JNTUH Electromagnetic Field Study Materials

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Lecture Notes

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Electromagnetic Field

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Contents Contents

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1 Fundamental concepts

1 3

Maxwell’s Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3

1.2

Phasors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

5

1.3

Constitutive relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

9

1.4

Boundary conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

11

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1.1

14

Plane waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

14

2.2

Cylindrical waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

20

2.3

Spherical waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

21

2.4

Waves in non homogeneous media . . . . . . . . . . . . . . . . . . . . . . . . . . .

22

2.5

Propagation in good conductors . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

23

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2 Waves in homogeneous media

3 Radiation in free space

28

Green’s functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

28

3.2

Elementary dipoles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

34

3.3

Radiation of generic sources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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3.1

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4 Antennas 4.1

Antenna parameters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

49

4.1.1

Input impedance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

50

4.1.2

Radiation pattern, Directivity and Gain . . . . . . . . . . . . . . . . . . . .

51

4.1.3

Effective area, effective height . . . . . . . . . . . . . . . . . . . . . . . . . .

54

4.2

Friis transmission formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

57

4.3

Examples of simple antennas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

59

4.3.1

Wire antennas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

60

4.3.2

Aperture antennas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

67

1

CONTENTS

5 Waveguides

76

5.1

Waveguide modes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

77

5.2

Equivalent transmission lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

80

5.3

Rectangular waveguide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

86

5.3.1

Design of a single mode rectangular waveguide . . . . . . . . . . . . . . . .

92

5.3.2

Tunneling effects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

94

5.3.3

Irises and waveguide discontinuities . . . . . . . . . . . . . . . . . . . . . . . 100

A Mathematical Basics

1 1

A.2 Calculus of vector fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

10

A.3 Multidimensional Dirac delta functions . . . . . . . . . . . . . . . . . . . . . . . . .

17

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A.1 Coordinate systems and algebra of vector fields . . . . . . . . . . . . . . . . . . . .

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B Solved Exercises

20 20

B.2 Plane Waves

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

23

B.3 Antennas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

28

B.4 Waveguides . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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B.1 Polarization and Phasors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Bibliography

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45

Chapter 1

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Maxwell’s Equations

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1.1

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Fundamental concepts

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All electromagnetic phenomena of interest in this course can be modeled by means of Maxwell’s equations  ∂    ∇ × E(r,t) = − B(r,t) − M(r,t) ∂t (1.1)  ∂   ∇ × H(r,t) = D(r,t) + J (r,t) ∂t Let us review the meaning of the symbols and the relevant measurement units.

electric field

H(r,t)

magnetic field

D(r,t)

electric induction

C/m2

magnetic induction

Wb/m2

electric current density (source)

A/m2

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J (r,t)

A/m

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B(r,t)

V/m

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E(r,t)

M(r,t)

magnetic current density (source)

[V/m2 ]

Customarily, only electric currents are introduced; it is in particular stated that magnetic charges and currents do not exist. However, it will be seen in later chapters, that the introduction of fictitious magnetic currents has some advantages: • The radiation of some antennas, such as loops or horns, is easily obtained • Maxwell’s equations are more symmetric 3

4 • (surface) magnetic currents are necessary for the formulation of the equivalence theorem, a fundamental tool for the rigorous modelling of antennas In circuit theory, one has two types of ideal generators, i.e. current and voltage ones: likewise, in electromagnetism one introduces two types of sources.

Concerning the symmetry of Maxwell’s equations, we cite the principle of duality: performing the exchanges E ↔H

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B ↔ −D J ↔ −M Maxwell’s equations transform into each others.

∂ρ(r,t) =0 ∂t

(1.2)

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∇ · J (r,t) +

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Experiments show that the electric charge is a conserved quantity. This implies that electric current density and electric charge (volume density are related by a continuity equation

∇ · B(r,t) = ρm (r,t)

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By analogy, we assume that also magnetic charges are conserved, so that a similar continuity equation must be satisfied: ∂ρm (r,t) =0 (1.3) ∇ · M(r,t) + ∂t It can be proved that eqs.(1.1), (1.2) (1.3) imply the well known divergence equations

∇ · D(r,t) = ρ(r,t)

(1.4)

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Some authors prefer to assume eqs.(1.1), 1.4) as fundamental equations and obtain the conservation of charge (1.2) (1.3) as a consequence.

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Maxwell’s equations can be written in differential form as above, so that they are assumed to hold in every point of a domain, or in integral form, so that they refer to a finite volume. The integral form can be obtained by integrating eq.(1.1) over an open surface Σo with boundary Γ and applying Stokes theorem: Z Z I d B · ν̂ dΣo − M · ν̂ dΣo E · τ̂ ds = − dt Σo Σo Γ (1.5) Z Z I d D · ν̂ dΣo + J · ν̂ dΣo H · τ̂ ds = dt Σo Σo Γ

In words, the first equation says that the line integral of the electric field, i.e. the sum of all voltage drops along a closed loop, equals the time rate of change of the magnetic induction flux plus the total magnetic current. The second equation says that the line integral of the magnetic field along a closed loop equals the time rate of change of the electric induction flux plus the total electric current.

Then we integrate eq.(1.4) over a volume V with surface Σ and apply the divergence theorem: I

Z B · ν̂ dΣ =

Σ

I ρm dV

V

Z D · ν̂ dΣ =

Σ

ρ dV V

(1.6)

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Closed surface Σ for the application of the divergence theorem.

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Figure 1.2.

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Figure 1.1. Open surface Σo for the application of Stokes theorem. Notice that the orientations of ν̂ and τ̂ are related by the right-hand-rule: if the thumb points in the direction of ν̂, the fingers point in that of τ̂ .

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which is the usual formulation of Gauss theorem.

(1.7)

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The same procedure on eq.(1.2) produces: Z I d ρ dV = 0 J · ν̂ dΣ + dt V Σ

This says that the total current out of a volume equals the time rate of decrease of the internal charge.

1.2

Phasors

Field variables can have any time dependence but a particularly important one is the so called time-harmonic regime. Consider a general time-harmonic electric field in a particular point: E(t) = Ex0 cos(ω0 t + ϕx )x̂ + Ey0 cos(ω0 t + ϕy )ŷ + Ez0 cos(ω0 t + ϕz )ẑ

6 The three cartesian components are sinusoidal functions of time with different amplitudes and phases, but the same frequency. This equation can be transformed in the following way. n o n o n o E(t) = R Ex0 ej(ω0 t+ϕx ) x̂ + R Ey0 ej(ω0 t+ϕy ) ŷ + R Ez0 ej(ω0 t+ϕz ) ẑ ©¡ ¢ ª (1.8) = R Ex0 ejϕx x̂ + Ey0 ejϕy ŷ + Ez0 ejϕz ẑ ejω0 t © jω0 t ª =R Ee The complex vector E is called the phasor of the time-harmonic vector E(t) and is measured in the same units. It can be decomposed into real and imaginary parts:

with

E0 = Ex0 cos ϕx x̂ + Ey0 cos ϕy ŷ + Ez0 cos ϕz ẑ

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E00 = Ex0 sin ϕx x̂ + Ey0 sin ϕy ŷ + Ez0 sin ϕz ẑ

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E = E0 + jE00

Consider again eq.(1.8):

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© ª E(t) = R E ejω0 t © ª = R (E0 + jE00 ) ejω0 t = R {(E0 + jE00 ) (cos ω0 t + j sin ω0 t)} = E0 cos ω0 t − E00 sin ω0 t

E"

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We have obtained a representation of the time-harmonic electric field as the sum of two vectors, with arbitrary directions, in time quadrature one with respect to the other: in other words both these vectors are sinusoidal functions of time, with the same frequency but with a relative delay of a quarter of a period. This representation is useful to study the polarization of the time-harmonic vector, i.e. the form of the curve traced by the vector E(t) as a function of time. It can be shown

E'

E(t )

Figure 1.3.

Elliptically polarized field

that, in general, this curve is an ellipse inscribed in a parallelogram that has E0 and E00 as half medians, as shown in Fig. 1.3. We see easily from the previous equation that

7 • for t = 0, E = E0 • for t = T /4, E = −E00 • for t = T /2, E = −E0 • for t = 3T /4, E = E00 where T = 2π/ω is the period. Hence the sense of rotation is from E0 to −E00 . In this case the field is said to be elliptically polarized.

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There are particular cases. When E0 and E00 are parallel or one of the two is zero, the parallelogram degenerates into a line and the polarization is linear. The two cases can be condensed in the single condition (cross product, i.e. vector product): E0 × E00 = 0

|E0 | = |E00 |

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The other particular case is that in which

E0 · E00 = 0

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The parallelogram becomes a square and the ellipse a circle: the field is circularly polarized.

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The plane defined by the two vectors E0 and E00 is called polarization plane. Suppose we introduce a cartesian reference in this plane with the z axis orthogonal to it. The phasor E has only x and y components, E = Ex x̂ + Ey ŷ

or

where Ex and Ey are complex numbers. This means that the original time-harmonic field is represented as the sum of two sinusoidally varying orthogonal vectors, with arbitrary magnitudes and phases. On this basis, the type of polarization is ascertained with the following rules:

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• if the phase difference between the two components δ = arg Ey − arg Ex is 0 or π the polarization is linear, along a line that forms an angle ψ = arctan(|Ey |/|Ex |) (if δ = 0) or ψ = − arctan(|Ey |/|Ex |) (if δ = π)

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• if δ = ±π/2 and |Ey | = |Ex | the polarization is circular, clockwise if δ = π/2, counterclockwise if δ = −π/2 • in the other cases, the polarization is elliptical

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To illustrate these concepts consider the following example.

Example The phasor of a magnetic field is H = (1 + j)x̂ + (1 − 3j)ŷ. Determine the type of polarization, write the expression of the time varying field H(t) and draw the polarization curve defined by this vector.

Solution Find real and imaginary part of the phasor H0 = x̂ + ŷ

H00 = x̂ − 3ŷ

8 Compute H0 × H00 = (x̂ + ŷ) × (x̂ − 3ŷ) = (−3 − 1)ẑ 6= 0 H0 · H00 = (x̂ + ŷ) · (x̂ − 3ŷ) = 1 − 3 = −2 6= 0 The polarization is neither linear nor circular, hence it is elliptical counterclockwise (H(t) goes from H0 to −H00 ).

The time varying field is H(t) = (x̂ + ŷ) cos ω0 t − (x̂ − 3ŷ) sin ω0 t

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The plot is shown in Fig. 1.4 y 4

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−2

−3

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Polarization curve defined by H(t) above

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Figure 1.4.

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The time-harmonic regime is important because of the property

ª ª © d d © E(r,t) = R E(r) ejω0 t = R jω0 E(r) ejω0 t dt dt

(1.9)

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so that time derivatives become algebraic operations. If we substitute the representation (1.8) into (1.1) we obtain, after canceling common factors exp(jω0 t): ∇ × E(r)

=

−jω0 B(r) − M(r) (1.10)

∇ × H(r) =

jω0 D(r) + J(r)

If the time dependence is of general type, eq. (1.8) is generalized by the spectral representation (inverse Fourier transform) Z ∞ 1 E(r,ω) ejωt dω (1.11) E(r,t) = 2π −∞

9 In words, a generic time varying electric field is represented as a sum of an infinite number of harmonic components of all frequencies and amplitude E(r,ω)dω/(2π), where Z ∞ E(r,ω) = E(r,t) e−jωt dt (1.12) −∞

(direct Fourier transform). E(r,ω) is a spectral density of electric field, hence it is measured in V/(m Hz). Due to the fact that E(r,t) is real, E(r, − ω) = E∗ (r,ω), so that the previous equation can also be written ¾ ½ Z ∞ 1 jωt E(r,ω) e dω E(r,t) = 2R 2π 0

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in terms of positive frequencies only.

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The importance of the spectral representation is related to the property Z ∞ Z ∞ 1 d d 1 jωE(r,ω) ejωt dω E(r,ω) ejωt dω = E(r,t) = 2π −∞ dt 2π −∞ dt

=

(1.13)

jωD(r,ω) + J(r,ω)

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∇ × H(r,ω) =

−jωB(r,ω) − M(r,ω)

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∇ × E(r,ω)

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If we take the Fourier transform of (1.1), we get

Constitutive relations

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1.3

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While ω0 is a specific frequency value, (1.13) must be satisfied for all frequencies. We refer to these system as Maxwell’s equations in the frequency domain. The variables will be interpreted as phasors or Fourier transforms, depending on the application.

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Clearly Maxwell’s equations as written in the previous section form an underdetermined system: indeed there are only two equations but four unknowns, the two fields and the two inductions. It is necessary to introduce the constitutive relations, i.e. the equations linking the inductions to the fields. From a general point of view, matter becomes polarized when it is introduced into a region in which there is an electromagnetic field, that is, the electric charges at molecular and atomic level are set in motion by the field and produce an additional field that is summed to the original one. The inductions describe the electromagnetic behaviour of matter.

The simplest case is that of free space in which B(r,t)

=

µ0 H(r,t) (1.14)

D(r,t) =

²0 E(r,t)

where ²0 , dielectric permittivity, and µ0 magnetic permeability, have the values µ0 ²0

= 4π · 10−7 H/m 1 1 · 10−9 ≈ = 36π µ0 c2

F/m

10 and the speed of light in free space c has the value c = 2.99792458 · 108

m/s.

In the case of linear, isotropic, dielectrics, the constitutive relations (1.14) are substituted by B(r,ω)

=

µ(ω) H(r,ω)

D(r,ω) =

ε(ω) E(r,ω)

(1.15) where µ0 µr (ω) ²0 ²r (ω)

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µ(ω) = ²(ω) =

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and µr (ω), ²r (ω) (pure numbers) are the relative permittivity and permeabilities. All non ferromagnetic materials have values of µr very close to 1. Notice that since molecular and atomic charges have some inertia, they cannot respond instantaneously to the applied field, so that the response depends on the time rate of change of the excitation. The description of such a mechanism is best performed in the frequency domain, where ε(ω) and µ(ω) play the role of transfer functions.

The fact that they depend on frequency is called dispersion: hence free space is non dispersive. In general ε(ω) and µ(ω) are complex: µ = µ0 − jµ00

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ε = ε0 − jε00

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It can be shown that the volume density of dissipated power in a medium is related to their imaginary part 1 1 Pdiss = ε00 |E|2 + µ00 |H|2 2 2 Notice that ε00 , µ00 are positive in a passive medium because of the phasor time convention exp(jω0 t). Some authors use the convention exp(−jω0 t): in this case passive media have negative ε00 , µ00 . Clearly with the time convention we adopt, negative ε00 , µ00 characterize active media, such as laser media.

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When the dielectric contains free charges, the presence of an electric field E(r,ω) gives rise to a conduction current density Jc (r,ω): Jc (r,ω) = γ(ω) E(r,ω)

where γ(ω) is the conductivity of the dielectric, measured in S/m. The previous equation is the microscopic form of Ohm’s law of circuit theory.

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The conduction current enters into the second Maxwell’s equation (1.13), which becomes ∇ × H = jωεE + γE + J

where the term J is the (independent) source. It is customary to incorporate the conduction current into the polarization current by means of an equivalent permittivity. Indeed we can write ³ γ´ E = jωεeq E jωεE + γE = jω ε − j ω

with εeq = ε0 − j(ε00 + γ/ω). In practice the subscript eq is always dropped: the imaginary part of ε takes into account all loss mechanisms, both those due to molecular and atomic transitions and those due to Joule effect.

11 In the case of low loss dielectrics one often introduces the loss tangent tan δ =

so that we can write

ε00 ε0

ε = ε0 (1 − j tan δ)

Values of tan δ ' 10−3 : 10−4 characterize low loss dielectrics.

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It is interesting to note that for fundamental physical reasons, there is relationship between the real and the imaginary part of the dielectric permittivity and of the magnetic permeability.

Indeed, in the case of the permittivity, just as a consequence of causality, ε0 (ω) − ε0 and ε00 (ω) are Hilbert transforms of each other, that is Z ∞ 00 ε (α) 1 dα ε0 (ω) − ε0 = P π −∞ α − ω Z ∞ 0 ε (α) − ε0 1 00 dα ε (ω) = P α−ω π −∞

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These equations are called Kramers-Krönig relations. The symbol P denotes the Cauchy principal value of the integral, that is, for f (0) 6= 0 ¾ ½Z −a Z ∞ Z ∞ f (x) f (x) f (x) dx dx + dx = lim P a→0 x x a −∞ −∞ x

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The constitutive equations (1.15) imply that the inductions are parallel to the applied fields, which is true in the case of isotropic media but not in the case of crystals. These media are said to be anisotropic and are characterized by a regular periodical arrangement of their atoms. In this case the permittivity constitutive equations must be written in matrix form:      Dx εxx εxy εxz Ex  Dy  =  εyx εyy εyz   Ey  Dz εzx εzy εzz Ez

1.4

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In the case of an isotropic dielectric, the matrix is a multiple of the identity: ε = εI.

Boundary conditions

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Maxwell’s equations (1.13) are partial differential equations (PDE), valid in every point of a given domain, which can be finite or infinite. If it is finite, we must supply information about the nature of the material that forms the boundary. In mathematical terms, we must specify the boundary conditions, i.e. the values of the state variables on the boundary.

Often the boundary is a perfect electric conductor (PEC), that is a material with infinite conductivity. Note that copper is such a good conductor that up to microwave frequencies it can be modeled as a PEC. If the conductivity is infinite, the electric field must vanish everywhere in the volume of a PEC, otherwise the conduction current would be infinite. The first Maxwell’s equation shows that also the magnetic field is identically zero, provided the frequency is not zero.

Since we are essentially interested in time-varying fields, we conclude that in a PEC both fields vanish identically. At the surface, since the conduction current cannot have a normal component,

12 only the tangential component of the electric field must be zero. If ν̂ is the unit normal at the boundary, this condition can be written ν̂ × E = 0

on the boundary

(1.16)

Indeed, ν̂ × E is tangential to the boundary, as shown in Fig. 1.5.

ν̂

ν̂

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E tg

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Figure 1.5. Boundary condition at the surface of a perfect electric conductor. Relationship between the tangential component of the electric field Etg and ν̂ × E

ν̂ × (H(rΣ+ ) − H(rΣ− )) = 0

or

ν̂ · (B(rΣ+ ) − B(rΣ− )) = 0

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Sometimes the permittivity or the permeability change abruptly crossing a surface in the domain. By applying some integral theorems to Maxwell’s equations, it can be shown that the following continuity conditions hold ν̂ × (E(rΣ+ ) − E(rΣ− )) = 0

(1.17)

ν̂ · (D(rΣ+ ) − D(rΣ− )) = 0

(1.18)

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where ν̂ is the normal to the surface and rΣ± are infinitely close points, lying on opposite sides of the surface, as shown in Fig. 1.6. It can also be proved that if a surface current Js or Ms exist, then the fields are discontinuous

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ν̂

rΣ−

ε 2 µ2

rΣ+

Σ

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ε1 µ1

Figure 1.6.

Boundary conditions at the surface of separation between different dielectric media

ν̂ × (H(rΣ+ ) − H(rΣ− )) = Js

ν̂ × (E(rΣ+ ) − E(rΣ− )) = Ms

Since it can be proved that when a PEC is present in a magnetic field, the induced current flows only on the surface of it and the magnetic field is identically zero in the PEC, we can write ν̂ × H(rΣ+ ) = Js

(1.19)

13 Notice that this is an equation that does not force a condition on H but allows the computation of Js . Hence the boundary condition to be enforced at the surface of a PEC is only (1.16). Notice also that the units of a surface electric current are A/m and those of a surface magnetic current V/m.

This is necessary for the validity of the previous equations, but it is also obvious for geometrical reasons, as Fig. 1.7 shows. ν̂ γ

Js

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Figure 1.7. Surface electric current induced on a perfect electric conductor (PEC). γ is a curve lying on the PEC surface. Js is a current density per unit length measured along γ (A/m)

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Sometimes an approximate boundary condition of impedance type is used: it is a linear relation between the tangential electric and magnetic fields, called also a Leontovich boundary condition, that can be written ν̂ × E(rΣ+ ) = Zs (ν̂ × H(rΣ+ )) × ν̂ (1.20)

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in terms of a suitable surface impedance. This condition is typically applied when the boundary is a real metal and one desires a more accurate model than just a PEC. The double vector product on the right hand side makes the tangential electric and magnetic field orthogonal on the surface.

If the surface is not smooth but has a sub-wavelength wire structure, the surface impedance is not a scalar but a tensor (matrix).

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If the domain is infinite and sources are all at finite distance from the origin, the only necessary boundary condition is that the field is only outgoing at large distance from the origin.

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Sometimes the geometry of dielectric and metal bodies possesses sharp edges or sharp vertices (e.g. plates, cubes, cones), as shown in Fig. 1.8. In this case some field components can become infinite at the geometric singularity: however the fields must always be locally square integrable.

In physical terms this condition means that the electromagnetic energy contained in a finite neighborhood of the singularity must always be finite. Apart from these cases of true singularities of the geometry, fields are always regular, i.e. differentiable. This is to be noted in particular when the geometry singularity is only apparent because it is related to the particular coordinate system.

For example if we use cylindrical coordinates in a homogeneous medium the fields must be regular in the origin even if the coordinates have a singularity there.

,ĂůĨƉůĂŶĞ

tĞĚŐĞ

Figure 1.8.

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Chapter 2

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Waves in homogeneous media

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At this point we have finished the preliminaries. We have decided to use the electric and magnetic field as state variables and we have the relevant equations:  ∇ × E = −jωµH − M    ∇ × H = jωεE + J (2.1)    + boundary conditions

2.1

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where for brevity we have dropped the dependence of all variables on r, ω, but it is understood.

The line “boundary conditions” contains the form of the domain where we want to compute the fields created by the given sources J,M and information on the nature of the material that forms the boundary. In general permittivity and permeability are functions of r and provide information on the shape of the bodies in the domain and on their nature. In this way the problem is well posed and it can be shown it has a unique solution, provided there is at least a region in the domain where energy is dissipated. Needless to say the problem (2.1) can be very complicated and can be solved only by approximate numerical techniques. For this reason it is useful to proceed by small steps, by analyzing first a very idealized problem that is so simple to allow an analytical solution.

Plane waves

w w

Let us start by assuming that the domain of interest is infinite and the medium is homogeneous and lossless, so that ε, µ are real constants. The only boundary condition to enforce is that the field is regular everywhere, in particular at infinity.

Moreover we assume that sources are identically zero. This is, at first sight, a strange assumption since it would seem to imply that also the fields must be identically zero! However, if there are no sources, Maxwell’s equations become a homogeneous system of differential equations: it is well known that homogeneous differential equations have nontrivial solutions. Let us review some examples: •

d f (x) = 0, x ∈ R ⇒ f (x) = const dx

14

15 • Harmonic oscillator ¶ µ 2 d 2 + ω 0 f (t) = 0, t ∈ R ⇒ f (t) = A cos ω0 t + B sin ω0 t dt2

• Transmission line  d   − V = dz   −dI = dz

 + −jkz + V0− e+jkz  V (z) = V0 e

jω L I

, z∈R⇒ jω C V



I(z)

=

Y∞ V0+ e−jkz − Y∞ V0− e+jkz

om

These are actually generalizations of the simple case of a linear system of algebraic equations Ax=0

∇×E

=

−jωµH jωεE

(2.2)

ru

∇×H =

m

So the problem we want to solve is (

.c

which has nontrivial solutions if the matrix A is non invertible.

ld

fo

Equations (2.2) are written in a coordinate-free language. However, in order to solve them it is necessary to select a coordinate system. Several choices are at our disposal, the more common being cartesian, cylindrical and spherical coordinates. The corresponding solutions of (2.2) will be plane, cylindrical and spherical waves, respectively. The simplest case is the first and we start with that.

or

Recalling the expression of the ∇ operator in cartesian coordinates

w

∇ = x̂

∂ ∂ ∂ + ẑ + ŷ ∂z ∂y ∂x

w .jn

tu

and that the medium is homogeneous, it is clear that (2.2) is a linear system of constant coefficient equations. On the basis of the experience with ordinary differential equations, we can expect that the solution is of exponential type, hence we assume tentatively E(r) = E0 exp(−jkx x) exp(−jky y) exp(−jkz z)

(2.3)

w w

and likewise for the magnetic field, where E0 and kx ,ky ,kz are constants to be found. The constants kx ,ky ,kz have dimensions rad/m and are wavenumbers along the three coordinate axes. It is convenient to work with a vector formalism, even if the coordinate system is fixed. By recalling that r = xx̂ + yŷ + zẑ and defining the wavevector k = kx x̂ + ky ŷ + kz ẑ, the assumed form of the solution is E(r) = E0 exp(−jk · r) H(r) = H0 exp(−jk · r) (2.4) Before substituting it into (2.2) it is useful to compute µ ¶ ∂ ∂ ∂ ∇ exp(−jk · r) = x̂ exp(−jkx x) exp(−jky y) exp(−jkz z) + ẑ + ŷ ∂z ∂y ∂x

= (−jkx x̂ − jky ŷ − jkz ẑ) exp(−jkx x) exp(−jky y) exp(−jkz z) = −jk exp(−jk · r)

16

Moreover we recall the identity of vector calculus ∇ × (A(r)φ(r)) = φ(r)∇ × A(r) + ∇φ(r) × A(r) so that the substitution of (2.4) into (2.2) yields −jk × E0 exp(−jk · r)

=

−jωµH0 exp(−jk · r)

−jk × H0 exp(−jk · r) =

jωεE0 exp(−jk · r)

−jk × E0

=

−jωµH0

−jk × H0

=

jωεE0

om

Canceling common factors

.c

Note that the fact that these equations do not contain the variable r any longer confirms the correctness of the assumption (2.4).

ru

k × E0 ωµ

(2.5)

fo

H0 =

m

Recalling the properties of vector products we learn that E0 , H0 , k form a righthanded triple of mutually orthogonal vectors. Next, to proceed, we eliminate H0 between the two equations. To this end, we solve the first equation with respect to H0 :

and substitute into the second one

or

The double vector product can be expanded

ld

k × (k × E0 ) + ω 2 µεE0 = 0

w

k (k · E0 ) − (k · k) E0 + ω 2 µεE0 = 0

w .jn

tu

The first term is zero because of the orthogonality of k and E0 , hence ¡ ¢ k · k − ω 2 µε E0 = 0

(2.6)

We are interested in nontrivial solutions of this equation, so that the following condition must hold k · k = ω 2 µε

(2.7)

w w

A relationship between frequency and wavenumbers is called in general a dispersion relation. We can read it from right to left or vice versa: in the first instance it tells us what the frequency must be so that the field distribution (2.4) with a specific wavevector k is a solution of Maxwell’s equations. From this point of view, the value of ω can be considered to be a resonance frequency of the structure. Notice that the requirement that the solution be regular everywhere (in particular at infinity) forces the wavector to be real. Apart from this condition there is no constrain on the possible values of the wavenumbers, hence the resonance frequencies of the system are infinite in number and even distributed continuously. We start seeing a property that characterizes all field problems. Whereas lumped element circuits have a finite number of resonances, distributed systems always have an infinite number of them. Moreover, if the structure has finite size its resonances are denumerably infinite: this means that they can be labeled with integers ω1 , . . . ,ωn , . . .. If the structure, as in this case, has infinite size, then the resonances should be labeled with a continuous

17 variable. To simplify the notation, we omit this labeling variable and remember that ω can take every real value.

The dispersion equation can also be read from left to right: in this case the frequency is considered fixed and we look for the wavevectors that satisfy eq.(2.7). It is convenient to introduce a unit vector ŝ, called direction of propagation, directed along k, so that k = kŝ; then the dispersion relation becomes √ k = ω εµ

(2.8)

Clearly the direction ŝ can be whatever, only the wavenumber k is specified. In other words for any given frequency there are an infinite number of waves with arbitrary directions of propagation.

m

.c

om

Considering again eq.(2.6), we see that if the dispersion relation is satisfied, the vector E0 can be arbitrarily chosen, provided it is orthogonal to ŝ. For a given ŝ, there are two linearly independent waves, but they are degenerate, because they have the same value of the wavenumber k. The corresponding magnetic field is obtained by the impedance relation (2.5): r √ ω εµŝ × E0 ε ŝ × E0 = Y ŝ × E0 (2.9) = H0 = µ ωµ

In conclusion a solution of the problem (2.2) is

fo

ru

where the wave admittance has been introduced. Its inverse is the wave impedance Z = 1/Y . In free space it has the value r µ0 ≈ 120π ≈ 377 Ω Z0 = ε0

tu

w

or

ld

Eŝ (r) = E0 exp(−jkŝ · r) (2.10) Hŝ (r) = H0 exp(−jkŝ · r) √ where ŝ is the direction of propagation, k = ω εµ, H0 = Y ŝ × E0 and E0 , H0 , ŝ are mutually orthogonal: electromagnetic waves are transverse. Since the problem is linear, the general solution of (2.2) can be written as a linear combination of waves of the type (2.10) with all possible directions ŝ.

w .jn

Wavefronts are defined to be the surfaces on which the phase Φ(r) of the wave is constant. In this case Φ(r) = −kŝ · r = constant is the equation of a family of planes perpendicular to ŝ: hence the fields (2.10) are called plane waves because the wavefronts are planes.

w w

Let us now study the polarization of plane waves. This requires going back to time domain via eq.(1.8) E(r,t) = R {E0 exp(−jkŝ · r) exp(jω0 t)} = E0 0 cos (ω0 t − kŝ · r) − E00 0 sin (ω0 t − kŝ · r)

(2.11)

We see clearly from this equation that the type of polarization in every point of space is specified

by E0 , which is the electric field in the origin. What changes from point to point is the time evolution, due to the propagation delay. The constant phase surfaces of the time varying field are ω0 t − kŝ · r = const, from which we find ω0 t − const ŝ · r = k

18

û

ǁĂǀĞĨƌŽŶƚƐ

ŝ W

E

r

ĐŽŶƐƚсϬ

H

om

Figure 2.1. Wavefronts of a plane wave with direction of propagation ŝ. They move at the phase velocity along ŝ. The vector û denotes an arbitrary direction

c 1 ω0 ω0 = √ =√ =√ εr µr εµ ω0 εµ k

(2.12)

ru

vph =

m

.c

This means that the wavefronts are not fixed but are moving. Indeed, consider a specific wavefront, say the zero-phase one, i.e. the one for which const=0, as shown in Fig.2.1; the vector r that denotes its points is such that its projection on ŝ increases linearly with time. In other words, the plane moves as a whole with speed

fo

This velocity is called phase velocity of the wave, because it has been defined by means of the constant phase surfaces. Now consider an arbitrary straight line with direction û, an let P be its intersection with the zero-phase wavefront. The velocity of P is

c 1 √ û · ŝ εr µr

ld

vph (û) =

tu

w

or

√ Clearly, for all directions û 6= ŝ this velocity is larger than c/ εr µr . Notice, however, that even when this velocity is larger than the speed of light in empty space, the theory of relativity is not violated. Indeed no matter or energy is moving in the direction of û, but only a mathematical point, e.g. a maximum or a node of the oscillation. This concept is at the basis of the fact that the phase velocity in a waveguide is always greater that the speed of light in vacuum.

w .jn

Fig.2.2 shows the fields of a linearly polarized plane wave propagating in the ŝ = ẑ direction: remember that the electric and magnetic field are orthogonal. For clarity, the field vectors have been drawn only for a number of points on the z axis, even if they are defined in every point of space.

w w

Wave propagation is always associated to energy flow. The Poynting vector has the meaning of power density (per unit surface) associated to the wave. Let us compute the Poynting vector S in the case of a plane wave: 1 1 (E × H∗ ) = (E0 exp(−jkŝ · r) × H0 ∗ exp(jkŝ · r)) 2 2 1 1 = (E0 × H0 ∗ ) = (E0 × (ŝ × E0 ∗ )Y ∗ ) 2 2 1 |E0 |2 = ŝ 2 Z

S=

(2.13)

where we used the impedance relation (2.9) and the property E0 × (ŝ × E0 ∗ ) = |E0 |2 ŝ − (E0 · ŝ) E∗0 = |E0 |2 ŝ

(2.14)

19

dž

E( z, t ) LJ

H( z , t )

Snapshot of a linearly polarized plane wave propagating in the ŝ = ẑ direction

m

Figure 2.2.

.c

om

nj

or

ld

fo

ru

because of the orthogonality between E0 and ŝ. Note that the Poynting vector is real, hence no reactive power is associated to plane waves in a lossless medium. We see that the active power density (magnitude of S) associated to a plane wave is constant: this implies that the total power, obtained by integration over the whole wavefront, is infinite. Hence, a single plane wave is not a physically realizable field. This property, however, does not destroy the usefulness of the concept. Indeed, since Maxwell’s equations are linear, the superposition principle holds and linear combinations of plane waves are also solutions. It turns out that a continuous sum (integral) of plane waves not necessarily has infinite power: indeed all physically realizable fields can always be represented as integrals of plane waves.

tu

w

We recall that to each plane wave not only a power flow is associated, but also a flow of linear momentum and of angular momentum. In particular the linear momentum flow, which has direction ŝ, is responsible for the radiation pressure, that explains, for instance, the shape of comet tails and has been considered as a possible “engine” for interplanetary travels.

w .jn

The properties of plane waves do not change much if the dielectric is lossy. In this case the permittivity is complex and the dispersion relation (2.8) becomes p α≥0 k = ω (ε0 − jε00 )µ = β − jα,

w w

where β is the true phase constant, measured in rad/m and α is the attenuation constant, measured in Nepers/m. The electric field, for instance, obeys the propagation law E(r) = E0 exp(−jβŝ · r) exp(−αŝ · r)

(2.15)

Clearly the magnitude of the field is no longer constant in space and the wavefronts are also surfaces of constant field magnitude. Obviously, it must be remarked that the value of the phase velocity cannot be computed by (2.12), but it is given by vph =

and the wavelength is λ=

ω β

vph 2π = f β

20 Indeed, both the phase velocity and the wavelength are defined on the basis of the phase of the wave and β is exactly the phase rate-of-change, measured, as said above, in rad/m.

Let us compute the power flow.

1 1 (E × H∗ ) = (E0 exp(−jkŝ · r) × H0 ∗ exp(jk ∗ ŝ · r)) 2 2 1 1 = (E0 × (ŝ × E0 ∗ )Y ∗ exp(−2αŝ · r)) = Y ∗ |E0 |2 exp(−2αŝ · r)ŝ 2 2

S=

(2.16)

where we have used again the property (2.14). In this case the Poynting vector is complex. The active power density of the wave decreases during propagation because part of it is transferred to the dielectric in the form of heat.

Cylindrical waves

fo

2.2

ru

m

.c

om

All the plane waves considered up to now are called uniform because their propagation direction ŝ is real. If we go through all the steps of the derivation, we realize that even if ŝ is complex (whatever this means!) the expression (2.10) is a valid solution of Maxwell’s equations, although only in a halfspace. Such a generalization leading to non uniform plane waves is required when solving a scattering problem where a plane wave is incident on the interface separating two dielectrics. It is to be remarked that also the plane wave (2.15) is a valid solution only in a halfspace.

Indeed, if ŝ · r → −∞, the field diverges, which is not physically acceptable.

w

or

ld

To solve eqs.(2.2) it is also possible to use a cylindrical coordinate system instead of a cartesian one. The mathematics is considerably more complicated in this case. The reason is that the unit vectors of the cylindrical coordinate system are not constant but change from point to point. As a consequence, the expression of the differential operators is no longer with constant coefficients and the solutions are no longer of exponential type, but are expressed in terms of Bessel functions.

These special functions of mathematical physics were actually introduced, along with many others, in order to solve the wave equation.

w .jn

tu

If a cylindrical or spherical coordinate system is used, Maxwell’s equations (2.2) are not attacked directly but are first transformed into a single second order equation. We write them again for convenience ( ∇ × E = −jωµH ∇×H =

jωεE

w w

Since it is a system of equations, we can eliminate one of the two unknowns. We solve the first equation with respect to H and substitute in the second  ∇×E   =  H −jωµ    ∇ × (∇ × E) = ω 2 εµE

As expected, the second equation contains only the electric field: the price to pay for it is that it is second order in the space derivatives; it is called the curl-curl equation. However its form can be simplified recalling the identity ∇ × (∇ × E) = ∇(∇ · E) − ∇2 E = −∇2 E

21 where we have used the fact that (2.2) do not have sources, hence also the charge density ρ(r) is zero and the electric field has zero divergence, ∇ · E = ρ = 0. We obtain in this way the vector Helmholtz equation ∇2 E + ω 2 εµE = 0 (2.17) Even if it is written in coordinate-free language, its meaning is easily understood in cartesian coordinates only, where E(r) = Ex (r)x̂ + Ey (r)ŷ + Ez (r) Since the unit vectors are not function of r, each cartesian component of the electric field satisfies Helmholtz equation, which then becomes scalar:

om

∇2 ψ + ω 2 εµψ = 0

m

(2) (kρ ρ)e−jkz z e−jmϕ ψ(ρ,ϕ,z) = ψ0 Hm

.c

where ψ(r) denotes any component of E . It is interesting to note that even if we are using cartesian components to represent the electric field, we are not forced to use necessarily cartesian coordinates to specify the observation point, i.e. the components of r. By using the classical method of the separation of variables in cylindrical coordinates, we can find

(2.18)

or

ld

fo

ru

where m = 0, ± 1, ± 2, . . ., kρ ∈ [0,∞) and kz identify the various outgoing cylindrical waves. These (2) three parameters play the role of kx , ky , kz in the case of plane waves. The function Hm (kρ ρ) is a Hankel function of second kind and order m. Its asymptotic expansion is s h 2 π π i (2) Hm (kρ ρ) ∼ exp −j(kρ ρ − m − ) 4 2 πkρ ρ

The dispersion relation is

w

kz =

q ω 2 εµ − kρ2

tu

Notice that all three components have this form, but the values of ψ0 for each must be interrelated so that the resulting vectors E and H satisfy Maxwell’s equations.

Spherical waves

w w

2.3

w .jn

We are not going to describe in detail the properties of these waves. To explain the name, it is enough to say that the wavefronts are cylinders having ẑ as axis, at least in the case m = 0 and √ kρ = ω εµ.

The case of spherical waves is similar, from a certain point view, to that of cylindrical waves. Again the mathematics is fairly complicated and new special functions are introduced. In this case the scalar Helmholtz equation is solved in spherical coordinates and the result is (2)

ψ(r,ϑ,ϕ) = ψ0 hl (kr)Plm (cos ϑ)e−jmϕ where l = 0,1,2, . . . and −l ≤ m ≤ l identify the various outgoing spherical waves. The functions (2) hl (kρ) are spherical Hankel functions of second kind and order l, whereas Plm (cos ϑ) are associated Legendre polynomials of degree l and order m. Again, the various solutions for the three cartesian

22

ε1 µ1 r

ε2 µ2

ε3 µ3

t ε4

µ4

i

om

Figure 2.3. Scattering from a stratified dielectric: i, incident wave; r reflected wave; t, transmitted wave. For clarity, the couple of plane waves existing in each of the internal layers has not been indicated

m

.c

components must be related so that the resulting E and H satisfy Maxwell’s equations. The asymptotic expansion of the spherical Hankel functions is h π π i 1 (2) hl (kr) ∼ exp −j(kr − m − ) 4 2 kr

Waves in non homogeneous media

fo

2.4

ru

Hence the wavefronts are spheres with center in the origin and this justifies the name.

w .jn

tu

w

or

ld

The case that has been considered, namely a homogeneous medium filling the whole space is highly idealized. In a realistic situation, ε(r), µ(r) are not constant and obviously the plane waves (2.10) are not solution of Maxwell’s equations (2.2). In order to consider a simple case, let us assume that the medium is piecewise homogeneous and that the interfaces between the different materials are planar: the structure is called a stratified dielectric. In the left half space an incident plane wave is assumed. In each layer, plane waves are solutions of (2.2), but the continuity conditions (1.17) must be obeyed. It can be proved that in each one of the internal layers two plane waves are present, one forward (incident on the following interface) the other backward (reflected from the following interface); in the right half space only one, because the medium extends to infinity and there is no other interface. All these plane waves have the same transverse (to z) component of the wavevector and their amplitudes can be easily determined so that the continuity conditions are satisfied. The single wave in the fourth medium is the transmitted field, the second one in the first medium is the reflected field, as sketched in Fig. 2.3.

w w

If the interfaces are not planar, the problem becomes much more difficult. Consider, for example, the case of Fig. 2.4, where a plane wave is incident on a cylinder with parameters ε2 , µ2 , embedded in a homogeneous medium with parameters ε1 , µ1 . It can be shown that the continuity conditions require that an infinite number of plane waves are excited, each one with the right amplitude. Collectively. these are called scattered waves. Hence the difficulty of the problem stems from the necessity to solve a linear system with an infinite number of unknowns.

If the medium is not even piecewise homogeneous but arbitrarily inhomogeneous, no analytical solution is at our disposal. It is, however, to be mentioned that when the variations of ε(r), µ(r) are small on the wavelength scale, a well known approximate method can be used, i.e. Geometrical Optics. Whereas the plane waves discussed up to now can be defined global plane waves since each one is defined over the whole space, the elementary geometrical optics field is a local plane wave.

For instance, a spherical wave in free space can be approximated by a collection of local plane

23

s

ε2 µ2

s

ε1 µ1

s

s

i

om

Figure 2.4. Scattering from a non planar interface: i, incident wave; s, scattered waves. For clarity, the plane waves existing inside the cylinder have not been indicated

m

.c

waves because its wavefront (a sphere) can be approximated locally by the relevant tangent plane.

The k vectors of these local plane waves define a vector field, whose field lines are the geometrical optics rays. It turns out that rays are also the field lines of the Poynting vector field: hence a plot of the rays provides information about the power flow.

Propagation in good conductors

or

2.5

ld

fo

ru

Geometrical optics is a very powerful technique, but sometimes yields definitely wrong results.

This happens when rays cross in a point or along a line, because in this case it predicts a field of infinite intensity. These singularities are called caustics and the focus of a converging lens is an example: in such a point the electromagnetic field can be very large but is certainly finite. Hence geometrical optics can be safely used only away from caustics.

w .jn

tu

w

Apart from the case of optical fibers, guided wave propagation is possible in structures containing metal conductors. Examples are coaxial cables, parallel wire transmission lines, microstrip lines, waveguides with any cross section. Since the metals used in the applications (such as copper) are characterized by a very large conductivity, in a first approximation they can be considered to be perfect conductors (PEC), an assumption that greatly simplifies the study. However, in order to build more accurate numerical models of real devices, it is necessary to take into account the finite conductivity of real metals. In this section we consider the propagation of plane waves in good conductors, in order to draw some conclusions pertaining to transmission systems.

w w

Metals are characterized by so a large conductivity that the displacement currents can be safely neglected with respect to the conduction currents, so that some simplifications in the general formulas of Section 2.1 are possible. Starting with the wavenumber, p p km = ω (ε − jγ/ω)µ ≈ ω (−jγ/ω)µ

if

Recalling that

γ À1 ωε

(good conductor) p 1−j −j = ± √ 2

24 and that Imk ≤ 0 for a passive medium, we find

1−j 1−j√ ωγµ = km = √ δ 2

(2.19)

where we have introduced the so called skin depth r 2 δ= ωµγ

(2.20)

om

which, of course, should not be confused with the loss angle, introduced in Section 1.3, indicated with the same symbol. This relation can also be written r p 1 fδ = = const πµγ

m

.c

where f is the frequency and the constant depends only on the material. For instance, in the case of copper, γ = 5.8 · 107 S/m and µ = µ0 = 4π · 10−7 H/m, hence p √ f δ = 0.0661 Hzm (2.21)

ru

The reason for the name will be explained below.

ld

fo

The wave impedance is computed with the same approximation: s r r r 1 + j ωµ jµω µ µ = √ Zm = = ≈ γ γ −jγ/ω ε − jγ/ω 2

or

that is

Zm = Rs (1 + j)

(2.22)

w w

w .jn

for which we can write

tu

w

where we have introduced the surface resistance r 1 ωµ = Rs = γδ 2γ

R √s = f

r

πµ = const γ

x E

i

Et

Hi Hr

Ht

Er

ŵĞƚĂů

ĨƌĞĞ ƐƉĂĐĞ

z

Figure 2.5. Good conductor in a plane wave field. In the free space region both an incident and a reflected wave exist, in the metal only the transmitted one. The wavevector of the transmitted wave is drawn dashed, to indicate that it is complex.

25 Again, in the case of copper,

√ R √ s = 2.6090 · 10−7 Ω/ Hz f

Consider now a (highly idealized) conductor in the form of a half space, which faces free space, with a linearly plane wave incident normally on it, as shown in Fig. 2.5. The tangential electric and magnetic fields are continuous at the interface, then the ratio of their magnitudes is the same in z = 0− and in z = 0+ . But in z = 0+ this ratio is Zm by definition, so we can easily understand that the expressions of the electric fields are Ei = E0 e−jk0 z x̂ Er = ΓE0 ejk0 z x̂

Γ=

Zm − Z0 Zm + Z0

.c

where the reflection coefficient is

om

Et = (1 + Γ)E0 e−jkm z x̂

m

Since |Zm | ¿ Z0 , Γ is close to −1. Indeed,

from which

fo

s

ru

s 2Zm πε0 f 2Zm = 2(1 + j) ≈ 1+Γ= γ Z0 Zm + Z0

ld

Γ ≈ −1 + 2(1 + j)

In the case of copper,

πε0 f γ

or

Γ ≈ −1 + 2(1 + j) · 6.9252 · 10−10

p f

tu

w

(frequency in Hz). We can also say that the metal enforces an impedance type boundary condition (see (1.20)) with Zm as surface impedance.

The magnetic field is

w .jn

Hi = Y0 E0 e−jk0 z ŷ Hr = −Y0 ΓE0 ejk0 z ŷ t

H = Y0 (1 − Γ)E0 e

−jkm z

(2.23) x̂ ≈ 2Y0 E0 e

−jkm z

ŷ

w w

Note that the total magnetic field at the interface is approximately twice the incident one because Γ is very close to −1.

The electric field in the metal produces a conduction current in the x̂ direction p Jc = γEt = 2(1 + j) πε0 f γE0 e−jkm z x̂

In the case of copper, this becomes p Jc = 2(1 + j) · 0.402 f E0 e−jz/δ e−z/δ x̂

(frequency in Hz) where we have used (2.19). The magnitude of this current density is maximum at the interface and then decays exponentially in the metal. At a depth z = δ, it has reduced by a factor 1/e = 0.368. Eq.(2.21) allows a simple computation of δ for various frequencies, reported in

26

Skin depth for copper at various frequencies

Frequency

Skin depth

50 Hz 1 kHz 1 MHz 1 GHz

9.3 mm 2.1 mm 66.1 µm 2.1 µm

om

Table 2.1.

fo

ru

m

.c

Table 2.1. We see clearly that as the frequency increases, the current density remains appreciable only in a very thin layer close to the metal surface, which can be considered as its “skin”. Even if this analysis strictly refers to a metal half space, we can use it to draw qualitative conclusions in the case of finite thickness conductors or even round conductors, provided the thickness is much larger than the skin depth. At the power frequency of 50Hz, the skin depth is so large that the current has a uniform distribution in ordinary wires. At the frequency of 1MHz, instead, most of the conductor copper is not used. At microwave frequencies, a few microns of copper deposited on an insulator perform as an excellent conductor.

The consequence of the skin depth change with frequency is that the resistance of a conductor is an increasing function of frequency: indeed, the “effective” cross-section of the conductor decreases as the frequency increases. This phenomenon is generally called skin effect.

w

or

ld

Let us compute the impedance of the structure of Fig. 2.5, viewed as a current carrying conductor. Since the fields and the current density does not depend on y, we consider a strip of unit length in this direction. We compute first the total current I, flowing in the x̂ direction, per unit length along y: Z ∞ Z ∞ Jc0 δ I= Jc (x,z) · x̂dz = Jc0 e−j(1−j)z/δ dz = (2.24) 1+j 0 0

w .jn

tu

Notice that the dimensions of I are correctly A/m, since Jc0 is a surface current density with value p (2.25) Jc0 = 2(1 + j) πε0 f γE0

w w

Next, consider a unit length in the x̂ direction of this conductor and compute the potential difference along this length by integrating the electric field Ex along the x axis (y = 0, z = 0). Note that Ex does not depend on x, hence Ex itself coincides numerically with this potential difference.

Finally, the impedance per unit width in the y direction and unit length in the x direction is given by Jc0 /γ 1+j Ex (0,0) = = Zm = Zpul = γδ Jc0 δ/(1 + j) I

where we used (2.22). In conclusion we have this remarkable result: the impedance seen by a current flowing through a square of unit sides coincides with the wave impedance in the metal.

Notice that, apart from the similarity in the symbols, Zm =

Ex Hy

hence it is a completely different concept. Moreover, since the conductor we are considering has unit width in the y direction, unit length in the x direction (and infinite thickness in the z direction)

27 the previous analysis shows why often the value of the surface resistance Rs is expressed in Ω/¤ (read Ohm per square).

As a final remark, we note that the material becomes a perfect conductor when γ → ∞. In these conditions, the skin depth vanishes and the value of the current density at the interface tends to infinity, according to (2.25). Nevertheless, we see from (2.24) that the total current is finite and its value, independent of γ is r ε0 Jc0 δ E0 I= =2 µ0 1+j

om

where (2.20) and (2.25) have been used. This means that in a perfect conductor the current density can be written r ε0 E0 δ(z) = Jσ δ(z) Jc (x,z) = 2 µ0

.c

On the other hand, from (2.23) we see that in the limit γ → ∞ the magnitude of the total magnetic field at the z = 0− interface coincides with Jσ . Taking the directions of the vectors into account, we conclude that if a perfect conductor is immersed in an electromagnetic field, on its surface a current density Jσ (A/m) appears, such that

m

Jσ = ν̂ × H

ru

where ν̂ is the normal to the PEC surface, pointing toward free space. In practice, this is the proof of Eq.(1.19).

w

The results are the following:

or

ld

fo

Another example that we consider now is that of sea water: because of the salt contained in it, the conductivity is γ = 5 S/m, whereas the relative permittivity, up to the microwave region, does not change very much and will be taken to be εr = 80. We compute the complex wavenumber and the attenuation constant by the general equation p km = ω (ε − jγ/ω)µ

tu

• At f = 100Hz, γ/(2πf ε0 εr ) = 1.1234 · 107 , so sea water behaves as a good conductor;

w .jn

k = (4.4429 − j4.4429) · 10−2 m−1

α = 0.3859dB/m

• At f = 10kHz, γ/(2πf ε0 εr ) = 1.1234 · 105 , so sea water behaves as a good conductor;

w w

k = (0.4429 − j0.4429)m−1

α = 3.8590dB/m

• At f = 1GHz, γ/(2πf ε0 εr ) = 1.1234, so the displacement currents cannot be neglected; k = (209.7536 − j94.1066)m−1

α = 817.3998dB/m

• At f = 10GHz, γ/(2πf ε0 εr ) = 0.1123, so the displacement currents cannot be neglected; k = (1877.5266 − j105.1341)m−1

α = 913.1833dB/m

Obviously, at microwave frequency, the attenuation of sea water precludes the possibility of communicating with submarines during subsurface navigation. This becomes possible at low frequencies, where, however, the available bandwidth is very narrow.

Chapter 3

om

Radiation in free space

ru

m

.c

The fundamental problem in electromagnetics is computing the fields created by a specified set of sources in a given region of space. This means that the functions ε(r), µ(r) are assigned, as well as the form of the region boundary and the material of which it consists. Then the sources are specified in terms of electric and magnetic current densities J(r), M(r).

Green’s functions

or

3.1

ld

fo

In order to understand the basic mechanism of radiation, it is convenient to consider first a highly idealized problem, wherein the sources radiate in an infinite homogeneous medium. Later we will see how to apply the results of this chapter to the real antenna problem.

w

The radiation problem is mathematically formulated as ( ∇ × E = −jωµ0 H − M

tu

∇×H

(3.1)

= jωε0 E + J

w w

w .jn

in an infinite homogeneous domain that we assume to be free space. These equations are linear with constant coefficients and the independent variable is r. We can interpret them as the equations of a Linear Space Invariant system (LSI), where the source currents play the role of input and the radiated fields that of output, see Fig. 3.1. The box represents a system with two inputs and two outputs.

G EJ

J(r)

G HJ M(r)

Figure 3.1.

E(r) G EM

G HM

H(r)

Linear system view of the radiation phenomenon

28

29 LSI systems are clearly a multidimensional generalization of Linear Time Invariant (LTI) systems. Let us review briefly the properties of the latter. LTI systems, as shown in Fig. 3.2 are completely characterized in time domain by their impulse response h(t), that is the output that is obtained when the input is a Dirac delta function δ(t). An arbitrary (continuous) input x(t) can be represented as a linear combination of pulses thanks to the sifting property of the delta function Z ∞ x(t) = x(t0 )δ(t − t0 )dt0 −∞

Because of linearity, the response y(t) to x(t) can be found by convolution Z ∞ y(t) = h(t) ∗ x(t) = h(t − t0 )x(t0 )dt0

om

−∞

m

.c

Alternatively, an LTI system can be characterized by its transfer function: when the input is x(t) = exp(jωt), the output is proportional to it and the constant of proportionality is, by definition, H(ω), so that y(t) = H(ω) exp(jωt). It can be proved that the impulse response and the transfer function of a system are related by a Fourier transform Z ∞ H(ω) = h(t) exp(−jωt)dt

X (ω )

Y (ω )

ld

H (ω )

Time domain and frequency domain description of an LTI system

or

Figure 3.2.

y (t )

fo

h(t )

x(t )

ru

−∞

w w

w .jn

tu

w

As a preparation for (3.1), let us consider the simpler case of an infinite, uniform transmission line excited by a distribution of voltage and current generators, vs (z) and is (z), as shown in Fig. 3.3.

Since these generators are distributed continuously, vs (z) and is (z) are densities per unit length of generators described, as usual, in terms of their open circuit voltage (V/m) and short circuit current (A/m), respectively. The differential equations of the system are  dV  − = jωLI + vs dz (3.2)   − dI = jωCV + i s dz

vs ( z )

+ + + + is ( z )

Figure 3.3.

Infinite uniform transmission line with distributed voltage and current generators.

30

GVis

is ( z )

V ( z)

GVvs

GIis vs ( z )

Linear system view of the transmission line with distributed generators.

om

Figure 3.4.

I ( z)

GIvs

fo

ru

m

.c

Generally, transmission lines are excited at one end by a generator that acts as a transmitter.

The model shown in Fig. 3.3 refers to a situation of electromagnetic compatibility, where a line is excited by an electromagnetic wave that couples to the line along a certain segment of it. It is easy to recognize that this is the one dimensional analogue of Maxwell’s equations (3.1). The problem is again LSI and can be schematized as in Fig. 3.4. Hence, as suggested by this picture, the solution can be expressed as Z ∞ Z ∞ 0 0 0 V (z) = Z∞ GV is (z − z )is (z )dz + GV vs (z − z 0 )vs (z 0 )dz 0 −∞ −∞ Z ∞ Z ∞ 0 0 0 I(z) = GIis (z − z )is (z )dz + Y∞ GIvs (z − z 0 )vs (z 0 )dz 0 −∞

−∞

or

ld

The system here has two inputs and two outputs: each output depends on both inputs, so that in practice there are four Green’s functions, each one a pure number. They can be obtained by applying the spatial Fourier transform to the system equations (3.2). However, by the very definition of Green’s function

w

• GV is (z) is the voltage wave V (z)/Z∞ created by a unit amplitude current generator located in z = 0

w .jn

tu

• GV vs (z) is the voltage wave V (z) created by a unit amplitude voltage generator located in z=0 • GIis (z) is the current wave I(z) created by a unit amplitude current generator located in z=0

w w

• GIvs (z) is the current wave I(z)Y∞ created by a unit amplitude voltage generator located in z=0 so that they can be found by simple circuit theory, just recalling that the input impedance of an infinitely long line is Z∞ . The resulting expressions are 1 GV is (z) = GIvs (z) = − e−jk|z| 2 1 GV vs (z) = GIis (z) = − sgn(z)e−jk|z| 2

where sgn(z) is the sign function ½ sgn(z) =

1 if z > 0 −1 if z < 0

31 As another preparatory example before tackling (3.1), let us consider the case of sound waves.

It can be shown that the excess pressure p(r) with respect to the background pressure satisfies the scalar Helmholtz equation ¶ µ ω2 2 ∇ + 2 p(r) = −S(r) (3.3) Vs

−∞

−∞

.c

om

where Vs is the sound velocity and S(r) is a source term. This equation corresponds to the picture of Fig. 3.5, where S(r) is the input and p(r) the output. In this case the system has only one input and one output but it is multidimensional, since both depend on the three independent variables x, y, z. In perfect analogy with LTI systems, LSI systems are completely characterized in space domain by their “impulse response” G(r), which is traditionally called Green’s function.

This is the output of the system when the input is a point source located at the origin of the coordinate system, which can be represented mathematically by a three-dimensional Dirac delta function S(r) = δ(r) = δ(x)δ(y)δ(z). The fundamental property of this multidimensional Dirac δ function is Z Z ∞Z ∞Z ∞ δ(r)dr = δ(x)δ(y)δ(z)dx dy dz = 1 −∞

ru

m

When the input is an arbitrary function, the output is found by (three-dimensional) convolution Z p(r) = G(r − r0 )S(r0 )dr0

w .jn

tu

w

or

ld

fo

Alternatively, an LSI system can be characterized in the spectral domain. When the input is a harmonic function of x, y, z, that is S(r) = exp(−j(kx x + ky y + kz z)) = exp(−jk · r), the output is proportional to it and the coefficient of proportionality is, by definition, the transfer function H(k). Again, transfer function and Green’s function of the same system are related by a Fourier transform: however, in this case, it is triple, since it operates on the three variables x, y, z. The couple of inverse and direct 3-D Fourier transform is given by Z 1 H(k) exp(−jk · r)dk G(r) = (2π)3 Z (3.4) H(k) = G(r) exp(jk · r)dr

where dk = dkx dky dkz . It can be shown that in the case of free space, the transfer function is H(k) =

1 − ω 2 /Vs2

(3.5)

exp(−jk0 r) 4πr

(3.6)

k2

w w

and the corresponding Green’s function is

S (r )

Figure 3.5.

G(r) =

G (r )

p (r )

S (k )

H (k )

p(k )

Space domain and spatial frequency domain description of the sound radiation phenomenon

32 with k0 = ω/Vs denoting the wavenumber. This expression describes a diverging spherical wave.

Indeed, the constant phase surfaces are obviously r = const, a series of concentric spheres with center in the origin. Moreover, assuming that the source is harmonic with frequency ω0 , the expression of the Green’s function in time domain is ½ ¾ exp(−jk0 r) cos(ω0 t − k0 r) exp(jω0 t) = g(r,t) = R 4πr 4πr

from which it is evident that the phase velocity is Vph = ω0 /k0 = Vs > 0

ρ(r) ε

1 εk 2

1 4πεr

ld

G(r) =

fo

and the corresponding Green’s function (from (3.6))

ru

H(k) =

m

The transfer function associated to this equation is (from (3.5))

.c

∇2 V (r) = −

om

As another well known example of LSI system, let the frequency ω go to zero in (3.3), so that the Helmholtz equation becomes Poisson equation. This, for example, relates the electric potential V (r) to a charge distribution ρ(r), which acts as its source:

or

We recognize immediately this expression as the potential generated by a point charge q = 1 C in a dielectric with permittivity ε.

w .jn

tu

w

We are ready now for Maxwell’s equations (3.1), which are still more complicated because in addition to being multidimensional and multiple input/output, they are vector equations: this means that the output is a vector that is not necessarily parallel to the input. This implies that each of the four Green’s function is not a scalar but a linear operator (a tensor ), which, in a basis, is represented by a 3 × 3 matrix. This means that, differently from the case of sound waves, the Green’s function is not directly the field radiated by a point source. The source is really a point but is also a vector, which can have all possible orientations. From a certain point of view, we can say that the Green’s tensor yields the field radiated in a given point by a point source in the origin with all possible orientations. This concept will be better clarified in Section 3.2.

w w

In coordinate-free language Z Z 0 0 0 E(r) = −jωµ0 GEJ (r − r ) · J(r )dr − GEM (r − r0 ) · M(r0 )dr0 Z Z H(r) = GHJ (r − r0 ) · J(r0 )dr0 − jωε0 GHM (r − r0 ) · M(r0 )dr0

(3.7)

To check the dimensions of the various Green’s functions, it is useful to note that ωµ0 = k0 Z0

ωε0 = k0 Y0 = k0 /Z0

Hence we recognize that GEJ and GHM are measured in m−1 , GEM and GHJ in m−2 . The explicit expressions of the various dyadic Green’s functions can be obtained by applying the Fourier

33 transform technique to (3.1). The most appropriate coordinate system that can be used to show the result is the spherical one, because the source is a point. It can be shown that the matrices representing the Green’s functions are:   A 0 0 exp(−jk0 r) GEJ (r,ϑ,ϕ) = GHM (r,ϑ,ϕ) ↔  0 B 0  4πr 0 0 B   (3.8) 0 0 0 exp(−jk r) 0 GEM (r,ϑ,ϕ) = GHJ (r,ϑ,ϕ) ↔ −jk0  0 0 −C  4πr 0 C 0

om

√ where the wavenumber is k0 = ω ε0 µ0 . Note that the row and column indices are r̂, ϑ̂, ϕ̂, respectively. In dyadic form

m

.c

h i exp(−jk r) 0 GEJ (r,ϑ,ϕ) = GHM (r,ϑ,ϕ) = Ar̂r̂ + B ϑ̂ϑ̂ + B ϕ̂ϕ̂ 4πr h i exp(−jk r) 0 GEM (r,ϑ,ϕ) = GHJ (r,ϑ,ϕ) = −jk0 C ϕ̂ϑ̂ − ϑ̂ϕ̂ 4πr

where

or

ld

fo

ru

¶ µ 1 1 + A=2 j k0 r (k0 r)2 1 1 1 − B =1− A=1−j k0 r (k0 r)2 2 1 C =1−j k0 r

w

Consistently with the fact that the source is a point, the Green’s function does not depend on the angular variables.

tu

The behavior of the three coefficients at large distance from the source, k0 r → ∞ (i.e. r À λ)

w w

w .jn

is

µ A=O

1 k0 r

¶

B→1 C→1

In this region, usually called far field region, the expressions of the Green’s functions simplify and become exp(−jk0 r) 4πr exp(−jk0 r) GEM (r,ϑ,ϕ) = GHJ (r,ϑ,ϕ) ∼ −jk0 r̂ × Itr 4πr GEJ (r,ϑ,ϕ) = GHM (r,ϑ,ϕ) ∼ Itr

(3.9)

where Itr is the transverse to r identity dyadic, see Appendix. This operator, when applied to an arbitrary vector, produces as a result the projection of the vector on the plane perpendicular to r.

The operator r̂ × Itr adds a further 90◦ counterclockwise turn around r.

34 Conversely, in the near field region, k0 r → 0 (i.e. r ¿ λ), the coefficients become 2 (k0 r)2 1 B∼− (k0 r)2 1 C ∼ −j k0 r A∼

3.2

Elementary dipoles

m

.c

om

As discussed previously, the Green’s function is the basic tool for the computation of the field radiated by any source by means of eq. (3.7). However, it is convenient to start with the simplest one, i.e. a point source, and this will help in understanding the properties of the Green’s functions.

Consider first an elementary source of electric type located at the origin of the coordinate system, modeled by the current distribution J(r) = Me δ(r)

∞

xn Wξ (x)dx

ld

mn = E{ξ n } =

fo

ru

The vector Me is called the electric dipole moment of the current distribution and is measured in Am (recall that the dimensions of the three dimensional δ function are m−3 ). An arbitrary current distribution can be characterized by its moments. This concept is used also in the theory of probability: if ξ is a random variable with density function Wξ (x), moments of all orders can be defined by Z −∞

tu

w

or

where E{ } denotes the expectation value. In the case of the current distribution, the role of Wξ (x) is played by J(r), but the situation is more complicated because its vector nature implies that the moments beyond the first are tensors. The first moment (dipole moment) is a vector, defined by Z Me =

J(r)dr

(3.10)

w w

w .jn

In the case of the point source introduced above, thanks to the properties of the delta function, the previous equation becomes an identity and we understand the reason for the name of the coefficient.

From a practical point of view, we can imagine to obtain this source by a limiting process, starting from a rectilinear current I, whose length l is progressively reduced without changing the aspect ratio (diameter/length of the wire), while, at the same time, the current is increased, so that the value of the integral (the dipole moment Il) remains constant.

Introducing the dipole current into (3.7), we find that the radiated fields are given by Z E(r) = −jωµ0 GEJ (r − r0 ) · Me δ(r0 )dr0 = −jωµ0 GEJ (r) · Me Z H(r) = GHJ (r − r0 ) · Me δ(r0 )dr0 = GHJ (r) · Me

(3.11)

Since we know the expressions of the matrices representing the Green’s functions in the spherical basis, it is necessary to express the vector Me in the same basis. Let us assume that the polar axis of the coordinate system is in the direction of Me , i.e. assume Me = Me ẑ. Note that this step is allowed because the Green’s function does not depend on the angular variables, as a

35 consequence of the isotropy of free space. It is to be remarked, as a general rule, that the Green’s function depends only on the structure and, hence, shares its symmetries. This choice guarantees the simplest description of the radiated field. Since the radiated field must have the direction of Me as a symmetry axis, orienting the polar axis of the coordinate system in this direction allows the expressions to be independent on ϕ.

Me = (Me · r̂)r̂ + (Me · ϑ̂)ϑ̂ + (Me · ϕ̂)ϕ̂ (3.12)

= Me (ẑ · r̂)r̂ + Me (ẑ · ϑ̂)ϑ̂ + Me (ẑ · ϕ̂)ϕ̂ = Me cos ϑr̂ − Me sin ϑϑ̂

m

.c

om

where we have exploited (A.12). Now recalling the expression of the Green’s function (3.8), we obtain    A 0 0 cos ϑ exp(−jk0 r)  0 B 0   − sin ϑ  Me E(r) = −jωµ0 4πr 0 0 B 0 (3.13) ´ Z0 Me exp(−jk0 r) ³ A cos ϑr̂ − B sin ϑϑ̂ = −j 2rλ

where use has been made of

tu

w

or

ld

fo

ru

2π Z0 λ and Z0 is the wave impedance. Concerning the meaning of (3.12), it is to be remarked that the matrix (3.8) represents the Green’s function in the spherical basis consisting of the unit vectors r̂, ϑ̂, ϕ̂ defined in the observation point r. Hence, even if the source is located in the origin, its components are evaluated in the basis associated to the point r.

We can proceed similarly for the magnetic field:    0 cos ϑ 0 0 exp(−jk0 r)  0 0 −C   − sin ϑ  H(r) = −jk0 4πr 0 C 0 0 (3.14) Me exp(−jk0 r) =j C sin ϑϕ̂ 2rλ ωµ0 = k0 Z0 =

w w

w .jn

In conclusion, the electromagnetic field radiated by an electric dipole has the following expression ¶ ¸ ¶ µ · µ 1 1 1 1 Z0 Me exp(−jk0 r) sin ϑ ϑ̂ − cos ϑr̂ + − 1 − j 2 j E(r) = −j k0 r (k0 r)2 k0 r (k0 r)2 2rλ

H(r) = j

Me exp(−jk0 r) 2rλ

µ

1−j

1 k0 r

(3.15)

¶ sin ϑϕ̂

This wave has two components of electric field and only one of magnetic field. Imagine a geographical system of coordinates such that the direction of the dipole moment defines the direction of the earth axis. The angle ϑ is the colatitude (= 90◦ −latitude), the angle ϕ is the longitude.

Then the electric field is contained in the meridian planes and the magnetic field is tangent to the parallels. This type of wave is called TM (Transverse Magnetic) since the magnetic field has no radial component. We recognize also that the radial component of the electric field is dominant close to the source, but negligible with respect to the others at large distance. Here the wave is essentially TEM, since neither field has a (significant) radial component.

36 Let us compute the energy budget by means of the Poynting theorem. The Poynting vector is S = E × H∗ =

i Z0 Me2 h 2 ∗ ∗ BC sin ϑr̂ + AC sin ϑ cos ϑ ϑ̂ 4r2 λ2

om

Compute the components ¸ ¸· · 1 1 1 1 1 1 1 1 ∗ −j + +j − =1−j − BC = 1 − j 1+j (k0 r)3 k0 r (k0 r)2 k0 r (k0 r)2 k0 r k0 r (k0 r)2 1 =1−j (k0 r)3 ¸ ¸· · 2 2 2 2 2 2 1 +j − + =j + AC ∗ = j 1 + j (k0 r)3 (k0 r)2 k0 r (k0 r)2 k0 r k0 r (k0 r)2 ¸ · 2 2 + =j k0 r (k0 r)3

m

.c

Substituting in the previous equation we get ¶ ¸ ¶ ·µ µ 2 1 2 Z0 Me2 2 sin ϑ cos ϑϑ̂ + sin ϑr̂ + j 1−j S= 2 2 k0 r (k0 r)3 (k0 r)3 4r λ

ru

According to Poynting theorem, the surface density of active power flow is

fo

1 dP = R{S · ν̂} 2 dΣ

(3.16)

w .jn

tu

w

or

ld

where ν̂ is the normal to the surface element. In order to compute the total radiated active power, we have to evaluate the flux of the Poynting vector across a closed surface surrounding the source.

For maximum simplicity we choose a sphere of radius r: Z Z Z 1 1 2π π Z0 Me2 sin2 ϑ r2 sin ϑ dθdϕ Prad = R S · ν̂ dΣ = 2 2 2 0 2 0 4r λ Z π 1 Z0 Me2 (3.17) 2π sin3 ϑ dθ = 2 4λ2 0 1 Z0 Me2 2π = 2 3λ2

Here we have used the following facts

w w

• the normal to the spherical surface is ν̂ = r̂ • the area element in spherical coordinates is dΣ = r2 sin ϑ dθdϕ • the integrand does not depend on ϕ, so the ϕ integration yields the 2π factor • the ϑ integration yields

Z 0

π

sin3 ϑ dθ =

4 3

The factor 1/2 has been left explicit to make it clear that Me is a peak value, that is the time domain dipole moment is Me (t) = Me cos(ω0 t). If, on the contrary Me is an effective value, the factor 1/2 has to be dropped.

37 We notice that the total radiated power does not depend on the radius of the sphere chosen to compute it. Algebraically this is the result of the cancellation between the r2 factor in the denominator of the Poynting vector and the one in the area element dΣ. To get a more physical explanation, consider the fluxes through two concentric spheres of different radii: if they were different, power would be lost or generated in the shell, which is impossible by conservation of energy in a lossless medium.

om

In (3.17) we took the real part of the integral. The imaginary part is the reactive power Z Z Z 1 1 2π π Z0 Me2 1 sin2 ϑ r2 sin ϑ dθdϕ Q = I S · ν̂ dΣ = − 3 2 2 2 0 2 0 4r λ (k0 r) Z π 1 Z0 Me2 1 2π sin3 ϑ dθ =− 2 4r2 λ2 (k0 r)3 0 1 Z0 Me2 2π 1 =− 2 3λ2 (k0 r)3

ru

m

.c

It is reasonable that Q depends on r: the reactive power is the energy that twice per period is exchanged between generator and load, hence crosses the spherical surface of radius r. Moreover, it may be remarked that in the ϑ direction, the structure is closed as a (virtual) cavity. This explains why the ϑ component of the Poynting vector is pure imaginary. Indeed, a real part of Sϑ would imply a steady energy flow in that direction; however, this is impossible, since the angular domain is finite (0 ≤ ϑ ≤ π/2) and the dielectric is lossless.

fo

It is useful to explicitly indicate the dominant components close to the source and far from it.

In the far field region, r À λ

ld

Z0 Me exp(−jk0 r) sin ϑϑ̂ 2rλ

or

E(r) ∼ j

Me exp(−jk0 r) sin ϑϕ̂ 2rλ

w

H(r) ∼ j

(3.18)

w .jn

tu

We see that the fields tend to be linearly polarized, orthogonal and proportional to each other and also orthogonal to the radial direction. These properties are summarized in the impedance relation 1 r̂ × E(r) Z0 E(r) ∼ Z0 H(r) × r̂

H(r) ∼

w w

The radiated field is a spherical wave (because of the factor exp(−jk0 r)), diverging from the origin √ with phase velocity c = ω/k0 = 1/ ε0 µ0 . Indeed, the time varying electric field is given by

E(r,t) = R{E(r)ejωt } Z0 Me sin ϑ sin(ω0 t − k0 r)ϑ̂ ∼− 2rλ

Note that there is no contradiction between the statement that the radiated field is a spherical wave and the fact that this field has the sin ϑ dependence: the former refers to the constant phase surfaces, the latter to a magnitude factor.

The field amplitudes decay as 1/r. This behavior is strictly connected with the principle of conservation of energy. Indeed, it is easy to see that the total radiated active power computed in

38 (3.17) is due only to these 1/r components.

The active power density per unit surface dP/dΣ, at any distance, (dΣ orthogonal to r̂) is given, according to (3.16), by 1 Z0 Me2 dP sin2 ϑ (3.19) = 2 4r2 λ2 dΣ Thus, in the far field, 1 |E|2 dP ∼ 2 Z0 dΣ

.c

Me 1 sin ϑϕ̂ 2rλ k0 r

m

H(r) ∼

om

exactly as in the case of a plane wave, see (2.13). This fact should not be surprising, since a spherical wave can be locally approximated by a plane wave with direction of propagation coincident with the radial direction. In the near field region r ¿ λ ¸ · 1 2 Z0 Me sin ϑ ϑ̂ cos ϑr̂ + E(r) ∼ −j (k0 r)2 2rλ (k0 r)2 (3.20)

which may be reduced to

ru

Me sin ϑϕ̂ 4πr2

(3.21)

ld

H(r) ∼

i Z0 Me λ h 2 cos ϑr̂ + sin ϑϑ̂ 2 3 8π r

fo

E(r) ∼ −j

tu

w

or

We notice that the exponential has been dropped, since its value, for r ¿ λ is essentially 1. Clearly, very close to the source, the propagation delay is negligible. We can recognize that the electric field coincides with that of a static dipole, whose moment, however, is a sinusoidal function of time. Likewise, the magnetic field coincides with that created, in accordance with the Biot-Savart law, by an infinitesimal current element, where the current is a sinusoidal function of time. This regime is called quasi-static.

w w

w .jn

Fig. 3.6 shows a sketch of the electric field lines in a plane ϕ = const; close to the source the field line behavior is similar to that of a static dipole, further away, entering in the radiation region, they are completely different. Note that they are closed: we are accustomed to magnetic field line being closed because ∇ · B = 0, i.e. B is solenoidal. Actually in this case also the electric field is (almost) everywhere solenoidal. Recall the divergence equation ∇·E=

ρ =0 ε

for r 6= 0

because the source is point like and located in the origin.

Note that the image shown in the figure refers to a specific time; as time passes the contours move outward radially. Fig. 3.7 shows the traditional way to plot the ϑ component of the electric field in the far field region, in a plane ϕ =const. It is a polar plot and the curves have equation ρ = sin ϑ (with ρ denoting the length of OP ), hence they are circles.

The other elementary source that we describe now is the dual of the electric dipole, i.e. a magnetic dipole. We consider then formally a point source of magnetic current, M(r) = Mm δ(r)

Dipole electric field lines

z

ld

Figure 3.6.

fo

ru

m

.c

om

39

P

O

1

x

tu

w

or

ϑ

w .jn

Figure 3.7. Dipole far field. Normalized polar plot of Eϑ and Hϕ at large distance. The maximum field value is reached on the equator ϑ = π/2 and is normalized to 1. For any ϑ the length of the segment OP is proportional to the value of Eϑ

w w

The vector Mm is called the magnetic dipole moment of the current distribution and is defined as Z Mm = M(r)dr Its units are Vm. It is well known that magnetic currents do not exist, as flow of magnetic charges, but can be introduced formally to describe electric currents circulating in closed loops. Indeed, this elementary source can be imagined to be obtained with a limiting procedure from a loop, enclosing an area S, on which the current I is circulating. The “equivalent” magnetic current has a direction perpendicular the plane of the loop as indicated by the thumb of the right hand when the fingers are aligned along the electric current flow. The point source is obtained by shrinking the loop and increasing the current so that the magnetic dipole moment remains constant. It can

40 be shown that the magnetic dipole moment of a small loop is Mm = jωµSI

om

Notice that since ωµ = k0 Z0 , this quantity is measured in Ω/m and Mm turns out with the right dimensions. In order to compute the fields radiated by such a source, we substitute its expression in (3.7) and obtain Z E(r) = − GEM (r − r0 ) · Mm δ(r0 )dr0 = −GEM (r) · Mm Z (3.22) 0 0 0 H(r) = −jω²0 GHM (r − r ) · Mm δ(r )dr = GHM (r) · Mm

m

.c

As in the case of the electric dipole, we choose a spherical coordinate system with the polar axis aligned along Mm , which then takes the form Mm = Mm ẑ. Recalling (3.8), (3.12) and computing the scalar products we obtain ¶ µ Mm exp(−jk0 r) 1 sin ϑϕ̂ E(r) = −j 1−j k0 r 2rλ

(3.23)

fo

ru

¶ ¸ ¶ µ · µ 1 1 Y0 Mm exp(−jk0 r) 1 1 sin ϑ ϑ̂ − cos ϑr̂ + H(r) = −j − 1 − j 2 j k0 r (k0 r)2 k0 r (k0 r)2 2rλ

or

ld

If we compare these expressions with those of the fields radiated by an electric dipole (3.15), we see that they (obviously!) satisfy the principle of duality, in the sense that the expressions of electric and magnetic fields are exchanged in the two cases. This wave is of T E type since the electric field has no radial component and tends to become T EM in the far field region.

The Poynting vector associated to these fields is

w

i 2 h Y0 Mm 2 ∗ ∗ B C sin ϑr̂ + A C sin ϑ cos ϑ ϑ̂ 4r2 λ2

tu

S = E × H∗ =

w .jn

which is just the complex conjugate of the corresponding expression for the electric dipole. Hence the result is written down by inspection ¶ ¸ ¶ ·µ µ 2 2 2 1 Y0 Mm 2 sin ϑ cos ϑ ϑ̂ + sin ϑr̂ 1 − j + j S= k0 r (k0 r)3 (k0 r)3 4r2 λ2

w w

The total radiated active power is given by the real part of the flux of S through a sphere concentric with the source and is Z 2 2π 1 1 Y0 Mm (3.24) Prad = R S · ν̂ dΣ = 2 3λ2 2

For completeness we report the expression of the reactive power Z 2 2π 1 1 1 Y0 Mm Q = I S · ν̂ dΣ = 2 (k0 r)3 2 3λ 2

which has the opposite sign with respect to the case of the electric dipole, since this is a magnetic source. This must not be surprising: also in circuit theory the reactive power in inductors and capacitors has opposite signs.

41 It is useful to explicitly indicate the dominant components close to the source and far from it.

In the far field region, r À λ E(r) ∼ −j

Mm exp(−jk0 r) sin ϑϕ̂ 2rλ

(3.25) H(r) ∼ j

Y0 Mm exp(−jk0 r) sin ϑϑ̂ 2rλ

The impedance relation is the same as the one for the electric dipole fields: 1 r̂ × E(r) Z0 E(r) ∼ Z0 H(r) × r̂

om

H(r) ∼

Mm 1 sin ϑϕ̂ 2rλ k0 r

m

E(r) ∼ −

.c

In the near field region r ¿ λ

(3.26)

which may be reduced to

Mm sin ϑϕ̂ 4πr2

or

E(r) ∼ −

ld

fo

ru

¸ · Y0 Mm 1 2 sin ϑ ϑ̂ cos ϑr̂ H(r) ∼ −j + (k0 r)2 (k0 r)2 2rλ

(3.27)

tu

w

i Y0 Mm λ h 2 H(r) ∼ −j cos ϑr̂ + sin ϑ ϑ̂ 8π 2 r3

3.3

w .jn

Obviously, Figs. 3.6 and Fig. 3.7 can be used also for magnetic dipoles.

Radiation of generic sources

w w

After the analysis of the two elementary sources, we can discuss the radiation properties of arbitrary distributions of electric and magnetic currents. Actually, the solution of this problem has already been given in (3.7): the integrals are generally to be evaluated by numerical techniques and the task is absolutely non trivial, since the integrands are highly oscillatory due to the presence of the exponential exp(−jk0 r). However, when the point where the field is to be computed is far from the source, it is possible to carry out a number of approximations that lead to a closed form evaluation of the radiation integrals in a number of interesting cases. In this section we discuss this approximation and find its domain of validity, obtaining a generalization of the concept of far field region already introduced in the previous section.

Let us make reference to Fig. 3.8, which shows the geometry of the problem. The field radiated in the observation point P (often called field point) is given by (3.7) that we repeat here for ease

42 d

P r'

r

O J, M

of reference

Z

om

Figure 3.8. Radiation of an arbitrary source. Geometry for the approximate evaluation of the radiation integral. P is the field point, r0 denotes a generic point of the current distribution and the vector there represents the elementary dipole J(r0 )dV

Z 0

0

E(r) = −jωµ0 GEJ (r − r ) · J(r )dr − GEM (r − r0 ) · M(r0 )dr0 Z Z H(r) = GHJ (r − r0 ) · J(r0 )dr0 − jωε0 GHM (r − r0 ) · M(r0 )dr0

(3.28)

m

.c

0

or

ld

fo

ru

The meaning of these equations is very clear: the given source distribution is described as a collection of elementary electric (J(r0 )dV ) and magnetic (M(r0 )dV ) dipoles located in the source point r0 , the field radiated by each of them is computed, as in the previous section, and the partial results are summed. In other words, it is just an application of the superposition principle implied by the linearity of Maxwell’s equations. The position of P is specified by its spherical coordinates r, ϑ, ϕ with respect to a (global) coordinate system with origin in O. Let us introduce the relative position of P with respect to the source point r0 d = r − r0

w w

w .jn

tu

w

and assume that k0 d À 1, so that the asymptotic expressions of the Green’s functions (3.9) can be used: exp(−jk0 d) GEJ (d,ϑd ,ϕd ) = GHM (d,ϑd ,ϕd ) ∼ Itd 4πd (3.29) exp(−jk0 d) GEM (d,ϑd ,ϕd ) = GHJ (d,ϑd ,ϕd ) ∼ −jk0 d̂ × Itd 4πd where d, ϑd , ϕd denote the spherical coordinates of P with respect to the source point r0 . Now suppose, as it is usual in a telecommunication application, that the field point is so far away that the angle subtended by the source, when observed from the field point, is negligible. This means that the vectors d and r can be considered to be parallel, so that the various local spherical coordinate systems can be collapsed in the global one with origin in O, so that ϑd ≈ ϑ and ϕd ≈ ϕ.

Of course d is different from r and can be approximated as follows. s µ 0 ¶2 p p √ r̂ · r0 r −2 d = d · d = (r − r0 ) · (r − r0 ) = r2 − 2r · r0 + r02 = r 1 + r r

The observation point P is assumed to be at large distance, so r0 /r ¿ 1 and the square root can be expanded in Taylor series (binomial theorem) √ 1 1 1 + x = 1 + x − x2 + . . .

8 2

43 By letting x=

µ 0 ¶2 r̂ · r0 r −2 r r

we obtain, by keeping only the terms up to the second degree in r0 /r # " µ ¶2 µ 0 ¶2 1 1 r0 r̂ · r0 r + . . .] + (r̂ · r̂0 ) + d=r 1− r 2 2 r r

om

where the last term is the lowest degree one coming from the term (1/8)x2 . Factoring common terms yields i r02 h 1 − (r̂ · r̂0 )2 + . . .

(3.30) d = r − r̂ · r0 + 2r

ru

m

.c

This infinite expansion can be truncated at any point: more terms retained means more accurate results but also greater difficulty in the evaluation of the radiation integral. The relative distance d appears in two places in the Green’s function, in the denominator and in the phase exponential.

To appreciate that the accuracy required in the two cases is different, let D be the diameter of the smallest sphere containing the source. Then, as the source point sweeps the source volume ¯ ¯ ¯ ¯d D D ¯ (3.31) − 1¯¯ ≤ i.e. |d − r| ≤ , ¯ 2r r 2

D 2

with

or

d=r+ξ

ld

fo

Then substituting d with r (i.e. truncating the Taylor expansion at the first term) in the denominator produces an error that can be made as small as one wishes by assuming r large enough.

Concerning the exponential, (3.31) implies that

w

but then

|ξ| ≤ 1

tu

exp[−jk0 d] = exp[−jk0 r] exp[−jk0 ξ

D ] 2

w w

w .jn

and approximating d with r would imply an error that does not depend on r but only on the electrical size of the source, so that it could not be made small by increasing r. In conclusion, the phase term requires that in (3.30) at least the first two terms are used. The denominator is not critical and we can take d ≈ r there. In conclusion, we use: d ≈ r − r̂ · r0 d≈r

in the exponential in the denominator

In this case we say that the radiation integral is computed in the Fraunhofer approximation. If also the quadratic term in (3.30) is used in the exponential, we have the Fresnel approximation , which clearly is more accurate or it has a wider domain of applicability.

Consider first the Fraunhofer approximation of which we establish the domain of applicability.

A sufficient condition for the validity of the approximation is that the phase error produced by neglecting the quadratic term in (3.30) is small. In the worst case ( r̂ · r0 = 0) the phase error is ∆ϕ = k0

r02 2r

44 and reaches the maximum value ∆ϕmax = k0

D2 D2 =π 4rλ 8r

0 for r0 = rmax = D/2. A this point one could imagine that the requirement ∆ϕ ¿ 1 is enforced.

However, the distance rmin that would result would be very large and traditionally the condition

∆ϕmax ≤

π 8

m

.c

om

is assumed instead. This condition yields for the domain of applicability of the Fraunhofer approximation D2 r ≥ rmin = 2 λ The procedure seems to be a bit strange, but one must remember that the integrand is an oscillatory function so that the sufficient condition ∆ϕ ¿ 1 is not necessary. The distance rmin is called far field distance or Fresnel distance: indeed it is the boundary between the far field region (also called Fraunhofer region or radiation region) and the Fresnel region, which is closer to the source even if not adjacent to it.

ru

Assuming the field point P is in the far field region of the source, the expressions of the Green’s functions (3.29) become

(3.32)

ld

fo

exp(−jk0 r) exp(jk0 r0 · r̂) 4πr exp(−jk0 r) GEM (d,ϑ,ϕ) = GHJ (d,ϑ,ϕ) ∼ −jk0 r̂ × Itr exp(jk0 r0 · r̂) 4πr GEJ (d,ϑ,ϕ) = GHM (d,ϑ,ϕ) ∼ Itr

w w

w .jn

tu

w

or

Remember that here two approximations have been performed: first r À λ and then r ≥ 2D2 /λ and, clearly, both must be satisfied. If these expressions are substituted in (3.28) we obtain the far field expressions to be used for an arbitrary source: Z exp(−jk0 r) Itr · J(r0 ) exp(jk0 r0 · r̂)dr0 + E(r) ∼ −jωµ0 4πr Z exp(−jk0 r) + jk0 r̂ × Itr · M(r0 ) exp(jk0 r0 · r̂)dr0 4πr

Z exp(−jk0 r) H(r) ∼ −jk0 r̂ × Itr · J(r0 ) exp(jk0 r0 · r̂)dr0 + 4πr Z exp(−jk0 r) Itr · M(r0 ) exp(jk0 r0 · r̂)dr0 − jωε0 4πr

These bulky expressions can be rewritten in a simpler form, very similar to the one in (3.18) pertaining to an elementary dipole: E(r) ∼ −j

Z0 exp(−jk0 r) Pe (r̂) 2rλ

(3.33) H(r) ∼ −j

exp(−jk0 r) r̂ × Pe (r̂) 2rλ

45 where the generalized electric dipole moment Pe (r̂) is defined by Z Z 0 0 0 Pe (r̂) = Itr · J(r ) exp(jk0 r · r̂)dr − Y0 r̂ × Itr · M(r0 ) exp(jk0 r0 · r̂)dr0

(3.34)

The units of Pe (r̂) are Am and, on comparing this expression with (3.10), we realize that the generalization consists in the fact that • it depends on integrals of both electric and magnetic currents

om

• the integrals contain also the exponential factor, so that the generalized electric dipole moment depends on r̂. Moreover, due to the transverse identity Itr , Pe (r̂) · r̂ = 0, i.e. the generalized dipole moment is always orthogonal to r̂.

In the case of an electric dipole J(r) = Me ẑδ(r)

ru

and in the case of a magnetic dipole

m

Pe (r̂) = −Me sin ϑϑ̂

.c

we find (from (3.18) more easily than from the definition) that

(from (3.25) rather than from the definition)

fo

M(r) = Mm ẑδ(r)

ld

Pe (r̂) = Y0 Mm sin ϑϕ̂

w

or

It is interesting to note that, on recalling the definition of the triple Fourier transform of a function of three space variables (3.4), the definition of the generalized electric dipole moment (3.34) can be rewritten as

tu

Pe (r̂) = Itr · F3 {J(r)}|k=k0 r̂ − Y0 r̂ × Itr · F3 {M(r)}|k=k0 r̂

(3.35)

w w

w .jn

from which we see that the radiated field is related to the Fourier transform of the source distribution. In particular, the spectral component that contributes in a given observation direction r̂ is 2π k= r̂ λ This observation is important because it allows to derive some general properties of the radiated field by basic theorems of the Fourier transform: for instance, a consequence of the uncertainty principle of Fourier transforms is that a directive antenna must be large in terms of wavelengths.

From (3.33) we can derive some general properties of the field radiated by an arbitrary source in the far field region: • whatever the source, the radiated field is a spherical wave, as indicated by the presence of the exponential exp(−jk0 r). It can be shown that close to the source the wavefronts have a shape that more or less matches that of the source, but the propagation phenomenon modifies them so that they transform smoothly into spheres.

• the magnitude of electric and magnetic field decays as 1/r

46

Peϕ (ϑ , ϕ )

ϑ

fo ld

ϕ

ru

m

.c

om

Peϑ (ϑ , ϕ )

or

Figure 3.9. Polar plot of Pe (r̂) for a fairly directive antenna. For the particular direction shown, the spherical components of Pe are shown. Obviously Per ≡ 0

w

.

w w

w .jn

tu

• Pe (r̂) for any source with a finite size in terms of wavelength is a function of ϑ, ϕ, which is more rapidly varying than the sin ϑ typical of a point dipole. Indeed, the transform of the Dirac delta function is a constant and the sin ϑ originates from the dot product Itr ·ẑ. Fig. 3.9 shows a polar plot of Pe (r̂) for a fairly directive antenna. The plot is a surface constructed as follows. For each direction specified by spherical angles ϑ, ϕ, as the one shown, a point of the surface is identified: its distance r from the origin is proportional to the magnitude of Pe (ϑ,ϕ). For the particular direction shown, also the spherical components Peϑ , Peϕ are shown. Obviously the radial component is identically zero. It must be clear that the two vectors shown are not tangent to the surface shown.

The plot shows that the magnitude of Pe (r̂) (and then also of the radiated electric field) is maximum in the direction of the z axis, which appears to be a symmetry axis of the antenna: |Pe | does not depend on ϕ. One of the most striking characteristics of the plot is that it shows that there are many directions for which |Pe | vanishes, so that the antenna does not radiate any field in those directions. The explanation is simple. As (3.34) shows, Pe (r̂) is given by an integral, that is by a sum of many infinitesimal contributions (J(r0 )dV ), each one weighted by a phase factor (exp(jk0 r0 ·r̂)) that depends on the the observation direction.

For some directions (as ẑ in the figure) the various contribution sum in phase and the result is large (constructive interference). For other directions (ϑ = const. in the figure), the contributions are out of phase and the end result is zero because of destructive interference.

47 In other words, the zeros arise always because of cancellation. It is to be borne in mind that this characteristic behavior is possible only because we assume that the source is time harmonic, with a well defined frequency. In the case of a thermal source of light, i.e. a lamp or a flame, the source is absolutely not time harmonic, but can be better described as a white noise source: hence no constructive or destructive interference is possible and the plot is in the form of a kind of sphere. Indeed, consider a certain direction: even if for one frequency the various contributions give rise to cancellation, for other frequencies they do not and the total field is never zero. This is the reason why two lamps on a table produce a uniform brighter illumination of it. If the two lamps were ideally single frequency, characteristic dark fringes (interference fringes) would appear

om

• the Poynting vector has the following expression Z0 Pe (r̂) × (r̂ × P∗e (r̂)) 4r2 λ2 © ª |Pe (r̂)|2 r̂ − (r̂ · Pe (r̂)) P∗e (r̂)

S(r) = E(r) × H∗ (r) ∼

(3.36)

m

.c

Z0 4r2 λ2 Z0 ∼ 2 2 |Pe (r̂)|2 r̂ 4r λ

=

Z

Z

ld

1 R 2

S · r̂ dΣ =

1 Z0 2 4λ2

or

Prad =

fo

ru

because the generalized dipole moment is transverse to r̂. Since S is real and its field lines are radial, active power flows in the radial direction. If we want to compute the total radiated power, we have to compute the flux of the Poynting vector across any closed surface surrounding the source. It is obviously convenient to choose a sphere with large enough radius r, so that the previous expression can be used

Z

2π

π

dϕ

0

|Pe (ϑ, ϕ)|2 sin ϑdϑ

(3.37)

0

w .jn

tu

w

where we have used dΣ = r2 sin ϑdϑdϕ. As in the case of an elementary dipole, the total power leaving the sphere must be independent of its radius, so that we can simply talk about the total power radiated by the source. Recalling the definition of solid angle dΩ = sin ϑdϑdϕ, we can introduce the concept of power density per unit solid angle (W/steradiant) or radiation intensity 1 Z0 dPrad |Pe (ϑ, ϕ)|2 = (3.38) 2 4λ2 dΩ

w w

Alternatively, we can introduce the power density per unit surface (oriented as the tangent plane to the sphere of radius r through the observation point) by

1 Z0 dPrad |Pe (ϑ, ϕ)|2 = 2 4λ2 r2 dΣ

(3.39)

If we compare this expression with (3.33), we realize that

1 |E|2 dPrad = 2 Z0 dΣ

(3.40)

which is the same expression as (2.13) for plane waves, as we pointed out in the case of the dipole fields. Then Pe (r̂) contains the information about the radiation pattern, i.e. the way in which radiated power is distributed according to the direction. Moreover, it determines

48 also the polarization. Indeed, it can be expressed in the spherical basis relative to any observation direction as: Pe (ϑ, ϕ) = Peϑ (ϑ, ϕ)ϑ̂ + Peϕ (ϑ, ϕ)ϕ̂ as shown in Fig. 3.9 – If Peϑ and Peϕ have the same phase, the polarization is linear.

– If Peϑ and Peϕ have the same magnitude and are in phase quadrature, the polarization is circular.

– In all other cases, the polarization is elliptical

Pe (r̂) = Itr · Me = −Me sin ϑϑ̂

.c

om

It is finally to be remarked that any source that is significantly smaller than the wavelength behaves as an elementary dipole, independently of the details of its actual shape. Indeed, in these conditions, for all the points of the source the exponential in (3.34) is essentially equal to one and

w w

w .jn

tu

w

or

ld

fo

ru

m

as for an elementary dipole. As a consequence, we cannot deduce the shape of a small antenna from measurements of the radiated field carried out in the far field region. Alternatively, small details of a large source cannot be inferred from measurements of the radiated field. This property explains why no microscope can show details of an object that are smaller than a wavelength.

Chapter 4

.c

om

Antennas

ru

m

Antennas are fundamental components in wireless telecommunication systems. Information transfer between transmitter and receiver is carried out by free space propagation and antennas provide the necessary coupling between circuits and open space. Even if at sufficiently high frequency any circuit radiates electromagnetic energy, antennas are designed to do it in a particularly efficient way.

w .jn

tu

w

or

ld

fo

Depending on the frequency range and on the applications, antennas may have very different forms. If the frequency of operation is very small (and consequently the wavelength is very large) ordinary antennas are forced to be electrically small for size limitation reasons. Usually they are in the form of dipoles or loops constructed with metal wires or rods. They radiate energy all around them and are in general inefficient. At intermediate frequencies, antennas can have a size comparable with the wavelength: widespread is the use of Îť/2 dipoles or Îť/4 monopoles; loops with one wavelength circumference are also used. These antennas are still not directive but their efficiency is larger than that of small antennas. Directive antennas, i.e. such that can radiate electromagnetic energy essentially in one direction, must be electrically large, hence are practicable only at very high frequency. Typical examples are reflector antennas, which exploit the same working principles of optical telescopes. Antennas for satellite telecommunication belong generally to this class. Similar properties have antenna arrays: they consist of electrically large periodic arrangements of small antennas (half-wave dipoles for example).

w w

In this chapter the main types of antennas will be described, but first the parameters necessary to quantify their performance must be introduced

4.1

Antenna parameters

Antennas are used in transmit and in receive mode. Even if their behaviors in the two operating modes are strictly related, they are characterized by different parameters.

Transmitting antennas behave as loads for their feeding transmission line, hence are characterized by an input impedance or a reflection coefficient. Moreover the power density radiated in the various directions is different: parameters are necessary to describe this property.

Receiving antennas behave as generators as seen by the relevant receivers. Hence they can be characterized by an internal impedance and by an open circuit voltage in a TheĚ venin equivalent 49

50 circuit. Also, their sensitivity to waves incident from various directions is different and has to be quantified.

4.1.1

Input impedance

Consider an antenna with a couple of terminals, such as the one on the left in Fig. 4.1 The input

om

Ia

.c

Va

ru

m

Γin

fo

Figure 4.1. Left: wire antenna (dipole) with a couple of terminals at which voltage and current can be defined. Right: horn antenna connected to a waveguide; in this case only the reflection coefficient for the fundamental mode can be defined

ld

impedance is obviously defined as

or

Zin = Rin + jXin =

Va Ia

w .jn

tu

w

The presence of a real part means that the active power (Pin ) is absorbed from the feeding line: part of this is radiated away (Prad ) and the rest is dissipated into heat due to the limited metal conductivity (Ploss ). The imaginary part of the input impedance is related to the energy stored in the reactive fields in close proximity of the antenna structure. We can define the antenna radiation efficiency by Prad (4.1) η= Pin

w w

Obviously η ≤ 1. Microwave antennas have very high efficiency, η ' 1, but at a frequency of 100kHz it can be as low as η = 0.1.

Since Pin = 21 Rin |Ia |2 , it is customary to introduce the antenna radiation resistance as

Rrad =

2Prad |Ia |2

and the efficiency can also be written as η=

Rrad Rin

Clearly the radiation resistance is an equivalent resistance and has nothing to do with the Joule effect. Notice that there is no contradiction in the fact that radiation in a lossless medium is

51 described by a resistance: indeed radiated power is irreversibly transferred from the generator to infinity, just as electrical power flowing in a copper conductor is irreversibly transformed from electrical to thermal nature. Notice also that there is no “radiation reactance”, because this is a contradiction in terms. Indeed, radiation is related to a steady (one directional) energy flow, whereas the input reactance is associated to the energy exchanged twice per period back and forth between the antenna and the surrounding space. It turns out that the antenna behaves as a circuit with many resonance frequencies, in correspondence of which the input reactance vanishes.

It is to be remarked that there is a relationship between the generalized dipole moment Pe (r̂) and the radiation resistance Rrad , which originates from (3.37): 2Z0 4λ2 |Ia |2

Z

Z

2π

π

dϕ 0

|Pe (ϑ, ϕ)|2 sin ϑdϑ

0

(4.2)

om

Rrad =

Radiation pattern, Directivity and Gain

ru

4.1.2

m

.c

In the case of the horn antenna shown in the right part of Fig. 4.1 the definition of an input impedance is conventional, whereas the input reflection coefficient (for the fundamental mode) is well defined and similar considerations can be carried out.

w w

w .jn

tu

w

or

ld

fo

Antennas radiate the power they get from the transmitter with different density per unit solid angle in the various directions . The radiation pattern is a polar plot of this density and has the form of a surface as shown in Fig. 4.2: for each direction ϑ, ϕ, a point is defined at a distance r from the origin proportional to dPrad /dΩ. The set of all these points has the characteristic shape shown in the figure. The antenna is moderately directive and exhibits a major lobe (or main lobe) and a number of minor lobes (side lobes) Associated to the antenna is a spherical coordinate

Figure 4.2.

Coordinate system for antenna analysis

52 system that is the most appropriate to describe the field radiated in the Fraunhofer region (far field region). A sphere is shown with radius greater than the Fresnel distance and an elementary patch on it dΣ. Since dΣ = r2 dΩ, we can say that the radiation pattern is a plot of dPrad /dΣ as the patch dΣ sweeps the spherical surface. The figure shows also the three basis vectors in a particular observation point.

w w

w .jn

tu

w

or

ld

fo

ru

m

.c

om

Often two-dimensional patterns are used to describe the characteristics of an antenna, instead of the three-dimensional one. An example is shown in Fig. 4.3. The patterns are always normalized

Figure 4.3. Normalized two-dimensional patterns. (a) is a polar plot of the field magnitude in linear scale. (b) is a polar plot of the power density in linear scale. (c) is a polar plot of the power density in dB. In each plot the Half Power Beam Width (HPBW) is indicated. The lobes are marked alternatively by a plus and a minus to indicate that the field has a phase 0 and π rad, respectively

53 to the maximum. Sometimes field patterns instead of power patterns are used: to appreciate the relationship between the two it is to be remembered that the power density is proportional to the square of the field. Thus in a power pattern using linear scales, sidelobes become less visible. On the contrary, a dB scale makes sidelobes more visible. It is clear that the normalized field pattern coincides, apart for a scale factor, with the polar plot of Pe (ϑ, ϕ), shown in Fig. 3.9.

An important characteristic of a pattern is the width of the main lobe. Usually the Full Half Power Beam Width (FHPBW) is used for this purpose. The main lobe may be rotationally symmetric or not. In the latter case two FHPBW must be specified, relative to two orthogonal planes. If the field radiated along the axis of the main beam is linearly polarized, these planes are: • the E-plane, defined by the main lobe axis and the E vector

om

• the H-plane, defined by the main lobe axis and the H vector

or

ld

fo

ru

m

.c

The directivity of an antenna is a function of ϑ, ϕ defined as the ratio between the radiation intensity for that direction and the average of the radiation intensity over all directions. The average is simply obtained as the ratio between the total radiated power and the total solid angle.

Hence: ¯ dPrad ¯¯ dΩ ¯(ϑ,ϕ) (4.3) d(ϑ,ϕ) = Prad 4π The denominator can also be interpreted as the radiation intensity that would be produced by a (hypothetical) isotropic radiator, radiating the same total power as the actual antenna. The directivity is a pure number and its polar plot coincides with the power pattern, apart for a scale factor. The maximum value of the directivity function, i.e. the one corresponding to the main lobe direction, is called simply the antenna directivity, typically expressed in dB:

w

D = 10 log10 max d(ϑ,ϕ)

w .jn

tu

A property of the directivity function is that its average value over all directions is one. Indeed Z Z dPrad 1 1 1 1 1 Prad = 1 dΩ = d(r̂)dΩ = P 4π Prad dΩ 4π rad 4π 4π 4π

w w

A small antenna has in general low directivity. By definition, an isotropic radiator, has unit directivity. It can be proved that such an antenna is not physically realizable. Using (3.19) we find that elementary dipoles have directivity D = 3/2. Because of the previous property, an antenna with large directivity has necessarily a narrow main lobe and viceversa.

Closely related to the directivity is the gain of an antenna. The definition is ¯ dPrad ¯¯ dΩ ¯(ϑ,ϕ) g(ϑ,ϕ) = Pin 4π

This means that the radiation intensity in a given direction is compared with the radiation intensity that would be produced by a lossless isotropic radiator fed with the same input power as the actual antenna. Hence the plots of both the directivity and gain functions are normalized power patterns,

54 the difference being just the normalization factor. The directivity is a purely geometrical parameter, whereas gain takes also antenna efficiency into account. Recalling (4.1), we find that g(ϑ,ϕ) = ηd(ϑ,ϕ)

om

Sometimes gain and directivity are defined as ratios of power densities per unit surface, produced at the same distance r by the actual antenna and by the isotropic radiator: ¯ dPrad ¯¯ dΣ ¯(ϑ,ϕ) g(ϑ,ϕ) = (4.4) Pin 4πr2 As for the directivity, when a single number is indicated as the gain of an antenna, it is meant to be the maximum gain, usually expressed in dB:

G = 10 log10 max g(ϑ,ϕ)

32400 Θ1d Θ2d

m

D'

.c

A useful approximate formula allows to estimate the maximum directivity. If the antenna is very directive, and the side lobes can be neglected,

Effective area, effective height

fo

4.1.3

ru

where Θ1d , Θ2d are the FHPBW (in degrees) in two orthogonal planes

w .jn

tu

w

or

ld

An antenna in receive mode behaves as a generator for the receiver, as shown in Fig. 4.4. The wave incident on it is generally a spherical wave, but since it is produced by a very far away transmitting antenna, it can be approximated by a plane wave in the neighborhood of the receiving antenna.

It turns out that if the polarization of this plane wave is changed while the incidence direction is kept fixed, the received power changes. In other words, the sensitivity of the receiving antenna depends also on the polarization of the incident field. This concept may seem obvious in the case of the wire antenna. Since the current on it can flow only in the direction of the wire and since this current is produced by the incident electric field that applies forces on the electrons in the metal and sets them in oscillation, it is to be expected that the antenna response is maximum when the electric field is parallel to the wire. The aperture antenna is apparently symmetrical, hence this property is not so clear. The explanation is that even though the feeding waveguide is circular, only one well defined polarization is used, which is fixed by the waveguide-coaxial cable transition, not shown in the figure.

w w

The internal impedance of the equivalent generator is denoted by Zg . It can be proved, by means of the theorem of reciprocity, that Zg coincides with the input impedance Zin of the same antenna when used in transmit mode. The generator itself can be characterized by means of the available power or by means of the open circuit voltage Vg . As a consequence, two antenna parameters can be defined.

The effective area (or equivalent area) in a particular direction is the ratio of the available power at the output terminals to the power flux density (per unit surface) of a plane wave incident from that direction, the incident wave being polarization matched to the antenna: Aeq (ϑ,ϕ) =

max Pavail ¯ dPinc ¯¯ dΣ ¯(ϑ,ϕ)

(4.5)

55

Ei +

(ϑ , ϕ )

Vg

Zg

(a)

om

(c)

.c

Ei

(ϑ , ϕ )

ru

m

(b )

fo

Figure 4.4. Receiving antennas and their equivalent circuit. (a) Wire antenna with an incident plane wave coming from direction (ϑ,ϕ). (b) Horn, as an example of an aperture antenna, with an incident plane wave. (c) Thévenin equivalent circuit of both types of antennas

w

or

ld

where it is understood that the maximization regards only the polarization of the incident wave.

The effective length (or effective height) in a particular direction is the ratio of the open circuit voltage at the output terminals to the magnitude of the electric field of a plane wave incident from that direction, the incident wave being polarization matched to the antenna:

tu

hef f (ϑ,ϕ) =

max Vg |Ei ||(ϑ,ϕ)

w .jn

Note that the effective height is a vector quantity: its direction is that of the incident electric field that is polarization matched to the antenna. Hence the effective height defines the polarization of the antenna in receive mode.

w w

As a matter of principle, both parameters can be used to describe both types of antennas.

However, it is clear the equivalent area is concept is more appropriate for aperture antennas and the effective height is more appropriate to wire antennas. Actually it is customary to define an aperture efficiency ν max Aeq (ϑ,ϕ) νa = Ag

where Ag is the area of the geometrical aperture and the maximization is performed with respect to the direction of the incoming plane wave. Likewise the length efficiency of a wire antenna is defined by max hef f (ϑ,ϕ) νl = `g

where `g is the length of the antenna.

56 In general the theorem of reciprocity allows to establish a relationship between the receive mode and transmit mode properties of a same antenna. It can be shown that the effective area is connected to the gain by λ2 g(ϑ,ϕ) Aeq (ϑ,ϕ) = (4.6) 4π and that the effective height is connected to the generalized electric dipole moment by

1 Pe (ϑ,ϕ) Ia

hef f (ϑ,ϕ) =

(4.7)

om

From this we learn that the polarization of an antenna in receive mode, relative to a particular direction, coincides with the polarization of the electric field radiated in that direction by the same antenna in transmit mode.

m

Vg = hef f (ϑ,ϕ)|Ei |p̂T X · p̂RX

.c

From an application point of view, it is useful to compute the equivalent generator in the case the incident wave is not polarization matched. If p̂T X , p̂RX are the polarization unit vectors of the incident wave and of the receiving antenna, respectively, it can be shown that

¯ dPinc ¯¯ 2 |p̂T X · p̂RX | dΣ ¯(ϑ,ϕ)

(4.9)

fo

Pavail = Aeq (ϑ,ϕ)

ru

and

(4.8)

ld

Notice that the polarization of the incident wave has been denoted by p̂T X , because this is exactly the definition of the polarization of the transmitting antenna that produces this wave.

or

Several relationships exist among the various parameters introduced up to now.

Z

w

• Radiation resistance From (4.2)

tu

Rrad

2Z0 = 4λ2

Z

2π

dϕ 0

π

|hef f (ϑ, ϕ)|2 sin ϑdϑ

(4.10)

0

w w

w .jn

• Effective height and gain.

Recall the definition of gain (4.4) and (3.40)

¯ ¯ |E|2 ¯¯ dPrad ¯¯ 2Z0 ¯(ϑ,ϕ) dΣ ¯(ϑ,ϕ) = g(ϑ,ϕ) = Pin Pin 2 4πr2 4πr

(4.11)

Moreover, from (3.39) and (4.7)

1 Z0 |Ia |2 dPrad |hef f (ϑ, ϕ)|2 = 2 4λ2 r2 dΣ

and Pin = 21 Rin |Ia |2 . Substituting we get

g(ϑ,ϕ) =

1 πZ0 |hef f (ϑ, ϕ)|2 λ2 Rin

(4.12)

57 • Equivalent area and effective height.

From (4.9) the available power is Pavail = Aeq (ϑ,ϕ)

1 |Ei |2 2 |p̂T X · p̂RX | 2 Z0

From the equivalent circuit of Fig. 4.4, we have Pavail =

1 |hef f (ϑ,ϕ)|2 |Ei |2 |p̂T X · p̂RX |2 1 |Vg |2 = 4Rin 2 2 4Rg

From the comparison between the two, the desired relationship follows Z0 |hef f (ϑ,ϕ)|2 4Rin

om

Aeq (ϑ,ϕ) =

.c

This relationship could also be derived by direct application of (4.6)on (4.12).

fo

ru

m

• Directivity and effective height.

From (4.3), (3.38) and (4.7), we get ¯ dPrad ¯¯ 1 Z0 |Ia |2 |hef f (r̂)|2 ¯ dΩ r̂ 2 4λ 2 = d(r̂) = Z Prad 1 Z0 |Ia |2 1 |hef f (r̂)|2 dΩ 4π 2 4λ2 4π

ld

and cancelling common factors

|h (r̂)|2 Z ef f 1 |hef f (r̂)|2 dΩ 4π

tu

• Radiated electric field.

From (3.33)

w

or

d(r̂) =

w .jn

E(r) ∼ −j

Z0 Ia exp(−jk0 r) |hef f (r̂)| p̂T X 2rλ

Now invert (4.12)

Z0 Ia exp(−jk0 r) E(r) ∼ −j λ 2rλ

r

Rin g(r̂) p̂T X πZ0

w w

and use again Pin = 21 Rin |Ia |2 to find

4.2

E(r) ∼ −j

Z0 exp(−jk0 r) 2r

r

2Pin g(r̂) p̂T X πZ0

Friis transmission formula

When designing a radio link the fundamental problem is to compute the received power knowing the transmitted power, the characteristics of the antennas and their distance. The solution is given

by Friis formula, that we derive.

58

Pin

Pavail

GTX pˆ TX

GRX pˆ RX

R Figure 4.5.

Geometry for the Friis transmission formula

dPinc 2 |p̂T X · p̂RX | dΣ

m

Pavail = Aeq

.c

om

Consider a transmitting antenna with gain GT X in the direction of the link and polarization unit vector p̂T X , fed with the power Pin , as shown in Fig. 4.5. Let GRX , p̂RX , the corresponding parameters of the receiving antenna, placed at a distance R from the transmitting one. This distance must be large enough so that the two antennas are one in the far field region of the other.

This condition is always satisfied in all common situations. In the cases where the relative distance is small, Friis formula must be substituted by a more complex one.

From the definition of equivalent area (4.5) we can relate the available power at the output port of the receiving antenna to the incident power flux density

fo

ru

Now the incident power flux density is created by the transmitting antenna and we can relate it to the input power to the transmitting antenna via the definition of gain (4.4)

ld

dPrad dPinc Pin = GT X = 4πR2 dΣ dΣ

hence

Pin 2 |p̂T X · p̂RX | 4πR2 It is convenient to express the effective area of the receiving antenna in terms of its gain by (4.6), so that the Friis formula reads

w

or

Pavail = Aeq GT X

(4.13)

w .jn

tu

GT X GRX 2 Pavail = Pin µ ¶2 |p̂T X · p̂RX | 4πR λ

w w

This result is remarkably simple and one could be perhaps kind of astonished, after having seen the complicated equations yielding the field radiated by an antenna, e.g. (3.33) and (3.34). The explanation, however, is straightforward: all the complications are hidden in the gains of the antennas (see (4.12)), whose definition was exactly conceived with a view to a simple form of the Friis transmission formula!

However, it must be remarked that the gain of an antenna can be measured fairly easily as suggested by (4.13). First two identical antennas are placed one in front of the other at a convenient distance. Then the received power is maximized by changing the polarization of the transmitting antenna: in the case of linearly polarization, this requires just a rotation of the antenna around its axis. Finally, from a measurement of the input power to the T X antenna and of the received power, it is simple matter to obtain the maximum gain of each antenna. If this operation is repeated for all frequencies in a band, a calibrated reference antenna has been obtained. This can now be used to measure another antenna, whatever its characteristics, by the same technique. Also, if the reference antenna is kept fixed while the antenna under test is rotated, as shown in Fig. 4.6, its

59

ϑ ƌĞĨĞƌĞŶĐĞ ĂŶƚĞŶŶĂ

ĂŶƚĞŶŶĂ

ƵŶĚĞƌ ƚĞƐƚ

Figure 4.6. Measurement of the radiation pattern of an antenna under test by means of a calibrated reference antenna

Examples of simple antennas

.c

4.3

om

gain function is measured for any direction ϑ, ϕ. Of course, the spherical coordinate system is attached to the rotating antenna, so that the line joining the two antennas has the direction ϑ, ϕ in this reference.

w w

w .jn

tu

w

or

ld

fo

ru

m

We describe in the following a few types of antennas, commonly used in the applications. The structures we describe are made of metal, whose conductivity is so high that, as far as the computation of the radiation pattern is concerned, can be considered infinite. Wire antennas have a pair of terminals that are connected to a transmission line. When the antenna is in operation, it absorbs a current Ia from the transmission line, which then gives rise to a current distribution J on the surface of all the parts of the antenna. This current is the source of the radiated field and Eq.(3.34) allows us to compute it. We face two problems here, however. When (3.34) was derived, it was assumed that the current distribution J(r) was radiating in free space and not in presence of the metal structure. Indeed, the Green’s function we have found in section 3.1 is that of free space. The second problem is that the distribution J(r) is not known, because it is not under our direct control. When we solved the radiation problem, we assumed to know the current distribution and, on the basis of that, we were able to compute the radiated field. In the antenna case, the only thing that we can control is Ia but nothing else. Hence the question arises: according to what rule does the current distribute on the parts of the antenna? The answer lies in the boundary conditions that we introduced in (1.16): the current distribution has such a form that the radiated electric field has zero tangential component on the antenna surface. In this way all the electromagnetic field components in the perfect conductor are zero, as they should be. It is clear that this prescription does not help us very much in finding the expression of J(r), since mathematically it entails the solution of a complicated integral equation! However, we have the answer to the first question we asked above. Indeed, if the electromagnetic field inside a volume is zero, we can freely change the constitutive relations of the material in that volume. A similar situation exists in circuit theory. Suppose that in a circuit the element ZL is connected between nodes A and B and suppose it is known that both the current IA and the voltage VAB are zero.

Ohm’s law VAB = ZL IA becomes the identity 0 = ZL · 0, which is satisfied for any ZL . Hence the value of this impedance can be changed and chosen at will, without affecting the rest of the circuit.

Similarly, in the electromagnetic case, if both the electric and magnetic fields in the metal are zero, the metal can be substituted with any other dielectric, without any consequences in the rest of the structure. The only sensible choice in this case is to substitute the metal with free space, so that the medium is altogether homogeneous and the free space Green’s functions (3.8) can be used safely to compute the radiated field everywhere and in particular in the far field region.

To better appreciate what we have found, consider another case, where a given source J(r), for instance an electric dipole, is supposed to radiate in presence of a metal sphere, placed at a certain

60 distance. In this case the metal cannot be substituted with free space, because the given source is arbitrary and does not produce zero field in the region occupied by the metal. Nevertheless, if the metal is a perfect conductor, the field there is zero, but then it is the Green’s function that takes care of that. In conclusion the Green’s function to be used in this case is not the free space one and Z E(r) = −jωµ G(r,r0 ) · J(r0 )dr0

.c

om

Note that this is not a convolution integral, because the Green’s function G(r,r0 ) depends on the coordinates of both the source r0 and the field point r separately and not only on their difference r − r0 : this is an obvious consequence of the fact that the system is no longer space invariant, i.e. free space containing the metal sphere is no longer homogeneous.

Another way to look at this problem is to fix the attention on the current distribution JΣ that is induced on the surface of the sphere. The form of it is such that the total electric field radiated by J(r) and JΣ has zero tangential component on the PEC sphere. Then, as before, since the total E field in the metal sphere is zero due to the current distribution, we can substitute the metal with free space and use the free space Green’s function. In conclusion we have two alternatives:

m

• use only the independent current distribution, but use a complicated Green’s function (generally not known explicitely) to enforce the vanishing E field in the metal

fo

ru

• introduce also the unknown induced current distribution, but use the free space Green’s function to compute the radiated field In the following applications we will always use the second approach.

Wire antennas

or

4.3.1

ld

These simple considerations are formulated in a more rigorous form by the equivalence theorem.

tu

w

Consider an antenna made of a metal wire of small diameter and with a shape described by the parametric equation r = rγ (s), where s is the arc length parameter. Let I(s) be the electric current distribution on it (measured in A). The electric current density can be written

s

w w

w .jn

ŝ

Figure 4.7.

I (s)

P rγ ( s)

r O

Wire antenna of shape r = rγ (s). P is the field point

J(r) = I(s)δ(r − rγ (s))ŝ where ŝ is the unit vector tangent to the curve, δ is a 2D Dirac delta function with support on the curve r = rγ (s). Its role is to express mathematically the statement that the current flows on a wire of negligible diameter. The magnetic current for this source is obviously zero. This is the

61 z

P r

2

ϑ

Ia y

Va x

−

Linear antenna. (a) Actual shape. (b) Computational model

.c

Figure 4.8.

(b )

om

(a)

2

m

current that was called JΣ above. Rigorously, its expression should be found by solving an integral equation, but sometimes it can be estimated with reasonable accuracy.

fo

ru

If we want to compute the electromagnetic field radiated in the far field region, it is enough to evaluate the generalized electric dipole moment of this source. According to the definition (3.34): Z Z sb 0 0 Pe (r̂) = Itr · I(s)δ(r − rγ (s))ŝ exp(jk0 r̂ · r )dr = Itr · I(s)ŝ exp(jk0 r̂ · rγ (s))ds (4.14) sa

w

or

ld

The integration limits define the beginning and the end of the wire. This is a very general formula, which can be applied for any shape of the wire and any current distribution on it. In general the integral has to be evaluated by numerical quadrature, but there are some cases in which a closed form result is obtained.

tu

Short dipole

w .jn

As a first example we consider a rectilinear antenna of length `, as shown in Fig. 4.8. The current distribution is I(z) can be assumed to be constant if ` ¿ λ, say ` ≤ λ/10. This simple antenna is called short dipole or Hertzian dipole. Sometimes two small spheres or plates are added to the wire ends to help creating a constant current distribution. In this case

w w

s=z

Hence

ŝ = ẑ

Z

`/2

Pe (r̂) = Itr · ẑ −`/2

r̂ · rγ (s) = z cos ϑ

I(z) = Ia Z

Ia ejk0 z cos ϑ dz = −ϑ̂ sin ϑIa

`/2

ejk0 z cos ϑ dz

−`/2

Here we have used Itr · ẑ = (ϑ̂ϑ̂ + ϕ̂ϕ̂) · ẑ = −ϑ̂ sin ϑ recalling (A.4). The integral is Z

`/2

jβz

e −`/2

¯`/2 sin(β`/2)) 2j sin(β`/2)) ejβ`/2 − e−jβ`/2 ejβz ¯¯ =` = = dz = ¯ (β`/2) jβ jβ jβ −`/2

62 Fe(x) 1 0.8 0.6 0.4

om

0.2 0

2

4

6

8

m

−0.4 0

10

ru

x

Plot of the function FE (x)

fo

Figure 4.9.

.c

−0.2

ld

It is useful to introduce the function FE (x), shown in Fig. 4.9: sin x x

(4.15)

or

FE (x) =

w

Note that FE (x) is even and vanishes for x = nπ, n = 1,2, . . .. Moreover the side lobes are about 13.4 dB below the maximum.

In conclusion, the required dipole moment is

tu

sin(k0 ` cos ϑ/2)) = −ϑ̂ sin ϑMe FE (k0 ` cos ϑ/2) (k0 ` cos ϑ/2)

w .jn

Pe (r̂) = −ϑ̂ sin ϑIa `

w w

where we have introduced the dipole moment Me = Ia `. The maximum value of the argument of FE is π`/λ, which is very small, so that FE ' 1 and the Hertzian dipole has the same pattern as the elementary dipole. Notice that the previous formula can be used for any value of `, even greater than λ. However, if the antenna is long, the current I(s) cannot be assumed to be constant and the previous formulas are no longer applicable. We list here the main properties: • The effective height (4.7) is hef f (ϑ,ϕ) = −` sin ϑ ϑ̂

• The radiation resistance is, see (4.10) and (3.17) Rrad =

2π 2Prad = Z0 2 3 |Ia |

µ ¶2 ` λ

For example, if ` = λ/20, the radiation resistance is Rrad ' 2Ω. This means that a fairly large current Ia ' 10 is necessary to radiate a power Prad = 100W.

63 If we want to estimate the efficiency of short dipoles we have to compute the dissipated power. It can be proved that the resistance per unit length of a wire with radius rw , length ` and conductivity γ is 1 1 Rdiss = γ 2πrw δ

om

where the denominator of the second fraction is an estimate of the area of the annulus with radii rw and rw − δ, with δ denoting the skin depth (see(2.20)), in the assumption of well developed skin effect δ/rw ¿ 1. Consider for example a dipole with length ` = 1 m for frequencies in the range between 100kHz and 30MHz, for which `/λ varies between 3.3 · 10−4 and 0.1, so that the dipole is always short. Fig. 4.10 shows plots of radiation resistance Rrad and of wire resistance Rdiss in the case the wire radius is rw = 1mm. Notice that in this frequency range, the skin depth δ varies between 0.2 and 0.01 mm. Fig. 4.11 shows a plot Radiation and loss resistance [Ω]

.c

2.5

m

2 (a)

ru

1.5

fo

1

0.5

0

0.04

0.06

or

0.02

ld

(b)

0.08

0.1

Plot of Rrad (a) and Rdiss (b) in the case of a short dipole with radius rw = 1 mm

tu

Figure 4.10.

w

L/λ

w .jn

of the efficiency for the same short dipole, when the wire radius is rw = 1, 3, 5 mm. We see clearly that the the efficiency η of very short dipoles can be very low, but is almost 1 for dipoles about λ/10.

w w

• The directivity is the same as that of the elementary dipole: recalling (4.3),(3.19), (3.17), we find ¯ dPrad ¯¯ Z0 Me2 sin2 ϑ dΩ ¯(ϑ,ϕ) 2 3 4λ = = sin2 ϑ d(ϑ,ϕ) = 2 Prad 2 1 Z0 Me 2π 1 4π 2 3λ2 4π

Hence the maximum directivity is D = 1.5 and the maximum gain G = 1.5η.

Half-wavelength dipole We consider now a dipole whose length L is comparable with the wavelength. In this case we cannot suppose that the current I(s) is constant. It can be shown that a good approximation of

64

efficiency 1 (c) 0.8

0.6

(b)

(a)

0.4

0.04 0.06 L/lambda

0.08

0.1

the current distribution is Ia sin k0 sin k0 L/2

µ

L − |z| 2

¶ for

−

L L ≤z≤ 2 2

fo

I(z) =

m

Plot of the short dipole efficiency. (a) rw = 1 mm, (b) rw = 3 mm, (c) rw = 5 mm

ru

Figure 4.11.

0.02

.c

0 0

om

0.2

or

tu

0.5

w

1

ld

This current distribution is shown in Fig. 4.12 for two lengths of the dipole, i.e. for L = λ/2 and 3λ/2: if the geometrical length of the wire is the same in the two cases, the frequency in the second case is three times the one in the first case. Note that the current is zero at the dipole ends, since

w .jn

0

w w

−0.5

−1 −0.5

0 z/L

0.5

Figure 4.12. Plot of the current distribution on a dipole λ/2 and 3λ/2 long. A sketch of the dipole is superimposed on the current plot

no plates (with the goal of providing capacitance) are present there, as in the case of the Hertzian dipole. It can be shown that the generalized dipole moment, computed by (4.14) is Pe (ϑ,ϕ) = −λIa

cos(k0 L cos ϑ/2) − cos(k0 L/2) ϑ̂ π sin ϑ

65 FH(x) 1.2 1 0.8 0.6

om

0.4 0.2

2

4

6

8

m

−0.2 0

10

ru

x

Plot of the function FH (x)

fo

Figure 4.13.

.c

0

ld

The case of particular interest in the applications is that of L = λ/2, i.e. the half wave dipole.

The relevant generalized dipole moment is cos(π cos ϑ/2) ϑ̂ π sin2 ϑ

or

Pe (ϑ,ϕ) = −λIa sin ϑ

that we write as

w .jn

tu

w

´ ³π 2 cos ϑ ϑ̂ Pe (ϑ,ϕ) = −λIa sin ϑ FH 2 π where the function FH (x) is defined by FH (x) =

cos x µ ¶2 2x 1− π

w w

and its plot is shown in Fig. 4.13 Note that FH (x) is even and vanishes for x = n = 1,2, . . .. Also, the side lobes are about 23 dB below the maximum.

(4.16)

(2n + 1)π , 2

The maximum value of the argument of FH is π/2, reached for ϑ = 0. The pattern turns out to be slightly more directive than that of the elementary dipole, as shown in Fig. 4.14 The main properties of the half-wavelength dipole are • Maximum directivity D = 1.643, to be compared with that of the elementary dipole D = 1.5.

• Full Half Power Beamwidth =78◦ , to be compared with 90◦ of the elementary dipole.

• Radiation resistance Rrad ' 73 Ω. This is the most important property of half-wavelength dipole antennas. Indeed, matching to transmission lines that typically have characteristic

66

90

1.5 60

120

1 30

150 0.5

0

210

240

.c

330

om

180

m

300

ru

270

fo

Figure 4.14. Directivity pattern of the λ/2 dipole (solid line) and of the elementary dipole (dashed line). The maximum directivity is D = 1.643 for the λ/2 dipole and D = 1.5 for the elementary dipole.

tu

w

or

ld

impedance Z∞ = 50 Ω or Z∞ = 75 Ω is very simple. Moreover, to radiate a power Prad = 100 W requires only a current Ia = 1.65 A instead of 10 A, as in the case of the λ/20 dipole previously examined. Hence, the efficiency is higher. Finally, note that if the radiation resistance of the half-wavelength dipole is computed according to the equation derived for short dipoles, the value Rrad = 197 Ω is obtained. The difference is due to the different current distribution I(z) assumed in the two cases.

w w

w .jn

It is to be remarked that the dipole antenna requires a balanced feeding transmission line, such as a twin-lead one (two-wire transmission line). A coaxial cable is an inherently unbalanced line: if the inner conductor is connected to a dipole arm and the outer to the other, the result is that an unwanted current flows to ground on the outer surface of the outer conductor, with a degradation of the antenna performance. To avoid this phenomenon, a balun (balanced to unbalanced) must be placed between the coax and the dipole. ŵŽŶŽƉŽůĞ

L=

λ 4 ŐƌŽƵŶĚ ƉůĂŶĞ

ĐŽĂdž

Figure 4.15. Half-wavelength dipole realized by a λ/4 monopole on a metal ground plane and fed by a coaxial cable, whose outer conductor is bonded to the plane.

67 Sometimes a half-wavelength dipole is realized in the form of a λ/4 long monopole, mounted perpendicularly to a metal ground plane, as sketched in Fig 4.15. If this ground plane were infinite, image theorem would guarantee the perfect equivalence. In practice some degradations occur.

Loop antenna

om

Loop antennas are generally classified into two groups, electrically small or large depending on whether the circumference is smaller than λ/10 or larger, up to one wavelength. If the loop is small, its actual shape plays no role: circular, square or elliptical loops have the same radiation characteristic of an elementary magnetic dipole, directed perpendicularly to the loop plane.

Moreover, radiation is maximum in the loop plane. As the circumference is increased toward one wavelength, the maximum of the pattern shifts from the plane of the loop to the direction of the axis.

.c

Small loops are very poor radiators, because of their very low radiation resistance. Radiation resistance can be increased up to a value comparable to that of typical transmission lines by increasing the number of turns or by winding the loops on a ferrite core.

or

ld

fo

ru

m

Concerning the equivalence between small loops and magnetic dipoles, it can be shown that the magnetic dipole moment of a loop embracing a surface S, on which the current distribution is constant and equal to Ia , is 2π Z0 SIa Mm = jωµSIa = λ Then the radiated fields can be computed by means of (3.23). As it was explained in detail there, the radiation pattern of the magnetic dipole is the same as that of the electric one. The difference is in the fields: the magnetic dipole radiates a magnetic field with ϑ (dominant in far field) and r components, and an electric field with only a ϕ component, parallel to the loop plane.

The radiation resistance is computed, according to the definition, by means of (3.24):

w

µ

kS λ

¶2

µ = 20π

2

C λ

¶4

tu

Rrad

2π 2Prad = Z0 = 2 3 |Ia |

w .jn

If the loop antenna is made with N turns, the previous radiation resistance has to be multiplied

by N 2 . Indeed, the N -turns coil with a current Ia is equivalent to a single coil with current N Ia , which radiates a power N 2 larger than the single turn one.

As an example, an antenna made of N = 100 turns with radius a = 1 cm at f = 300 MHz has a radiation resistance Rrad = 30.7 Ω, whereas the single coil one has only Rrad = 30.7 · 10−4 Ω!

w w

The maximum directivity of the small loop is the same as that of the elementary dipole, i.e.

D = 1.5.

4.3.2

Aperture antennas

Aperture antennas are completely different from wire antennas. They are still made of metal parts, and we can consider the currents flowing on them as responsible for the radiated fields.

However, these antennas are characterized by the presence of apertures, as in the case of horns and paraboloids, as shown in Fig. 4.16 and Fig. 4.17. Actually, a paraboloid is just a mirror and typically is used in conjunction with a horn, placed in the focus, that acts as a feed, as sketched in Fig.4.17. In the case of aperture antennas, one can estimate the fields on the aperture with

68

Examples of horn antennas, connected to rectangular and circular waveguides.

w

or

&ĞĞĚ ŚŽƌŶ

ld

ĚŝƐŚ

fo

ru

m

.c

ĂƉĞƌƚƵƌĞ

om

Figure 4.16.

w .jn

tu

Figure 4.17. Front fed paraboloid. The aperture is the circle defined by the dish rim and the plane Π is to be considered in contact with it.

w w

reasonable accuracy, while the currents on the metal parts can be very difficult to guess. By means of the equivalence theorem it is possible to represent the aperture fields in terms of equivalent sources, which are responsible for the radiated fields.

Rectangular horn antenna Horns are always fed by a waveguide of convenient cross section, generally rectangular or circular.

Waveguides are used invariably in single mode operation, so that their cross section has a size less than one wavelength. For this reason a truncated waveguide is a very poor radiator: • the mode of propagation approaching the waveguide end suffers a fairly large reflection coefficient • the radiation pattern is very wide

69 In practice a horn carries out a slow transformation of the waveguide cross section into a larger one, so that both problems are solved at the same time. In the case of a pyramidal horn fed by a rectangular waveguide the aperture electric and magnetic fields are assumed to be ³ πx ´ E(x,y) = E0 cos ŷ A ³ πx ´ E0 1 cos H(x,y) = x̂ ẑ × E(x,y) = − (4.17) A Z0 Z0 B B A A ≤y≤ and − for − ≤ x ≤ 2 2 2 2

m

.c

om

where the aperture is rectangular with sides A and B, as shown in Fig. 4.18. This field distribution is real because the phase error has been neglected for simplicity. This phase error takes into account the fact that the wavefront propagating in the horn is spherical instead of plane, as assumed here.

The electric field distribution is that of the fundamental mode of the feed rectangular waveguide, mapped smoothly onto the aperture. The magnetic field would be also that of the fundamental mode if the wave impedance were substituted by the modal impedance. The use of Z0 is more accurate for apertures with size greater than 2-3 λ. A plot of the normalized field distribution is shown in Fig. 4.19 It can be shown that the generalized dipole moment of this horn has the

ld

fo

B

ru

y

x

Rectangular horn aperture, with sides A and B.

w

Figure 4.18.

or

A

tu

expression (valid only in the front half space z ≥ 0, i.e. 0 ≤ ϑ ≤ π/2))

w .jn

Pe (ϑ,ϕ) = −2Y0 E0 AB cos2

ϑ 2 FH (ηx ) FE (ηy ) p̂ 2 π

with

πA sin ϑ cos ϕ λ πB sin ϑ sin ϕ ηy = λ

w w

ηx =

where the functions FE and FH were introduced in (4.15) and (4.16) and the polarization unit vector p̂ is given by p̂ = sin ϕ ϑ̂ + cos ϕ ϕ̂ Fig. 4.20 helps in understanding the definition of p̂. The rectangle is the horn aperture, seen from a point on the z axis. The circle indicates a cone of directions with the same ϑ (assumed to be very small) and different ϕ: in each point the local p̂ is shown. It appears to be parallel to ŷ, but actually is tangent to the sphere through the observation point. figura diagramma horn polarizzzione 3D bis This is more clearly indicated in Figs. 4.21,4.22 from two different views.

70

1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2

0.5 0.5 0

Figure 4.19.

−0.5

−0.5

x/A

m

y/B

.c

0

om

0.1

Horn normalized aperture field distribution.

tu

w

fo

or

ld

p̂

ru

y

ϑ = const

Polarization vector p̂ for small ϑ.

w .jn

Figure 4.20.

x

The plane x,z (ϕ = 0,π) is called H plane because it contains the H field; similarly, the plane y,z (ϕ = π/2,3π/2) is called E plane because it contains the E field.

w w

An example of 3D field pattern for a horn with size A = 2λ, B = λ is shown in Figs. 4.23. For better clarity a cartesian diagram is used, hence the z coordinate (vertical axis) shows |Pe (ϑ,ϕ)|.

The angles ϑ, ϕ define a unit vector ŝ and its x and y components are used as coordinates in the horizontal plane: sx = sin ϑ cos ϕ and sy = sin ϑ sin ϕ. The polar coordinates in this plane are sρ =

q s2x + s2y = sin ϑ and

ϕ

As a consequence, the limiting circle with unit radius in the horizontal plane corresponds to ϑ = π/2.

Fig. 4.24 shows the field pattern of the same horn in the standard polar diagram. The aperture is also shown.

An example of a more directive horn, with A = 4λ and B = 4λ is shown in Figs. 4.25, 4.26.

Polarization vector field p̂ drawn on the half sphere of directions.

w w

w .jn

tu

w

or

ld

fo

ru

Figure 4.21.

m

.c

om

71

Figure 4.22.

Polarization vector field p̂ drawn on the half sphere of directions.

Two dimensional plots are easier to use. Figs. 4.27, 4.28 show the plots of the functions FE (ηy ), FH (ηx ) in dB scale. The corresponding plots in linear scale are shown in Figs. 4.29, 4.30.

The E plane pattern has a number of side-lobes that increases with B/λ: indeed, the maximum value of ηy is πB/λ. The first side-lobe is near ηy = 1.5π and its value is -13.3 dB. As for the H plane, the maximum value of ηx is πA/λ. The first side-lobe is near ηx = 2 and has a value of -23 dB. The reason why the H plane side-lobes are lower than the E plane ones is due to the fact that the field distribution is tapered in the x direction but not in the y direction, see Fig. 4.19. As

72

Rectangular horn: A/λ=2, B/λ=1

E plane φ=π/2

H plane φ=0

1

0.8

0.6

0.4

0 1

1

0.5 0.5

.c

0 0 −0.5

−0.5 −1

m

−1

sinθ sinφ

ru

sinθ cosφ

Rectangular horn field pattern, cartesian diagram, A = 2λ and B = λ.

fo

Figure 4.23.

om

0.2

H plane

B/λ A/λ

w w

w .jn

tu

w

or

E plane

ld

Rectangular horn: A/λ=2, B/λ=1

Figure 4.24.

Rectangular horn field pattern, standard polar diagram, A = 2λ and B = λ.

a consequence, the beamwidth is larger in the H plane than in the E plane. If the main lobe beamwidth (between the zeros) is to be the same in the E and H planes in order to have a rotationally symmetrical main lobe, the condition B = (2/3)A must be enforced.

73

Rectangular horn: A/λ=4, B/λ=4

E plane φ=π/2

H plane φ=0

1

0.8

0.6

0.4

0 1

1

0.5 0.5

.c

0 0 −0.5

−0.5 −1

−1

m

sinθ sinφ

ru

sinθ cosφ

Rectangular horn field pattern, cartesian diagram, A = 4λ and B = 4λ.

fo

Figure 4.25.

om

0.2

H plane

w .jn

tu

w

or

E plane

ld

Rectangular horn: A/λ=4, B/λ=4

B/λ

w w

A/λ

Figure 4.26.

Rectangular horn field pattern, standard polar diagram, A = 4λ and B = 4λ.

Finally, it can be shown that the aperture efficiency of horns is νa =

Aeq = 0.8 Ag

74

FE(ηy) 0

−5

−10

dB

−15

−20

om

−25

−30

1

1.5

2 ηy/π

2.5

3

3.5

4

Horn E plane field pattern: plot of the function FE (ηy ).

fo

Figure 4.27.

0.5

m

0

ru

−40

.c

−35

FH(ηx)

ld

0

or

−5

w

−10

tu

−20

w .jn

dB

−15

−25

w w

−30

−35

−40

0

Figure 4.28.

0.5

1

1.5

2 η /π

2.5

3

3.5

4

x

Horn H plane field pattern: plot of the function FH (ηx ).

In this way, if the size A, B of the horn is known, the maximum gain can be easily computed from Aeq = νa AB

G=

4π Aeq λ2

75

F (η ) E y 1

0.9

0.8

0.7

0.6

0.5

0.4

om

0.3

0.2

0

0

0.5

1

1.5

2

2.5

3.5

4

ru

Horn E plane field pattern: plot of the function FE (ηy ).

fo

Figure 4.29.

3

m

ηy/π

.c

0.1

F (η ) H x

ld

1

or

0.9

0.8

w

0.7

tu

0.6

0.5

w .jn

0.4

0.3

w w

0.2

0.1

0

0

Figure 4.30.

0.5

1

1.5

2

ηx/π

2.5

3

3.5

4

Horn H plane field pattern: plot of the function FH (ηx ).

Also, losses are very small and gain and directivity coincide.

Chapter 5

m

.c

om

Waveguides

w .jn

tu

w

or

ld

fo

ru

Electromagnetic waves have the natural tendency to propagate in all directions away from the point where they are generated. The fields created by a dipole in free space are a good example.

For some applications, it is desirable that waves propagate in only one direction, say z, while the field is appreciable only in a neighborhood of it. This behavior can be obtained by generating the electromagnetic wave inside a metal tube, called waveguide. The side walls have very high (if not infinite) conductivity and act as mirrors that prevent propagation in the transverse direction. The cross section of the tube is arbitrary, but generally it is rectangular, circular or in the form of a ridge, as shown in Fig. 5.1. Sometimes the cross section is simply connected, sometimes doubly or multiply connected: the coaxial cable belongs to this latter class.

w w

Figure 5.1. Various waveguide types. From left to right: rectangular, cylindrical, coaxial and ridge waveguide

Waveguides are not always in the form of hollow tubes. The microstrip is a different kind of waveguide, which appears to be open, without sidewalls. the field confinement is due to the dielectric inhomogeneity of the cross section. Optical fibers are another example of waveguides without sidewalls. The confinement here is related to the total reflection at the interface between two different dielectrics.

In these notes we focus on the hollow tube waveguides.

76

77

5.1

Waveguide modes

Fig. 5.2 shows an example of rectangular waveguide containing a source, in the form of a small dipole, connected to the inner conductor of a coaxial cable through a hole in the wide side of the waveguide. The problem that we want to solve is to compute the fields radiated by this source in the waveguide. For simplicity, we assume that the walls are perfect conductors and that the waveguide is filled with a homogeneous dielectric with permittivity ε and permeability µ. The mathematical formulation of the problem is the same as (2.1), that we repeat here for convenience

z

om

y

fo

ru

m

.c

x

= =

−jωµH − M jωεE + J on the metal

w

or

   ∇×E ∇×H   Etg = 0

ld

Figure 5.2. Waveguide with rectangular cross section. The field in the waveguide is radiated by a kind of small dipole, connected to a coaxial cable through a hole in the wide side of the waveguide

(5.1)

w .jn

tu

As we already did in Chapter 2, we solve first the problem without sources, in order to find the possible fields that can exist in the waveguide. They must satisfy • Maxwell’s equations

w w

• the boundary conditions that is

   ∇×E ∇×H   Etg = 0

= −jωµH = jωεE on the metal

(5.2)

The problem is homogeneous and its solutions are called modes of propagation. When they are known, we solve the radiation problem by computing the excitation coefficient of each of them due to the dipole antenna.

The z axis is a symmetry axis of the system: the system is LSz I, i.e. Linear z-invariant. In other words the coefficients of the equations do not depend on z, hence we expect the solutions

78 to depend on z by means of exponentials that we denote by exp(−jkz z). This means that the solution of (5.2) can be written as Et (ρ) = V (z)e(ρ) Ez (ρ) = I(z)Zt ez (ρ) Ht (ρ) = I(z)h(ρ) Hz (ρ) = V (z)Yt hz (ρ)

(5.3)

where ρ denotes the transverse plane coordinates: a generic point P has coordinates (ρ, z). The functions V (z), I(z) are V (z) = V0+ e−jkz z + V0− ejkz z

om

I(z) = Yt V0+ e−jkz z − Yt V0− ejkz z

(5.4)

fo

E = Et + Ez ẑ

ru

m

.c

We recognize that the z dependence of the field is indeed in the form of complex exponentials that represent traveling waves. Moreover the form of these equations is the same as the ones describing the electrical state of a transmission line. In this case, however, there is no real transmission line but only an “equivalent” one: the z dependence of each propagation mode can be studied by means of a modal transmission line with suitable propagation constant kz and characteristic (or modal) impedance Zt , where the t subscript stands for transverse (to z). Actually, we notice that the presence of the z symmetry axis makes it convenient to split the fields into transverse (Et , Ht ) and longitudinal components (Ez , Hz ): (5.5)

ld

H = Ht + Hz ẑ

w .jn

tu

w

or

Both the transverse and the longitudinal fields are written in the form of products of normalized functions of the transverse coordinates, called modal functions, and of functions of z (modal voltage and current) that can be considered as coefficients.

It is to be remarked that the transverse components play a major role: indeed the power flow budget is evaluated, as always, by means of the flux of the Poynting’s vector S. Obviously, power can flow only in the z direction, so that only the z component of S has to be computed: it is easy to realize that only the transverse field components yield a contribution Z 1 P = R (Et × H∗t ) · ẑ dΣ 2 Σ

w w

A mode of propagation is characterized by the set of modal functions {e, ez , h, hz } and a number kt called transverse wavenumber. The modal functions are not independent, however. For this type of waveguide (PEC walls and transversally homogeneous dielectric) it turns out that the modes of propagation belong two sets: T E modes, identified by a vanishing longitudinal electric field, ez (ρ) ≡ 0 over the complete cross section Σ and T M modes, with identically zero longitudinal magnetic field hz (ρ) ≡ 0. Each family consists of an infinite number of elements, that we label by an index i.

In particular, T M modes, denoted by a prime have the following characteristics • h0iz (ρ) ≡ 0,

• e0 i (ρ) = −

e0iz (ρ) =

∇t Φi (ρ) 0 kti

0 kti Φi (ρ) jkzi

79 • h0 i (ρ) = ẑ × e0 i (ρ) 0 • kti

T E modes are denoted by a double prime and have the properties • e00iz (ρ) ≡ 0,

• h00 i (ρ) = −

h00iz (ρ) =

00 kti Ψi (ρ) jkzi

∇t Ψi (ρ) 00 kti

om

• e00 i (ρ) = h00 i (ρ) × ẑ 00 • kti

ru

m

.c

When no prime or double prime is used, it is understood that the statement is applicable to both T M and T E modes.

0 We see that all the T M mode functions are derived from the couple {Φi (ρ), kti } and all T E 00 ones are derived from the couple {Ψi (ρ), kti }, where the generating functions Φi (ρ), Ψi (ρ) are proportional to the longitudinal field components. The T M generating functions are solutions of the two-dimensional Dirichlet problem for the Helmholtz equation:

fo

02 (∇2t + kti )Φi (ρ) = 0 Φi (ργ ) = 0

(5.6)

or

ld

We recognize the functions Φi (ρ) to be eigenfunctions of the Dirichlet transverse laplacian with 02 eigenvalue −kti .

w

Similarly, the T E generating functions are solutions of the two-dimensional Neumann problem for the Helmholtz equation:

w .jn

tu

002 (∇2t + kti )Ψi (ρ) = 0 ¯ ∂Ψi (ρ) ¯¯ =0 ∂ν ¯

(5.7)

ργ

w w

It is interesting to note that the generating functions Φi (ρ) can be interpreted as the oscillation modes of a drum, i.e. of an elastic membrane with the rim fixed on rigid ring with the same shape as the waveguide cross section. Analogously, the generating functions Ψi (ρ) can be interpreted as the oscillation modes of the surface of a liquid in a box with the same cross section as the waveguide.

If the waveguide cross section is not simply connected, as in the case of the coaxial cable, it can be shown that T M modes exist with zero transverse wavenumber. Such TM modes are special, because beyond having hz (ρ) ≡ 0, they have also ez (ρ) ≡ 0, so that they are called T EM modes.

This means that a coaxial cable has a T EM mode and an infinite number of T M and T E modes.

When one describes a coaxial cable as a “transmission line”, only the T EM mode is studied.

Indeed, it is always stated that “a coax can be modeled as a transmission line up to a certain maximum frequency that depends on the size of the cross section”. Indeed, for higher frequencies, higher order modes are no longer cut-off and start propagating (see next section).

0 00 The transverse wavenumbers kti , kti are real and positive and form an unbounded sequence.

80 It can be shown that the generating functions are orthogonal to each other. Defining the inner product of two functions by Z < f,g >= f (ρ)g ∗ (ρ)dΣ (5.8) Σ

the orthonormality conditions are < Φi (ρ),Φj (ρ) >= δij and < Ψi (ρ),Ψj (ρ) >= δij

om

where δij is the Kronecker delta. As a consequence, also the vector modal functions are orthonormal Z 0 0 0 0 < ei ,ej >=< ei ,hj × ẑ >= e0i × h0∗ j · ẑ dΣ = δij Σ

Z < e00i ,e00j >=< e00i ,h00j × ẑ >=

Σ

e00i × h00∗ j · ẑ dΣ = δij

Σ

e0i × h00∗ j · ẑ dΣ = 0

m

< e0i ,e00j >=< e0i ,h00j × ẑ >=

.c

Z

or

1 X 0 0∗ R (Vi Ii + Vi00 Ii00∗ ) 2 i

w

=

ld

fo

ru

A direct consequence of these orthonormality properties is that the total power carried by a number of modes is the sum of the individual powers carried by each mode: Z 1 P = R (Et × H∗t ) · ẑ dΣ 2 Σ  !  Z ÃX X X X 1  dΣ Vi0 e0i + Vi00 e00i ×  Ij0∗ h0∗ Ij00∗ h00∗ = R j + j 2 Σ i i j j

tu

Notice that this property is absolutely not trivial and does not depend on the linearity of Maxwell’s equations, since power depends quadratically on the fields. We can view this relation as a statement of a Pythagora’s theorem in a Hilbert space with an infinite number of dimensions.

w w

w .jn

In conclusion, propagation modes in a uniform waveguide are completely independent. If many of them are simultaneously present in a waveguide, each one has a completely autonomous evolution, also form the energy point of view. As soon as an obstacle or a discontinuity is introduced in the waveguide, such as a screw, a post, an iris, or a change of cross section, new boundary conditions must be satisfied on the obstacle and this causes mode coupling.

5.2

Equivalent transmission lines

In this section we focus on the evolution of modes along the waveguide: in other words, we study the modal voltages and currents. It was said above that each propagation mode is characterized 0 00 by a number of modal functions and a constant kti , kti called transverse wavenumber. This wavenumber determines the longitudinal evolution of the mode, since it can be shown that the longitudinal propagation constant kzi is given by q 2 (5.9) kzi = k 2 − kti

81 for both T M and T E modes, where k 2 = ω 2 εµ is the wave number in the medium that fills the waveguide. Moreover, the modal impedances can be shown to be 0 Zti =

kzi ωε

for T M modes

00 Zti =

ωµ kzi

for T E modes

om

Depending on the frequency of operation, the evolution of modes can be totally different. Consider a specific mode i. It is clear from (5.9) that at very high frequency kzi is real, whereas a very low frequencies it is imaginary. In particular, we define the critical frequency (or cut-off frequency) of each mode kti (5.10) fci = √ 2π εµ

m

• if f ≥ fci the mode is above cut-off : kzi and Zti are real

.c

In terms of it, for both T M and T E modes

ru

• if f ≤ fci the mode is below cut-off : kzi and Zti are pure imaginary

ld

fo

0 00 At the cut-off frequency, kzi = 0 and Zti = 0 and Zti → ∞.

The mode with the smallest critical frequency is called fundamental mode.

Let fc0i be the critical frequency of mode i in an empty waveguide. As it is clear from the definition, the presence of the dielectric makes the critical frequency smaller:

w

or

fc0i fci = √ εr µr

z

w .jn

tu

0

w w

ĐŽĂdž

Figure 5.3.

Infinitely long waveguide with a source that excites a mode above cut-off

In order to clearly understand the characteristics of the two regimes, suppose that the waveguide is infinitely long, as shown in Fig. 5.3, so that only the forward wave is excited by the source: Vi (z) = Vi0+ e−jkzi z Ii (z) = Yti Vi0+ e−jkzi z for z ≥ 0

82 1. Suppose f > fci . Let us compute the time evolution of this wave. vi (z,t) = R{Vi (z)ejωt } = |Vi0+ | cos(ωt − kzi z + arg(Vi0+ )) ii (z,t) = R{Ii (z)ejωt } = Yti |Vi0+ | cos(ωt − kzi z + arg(Vi0+ )) Since kzi is real, the mode is propagating with phase velocity √ c/ εr µr ω ω s = vph = = s µ ¶2 µ ¶2 kzi fci kti 1− k 1− f k

om

where we have used (5.10) and the fact that

c ω ω =v = √ =√ εr µr ω εµ k

or

ld

fo

ru

m

.c

Note that this last quantity is the phase velocity v of a plane wave in the medium that fills the waveguide. Note also that if the waveguide is empty, εr = 1 and the phase velocity of the mode is greater than the speed of light c. However, the theory of relativity is not violated: indeed, only geometrical points (wave nodes or wave crests) move at the velocity vph and no mass or energy or information. See the analogous discussion concerning plane waves (2.12).

The previous formula can also be written in a slightly different for in terms of the critical frequency of the mode in an empty waveguide: √ c/ εr µr c vph = s ¶2 µ µ ¶2 = s fc0i fci εr µr − 1− f f

w w

w .jn

tu

w

We remark that the mode phase velocity depends on frequency, hence propagation in a waveguide is dispersive. When the field in the waveguide is not monochromatic, the various frequency components move at different velocity and the signal suffers distortions. If however the signal is narrow band, as in the case of an amplitude modulated pulse of much longer duration than the period of the carrier, the distortion manifests itself in the fact that the envelope appears to travel at a different velocity than the carrier: the former travels at the group velocity (vg ), the latter at the phase velocity (vph ). The group velocity can be shown to be given by 1 vg = d R{kzi } dω Carrying out the derivative, we find s s ¶2 µ µ ¶2 c fc0i fci c = 1− 1− vg = √ f εr µr f εr µr

Note that since the envelope does not changes shape, the difference in velocities is conventionally not considered a real distortion.

It can also be shown that the energy of the mode travels at the group velocity. We see that the group velocity is always smaller than c and this is very important for the requirements of the theory of relativity.

83 Finally, it is to be remarked that a simple relationship exists between phase and group velocity: c2 vg vph = εr µr

The wavelength on the equivalent transmission line is called guided wavelength and is defined as 2π λ λgi = =s µ ¶2 kzi fci 1− f

λ=

√ c/ εr µr λ0 2π =√ = εr µr f k

om

where

The modal impedance of T E modes is given by

ru

1−

λ0 ¶2 ¶2 = s µ fc0i fci εr µr − f f

m

λ µ

λgi = s

.c

is the plane wave wavelength in the dielectric that fills the waveguide and λ0 the corresponding one in empty space. Obviously λgi > λ. Also this formula can be written in terms of fc0i :

r

fo

µr εr Z0 µr ωµ ωµ 00 =s =s Zti = = s ¶2 ¶ ¶ µ µ µ 2 2 kzi fc0i fci kti εr µr − 1− k 1− f f k

It is useful to recall that

or

ld

Z0

r

w

Z=

µ = Z0 ε

r

µr εr

tu

is the plane wave impedance in the dielectric that fills the waveguide.

w w

w .jn

The modal impedance of T M modes is given by s µ ¶2 kti s s k 1− ¶2 µ µ ¶2 r k Z fc0i f µ k 0 ci r zi 0 = εr µr − 1− Zti = = Z0 = f εr f εr ωε ωε

Since the characteristic impedance Zti is real, the wave is characterized by the power flow P =

1 |Vi0+ |2 2 Zti

2. Suppose f < fci . In this case kzi = −j|kzi | is imaginary. The negative sign of the propagation constant is related, as always, to our time convention exp(jω0 t) for phasors. The time evolution of the modal voltage and current wave is vi (z,t) = R{Vi (z)ejωt } = |Vi0+ |e−|kzi z | cos(ωt + arg(Vi0+ )) ii (z,t) = R{Ii (z)ejωt } = |Yti ||Vi0+ ||e−|kzi z | cos(ωt + arg(Vi0+ ) ± 90◦ )

84

z

0

ĐŽĂdž

Figure 5.4.

Infinitely long waveguide with a source that excites a mode below cut-off

.c

om

Also the modal admittance is imaginary, but its sign depends on the polarization: negative for T E modes and positive for T M . Fig. 5.4 shows clearly that the voltage and current waves are evanescent. Since the phase of the wave does not depend on z, phase velocity, group velocity and guided wavelength are not defined. The decay constant, i.e. the length necessary for the amplitude to reach the level 1/e of the starting value, is 1 0.9

0.7

m

0.8

1/e

ru

0.6 0.5 0.4

fo

0.3 0.2 0.1

ld

4

6

8

10

12

14

Decay constant of a mode below cut-off

tu

w

Figure 5.5.

Li 2

or

0 0

Li =

1 |kzi |

w w

w .jn

and is shown in Fig. 5.5. We see also that voltage and current are in phase quadrature, hence no active power flow is associated to a purely forward evanescent wave. The situation is different when both forward and backward wave are present in the same region, as we will explain later. Nevertheless, an evanescent wave plays an important role in the energy budget, since it stores energy. Since the wave does not move, but remains attached to the source, it does not give rise to any power flow. The wave oscillates in time and exchanges its energy with the source twice per period.

If we look at the equations derived above, we see that the various modal parameters have two different frequency behaviors. In particular s µ ¶2 0 Zti vg fci kzi = 1− = = f Z v k

and

1 Z 00 vph λg = ti = s = µ ¶2 Z v λ fci 1− f

85 The symbols λ, v, Z, denote the values of wavelength, phase velocity and wave impedance in the dielectric that fills the waveguide. The two behaviors are plotted in Fig. 5.6 It is evident that the 2

1.5 λg/λ=vph/v=Z’’t/Z 1

om

kz/k=vg/v=Z’t/Z

0 0

2

3

f/fc

4

5

m

1

.c

0.5

ru

Figure 5.6. Dispersion curves of various parameters of a mode above cut-off. The two vertical dash– dotted lines indicate the standard operating band of a rectangular waveguide in the fundamental mode. Note that the first higher order mode goes above cut-off at f /fc = 2.

or

ld

fo

waveguide is very dispersive close to cut-off. However, it is not possible to use it in the frequency band where the curves are almost flat, since the guide is not single mode there. Actually, assuming that the curves refer to the fundamental mode of a rectangular waveguide, the vertical lines are the limits of the standard operating range. Indeed, the first higher order mode goes above cut-off at f /fc = 2.

w .jn

tu

w

In general a source excites all the modes of a waveguide, the amount of excitation depending on the geometry of the source. At a given frequency, a finite number of excited modes are above cut-off and, by carrying energy away from the source, is responsible for the radiation phenomenon.

The value of the radiation resistance of the dipole is related to them. At the same time, an infinite number of excited modes are below cut-off: their effect is important only in the neighborhood of the source, where they describe the reactive field. Moreover, evanescent modes are responsible for the imaginary part of the input impedance of the dipole.

w w

A waveguide is said to be single mode if at the operation frequency only the fundamental mode is above cut-off. At microwave frequency, waveguides are essentially always single mode, in order to avoid interference effects between the various modes above cut-off.

The sources present in the waveguide can be represented in circuit terms by means of voltage and current generators to be inserted on the modal equivalent lines, as shown in Fig. 5.7. If the source is described by the distributions J(r), M(r), it can be shown that the corresponding generators are vi (z) =< Mt ,hi > + < Jz ,Zti ezi > ii (z) =< Jt ,ei > + < Mz ,Yti hzi >

(5.11)

The inner products, defined in (5.8), are integrals over the waveguide cross section, hence, the generators are distributed on the line since their strength is a function of z.

86

5.3

Rectangular waveguide

The simplest waveguide to analyze is the one with rectangular cross section, see Fig. 5.8 The cross section is simply connected, hence the modes of this waveguide are T M and T E but not T EM . As for T M modes, the solution of (5.6), obtained by the classical method of separation of variables, is

³ mπ ´ ³ nπ ´ 2 sin Φmn (x,y) = √ sin b a ab r³ ´ ´ ³ nπ 2 mπ 2 0 + ktmn = b a

m, n = 1,2,3, . . .

(5.12)

fo

ru

m

.c

om

This is the generating function of T M modes and is proportional to the longitudinal component of the electric field ez . We see that the index i used in the previous section to label the mode functions is actually a double index i ↔ (m, n). Fig. 5.9 and Fig. 5.10 show the plots of two generating functions that help in understanding the meaning of the integer labels: m is the number of hills and valleys along the x direction, whereas n refers to the y direction. The contour lines plotted on the bottom plane serve two purposes: first, they are constant height lines and help in the comprehension of the 3D plot, second they are the field lines of the transverse magnetic field hmn .

Indeed, according to the equations of Section 5.1, e0 i (ρ) is proportional to the transverse gradient of Φ(ρ), hence it is orthogonal to the contours. However, h0 i (ρ) is orthogonal to e0 i (ρ), hence it is tangent to the contour lines. Note that the mode is T M , hence h0zi (ρ) ≡ 0 and h0 i (ρ) is the total magnetic field. It is no surprise that the contour lines are closed: this is a consequence of the fact that the magnetic field is solenoidal. The field lines of e0 i (ρ) are not drawn, but can be easily imagined as the orthogonal trajectories to the contour lines shown. See, however, Fig. 5.13.

but not m = n = 0

(5.13)

tu

w

or

ld

Concerning T E modes, the solution of (5.7) is r ³ nπ ´ ³ mπ ´ ²m ²n m, n = 0,1,2, . . . cos cos Ψmn (x,y) = b a ab r³ mπ ´2 ³ nπ ´2 00 + ktmn = b a

w .jn

where ²m is the Neumann symbol :

²0 = 1 ²m = 2 for m 6= 0

w w

Fig. 5.11 and Fig. 5.12 show the plots of the generating functions of two T E modes. Again, we see clearly that the indices m, n count the hills and valleys in the x and y directions, respectively.

vi ( z )

н

k zi ii ( z ) Z ti

Figure 5.7. Distributed generators on the modal transmission line corresponding to mode i, characterized by propagation constant kzi and characteristic impedance Zti .

87

y b

a

z

Rectangular waveguide with sides a, b, a ≥ b.

.c

om

Figure 5.8.

x

m

1

ru

0.5

fo

0

−1 1

0

2 x

0

Rectangular waveguide (a = 3, b = 1): generating function Φ11 (x,y) of mode T M11

w .jn

Figure 5.9.

1

3

tu

w

0.5 y

or

ld

−0.5

w w

When one of the labels is zero, the function is constant in the corresponding direction, as in the case of Fig. 5.11, referring to the fundamental mode. The contour lines are the field lines of e00 i (ρ).

Indeed, h00 i (ρ) is proportional to the gradient of Ψ(ρ) and is then orthogonal to the level lines.

Moreover, e00 i (ρ) is orthogonal to h00 i (ρ), hence tangent to the contour lines. The contours are closed curves because the transverse electric field (that coincides with the total one, since the mode is T E) is solenoidal: indeed no charges are present since modes are solutions in absence of sources.

The field lines of h00 i (ρ) are not shown here, but it is easy to imagine them, orthogonal to the level lines, see also Fig. 5.14.

The mode functions ei (ρ) and hi (ρ) can be obtained by computing the derivatives of the generating functions, according to the general formulas of the previous section. We limit ourselves here to write the mode functions of the fundamental mode T E10 :

88

1 0.5 0

om

−0.5 −1 −1.5 1y 0.5

x

m

1

ru

0 0

Rectangular waveguide (a = 3, b = 1): generating function Φ32 (x,y) of mode T M32

or

ld

fo

Figure 5.10.

2

.c

3

w

1

tu

0.5

w .jn

0

−0.5

3

w w

−1 1

Figure 5.11.

y

2 x

0.5

1 0 0

Rectangular waveguide (a = 3, b = 1): generating function Ψ10 (x,y) of mode T E10

r

³ πx ´ 2 sin ŷ =− a ab r ³ πx ´ 2 sin x̂ h00 10 (x,y) = a ab r ³ πx ´ 2 −j π cos hz10 (x,y) = a kz10 a ab e

00

10 (x,y)

(5.14)

89

1 0.5 0

om

−0.5 −1 −1.5 1 y0.5

x

m

1

ru

0 0

Rectangular waveguide (a = 3, b = 1): generating function Ψ32 (x,y) of mode T E32

fo

Figure 5.12.

2

.c

3

w

or

ld

Notice that neither the mode functions nor the transverse wavenumbers depend on the dielectric that fills the waveguide(provided the dielectric is homogeneous). Fig. 5.13 shows the field lines of the first T M modes. Note that the in the cross section the electric field lines are orthogonal to the magnetic ones. Moreover, on the lateral surface, the magnetic field lines are orthogonal to the field lines of the induced current, according to (1.19). Fig. 5.14 shows, in a similar manner, plots of the field lines of the first T E modes.

w w

w .jn

tu

Concerning the transverse wavenumbers, we see that they are given by the same equation for T M and T E modes: the difference is just in the allowed values of the indices. There is a useful geometrical representation of transverse wavenumbers that allows to determine graphically the modes that are below or above cut-off. Looking at the defining equations, one has the idea to arrange the points representing modes in a regular lattice as shown in Fig. 5.15, which is based on the use of Pythagora’s theorem. The distance of the representative point of a given mode from the origin is just its transverse wavenumber. The plane in which the points are drawn is a spectral plane of spatial frequencies, i.e. wavenumbers, ξ conjugate to x and η conjugate to y. Now, suppose that we want to find which modes are above cut-off at a given frequency. It is √ enough to draw a circle with center in the origin and radius equal to k = ω εµ: the modes whose representative points are inside the circle have real kzi and then are above cutoff.

As explained in the previous section, below cut-off modes are characterized by evanescent fields that remain attached to their sources and do not provide active power transfer (excluding tunnel effects to be discussed later). Their importance is lower and lower as the magnitude of their imaginary propagation constant kzi increases. Then it is useful to arrange modes in order of increasing kti . Fig. 5.15 shows clearly that the points closest to the origin are those corresponding to • T E10 , T E20 , T E01 if b < a/2

w w

w .jn

tu

w

or

ld

fo

ru

m

.c

om

90

Figure 5.13. Field lines of the first T M mode functions. The symbol Emn is equivalent to T Mmn .

1. is the cross-sectional view, 2. the longitudinal view and 3. the (opened) surface view. The critical wavelengths are also indicated

â&#x20AC;˘ T E10 , T E01 , T E20 if b > a/2 Only T E modes need be considered since T M points are not on the coordinate axes. The first mode

w w

w .jn

tu

w

or

ld

fo

ru

m

.c

om

91

Figure 5.14. Field lines of the first T E mode functions. The symbol Hmn is equivalent to T Emn .

1. is the cross-sectional view, 2. the longitudinal view and 3. the (opened) surface view. The critical wavelengths are also indicated

is T E10 that is then the fundamental mode of the rectangular waveguide. Different waveguides have different fundamental mode: T E11 for the circular waveguide and T EM for the coaxial cable.

The first higher order mode is different according to the aspect ratio of the cross section. The

92 critical frequencies of these modes in an empty waveguide, computed by (5.10) are • fc10 =

c c c , fc20 = , fc01 = 2b a 2a

It is clear that the bandwidth over which the guide is single mode is maximum if the mode sequence is T E10 , T E20 , T E01 ; in this case it equals an octave. If b > a/2 the single mode bandwidth reduces, down to zero in the case of square waveguides: in this extreme case, the T E10 and T E01 modes are degenerate, i.e. they have the same critical frequency. Mode degeneracy is a very common phenomenon in rectangular waveguides, due to their high degree of symmetry.

Design of a single mode rectangular waveguide

ru

5.3.1

m

.c

om

It was said before that waveguides for microwave applications are almost invariably single mode.

If, on the contrary, the sides are very large with respect to λ, counting the modes above cut-off can be not straightforward. However, it is not difficult to estimate this number Nabove . Indeed, the area of the quarter of circle with radius k is πk 2 /4; the density of points is 2 (T E and T M ) for each rectangle with sides π/a × π/b, ignoring that T M modes have labels different from zero.

Then (ka)(kb) πk 2 π 2 ab πk 2 ab = 2π = =2 / Nabove = 2 2 λλ 2π 4 π 4 ab

w

×

η

ו

ו

ו

ו

ו

ו ( 3,1)

ו

ו

w .jn

tu

2π b

or

ld

fo

A typical design problem is that of finding the dimensions of a rectangular waveguide so that it is single mode over the bandwidth [fmin ,fmax ], with fmax ≤ 2fmin , of course. The specifications are generally completed by asking that the first higher order mode has an attenuation greater than a minimum value αlim over the complete operative band and that the waveguide can carry a maximum power Plim without dielectric breakdown.

The operating conditions are indicated in Fig. 5.16:

π

w w

b

×

kt 31

k (1,0 )×

π

a

×

×

×

2π a

3π a

4π a

ξ

Figure 5.15. Mode scheme of the rectangular waveguide. The transverse wavenumber of modes T M32 and T E32 is displayed. At the indicate frequency, only the T E10 mode (fundamental mode) is above cut-off

93

f

0

f min

f c10

f max

f c 20

Figure 5.16. Operating conditions for a single mode waveguide. The grey region between fmin and fmax is the required single-mode band of operation. The square parentheses denote the available single-mode band of the waveguide. If a is modified, the position of the parentheses changes, with the constant ratio fc20 = 2fc10

c 2a c = a

fmax ≤ fc20

2fmin

≤a≤

c

fmax

(5.15)

m

c

.c

from which we find the range of values of a that satisfy the specification

om

fmin ≥ fc10 =

It is minimum at fmax

tu

w

or

ld

fo

ru

If a is chosen close to amin , the single-mode band of the waveguide, denoted by the square parentheses in the figure, shifts to the right. In these conditions, the mode T E20 has the maximum attenuation over [fmin ,fmax ], but in the first part of this band the propagation is highly dispersive, as Fig. 5.6 clearly indicates.

If, on the contrary, a is chosen close to amax , the single-mode band of the waveguide shifts to the left. In these conditions, the dispersion effects are minimized, but the attenuation of mode T E20 is insufficient, in particular for frequencies close to fmax . Actually, there is a specification concerning this point. Compute the attenuation of mode T E20 : sµ ¶2 q fc20 2 2 −1 α20 = kt20 − k = k f

2πfmax = c

w .jn

α20min

sµ

fc20 fmax

¶2 −1

w w

By requiring α20min ≥ αlim , we find a constraint on a (contained in fc20 )

sµ

a≤

fmax

c

cαlim 2πfmax

(5.16)

¶2 +1

The third specification, that dispersion effects are minimized, forces the ≤ sign to be substituted by an equal in the previous equation. Assuming that the problem proposed has a solution, i.e. that this value lies in the range (5.15), the width of the waveguide has been found.

The fourth specification, concerning the power at the breakdown limit, affects only the height b.

The active power flow on the T E10 modal line is P =

1 |Vmax |2 1 1 |V + |2 = 2 Zt10 S 2 Zt10

94 assuming that the modal line is mismatched, the VSWR has the value S, and Vmax is the maximum voltage along the line.

By recalling (5.14), the maximum value of the electric field is reached at x = a/2, so that r 2 Vmax Emax = ab

Substituting in the previous expression, we get P =

2 1 1 ab Emax 2 2 Zt10 S

(5.17)

m

.c

om

This is the power in the waveguide for a given value of Emax . Obviously, S = 1 if the line is matched. If we solve the previous equation with respect to Emax for a given power Plim , we see that the maximum field is proportional to the modal impedance. Hence, recalling the plot of Fig. 5.6, we conclude that the condition Emax ≤ R (R denoting the dielectric rigidity) has to be enforced at f = fmin , where the impedance is maximum. The value of a to be used in the previous equation is that found by (5.16). Moreover, it is to be remarked that the condition on the power at the breakdown limit must be consistent with b ≤ a/2, on which the determination of a is based.

fo

ru

Alternatively, we can compute the power at the discharge limit Pdisch by substituting in (5.17) Emax with the dielectric rigidity of air R = 20 kV/cm. Then Pdisch must be enforced to be greater than the desired power Plim . Pdisch depends on frequency through Zt10 : the minimum value of Pdisch is that at f = fmin , where Zt10 is maximum. Hence the condition to be guaranteed is

ld

Pdisch (fmin ) ≥ Plim

w

or

In order to make an example with numerical values, we can consider the WR90 standard rectangular waveguide for X band, whose data are

tu

• a = 0.9 inches=2.286 cm; b = 0.4 inches=1.016 cm • fmin = 8.0 GHz, fmax = 12.4 GHz

w .jn

• αlim = 7.8 dB/cm

• Plim = 201.6 kW, in matched conditions, S = 1.

w w

We can add that the attenuation of the T E20 mode is α = 18.9 dB/cm at f = fmin .

5.3.2

Tunneling effects

It was shown in section 5.2 that the modes below cut-off do not carry active power. Actually, this is not true if in a waveguide region both a forward and a backward below cut-off mode exist.

This phenomenon is called tunneling because of the analogy with the tunnel effect in quantum mechanics.

Let us consider a rectangular waveguide filled completely with dielectric apart from the region √ AB, as shown in Fig.5.17(a). Let fc0 and fcd = fc0 / εr be the critical frequencies of the T E10

95

εr

TE10

−

( b)

A+

−

z

B+

Z∞0 kz 0

Z ∞d k zd −

A+

Z ∞d k zd

−

B+

om

εr

.c

(a)

ru

m

Figure 5.17. Tunnel effect in an inhomogeneously filled rectangular waveguide. The mode T E10 is above cut-off in the dielectric and below cut-off in the air region. (a) Actual structure. (b) Modal equivalent transmission line circuit

ld

fo

mode in the empty and in the filled waveguide, respectively. Assume that the the frequency f of the incident mode is fcd < f < fc0

w .jn

tu

w

or

so that the mode is above cut-off in the filled waveguide and below in the empty one. The problem is to compute the field everywhere. First of all, the point to address what happens at the interface in A. Two are the alternatives: either the interface gives rise to mode coupling or not. The conditions to be enforced, according to (1.17), are the continuity of the tangential components of electric and magnetic field across the interface. Since the interface is a plane perpendicular to ẑ, the tangential components coincide with the transverse ones. Hence, the condition that must hold is Et (ρ,zA− ) = Et (ρ,zA+ ) Ht (ρ,zA− ) = Ht (ρ,zA+ )

w w

The field in a waveguide is a superposition of modes of propagation of the type (5.3), then X

Vi (zA− )ei (ρ) =

i

X i

X

Vi (zA+ )ei (ρ)

i

Ii (zA− )hi (ρ) =

X

Ii (zA+ )hi (ρ)

i

Eqs.(5.12), (5.13) show that the mode functions do not depend on the material that possibly fills the waveguide, then the previous equations lead to Vi (zA− ) = Vi (zA+ ) Ii (zA− ) = Ii (zA+ ) ∀i

96

kz0 = −jα

Z∞0 = jX∞0

m

Notice that the mode is below cut-off in the empty waveguide, hence

.c

om

This equation shows clearly that the dielectric interface does not produce mode coupling: each mode can be analyzed independently from the others. Notice that if the interface is not planar or not orthogonal ẑ, it produces mode coupling. This means that even if only the T E10 mode is incident on it, all the other modes (in general) are excited, with the right amplitudes, so that the appropriate continuity conditions are satisfied. In this case, since only the T E10 mode is incident, no other mode is generated and the modal equivalent circuit is the one shown in the lower part of Fig. 5.17.

The parameters of the modal lines are s s ¶2 ¶2 µ µ fc0 fc0 kzd = k0 εr − kz0 = k0 1 − f f Z0 Z0 Z∞d = s Z∞0 = s ¶2 ¶2 µ µ fc0 fc0 εr − 1− f f

fo

ru

with α ≥ 0, as required by the usual convention, and X∞0 ≥ 0 because Zt00 = ωµ/kz . If the incident electric field is known, we can compute the incident voltage in A, VA+− . Next we compute the reflection coefficient in A− . Start with

Z∞d ZB + = −j X∞0 Z∞0 z − −1 = B with |ΓB − | = 1 zB − + 1 = ΓB − exp(−2jkz0 LAB ) = ΓB − exp(−2αLAB )

w

Γ A+

or

ΓB −

ld

zB − =

w .jn

tu

The trajectory on the Smith chart of the reflection coefficient when the observation point moves on a transmission line with imaginary propagation constant is a radial line instead of a circumference.

Hence ΓA+ < 1. Next 1 + Γ A+ = rA+ − jxA+ zA+ = 1 − ΓA+

with rA+ > 0 and xA+ > 0 strictly. Compute zA−

w w

zA− = zA+

X∞0 X∞0 X∞0 Z∞0 + jrA+ = xA+ = (rA+ − jxA+ )j Z∞d Z∞d Z∞d Z∞d

Note that the real part of zA− (as well as the imaginary part) is strictly positive, hence the point ΓA− is strictly inside the unit circle, ΓA− < 1. As a consequence, the active power flowing beyond the point A is ¢ 1 |VA+− |2 ¡ 1 − |ΓA− |2 > 0 PA− = 2 Z∞d

Now, this means that there is an active power flow through the part AB of the waveguide, although the T E10 mode is below cut-off. The paradox is easily solved if one takes into account that in the part AB both the forward and the backward waves exist (because of the discontinuity in B). The

97 formula to be used for the computation of the active power on a line with imaginary characteristic impedance Z∞0 = jX∞0 can be shown to be P (z) =

1 1 Im{V + (z)V −∗ (z)} R{V (z)I ∗ (z)} = X∞0 2

(5.18)

om

We see that although a single evanescent wave does not give rise to any active power flow, the presence of a forward and a backward one does: one can say that the two evanescent waves “cooperate” in order to produce the power flow.

It is interesting to examine what happens when the thickness LAB → ∞, so that the structure contains only the interface in A. From the previous equations we see easily that, in this limit, it is ΓA+ = 0, so that rA+ = 1, xA+ = 0 and rA− = 0 with the consequence that |ΓA− | = 1 and PA− = 0. This agrees with (5.18): if the interface in B is absent, the backward wave is identically zero.

0.4

.c

We can see numerical data about this phenomenon in Fig. 5.18, which shows an explanation of the tunnel effect in terms of phasor diagrams. The waveguide is the standard WR90 guide,

m

V+

0.3

V

ru

0.2

I−

0

ld

−0.1

V−

or

−0.2

w

−0.3

tu

−0.4

−0.5

fo

0.1

0

0.2

I 0.4

0.6

0.8

w .jn

−0.2

I+

w w

Figure 5.18. Phasor diagram illustration of tunnel effect at the frequency f = 5 GHz. The vectors represent, in the complex plane, the phasors of voltage and current in a point halfway between A and B. Note that the phase difference between V and I is less than π/2, implying an active power flow.

with a = 0.9 in = 2.286 cm and b = 0.4 in = 1.016 cm. The dielectric has a relative permittivity εr = 4, the thickness of the empty guide is LAB = 2 cm. The critical frequencies of the T E10 mode are fc0 = 6.56 GHz and fcd = 3.28 GHz and the frequency of the incident wave is f = 5 GHz, so that the mode is below cut-off in the empty guide. Voltages and currents are evaluated in a point M halfway between A and B. We see that the forward voltage is in phase quadrature with the forward current, consistently with the fact that the characteristic impedance is positive imaginary: indeed, the voltage is leading. Also, the backward voltage is in phase quadrature with the backward current, but the current leads now. As a result, the total voltage and current are not in phase quadrature, implying an active power flow. Since the phase difference between V and I is less than π/2, the power flow is positive. Obviously, this phase difference is the phase of the

98

0.2

V+

0.15 0.1 0.05

I−

0 −0.05

V

−0.1 −0.15

I+

−0.25

I

V−

−0.1 −0.05

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

.c

−0.3

om

−0.2

ru

m

Figure 5.19. Phasor diagram illustration of tunnel effect at the frequency f = 5 GHz. The vectors represent, in the complex plane, the phasors of voltage and current in B − .

Note that VB − and IB − are in phase.

ld

fo

− + local impedance in M and the phase difference between VM and VM is the phase of the reflection coefficient ΓM .

Fig. 5.19 is a similar plot that refers to the point B − . We notice in this case that the total V and

or

S11

w

1

w .jn

0.6

tu

0.8

w w

0.4

0.2

0 2

fcd 4

6 fc0 8 10 frequency (GHz)

12

14

16

Figure 5.20. Tunnel effect in an inhomogeneously filled rectangular waveguide. The vertical dashed lines denote the critical frequencies in the empty and in the filled waveguide. Plot of the magnitude of the scattering parameter S11 (f ) vs. frequency

I are even in phase. The reason is that they coincide with VB + and IB + , which are certainly in

99 S21 1

0.8

0.6

0.2

fcd 4

6 fc0 8 10 frequency (GHz)

12

14

16

.c

0 2

om

0.4

ru

m

Figure 5.21. Tunnel effect in an inhomogeneously filled rectangular waveguide. The vertical dashed lines denote the critical frequencies in the empty and in the filled waveguide. Plot of the magnitude of the scattering parameter S21 (f ) vs. frequency

fo

phase because the wave there is purely forward and above cut-off. See also Fig. 5.22 and Fig. 5.23.

or

ld

The frequency response of the structure is illustrated in Fig. 5.20, and Fig. 5.21, which show plots of the scattering parameters S11 and S21 versus frequency at the reference planes A− and B + . We note first of all that the two curves are complementary: since the structure is lossless, the scattering matrix is unitary, so that

w

|S11 (f )|2 + |S21 (f )|2 = 1

w w

w .jn

tu

As long as fcd < f < fc0 , the T E10 mode is below cut-off in the region AB and the reflection coefficient is large but not of unit magnitude. Correspondingly, the transmission coefficient S21 is not zero, indicating that a power flow across the region AB exists. Next, as the frequency is increased, the T E10 mode is everywhere above cut-off. The transmission coefficient increases steadily up to unit magnitude at a frequency of about 10 GHz, while the reflection coefficient is zero at the same frequency. We can check that at this frequency the thickness LAB is exactly half of the guided wavelength. This means that ΓA+ = ΓB − , a complete turn on the Smith chart, and ZA− = ZB + . The consequence is obviously a zero reflection coefficient.

When the frequency is still increased, the reflection coefficient increases, again reaching a maximum at the frequency for which LAB = (3/4)λg0 . This behavior continues forever, with S11 = 0 at the frequencies for which LAB = nλg0 /2 for any integer n and local maxima at the frequencies for which LAB = (2n + 1)λg0 /4. The local maxima are different but tend to have the same height for large frequency. This is due to the fact that the modal impedances are functions of frequency but for large frequency, f À fc10 Z0 Z∞0 → Z0 Z∞0 → √ εr

It is useful to examine the plots of the T E10 modal voltage and current. Fig. 5.22 and Fig. 5.23 refer to a frequency f = 5 GHz, at which the mode is below cut-off in the empty waveguide.

100

om

Both voltage and current are evanescent between A and B: even if it is not so evident, note that the curve is a combination of a decaying (the forward wave) and a growing (the backward wave) exponential. The VSWR to the left of A is very large, because of the high reflection coefficient.

This causes the small minima. The plot is flat to the right of B because the transmission line is matched and only the forward wave exists there.

If the frequency increases beyond fc0 , the T E10 mode goes above cut-off also in the empty region.

Fig. 5.24 and Fig. 5.25 show the voltage and current plots for f = 9 GHz. The reduction in the value of S11 shows up in the reduction of the VSWR to the left of A and in the increase of the transmitted wave to the right of B. It is also evident that the mode is above cut-off in the AB region since the curve is oscillatory there.

Fig. 5.26 and Fig. 5.27 show the voltage and current plots for f = fr = 9.965 GHz, the frequency for which the length AB is exactly half of the guided wavelength λg0 . The reflection coefficient in A is zero, so that the VSWR is zero to the left of A and the plot is flat. Moreover, the magnitude of the transmission coefficient is one. The empty waveguide behaves as a resonator with resonance frequency fr , i.e. as band-pass filter, as it is evident from the |S21 | plot in Fig.5.21

fo

ru

|V(z)| 2

m

.c

The phenomenon that we have described is called tunnel effect because it is similar (and is described by the same mathematics) to the one of quantum mechanics concerning the transmission of electrons through potential barriers.

or

ld

1.5

tu

w .jn

0.5

w

1

0 −0.04

−0.02

0 0.02 A z (m) B

0.04

0.06

w w

Figure 5.22. Tunnel effect in an inhomogeneously filled rectangular waveguide. The vertical dashed lines denote the interfaces in A and B between the empty and the filled waveguide. Plot of the magnitude of the T E10 modal voltage vs. z at the frequency f = 5 GHz.

5.3.3

Irises and waveguide discontinuities

Modes in a perfect, uniform waveguide are independent: each of them can exist alone in the waveguide since it satisfies Maxwell’s equations and all boundary conditions. If an obstacle is introduced, or a change in the cross section, new boundary conditions arise, which were not taken into account in the definition of the modes. In general then, all waveguide modes are excited

101 Z∞ d|I(z)| 2

1.5

1

−0.02

0 0.02 A z (m) B

0.04

0.06

.c

0 −0.04

om

0.5

fo

ru

m

Figure 5.23. Tunnel effect in an inhomogeneously filled rectangular waveguide. The vertical dashed lines denote the interfaces in A and B between the empty and the filled waveguide. Plot of the magnitude of the T E10 modal current vs. z at the frequency f = 5 GHz. The current is multiplied by the modal impedance for viewing convenience |V(z)|

ld

2.5

or

2

tu

w .jn

1

w

1.5

w w

0.5

0 −0.02

−0.01

0 A

0.01 z (m)

0.02 B

0.03

0.04

Figure 5.24. Tunnel effect in an inhomogeneously filled rectangular waveguide. The vertical dashed lines denote the interfaces in A and B between the empty and the filled waveguide. Plot of the magnitude of the T E10 modal voltage vs. z at the frequency f = 9 GHz

by the discontinuity with suitable coefficients, such that the total electric and magnetic field, consisting of incident and scattered waves for all the modes, satisfy the new boundary conditions.

As a example, consider an iris in a rectangular waveguide, as shown in Fig. 5.28. Irises are metal windows, perpendicular to the guide axis. The thickness of irises is often small and can be neglected

102 Z∞ d|I(z)| 2.5

2

1.5

0.5

−0.01

0 A

0.01 z (m)

0.02 B

0.03

0.04

.c

0 −0.02

om

1

fo

ru

m

Figure 5.25. Tunnel effect in an inhomogeneously filled rectangular waveguide. The vertical dashed lines denote the interfaces in A and B between the empty and the filled waveguide. Plot of the magnitude of the T E10 modal current vs. z at the frequency f = 9 GHz. The current is multiplied by the modal impedance for viewing convenience |V(z)|

ld

3

or

2.5

w

2

w .jn

1

tu

1.5

w w

0.5

0 −0.02

−0.01

0 A

0.01 z (m)

0.02 B

0.03

0.04

Figure 5.26. Tunnel effect in an inhomogeneously filled rectangular waveguide. The vertical dashed lines denote the interfaces in A and B between the empty and the filled waveguide. Plot of the magnitude of the T E10 modal voltage vs. z at the frequency for which S11 = 0.

as a first approximation. Suppose that the fundamental mode T E10 is incident from the left. The boundary conditions to be satisfied on the iris are that the tangential electric field is zero on the metal parts of the iris and different from zero on the aperture. In this case the tangential field is the transverse field. If we recall the electric field distribution of T E10 , shown in Fig. 5.14, we

103 Z∞ d|I(z)| 2.5

2

1.5

0.5

−0.01

0 A

0.01 z (m)

0.02 B

0.03

0.04

.c

0 −0.02

om

1

fo

ru

m

Figure 5.27. Tunnel effect in an inhomogeneously filled rectangular waveguide. The vertical dashed lines denote the interfaces in A and B between the empty and the filled waveguide. Plot of the magnitude of the T E10 modal current vs. z at the frequency for which S11 = 0. The current is multiplied by the modal impedance for viewing convenience

jB

TE10 ŵŽĚĂů ůŝŶĞ

tu

kz 0 , Z∞ 0

z

w

A

or

ld

y

x

w .jn

A

w w

Figure 5.28. Example of an iris in a rectangular waveguide. Upper left: longitudinal view. Upper right: cross section view. Lower left: equivalent circuit for the fundamental mode (B > 0, capacitive iris).

realize that it is impossible to satisfy these boundary conditions with just the incident reflected and transmitted waves of this mode. Indeed, if the total Ey field must be zero on the the metal, certainly Eyr = −Eyi , that is the reflection coefficient must be Γ = −1. But the modal fields do not depend on y (the second subscript is n = 0), hence the total Ey field is zero also on the aperture and the transmitted field should be zero. This would imply that the iris behaves as a short circuit plate and this is contrary to experimental evidence: indeed, although the reflection coefficient can be high when the aperture is small, power is always transmitted through the hole. In conclusion, the behavior is more complicated and an infinite number of higher order modes are excited in order to satisfy the boundary conditions. In this way the reflected wave of T E10 need not be opposite to the incident wave and the reflection coefficient is smaller than one.

104

yA− = jb + 1

fo

−jb yA− − 1 = 2 + jb yA− + 1

2 −jb = 2 + jb 2 + jb

ld

S11 = ΓA− = −

ru

YA− = jB + Y∞0

m

.c

om

If the iris thickness is negligible, its fundamental mode equivalent circuit is an admittance connected in parallel on the T E10 modal line. Indeed, the transverse electric field must be continuous on the complete cross section: equal to zero on the metal and different from zero (but continuous) on the aperture. Hence the modal voltage is continuous, V (A− ) = V (A+ ), and the equivalent circuit is a load in parallel. Note that this load takes into account only the effect the iris has on higher order modes. Indeed, this load is connected on the T E10 modal line and the iris effects on the propagation of this mode are taken into account directly by transmission line theory. Generally the waveguide is operated in single mode conditions, so that all higher order modes are below cut-off.

As well known, they do not contribute active power propagation but store electromagnetic energy.

The admittance that represents the iris on the T E10 modal line must describe this behavior, hence it is a pure susceptance jB. As an example of this procedure, we compute the scattering matrix of the iris for the fundamental mode, assuming that the susceptance B is known. Assume that the reference planes are in A− and in A+ and the reference impedances at both ports coincide with the modal impedance. Let b = BZ∞0 be the normalized susceptance (not to be confused with the small side of the waveguide!!). To compute S11 and S21 , close port 2 on the reference impedance and compute reflection and transmission coefficients. They are

or

S21 = 1 + ΓA− = 1 +

w w

w .jn

tu

w

Then S12 = S21 because of reciprocity and S22 = S11 because of the symmetry of the structure.

The problem is the determination of the susceptance, which is a difficult one and requires the solution of an integral equation. Fig. 5.29 shows plots of the normalized susceptance of a symmetrical iris. It can be shown that when the edges are perpendicular to the electric field lines, B > 0 and the iris is said to be capacitive. Two sets of curves are shown for better reading accuracy: the right set uses the vertical axis on the left and is to be used for large apertures (d/b ' 1), the left one uses the axis on the right that measures essentially the inverse of the normalized susceptance.

Note that when the aperture is small, it resembles a short circuit, so the susceptance is large. If the edges of the iris are parallel to the electric field lines, the iris is said to be inductive, because its susceptance is negative, B < 0. Fig. 5.30 shows the normalized reactance of a symmetrical inductive iris. Again, if the aperture is small, the reactance is small, i.e. the susceptance is large as that of a short circuit.

It is interesting to compare the susceptance values of a capacitive and an inductive iris with the same aperture: the former is much smaller than the latter. As a numerical example, consider a WR90 (a = 2.286 cm, b = 1.016 cm) waveguide at f = 12 GHz. An inductive iris with an aperture of d = 0.3 cm has a normalized susceptance B/Y∞ ' −32, whereas a capacitive iris with the same aperture has a susceptance B/Y∞ ' 1.16.

Alternatively, consider a capacitive iris with a very small aperture, d = 0.5 mm, which has a susceptance B/Y∞ = 4.3. An inductive iris with B/Y∞ = −4.3 has a much wider aperture, d = 0.68 cm.

Such a susceptance produces |S11 | = 0.9067

|S21 | = 0.4217

105

om

Ä&#x161;

w w

w .jn

tu

w

or

ld

fo

ru

m

.c

Ä?

Figure 5.29. Normalized susceptance of a symmetrical capacitive iris. wavelength and Y0 is the modal admittance.

Îťg is the guided

A possible explanation of this phenomenon is related to the induced currents on the iris plates, which are parallel to the electric field since they are perpendicular to H (see (1.19)), which in turn is perpendicular to E. These induced currents radiate the scattered field but are disturbed

Ă

w w

w .jn

tu

w

Ě

or

ld

fo

ru

m

.c

om

106

Figure 5.30. Normalized reactance of a symmetrical inductive iris. λg is the guided wavelength and Z0 is the modal impedance. The curves are parametrized by a/λ where λ = c/f is the free space wavelength.

(interrupted) by the aperture in the case of capacitive iris so that their effect is small. In the case of inductive iris they are not interrupted and are then very efficient.

107

w w

w .jn

tu

w

or

ld

fo

ru

m

.c

om

Irises are reactive circuit elements and can be used to build matching devices: in practice, they substitute stubs in waveguide technology. Moreover they can be used to build filters. A simple example consists of two irises separated by a distance of about Îťg /2. They form a cavity, coupled to the waveguide by means of the apertures: For the center frequency the structure is transparent, otherwise it is highly reflecting, so that it behaves as a bandpass filter. Cavities are the equivalent of LC resonators in waveguide technology.

Appendix A

.c

Coordinate systems and algebra of vector fields

m

A.1

om

Mathematical Basics

w

or

ld

fo

ru

The position of a point in space is specified by means of a coordinate system, that is a one to one correspondence between points in space and triples of real numbers. The most common are cartesian, cylindrical and spherical ones. It is to be remarked that all coordinate systems are equivalent: a specific one is chosen according to the geometry of the problem at hand. If the domain has a symmetry, it is convenient that the coordinate system has the same symmetry. For example, to describe the position of points on the surface of the Earth, a spherical system is more convenient than a cartesian one. To describe the magnetic field created by a straight wire carrying an electric current, a cylindrical system is better suited than a cartesian one. To describe the position of objects in a room, a cartesian system is more appropriate than a spherical one. Fig. A.1 shows the main characteristics of the cylindrical reference system. Fig. A.2 describes the spherical reference system.

w .jn

tu

The relationship between cylindrical (ρ, ϕ, z) and cartesian coordinates is x = ρ cos ϕ y = ρ sin ϕ

(A.1)

w w

and that between spherical (r, ϑ, ϕ) and cartesian coordinates is x = r sin ϑ cos ϕ y = r sin ϑ sin ϕ z = r cos ϑ

(A.2)

Every coordinate system has a set of three fundamental unit vectors, that will be denoted by carets. The expressions of the cylindrical unit vectors in the cartesian basis are ρ̂ = cos ϕx̂ + sin ϕŷ ϕ̂ = − sin ϕx̂ + cos ϕŷ

(A.3)

Clearly the unit vectors are different in each point. The origin is a singular point of the system of coordinates, since the unit vectors are not defined there. The corresponding expressions for the 1

w .jn

tu

w

or

ld

fo

ru

m

.c

om

2

w w

Figure A.1. Cylindrical coordinate reference system. (a) The three mutually perpendicular surfaces defining the position of a point. (b) The three unit vectors in the point P ; in the text ρ̂, ϕ̂, ẑ are used in place of aρ , aφ , az . (c) The elementary volume dV at P : the sides are dρ, ρdφ, dz.

w w

w .jn

tu

w

or

ld

fo

ru

m

.c

om

3

Figure A.2. Spherical coordinate reference system. (a) Definition of the spherical coordinates. (b) The three mutually perpendicular surfaces defining the position of a point. (c) The three unit vectors in the point P ; in the text r̂, ϑ̂, ϕ̂ are used in place of ar , aθ , aφ . (d ) The elementary volume dV at P : the sides are dr, rdϑ, r sin ϑdϕ.

4 spherical system are r̂ = sin ϑ cos ϕx̂ + sin ϑ sin ϕŷ + cos ϑẑ (A.4)

ϑ̂ = cos ϑ cos ϕx̂ + cos ϑ sin ϕŷ − sin ϑẑ ϕ̂ = − sin ϕx̂ + cos ϕŷ

and again in each point the basis of the unit vectors is different. As in the previous case, the origin is a singular point. The cartesian system is different in this respect: indeed, the basis of unit vectors is the same in each point, x̂,ŷ,ẑ. Moreover, no point is singular.

om

Suppose that a certain vector field E(r) is given in a certain region of space. This means that in every point r the vector E(r) is defined: we can imagine it as an arrow with the tail in the point r. This is the abstract picture (coordinate-free). According to the system of coordinates, in the point r a basis of unit vectors is defined and three numbers are associated to the vector, i.e. its components:

.c

E(r) = Ex x̂ + Ey ŷ + Ez ẑ = Eρ ρ̂ + Eϕ ϕ̂ + Ez ẑ = Er r̂ + Eϑ ϑ̂ + Eϕ ϕ̂

(A.5)



 Eρ ¡  Eϕ  = r̂ ϑ̂ Ez

ld

Eq.(A.5) can be conveniently written as   Ex ¡ ¢ ¡ E(r) = x̂ ŷ ẑ  Ey  = ρ̂ ϕ̂ Ez

fo

ru

m

Components can also be arranged in a 3 × 1 matrix (column vector) so that the previous equations can also be written as       Ex Eρ Er E(r) ↔  Ey  E(r) ↔  Eϕ  E(r) ↔  Eϑ  (A.6) Ez Ez Eϕ  ϕ̂

or

ẑ

¢

¢

 Er  Eϑ  Eϕ

(A.7)

w

Notice that the row vector contains the unit vectors instead of numbers.

ρ̂

ϕ̂ ẑ

w .jn

¡

tu

Also eq.(A.3) can be written in matrix form ¢

=

¡

x̂

 ŷ

ẑ

¢

cos ϕ  sin ϕ 0

− sin ϕ cos ϕ 0

 0 0  1

(A.8)

As for the spherical system r̂ ϑ̂

w w

¡

ϕ̂

¢

=

¡

 x̂

ŷ

ẑ

¢

sin ϑ cos ϕ  sin ϑ sin ϕ cos ϑ

cos ϑ cos ϕ cos ϑ sin ϕ − sin ϑ

 − sin ϕ cos ϕ  0

(A.9)

In this way it is a simple matter to find the relationship between cylindrical and cartesian components of a vector. From (A.7) and (A.8)     Ex Eρ ¡ ¢ ¡ ¢ E(r) = x̂ ŷ ẑ  Ey  = ρ̂ ϕ̂ ẑ  Eϕ  Ez Ez    (A.10) cos ϕ − sin ϕ 0 Eρ ¡ ¢ = x̂ ŷ ẑ  sin ϕ cos ϕ 0   Eϕ  0 0 1 Ez

5 from which we get, on comparing the second and the fourth term of this chain of equalities     Ex Eρ  Ey  = P  Eϕ  CaCy Ez Ez

where the change of basis matrix is



P CaCy

The inverse transformation is

cos ϕ =  sin ϕ 0

− sin ϕ cos ϕ 0

 0 0  1

   Ex Eρ  Ey   Eϕ  = P CyCa Ez Ez

where

 sin ϕ 0 cos ϕ 0  0 1



.c

cos ϕ =  − sin ϕ 0

m

P CyCa = P −1 = P TCaCy CaCy

om



ru

Since both the cartesian and the cylindrical basis are orthonormal, the change of basis matrix is unitary, hence the inverse of it coincides with its transpose.

w

or

ld

fo

The transformation of the components of a vector between the spherical and cartesian bases is obtained analogously. From (A.7) and (A.9)     Ex Er ¡ ¢ ¡ ¢ E(r) = x̂ ŷ ẑ  Ey  = r̂ ϑ̂ ϕ̂  Eϑ  Ez Eϕ    (A.11) sin ϑ cos ϕ cos ϑ cos ϕ − sin ϕ Er ¡ ¢ = x̂ ŷ ẑ  sin ϑ sin ϕ cos ϑ sin ϕ cos ϕ   Eϑ  cos ϑ − sin ϑ 0 Eϕ

w .jn

tu

from which we get, on comparing the second and the fourth term of this chain of equalities     Ex Er  Ey  = P  Eϑ  CaS Ez Eϕ

w w

where the change of basis matrix is  P CaS

The inverse transformation is

sin ϑ cos ϕ cos ϑ cos ϕ =  sin ϑ sin ϕ cos ϑ sin ϕ cos ϑ − sin ϑ

 − sin ϕ cos ϕ  0

   Ex Er  Ey   Eϑ  = P SCa Ez Eϕ 

where



P SCa = P −1 = P TCaS CaS

sin ϑ cos ϕ =  cos ϑ cos ϕ − sin ϕ

sin ϑ sin ϕ cos ϑ sin ϕ cos ϕ

 cos ϑ − sin ϑ  0

(A.12)

6 Vectors can be multiplied by a scalar, i.e. a number: the result is a vector with the same direction.

Two vectors can be multiplied in two ways. The scalar product of two arbitrary vectors is a scalar, the vector product is a vector. In the former case A · B = |A||B| cos α = Ax Bx + Ay By + Az Bz = Aρ Bρ + Aϕ Bϕ + Az Bz = Ar Br + Aϑ Bϑ + Aϕ Bϕ

(A.13)

om

where ϑ is the angle between the two vectors. The scalar product of orthogonal vectors is obviously zero. The first definition is coordinate-free. The other expressions are so simple because the bases are orthonormal. In matrix form   Bx ¡ ¢ A · B = Ax Ay Az  By  Bz Indeed, recalling (A.7)

Ax

m

Ay

(A.14)

Ay

fo

=

¡

Ax

¢

ru

A·B=

   x̂ Bx ¡ ¢ Az  ŷ  · x̂ ŷ ẑ  By  ẑ Bz    x̂ · x̂ x̂ · ŷ x̂ · ẑ Bx ¢ Az  ŷ · x̂ ŷ · ŷ ŷ · ẑ   By  ẑ · x̂ ẑ · ŷ ẑ · ẑ Bz

.c



¡

ld

and the central 3 × 3 matrix is the identity matrix since the cartesian basis is orthonormal.

or

Two vectors can also be multiplied so that the result is a vector A × B (or, more appropriately, A ∧ B) : |A × B| = |A||B| sin α (A.15)

w .jn

tu

w

and the direction is orthogonal to the plane on which the two vectors lie and points in the direction of the “right-hand-rule”. The vector product of parallel vectors is obviously zero. The magnitude of the vector product has the geometrical meaning of surface of the parallelogram with sides A and B . The easiest way to introduce it is by specifying the vector products of unit basis vectors:       x̂ x̂ × x̂ x̂ × ŷ x̂ × ẑ 0 ẑ −ŷ ¡ ¢  ŷ  × x̂ ŷ ẑ =  ŷ × x̂ ŷ × ŷ ŷ × ẑ  =  −ẑ 0 x̂  (A.16) ẑ ẑ × x̂ ẑ × ŷ ẑ × ẑ ŷ −x̂ 0

w w

from which we see very clearly that the vector product is anticommutative. Recalling (A.7) again     x̂ Bx ¡ ¢ ¡ ¢ A × B = Ax Ay Az  ŷ  × x̂ ŷ ẑ  By  ẑ Bz     0 ẑ −ŷ Bx ¡ ¢ x̂   By  = Ax Ay Az  −ẑ 0 (A.17) ŷ −x̂ 0 Bz   x̂ ŷ ẑ = det  Ax Ay Az  Bz By Bz

7 The last expression is the traditional one (although only mnemonic!), where the determinant must be expanded using the elements of the first line.

Similar expressions hold for the cylindrical and cylindrical system    ρ̂ ρ̂ × ρ̂ ρ̂ × ϕ̂ ¡ ¢  ϕ̂  × ρ̂ ϕ̂ ẑ =  ϕ̂ × ρ̂ ϕ̂ × ϕ̂ ẑ ẑ × ρ̂ ẑ × ϕ̂

  ρ̂ × ẑ 0 ϕ̂ × ẑ  =  −ẑ ẑ × ẑ ϕ̂

 ẑ −ϕ̂ 0 ρ̂  −ρ̂ 0

(A.18)

   r̂ × ϕ̂ 0 ϕ̂ −ϑ̂ r̂  ϑ̂ × ϕ̂  =  −ϕ̂ 0 ϑ̂ −r̂ 0 ϕ̂ × ϕ̂

(A.19)

om

and for the spherical one    r̂ r̂ × r̂ r̂ × ϑ̂ ¡ ¢  ϑ̂  × r̂ ϑ̂ ϕ̂ =  ϑ̂ × r̂ ϑ̂ × ϑ̂ ẑ ϕ̂ × r̂ ϕ̂ × ϑ̂

spherical coordinates. In particular, for the

Correct order of unit vectors

fo

Figure A.3.

ru

m

.c

These definitions can be easily remembered by the method of Fig. A.3 A useful identity to know

ld

is the one that allows to expand a double vector product

or

A × (B × C) = (A · C) B − (A · B) C

w

Vector and scalar product can be combined in the mixed product A · (B × C). This number is the volume of the (skewed) parallelepiped with edges defined by A, B and C; it is clearly zero if the three vectors lie in the same plane.

w .jn

tu

An important concept associated to vector fields is that of field line. A field line is a line such that its tangent at any point r is parallel to the vector field E(r) in that point. They are useful tools for visualizing vector fields. If the vector field is regular, for each point only a field line passes.

w w

Sometimes in a region two vector fields are given and there is functional dependence of one on the other. The simplest relationship that can exist is a linear one, so that the principle of superposition holds. An example in electromagnetism is the relationship existing between the electric field and the electric induction in a dielectric. Or the one between an electric current and the electric and magnetic fields it produces. Referring to the former example, we can write in abstract form D = L{E} However just as in the case of vectors, it is generally useful to introduce a basis to carry out computations. Using a cartesian basis, the previous equation becomes, because of linearity D = Dx x̂ + Dy ŷ + Dz ẑ = Ex L{x̂} + Ey L{ŷ} + Ez L{ẑ} In matrix form



   Dx Ex ¡ ¢  Dy  = L{x̂} L{ŷ} L{ẑ}  Ey  Dz Ez

8 Now L{x̂} is a (abstract) vector with three components along the three axes that can be collected in a column vector. The same can be done for the other two elements so that a 3 × 3 matrix arises and the previous equation can be written, with a more physical notation      Dx εxx εxy εxz Ex  Dy  =  εyx εyy εyz   Ey  (A.20) Dz εzx εzy εzz Ez where



 εxx L{x̂} =  εyx  εzx



 εxy L{ŷ} =  εyy  εzy



 εxz L{ẑ} =  εyz  εzz

.c

om

The row index denotes the component, the column index the unit vector. The matrix ε represents the abstract linear operator L in the cartesian basis:   εxx εxy εxz L ↔  εyx εyy εyz  = ε εzx εzy εzz

m

and eq.(A.20) can be rephrased as D = εE

ru

Following in the reverse direction the steps leading from (A.5) to (A.6), this equation can be written

fo

L = εxx x̂x̂ + εxy x̂ŷ + εxz x̂ẑ + εyx ŷx̂ + εyy ŷŷ + εyz ŷẑ + εzx ẑx̂ + εzy ẑŷ + εzz ẑẑ

(A.21)

or

ld

Symbols such as x̂ŷ are called (unit) dyads and play just the role of placeholders to denote a specific row and column. However, it is simple to give rules to operate with scalar products on dyads so as to reproduce the results that could be obtained by the matrix formalism. The rules are the following def

E · x̂ŷ = (E · x̂) ŷ = Ex ŷ

w

(A.22)

def

tu

x̂ŷ · E = x̂ (ŷ · E) = Ey x̂

Eq.(A.21) is called the dyadic form of the linear operator L.

w w

w .jn

A dyad must not necessarily be formed by two unit vectors. If A, B are arbitrary vectors, we form the dyad AB with the rule of operation def

E · AB = (E · A) B

(A.23)

def

AB · E = A (B · E)

If we want to obtain the matrix that represents the abstract operator L = AB in the cartesian basis, we can use a distributive law of the “side-by-side-placement” of two vectors L = AB = (Ax x̂ + Ay ŷ + Az ẑ) (Bx x̂ + By ŷ + Bz ẑ) = Ax Bx x̂x̂ + Ax By x̂ŷ + Ax Bz x̂ẑ + Ay Bx ŷx̂ + . . . + Az Bz ẑẑ

The matrix form can be obtained immediately     Ax Bx Ax By Ax Bz Ax ¡ L ↔  Ay Bx Ay By Ay Bz  =  Ay  Bx Az Bx Az By Az Bz Az

By

Bz

¢

(A.24)

9 It is possible to introduce the vector product of dyads and vectors: it is just enough to extend (A.23): def

E × AB = (E × A) B

(A.25)

def

AB × E = A (B × E) Useful linear operators are • the identity I, such that I·A=A·I=A



 0 0  1

ru

and in spherical

0 1 0

m

1 I = ρ̂ρ̂ + ϕ̂ϕ̂ + ẑẑ ↔  0 0

om

in cylindrical

coordinates are  0 0  1

.c

for any vector A. Its dyadic and matrix form in cartesian  1 0 I = x̂x̂ + ŷŷ + ẑẑ ↔  0 1 0 0



fo

1  0 I = r̂r̂ + ϑ̂ϑ̂ + ϕ̂ϕ̂ ↔ 0

0 1 0

 0 0  1

or

ld

• the identity transverse to ẑ, i.e. the projection operator onto the x,y plane, such that Itz · A = A − Az ẑ 1 0 = x̂x̂ + ŷŷ ↔  0 1 0 0

 0 0  0

tu

w

Itz



w w

w .jn

• the identity transverse to r̂, i.e. the projection operator onto the tangent plane to the sphere r =const., such that Itr · A = A − Ar r̂ 

Itr

0 = ϑ̂ϑ̂ + ϕ̂ϕ̂ ↔  0 0

0 1 0

 0 0  1

Consider again the vector product of two vectors A × B = C. Since the vector C depends linearly on B, we can say that it is the result of the application of a linear operator on B. This operator can be found by recalling the second line of (A.17)    0 ẑ −ŷ Bx ¡ ¢ x̂   By  A × B = Ax Ay Az  −ẑ 0 ŷ −x̂ 0 Bz    0 −Az Ay Bx ¡ ¢ 0 −Ax   By  = −Ay ẑ + Az ŷ Ax ẑ − Az x̂ −Ax ŷ + Ay x̂ =  Az −Ay Ax 0 Bz (A.26)

10 First we have carried out the multiplication of the row vector times the (formal) matrix, obtaining a row vector whose elements are abstract vectors. Their components can be interpreted as columns of the successive matrix. Alternatively, A×B=A×I·B so that the matrix we are looking for is the one representing A × I, i.e.

A × I = (A × x̂) x̂ + (A × ŷ) ŷ + (A × ẑ) ẑ

(A × x̂) = (Ax x̂ + Ay ŷ + Az ẑ) × x̂ = −Ay ẑ + Az ŷ (A × ŷ) = (Ax x̂ + Ay ŷ + Az ẑ) × x̂ = Ax ẑ − Az x̂ (A × ẑ) = (Ax x̂ + Ay ŷ + Az ẑ) × ẑ = −Ax ŷ + Ay x̂

(A.27)

.c

Substituting in the previous equation we get

om

Then

m

A × I = (−Ay ẑ + Az ŷ)x̂ + (Ax ẑ − Az x̂)ŷ + (−Ax ŷ + Ay x̂)ẑ = −Ay ẑx̂ + Az ŷx̂ + Ax ẑŷ − Az x̂ŷ − Ax ŷẑ + Ay x̂ẑ 0 A × I ↔  Az −Ay

−Az 0 Ax

 Ay −Ax  0

ru



fo

In conclusion

w

or

ld

As a direct application of this result, we consider another useful operator to be added to the previous list. It is r̂ × I = r̂ × Itr :   0 0 0 r̂ × I ↔  0 0 −1  ↔ ϕ̂ϑ̂ − ϑ̂ϕ̂ 0 1 0

A.2

w .jn

tu

where we have used the spherical basis and clearly r̂ ↔ (1 0 0)T in this basis.

Calculus of vector fields

w w

In this section we review the basic concepts of calculus, i.e. derivatives and integrals, applied to vector fields. Let us start with a scalar field h(x,y), where h is a smooth function of only two variables for simplicity. The partial derivatives ∂h ∂x

∂h ∂y

evaluated in the point P ↔ (x0 ,y0 ) have the meaning of local rate of change of h for small increments of the coordinates dx, dy in the x, y directions around point P , respectively. A quantity of interest is the directional derivative of h in the direction ŝ, which has the meaning of local rate of change of h for small displacements in the ŝ direction. According to the chain rule,

∂h dx ∂h dy ∂h + = ∂y ds ∂x ds ∂s

11

y

sy

ŝ

dy

ds

α sx

dx

Direction along which the directional derivative is computed

om

Figure A.4.

x

dx = cos α = sx ds

ru

∂h ∂h ∂h sy sx + = ∂y ∂x ∂s

m

so that

dy = sin α = sy ds

.c

With reference to Fig. A.4, we find

∂h ∂h ŷ x̂ + ∂y ∂x

ld

gradh = ∇h =

fo

This formula can be considered as a scalar product of the vector

w .jn

tu

w

or

called gradient of h, times the unit vector ŝ. The gradient of a scalar function is a vector field.

The magnitude of the gradient is the maximum rate of change of h when the direction ŝ is allowed to vary. The direction in which his happens is the direction of the gradient. If h is the height of the ground above the sea level, so that h(x,y) can be interpreted as the local height of a hill, the direction of the gradient is that of the steepest slope. Note, however, that the gradient is a vector belonging to the x, y plane and it is not tangent to the surface. When ŝ is orthogonal to the gradient, the directional derivative is zero: in that direction h is not changing, hence the gradient is always orthogonal to contour lines (lines on which h= const). Fig. A.5 shows an example of a scalar field h(x,y). Fig. A.6 shows a contour plot of the same function with the vector field gradh, computed numerically by finite differences. Note that the gradient is zero in the extrema of the graph of h, i.e. maxima, minima and saddle points. Also, the arrows are orthogonal to the contours.

w w

If the scalar field h(x,y,z) depends on the three space coordinates, the gradient gradh is defined

as

gradh = ∇h =

∂h ∂h ∂h ẑ ŷ + x̂ + ∂z ∂y ∂x

(A.28)

If the scalar field is given in a non cartesian system of coordinates the expression of the gradient are more complicated, essentially because the unit vectors are not constant. It can be proved that • in cylindrical coordinates gradh(ρ,ϕ,z) = ∇h =

∂h 1 ∂h ∂h ẑ ϕ̂ + ρ̂ + ∂z ρ ∂ϕ ∂ρ

(A.29)

12

z=h(x,y)

10 8 6 4 2 0 −2

om

−4 −6 −8 3

z y

x

2

3

1

.c

2

0

1

0

−1

−1

−2

−2

m

Scalar field z = h(x,y)

ld

fo

Figure A.5.

−3

ru

−3

grad h(x,y) and contours

or

2

w

1.5

y

w .jn

0.5

tu

1

0

w w

−0.5

−1

−1.5

−2 −2

−1.5

Figure A.6.

−1

−0.5

0 x

0.5

1

1.5

Contours of z = h(x,y) and vector field gradh(x,y)

2

Closed surface Σ for the definition of the flux of a vector field.

.c

Figure A.7.

om

13

• in spherical coordinates

m

1 ∂h 1 ∂h ∂h ϕ̂ ϑ̂ + r̂ + r sin ϑ ∂ϕ r ∂ϑ ∂r

(A.30)

ru

gradh(r,ϑ,ϕ) = ∇h =

ld

fo

We turn now to a vector field A(r) and define its flux across a surface, which may be either open or closed. Fig. A.7 shows a closed surface Σ enclosing the volume V , with the outward normal unit vector ν̂. The outward flux of A through Σ is defined by the scalar I ΦΣ (A) = A · ν̂dΣ Recalling Gauss Theorem

I

w

Z

or

Σ

∇ · AdV =

A · ν̂dΣ

(A.31)

Σ

tu

V

w .jn

we can interpret this equation as defining an “average divergence” I Z 1 1 A · ν̂dΣ ∇ · AdV = average{∇ · A} = V Σ V V

w w

In the limit of the volume reducing to a point we obtain the divergence as flux per unit volume across a small closed surface surrounding the point. It can be shown that the explicit expressions for the divergence of a vector field are • in cartesian coordinates

∂Az ∂Ay ∂Ax + + ∂z ∂y ∂x

(A.32)

∂Az 1 ∂Aϕ 1 ∂ + (ρAρ ) + ∂z ρ ∂ϕ ρ ∂ρ

(A.33)

div A = ∇ · A =

• in cylindrical coordinates div A = ∇ · A =

14 • in spherical coordinates div A = ∇ · A =

∂ 1 ∂ 2 1 1 ∂Aϕ (sin ϑAϑ ) + (r Ar ) + r sin ϑ ∂ϕ r sin ϑ ∂ϑ r2 ∂r

(A.34)

om

In conclusion, the divergence of a vector field is a particular combination of derivatives of the field components, whose meaning is clarified by Gauss theorem. Consider a flow of charges across a closed surface so that A = J = ρv where v is the velocity field and ρ the volume charge density.

The units of J are A/m2 , hence it is a current density. The flux ΦΣ (J) has the units of A and it yields the amount of charge coming out every second from the total volume V , i.e. the total current. The source of this outflow is the divergence of J, which tells us how much charge per unit volume is coming out every second from each point of V . The flux of J provides a global information, the divergence of J a local one.

m

.c

Finally it is useful to write Gauss theorem (A.31) for the one dimensional domain [a,b], with boundary consisting of the points x = a (with outward normal −x̂) and x = b (with outward normal x̂) Z b ∂Ax dx = Ax (b) − Ax (a) a ∂x

Clearly, this is just the fundamental formula of integral calculus.

Γ

fo

ru

Another type of integral that we can form with a vector field is a line integral Z A · τ̂ ds

Γ

w

Σo

or

ld

where the curve Γ may be either open or closed. Assume it is closed, so that the integral is called the circulation of A. Stokes theorem states that Z I ∇ × A · ν̂dΣo = A · τ̂ ds (A.35)

w w

w .jn

tu

where Σo is an open surface with Γ as boundary; the orientation of the tangent vector τ̂ is related

Figure A.8.

Geometry for the application of Stokes theorem

to that of ν̂ by the right-hand-rule, as shown in Fig. A.8. For the purpose of the definition, it is convenient to limit the generality of the theorem and consider a planar curve Γ so that Σo can

15 be taken as the part of plane inside Γ, with constant normal vector ν̂. This equation allows us to define the ν̂ component of an “average curl ” I Z 1 1 A · τ̂ ds ∇ × A · ν̂dΣo = average{∇ × A} · ν̂ = Σo Γ Σo Σo

Obviously, by choosing three loops with linearly independent normals we can define completely the “total amount of rotation” or average curl. By letting the size of the loops go to zero, we can define the curl in a point as circulation per unit area in the neighborhood of a single point. By carrying out this prescription in different systems of coordinates, it can be proved that the explicit expressions of the curl of a vector field are • in cartesian coordinates

(A.36)

.c

om

curl A = ∇ × A = ¶ ¶ µ ¶ µ µ ∂Ax ∂Ay ∂Az ∂Ax ∂Ay ∂Az ẑ − ŷ + − x̂ + − ∂y ∂x ∂x ∂z ∂z ∂y

• in cylindrical coordinates

(A.37)

ru

m

curl A = ∇ × A = µ ¶ ¶ µ ¶ µ 1 ∂Az 1 ∂Aρ 1 ∂ ∂Az ∂Aρ ∂Aϕ ẑ (ρAϕ ) − ϕ̂ + − ρ̂ + − ρ ∂ϕ ρ ∂ρ ∂ρ ∂z ∂z ρ ∂ϕ

fo

• in spherical coordinates

(A.38)

w

or

ld

curl A = ∇ × A = µ µ ¶ ¶ 1 ∂Ar ∂ ∂ 1 ∂Aϑ 1 (rAϕ ) ϑ̂+ − r̂ + (Aϕ sin ϑ) − ∂r r sin ϑ ∂ϕ ∂ϕ r sin ϑ ∂ϑ µ ¶ 1 ∂ ∂Ar ϕ̂ (rAϑ ) − ∂ϑ r ∂r

w .jn

tu

If we examine the expressions of gradient, divergence and curl in cartesian coordinates we can identify a kind of formal vector ∇ defined by ∇ = x̂

∂ ∂ ∂ + ẑ + ŷ ∂z ∂y ∂x

(A.39)

w w

such that the various differential operators can be imagined to be formed by means of standard product, scalar product and vector product of ∇ times a scalar field or a vector field. It is to be remarked that such an interpretation (due to the American physicist W. Gibbs 1839-1903) is possible only in cartesian coordinates. Hence, for instance in spherical coordinates, even if the curl of A is written usually as ∇ × A, this expression has to be taken as a single symbol (whose meaning is given by (A.2)) and not interpreted as the product of two factors.

First order differential operators can combined to form second order operators.

• The gradient of a scalar field is a vector, so that we can compute the curl and divergence of it. However ∇ × ∇h = 0 identically for any smooth h(r). To remember the property, we can note that the vector product of two equal vectors is zero.

16 • The divergence of a gradient produces the Laplace operator or laplacian ∇ · ∇h = ∇2 h = 4h We recall only the expression of the laplacian in cartesian coordinates ∇2 h =

∂2h ∂2h ∂2h + 2 + 2 ∂z ∂y ∂x2

• The divergence of a vector field A(r) is a scalar, hence its gradient can be computed

om

∇(∇ · A)

.c

• The curl of a vector field A(r) is a vector, hence its divergence and curl can be computed.

However ∇ · (∇ × A) = 0

m

identically for any smooth A(r). To remember the property, we can note that the triple mixed product with two equal factors is zero.

ru

• As for the curl of the curl, the following identity is to be noted

fo

∇ × (∇ × A(r)) = ∇(∇ · A) − ∇2 A

or

ld

Often multidimensional integrals have to be computed. It is useful to remember from Fig. A.2 that the elementary volume in spherical coordinates is a cube with edges of size dr (along r̂), rdϑ (along ϑ̂) and r sin ϑdϕ (along ϕ̂), thus

w

dV = r2 sin ϑdrdϑdϕ

dΣ = r2 sin ϑdϑdϕ

w .jn

tu

Concerning surface integrals in spherical coordinates, the elementary patch is a square with sides rdϑ (along ϑ̂) and r sin ϑdϕ (along ϕ̂), thus

Connected with this is the concept of solid angle, displayed in Fig. A.9

w w

The natural measurement unit of a plane angle is radian (rad). A one radian angle is the one with the vertex at the center of a circle of radius r that subtends an arc whose length is equal to the radius r. Thus the radian measure of an angle is equal to the ratio between the length of the subtended arc and the radius: for instance, the measure of a full round angle is 2π radians.

Similarly, the solid angle, Ω, is the two-dimensional angle in three-dimensional space that an object subtends at a point; it is measured in steradians (sr). A one steradian solid angle is the one with the vertex at the center of a sphere of radius r that subtends a patch whose area is equal to r2 . Since the surface of a sphere is 4πr 2 , the measure of the total solid angle around a point is 4π steradians. As another example, the solid angle defined by x ≥ 0, y ≥ 0, z ≥ 0 has a measure of 4π/8 = π/2 steradians. The elementary solid angle subtended by a patch of area dΣ is, in spherical coordinates dΣ dΩ = 2 = sin ϑdϑdϕ r

Multidimensional Dirac delta functions

or

A.3

Definition of radian for a plane angle and of steradian for a solid angle

ld

Figure A.9.

fo

ru

m

.c

om

17

w

The one dimensional Dirac δ function is a distribution defined by the property Z ∞ f (x)δ(x − x0 )dx = f (x0 )

tu

−∞

w .jn

for any function f (x) continuous in x = x0 . If f (x) ≡ 1 identically, this becomes Z ∞ δ(x − x0 )dx = 1 −∞

w w

The support of δ(x − x0 ) is the point x = x0 .

Suppose now that f (x,y,z) is a function of three variables and suppose that we want to compute Z ∞Z ∞Z ∞ I= f (x,y,z)δ(x − x0 )dxdydz −∞

−∞

−∞

Integrating first with respect to x leads to Z ∞Z I= −∞

∞

f (x0 ,y,z)dydz

−∞

The support of δ(x − x0 ) is still x = x0 , which in this case is not a single point but a plane.

Physically, such a δ function can describe a surface charge. The value of the integral has been

18 obtained by evaluating the function f on the support of the δ function and then integrating along it.

This concept can be generalized if we consider a δ function with support on a smooth surface of R3 with parametric equation r = rΣ (u,v). Consider the integral Z I = f (r)δ(r − rΣ )dr Again its value is obtained by evaluating f on the support rΣ of the δ function and then integrating along it ° ° Z Z Z ° ∂rΣ ∂rΣ ° ° dudv × I = f (rΣ )dΣ = f (rΣ (u,v)) ° ° ∂u ∂v °

where

om

° ° ° ∂rΣ ∂rΣ ° ° ° dudv dΣ = ° × ∂v ° ∂u

.c

is the elementary area on the surface Σ. Note that the δ function can be described as one dimensional here and its support is two dimensional.

ru

m

Consider now a δ function with one dimensional support consisting of the straight line parallel to the z axis with equation x = x0 y = y0

−∞

−∞

or

−∞

ld

fo

Physically, it can describe a line charge. Mathematically, its expression is δ(x − x0 )δ(y − y0 ), hence it can be defined as two dimensional, because it is the product of two one dimensional δ functions.

We can be interested in computing the integral Z ∞Z ∞Z ∞ I= f (x,y,z)δ(x − x0 )δ(y − y0 )dxdydz

−∞

tu

w

Its value is obviously the integral of f along that line Z ∞ I= f (x0 ,y0 ,z)dz

w .jn

A generalization of this result is obtained if the support of the δ function is a smooth curve γ ∈ R3 with parametric equation r = rγ (s). Suppose we must evaluate the integral Z I = f (r)δ(r − rγ )dr

w w

Once again, its value is obtained by evaluating f on γ and integrating along it ° ° Z Z ° drγ ° ° ds ° I = f (rγ )d` = f (rγ (s)) ° ds °

where

° ° ° drγ ° ° ds ° d` = ° ds °

is the line element on γ.

The last example to consider is that of a three-dimensional δ function with support on the point x = x0 y = y0 z = z0

19 which can represent a point charge. Next, consider the integral Z I = f (r)δ(r − r0 )dr Its value is I = f (r0 ) We can say that this result is obtained by evaluating the function f on the support of the δ function and then integrating along it. However, since this support is zero-dimensional, this last integral disappears.

w w

w .jn

tu

w

or

ld

fo

ru

m

.c

om

In conclusion, all the integrals can be evaluated by the same rule: evaluate the integrand on the support of the δ function and integrate along it. Notice that the sum of the dimensionality of the δ function and of its support have constant sum 3, the dimensionality of ambient space.

Appendix B

.c

Polarization and Phasors

m

B.1

om

Solved Exercises

fo

ru

Exercise n. 1 A time harmonic electric field is E(t) = cos ω0 t x̂ + sin ω0 t ŷ. Determine the type of polarization, draw the polarization curve defined by this vector and write the expression of the phasor E. Solution Write

or

ld

© ª © ª E(t) = R x̂ ejω0 t + R −jŷ ejω0 t © ª = R (x̂ − jŷ) ejω0 t

w

Hence the phasor is E = x̂ − jŷ. Compute

tu

E0 × E00 = x̂ × (−ŷ) = −ẑ 6= 0

w .jn

E0 · E00 = x̂ · (−ŷ) = 0 and

|E0 | = |E00 |(= 1)

w w

The polarization is circular counterclockwise. The plot of the polarization curve is shown in Fig. B.1.

Alternatively, the polarization is circular because Ex = 1

Ey = −j = e−jπ/2

δ = −π/2

|Ex | = |Ey | = 1 ¥

Exercise n. 2 √ √ A time harmonic electric field is E(t) = 2 cos(ω0 t + 45◦ ) (x̂ + ŷ) + 2 sin(ω0 t + 45◦ ) (x̂ − ŷ).

Determine the type of polarization, draw the polarization curve defined by this vector and write the expression of the phasor E.

20

21

y 2

1.5

1

0.5

0 x −0.5

−1

−2 −2

−0.5

0

0.5

1

1.5

2

Polarization curve defined by E(t) above

m

Figure B.1.

−1

.c

−1.5

om

−1.5

Solution Write

ld

fo

ru

o o n n √ √ ◦ ◦ E(t) = R (x̂ + ŷ) 2ej45 ejω0 t + R −j(x̂ − ŷ) 2ej45 ejω0 t © ª = R {(1 + j)(x̂ + ŷ) − j(1 + j)(x̂ − ŷ) ejω0 t } © ª = R {(1 + j)(x̂ + ŷ) + (1 − j)(x̂ − ŷ) ejω0 t }

or

Hence the phasor is E = 2x̂ + j2ŷ. Compute

E0 × E00 = 2x̂ × (2ŷ) = 4ẑ 6= 0

w

E0 · E00 = 2x̂ · (2ŷ) = 0 and

|E0 | = |E00 |(= 2)

Ex = 2

Ey = j = ejπ/2

δ = π/2

|Ex | = |Ey | = 2 ¥

w w

w .jn

tu

The polarization is circular clockwise. The plot of the polarization curve is shown in Fig. B.2 Alternatively, the polarization is circular because

Exercise n. 3 The phasor of a magnetic field is H = (1 + j)x̂ + 2(1 + j)ŷ. Determine the type of polarization, write the expression of the time varying field H(t) and draw the polarization curve defined by this vector.

Solution Find real and imaginary part of the phasor H0 = x̂ + 2ŷ

H00 = x̂ + 2ŷ

22

y 4

3

2

1 x 0

−1

−2

−2

−1

0

1

2

3

Figure B.2.

Polarization curve defined by E(t) above y

ru

4

3

fo

2

0

or

−1

w

−2

tu

−3

−4 −4

w .jn

−3

Figure B.3.

w w

x

ld

1

Compute

4

.c

−3

m

−4 −4

om

−3

−2

−1

0

1

2

3

4

Polarization curve defined by H(t) above

H0 × H00 = (x̂ + 2ŷ) × (x̂ + 2ŷ) = 0

The polarization is linear.

The time varying field is shown in Fig. B.3 H(t) = (x̂ + 2ŷ)(cos ω0 t − sin ω0 t) √ = (x̂ + 2ŷ) 2 cos(ω0 t + 45◦ )

The plot is shown in Fig. B.3 ¥

23

B.2

Plane Waves

Exercise n. 1 Consider a plane wave in a dielectric with εr = 4, µr = 1, at the frequency f = 10GHz. The electric field in the origin is E(0) = 2x̂ − jŷ + ẑ V/m 1. Compute the phase velocity vph , the wavelength λ, the wave impedance Z, the wavenumber k in deg/cm, the power density dP/dΣ.

om

2. Find the polarization of the electric field E0 . Find the direction of propagation ŝ knowing that the phase of the wave decreases in the z direction. √ 3. Compute the magnetic field in the point P : (2,2,2)T 5 cm at the time t = T /4

c c 1 = = 1.5 · 108 m/s vph = √ = √ 2 εr µr εµ

m

Solution Phase velocity:

.c

Use the approximate values c = 3 · 108 m/s and Z0 = 377Ω.

ru

Wavelength

r

Wavenumber:

r

Z0 µr = 188.5Ω = 2 εr

or

Z=

µ = Z0 ε

ld

Wave impedance:

fo

1.5 · 108 vph = 1.5 · 10−2 m = 1.0 · 1010 f

λ=

6.2832 2π = 4.1888 rad/cm = 1.5 · 10−2 λ 360◦ = 240.0 deg/cm = λ

w .jn

Power density

tu

w

k=

1 (4 + 1 + 1) 1 |E0x |2 + |E0y |2 + |E0z |2 1 |E0 |2 dP 2 = 0.01591 W/m = = = 2 188.5 188.5 2 2 Z dΣ

w w

Polarization: The real and imaginary part of E(0) are E00 = 2x̂ + ẑ

E000 = −ŷ

They are not parallel, hence the polarization is not linear. Moreover E00 · E000 = 0

but

|E00 | = 6 |E000 |

hence the polarization is not circular. In conclusion, it is elliptical.

Direction of propagation: Since the electric field is perpendicular to ŝ, we have ŝ = ±

(2x̂ + ẑ) × (−ŷ) −2ẑ + x̂ E00 × E000 =± =± √ |(2x̂ + ẑ) × (−ŷ)| |E00 × E000 | 5

24

The phase factor is exp[−j(kx x + ky y + kz z)] If the phase is decreasing in the z direction, then kz = ksz > 0, hence 2ẑ − x̂ ŝ = √ 5 Phasor of magnetic field in the origin ¶ ¶ µ µ 1 5 2 2ẑ − x̂ √ × (2x̂ − jŷ + ẑ) = Y j √ x̂ + √ ŷ + j √ ẑ H0 = Y ŝ × E0 = Y 5 5 5 5

.c

om

Phasor of magnetic field in P : ¶ µ ¸ ¸ · · √ 2π 2ẑ − x̂ 2π √ · (2x̂ + 2ŷ + 2ẑ) 5 = H(P ) = H0 exp −j ŝ · rP = H0 exp −j λ λ 5

¸ · ¸ √ 2π · 2 cm 2π 1 √ (2 · 2 − 2) 5 = H0 exp −j = H0 exp −j = H0 exp [−j 120◦ ] 1.5 cm λ 5

m

·

ru

Magnetic field in P at t = T /4

w

or

ld

fo

n o © ª ◦ ◦ H(rP ,t) = R H(rP )ejω0 t t=T /4 = R H0 ej(90 −120 ) Ã √ !

3 5 1 2 √ ŷ + √ x̂ + √ ẑ = = H00 cos(−30◦ ) − H000 sin(−30◦ ) = Y 2 5 2 5 2 5 Ã ! √ 1 15 1 1 √ x̂ + = ŷ + √ ẑ 2 188.5 2 5 5

w .jn

tu

¥

Exercise n. 2 Consider a plane wave propagating in the z direction in a dielectric with εr = 4,µr = 1 and γ = 0.01 S/m at the frequency f = 1.0 GHz and E0 = x̂.

w w

1. Compute the wavenumber k, the phase velocity vph , the wavelength λ 2. Compute the wave impedance Z, the active power density in the origin dP/dΣ, the attenuation αdB in dB/m Use c = 2.99792458 · 108 m/s, ε0 = 8.854 · 10−12 F/m.

Solution Wavenumber: r³ k=ω

γ´ √ µ0 = ω ε0 µ0 ε0 εr − j ω

r r γ γ εr − j = k0 εr − j ωε0 ωε0

25 The free space wavelength and wavenumber are λ0 =

c = 0.29979 m f

k0 =

2π = 20.9585 rad/m λ

Then k = 20.9585

p 4 − j 0.1798 = 20.9585(2.0005 − j 0.0449) = 41.9275 − j 0.9416 m−1

The real and imaginary part are

vph =

ω = 1.4986 · 108 m/s β

Wavelength λ=

vph = 0.1499 m f

.c

Phase velocity:

α = 0.9416 Np/m

om

β = 41.9275 rad/m

fo

ru

m

Wave impedance: r r r µr µ0 µ = Z0 (0.4996 + j 0.0112) = 188.2227 + j 4.2270 Ω Z= = ε0 εr − jγ/(ωε0 ) ε0 εr − jγ/ω

Wave admittance

ld

1 1 = = (5.3102 − j 0.1192) · 10−3 S 188.2227 + j 4.2270 Z

Notice that

1 1 = 5.3129 · 10−3 S = 188.2227 R{Z}

w

R{Y } = 6

or

Y =

w .jn

tu

In this case the difference is small because the phase of Z is small but becomes enormous when this phase approaches π/2.

Active power density in the origin

1 1 dP 2 = R{Y }|E0 |2 = 5.3102 · 10−3 (|E0x |2 + |E0y |2 + |E0z |2 ) = 2.6551 · 10−3 W/m 2 2 dΣ

w w

Attenuation:

α = Im{k} = 0.9416 Np/m

αdB = α 20 log10 e = α 8.68589 = 8.1785 dB/m ¥

Exercise n. 3 Consider a plane wave propagating in free space at the frequency f = 5 GHz. The electric field in the points of the plane z = 0 has the value E(x,y,z)|z=0 = E0 exp[−j(ξx + ηy)]

∀x, y

26 with ξ = k0 /5 and η = k0 /2 and it is known that this wave is a TE field, i.e. E0 has no zcomponent and that it carries the power density dP/dΣ > 0 through the plane z = 0.

Compute the direction of propagation ŝ and the spherical angles that define this direction. Compute also the fields in the origin E0 , H0 .

Solution The propagation factor of a plane wave is exp[−jk · r] = exp[−j(kx x + ky y + kz z)]

om

hence, by inspection, we find kx = ξ and ky = η. From the dispersion relation kx2 + ky2 + kz2 = ω 2 ε0 µ0 = k02

kz =

k02

−

kx2

−

ky2

r √ q 1 1 71 2 2 2 − = k0 = k0 − ξ − η = k0 1 − 10 25 4

m

q

.c

it follows that

ld

fo

ru

The sign of the square root is taken to be positive because power is flowing toward the region z > 0, due to the fact that dP/dΣ > 0 through a dΣ belonging to the plane z = 0. From this √ 1 71 1 k = x̂ + ŷ + ẑ ŝ = 10 2 5 k0

or

The spherical angles that identify the direction ŝ are found by recalling (A.2) and noting that |ŝ| = 1

From this we find

tu

w

sx = sin ϑ cos ϕ sy = sin ϑ sin ϕ sz = cos ϑ

w .jn

√ 71 = 32.5827 deg ϑ = arccos sz = arccos 10 ³ s ´ ³ s ´ x y = 68.1986 deg = arcsin ϕ = arccos sin ϑ sin ϑ

w w

The electric field in the origin, with magnitude r E0 =

2Z0

dP dΣ

must be perpendicular to ŝ because it is a plane wave and perpendicular to ẑ because required in the text. Then E0 = E 0

k × ẑ ŝ × ẑ (ηx̂ − ξŷ) (ξx̂ + ηŷ + kz ẑ) × ẑ = E0 = E0 p = E0 |k × ẑ| |k × ẑ| |ŝ × ẑ| ξ2 + η2

Thanks to the denominator, the vector multiplying E0 has unit magnitude.

27 The magnetic field is computed by the impedance relation H0 = Y0 ŝ × E0 =

=

k0

Y E p0 0 ξ2 + η2

Y0 E0 Y0 (ηx̂ − ξŷ) (ξx̂ + ηŷ + kz ẑ) × p k × E0 = k0 k0 ξ 2 + η2 ¡ ¢ kz (ξx̂ + ηŷ) − (ξ 2 + η 2 )ẑ

Exercise n. 4 Consider a plane wave field propagating in free space. The electric field is E(x,y,z) = E0 x̂ cos kz

m

.c

Compute the magnetic field.

om

¥

ru

Solution The general relation between electric and magnetic fields is the first Maxwell equation:

fo

1 ∇×E jωµ

ld

H=−

In the case of a plane wave this equation simplifies in the impedance relation

or

1 k × E = Y ŝ × E ωµ

w

H=

tu

The given electric field is not a plane wave, but is the sum of two counter-propagating ones, as it can be seen by Euler’s formula

w .jn

E(x,y,z) = E0 x̂ cos kz =

E0 −jkz E0 jkz + x̂e x̂e 2 2

w w

The direction of propagation of the first wave is ŝ = ẑ, that of the second ŝ = −ẑ. Now we apply the impedance relation to each wave, using the appropriate unit vector ŝ:

E0 E0 ẑ × x̂e−jkz + Y0 (−ẑ) × x̂ejkz 2 2 E0 jkz E0 −jkz − Y0 ŷe = Y0 ŷe 2 2 = −jY0 E0 ŷ sin kz

H(x,y,z) = Y0

28

B.3

Antennas

Exercise n. 1 Consider a horn antenna whose rectangular aperture has size A = 10 cm, B = 4 cm, operated at the frequency f =20GHz. Compute 1. the maximum gain G in natural units and in dB. 2. the full width of the main lobe between the zeros in the E and H plane 3. the direction and the level of the first sidelobe in the E and H plane

om

4. the magnitude of the radiated electric and magnetic field on axis, at the distance R = 5 km, when the horn has an input power Pin = 10 W.

.c

Solution 1. The geometrical area of the aperture is Ag = AB = 40 cm2 .

3. The wavelength is λ = c/f = 3. · 108 /20 = 1.5 cm

ru

m

2. The aperture efficiency of a rectangular horn is νa = 0.8, hence the effective area is Aeq = νa Ag = 0.8 · 40 = 32 cm2

ld

fo

4. The maximum gain is G = (4π/λ2 )Aeq = 179 = 22.5 dB. The directivity is the same: antennas with size comparable to wavelength have ohmic efficiency very close to one.

w .jn

tu

w

or

5. Compute the far field distance rf f = 2D2 /λ. The characteristic size D is the diagonal of the rectangle D = 10.77 cm and rf f = 1.54 m. Since the observation distance is R À rf f the concept of gain can be used. Hence, we compute the power density per unit surface in the observation point: Pin dPrad =G = 5.6977 · 10−6 4πR2 dΣ From this we compute the electric field via r dPrad = 65.5 mV/m |E| = 2Z0 dΣ

w w

and the magnetic field from the impedance relation |H| =

|E| = 173 µA/m Z0

¥

Exercise n. 2 Consider a radio link between a paraboloid and a half wavelength dipole antenna. The paraboloid has a diameter D = 1 m, is operated at the frequency f = 10 GHz, radiates a circularly polarized field and its input power is Pin = 0.1 W. The receiving dipole has ohmic efficiency η = 1, is located

29 x

x

α

α

y

x

z

z

y

(b)

Figure B.4.

om

(a) Radio link scheme.

m

.c

on axis at the distance R = 20 km and perpendicular to the link direction. However it is not vertical, but forms the angle α with the vertical, as shown in Fig. B.4. Compute the received power as a function of the angle α.

ru

Solution

fo

• The geometrical area of the paraboloid aperture is Ag = πD2 /4 = 0.7854 m2 .

ld

• The aperture efficiency of a paraboloid is νa = 0.6, hence the effective area is Aeq = νa Ag = 0.6 · 0.7854 = 0.4712 m2

or

• The wavelength is λ = c/f = 3. · 108 /10 = 3.0 cm

w

• The maximum gain is G = (4π/λ2 )Aeq = 6580 = 38.2 dB. The directivity is the same: antennas with size comparable to wavelength have ohmic efficiency very close to one.

w .jn

tu

• Compute the far field distance rf f = 2D2 /λ. The characteristic size D is the diameter of the aperture and rf f = 66.67 m. Since the observation distance is R À rf f the Friis equation can be used.

• The maximum gain of the dipole (equal to the directivity since the ohmic efficiency is η = 1) is G = D = 1.643

w w

• The polarization of the transmitting antenna is circular. The electric field lies in the x, y plane and the polarization vector can be written p̂T X = c(x̂ ± jŷ)

The double sign is necessary because the text does not specify whether the polarization is clockwise or counterclockwise. The constant c has the value that makes p̂T X a unit vector: |p̂T X |2 = p̂T X · p̂∗T X = c(x̂ ± jŷ) · c∗ (x̂ ∓ jŷ) = = |c|2 (x̂ · x̂ ∓ jx̂ · ŷ ± jŷ · x̂ + ŷ · ŷ) = |c|2 (1 + 1) = 2|c|2 Alternatively, |p̂T X |2 = |(p̂T X )x |2 + |(p̂T X )y |2 = |c|2 + |c|2 = 2|c|2

30 √ Hence c = 1/ 2 and

1 p̂T X = √ (x̂ ± jŷ) 2

• The polarization of the receiving antenna is, by definition, the polarization of the E field radiated in the direction of the paraboloid if the dipole were used as transmitting antenna.

The link direction coincides with the one of maximum radiation and the E field is parallel to the antenna. Hence p̂RX = ±(cos αx̂ − sin αŷ)

om

• Compute the polarization matching factor ¯2 ¯ ¯ ¯ 1 1 2 ¯ p = |p̂T X · p̂RX | = ¯± √ (cos α ∓ j sin α)¯¯ = 2 2

.c

We see that p is independent of α and has the constant value of 1/2: a 3 dB loss, with respect to the case of polarization matching.

m

• The available power at the output terminals of the receiving antennas is given by Friis formula

¥

ld

fo

ru

6580 · 1.643 1 GT X GRX 2 −12 W Pavail = Pin µ ¶2 |p̂T X · p̂RX | = 0.1 µ ¶2 = 7.7019 · 10 2 4πR 4π · 20 · 103 λ 3 · 10−2

w

or

Exercise n. 3 Consider a radio link between two identical paraboloids. The paraboloids have clockwise circular polarization. Compute the polarization matching factor.

tu

Solution

w .jn

Obviously, the result must be 1, since two identical antennas are polarization matched by definition.

The computation is, however, a little tricky.

Consider the transmitting antenna. Then 1 loc p̂T X = √ (x̂loc TX + jŷTX ) 2

w w

With reference to Fig. B.5, we realize that the previous phasor describes really a clockwise polarization if observed from a point in the far field of the TX antenna. The axes have been identified with a T X subscript and a loc (as local) superscript because they are to be considered as attached to the transmitting antenna.

Now take this antenna, together with the reference system, and rotate it 180◦ around the xloc T X axis in order to make it coincident with the receiving antenna. The reference system now is different from the original one and is indicated with the RX subscript. Clearly, the field radiated by the receiving antenna if it were connected to a generator would have still clockwise circular polarization, as observed by a point in the far field of this RX antenna and its expression would be 1 loc p̂RX = √ (x̂loc RX + jŷRX ) 2

31 loc xTX

xloc RX

loc yRX

ETX

Figure B.5.

loc zTX

Radio link scheme with two identical antennas.

om

loc yTX

Note that loc x̂loc RX = x̂TX

.c

loc loc ŷRX = −ŷTX

m

loc ẑloc RX = −ẑTX

fo

ru

Then we compute the polarization matching factor ¯2 ¯ ¯ ¯ 1 1 1 loc loc loc ¯ loc 2 ¯ p = |p̂T X · p̂RX | = ¯ √ (x̂T X + jŷT X ) · √ (x̂RX + jŷRX )¯ = (1 + j · j · (−1)) = 1 2 2 2

¥

or

ld

as it should be.

w .jn

tu

w

Exercise n. 4 A Hertzian dipole of length L = 2 m is fed by a transmission line with characteristic impedance Z∞ = 50 Ω. The dipole has an ohmic efficiency η = 0.8 and an input capacitance C = 30 pF. Suppose that this dipole is required to radiate a total power Prad = 100 W at the frequency f = 10 MHz. Compute: 1. the voltage that must be applied at the dipole input

w w

2. the reflection coefficient at the dipole input and the incident power on the line so that Prad = 100 W. Compute also the maximum and minimum voltage on the line and the VSWR. Solution

1. The wavelength is λ = 3 · 108 /107 = 30 m The radiation resistance is µ ¶2 2π L = 3.5 Ω Rrad = Z0 3 λ

The input resistance is Rin =

Rrad = 4.4 Ω η

32 The input impedance is Zin = Rin − j

1 = 4.4 − j 530.5 Ω 2πf C

The input current Ia is such that the radiated power is Prad =

Hence

r |Ia | =

1 Prad |Ia |2 2

2Prad = Rrad

r

2 · 100 = 7.55 A 3.5

om

The voltage Va to be applied is

2. The input reflection coefficient is

m

◦ Zin − Z∞ = 0.9985 e−j10.8 Zin + Z∞

ru

Γa =

.c

|Va | = |Zin ||Ia | = 4005 V

The input power is Prad /η = 125 W. The necessary incident power is

+ 2 1 |V | 2 Z∞ ,

fo

hence

or

On the other hand P + =

Pin = 40.3 kW 1 − |Γa |2

ld

P+ =

p 2Z∞ P + = 2008 V

w

|V + | =

tu

The maximum voltage is then

Vmax = |V + |(1 + |Γa |) = 4013 V

w .jn

and the minimum

Vmin = |V + |(1 − |Γa |) = 3.1 V

w w

The VSWR is

V SW R =

1 + |Γa | = 1291 1 − |Γa |

Obviously, using this antenna in these conditions is absolutely ridiculous and a matching device is to be used.

¥

Exercise n. 5 Find the sizes A, B of a rectangular horn with gain G = 19 dB at the frequency f = 12 GHz and

33 symmetrical main lobe, as measured between the zeros, i.e. with equal beamwidth in the E and H planes. Compute also the full beamwidth.

Solution The first zero in the E plane is located at ηy = π, from which λ B

sin ϑE z1 =

Similarly, the first zero in the H plane is located at ηx = 3π/2, from which

2A2 3

The maximum effective area is Aeq = νa Ag = νa

The maximum gain is

8πνa 4π Aeq = 3 λ2

from which we get

r

3 G 8πνa

ld

A = λ

µ ¶2 A λ

fo

G=

2A2 3

.c

Ag = AB =

m

E If ϑH z1 = ϑz1 , then B = 2A/3.

The geometrical area is

om

3λ 2A

ru

sin ϑH z1 =

The required gain is GdB = 19 dB, i.e.

or

G = 10GdB /10 = 79.43

w

We find then A = 3.44λ = 8.600 cm, since the wavelength is

3 · 108 c = 2.5 · 10−2 m = 12 · 109 f

tu

λ=

w .jn

Moreover B = 5.733 cm and

ϑE z1 = arcsin

µ

λ B

¶ = arcsin(0.436) = 25.852◦

w w

The full beamwidth (between the zeros) is H ◦ BW = 2ϑE z1 = 2ϑz1 = 51.704

¥

Exercise n. 6 Consider a paraboloid antenna with the following radiation pattern at the frequency f = 5 GHz: π g(ϑ,ϕ) = G0 cosα ϑ for ϑ ≤ 2 π =0 for ϑ > 2

34 The full half-power beam-width is FHPBW=10◦ .

Compute G0 and estimate the antenna diameter.

Solution First find α from the FHPBW. Set ϑ−3dB =FHPBW/2=5◦ : G0 = G0 cosα ϑ−3dB 2

Take logarithms

−3 = 10 log10 (cosα ϑ−3dB ) −3 = 181.18 10 log10 (cos ϑ−3dB )

Next recall the normalization condition Z 2π Z π2 1 g(ϑ,ϕ) sin ϑdϑdϕ = 1 4π 0 0

om

α=

.c

Hence

0

Z

π 2

g(ϑ,ϕ) sin ϑdϑdϕ =

0

Set u = cos ϑ G0 2

Z

0

Z

1 4π

Z

2π

0

G0 2

uα d(−u) =

1

π 2

ru

2π

G0 cosα ϑ sin ϑdϑdϕ =

0

Z 0

1

G0 2π 4π

Z

π 2

cosα ϑ sin ϑdϑ

0

fo

Z

uα du =

¯1 G0 u(α+1) ¯¯ G0 = 2(α + 1) 2 α + 1 ¯0

ld

1 4π

m

Compute the integral, noting that g(ϑ,ϕ) does not depend on ϕ, so that the ϕ integral equals 2π:

or

In conclusion, the normalization condition yields

w

G0 = 2(α + 1) = 364.36

w w

w .jn

tu

that is (G0 )dB = 25.61 dB. To find the diameter of the antenna, compute the maximum effective area λ2 G0 = 1043.8 cm2 Aeq = 4π since λ = c/f = 6 cm. The aperture efficiency of paraboloids is νa = 0.6, so

Ag =

Aeq = 1739.7 cm2 νa

and the required antenna diameter is r D=

4Ag = 47.06 cm π

Fig. B.6 shows the radiation pattern in cartesian coordinates, Fig. B.7 shows the same pattern in polar coordinates.

35

400

350

300

250

200

om

150

100

−80

−60

−40

−20

0 ϑ

20

40

60

80

Figure B.6.

or

ld

fo

ru

Radiation pattern in cartesian coordinates

w

90

400 60

120

tu

300

150

30

w .jn

200

100

180

w w

100

m

0 −100

.c

50

0

210

330

240

300 270

Figure B.7.

Polar radiation pattern

36

B.4

Waveguides

Exercise n. 1 Find the dimensions of a rectangular waveguide with the following specifications: .

1. it is single mode in the band between fmin = 10 GHz and fmax = 15 GHz 2. the attenuation of the first higher order mode must be at least αlim = 8 dB/cm for the whole band [fmin ,fmax ] 3. dispersion must be minimized

om

4. the waveguide must carry up to PT = 150 kW without discharge when it is matched If the waveguide designed above is connected to a load with reflection coefficient |ΓL | = 0.2, what is the maximum power that the guide can carry without discharge?

from which we find

c

≤a≤

c

fmax

ld

2fmin

fo

ru

m

.c

Solution Certainly we choose b < a/2 so that the first higher order mode is T E20 . Next find a.

Enforce c fmin ≥ fc10 = 2a c fmax ≤ fc20 = a

that is

or

1.5 ≤ a ≤ 2 cm

w

Compute the attenuation of mode T E20 :

w .jn

It is minimum at fmax

tu

α20

q 2 − k2 = k = kt20

α20min

2πfmax = c

sµ

sµ

fc20 f

fc20 fmax

¶2 −1

¶2 −1

w w

By requiring α20min ≥ αlim , we find a constraint on a (contained in fc20 )

sµ

a≤

fmax

c

cαlim 2πfmax

= 1.919 cm

¶2 +1

Note that the value of αlim given in dB must be converted in Neper, by diving it by 8.6859.

Dispersion is minimized if a = 1.919 cm is chosen. This value is inside the range 1.5 ≤ a ≤ 2 cm, hence it is acceptable. Then the maximum value for b is bmax = a/2 = 0.960 cm Now find b on the basis of the power to be transmitted PT . The active power flow on the T E10 modal line is 1 |V + |2 Pwg = 2 Zt10

37

because the modal line is matched.

The maximum value of the electric field is reached at x = a/2, so that r 2 + V Emax = ab Substituting in the previous expression, we find that the power in the guide for a given value of Emax is 2 1 ab Emax Pwg = 2 2 Zt10

om

Now, setting Emax = R = 20 kV/cm the dielectric rigidity of air, we obtain the power at the onset of discharge Pdisch : 1 ab R2 Pdisch = 2 2 Zt10

ru

R2 1 ab ≥ PT 2 2 Zt10 (fmin )

m

.c

Note that this power is a function of frequency because Zt10 is. At f = fmin the modal impedance is maximum, so Pdisch is minimum: this is the most critical situation. If the condition Pdisch (fmin ) ≥ PT is satisfied, Pdisch (f ) ≥ PT for all f . Hence enforce

from which we find

fo

4Zt10 (fmin )PT = 0.472 cm aR2 In conclusion we find 0.472 ≤ b ≤ 0.960 cm; any value of b in this range is acceptable.

Suppose b = 0.472 cm is chosen. Then the discharge power is exactly Pdisch = PT = 150 kW. If the guide is mismatched, the discharge power is reduced by the factor 1/S,

or

ld

b≥

w

Pdisch =

w .jn

tu

with

S=

1 ab R2 1 2 2 Zt10 S

1 + |ΓL | = 1.5 1 − |ΓL |

w w

so the discharge power is Pdisch = 150/1.5 = 100 kW. In conclusion, the guide previously designed can carry only a maximum power of 100 kW, in mismatched conditions, if discharge is to be avoided. ¥

Exercise n. 2 The rectangular waveguide WR90 of Fig. B.8 is connected to a load that has a reflection coefficient ΓL = 0.4 exp(−j45◦ ) for the fundamental mode T E10 , at the frequency f = 12 GHz. Design a small piece of dielectric to be inserted at a convenient position in the guide so that matching is obtained. Then draw the voltage plot |V (z)| with V inc = 1 V. Solution Compute first the parameters of the modal lines. The waveguide is WR90, with a = 0.9 in=0.9 ·

38

εr A

kz 0 , Z ∞ 0

B

ΓL

k zd , Z ∞d

ΓL

kz 0 , Z ∞ 0

A

C

om

B

2.54 cm. and fc0 = c/(2a) = 6.56 GHz. Recall

1−

kz0 = k0

fc0 f

s

¶2

kzd = k0

Z0 ¶2 µ fc0 1− f

µ

ru

µ

εr −

fc0 f

¶2

Z0 ¶2 µ fc0 εr − f

fo

s

m

.c

Figure B.8. Matching of a load by means of a piece of dielectric and equivalent modal circuit for the T E10 mode

Z∞d = s

or

ld

Z∞0 = s

w

which yield

λg0 = 2.986 cm

tu

kz0 = 2.1043 rad/cm Z∞0 = 450.2789 Ω

Z∞d =

p RA RB

w w

w .jn

The matching device is a λ/4 line. It must be positioned in such a way that ZB is real ZB = RB .

Then the characteristic impedance of the line is computed according to

where RA = Z∞0 , the desired input impedance. To find RB draw on the Smith chart the circle through ΓL with center in the origin. The intersections with the real axis are rm =

1 − |ΓL | 1 + |ΓL |

and

rM =

1 1 + |ΓL | = rm 1 − |ΓL |

Which one of the two must be selected? Let r be one of the two. Then √ Z∞d = Z∞0 r

39

om

From the equations above, Z∞d ≤ Z∞0 , hence r ≤ 1 so the point rm must be chosen. Now we can compute εr . √ Z∞d = Z∞0 r √ Z0 Z0 s ¶2 rm ¶2 = s µ µ fc0 fc0 1− εr − f f s s ¶2 ¶2 µ µ 1 fc0 fc0 =√ 1− εr − f rm f à ¶2 ¶2 ! µ µ fc0 fc0 1 + 1− εr = f f rm

fo

µ ¶T L ` = 0. λ B+

m

This is exactly the value of LBC /λg0 , since

ru

µ ¶T L ` = 0.1875 λ C

.c

We find rm = 0.4286 and εr = 1.9347. The phase of ΓC is −45◦ : on the Smith chart, we read

ld

Hence

or

LBC = 0.1875λg0 = 0.5599 cm

w

Moreover λgd = 1.9548 cm and LAB = λgd /4 = 0.4887 cm. If we choose rM instead of rm , we obtain εr ≤ 1, which is not realizable.

w w

Next

w .jn

tu

In order to draw the voltage plot, i.e. the electric field plot in the waveguide, we mark some points on the Smith chart of Fig. B.9. From ΓL turn clockwise (Toward Generator) up to zB + = rm = 0.4286. Then Z∞0 = 0.6547 zB − = zB + Z∞d

zA+ =

1

zB −

and

zA− = zA+

= 1.5275 Z∞d =1 Z∞0

as it should be.

To draw the plot, note that zA− = 1 implies that |V (z)| is flat to the left of A since there is no backward wave. Moreover in A+ there is a maximum of voltage and in B − a minimum, since the corresponding points are on the positive and negative part of the real axis, respectively. Then VA− = 1 V. The VSWR on AB is S=

1 + |ΓB − | 1 + |ΓA+ | = 1.5275 = 1 − |ΓB − | 1 − |ΓA+ |

40

zB−

z

+

=r

m

m

.c

om

B

rM

zA−

Smith Chart plot relative to the design of the matching device and to the voltage plot

ld

Figure B.9.

fo

ru

ΓL

zA+

w

or

This number, of course, coincides with zA+ . Then

Vmax = 0.6547 S

tu

VB − = Vmin =

w w

w .jn

Of course VB − = VB + . To draw the part relative to BC, note that VC = VC+ (1 + ΓL ) = VB++ e−jkz0 LBC (1 + ΓL ) 1 + ΓB − VB++ = VB+− 1 + ΓB + + + −jkzd LAB = −jVA++ VB − = VA + e 1 + Γ A− VA++ = VA+− 1 + ΓA+

We find |VC | = 1.4333 V. Alternatively and more rapidly, we can use the conservation of energy to compute |V (z)|. Since the structure is lossless, the net power is the same everywhere in the circuit. In general

1 1 |V inc |2 = |V (z)|2 R{Y (z)} 2 2 Z∞0

41 The left hand side is the net power in A− , because the reflected power is zero. In particular

|V inc |2 |V inc |2 = R{yL } Z∞0 R{YL } inc 2 |V inc |2 |V inc |2 |V | = = |VB |2 = Z∞0 Y∞d R{yB − } R{yB + } Z∞0 R{YB } |VC |2 =

The resulting plot is shown in Fig. B.10 |V(z)| 1.5

om

C

1.4

.c

1.3

1.1

A

ru

1

fo

0.9

ld

0.8

−2

−1.5

w

−2.5

or

0.7

−3

m

1.2

−1

−0.5 z (cm)

B

0.5

1

1.5

Voltage plot

¥

w w

w .jn

tu

Figure B.10.

0

Exercise n. 3 Consider the waveguide structure of Fig. B.11. It consists of a semi-infinite rectangular waveguide, short circuited on the left and infinite on the right, excited by a dipole antenna in B. The guide is empty up to C and then it is completely filled with a dielectric with relative permittivity εr = 4.

The guide cross section has dimensions a = 5 cm, b = 2 cm. The frequency of the source is f = 4 GHz. Moreover LAB = 2.5 cm and LBC = 2 cm. The dipole can be represented on the fundamental mode equivalent line by a current generator is = 0.01 A. Compute the power that the source radiates beyond C and select LAB so as to maximize it.

Solution Compute first the parameters of the modal lines. The waveguide has a = 5 cm, hence the critical

42 frequency of the T E10 mode is fc0 = c/(2a) = 3 GHz. Recall s

µ 1−

kz0 = k0

fc0 f

s

¶2 kzd = k0

µ

εr −

fc0 f

¶2

Z0 ¶2 µ fc0 εr − f

Z0 ¶2 µ fc0 1− f

Z∞d = s

Z∞0 = s

.c

λg0 = 0.1134 m kz0 = 55.4125 rad/m kzd = 155.3245 rad/m λgd = 0.0405 m Z∞0 = 569.9704 Ω Z∞d = 203.3387 Ω

ru

C

B

C

tu

Waveguide structure with a dipole source and modal circuit for the fundamental mode

w w

w .jn

Figure B.11.

k zd , Z ∞d

w

A

kz 0 , Z ∞ 0

or

is

ld

fo

B

m

εr A

om

which yield

VB

ZB

−

is

ZB

+

B− B+ Figure B.12.

Loads seen by the current generator

Then compute the load seen by the generator. Start with that looking to the right. The line to the right of C is infinitely long, hence ZC = Z∞d . Compute zC − =

Z∞d = 0.3568 Z∞0

43 zC − − 1 = −0.4741 zC − + 1

ΓC − =

ΓB + = ΓC − e−j2kz0 LBC = 0.4741ej53.0039 zB + =

and finally

◦

1 + ΓB + = 1.1850 + j1.1577 1 − ΓB +

→ − Z B + = zB + Z∞0 = 675.43 + j659.83 Ω

These values have been obtained by means of a computer, but can be obtained, with less accuracy, by the Smith chart. Similarly for the load looking to the left.

m

and finally

.c

◦ ← − ← − Γ B − = Γ A e−j2kz0 LAB = ej21.2549 ← − 1 + Γ B− ← − z B− = = j5.3293 ← − 1 − Γ B−

om

← − Γ A = −1

ru

← − − Z B− = ← z B − Z∞0 = j3037.6 Ω

fo

The original modal circuit can be lumped into the one of Fig. B.12. The voltage at the generator terminals is is = −4.3934 − j6.2010 V (B.1) VB − = VB + = − ← − − → Y B− + Y B+

or

ld

In order to compute the power radiated beyond C we can apply the principle of energy conservation since the structure is lossless. Then

− → 1 |V + |2 R{ Y B + } = 0.0219 W 2 B

w

PC = PB + =

w .jn

tu

→ − since Y B + = (7.5756 − j7.4007) · 10−4 S. If we want a more detailed computation, that yields also the phase of voltage and current, we can proceed step by step across the discontinuities: VB++ =

VB + = −4.4530 − j3.5127 V 1 + ΓB +

w w

VC+− = VB++ e−jkz0 LBC = −5.1307 + j2.4176 V VC++ = VC+−

1 + ΓC − = −2.6982 + j1.2714 V 1 + ΓC +

where ΓC + = 0, of course. We can check that PC =

1 |VC++ | = 0.0219 W 2 Z∞d

exactly as found above by the energy method.

If we compute the power radiated to the left, we should find zero, since the waveguide is shorted in A. Indeed ← ← − − 1 P B − = |VB−− |R{ Y B − } = 0 2

44 ← − because Y B − is pure imaginary.

If we want to maximize the radiated power, it is necessary to maximize the magnitude of VB .

Since the numerator of (B.1) is fixed, we must minimize the magnitude of the denominator. Note − → ← − that Y B + is fixed and that Y B − is pure imaginary and its value depends on the length LAB . → Clearly, in these conditions, the denominator has minimum amplitude if it is real. Now, − y B+ = ← − 0.4318 − j0.4218, so that we must enforce y B − = j0.4218, which requires LAB /λg0 = 0.3135 and LAB = 3.5551 cm.

w w

w .jn

tu

w

or

ld

fo

ru

m

.c

om

In ordinary applications, the waveguide is matched when looking to the right of the source, so − − that → y B + = 1 and the optimum situation is reached if ← y B − = 0, which implies LAB = λg0 /4.

45

w w

om

.c

m

ru

fo

ld

or

w

tu

w .jn