2 Dipole radiator near perfect conductorAn interesting particular case of a two-element antenna array corresponds to the situ-ation when a single dipole is located in the vicinity of a perfect conductor at z = 0,say—see the figure above—such as the earth surface. In this case the method of imagescan be employed to work out the radiation field. To this end, we first derive an expres-sion for the far-field of a dipole located off the origin, above the conducting plane. Forsimplicity, we assume that the dipole is oriented normal to the plane and take our z-axis along the dipole direction so that p = pez and assume that the dipole lies on thez-axis, implying that its position radius-vector can be written as r0 = z0ez . Next, itfollows at once from Eq. (3.113) that the field far way from a radiated dipole located atthe origin can be expressed as

E0 = −ω2p sin θ

4πcη0

(ejkr

r

)eθ, (4.57)

If one shifts the dipole position to the point r0, the electric field will have the sameform in the coordinate frame with the origin shifted to r0. Hence, the far-field of thedipole reads

E0 = −ω2p sin θ

4πcη0

(ejk|r−z0ez|

|r− z0ez|

)eθ. (4.58)

We now employ a method of images by replacing the perfect conductor with animage dipole, located below the plane. The total electric field in the upper half spacewill then read

E = −ω2 sin θ

4πcη0

(peθ

ejk|r−z0ez|

|r− z0ez|+ p′e′θ

ejk|r−z′0ez|

|r− z′0ez|

), (4.59)

where p′ and z′0 are the dipole moment and position of the image. As the plane isa perfect conductor, the tangential component of the electric field must be zero at theplane z = 0, implying that

(ez ×E)|z=0 = 0. (4.60)

69

Figure 4.5: Illustrating radio wave transmission geometry near the earth surface.

Figure 4.6: Illustrating the image dipole orientation.

By symmetry, we choose p′ = p and z′0 = −z0. As is seen in Fig. 4.7, on the planez = 0,

(ez × eθ)|z=0 = −(ez × e′θ)|z=0. (4.61)

It then follows from Eq. (4.59) that the field in the upper half plane, z ≥ 0, is given bythe expression

E = −ω2p sin θ

4πcη0

(eθejk|r−z0ez|

|r− z0ez|+ e′θ

ejk|r−z′0ez|

|r− z′0ez|

); (4.62)

where

|r− z0ez| =√ρ2 + (z − z0)2, |r− z′0ez| =

√ρ2 + (z + z0)2. (4.63)

Momentary inspection of Eqs. (4.61) through (4.63) convinces one that the boundarycondition, Eq. (4.60) is indeed satisfied by this choice, thereby giving an acceptable so-

70

lution to the problem of determining the elemental dipole radiation field in the vicinityof a perfect conductor.

A useful application of this result is to radio transmission near the earth surface.If a transmitter antenna size is much smaller than the distance between the transmitterand receiver—which is almost always the case—the former can be viewed as a dipoleantenna. The receiver signal is then comprised of a line-of-sight propagating wavecontribution and that resulted from a single reflection from the earth surface, see Fig.4.5 . As the latter is a perfect conductor, it can be replaced by an image dipole, and theradiation at the receiver is given by the two-element array of Eq. (4.61).

4.3.3 Multi-element arrays

Figure 4.7: Illustrating a uniform linear antenna array.

Let us now consider the case of N identical, nearly isotropic radiators placed alonga straight line which we assume to coincide with the z-axis, say. If each individualradiators have identical excitation coefficients, this is known as a uniform linear ar-ray. We assume that the radiators are equally spaced, with the adjacent radiators beingspaced a distance d apart so that their phase difference is

ψ = ψj+1 − ψj = kd cos θ + ξ. (4.64)

Thus, the antenna array factor is given by

AF =∣∣∣ 1N

∑N−1s=0 e−jsψ

∣∣∣ =1

N

∣∣∣∣1− e−jNψ1− e−jψ

∣∣∣∣ =1

N

∣∣∣∣e−jψN/2(ejψN/2 − e−jψN/2)

e−jψ/2(ejψ/2 − e−jψ/2)

∣∣∣∣ ,(4.65)

where A0 = const is an (nearly) isotropic contribution due to an individual radiatorand we used the known expression for the geometric series sum,

∑N−1n=0 x

n = (1 −xN )/(1− x). Using the Euler formula,

sinx =(ejx − e−jx)

2j, (4.66)

we can transform Eq. (4.65) as

AF =

∣∣∣∣ sinNψ/2N sinψ/2

∣∣∣∣ . (4.67)

71

We can now examine the antenna array factor in detail.

1. Maxima & main-beam direction.

sinψm = 0, =⇒ ψm = πm, m = 0,±1,±2, . . . (4.68)

Of special importance is the main-beam direction, the first maximum corre-sponding to ψ = 0, i,e., as follows from Eq. (4.52)

cos θ0 = − ξ

kd. (4.69)

We can distinguish two especially relevant cases

• Broadside array.—In this case, the radiation maximum takes place in thedirection perpendicular to the array, θ0 = ±π/2, just like the case of asimple dipole. It follows at once from Eq. (4.69) that it requires

ξ = 0, (4.70)

i.,e., all the array elements must radiate in phase.

• End-fire array.—In this case, the radiation maximum occurs along the arrayaxis, θ0 = 0, implying that there exists a phase difference of

ξ = −kd, (4.71)

between adjacent radiators.

2. Null locations. The null locations occur at

sinNψn/2 = 0, =⇒ ψn = 2πn/N, n = ±1,±2, . . . , (4.72)

which are obviously different for broadside and end-fire arrays because of thedifferent phase differences, ξ implicit in ψ.

Figure 4.8: Illustrating the main beam width concept.

72

3. Main beam width. The main beam width is determined by the first null, that isby

Nψ1/2 = ±π, =⇒ ψ1 = ±2π/N. (4.73)

Again we shall distinguish broadside and end-fire arrays because the adjacentelement phase difference is different for the two.

• Broadside array.—For a broadside array,

ξ = 0, θ0 = π/2, (4.74)

It follows that the main lobe width, whose angular width is determined by

2∆θ = 2(θ1 − θ0) (4.75)

can be calculated using Eq. (4.64). Specifically,

kd cos θ1 = kd cos(θ0 + ∆θ) = ψ1. (4.76)

It follows at once from Eqs. (4.73) through (4.76) that

kd cos θ1 = kd cos(π/2 + ∆θ) = −kd sin ∆θ = −2π/N. (4.77)

or

∆θ = sin−1

(λ

Nd

)' λ

Nd, (4.78)

where we assumed that the array length L ' Nd is much longer than theradiation wavelength λ.

• End-fire array.—In this case,

ξ = −kd, θ0 = 0, (4.79)

and the similar reasoning yields

kd(cos θ1 − 1) = kd(cos ∆θ − 1) = −2π/N. (4.80)

Next, under the same approximation, L λ, translating to ∆θ 1, wecan use the Taylor expansion

cos ∆θ ' 1− (∆θ)2/2. (4.81)

Combining Eqs. (4.73), (4.80), and (4.81) results in

∆θ '√

2λ

Nd. (4.82)

Exercise 4.3. Determine the directivity of a closely spaced, kd 1, broadsideuniform linear antenna array.Solution.—By definition of a broadside array, ξ = 0, implying that ψ = kd cos θ. For

73

a closely spaced array, ψ 1. One can then make an approximation sinψ/2 ' ψ/2in Eq. (4.67) transforming it into

AF =sin(Nψ/2)

Nψ/2= sinc(Nψ/2). (4.83)

It then follows at once on substituting from Eq. (4.83) into Eqs. (3.162) and (3.169)that

Dar =4π∫ 2π

0dφ∫ π

0dθ sin θ sinc2( 1

2Nkd cos θ), (4.84)

or performing the trivial φ-integration and cancelling out the common factors,

Dar =2∫ π

0dθ sin θ sinc2( 1

2Nkd cos θ), (4.85)

Changing the integration variable in Eq. (4.85) to x = 12Nkd cos θ, we obtain

Dar =Nkd∫ Nkd

−Nkd dx sinc2(x), (4.86)

and assuming, as usual, a long array L ' Nd λ, implying that Nkd 1, we canreplace the integration limits by ±∞ and use the table integral,∫ ∞

−∞dx sinc2(x) = π. (4.87)

Combining Eqs. (4.87) and (4.86), we end up with the result

Dar = Nkd/π = 2Nd/λ ' 2L/λ, (4.88)

where L is the array length.

4.4 Non-uniform antenna array synthesisWe now consider the array design or synthesis whereby one is required to design anantenna array given certain specifications. For instance, one might be interested tohave linear antenna arrays with prescribed positions of radiation minima (nulls) or aprescribed array factor AF in the far zone. The former can be accomplished using arelatively simple technique due to Schelkunov, whereas the latter might require Fouriersynthesis or sampling to achieve the objective. One may also be interested in antennapatterns with no sidelobes which is realized by using binomial distribution of the arrayelement excitation coefficients. We proceed by discussing binomial arrays first.

74

4.4.1 Binomial arraysSo far we have discussed uniform linear arrays of nearly isotropic scalar radiators. Ingeneral, we can consider non-uniform distributions of elemental radiators. The corre-sponding array factor can be expressed as

AF (er) =

N−1∑n=0

AnNejnψ, (4.89)

where er = r/r is a unit vector in the radial direction, ψ is given by Eq. (4.64) for theindividual radiators placed along the z-axis; in a general situation ψ is given by

ψ = kd(er · el) + ξ, (4.90)

where el is a unit vector along the linear antenna array. In particular, the so-calledbinomial arrays have the amplitudes given by binomial coefficients,

AnN =N !

n!(N − n)!, (4.91)

where the factorial of a non-negative real number n is defined as n! ≡ 1×2×3 . . .×n;0! ≡ 1. The binomial coefficients arise as a result of the binomial expansion as

(p+ q)N =

N∑n=0

AnN pnqN−n. (4.92)

The binomial arrays are known to have suppressed sidelobes at the expense of havinglarger main beam widths than their uniform cousins.

4.4.2 Schelkunoff polynomial methodSchelkounoff method makes it possible to design antenna arrays with radiation nulls atprescribed positions. To see how it works, let us consider a linear array factor for thecase all elements are along the z-axis, for simplicity. The resulting array factor reads,

AF =

N−1∑n=0

anejnψ, (4.93)

where

ψ = kd cos θ + ξ =2π

λd cos θ + ξ . (4.94)

Notice that each phasor ejψ can be viewed as a unimodular complex number,

z = ejψ = ej(kd cos θ+ξ). (4.95)

Hence it can be visualized as being located on a unit circle in a complex plane. The AFis then a complex polynomial,

AF =

N−1∑n=0

anzn = a0 + a1z + . . . aN−1z

N−1. (4.96)

75

According to the fundamental theorem of complex algebra, AF has exactly N − 1—ingenerally complex—roots and it can be expressed as

AF = aN−1(z − z1)(z − z2)(z − z3) . . . (z − zN−1) . (4.97)

It follows from Eq. (4.94) that as 0 ≤ θ ≤ π, the argument ψ lies in the visiblerange

−2π

λd+ ξ ≤ ψ ≤ 2π

λd+ ξ . (4.98)

Eq. (4.98) gives a visible range of angles; no zero zs = ejψs outside this range corre-sponds to a physical null of the array radiation pattern for a given d. For instance, ford = λ/2 and no progressive phase shift between the adjacent aray elements, ξ = 0,−π ≤ ψ ≤ π, i. e., the visible range covers the whole circle; for d ≤ λ/2 the visiblerange is shorter. We can now illustrate by examples how non-uniform arrays with de-sired radiation null locations can be designed.Exercise 4.4. Design a linear array with an element spacing d = λ/4 and ξ = 0such that it has zeroes at θ = 0, π/2 and π. Determine the required number ofelements and their excitation coefficients.Solution.—In this case, ψ = (π/2) cos θ and the visible range spans half a circle,−π/2 ≤ ψ ≤ π/2. The desired zeros at θ = 0, π/2 and π correspond to the ψ valuesas ψ1 = π/2, ⇒ z1 = j, ψ2 = 0 ⇒ z2 = 1, and ψ3 = π ⇒ z3 = −j. Hence, thearray factor reads

(z − 1)(z − j)(z + j) = (z − 1)(z2 + 1) = z3 − z2 + z − 1.

We can read off the excitation coefficients now: a0 = −1, a1 = 1, a2 = −1, anda3 = 1.Exercise 4.5. Design a linear array with an element spacing d = 3λ/8 and ξ = 0such that it has zeroes at θ = 0, π/2 and π.a) Determine the required number of elements and their excitation coefficients.b) For the same inter-element spacing, find the range of progressive phase shifts ξsuch that the null at θ = π/2 no longer lies within the visible range, and hence noradiation pattern null occurs in this direction.Solution.—a) In this case, ψ = (3π/4) cos θ and the visible range spans the arc,−3π/4 ≤ ψ ≤ 3π/4. The desired nulls occur in the directions

ψ1 = 3π/4, , ψ2 = 0, ψ3 = −3π/4.

These zeroes have the complex representation on the unit circle,

z1 = e3jπ/4, , z2 = 1, z3 = e−3jπ/4.

It follows that

AF = (z− 1)(z− e3jπ/4)(z− e−3jπ/4) = (z− 1)(z2− e3jπ/2) = z3 − z2 + jz − j.

76

Thus, a0 = −j, a1 = j, a2 = −1, and a3 = 1.b) Since as ξ = 0, −3π/4 ≤ ψ ≤ 3π/4, it is clear that ψ = 0 falls outside the originalvisible range, provided that

|ξ| ≥ 3π/4.

4.4.3 Fourier series methodSuppose we want to design a linear array with a given array factor AF (ψ). The taskcan be easily accomplished if the array factor is periodic with a period of 2π. Thiswould correspond to the element separation such that

2kd = 2π =⇒ d = λ/2. (4.99)

Assuming an odd number of elements to the array such that the array has a non-zero“dc” contribution corresponding to a0, the array factor reads

AF (θ) = AF (ψ) =

M∑m=−M

amejmψ, (4.100)

where the phase isψ = kd cos θ + ξ. (4.101)

The array elements are placed at the positions

zm = md. (4.102)

If d = λ/2, Eq. (5.1) is a Fourier transform with a period of 2π. A simple Fourierinversion gives the desired array element excitation coefficients viz.,

am =

∫ 2π

0

dψ

2πAF (ψ)e−jmψ. (4.103)

4.4.4 Woodward-Lawson methodSuppose now we look to design an array with a specified array factor AFs and an ar-bitrary inter-element spacing d. Woodward and Lawson developed a sampling methodfor arrays of arbitrary inter-element spacings based on sampling an array with uniformarray samples, each having the angular dependence given by Eq. (4.67) such that

fm(θ) = bmsin[N2 kd(cos θ − cos θm)

]sin[

12kd(cos θ − cos θm)

] , (4.104)

where bm is an excitation weight of a given sample and l = Nd; for this design only,the array length l includes a distance d/2 beyond each element end. The total arrayfactor of 2M + 1 sampling terms is given by the expression

AF (θ) =

M∑m=−M

bmsin[N2 kd(cos θ − cos θm)

]sin[

12kd(cos θ − cos θm)

] , (4.105)

77

where the excitation coefficients bm of the array sample elements are equal to thespecified array factor AFs at a given sample point, that is

bm = AFs(θ = θm). (4.106)

The sample points are chosen uniformly such that

cos θm = mλ/l, m = 0,±1,±2, . . . (4.107)

The normalized excitation coefficient of each array element, yielding a desired angularpattern in the far zone, is then given by the expression

an(z) =1

N

M∑m=−M

bme−jkzn cos θm , (4.108)

where zn indicates the nth element position of the sough array, symmetrically locatedwith respect to the geometrical center of the array.

78

Chapter 5

Beyond linear antannas

5.1 Aperture antenna and angular spectrumLet us consider an aperture of an arbitrary shape surrounded by a perfect conductor.The aperture can be a waveguide opening and thus serves as an aperture antenna forelectromagnetic waves propagating in the waveguide and radiating out into a half-spacez ≥ 0. The time-harmonic electric field

E(r, t) = E(r)e−jωt, (5.1)

anywhere in the upper half-space can be expressed as a superposition of plane wavesas

E(r) =

∫∫ ∞−∞

dkx dky ejk·r E(kx, ky), (5.2)

where E(kx, ky) is known as the angular spectrum of plane waves and each plane wavehas the wave number satisfying the dispersion relation

k2 = k2x + k2

y + k2z = ω2/c2, (5.3)

in free space. It follows at once from Eq. (5.3) that if we choose the in-plane com-ponents, kx, ky of the wave vector k as independent variables, the axial component isgiven by

kz =√ω2/c2 − k2

x − k2y. (5.4)

As will become clear below, for plane waves propagating away from the aperture planeinto the upper half-space z ≥ 0, we must choose the positive root in Eq. (5.4).

Next, it is convenient to introduce the dimensionless wave vector components as

kx = pk, ky = qk, kz = mk, (5.5)

where k = ω/c and we introduced the notation

m =

√1− p2 − q2, p2 + q2 ≤ 1;

j√p2 + q2 − 1, p2 + q2 ≥ 1.

(5.6)

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We note that the plane waves corresponding to p2 +q2 ≤ 1 are ordinary homogeneousplane waves with the complex exponential space dependence, ejk(px+qy+mz), whereasthe waves associated with p2 + q2 ≥ 1 decay exponentially in the upper half spaceaway from the aperture plane z = 0 such that their spatial dependence is given byejk(px+qy)e−kmz for any z ≥ 0. Such waves are known as evanescent waves andthey do not contribute to the radiation field because they decay very fast and make anegligible contribution in the far zone where kz 1. As we are primarily interestedin the far zone behaviour, we neglect evanescent waves and re-write the electric fieldin Eq. (5.2) as

E(r) = k2

∫∫p2+q2≤1

dpdq ejk(px+qy+mz) E(p, q). (5.7)

Further, to determine the radiated electric field anywhere in the upper half-space,we only need to know the angular spectrum. To find the latter, we look into the electricfield behaviour in the aperture plane z = 0. We assume that the electric field in thewaveguide is TE-polarized in the x-direction, say, and a characteristic aperture sizeis much greater than the wavelength of the field. In this approximation, the radiatedfield is strongly localized along the z-axis of the system and we can safely ignorepolarization effects hereafter, effectively replacing the vector electric field with a scalarone. It then follows that the electric field in the aperture is TE-polarized as well andis naturally equal to the waveguide field. Further, the tangential electric field outsidethe aperture is zero in the perfectly conducting plane. Therefore, the field in the z = 0plane is specified by

E(x, y, 0) = ex

E0(x, y), inside aperture;0, outside aperture. (5.8)

In the aperture plane, Eq. (5.2) can be written as

E(x, y, 0) = k2

∫∫ ∞−∞

dq dp ejk(px+qy) E(p, q). (5.9)

Eq. (5.9) can then be Fourier inverted to yield the angular spectrum in the form

E(p, q) =

∫ ∞−∞

dx

2π

∫ ∞−∞

dy

2πe−jk(px+qy)E(x, y, 0), (5.10)

which, in view of Eq. (5.8) can be rewritten as

E(p, q) =

∫∫aperture

dxdy

(2π)2e−jk(px+qy)E0(x, y) . (5.11)

We can now introduce the scaled spatial variables by the expressions

sx = x/r sy = y/r, sz = z/r, (5.12)

and write down Eq. (5.7), dropping the unit vector in the x-direction, as

E(r) = k2

∫∫p2+q2≤1

dpdq ejkr(psx+qsy+msz) E(p, q). (5.13)

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The radiated field can be found by doing the integral on the r.h.s. of Eq. (5.13) in thelimit kr 1. This integral is of the form

J(sx, sy, sz) =

∫∫dpdqF (p, q)ejνG(p,q,sx,sy,sz), ν →∞. (5.14)

Integrals of this type can be asymptotically evaluated using the method of stationaryphase, which, roughly speaking, takes into account the leading contribution to the in-tegral coming from the critical points where the phase of G is stationary, ∂pG|c =∂qG|c = 0. Hereafter in this section, we denote partial derivatives as

Gp ≡ ∂pG; Gq ≡ ∂qG, Gpp ≡ ∂2ppG, Gpq ≡ ∂2

pqG, Gqq ≡ ∂2qqG. (5.15)

The result reads

J(sx, sy, sz) '2πjγ

ν√|∆|

F (pc, qc)ejνG(pc,qc,sx,sy,sz). (5.16)

Here

γ =

+1, ∆ > 0,Σ > 0;−1, ∆ > 0,Σ < 0;−j, ∆ < 0,

(5.17)

where ∆ and Σ are defined by

∆ = (GppGqq −G2pq)c, Σ = (Gpp +Gqq)c. (5.18)

Here the subscript “c” implies that the partials are evaluated at critical points labeledby “c”.

On comparing Eqs. (5.13) and (5.14), we identify

G(p, q, sx, sy, sz) = psx + qsy +msz, ν = kr. (5.19)

Let us now determine the critical points of the radiated field. Taking partial derivativeswe obtain,

gp = sx +mpsz, mp = − p√1− p2 − q2

= −mp. (5.20)

Therefore,gp|c = (sx −

p

msz)c = 0. (5.21)

By the same token,gq|c = (sy −

q

msz)c = 0. (5.22)

Eqs. (5.21) and (5.22) yield for the critical point

pc/mc = sx/sz, qc/mc = sy/sz. (5.23)

Further,mc =

√1− p2

c − q2c , sz =

√1− s2

x − s2y. (5.24)

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We can infer at once from Eqs. (5.23) and (5.24) that

pc = sx, qc = sy, mc = sz. (5.25)

In other words, the key contribution to any direction in the far zone is given by theplane wave propagating along the line-of-sight which is known as a geometrical opticslimit.

Next, we evaluate the second derivatives of G:

Gpp = −sz(m+ p2/m

m2

)= −sz

m

(1 +

p2

m2

). (5.26)

By the same token,

Gqq = −szm

(1 +

q2

m2

), (5.27)

andGpq = psz

mq

m2= −sz

m

pq

m2. (5.28)

At the critical point we have,

Gpp|c = −(

1 +s2x

s2z

), Gqq|c = −

(1 +

s2y

s2z

), Gpq|c = −sxsy

s2z

. (5.29)

Thus, we can evaluate Σ and ∆ to be

Σ = −(

1 +1

s2z

)< 0, ∆ =

1

s2z

> 0. (5.30)

It then follows from Eqs. (5.17) and (5.30) that γ = −1. Finally, recalling Eqs. (5.12), (5.13),and (5.16), we can write down the answer for the radiated electric field as

E∞(r) ' −2πjk(zr

)E(xr,y

r

) (ejkrr

). (5.31)

Notice that since

z/r = cos θ, x/r = sin θ cosφ, y/r = sin θ sinφ, (5.32)

the far-zone electric field of Eq. (5.31) can also be expressed in the spherical coordi-nates as

E∞(r) ' −2πjk cos θ E(sin θ cosφ, sin θ sinφ)

(ejkr

r

). (5.33)

Looking at Eq. (5.31), we immediately recognize a far-zone outgoing sphericalwave dependence, with the angular distribution of the spherical wave amplitude be-ing determined by the angular spectrum of the electric field in the aperture plane,Eq. (5.11). Hence, Eqs. (5.11) and (5.31) give the radiated electric field of an aper-ture antenna in the far zone, given the electric field in the aperture.

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Exercise 5.1. Determine the far-zone electric field of a rectangular aperture ofthe width a and height b, assuming that the electric field E0 is uniform over theaperture opening, −a/2 ≤ x′ ≤ a/2, and −b/2 ≤ y′ ≤ b/2.Solution.—It follows at once from Eqs. (5.11) that the angular spectrum in the apertureplane is given by

E = E0

∫ a/2

−a/2dx′ e−jkx

′p

∫ b/2

−b/2dy′ e−jky

′q. (5.34)

Evaluating elementary integrals and substituting p = x/r = sin θ cosφ and q =sin θ sinφ, we arrive at the angular spectrum

E = abE0 sinc(X) sinc(Y ), (5.35)

whereX = 1

2ka sin θ cosφ, Y = 12kb sin θ sinφ. (5.36)

Combining Eqs. (5.35) with Eq. (5.33) we arrive at the final answer,

E∞ = −2πjkabE0 cos θ sinc(X) sinc(Y )

(ejkr

r

). (5.37)

5.2 Spherical nanoparticle as a simple nano-antennaAs we mentioned in the introduction to antenna theory, the nano-natennae serve asinterfaces between microscopic transmitters/receivers such as atoms and/or moleculesand macroscopic radiation. As an example, we displayed in the figure below a typicalmolecule probing (excitation) geometry. A small spherical nanoparticle, serving as anano-antenna, is attached to the tip of a tapered waveguide or fiber. An excited atom ormolecule radiates electromagnetic fields. As the nano-antenna is brought into the vicin-ity of the excited atom/molecule, the incoming field induces a time-dependent dipolemoment of the nanoparticle—as we we are showing below—and the latter radiates intospace. A part of the radiation is converted into propagating waves within the taper andis captured by a detector camera, for example. The nanoparticle radiation is found tobe strongly frequency sensitive. In particular, it can be strongly enhanced at the fre-quencies corresponding to localized surface plasmons excited in the nano-antenna aswe are discussing below. Alternatively, an atom/molecule probe can be excited by firstexciting the nano-sphere with the electromagnetic waves propagating in the taper. Ineither case, the nano-sphere provides an interface between propagating waves in the(macroscopic) taper and a microscopic transmitter/receiver.

Hereafter, we consider nanoparticle excitation by a linearly polarized plane electro-magnetic wave, which illustrates the simplest nano-particle excitation mechanism. Weassume the plane wave of the wavelength λ, propagating in a homogenous backgroundmedium with the dielectric permittivity εb is impinged on a small metal nanoparticle ofradius a, embedded in the medium. The dielectric permittivity of the particle ε(ω) is a

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Figure 5.1: Illustrating small spherical nanoparticle attached to a taper as a nano-antenna. Reproduced from P. Bharadwaj et. al, Adv. in Opt. and Photon., 1, 438-483,(2009).

Figure 5.2: Illustrating small spherical nanoparticle excitation by a linearly polarizedplane wave.

function of the incident light frequency ω. The geometry of the problem is indicated inthe figure above. We assume, for simplicity that the incident wave is linearly polarizedand choose the z-axis along the plane wave field such that the complex amplitude ofthe electric field can be written as

Ei = ezE0ej(kby−ωt). (5.38)

We assume the particle to be of the subwavelength size, ka 1, where kb = k√εb/ε0

is the wavenumber in the background medium and k = ω/c = 2π/λ is a free spacewavenumber. Under the circumstances, the spatial variation of the wave can be ne-glected in the interior of the particle because ejka ' 1. Hence, a complex amplitude of

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the field can be well approximated by a time-harmonic uniform field, i.e.,

Ei = ezE0e−jωt. (5.39)

Let us now estimate the orders of magnitude of the vector and scalar potential. Itfollows at once from the Lorentz gauge condition, (3.67) that ∇ ·A ∼ A/a ∼ kV/c,implying that A ∼ (a/λ)V/c, i.e, for nanoparticles, the scalar potential dominates andthe electric field can be determined from the quasi-static condition,

E(r) = −∇V (r). (5.40)

The time-harmonic potential and electric field can then simply given by

V (r, t) = V (r)e−jωt, E(r, t) = E(r)e−jωt. (5.41)

Further by estimating the order-of-magnitude values of the two terms on the l.s.d. ofEq. (3.69), we have ∇2V ∼ V/a2 and k2V ∼ V/λ2, implying that k2V ∇2V andthe scalar potential obeys the Laplace equation,

∇2V (r) = 0. (5.42)

Next, the time-independent part of the incident electric field can be generated fromthe potential,

Vi = −E0z = −E0r cos θ, (5.43)

where we converted to the spherical coordinates better reflecting the symmetry of thesystem. The problem of finding the electromagnetic fields scattered by the particlereduces to the determination of the scalar potential inside and outside the nanospherefrom Eq. (5.42), subject to the following boundary conditions

en12 × (E1 −E2)|r=a = 0, (5.44)

anden12 · (D1 −D2)|r=a = 0. (5.45)

Here we label the interior of the sphere as medium 1 and the exterior as medium 2. Itfollows that en12 = er, the constitutive relations read

D1 = ε(ω)E1; D2 = εbE2. (5.46)

Combining Eqs. (5.40) and (5.44) through (5.46), we arrive at

er ×∇V1|r=a = er ×∇V2|r=a, (5.47)

andε er · ∇V1|r=a = εb er · ∇V2|r=a. (5.48)

It follows at once from the last two equations that in the spherical coordinates, theboundary conditions are satisfied provided that

V1(a, θ, φ) = V2(a, θ, φ), (5.49)

85

andε ∂rV1|r=a = εb ∂rV2|r=a. (5.50)

Next, examining the boundary condition structure—see Eqs. (5.49) and (5.50)—aswell the the incident wave structure, Eq. (5.43), we conclude that (i) the system hasazimuthal symmetry, implying that we can seek the potential independent of φ and (ii),we can look for solutions in the form

V1 = f1(r) cos θ, (5.51)

andV2 = [−E0r + f2(r)] cos θ. (5.52)

The Laplace equation reduces to

1r2 ∂r(r

2 ∂rV1,2) + 1r2 sin θ∂θ(sin θ ∂θV1,2) = 0. (5.53)

On substituting from Eq. (5.51) into Eq. (5.53) and doing the derivatives with respect toθ, we observe that both terms on the r.h.s contain the same angular dependence, cos θwhich can be cancelled, leading to the equation for the radial profile as

d

dr

(r2 df1

dr

)− 2f1 = 0, (5.54)

or

r2 d2f1

dr2+ 2r

df1

dr− 2f1 = 0. (5.55)

We can seek a solution to Eq. (5.55) in the form of a polynomial as

f1(r) ∝ rs, (5.56)

On substituting from Eq. (5.56) into (5.55), we observe that the latter is satisfied pro-vided s satisfies the quadratic equation

s2 + s− 2 = 0, (5.57)

with the roots, s1 = 1 and s2 = −2. Hence a general solution reads,

f1(r) = C1r +C2

r2, (5.58)

where C1 and C2 are yet unknown constants. Notice that the potential must be finiteanywhere inside the sphere, including the origin, r = 0. It follows at once that C2 = 0,implying a general solution inside the sphere

V1 = C1r cos θ, (5.59)

where C1 is to be determined by the boundary conditions.

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Next, on substituting from Eq. (5.51) into Eq. (5.53) and doing the derivatives withrespect to θ, we observe that both terms on the r.h.s contain the same angular depen-dence, cos θ which can be cancelled again, leading to the equation for the radial profileas

d

dr

[(r2 df2

dr− E0

)]− (−E0r + f2) = 0, (5.60)

implying that

r2 d2f2

dr2+ 2r

df2

dr− 2f2 = 0, (5.61)

which is exactly the same equation as Eq. (5.55) which we have just solved. It followsthat a general solution can be written as

f2(r) = B1r +B2

r2, (5.62)

where B1 and B2 are constants to be determined. In particular, as r → ∞, V2 → 0for any realistic potentials, implying that B1 = 0. Thus, a general solution outside thesphere reads

V2 =

(−E0r +

B2

r2

)cos θ. (5.63)

On substituting from Eqs. (5.59) and (5.63) into Eq. (5.49) and (5.50), we obtain a setof linear equations for C1 and B2 as

C1a = −E0a+B2

a2, (5.64)

andεC1 = −εbE0 − εb

2B2

a3, (5.65)

which can be easily solved yielding

B2 = E0a3 ε(ω)− εbε(ω) + 2εb

; C1 = − 3εbε(ω) + 2εb

E0. (5.66)

Thus, the potential distribution reads

V (r, θ) =

− 3εbε(ω)+2εb

E0r cos θ, r ≤ a;

−E0r cos θ +(E0a

3

r2

)ε(ω)−εbε(ω)+2εb

cos θ, r ≥ a.(5.67)

Let us now analyze the electric field distribution. Inside the nanoparticle, it followsat once from Eqs. (5.40), (5.43), and (5.67) that the field is uniform along the z-axisand is given by

E1 =3εb

ε(ω) + 2εbE0 ez. (5.68)

Outside the sphere, the potential can be re-written in the form

V = −E0z +p · r

4πεbr3, (5.69)

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where the second contribution is recognized as the potential of a dipole with the dipolemoment

p = 4πεbE0a3

(ε− εbε+ 2εb

)ez. (5.70)

As the dipole moment is induced by an external field E0 = E0 ez , we can introducethe polarizability α(ω) viz.,

p(ω) = ε0α(ω)E0, (5.71)

implying that the polarizability is given by

α(ω) = 4πa3

(εbε0

) [ε(ω)− εbε(ω) + 2εb

]. (5.72)

Note that, in physical terms, the polarizability determines the strength of an individualdipole polarization. As a consequence, if we have Nv dipoles per unit volume, theelectric flux density is given by

D = ε0E0 + P = ε0E0 +Nvp = ε0(1 +Nvα)E0. (5.73)

Incidentally, it is obvious from Eq. (5.73) that α has units of volume (cubic meters).The electric field outside the sphere can be read off using Eq. (3.109) as

E2 = E0(er cos θ − eθ sin θ) +p

4πεbr3(2er cos θ + eθ sin θ), (5.74)

where we notice that the first term is just a uniform field along the z-axis because

ez = er cos θ − eθ sin θ. (5.75)

An important parameter characterizing this nanoantenna is the total scattering cross-section. It is defined as a ratio of the time-averaged total scattered power—which wecan be calculated through any spherical shell as it is independent of the shell radius aswe will show—to the time-averaged incident field intensity,

σtot =〈Ptot〉〈P0〉

. (5.76)

The incident intensity can be determined using Eq. (3.161) as

〈P0〉 = 12ηb|E0|2, (5.77)

where ηb = (µ0/εb)1/2 is a background medium impedance; the medium is assumedto be nonmagnetic, µb = µ0, at optical frequencies which are of chief interest here.Exercise 5.2. Determine the total scattering cross-section for light scattering by aspherical nanoparticle in the far zone.Solution.—The time-averaged total scattered power through a sphere of radius r isgiven by

〈Pscatt〉 =

∫dΩ r2〈P〉, (5.78)

88

where dΩ = dφdθ sin θ is a differential solid angle element and the scattered intensitycan be expressed as

〈P〉 = 12ηb|E|2. (5.79)

It follows at once from Eqs. (5.77), (5.78) and (5.79) that

σscatt =

∫dΩ r2|E|2

|E0|2. (5.80)

We can now work it out the total scattering cross-section in the far zone by recallingthat the dipole field there is given by Eq. (3.113). It then follows that

|E|2 =|p|2ω4

16π2c2r2η2

b sin2 θ. (5.81)

Combining Eqs. (5.79) and (5.81) with (5.80), and using Eq. (5.71), yields

σscatt =ω4|α(ω)|2

16π2c2εbµ0ε

20〈 sin2 θ〉, (5.82)

where we introduced〈 sin2 θ〉 =

∫dΩ sin2 θ = 8π/3, (5.83)

c.f., Eq. (3.167). Notice the total cross-section is independent of the distance from thesphere as asserted. Also, using Eq. (5.83) in Eq. (5.82) and noticing that µ0ε0 = 1/c2,leads to the final result

σscatt(ω) =k4

6πεb/ε0|α(ω)|2. (5.84)

To calculate the total scattering cross-section dependence on the incident wave fre-quency, we have to work out the polarizability magnitude of a metal nano-sphere. Tothis end, we notice that the permittivity of a metal at optical frequencies is a complexfunction,

ε(ω) = ε′(ω) + jε′′(ω); |ε′′(ω)| |ε′(ω)|, (5.85)

where the real part characterizes the dielectric polarizability and the imaginary oneaccounts for light absorption by the sphere. It follows at once from Eqs. (5.72) that

|α(ω)|2 = (4πa3)2

(εbε0

)2[ε′(ω)− εb]2

[ε′(ω) + 2εb]2 + ε′′2(ω); (5.86)

We can infer from Eqs. (5.86) and (5.84) that the polarizability and hence the scatteringcross-section has a sharp maximum around the resonant frequency determined by thecondition

ε′(ωSP) + 2εb = 0. (5.87)

The frequency ωSP corresponds to a surface plasmon, a standing electromagneticwave excited along the nano-particle circumference. Thus, the scattering cross-sectionof a spherical nanoparticle is strongly enhanced by the surface plasmon presence. Assuch a nano-particle—located at the tip of a taper waveguide—is placed in the vicinity

89

Figure 5.3: Normalized scattering cross-section of small silver and gold sphericalnanoparticles as function of the incident field wavelength for several background me-dia. Reproduced from L. Novotny and B. Hecht, Principles of Nano-Optics, (Cam-bridge U. Press, 2006).

of an atom or a molecular probe, the probe radiation can be resonantly enhanced. Thisis one of the nano-antenna functions. The scattering cross-section enhancement forsilver and gold nanoparticles in three different media—air, water and glass—is illus-trated in Fig. 4.11. The permittivity dependence on frequency was extracted from theexperimental data.

5.3 Nano-antenna applicationsThis lecture will be delivered in the PowerPoint format.

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