CONTENTS 1. 1.1 1.2 1.3
INTRODUCTION .............................................................................................. Description of an electromagnetic field ………………….….............................. Wave equation ………………...........................................................…............... Potentials ………….........................................................................….................
6 8 13 14
2. 2.1 2.2 2.3 2.4 2.5
ELECTROMAGNETIC WAVES IN FREE SPACE .................................… Solution of the wave equation ....................................…..................................... Propagation of a plane electromagnetic wave …………………………………. Wave polarization ……………………………………………………………… Cylindrical and spherical waves ……………………………………………….. Problems ………………………………………………………………………..
17 17 18 28 32 34
3. 3.1 3.2 3.3 3.4
WAVES ON A PLANE BOUNDARY ………………………………………. Perpendicular incidence of a plane wave to a plane boundary ………………… Perpendicular incidence of a plane wave to a layered medium ………………... Oblique incidence of a plane electromagnetic wave to a plane boundary ……... Problems ………………………………………………………………………..
35 35 42 45 54
4.
SOLUTION OF MAXWELL EQUATIONS AT VERY HIGH FREQUENCIES ………………………………………………………………. 56
5.
GUIDED WAVES …………………………………………………………….. 61
6. 6.1 6.2 6.2.1 6.2.2 6.2.3 6.3 6.4
TEM WAVES ON A TRANSMISSION LINE ………………………………. 65 Parameters of a TEM wave …………………………………………………….. 65 Transformation of the impedance along the line ………………………………. 71 An infinitely long line ………………………………………………………….. 71 A line of finite length …………………………………………………………… 72 A line terminated by a short cut and by an open end ……….………………….. 73 Smith chart ……………………………………………………………………… 76 Problems ……………………………………………...………………………… 87
7. 7.1 7.2 7.3 7.4
WAVEGUIDES WITH METALLIC WALLS ……………………………... Parallel plate waveguide ……………………………………………………….. Waveguide with a rectangular cross-section …………………………………… Waveguide with a circular cross-section ………………………………………. Problems ………………………………………………………………………...
88 88 94 103 107
8. 8.1 8.2 8.3
DIELECTRIC WAVEGUIDES ……………………………………………… Dielectric layers ………………………………………………………………… Dielectric cylinders ……………………………………………………………... Problems ………………………………………………………………………...
108 109 114 115
9. 9.1 9.2
RESONATORS ……………………………………………………………….. 116 Cavity resonators ……………………………………………………………….. 116 Problems ………………………………………………………………………... 120
10. 10.1 10.2 10.3 10.4 10.5 10.6
RADIATION ………………………………………………………………….. Elementary electric dipoles …………………………………………………….. Elementary magnetic dipoles …………………………………………………... Radiation of sources with dimensions comparable with the wavelength ………. Antenna parameters …………………………………………………………….. Antenna arrays ………………………………………………………………….. Receiving antennas ……………………………………………………………...
121 121 126 129 133 134 137
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10.7
Problems ……………………………………………………………………….. 140
11. 11.1 11.2
11.4 11.5
WAVE PROPAGATION IN NON-ISOTROPIC MEDIA ………………… Tensor of permeability of a magnetized ferrite ………………………………… Longitudinal propagation of a plane electromagnetic wave in a magnetized ferrite …………………………………………………………………………… Transversal propagation of a plane electromagnetic wave in a magnetized ferrite …………………………………………………………………………… Applications of non-reciprocal devices ………………………………………… Problems ………………………………………………………………………...
12. 12.1 12.2 12.2.1 12.2.2 12.2.3 12.2.4 12.3 12.4 12.4.1 12.4.2 12.4.3 12.4.4
APPLICATIONS OF ELECTROMAGNETIC FIELDS ………………….. 156 Introduction to microwave technology …………………………………………. 156 Antennas ………………………………………………………………………... 162 Wire antennas …………………………………………………………………... 164 Aperture antennas ………………………………………………………………. 165 Broadband antennas …………………………………………………………….. 167 Planar antennas …………………………………………………………………. 168 Propagation of electromagnetic waves in the atmosphere ……………………… 169 Optoelectronic ………………………………………………………………….. 173 Optical waveguides …………………………………………………………….. 173 Optical detectors ………………………………………………………………. 176 Optical amplifiers and sources …………………………………………………. 177 Optical modulators and sensors ………………………………………………… 178
13.
MATHEMATICAL APPENDIX …………………………………………….. 180
14.
BASIC PROBLEMS ………………………………………………………….. 191
15
LIST OF RECOMMENDED LITERATURE ………………………………. 194
11.3
141 142 146 151 154 155
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1. INTRODUCTION This textbook is aimed at students of the Faculty of Electrical Engineering, Czech Technical University taking a course in Waves and Transmission Lines. The textbook builds on basic knowledge of time varying electromagnetic fields gained from courses in physics and electromagnetic field theory. The textbook introduces all the basic knowledge that an electrical engineer specializing in radio engineering and telecommunications should have and that is necessary for further courses such as microwave engineering, antennas and propagation, optical communications, etc. Sequential mastering of wave theory contributes to the final objective of university studies, which is to enable graduates to do creative work. The course in Waves and Transmission Lines studies the theory and applications of classical electrodynamics. It is based on Maxwell’s equations. This course provides a basis for understanding the behavior of all high frequency electric circuits and transmission lines, starting from those applied to transmitting and receiving electric energy, processing signals, microwave circuits, optical fibers, and antennas. The main applications lie in wireless communications, radio engineering and optical systems. The text follows the classical approach to macroscopic electrodynamics. All quantities are assumed to be averaged over the material, which by its nature has a microscopic structure consisting of atoms. This confines the description of electromagnetic effects using macroscopic theory on the high frequency side, as the wavelength must be much longer than the dimensions of the atoms and molecules. This boundary lies in the range of ultraviolet light. Nevertheless, the spectrum of frequencies in which electromagnetic effects can be treated using this macroscopic theory is really huge – over 17 orders. And this whole spectrum really is used in a variety of different applications. Modern communication systems use electromagnetic waves with ever shorter wavelengths. The spectrum of electromagnetic waves is shown in Fig. 1.1. First, we review the basic relations from electromagnetic field theory and introduce potentials describing a time varying electromagnetic field. The concept of a plane electromagnetic wave is carefully reviewed. In addition, a cylindrical wave together with a spherical wave are briefly introduced. The behaviour of a plane electromagnetic wave on the boundary between two different materials is studied in detail. Here we will start treating the incidence of a wave perpendicular to the plane boundary and to a layered medium. Oblique incidence is studied in general, and then special effects such as total transmission and total reflection are treated. Specific aspects of solving Maxwell equations at very high frequencies are discussed separately. Transmission lines are designed to transmit guided waves. After introducing the general properties of guided waves the TEM wave is treated, and the transformation of impedances along a line is described. The Smith chart, a very effective graphical tool for analysis and design of high frequency circuits, is described and its basic applications are explained through particular problems. Waves propagating along waveguides with metallic walls of rectangular and circular cross-sections are studied. Dielectric waveguides are treated separately. They form the basis of optical fibers. Cavity resonators, unlike low frequency L-C resonant circuits, are able to resonate on an infinite row of resonant frequencies. Several kinds of such resonators are analyzed in the text. Attention is paid to problems of radiation of electromagnetic waves. This covers antenna theory. Elementary sources of an electromagnetic field, such as an electric dipole and a magnetic dipole, are studied and compared. Then radiation from sources with dimensions comparable with the wavelength is described. The basic antenna parameters are defined. The basic idea of antenna arrays is built up. Finally, receiving antennas are dealt with, and the effective antenna length and effective antenna surface are derived.
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Non-isotropic materials are introduced, and the tensor of permeability of a magnetized ferrite is calculated. The propagation of a plane electromagnetic wave in this ferrite material homogeneously filling an unbounded space is studied, in particular when the wave propagates both in the direction parallel with the magnetizing field and in the direction perpendicular to the magnetizing field. Some devices using non-isotropic materials are mentioned. frequency (Hz)
wavelength (m)
applications
classification
1018 10-9
1015 10-6
ultraviolet radiation
nm 360
violet 460 blue green 560 yellow 660 red
visible light infrared radiation
1012 -3
10
quasi optical waves
decimeter waves
m i radar, space investigation c r o w radar, satellite commun. a v e s radar, TV, navigation
very short waves
TV, FM radio, services
short waves
radio, services
medium waves
AM radio
millimeter waves centimeter waves
109 0
He-Ne laser 630 nm
760
10
-2
argon laser 490 nm
10 10 106
102 103 104
long waves
103
phone, audio
106
The spectrum of electromagnetic waves. Fig. 1.1 The applications of electromagnetic fields in particular branches of electrical engineering are briefly introduced. An introduction to microwave technology is given. Here scattering parameters are introduced. The paragraph on antennas represents a continuation of Chapter 10, introducing basic types of antennas. Particular mechanisms of wave propagation in the atmosphere are explained. The transmission formula and radar equations are derived. The basics of optoelectronics are presented. This involves a characterization of optical waveguides, detection and optical detectors, optical amplifiers, and the sources of optical
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radiation, namely lasers, and finally modulators of an optical beam and a short review of optical fiber sensors. The book has a mathematical appendix summarizing the necessary knowledge of mathematics. Basic problems in the form of questions are summarized at the end of the text. They help the students in preparing for their examinations. A list of suggested literature is given. The textbook treats time varying electromagnetic fields. Harmonic dependence on time is assumed throughout most of the text. Such fields are described using symbolic complex quantities called phasors, and time dependence in the form e jωt is assumed. These phasors are not marked by special symbols. When we need to emphasize an instantaneous value, we will mark it by showing dependence on time, e.g., E(t). Vectors will be marked by bold characters, e.g., E.
1.1 Description of an electromagnetic field The sources of an electromagnetic field are electric charge Q [C] and electric current I [A], which is nothing else than the flow of charge. Charge is often distributed continuously in a space, on a surface, or along a curve. It is convenient in this case to define the corresponding charge densities. The charge volume density is defined as the charge amount stored in a unit volume ∆q [C/m3] . ∆V →0 ∆V
ρ = lim
(1.1)
The charge surface density is defined by analogy ∆q [C/m2] . ∆S →0 ∆S
σ = lim
(1.2)
The charge linear density is defined as the charge stored along a line or a curve of unit length ∆q [C/m] , ∆l →0 ∆l
τ = lim
(1.3)
Electric current is created by a moving charge. This is defined as the passing charge per time interval ∆t ∆q [A=C/s] . ∆t → 0 ∆t
I = lim
(1.4)
It is useful to define current densities, which are vector quantities, as it is necessary to define the current flow direction. Current density is defined as ∆I i 0 [A/m2] , ∆S → 0 ∆S
J = lim
(1.5)
where i0 is a unit vector describing the current flow direction. It is sometimes useful to use the abstraction of a surface current passing along a surface, see Fig. 1.2. The linear density of
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this current is defined as ∆I i 0 [A/m] . ∆x → 0 ∆x
K = lim
(1.6)
The surface current and its density are an abstraction used to simplify the mathematical I i0 description of the current passing a conductor at a K high frequency. Due to the skin effect, current S flows through only a very thin layer under the ∆x material surface. In fact the surface current defined by (1.6) represents the finite current Fig. 1.2 passing a cross-section of zero value. This requires infinite material conductivity. Another widely used abstraction is a current filament representing a conductor of negligible crosssection (e.g., a line) carrying a finite current. The total electric current crossing a closed surface is related to the charge accumulated inside the volume surrounded by this surface by the continuity equation in an integral form
∫∫ J ⋅ dS + jωQ = 0 ,
(1.7)
S
or in a differential form div J + jω ρ = 0 .
(1.8)
where Q is the total charge accumulated in volume V with boundary S. The current density is related by Ohm’s law in the differential form to an electric field J=σE ,
(1. 9)
where σ is conductivity in S/m. It should be noted that in spite of the movement of the free electrons, a conductor passed by an electric current stays electrically neutral, as the charge of the electrons is compensated by the positive charge of the charged atomic lattice. The electric field is described by the vector of electric field intensity E, the unit of which is V/m. The magnetic field is described by the vector of magnetic field intensity H, the unit of which is A/m. These vectors are related to the induction vectors by material relations B=µH ,
(1.10)
D=εE ,
(1.11)
where ε = ε 0ε r and µ = µ 0 µ r are permittivity and permeability, respectively, and
ε 0 = 10−9 / 36π F/m and µ 0 = 4π 10−7 H/m are the permittivity and permeability of a vacuum, respectively. Vectors E and H are related by the set of Maxwell’s equations. Their differential form reads rot H = (σ + jωε )E + J S ,
(1.12)
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rot E = − jωµ H ,
(1.13)
div(ε E ) = ρ0 ,
(1.14)
div(µ H ) = 0 ,
(1.15)
where JS is a current supplied by an external independent source, ρ0 is the volume density of a free charge supplied by an external independent source. The differential form of Maxwell’s equations is valid only at those points where the field quantities n are continuous and are continuously differentiated functions of 1 ε1 µ1 σ1 position. They are not valid, for example, on a boundary between two different materials where the material parameters change ε2 µ2 σ2 step-wise, Fig. 1.3. For this reason we have to append 2 corresponding boundary conditions to these equations. We suppose that the boundary between two materials contains a free Fig. 1.3 electric charge with density σ0, and electric current K passing along this boundary. The boundary conditions in vector form can be expressed n ⋅ (ε1E1 − ε 2E 2 ) = σ 0 , n × (E1 − E 2 ) = 0 , n ⋅ (µ1H1 − µ 2 H 2 ) = 0 ,
(1.16)
n × (H1 − H 2 ) = K , n ⋅ (J1 − J 2 ) = 0 , n is the unit vector normal to the boundary. A scalar form using the normal and tangential components of vectors is
ε1En1 − ε 2 En 2 = σ ,
(1.17)
Et1 = Et 2 ,
(1.18)
µ1H n1 = µ 2 H n 2 ,
(1.19)
H t1 − H t 2 = K ,
(1.20)
J n1 = J n 2 .
(1.21)
Specially on the surface of an ideal conductor with conductivity σ → ∞ , and since the electric and magnetic fields are zero inside this conductor, we have
ε1En1 = σ 0 , Et1 = 0 , H n1 = 0 , H t1 = K .
(1.22)
The first Maxwell equation (1.12) has three terms on its right hand side. Term jωE
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represents the displacement current, term σE represents the conducting current which causes conducting losses in the material, see (1.9), and JS is the current supplied by an internal source. Equation (1.12) can be simplified by introducing a complex permittivity σ E + J S = jωε 0ε r (1 − j tg δ e ) E + J S = rot H = jωε 0ε r E + σE + J S = jωε 0 ε r − j ωε 0 jωε 0ε c E + J S where complex permittivity εc is defined
εc = εr − j
σ = ε r (1 − j tg δ e ) = ε '− jε ' ' , ωε 0
(1.23)
and term tg δ e =
σ , ωε 0ε r
(1.24)
is called the loss factor, or thea loss tangent, and δe is the loss angle. This loss factor is frequently used to define the losses in a material in spite of the fact that it is frequency dependent. The reason is that this loss factor, similarly as the imaginary part of permittivity ε’’, contains in practice not only the conducting losses, but also polarization and another kinds of losses. Similarly we can introduce complex permeability µc representing all kinds of magnetic losses
µ c = µ r (1 − j tg δ m ) = µ '− jµ ' ' .
(1.25)
In material relations (1.9) – (1.11) conductivity σ , permittivity ε and permeability µ represent the electric and magnetic properties of a material. In a linear material these parameters do not depend on field quantities, while in a nonlinear material they depend on E or H. These parameters can depend on space coordinates, which is the case of a nonhomogeneous material. In a homogeneous material σ, ε, and µ do not depend on coordinates. An isotropic material has parameters that are constant in all directions, ε, µ, σ are scalar quantities. Non-isotropic materials possess different behaviour in different directions. Their permittivity, permeability, or conductivity are tensor quantities that can be expressed by matrices. An example is provided by magnetized ferrite or magnetized plasma. E.g., equation (1.11) for magnetized plasma can be rewritten into D=ε E ,
(1.26)
which gives three particular scalar equations Dx = ε xx E x + ε xy E y + ε xz E z , D y = ε yx E x + ε yy E y + ε yz E z ,
(1.27)
Dz = ε zx E x + ε zy E y + ε zz E z .
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Vector D has a different direction from vector E. Permittivity is then not a scalar quantity, but a tensor quantity ε xx ε = ε yx ε zx
ε xy ε xz ε yy ε yz . ε zy ε zz
(1.28)
Similarly we can express the permeability and conductivity of an anisotropic material. The density of power transmitted by an electromagnetic wave is described by the complex Poynting vector. This vector is defined as S=
1 E × H * = S av + jQ , 2
(1.29)
where Sav is the average value of the transmitted power, which represents the density of active power S av =
[
]
1 Re E × H * . 2
(1.30)
Q is the density of reactive power Q=
[
]
1 Im E × H * . 2
(1.31)
Poynting’s theorem represents the balance of power in an electromagnetic system in volume V. It can be read by dividing power into active and reactive power PS = PJ + PR ,
(1.32)
(
)
QS = 2 ω Wmav − We av + QR ,
(1.33)
where PS and QS are the average values of the active and reactive power supplied by an external source. The active power supplied by an external source is PS =
{
}
1 Re E ⋅ J *S dV , 2 ∫∫∫ V
(1.34)
JS is the current supplied by this source. The active power is partly lost in materials, power PJ, and partly radiated outside of our volume, power PR. These quantities are PJ =
1 2 σ E dV , ∫∫∫ 2 V
(1.35)
{(
)}
PR = ∫∫ S av ⋅ dS = ∫∫ Re E × H * ⋅ dS , S
S
(1.36)
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where S is the surface surrounding investigated volume V. Similarly, for reactive power we get QS = −
{
}
1 Im E ⋅ J *S dV . ∫∫∫ 2 V
(1.37)
This power covers the difference between the average values of the energy of an electric field, Weav, and the energy of a magnetic field, Wmav, and is partly radiated outside, QR. These values are
{
}
QR =
1 Im E × H * ⋅ dS , 2 ∫∫ S
Wmav
µ H = ∫∫∫ 4 V
2
dV ,
(1.38)
Weav
ε E 2 dV . = ∫∫∫ 4 V
(1.39)
1.2 Wave equation The solution of an electromagnetic field by Maxwell’s equations (1.12) and (1.13) involves solving six scalar equations for three components of the electric field and for three components of the magnetic field. We can reduce this number of equations by extracting one of the two vectors from (1.12) and (1.13), as vectors E and H are mutually dependent. Let us start with equation (1.12). Let us apply operator rot to this equation rot rot H = (σ + jωε ) rot E + rot J S . Inserting for doubly applied operator rotation rotrotH = grad div H − ∆H (13.81) we get grad div H − ∆H = (σ + jωε ) rot E + rot J S . rotE can be expressed from Maxwell’s second equation (1.13) and Maxwell’s fourth equation (1.15) states that divH=0. Now we have − ∆H = − j ω µ (σ + jωε )H + rot J S . The term containing material parameters and a frequency in front of H on the right hand side is usually denoted k 2 = − j ω µ (σ + jωε ) = ω 2 µε − jωµσ ,
(1.40)
where k is a complex propagation constant. Using this simplification we get the wave equation for the vector of magnetic field intensity in the form ∆H + k 2 H = −rot J S .
(1.41)
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The derivation of the equation for E is similar. Let us apply operator rotation to (1.13), and we get rot rot E = − jωµ rot H .
Using (13.81), expressing divE from Maxwell’s third equation (1.14), and using the propagation constant k (1.40) we get ∆E + k 2 E = grad
ρ0 + jωµ J S . ε
(1.42)
Wave equations (1.41) and (1.42) are basic equations which describe an electromagnetic field in a space with general sources supplying current JS and charge ρ0. These equations are, however, not suitable for solving the field in areas with sources, as their right hand sides are rather complicated functions of source quantities JS and ρ0. Equations for potentials are preferably applied to solve these problems, which, e.g., involve radiation of antennas.
1.3 Potentials Stationary or static fields are described by simplified Maxwell’s equations rot E = 0 ,
(1.43)
div B = 0 .
(1.44)
The potentials for these fields can then be simply defined. The scalar potential is defined by E = − grad ϕ ,
(1.45)
and the vector potential by B = rot A .
(1.46)
These potentials are solutions of Poisson’s equations
ρ0 , ε
(1.47)
∆A = − µJ S .
(1.48)
∆ϕ = −
Equation (1.48) holds, supposing that the Lorentz calibration condition in the form div A = 0
(1.49)
is laid to the vector potential. The solution of these equations can be obtained by applying the method of superposition
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1
ϕ=
4πε ∫∫∫
A=
µ 4π
∫∫∫
ρ0
dV ,
(1.50)
JS dV . r
(1.51)
r
In a non-stationary field we have again equation (1.44) and the vector potential is, therefore, defined by (1.46). An electric field is a different story. It is now described instead of (1.43) by Maxwell’s equation (1.13), which can be rewritten into rot E = − jωB = − jω rot A ,
rot (E + jωA ) = 0 . This last equation tells us that the rotation of the vector in brackets equals zero, consequently this vector can be expressed by a scalar potential. The electric field is now expressed by both the scalar potential and the vector potential E = −grad ϕ − jω A .
(1.52)
Let us now derive equations analogous to (1.47) and (1.48) which describe the distribution of the vector and scalar potentials of a time varying electromagnetic field. First we insert (1.46) and (1.52) into Maxwell’s first equation (1.12) and we get rot rot A = ( jωµε + µσ )(− grad ϕ − jωA ) + µ J S ,
(
)
grad divA + ∆A = −( jωµε + µσ ) grad ϕ + A ω 2 µε − jωµσ + µ J S , ∆A + k 2 A = grad[divA + ( jωµε + µσ )ϕ ] − µ J S . This equation can be simplified supposing that the argument of the gradient is zero, which is the Lorentz calibration condition for a time varying electromagnetic field div A = −( jωµε + µσ )ϕ .
(1.53)
Finally the wave equation for the vector potential reads ∆A + k 2 A = − µJ S .
(1.54)
The equation for the scalar potential is derived in a similar way. We insert (1.52) into Maxwell’s third equation (1.14) − div grad ϕ − jω div A =
ρ0 . ε
div A is expressed by calibration condition (1.53), and we have
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∆ϕ − jω ( jωµε + µσ )ϕ = −
ρ0 , ε
which gives the final form of the wave equation for the scalar potential ∆ϕ + k 2ϕ = −
ρ0 . ε
(1.55)
Equations (1.54) and (1.55) are more suitable for solving an electromagnetic field, as the terms on their right hand sides are simple functions of source quantities JS and ρ0. They are frequently used to analyze and design antennas. Note that Poisson’s equations (1.47) and (1.48) represent limiting cases of general wave equations (1.54) and (1.55), assuming that the frequency is decreased to zero, which is equivalent to decreasing propagation constant k. Due to this, the solution of wave equations can be expected in a form corresponding to (1.50) and (1.51). Let us first solve equation (1.55) for the scalar potential in the time domain. A lossless medium is assumed to simplify understanding. Now the equivalent to (1.55) in the time domain is ∆ϕ + µε
∂ 2ϕ ∂t 2
=−
ρ0 . ε
The variations of an electromagnetic field excited at distance R from a source are delayed by the time which is necessary for the wave to travel at speed v along this distance R = r − r , , Fig. 1.4 ∆t =
R . v
(1.56)
ϕ
Consequently we get R ρ0 t − ρ 0 (t − ∆t ) 1 1 v ϕ (t ) = dV dV = ∫∫∫ ∫∫∫ 4πε V 4πε V R R
Applying phasors we get, assuming that k=ω/v
ϕ e jωt =
1 4πε
∫∫∫
ρ 0 e jω (t − R / v )
V
R
ϕ=
4πε
∫∫∫ V
dV
ρ0
r
r’
dV . 0 Fig. 1.4
The final form of the solution is 1
R
ρ 0 e − jkR R
dV .
(1.57)
Similarly for the vector potential we have
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A=
µ 4π
∫∫∫ V
J S e − jkR dV . R
(1.58)
Again we see that (1.57) and (1.58) pass to (1.50) and (1.51) when the frequency is reduced to zero, and, consequently, the propagation constant goes to zero, which causes a transition e − jkr → 1 . Equations (1.57) and (1.58) are frequently used to calculate the electromagnetic field excited by sources, e.g., by antennas in space or probes in waveguides. The electric field and the magnetic field are then calculated from the potentials, using (1.46) and (1.52). Prior to performing this calculation we usually have to determine the distribution of the electric charge and the electric current. This task is often not simple, see section 10.
2. ELECTROMAGNETIC WAVES IN FREE SPACE 2.1 Solution of the wave equation Let us solve wave equation (1.42) for an electric field. We will look for a solution which describes a plane electromagnetic wave propagating in an infinite space filled by a homogeneous material. Wave fronts, which are the surfaces of a constant field phase, are - in the case of a plane wave - planes perpendicular to the direction of propagation. Such a wave can be excited by a source located at infinity, which excites a field with a constant amplitude and phase within the whole infinite plane. So this wave is only an abstraction. We will study its behaviour and propagation, as this provides a basis for understanding all phenomena connected with the propagation of electromagnetic waves. Most real waves can be expressed as the superposition of a series of plane electromagnetic waves. Let us assume a free space filled by a homogeneous material with parameters ε, µ, and σ. The wave equation is solved as a homogeneous equation in the form (1.42), i. e., without any source. This means that the solution describes the possible particular waves, called modes or eigen waves, which can propagate in our space. We rewrite equation (1.42) into a form describing the i-th coordinate of the electric field ∂ 2 E i ∂ 2 Ei ∂ 2 E i + + + k 2 Ei = 0 , i=x, y, z ∂x 2 ∂y 2 ∂z 2
(2.1)
This equation can be solved by the method of separation of variables. Ei is expected in the form Ei = X ( x )Y ( y ) Z ( z ) .
(2.2)
Introducing (2.2) into (2.1) we get X '' Y '' Z '' + + + k2 = 0 , X Z Y
(2.3)
where
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X '' Y '' = − k x2 , = − k y2 , X Y
Z '' = − k z2 Z
(2.4)
and the propagation constant (1.40) can be written in the form
k 2 = ω 2 µε − jωµσ = k x2 + k y2 + k z2 .
(2.5)
Equation (2.3) is then separated into three equations. The equation for the x-coordinate is X '' + k x2 X = 0 ,
(2.6)
with its solution in the form of the two propagating plane waves
X = A e jk x x + B e -jk x x .
(2.7)
Combining electric field Ei (2.2) we get
(
)(
Ei = A e jk x x + B e -jk x x C e
jk y y
+ De
- jk y y
)(E e
jk z z
)
+ F e -jk z z .
(2.8)
Constants A to F can be determined from the boundary conditions. Multiplying the brackets in j (± k x ± k y ± k z ) (2.8) we get eight terms of the form e x y z . We confine the solution of the wave equation to one of the terms representing the particular plane wave with vector complex amplitude E0 E = E 0 e − jk ⋅r = E 0 e
(
− j k x x+k y y +k z z
)
,
(2.9)
where the scalar product k ⋅ r represents the projection of vector r into the direction of k. These vectors are r = x x0 + y y 0 + z z 0 , k = k x x 0 + k y y 0 + k z z 0 , k 2 = k x2 + k y2 + k z2 = ω 2 µε − jωµσ .
(2.9) describes the plane electromagnetic wave propagating in the direction determined by vector k.
2.2 Propagation of a plane electromagnetic wave A plane electromagnetic wave propagating in a general direction in an unbounded space filled by a homogeneous material is described by the electric field intensity (2.9) E = E0 e − jk ⋅r = E0 e − j (β − jα )⋅r = E0 e − α ⋅r e − jβ ⋅r = E0 e − α ⋅r e − jβ ⋅r e jφ 0 ,
(2.10)
where the generally complex propagation vector can be rewritten into a real part and an imaginary part k = β − jα .
(2.11)
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The instantaneous value of an electric field is
{
}
E(r, t ) = Im E e jωt = E0 e − α ⋅r sin (ωt − β ⋅ r + φ0 ) .
(2.12)
Term E 0 e − α⋅r represents the dependence of the wave constant phase amplitude on position r. From the definition of the propagation constant (2.5) it follows that both its real part β and its imaginary part α are positive numbers. ϕ β Exponential function e −α⋅r then decreases with increasing value of its exponent. The amplitude therefore decreases r cos ϕ Fig. 2.1 and the wave is attenuated in the case of non-zero α. Vector α is therefore known as an attenuation constant (vector), as it describes the measure of the wave attenuation. The wave described by (2.10) with a negative exponent therefore propagates in the positive direction – in the direction defined by the propagation vector. The argument of the sinus function ω t − β ⋅ r + φ 0 determines the dependence of the field phase on time and coordinates. For this reason β is called a phase constant (vector). The surfaces of a constant phase – wave fronts – are planes determined by the condition β ⋅ r = β r cos ϕ = const. We can see from Fig. 2.1 that the surface of a constant phase is a plane perpendicular to phase vector β. The surfaces of a constant amplitude are determined by α ⋅ r = const. , and they represent planes perpendicular to attenuation vector α. The name plane wave is derived from the shape of these surfaces. A uniform wave has surfaces of constant amplitude and of constant phase that are identical, as vectors α and β are parallel. A non-uniform wave has surfaces of the constant phase different from surfaces of the constant amplitude, as vectors α and β are not parallel. The phase velocity is the velocity of the propagation of planes of a constant phase. Let us follow the propagation of a plane with a constant phase φ which in a time increment ∆t moves by a distance ∆r
φ = ωt − βr = ω (t + ∆t ) − β (r + ∆r ) ⇒ 0 = ω∆t − β∆r , from this we can define the phase velocity v=
∆r ω = . ∆t β
(2.13)
Group velocity represents the velocity with which the wave transmits energy, or the velocity of the propagation of planes of a constant amplitude. We will derive the group velocity as the velocity of the propagation of the amplitude of the superposition of two waves with frequencies ω and ω+dω and with corresponding phase constants β and β+dβ and with amplitudes equal to one. The superposition of these waves is
E ( z, t ) = E1 ( z, t ) + E 2 ( z, t ) = sin (ωt − βz ) + sin[(ω + dω )t − (β + dβ )z ] = dω dβ dω t − dβ z = 2 cos sin ω + t − β + z 2 2 2
.
19 book - 2
The amplitude of the resulting field is described by function cosine. In a time increment ∆t it moves by a distance ∆z
φ = dωt − dβz = dω (t + ∆t ) − dβ ( z + ∆z ) ⇒ 0 = dω∆t − dβ∆z from this we can define the group velocity 1 ∆z dω = = . d β ∆t dβ dω
vg =
(2.14)
This velocity must in any case be lower than the velocity of light in a vacuum c. As we will see later, the phase and the group velocities are not functions of frequency in a lossless material – an ideal dielectric. This is not the case of a lossy material with nonzero conductivity, where waves with different frequencies propagate with different velocities. Each signal transmitting any information is represented by the spectrum of frequencies, and consequently each component of this signal propagates with a different velocity. The result is that the signal is distorted by passing the lossy material. This distortion is called dispersion and the material is called dispersive. So only the ideal dielectric is a non-dispersive material in which a passing signal is not distorted. The wavelength is defined as the least distance, measured in the direction of propagation, of two points with the same phase sin (ωt − βz ) = sin[ωt − β ( z − λ )] ⇒ βλ = 2π , from this we can get
λ=
2π
β
=
2π
ω
=
v , f
(2.15)
v where f =ω/2π is the frequency. In the preceding section we solved the wave equation for an electric field. The relation between the electric field and the magnetic field can be derived from Maxwell’s equations. We insert the solutions of the wave equation in the form E = E0 e − jk ⋅r
, H = H 0 e − jk ⋅ r ,
into Maxwell’s second equation (1.13). The rotation of vector E is (….)
(
)
(
)
rot E = rot E0 e − jk ⋅r = e − jk ⋅r rot E0 + grad e − jk ⋅r × E0 = − jk e − jk ⋅r × E0 = − jk × E0 e − jk ⋅r ,
as the rotation of constant vector E0 is zero. Inserting into (1.13) we get − jωµ H 0 e − jk ⋅r = − jk × E0 e − jk ⋅r
jωµ H 0 = jk × E0 .
(2.16)
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Inserting into Maxwell’s first equation (1.12) we get
(σ + jωε )E0 = − jk × H 0
.
(2.17)
It is evident from (2.16) and (2.17) that vectors E, H and k represent a right handed rotating set of three vectors, Fig. 2.2. Vectors E and H are perpendicular to each other and both are perpendicular to the direction of propagation determined by vector E k. For this reason a plane electromagnetic wave propagating in an unbounded space filled by a homogeneous material is called a k transversal electromagnetic wave, abbreviated as TEM. We rewrite vector k as k=kn, where vector n is a unit vector H determining the propagation direction. Now using the form of k (2.5) we get from (2.17) Fig. 2.2
E0 =
jωµ − jkn × H 0 = H ×n . σ + jωε σ + jωε 0
The second root in this equation has the unit [Ω] and therefore it is known as the wave impedance of the space Zw =
k ωµ jωµ = = , σ + jωε k σ + jωε
(2.18)
it is generally a complex number which defines the ratio of E over H including the phase shift between an electric field and a magnetic field. Consequently, the relations between an electric field and a magnetic field are
E0 = Z w H 0 × n , H0 =
(2.19)
1 n × E0 Zw
(2.20)
Let us now simplify the description of a plane electromagnetic wave by considering that it propagates in the z direction. So we have k = k z 0 and r=z. Vectors E and H can then be written as E = Ex x0 + E y y 0
, H = H x x0 + H y y 0 .
Inserting these vectors into (2.19) we get E x x 0 + E y y 0 = −Z w H x y 0 + Z w H y x 0 , using this equation we can define the wave impedance as
Zw =
Ey Ex . =− Hy Hx
(2.21)
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The complex propagation constant k was defined by (2.5). The phase constant and the attenuation constant can be calculated by inserting the values of the material parameters and frequency into (2.5). Alternatively α and β can be obtained from (2.5) in the form 2 1 σ β = ω µε 1+ + 1 , ωε 2
(2.22)
2 σ 1 . 1 + − 1 ωε 2
(2.23)
α = ω µε
The power transmitted by a plane electromagnetic wave is defined by the average value of Poynting’s vector S av =
(
E × z 0 × E *0 1 1 1 Re E × H * = Re E 0 e −αz e − jβz × H *0 e −αz e jβz = e − 2αz Re 0 2 2 2 Z w*
{
}
{
}
1 1 = e − 2αz Re * E 0 ⋅ E *0 z 0 − (E 0 ⋅ z 0 )E*0 2 Z w
[(
S av =
E0
)
) =
2 1 − 2αz E 0 Re * z 0 = e 2 Z w
]
2
2 Zw
e − 2αz cos ϕ z z 0 ,
(2.24)
where ϕz is the argument of wave impedance Z w = Z w e jϕ z (2.18). The power is transmitted in the z direction, i.e., in the direction of the wave propagation. Let us now discuss the propagation of plane electromagnetic waves in materials with limit parameters. We will at the same time show the distribution of the electromagnetic field of this wave in dependence on the time and space coordinates. An ideal dielectric is a material with zero conductivity. The propagation constant and the wave impedance are real in this material k 2 = ω 2 µε = β 2 Zw =
⇒ β = ω µε
, α =0 ,
µ0 120 π µ = = . ε ε 0ε r εr
(2.25) (2.26)
This means that the wave propagates with a constant amplitude, i.e., without losses, and such a material is therefore called a lossless material. The electric field and the magnetic field are in phase, as Zw is a real number. The electric field and the magnetic field and their instantaneous values are, assuming E parallel with the x axis and H parallel with the y axis and zero starting phase, E = E0 e − jβz x 0
, H = H 0 e − jβz y 0 =
E0 − jβz e y0 , Zw
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E( z , t ) = E 0 sin (ωt − βz ) x 0
, H(z, t ) =
E0 sin (ωt − βz ) y 0 . Zw
These functions are plotted in Fig. 2.3 in dependence on time t for constant z and in dependence on the z coordinate for constant t. x
z=0 m
ωt=π
x
E
E
t
H
z
H
y
y Fig. 2.3
A good conductor is a material in which we have ωε<<σ. The formulas for the propagation constant (2.5) and the wave impedance (2.18) can under this condition be simplified to ⇒ k = β − jα = − jωµσ =
k = − jωµσ
ωµσ
β =α =
Zw =
2
jωµ
σ
ωµσ 2
(1 − j ) ,
,
(2.27)
(2.28) π
ωµ ωµ j 4 = (1 + j ) = e σ 2σ
, ϕz =
π 4
.
(2.29)
In a well conducting material the electric and magnetic fields have a mutual phase shift equal to π/4, i.e., 45°. Generally the phase shift between E and H lies between 0° for a lossless material and 45° for a good conductor. Phase constant β and attenuation constant α have high values (2.28), and the wave is rapidly attenuated. A material is sometimes characterized by penetration depth δ. This quantity determines the length at which the wave amplitude decreases to 1/e (e=2.718281 is the basis of the natural logarithm), i.e., to a value of 36.8 % of its starting value. The value of the penetration depth can be derived e −αδ = e −1 ⇒ δ =
1
α
=
2
ωµσ
.
(2.30)
____________________________ Example 2.1: Determine the penetration depth of a plane electromagnetic wave into copper for frequencies of 50 Hz, 10 kHz, and 100 MHz. Applying formula (2.30) we get δ = 9.35 mm for 50 Hz
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δ = 0.66 mm δ = 6.6 µm
for 10 kHz for 100 MHz At high frequencies the field does not penetrate at all into the conducting material. ____________________________ The electric field and the magnetic field and their instantaneous values are, in the case of a lossy material E = E0 e −αz e − jβz x 0
, H=
E0 −αz − jβz − jϕ z e e e y0 , Zw
E( z , t ) = E 0 e −αz sin (ωt − βz ) x 0
, H(z, t ) =
E0 −αz e sin (ωt − βz − ϕ z ) y 0 . Zw
These functions are plotted in Fig. 2.4 in dependence on time t for constant z and in dependence on the z coordinate for constant t, and a general lossy material is assumed. The dependence on the z coordinate shows the attenuation of the wave according to function exp(αz). x
z=0 m
ωt=π
x
ϕz/ω E
E
t
H
exp(-αz)
H
y
z
y Fig. 2.4 A dielectric material with non-zero conductivity σ which fulfils the condition
σ<<ωε
(2.31)
can be called a lossy dielectric. The plane electromagnetic wave propagating in such a material can be treated as a wave propagating in the dielectric only in the case of propagation over a short distance, for which its attenuation can be neglected. A wave propagating over a long distance can be substantially attenuated, even in the case of a low attenuation constant. Under condition (2.31) we can accept the following simplification for the propagation constant k = ω 2 µε − jωµσ = ω µε 1 − j 1 µ k = ω µε − j σ , 2 ε
σ 1 σ ≈ ω µε 1 − j , ωε 2 ωε (2.32)
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β = ω µε
,
α=
σ 2
µ 60πσ = . ε εr
(2.33)
We see that the phase constant is the same as the phase constant of the wave propagating in a dielectric, but we have non-zero attenuation. Other parameters Zw, v, λ are determined using the same formulas as in the case of a wave propagating in a dielectric. ________________________________ Example 2.2: Calculate the instantaneous value of voltage received by a frame antenna located according to Fig. 2.5, due to the propagation of an electromagnetic wave z E = 5 . 10 − 2 sin 10 8 t − x 0 3
z , H = 1.33 . 10 − 4 sin 10 8 t − y 0 3
We will calculate the received voltage in two ways. Firstly, according to the definition of voltage. A closed loop is represented by an antenna perimeter, Fig. 2.5.
( )
π u (t ) = ∫ E ⋅ dl = ∫ E ⋅ dl1 + ∫ E ⋅ dl 2 = −π E1 + π E2 = π 5.10− 2 − sin 108 t + sin 108 t − 3 1 2 c We can rewrite this result to the final form
x
π
π u (t ) = −0.157 sin 10 8 t + . 3
E k
H
The second way must result in the same formula. We apply Faraday’s induction law u (t ) = −
c
E1
E2
dl1
dl2
π
y
dφ dt
z
Fig. 2.5
The magnetic flux passing the antenna area is π
φ = µ ∫∫ H ⋅ dS = µπ ∫ H dz = µπ 1.33 .10 S
−4
0
π
∫ sin10 0
8
z t − dz = 3
z = 157 .10 −11 cos10 8 t − − cos 10 8 t 3
(
u (t ) = −
)
dφ π π = 0.157 sin 10 8 t − − sin 10 8 t = −0.157 sin 10 8 t + . 3 3 dt
(
)
The results are identical. Tilting the antenna into plane y-z, voltage u=0, as vectors dS and H are perpendicular. In this way we can determine the direction of the wave propagation, i.e. the
25 book - 2
direction in which the transmitter is located. In this example we have the antenna dimensions 2π 2π = = 6π . comparable with the wavelength, which is λ = β 13 ____________________________________ Example 2.3: Calculate the effective value of the voltage induced in a frame antenna, see Fig. 2.6, due to the propagating plane electromagnetic wave defined in Example 2.2. In this case the antenna dimensions are much lower than the wavelength. We can therefore simplify the calculation of the magnetic flux. This can be done by supposing
,
x
e − jβz ≈ e − jβ ( z + a ) , i.e. e − jβa ≈ 1
i.e.
βa << 1 .
a E
φ = B S cosψ ,
a
k
H
The magnetic flux is
a=0.1 m z
y
where ψ is the angle between vectors B and dS. In our case this angle is 0°. The induced voltage is u=−
dφ dH = −µ0 S dt dt
, U = − j ω µ0 S H
Fig. 2.6
, U ef =
ωµ 0 S H U = = 1.18 µV . 2 2
This value was obtained using the wave parameters from the previous example. _______________________________________ Example 2.4: A plane electromagnetic wave propagates in the direction of the positive z axis in air. It is incident perpendicularly to an ideally conducting plane located at z=0. Determine the field distribution, which is the superposition of the incident and reflected waves supposing that an electric field reaches at point z = -1 m and at time t = 0 s its maximum value E ( z = −1, t = 0) = E max = 100 V/m . The frequency is 100 MHz. Calculate the power transmitted by this wave. An incident wave propagating in the positive z direction is E i = E0i e − jkz e jϕ x 0 , H i = H 0i e − jkz e jϕ y 0 . A reflected wave which propagates in the negative z direction is E r = E0r e jkz e jϕ x 0 , H r = H 0r e jkz e jϕ y 0 . The resulting field is the superposition of the incident and reflected waves E = E i + E r = E0i e − jkz e jϕ x 0 + E0r e jkz e jϕ x 0 H = H i + H r = H 0i e − jkz e jϕ y 0 + H 0 r e jkz e jϕ y 0
26 book - 2
The relation between the amplitudes of the incident and reflected waves can be determined from the boundary conditions. The electric field is parallel to the conducting plane, and due to (1.22) its value must therefore be zero E( z = 0 ) = E i ( z = 0 ) + E r ( z = 0 ) ⇒ E 0 i = − E 0 r . The relation between the amplitudes of a magnetic field follows from the orientation of the vectors of the plane wave. The vectors of both the incident wave and the reflected wave are mutually perpendicular, Fig. 2.7. Consequently we have
Ei Hi
kr ki
Hr
Er
Fig. 2.7 H 0i = H 0 r . Now we can write the field distributions of both phasors and instantaneous values
(
)
E = E0i e − jkz − e jkz e jϕ x 0 = −2 jE 0i sin (kz ) e jϕ x 0 = 2 E0i sin (kz ) e j (ϕ −π / 2 ) x 0 E(z , t ) = −2 E0i sin (kz ) sin (ωt ) x 0
(
)
H = H 0i e − jkz + e jkz e jϕ y 0 = 2 H 0i cos(kz ) e jϕ y 0 H ( z , t ) = 2 H 0i cos(kz ) sin (ωt ) y 0 We now have a new kind of electromagnetic wave. The dependence on the z coordinate and time is separated. This indicates that it is not a propagating wave but a standing wave. The distribution of the electric field and the magnetic field of a standing wave is plotted in Fig. 2.8. The electric field is shifted by 90 deg corresponding to the magnetic field. It is evident from Fig. 2.8 that using an excited standing wave we can measure the wavelength, as the distance between two adjacent minima or maxima is equal to one half of the wavelength. The power transmitted by the standing wave is determined by Poynting’s vector
E (z )
−λ
−λ
-3 λ/4
-λ/2
−λ/4
-3 λ/4
−λ/2
−λ/4
H (z )
Fig. 2.8
(
)
S av = Re(E × H ) = Re 2 E0i sin (kz ) e j (ϕ −π / 2 ) x 0 × 2 H 0i cos(kz ) e jϕ y 0 = 0
27 book - 2
The power transmitted by the standing wave is zero, which is why it is called standing. This is a natural result, as the incident wave transmits some power and no power is lost in the ideally conducting wall. Consequently, the whole power is reflected back in the form of a reflected wave. The total power, which is the vector sum of these two, must be zero. The numerical results are: k=
ω c
=
2π 3
m −1 ,
Z w = 120π Ω ,
at t = 0 s and z = -1 m we have
π 2π E max = 100 = 2 E 0i sin − sin (ϕ ) ⇒ E 0i = 57.8 V/m , ϕ = 2 3 Consequently we have
π E(z , t ) = 115.6 sin (kz )sin ω t − x 0 2 π H ( z , t ) = 30.6 cos(kz )sin ω t − y 0 2 _________________________________________
2.3 Wave polarization We showed that vectors E and H are mutually perpendicular and lie in a transversal plane perpendicular to the direction in which the wave propagates. Their actual position in this plane can, however, be quite general, and it can vary with time. The wave polarization determines the way in which the end point of vector E moves in the transversal plane. Let us find an equation which gives a general description of the behaviour of vector E in the plane perpendicular to the direction of wave propagation. Let us suppose that this direction is identical with the z axis and that we have a lossless material. Vector E is E = E xm sin (ωt − kz + φ x ) x 0 + E ym sin (ωt − kz + φ y )y 0 .
Let us simplify this formula by choosing the position z = φ x k . Then we have E x = E xm sin (ωt ) ,
(2.34)
E y = E ym sin (ωt + φ ) ,
(2.35)
where φ = φ y − φ x . We eliminate time from (2.34) and (2.35). From (2.34) we get
E cos(ωt ) = 1 − x E xm
E sin (ωt ) = x , E xm
2
,
28 book - 2
and (2.35) gives Ey E ym
= sin (ωt + φ ) = sin (ωt ) cos φ + cos(ωt )sin φ .
Inserting for sin (ωt ) and cos(ωt ) we get Ey E ym
2
E E = x cos φ + 1 − x sin φ E xm E xm
This formula gives the equation Ey E ym
2
2
Ey E − 2 Ex cos φ + x = sin 2 φ . E xm E ym E xm
Assuming general values Exm, Eym and φ this equation represents the equation of an ellipse, Fig. 2.9, in an arbitrary position in the coordinate system Ex and Ey. Consequently the end point of vector E moves along an ellipse, and the wave is an elliptically polarized wave. The position and the shape of this ellipse depend on values Exm, Eym and φ. A special case is φ = nπ , in which we have cos φ = ±1, and sin φ = 0 . (2.36) then gives the equation E y = ±Ex
Ey
E ym
Ey
Ex Fig. 2.9
,
E xm
(2.37)
Ey
Eym=0
(2.36)
Ey
Exm=0
Ex
Ex
n odd Ex
Ey
n even Ex
Fig. 2.10 which is the equation of a line, Fig. 2.10, and our wave is a linearly polarized wave. Another special case is φ = (2 n + 1)
π
2
, in which cos φ = 0 and sin φ = ±1 . (2.36) then gives the
equation 2
2
Ex Ey =1 , + E xm E ym
(2.38)
29 book - 2
of an ellipse in the basic position, Fig. Ey Ey 2.11. This wave has an elliptic Exm>Eym Exm
π E x = E 0 sin (ωt − kz ) , E y = E 0 sin ωt − kz ± = ± E 0 cos(ωt − kz ) , 2 E( z , t ) = E0 [sin (ωt − kz ) x 0 ± cos(ωt − kz ) y 0 ] ,
and consequently by E = E0 (x 0 ± j y 0 ) e − jkz .
A survey of basic types of polarization is given in Tab. 2.1. ________________________________ Example 2.5: Decompose the generally elliptical polarized wave into the superposition of the two circularly polarized waves, one with left-handed polarization, the second with righthanded polarization. We start solving a reciprocal problem in which we combine two circularly polarized waves with opposite orientation of the electric field vector rotation. The waves with lefthanded and right-handed polarization are
30 book - 2
φ (deg) Exm=0
0
45
90
135
180
225
270
315
Exm< Eym Exm= Eym Exm> Eym Eym=0 left-handed
right-handed
Tab. 2.1 E L = A (x 0 + jy 0 ) e − jkz ,
E R = B (x 0 − jy 0 ) e − jkz .
The superposition of these waves is E = E L + E R = [( A + B )x 0 + (B − A) jy 0 ]e − jkz .
In the case A=B the resulting wave is linearly polarized, for A>B it is elliptically polarized right-handed, for A
Comparing this formula with the previous formula we get the amplitudes of the particular waves with circular polarization C−D C+D . , B= 2 2 _________________________________________ A=
Example 2.6: Calculate the power transmitted by a wave with generally elliptical polarization. A wave with elliptical polarization can be written in the form E = Ax 0 e − jkz + By 0 e − jkz e − jϕ .
Poynting’s vector is
31 book - 2
S av =
(
[ (
)
)]
1 1 Re E × H * = Re E × z 0 × E * = 2 2Z w
[(
) (
)]
1 Re Ax 0 e − jkz + By 0 e − jkz e − jϕ × A* y 0 e jkz − B * x 0 e jkz e jϕ = 2Z w
[
]
1 Re A 2 + B 2 z 0 2Z w _________________________________________
2.4 Cylindrical and spherical waves We will solve the wave equation for a vector potential in a cylindrical coordinate system in the free space filled by a lossless material without a source. We will assume that the field does not depend on α and z. As a result, we obtain a wave which propagates from the z axis to infinity with surfaces of a constant amplitude and a constant phase in the shape of cylinders. Such a wave is an abstraction as a plane wave. The excitation of a cylindrical wave supposes an infinitely long line source with a passing current of a constant amplitude and a constant phase along this source. The homogeneous wave equation for the z component of a vector potential A = A(r)z0 is (1.54) ∆A + k 2 A = 0 . As the vector potential does not depend on α and z, we have ∂2 A ∂r
2
+
1 ∂A + k2A = 0 . r ∂r
(2.39)
This is a Bessel-type equation with a solution in the form of the sum of Bessel functions of zero order of the first and second kind, see mathematical appendix, A = C1 J 0 (kr ) + C 2 Y0 (kr ) = C1, H 01 (kr ) + C 2, H 02 (kr ) ,
(2.40)
where Hankel functions of the zero order and of the first and second kind are defined by Bessel functions, see mathematical appendix (13.44) and (13.45). It is evident from their asymptotical expression (13.46) and (13.47) that function H 02 represents a wave propagating as kr increases toward infinity, whereas H 01 represents a wave propagating from infinity towards the z axis. Assuming only a wave propagating from the z axis in an area of high kr, i.e., far distant from the axis, we have finally the cylindrical wave described by the form A=C
e − jkr . r
(2.41)
Why does the amplitude of this wave decrease as function 1 / r , as we have no losses? The decrease in the amplitude is caused by the spread of power to space. Let us calculate the density of the power transmitted by this wave
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S av
D2 1 = r0 , 2Z w r
where D is a constant and Zw is the wave impedance. The power transmitted through the surface of a cylinder with radius r and length l is P = S av 2πrl =
D2 2πl . 2Z w
(2.42)
This power is constant, so we have no losses and it confirms why the field amplitude decreases as 1 / r . For high kr and a small space angle we can approximate this wave by a plane wave. Let us now solve the wave equation in a spherical coordinate system, assuming that the vector potential depends only on distance from the origin r. We will solve a homogeneous equation, as we have a source-less space. The solution corresponds to a spherical wave. Such a wave can be excited by an elemental omni-directional source placed at the origin. We then solve the equation 1 1 2 ∂A 2 r +k A=0 . 2 ∂r ∂ r r
(2.43)
The solution of this equation corresponding to a wave propagating from the origin towards infinity is A=
C 2 H 1/ 2 (kr ) , r
(2.44)
using the formula for a Hankel function of order ½ and of the second kind (13.48), we get A=C
e − jkr . r
(2.45)
The amplitude of this spherical wave decreases due to the spread of power as 1/r. This wave, for high kr and within a small space angle, can be approximated by a plane wave. ____________________________ Example 2.7: An ideal omnidirectional antenna transmits a spherical wave. What power must the antenna transmit to get transmitted power density Sav=1 mW/m2 at distance r1=1 km. What power density corresponds to this at distance r2=1 m? Calculate the corresponding electric field amplitude at r2=1 m. Assume a lossless material. The total transmitted power is P (r1 = 1000 ) = ∫∫ S av dS = S av 4πr12 = 12560 W .
The power density at r2 is
33 book - 2
S av (r2 = 1) =
P = 1000 W/m 2 2 4πr2
As the power density is 2
S av
1 E = cos(ϕ z ) 2 Zw
we get E =
2 S av Z w
= 868.1 V/m cos(ϕ z ) _________________________________________
2.5 Problems 2.1 Write the solution of the wave equation describing a plane electromagnetic wave propagating in the free space with parameters εr=8, µr=1, σ=40 S/m. The frequency is 40 GHz. k = ω 2 µε − jωµσ = (3140 − j 2040 ) m −1 E = E0 e −2040 z e − j 3140 z
E(z , t ) = E0 e −2040 z sin (25.5.1010 t − 3140 z + ϕ )
2.2 The plane electromagnetic wave propagates in a free space filled by a material with parameters εr=4, µr=1, σ=0. The frequency is 10 MHz. The electric field intensity has the direction of the x axis and has a positive maximum equal to 10 V/m at z=0 and t=0. Calculate the electric and magnetic field and their instantaneous values and Poynting’s vector at z=6 m and t=10-8 s. E x = 10 e jπ / 2 e − j 0.42 z = 10 e − j 0.94 V/m
π E x ( z , t ) = 10 sin 3.14 ⋅ 107 t − 0.42 z + = −5.94 V/m 2 jπ / 2 − j 0.42 z − j 0.94 H y = 0.053 e = 0.053 e e A/m π H y ( z , t ) = 0.053 sin 3.14 ⋅ 107 t − 0.42 z + = −0.0315 A/m 2 2 S av = 0.26 z 0 W/m 2.3 A plane electromagnetic wave propagates in a free space filled by a material with parameters εr=80, µr=1, σ=0.002 S/m. The frequency is 500 kHz. The electric field intensity has the direction of the x axis and has a positive maximum equal to 50 V/m at z=0 and t=0. Calculate the electric and magnetic field and their instantaneous values and Poynting’s vector. E x = 50 e −0.039 z e jπ / 2 e − j 0.101z
π E x ( z , t ) = 50 e − 0.039 z sin ω 7t − 0.101z + 2
34 book - 2
H y = 1.37 e −0.039 z e j (π / 2 − 0.366) e − j 0.101z
π H y ( z , t ) = 1.37 e − 0.039 z sin ω 7t − 0.1012 z + − 0.366 2 −0.078 z 2 S av = 32 e z 0 W/m 2.4 Two plane electromagnetic waves propagate in the same direction in a lossy material. The first wave has the frequency 1 GHz and the corresponding attenuation constant is β1=78 m-1. The second wave has the frequency 3 GHz and the attenuation constant is β2=83 m-1. The amplitudes of these waves are equal at z=0. Determine the distance z in which E1=10E2. ln 10 z= = 0.46 mm β 2 − β1 2.5 A plane electromagnetic wave propagates in a free space filled by a material with parameters εr=2, µr=1, σ=0.01 S/m. The frequency is 9 MHz. Calculate the electric and magnetic field amplitudes at the point at which Sav=10 W/m2. S av 2 Z w E = = 48.8 V/m cos ϕ z H =
E
Zw
= 0.58 A/m
2.6 To measure a receiving antenna we need a plane electromagnetic wave. To measure with a sufficiently low error we can accept a change in field amplitude of 1% within a region 1 m in length. The wave is excited by an ideal omnidirectional antenna transmitting a cylindrical wave. Calculate the distance from the transmitting antenna at which we have to measure. r=99 m
3. WAVES ON A PLANE BOUNDARY 3.1 Perpendicular incidence of a plane wave to a plane boundary Let us solve the problem of the perpendicular incidence of a plane electromagnetic wave on a plane boundary between two different materials, Fig. 3.1. An incident wave carries some power. Part of this power is reflected back at the boundary in the form of a reflected wave. In the first material, this wave is superimposed on the incident wave. Part of the power is transmitted to the second material in the form of a transmitted wave. The incident wave is described by the quantities marked by index i, the reflected wave is marked by index r, and the wave transmitted to the second material is marked by index t. We assume that the orientation of the vectors is according to Fig. 3.1. Our task is to determine the relations between the amplitudes of particular waves. The propagation vectors of these waves are k i = k1z 0 , k t = k2z 0 ,
(3.1)
k r = − k1z 0 ,
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where k1 and k2 are the propagation constants of a plane wave in the first and second materials, respectively. The total field in the first material is the superposition of the incident and reflected waves
ε2, µ2, σ2
ε1, µ1, σ1 Ei
Et ki
kt
Hi
Ht
kr
0
Er
z
Hr
Fig. 3.1
E1 = E i + E r = x 0 Eix e − jk1z + x 0 E rx e jk1z , k H1 = Η i + H r = 1 z 0 × x 0 Eix e − jk1z − z 0 × x 0 E rx e jk1z =
ωµ 1
(
(
)
1 = y 0 Eix e − jk1z − E rx e jk1z Z1
)
.
In the second material we have only the transmitted wave E 2 = E t = x 0 Etx e -jk2 z ,
H2 = Ht =
k2
ωµ 2
y 0 Etx e -jk2 z =
1 y 0 Etx e -jk2 z Z2
Both the electric field and the magnetic field are parallel to the boundary placed at z = 0. These fields must fulfill the boundary conditions for the tangential components (1.18) and (1.121). So we have at z = 0 Eix + E rx = Etx , k1
µ1
(Eix − E rx ) =
(3.2) k2
µ2
Etx .
(3.3)
This set of equations has the solution E rx = Eix
µ 2 k1 − µ1k 2 , µ 2 k1 + µ1k 2
(3.4)
36 book - 3
Etx = Eix
2µ 2 k1 . µ 2 k1 + µ1k 2
(3.5)
Now we can define the reflection coefficient as R=
E rx µ 2 k1 − µ1k 2 Z 2 − Z1 , = = Eix µ 2 k1 + µ1k 2 Z 2 + Z1
(3.6)
and the transmission coefficient as T=
Etx 2µ 2 k1 2Z 2 , = = Eix µ 2 k1 + µ1k 2 Z 2 + Z1
(3.7)
where Z1 and Z2 are the wave impedances of the two materials (2.18). Values R and T are generally complex. Inserting R and T from (3.6) and (3.7) into the boundary condition (3.2) we get the relation between R and T in the form T = 1+ R .
(3.8)
Similarly as in (3.5) and (3.6), we can define the reflection and transmission coefficients for a magnetic field. The reader can compare the relations for the reflection and transmission coefficients for a plane electromagnetic wave, which is perpendicularly incident on the plane boundary of two different materials, with the incidence 2Z 2 Z − Z1 Z1 Z2 R= 2 T= of a wave in the Z 2 + Z1 Z 2 + Z1 transmission lines, see Fig. 3.2. Fig. 3.2 For a lossless dielectric, the wave impedance is defined by (2.26), and we have R=
ε r1 − ε r 2 n1 − n2 Z 2 − Z1 , = = Z 2 + Z1 ε r1 + ε r 2 n1 + n2
(3.9)
T=
2 ε r1 2Z 2 2n1 , = = Z 2 + Z1 ε r1 + ε r 2 n1 + n2
(3.10)
where n is the refractive index
n = εr .
(3.11)
When ε1<ε2 we have Z1>Z2 and consequently R<0, and the orientation of the electric field vectors is according to Fig. 3.3a. The electric field is reflected out of phase. When ε1>ε2 we have Z1
0 and the orientation of the electric field vectors is according to Fig. 3.3b. Then electric field is reflected in phase.
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Let us now calculate the power transmitted by the field in the first material and the power carried by the transmitted wave in the second material, assuming that there are no losses in the two materials. In the first material we get
Ei
Et
ki
Ei
kt
Et
ki
kt
Er kr
Er
kr b
a Fig. 3.3
(
)
[(
)]
)(
1 1 * − jk1z Re E1 × H1* = Re x 0 Eix e − jk1z + x 0 E rx e jk1z × y 0 H iy* e jk1z − y 0 H ry e = 2 2 1 * * − 2 jk1z e = Re Eix H iy* − E rx H ry − Eix H ry + E rx H iy* e 2 jk1z z 0 = 2 (3.12) 1 1 * − 2 jk1z e = Re Eix H iy* z 0 + Re − R Eix R * H iy* − Eix H ry + E rx H iy* e 2 jk1z z 0 = 2 2
S1av =
[ (
(
]
= S i av 1 − R
2
)
)
[
]
This is a logical result. The total power transmitted in the first material is equal to the power transmitted by the incident wave reduced by the power transmitted by the reflected wave. The power transmitted in the second material must be equal to the power transmitted in the first material, as no power can be lost in the boundary itself. Let us now assume that the second material is a perfect conductor with σ2 infinite and consequently Z2=0. There is a zero field in such a material, i.e., no field penetrates into this material. We get R = -1 and T = 0. The field in the first material is the superposition of the incident and reflected waves. This was already solved in Example 2.4. The total field has the character of a standing wave described, see Example 2.4, by the phasors of the electric field and the magnetic field E = −2 jE 0i sin (kz ) x 0 ,
(3.13)
H = 2 H 0i cos(kz ) y 0 ,
(3.14)
The instantaneous values of these fields are, see Example 2.4, E( z , t ) = −2 E0i sin (kz ) cos(ω t ) x 0 ,
(3.15)
H ( z , t ) = 2 H 0i cos(kz )sin (ω t ) y 0 .
(3.16)
This wave does not transmit any power. The distribution of its amplitudes is plotted in Fig. 2.8. As was mentioned in Example 2.4, the excitation of this standing wave is frequently used to measure the wavelength, or consequently the frequency. Let us now turn our attention again to the perpendicular incidence of a plane electromagnetic wave to the plane boundary between two dielectric materials. Let us analyze in greater detail the field in the first material, which is the superposition of the incident and reflected waves, Fig. 3.1. We suppose that ε1>ε2 and consequently R>0. The xcomponent of an electric field is
38 book - 3
[(
]
)
E1x = Eix + E rx = E0 e − jk1z + R E0 e jk1z = E0 R e − jk1z + e jk1z + (1 − R ) e − jk1z ,
[
]
E1x = E0 2 R cos(k1 z ) + (1 − R ) e − jk1z ,
(3.17)
Similarly for a magnetic field we get
[
]
H 1 y = H 0 − 2 jR sin (k1 z ) + (1 − R ) e − jk1z .
(3.18)
In the case ε1<ε2 the formulae for an electric field and for a magnetic field are exchanged. The first terms in field distributions (3.17) and (3.18) represent a standing wave with amplitude 2RE0 and 2RH0, whereas the second terms represent a wave traveling toward the boundary with amplitude E0(1-R) and H0(1-R) and continuing in the second material as the transmitted wave. Only the traveling part transmits power, it therefore verifies (3.12), and we get S1av =
(
)
1 2 E0 H 0 1 − R . 2
(3.19)
The distribution of the amplitude of an electric field of this standing wave is plotted in Fig. 3.4. We can identify the minima and maxima of the amplitude, with their mutual distance equal to λ/2. The same distribution of a voltage wave and a current wave on a transmission line can be measured, assuming that this wave is incident to the junction between two lines with different wave impedances, Fig. 3.2. Standing waves are described by a standing wave ratio p, abbreviated as SWR, defined by p=
λ/2
E Emax=E0(1+R)
Emin=E0(1-R) z Fig. 3.4
E max 1 + R = . E min 1 − R
(3.20)
This SWR is very often used in practical applications, as it can be measured simply. Consequently, the measurement of a reflection coefficient or of an impedance is transposed to the measurement of SWR. As a last case, we will study the perpendicular incidence of a plane wave to a well conducting material with conductivity σ. The case of the ideal conductor has been already solved in Example 2.4, where the standing wave with p = ∞ was introduced. We assume that the first material is air. So the wave impedance and the propagation constant for the two materials are k1 = ω ε 0 µ 0 , Z1 =
ωµ 2σ µ0 , k 2 = (1 − j ) ε0 2
, Z 2 = (1 + j )
ωµ 2 2σ
39 book - 3
The transmission coefficient is T=
2Z 2 = Z 2 + Z1
2
.
(3.21)
e − jk2 z ,
(3.22)
σ (1 − j ) 1+ 2ωε 2
The transmitted wave can be expressed Et = x 0
2 Eix
σ (1 − j ) 1+ 2ωε 2
This intensity enables us to calculate the total current passing a strip of the second material of width equal to l, Fig. 3.5, as it determines the current density (1.9). We get ∞
∞
0
0
ε0, µ0 Ei
ε2, µ2, σ2 Et
ki
J2
kt
l Hi
Ht
I 2 = l ∫ J 2 x dz = lσ ∫ Etx dz = ∞
= lσEtxm ∫ e 0
− jk2 z
z y
2σZ 2 l dz = H tym = jk 2
Fig. 3.5
= l H tym
The result is very simple I 2 = l H tym = l K t ,
(3.23)
the total current is proportional to the value of the magnetic field just below the boundary, which can be substituted by current density K. The power lost in the strip, Fig. 3.5, of length d is P=
( )
(
)
ωµ 2 1 1 1 Re U I 2* = d Re Etxm I 2* = d l H tym 2 2 2 2σ
2
=
1 S ρ hf H tym 2
2
,
(3.24)
where S=ld is the surface of the strip, and ρhf is the high frequency resistance of the surface of a well conducting material
ρ hf =
ωµ 2 . 2σ
(3.25)
Formula (3.24) assumes homogeneous distribution of the magnetic field along the boundary. The more general form is
40 book - 3
P=
1 ρ hf 2
∫∫ H t
2
dS .
(3.26)
S
So to calculate the power lost in a material we have to know only the distribution of the tangent component of the magnetic field just at the surface of this material. ______________________________ Example 3.1: A plane electromagnetic wave with amplitude Eim=300 V/m is incident from the air perpendicular to the surface of a conducting material with parameters εr=80, µ=µ0, σ=5 S/m, and the frequency is 500 kHz. Calculate: a) The phasors and the instantaneous values of the reflected and transmitted waves. b) The power passing the boundary. c) The penetration depth and the wavelengths in the air and in the material. d) The distance at which the field amplitude decreases to 1/100 of its amplitude at the boundary. The wave impedances and propagation constants are jωµ 0 ≅ 0.888 e jπ / 4 , jωε + σ
Z1 = 120π ,
Z2 =
k1 = ω ε 0 µ 0 = 0.0105 m −1 ,
k 2 = − jωµ 0σ = (1 − j )π m −1 .
The reflection and transmission coefficients are R=
Z 2 − Z1 ≅ −1 , Z 2 + Z1
T=
2 Z2 = 0.0047 e jπ / 4 . Z 2 + Z1
The phasors and instantaneous values of the electric and magnetic fields are E i = x 0 300 e − j 0.0105 z ,
H i = y 0 0.796 e − j 0.0105 z ,
E r = x 0 300 e jπ e j 0.0105 z ,
H r = y 0 0.796 e j 0.0105 z ,
E t = x 0 1.410 e −πz e − jπz e jπ / 4 ,
H t = y 0 1.596 e −πz e − jπz ,
E i ( z , t ) = x 0 300 cos(ω t − 0.0105 z ) ,
H i ( z , t ) = y 0 0.796 cos(ω t − 0.0105 z ) ,
E r ( z , t ) = x 0 300 cos(ω t + 0.0105 z + π ) ,
H r ( z , t ) = y 0 0.796 cos(ω t + 0.0105 z ) ,
E t ( z, t ) = x 0 1.41e −πz cos(ω t − πz + π / 4) , H t ( z, t ) = y 0 1.596 e −πz cos(ω t − πz ) .
The particular power densities are S iav
(
)
2
1 1 Ei = Re E i × H *i = z 0 = z 0 119.4 W/ m 2 , 2 2 Z1
41 book - 3
S rav =
S tav
(
)
1 Re E r × H *r = − z 0 119.4 W/ m 2 , 2
2 2 1 1 Et 1 Et * = Re E t × H t = z 0 Re = z0 cos(ϕ z ) = z 0 0.8 W/ m 2 . 2 2 Z2 2 Z2
(
)
The penetration depth is
δ =
1
β2
= 0.318 m .
The distance z100 at which E(z100)=E(z=0)/100 is calculated e − β 2 z100 = 0.01
⇒
z=
ln (100 )
β2
= 1.488 m .
______________________________________
I
a1 a R21 b2 2 b = b e − jk2d 2 R12 a1 1 2 b 2 b1 T21 b2 a1
a1 b1
II T12 a1
k1 I
k2 d2
k3 d3
II
k4
...
a
b
II k2
a 2 = a1e − jk2d 2
d2
b2
c
Fig. 3.6
3.2 Perpendicular incidence of a plane wave to a layered medium Let us assume a plane electromagnetic wave incident from a half space filled by material perpendicular to the system of layers of different materials, Fig. 3.6a. Our task is to describe the field distribution in all layers and the field penetrating through this structure. This problem can be effectively solved using transmission matrices. We derive separately the transmission matrix describing the behaviour of the waves on the boundary between two adjacent materials, Fig. 3.6b, and the transmission matrix of a single layer, Fig. 3.6c. An electromagnetic wave with the complex amplitude of an electric field a1 is incident from the left to the boundary between two different materials I and II, Fig. 3.6b. This wave is partially reflected and partially transmitted through the boundary. Similarly, a wave with amplitude b2 propagating from the right is reflected and transmitted. T12 is the transmission coefficient from material I to material II. T21 is the transmission coefficient from material II to material I. R12 represents the reflection from the left, R21 is the reflection coefficient from the side of material II. The wave traveling from the boundary to the left has amplitude b1, Fig. 3.6b b1 = R12 a1 + T12 b 2 .
(3.27)
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Amplitude a2 of the wave propagating from the boundary to the right is, Fig. 3.6b a 2 = R21 b 2 + T12 a1 .
(3.28)
From (3.27) and (3.28) we get, using R12 = -R21 and T12T21 – R12R21 = 1, a1 1 1 b = 1 T12 R12
R12 a 2 . 1 b 2
(3.29)
So the transmission matrix describing the boundary between the two materials is
[A I,II ] = T1
12
1 R 12
R12 . 1
(3.30)
Now we derive the transmission matrix of only a single layer, Fig. 3.6c. In this layer we have one wave propagating to the left with amplitude b2 and one wave propagating to the right with amplitude a1. These two waves passing the layer of thickness d2 can be expressed a 2 = a1e − jk2 d 2 ,
b1 = b 2 e − jk2 d 2 ,
we consequently get the transmission matrix a1 e jk 2 d 2 b = 1 0
a 2 , e − jk 2 d 2 b 2 0
(3.31)
So the transmission matrix describing the behaviour of the waves in the single layer is
[A 2 ] = e
jk2 d 2
0
. e − jk2 d 2 0
(3.32)
The total transmission matrix of the layered structure, Fig. 3.6a, is the product of particular matrices of type (3.30) and (3.31) a1 1 b = T 1 12
1 R 12
R12 e jk 2 d 2 1 0
1 1 e − jk 2 d 2 T23 R23 0
R23 1
e jk 3 d 3 0
a N . . . . (3.33) e − jk 3 d 3 b N 0
Structures of this kind, Fig. 3.6a, are frequently used namely in optical wavelength technology as an antireflection coating which can provide even frequency selective behaviour. These structures can serve as band pass filters, band stop filters, etc. They can be designed in a similar way as microwave filters composed of sections of transmission lines of different wave impedances. _______________________________ Example 3.2: Design an inter-layer placed between two different dielectric materials to get zero reflection, a so called antireflection layer.
43 book - 3
The structure is shown in Fig. 3.7. There is a dielectric layer of thickness d placed between two dielectric layers. Our task is to design the thickness and the permittivity of this inter-layer to get zero reflection, i.e., to get b1 = 0. The transmission matrix of our structure is according to (3.33) the product of three matrices a1 1 b = 1 T12
1 R 12
R12 e jk 2 d 1 0
0 e
− jk2 d
1 T23
1 R 23
a1
a3
ε1
b1
ε2 d
R23 a 3 1 0
ε3 b3=0
Fig. 3.7
From this set of equations we get
(
)
(
)
a1 =
1 e jk2d + R12 R23 e − jk 2d a 3 , T12T23
b1 =
1 R12 e jk2d + R23 e − jk2d a 3 . T12T23
The reflection coefficient is R=
R e jk 2d + R23 e − jk2d R12 + R23 e −2 jk2d b1 . = 12 = a1 e jk2d + R12 R23 e − jk2d 1 + R12 R23 e −2 jk2d
(
)
Inserting for particular reflection coefficients R12 and R23 from (3.9) we get
(n1 − n2 )(n2 + n3 ) + (n2 − n3 )(n1 + n2 ) e −2 jk2d R= (n1 + n2 )(n2 + n3 ) + (n1 − n2 )(n2 − n3 ) e −2 jk2d
.
For the next procedure we assume two possible simplifications relating refractive indices n1>n2>n3 or n1
⇒
2k 2 d = π + 2mπ ,
where m is an integer. As k2 = 2π/λ2, we get the value of the antireflection layer thickness d=
λ2 4
+m
λ2 2
.
(3.34)
The minimum inter-layer thickness is equal to a quarter of the wavelength of a wave in the second material. Now from the condition R = 0 we get
(n1 − n2 )(n2 + n3 ) − (n2 − n3 )(n1 + n2 ) = 0 , which gives us the condition of permittivity
44 book - 3
ε r 2 = ε r1 ε r 3
or
n2 = n1 n3 .
(3.35)
Results (3.34) and (3.35) are parameters of the well known quarter-wavelength transformer frequently used in microwave engineering. __________________________________
3.3 Oblique incidence of a plane electromagnetic wave to a plane boundary In the case of the oblique incidence of a plane electromagnetic wave to a plane boundary between two different materials x and we have to treat two cases Hr kr Ei separately. We have to differentiate the two Er cases of polarization of an incident wave. Hi ϕi ki We assume a wave with a linear kix polarization. In the first case, vector E is n kiz perpendicular to the plane of incidence, ϕi ϕr which is defined by normal vector n together with the propagation vector of an incident wave ki, Fig. 3.7. This case is marked as the incidence of a wave with y z horizontal or perpendicular polarization. This notification follows from the theory of ϕt wave propagation above the earth ground and, e.g., their incidence to the ionosphere. Ht Et The second case is called the incidence of a wave with vertical or parallel kt polarization, as vector E is parallel to the plane of incidence. Fig. 3.8 Horizontal polarization. In this case we assume the orientation of all vectors as defined in Fig. 3.8. Our task is to determine the relation between the amplitudes of the electric fields of all three waves. To do this we have to force an electric field and a magnetic field to fulfill the boundary conditions for tangential components at plane x = 0 (1.18) and (1.20), assuming that no current passes along the boundary. Let us first express the propagation vectors, Fig. 3.8, in their components k i = −k ix x 0 + k iz z 0 =
,
(3.36)
k r = k1 cos(ϕ r ) x 0 + k1 sin (ϕ r ) z 0 ,
(3.37)
k t = −k 2 cos(ϕ t ) x 0 + k 2 sin (ϕ t ) z 0 ,
(3.38)
= −k1 cos(ϕ i ) x 0 + k1 sin (ϕ i ) z 0
So the electric and magnetic fields of the incident wave can be expressed in the form Ei = y 0 Ei 0 e − jk i ⋅r = y 0 Ei 0 e jk1 cos(ϕi )x e − jk1 sin (ϕi )z ,
(3.39)
45 book - 3
Hi =
k i × Ei
ωµ1
=
Ei 0 (− cos(ϕi ) z 0 − sin (ϕi ) x 0 ) e jk1 cos(ϕi )x e − jk1 sin (ϕi )z , Z1
(3.40)
For the reflected and transmitted waves we have
Hr =
Ht =
E r = −y 0 E r 0 e − jk r ⋅r = −y 0 E r 0 e − jk1 cos(ϕ r )x e − jk1 sin (ϕ r )z ,
(3.41)
E t = y 0 Et 0 e − jk t ⋅r = y 0 Et 0 e jk2 cos(ϕt )x e − jk2 sin (ϕt )z ,
(3.42)
k r × Er
ωµ 1 k t × Et
ωµ 2
Er 0 (− cos(ϕ r ) z 0 + sin (ϕ r ) x 0 ) e − jk1 cos(ϕr )x e − jk1 sin (ϕr )z , Z1
=
=
Et 0 (− cos(ϕ t ) z 0 − sin (ϕ t ) x 0 )e jk2 cos(ϕt )x e − jk2 sin (ϕt )z , Z2
(3.43)
(3.44)
The boundary conditions at x = 0 are Eiy (0, z ) − E ry (0, z ) = Ety (0, z ) ,
(3.45)
H iz (0, z ) + H rz (0, z ) = H tz (0, z )
(3.46)
Inserting into (3.45) the expressions for the electric field from (3.39) and similar expressions for Er and Et (3.41) and (3.42) we get
Ei 0 e − jk1 sin (ϕi )z − E r 0 e − jk1 sin (ϕ r )z = Et 0 e − jk2 sin (ϕt )z .
(3.47)
This condition must be fulfilled for each coordinate z. This means that the exponents in all terms must be equal. This results in
k1 sin (ϕ i ) = k1 sin (ϕ r ) = k 2 sin (ϕ t ) This equation defines Snell’s first and second laws
ϕi = ϕ r ,
(3.48)
k1 sin (ϕ i ) = k 2 sin (ϕ t ) .
(3.49)
Snell’s first law tells us that the angle of incidence equals the angle of reflection. Snell’s second law can be used to calculate the angle of transmission. Boundary condition (3.47) now reduces to
Ei 0 − E r 0 = Et 0 .
(3.50)
Similarly, inserting (3.40) and analogous formulas for Hr and Ht (3.43) and (3.44) into (3.46), and applying Snell’s laws, we get
46 book - 3
− cos(ϕ i )
Ei 0 E E − cos(ϕ i ) r 0 = − cos(ϕ t ) t 0 . Z1 Z1 Z2
(3.51)
Reflection coefficient R⊥ and transmission coefficient T⊥ for the oblique incidence of the plane wave with horizontal polarization can be now derived from the set of equations (3.50) and (3.51)
R⊥ =
E r 0 Z1 cos(ϕ t ) − Z 2 cos(ϕ i ) µ1k 2 cos(ϕ t ) − µ 2 k1 cos(ϕ i ) = = , Ei 0 Z1 cos(ϕ t ) + Z 2 cos(ϕ i ) µ1k 2 cos(ϕ t ) + µ 2 k1 cos(ϕ i )
(3.52)
T⊥ =
Et 0 2Z 2 cos(ϕ i ) 2µ 2 k1 cos(ϕ i ) = 1 − R⊥ = = . Ei 0 Z1 cos(ϕ t ) + Z 2 cos(ϕ i ) µ1k 2 cos(ϕ t ) + µ 2 k1 cos(ϕ i )
(3.53)
Note that the derived formulas hold only for the vector orientation shown in Fig. 3.7. Vertical polarization. In this case we assume the orientation of all vectors as defined in Fig. 3.9. The procedure for calculating the reflection and transmission coefficients is similar as the procedure used in the case of horizontal polarization. We exchange the form of electric field vectors by the form of magnetic field vectors. The propagation vectors are defined by (3.36) to (3.38). The field vectors, Fig. 3.9, are
Ei Hi
x
Er
kr
ki
Hr n
ϕi ϕr y
z Et
ϕt Ht kt
Fig. 3.9 E i = Ei 0 (cos(ϕ i ) z 0 + sin (ϕ i ) x 0 ) e jk1 cos(ϕi )x e − jk1 sin (ϕi )z , Hi = y0
Ei 0 jk1 cos(ϕi )x − jk1 sin (ϕi )z e e , Z1
(3.55)
E r = E r 0 (− cos(ϕ r ) z 0 + sin (ϕ r ) x 0 ) e − jk1 cos(ϕ r )x e − jk1 sin (ϕ r )z , Hr = y0
(3.54)
Er 0 − jk1 cos(ϕ r )x − jk1 sin (ϕ r )z e e , Z1
E t = Et 0 (cos(ϕ t ) z 0 + sin (ϕ t ) x 0 ) e jk2 cos(ϕt )x e − jk2 sin (ϕt )z ,
(3.56) (3.57)
(3.58)
47 book - 3
Ht = y0
Et 0 jk2 cos(ϕt )x − jk2 sin (ϕt )z e e . Z2
(3.59)
Field distributions (3.54) to (3.59) are inserted into boundary conditions (1.18) and (1.20) assuming K = 0. We get using Snell’s laws the set of equations cos(ϕ i )(Ei 0 − E r 0 ) = cos(ϕ t ) Et 0 ,
(3.60)
(E i 0 + E r 0 )
(3.61)
Z1
=
Et 0 . Z2
The reflection and transmission coefficients can be calculated from these two equations
=
Z1 cos(ϕ i ) − Z 2 cos(ϕ t ) ε 2 k1 cos(ϕ i ) − ε 1k 2 cos(ϕ t ) = , Z1 cos(ϕ i ) + Z 2 cos(ϕ t ) ε 2 k1 cos(ϕ i ) + ε1k 2 cos(ϕ t )
(3.62)
T =
2Z 2 cos(ϕ i ) 2ε 1k 2 cos(ϕ i ) = . Z1 cos(ϕ i ) + Z 2 cos(ϕ t ) ε 2 k1 cos(ϕ i ) + ε1k 2 cos(ϕ t )
(3.63)
R
Angle ϕt can be expressed in formulas (3.52), (3.53) and (3.62), (3.63) by the angle of incidence k 2 cos(ϕ t ) = k 2 1 − sin (ϕ t ) = k 2 2
2
k sin (ϕ t ) = k 22 − k12 sin 2 (ϕ i ) . 1 − 1 k2
(3.64)
The reflection and transmission coefficients in the case of both vertical and horizontal polarization depend not only on the parameters of the material but also on the angle of incidence. In the case of materials with nonzero conductivity these coefficients are complex numbers. In many cases the two materials are dielectrics. Then using (3.64) and µ1 = µ2 = µ0 we can express the reflection coefficients in the form − R⊥ =
R
=
ε1 ε cos(ϕ i ) + 1 − 1 sin 2 (ϕ i ) ε2 ε2 , ε1 ε1 2 cos(ϕ i ) + 1 − sin (ϕ i ) ε2 ε2 ε2 ε cos(ϕ i ) − 1 − 1 sin 2 (ϕ i ) ε1 ε2 . ε2 ε1 2 cos(ϕ i ) + 1 − sin (ϕ i ) ε1 ε2
(3.65)
(3.66)
Reflection coefficients (3.65) and (3.66) are plotted in Fig. 3.10 assuming that ε1/ε2 = 3, and ε1/ε2 = 1/3. In the following text we will discuss some special cases of the oblique incidence of a plane electromagnetic wave to a plane boundary between two different materials.
48 book - 3
R
R
1 ε1/ε2=1/3
ε1/ε2=3
0.8
0.8
-
0.6
0.2 0
-
0.2
ϕB
+
30
R⊥
0.4 R
60
0
90
ϕc
+
0.6
R⊥
0.4
0
1
0
ϕB
R +
30
60
90
ϕi (°)
ϕi (°)
Fig. 3.10
the plane of ______________________________ incidence Example 3.3: A plane linearly polarized electromagnetic wave with amplitude Ei0 = 10 V/m is incident to a plane Ei║ boundary between two dielectric materials with permittivities εr1=1 and εr2=8. Its frequency is 30 MHz. The angle of Ei0 incidence is ϕi = 30°. The electric field contained with the Hi0 plane of incidence angle α0 = 45°. Calculate the electric and α0 magnetic fields and the transmitted power of the reflected wave and the transmitted wave. Calculate power transmitted Ei┴ kr through the boundary. An incident wave must be decomposed to two waves, the first one polarized perpendicular to the plane of Fig. 3.11 incidence, while the second wave has a parallel polarization, see Fig. 3.11, where the wave is observed in its direction of propagation. The two particular waves have amplitudes Ei ⊥ = Ei 0 sin (α 0 ) = 7.07 V/m Ei
= Ei 0 cos(α 0 ) = 7.07 V/m
The angle of transmission is
k1 sin (ϕ i ) = 0.177 k2
ϕ t = arcsin
⇒
ϕ t = 10.2°
The amplitudes of the electric field of the particular wave with perpendicular and parallel polarization are
E r ⊥ = Ei ⊥
µ1k 2 cos(ϕ t ) − µ 2 k1 cos(ϕ i ) = −0.53 Ei ⊥ = −3.75 V/m µ1k 2 cos(ϕ t ) + µ 2 k1 cos(ϕ i )
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Et ⊥ = Ei ⊥
2 µ 2 k1 cos(ϕ i ) = 0.47 Ei ⊥ = 3.32 V/m µ1k 2 cos(ϕ t ) + µ 2 k1 cos(ϕ i )
Er
= Ei
ε 2 k1 cos(ϕ i ) − ε1k 2 cos(ϕ t ) = 0.43 Ei ε 2 k1 cos(ϕ i ) + ε1k 2 cos(ϕ t )
Et
= Ei
2ε 1k 2 cos(ϕ i ) = 0.5 Ei ε 2 k1 cos(ϕ i ) + ε1k 2 cos(ϕ t )
Now we can draw diagrams showing the orientation of the electric field vectors in a reflected wave and in a transmitted wave, Fig. 3.12. The amplitudes of the reflected and transmitted waves and the angles of their declination from the plane of incidence, Fig. 3.12, are
= 3.54 V/m
reflected wave plane of incidence
transmitted wave plane of incidence
Er║
Et║
Et0 Er0 kr
αt
αr
Ht0
Et┴
E r 0 = E r2⊥ + E r2 = 4.8 V/m
Hr0
(a)
α r = arctg
= 3.05 V/m
Er ⊥ = −51° Er
kt
Er┴ (b) Fig. 3.12
Et 0 = Et2⊥ + Et2 = 4.9 V/m
α t = arctg
Et ⊥ = 43° Et
To calculate the amplitudes of a magnetic field we have to calculate the wave impedances Z1 =
µ0 = 120π Ω , ε0
H i 0 = 0.0265 A/m ,
Z2 =
µ0 = 133 Ω , ε 0ε r 2
H r 0 = 0.0127 A/m ,
H t 0 = 0.0367 A/m
The densities of the active power transmitted by particular waves are S iav =
1 Ei 0 H i 0 = 0.1325 W/m 2 , 2
S rav =
1 E r 0 H r 0 = 0.0304 W/m 2 , 2
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S tav =
1 Et 0 H t 0 = 0.09 W/m 2 . 2
To calculate the power transmitted through the boundary we have to decompose these powers to components parallel and perpendicular to the boundary, Fig. 3.13. The power transmitted through the boundary is S xav = S iav cos(ϕ i ) − S rav cos(ϕ i ) = S tav cos(ϕ t ) = 0.0886 W/m 2 , The power transmitted through the boundary must be of the same value when calculated from both sides. The power transmitted along the boundary is
ϕi
S z1 = S iav sin (ϕ i ) + S rav sin (ϕ i ) = 0.08145 W/m 2 ,
Siav
Srav
S z 2 = S tav sin (ϕ t ) = 0.0156 W/m 2 .
Stav
ϕt
The power transmitted in the direction parallel to the boundary has different values in the two materials. ____________________________________
Fig. 3.13
Total transmission. Fig. 3.10 shows that the reflection coefficient for the vertical polarization equals zero at a certain angle. This angle is called Brewster’s angle ϕB. This means that there is no reflection at Brewster’s angle, and the whole power is transmitted to the second material. Putting (3.66) equal to zero, we get
sin (ϕ B ) =
ε2 , ε1 + ε 2
or
tg (ϕ B ) =
ε2 . ε1
(3.67)
This angle is sometimes called Brewster’s polarization angle. This name is given because an incident wave with an elliptic polarization when it is incident under this angle makes a reflection of only the linearly polarized wave, as there is no reflection of the component with an electric field parallel to the plane of incidence. This effect is used in polarizing elements. Total reflection. The plot in Fig. 3.10 drawn for the case ε1>ε2 shows that the reflection coefficients for the waves of the two kinds of polarization are equal to one starting from a certain angle. This angle is called the critical angle ϕc. The value of this angle follows from Snell’s law (3.49). In the case ε1<ε2 we have so called refraction to a normal, Fig. 3.14a, in which ϕi>ϕt. In the case ε1>ε2 we have so called refraction from a normal, Fig. 3.14b, at which ϕi<ϕt. Now angle ϕt can be equal to 90° assuming that ϕi = ϕc. From (3.49) we get for ε1>ε2
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ϕi
ϕi
ε1
ε1
ε2
ε2
ϕt
ϕt a
b Fig. 3.14
sin (ϕ c ) =
ε2 . ε1
(3.68)
At angles ϕi > ϕc the refraction angle has a complex value and R = 1 , Fig. 3.10. This means that the whole power is reflected back. This of course does not mean that there is no field in the second material. Let us now investigate the field distribution created after total reflection on the boundary between two dielectric materials with permittivities ε1>ε2 and assuming ϕi > ϕc. We assume only a wave with a horizontal polarization to simplify the derived formulas. The results are also valid in the case of a wave with a vertical polarization. Angle ϕt has a complex value, and from (3.49) we have
sin (ϕ t ) =
ε1 sin (ϕ i ) > 1 , ε2
cos(ϕ t ) = 1 − sin 2 (ϕ t ) = j cos(ϕ t ) .
Now the field in the second material is, Fig. 3.8 and (3.42), E t = T⊥ Ei 0 y 0 e − jk2 [-x cos(ϕ t )+ z sin (ϕ t )] = T⊥ Ei 0 y 0 e
k2 x cos (ϕ t )
e − jk2 z sin (ϕ t ) .
(3.69)
This formula describes a so-called surface wave. The amplitude of this wave decreases exponentially in the direction into the second material and propagates in the direction of axis z along the boundary with the propagation constant k 2 z = k 2 sin (ϕ t ) = k1 sin (ϕ i ) .
(3.70)
The phase velocity of this wave is v pz =
ω k2z
=
ω
k 2 sin (ϕ t )
=
ω
k1 sin (ϕ i )
< v p2 .
(3.71)
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Due to this the surface wave is called a slow wave in the second material. This is the case of a non-uniform electromagnetic wave, as planes of a constant amplitude are parallel with the boundary, and at the same time planes of a constant phase are perpendicular to the boundary. The field distribution is plotted in Fig. 3.15. In the case of total reflection we have R =1 and consequently
x
λx =
vpz>v1 fast wave
2π k1 cos(ϕ i )
Ey
R = exp( jψ ) The field in the first material is the superposition of the incident and reflected waves. These waves are (3.39) and (3.41)
vpz
E i = y 0 Ei 0 e − jk1[− x cos(ϕi )+ z sin (ϕi )] ,
Fig. 3.15
E r = −y 0 E r 0 e − jk1[x cos(ϕi )+ z sin (ϕi )] ,
their superposition is
[
]
E1 = E i + E r = y 0 Ei 0 e jk1x cos (ϕi ) − R⊥ e − jk1x cos(ϕi ) e − jk1z sin (ϕi ) = = y 0 E i 0 e j [k1x cos(ϕi )−ψ / 2] − e − j [k1x cos(ϕi )−ψ / 2] e jψ / 2 e − jk1z sin (ϕi ) = . = y 2 jE sin[k x cos(ϕ ) − ψ / 2]e jψ / 2 e − jk1z sin (ϕi )
[
0
i0
]
1
(3.72)
i
The magnetic field can be derived in a similar way using distributions (3.40) and (3.43) H1 = H i + H r = 2
Ei 0 {− x 0 sin (ϕ i )sin[k1 x cos(ϕ i ) − ψ / 2] − Z1
− z 0 cos(ϕ i ) cos[k1 x cos(ϕ i ) − ψ / 2]}e jψ / 2 e − jk1z sin (ϕi ) .
(3.73)
(3.72) and (3.73) describe a wave which propagates in the z direction and has the character of a standing wave in the x direction. The distribution of the electric field is plotted in Fig. 3.15. The phase velocity of this wave v pz =
ω k1z
=
ω
k1 sin (ϕ i )
> v p1 =
ω k1
.
(3.74)
For this reason, the surface wave in the first material is known as a fast wave. Later we will explain the principle of a dielectric waveguide on the basis of the total reflection and of this surface wave.
53 book - 4
The oblique incidence of a plane wave to the surface of a lossy material. Let us calculate the distribution of the electric field in the second lossy material, assuming horizontal polarization. The transmitted wave is (3.42) E t = y 0T⊥ Ei 0 e − jk2 [− x cos(ϕt )+ z sin (ϕt )] .
The second material is a lossy material with nonzero conductivity, consequently its phase constant is a complex number k2 = β2 – jα2. From Snell’s law (3.49) it follows that sin(ϕt) is a complex number and cos(ϕt) = a + jb. Inserting these values into the formula for an electric field we get E t = y 0T⊥ Ei 0 e x (α 2a − β 2b )e − j [− x ( β 2a +α 2b )+ k1z sin (ϕt )] .
This formula describes a non-uniform electromagnetic wave. The planes of a constant amplitude are parallel with the boundary, Fig. 3.16, whereas the planes of a constant phase are determined by the equation
(3.75)
ϕi ε0, µ0 ε2, µ2, σ2 plane of a constant phase
x β 2 a + α 2b − k1z sin ϕ t
= const .
plane of a constant amplitude
ϕT
This gives the angle under which the wave propagates in the second material, Fig. 3.16, Fig. 3.16 tg (ϕ T ) = −
k1 sin (ϕ i ) . β 2 a + α 2b
(3.76)
Increasing the conductivity of the second material causes a rise of α2 and β2, which means that angle ϕT decreases to zero. Finally a wave in a well conducting material propagates perpendicular to the boundary, independently of the angle of incidence, and is a uniform wave which is of course attenuated very fast.
3.4 Problems 3.1 A plane electromagnetic wave is incident from the air to the plane surface of a dielectric. A reflection causes a standing wave in the air with standing wave ratio p=2.7. Calculate the permittivity εr of the dielectric and reflection coefficient R. There are two solutions: R=0.46, εr=0.138 R=-0.46, εr=7.2 3.2 What permittivity has a dielectric, the surface of which reflects at most 1% of the energy of a wave incident perpendicular. 0.67<εr<1.5
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3.3 A plane electromagnetic wave is incident from the air to the plane surface of sea water with parameters εr=81 and σ=5 S/m. The frequency is 10 MHz, and the amplitude of the wave is Eim=100 V/m. Calculate then electric field intensity at depth 1 m under the surface. Et (1) = 0.00175 V / m 3.4 Design an antireflection layer on the surface of a silicon photodiode for light with wavelength 0.8 µm. The permittivity of silicon is 12. d=(0.11+m0.22) µm εr2=3.46 3.5 Calculate the complex amplitude of a wave passing the layered structure from Fig. 3.6. The amplitude of the incident wave is Ei=10 V/m, the parameters of the structure are: εr1=1, εr2=4, εr3=6, d=8 cm, and the frequency is 10 GHz. a3=6e-j2π/3 V/m 3.6 Calculate the average value of Poynting’s vectors representing the power transmitted by the surface wave described by (3.69) and (3.72). 2
S 2 zav =
k 2 T⊥ Ei20 2ωµ 0
sin (ϕ t )
2 Ei20 ψ S1zav = sin (ϕ i )sin 2 k1 x cos(ϕ i ) − r Z1 2 There is no active power transmitted in the x direction. 3.7 Calculate the slant of an output glass window with permittivity εr=2.13, as a wave with a parallel polarization passes through it without losses, Fig. 3.17. The angle must be equal to Brewster’s angle. ϕ = ϕB = 55.6°
ϕ
εr
Fig. 3.17 3.8 A plane electromagnetic wave is incident to the plane boundary between two dielectrics with permittivities εr1=2.53, εr2=1. Calculate the minimum angle of incidence at which total reflection takes place. ϕic = 33° 3.9 A plane electromagnetic wave with perpendicular polarization is incident from the air to the plane surface of a material with parameters εr2=1.5, σ2=2.10-5 S/m. The frequency is f = 2 MHz, and the angle of incidence is ϕi = 60°. Calculate the direction of propagation of a wave transmitted into the second material, its attenuation and phase constants and phase velocity. ψT = 44° vp2 = 2.42.108 m/s β 2 = 0.0519 m-1
α 2 = 0.0382 m-1
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3.10 Calculate the angle of propagation of the wave transmitted into copper after the incidence of a plane electromagnetic wave from the air under angle ϕi = 45°. The frequency is 2 MHz the, and conductivity of copper is σ = 5.8.107 S/m. ψT = 0° 3.11 Calculate the phase velocity of the surface wave which is excited by the total reflection of a plane electromagnetic wave that is incident from a dielectric with permittivity εr1=4 to its boundary with the air. The angle of incidence is ϕi = 45°, and the frequency is 10 MHz. vpz = 2.16.108 m/s
4. SOLUTION OF MAXWELL EQUATIONS AT VERY HIGH FREQUENCIES We will assume the propagation of an electromagnetic wave in a generally nonhomogeneous lossless medium. The medium will be described by the distribution of the refractive index n( x, y, z ) = n(r ) = ε r (r ) . We assume a very high frequency, so that the wavelength is very short, much shorter than the distances at which the refractive index varies significantly. We assume the distribution of the electric field of a wave propagating in our medium in a form analogous to a plane electromagnetic wave
E = E m (r ) e − jk0φ (r ) ,
(4.1)
where vector function Em(r) determines the distribution of the field amplitude and function φ(r) determines the distribution of the phase, and k0 is the phase constant in the vacuum k 0 = ω µ 0ε 0 . The equation Em(r) = const defines the planes of a constant amplitude, and the equation φ(r) = const defines planes of a constant phase, known as wave-fronts. These two sets of planes are generally different. Nevertheless, we will show that a propagating wave has locally the same properties as a plane TEM wave propagating in a homogeneous space. Let us insert (4.1) and a similar formula for the magnetic field into Maxwell’s equations (1.12), (1.13) and (1.14)
(
)
(4.2)
(
)
(4.3)
rot H m e − jk0φ = jωε E m e − jk0φ , rot E m e − jk0φ = − jωµ H m e − jk0φ ,
(
)
div ε E m e − jk0φ = 0 .
(4.4)
Using formula (13.78) from the mathematical appendix describing the rotation applied to the product of two functions, we rewrite (4.2) e − jk0φ rot (H m ) − jk 0 e − jk0φ gradφ × H m = jωε E m e − jk0φ . Reducing the exponential function we get
56 book - 5
rot (H m ) − jk 0 gradφ × H m = jωε E m ,
(4.5)
and similarly from equation (4.3) rot (E m ) − jk 0 gradφ × E m = − jωµ H m .
(4.6)
Using formula (13.70) for the divergence of the product of two functions, we rewrite (4.4) to
(
)
E m ⋅ grad ε e − jk0φ + ε e − jk0φ div(E m ) = 0
[
]
E m ⋅ e − jk0φ grad(ε ) − jk 0 e − jk0φ ε gradφ = −ε e − jk0φ div(E m ) . Finally we have E m ⋅ [grad(ε ) − jk 0 ε gradφ ] = −ε div(E m ) .
(4.7)
Rewriting (4.5), (4.6) and (4.7) we get gradφ × H m +
ε0 1 ε r Em = rot (H m ) , µ0 jk 0
(4.8)
gradφ × E m −
ε0 1 Hm = rot (E m ) , µ0 jk 0
(4.9)
gradφ ⋅ E m =
1 gradε + div(E m ) . Em jk 0 ε
(4.10)
Now we use the assumption that we treat the field at very high frequencies. This means that 1 f → ∞ , and consequently k 0 → ∞ and → 0 . We can now set the right hand sides of jk 0 (4.8), (4.9) and (4.10) equal to zero. gradφ × H m +
ε0 ε r Em = 0 , µ0
(4.11)
gradφ × E m −
ε0 Hm = 0 , µ0
(4.12)
gradφ ⋅ E m = 0 .
(4.13)
As φ(r) = const determines the wave-fronts, and vector grad φ is perpendicular to these wavefronts we have a very important result. According to (4.13) vectors gradφ and Em are perpendicular. This means that vector Em is tangent to the wave-front, consequently it is perpendicular to local direction of the wave propagation. Inserting now from (4.12) for Hm to (4.11) we have
57 book - 5
ε0 ε gradφ × (gradφ × E m ) = − 0 ε r E m . µ0 µ0 Rewriting the double vector product (13.11) we have gradφ (gradφ ⋅ E m ) − E m (gradφ ⋅ gradφ ) = −ε r E m . Owing to (4.13) we have 2
gradφ
= ε r = n2 ,
(4.14)
this equation couples the wave-fronts with the distribution of the relative permittivity. We are thus able to determine the distribution of phase φ solving the partial differential equation 2
2
2
∂φ ∂φ ∂φ 2 + + = n . ∂x ∂y ∂z
(4.15)
The mutual relations between the electric and magnetic fields follow from (4.11) and (4.12) Em = -
µ 0ε 0 gradφ × H m , ε
(4.16)
Hm =
µ 0ε 0 gradφ × E m . µ
(4.17)
Using (4.14), vector gradφ can be expressed as grad φ = grad φ l 0 = nl 0 ,
(4.18)
where unit vector l0 determines the local direction of the wave propagation. (4.16) and (4.17) tell us that vectors Em and Hm are mutually perpendicular and are perpendicular to the direction of the wave propagation at each point, Fig. 4.1. This means that a wave propagating at very high frequencies in a generally nonhomogeneous medium behaves locally as a TEM wave propagating in a free space. The power transmitted by this wave is defined by Poynting’s vector. Locally this vector again corresponds to the Poynting vector of a plane TEM electromagnetic wave S av =
gradφ = nl0
Em
r Hm
Fig. 4.1
1 1 ε 2 Re[E m × H m ] = Em l0 . 2 2 µ
(4.19)
The power is transmitted in the l0 direction, i.e., perpendicular to the wave-fronts.
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The row of vectors l0 represents a ray. Let us derive the equation describing the ray. We will start with equation (4.18), where, according to Fig. 4.2, l0 = dr/ds. Deriving l0 (4.18) over s we have ray
d dr d n(r ) = grad[φ (r )] ds ds ds
φ
ds
Let us now pay attention to the right hand side part of this equation dx ∂ d grad φ = ∑ i ds i ds ∂xi
s dr
φ+dφ r
∂φ ∑ ∂x x j 0 , j j
r+dr 0
where xi is the i-th coordinate, i.e., x, y or z. Changing the order of the summation we get dx
∂φ
∂
∑ dsi ∂x ∑ ∂x i
i
= ∑ x j0 j
j
j
x j0 = ∑ x j0 j
Fig. 4.2
∂ ∂φ dxi ∂ dr = ∑ x j0 grad[φ (r )] ⋅ = ∑ ∂x j i ∂xi ds ∂x j ds j
∂ n(r ) l 0 ⋅ l 0 = grad[n(r )] ∂x j
Consequently we get d dr n(r ) = grad[n(r )] , ds ds
(4.20)
which is the equation describing the trajectory of the ray. The changes in the direction of the rays are caused by changes in the refraction index. Solving (4.20) we are able to find the rays and to represent the wave by rays. In the case of a homogeneous material we have n = const, consequently grad(n) = 0 and d dr n(r ) = 0 ds ds
=>
r = sa + b
where a and b are constant vectors. The ray represents a straight line. According to (4.19) the power transmitted by the wave propagates along rays. From this we can estimate the value of the field amplitude. Let us take an area dS1 on the wave-front φ1. The wave transmits power P1 through this area, Fig. 4.3. The rays starting at the perimeter of area dS1 circumscribe a tube of constant power and mark on wave-front φ2 area dS2 through which the same power P1 passes. This power is
P1
dS2
φ2 a tube of constant power P1
φ1
dS1 Fig. 4.3
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P1 =
1 ε 1 ε 2 2 E m1 dS1 = E m 2 dS 2 . 2 µ 2 µ
From this equation we get E m 2 = E m1
dS1 . dS 2
(4.21)
The amplitude of the electric field is higher where the area is smaller, which corresponds to the higher density of the rays. This technique using rays to describe waves is known as the geometric optic. The waves are described by a set of rays which propagate independently. Their trajectories are described by (4.20). If they fall on any boundary, they behave according to Snell’s laws. ________________________________ Example 4.1: Determine the trajectory of the ray which propagates at the plane y = 0 in a dx dielectric layer, see Fig. 4.4. The boundary conditions are x(z=0) = 0, = tg (ϑ0 ) ≈ ϑ0 . The dz refractive index depends on the x component
(
)
n( x ) = n0 1 − k x 2 , where constant k << 1 . This is known as the parabolic distribution of the refractive index. Assume that the ray changes its direction very little in the relation to the z axis, so we can put s = z. Such a ray is known as a par-axial ray. The trajectory of the ray is described by (4.20). This equation can be generally solved only numerically. Assuming a paraxial ray we have s ≈ z , and (4.20) now reads d dr dn ( x ) x0 , n( x ) = dz dz dx
z
ϑ0 x Fig. 4.4
r = xx0 + zz0 .
d dx dn( x ) n( x ) x 0 + z 0 = x0 dz dx dz as
dn = 0 we get dz d dx dn ( x ) , n( x ) = dz dz dx
=>
n( x )
d2x = −2 k n0 x => dz 2
2 k n0 x d2x =− 2 dz n0 1 − kx 2
(
)
Now we neglect kx2 comparing to 1 and we get
60 book - 5
d2x = −2 k x . dz 2 The solution of this equation is
x = A sin (qz ) + B cos(qz ) ,
q = 2k .
Applying the boundary conditions we get x(0) = 0
=>
B=0 dx dz
dx = A q cos(qz ) , dz x=
ϑ0 2k
(
= ϑ0
A=
z =0
ϑ0
π 2k
,
,
2k
z0
)
sin 2k z .
ϑ0
A part of the ray trajectory is shown in Fig. 4.5. The parameters of the trajectory are zp =
=>
z0 =
xmax
zp z
x Fig. 4.5 zp 2
,
x max =
ϑ0 2k
The ray is coupled to the area around the z axis, and it bends and returns back. For x < 0 the ray has a symmetric shape. __________________________________
5. GUIDED WAVES Up to now we have studied electromagnetic waves propagating in a free space filled by a homogeneous or non-homogeneous material. This is very important for the theory of wave propagation in the atmosphere, and is used in the design of communication systems, particularly the channels represented by transmitting and receiving antennas, and the space between them. On the other hand, there are waves the existence of which is based on the presence of a boundary between different materials. These boundaries can be of various shapes, depending on the required behaviour and the application, and they form a transmission line. We have studied the surface wave excited due to the incidence of a plane electromagnetic wave on a plane boundary of two dielectric materials under an angle greater than the critical angle. This surface wave is one from the examples of guided waves. A wide variety of transmission lines are used. Their geometry depends on many factors. The most important are the technology of the system or the circuit in which the line is applied, the frequency band and the transmitted power. Transmission lines can be generally categorized into three groups according to the form of the transmitted wave. In the first group, there are transmission lines which are able to transmit a TEM wave. Later we will see that such a transmission line must be able to transmit DC current, so it must consist of at least two
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separate conductors. Examples of these transmission lines are shown in Fig. 5.1a. The two transmission lines shown in Fig. 5.1b consist of two conductors, and are thus able to conduct DC current. These are planar lines with conducting strips located on a dielectric substrate. The electric field and the magnetic field have to fulfill boundary conditions on the surfaces of this substrate. Due to this, the longitudinal components of the fields are always present. The propagating wave is not a TEM wave. At low frequencies this wave can be treated as a TEM wave. Consequently it is known as a quasi TEM wave. The lines shown in Fig. 5.1c do not guide a TEM wave. The first two lines in Fig. 5.1c are known as waveguides as they guide a wave along their hollow center.
two-conductor transmission line
co-axial transmission line
parallel-plate waveguide
Fig. 5.1a Transmission lines with a TEM wave strip-line
microstrip line
coplanar waveguide dielectric
Fig. 5.1b Transmission lines with a quasi-TEM wave
waveguide with a rectangular cross-section
waveguide with a circular cross-section
optical fibre
Fig. 5.1c Transmission lines without a TEM wave To simplify the analysis of transmission lines we will assume lines without losses, i.e., dielectric materials with zero conductivity, and metals with infinite conductivity. The lines
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will be assumed to be infinitely long and longitudinally homogeneous. The particular materials will be homogeneous. The lines will be directed in the positive z-axis direction, which is thus equal to the direction of the wave propagation. The field distribution of a wave propagating along the transmission line in the z-axis direction will be calculated by solving the homogeneous wave equation (2.1). Therefore we will get only the waves, that can propagate along the line. These waves are known as the eigenmodes, and represent particular solutions of the wave equation. Let us apply the method of separation of variables for solving the wave equation. The longitudinal component of electric field Ez is assumed in the form E z ( x, y, z ) = E 0 z ( x, y )P( z ) .
(5.1)
This form of the field distribution is inserted into the wave equation (2.1) and is rewritten to ∆ T E0 z 1 d 2 P + + k2 = 0 E0 z P dz 2
(5.2)
where ∆ T represents Laplace’s operator calculated using derivatives over the transversal coordinates x and y. The first and second terms on the left hand side of (5.2) must to be equal to constants kp and kz, which represent the transversal propagation constant and the longitudinal propagation constant. In this way we decompose equation (5.2) into two equations for E0z and P
∆ T E0 z + k 2p E0 z = 0 ,
(5.3)
P ' '+ k z2 P = 0 .
(5.4)
The propagation constants kp and kz are coupled by k 2 = k p2 + k z2 .
(5.5)
Equation (5.4) has the solution describing the phase variation of the wave propagating along the line in the form P ( z ) = e − jk z z .
(5.6)
We omitted here the wave propagating in the negative z direction. All components of the electric and magnetic fields have the same character as Ez has. Consequently the derivatives of these components over z can be expressed simply. So for the i-th component of the electric field we have dE i = − jk z E i . dz
(5.7)
The character of the field described by (5.1) enables us to divide the modes propagating along the transmission line into two groups, and the modes in each group can be treated separately. These modes are the transversal electric (TE) modes and the transversal magnetic (TM) modes. The TE modes have Ez=0, and their field contains components: Ex,
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Ey, Hx, Hy, Hz, while in complement the TM modes have Hz=0, and their field contains components: Ex, Ey, Ez, Hx, Hy. The field distribution and the propagation constants of these modes are determined by solving the wave equation (5.3) for the longitudinal components of the electric and magnetic fields, and the transversal components are derived from the longitudinal components. We start from Maxwell’s first and second equations (1.12) and (1.13), putting σ=0 and Js=0. These two vector equations can be expressed in the scalar form ∂H z + jk z H y = jω ε E x , ∂y − jk z H x −
∂H y
∂H z = jω ε E y , ∂x
(5.9)
∂H x = jω ε E z , ∂y
+
∂x
(5.8)
(5.10)
∂E z + jk z E y = − jω µ H x , ∂y − jk z E x −
∂E y ∂x
(5.11)
∂E z = − jω µ H y , ∂x
(5.12)
∂E x = − jω µ H z . ∂y
+
(5.13)
From (5.12) we express Hy and insert it into (5.8). Now (5.8) contains only Ex, Ez, and Hz and in this way we get the dependence Ex on the longitudinal components of the electric and magnetic fields. Similarly we get the other transversal components. So we have Ex = −
k ∂E z jω µ ∂H z , + z 2 k p ∂y ω µ ∂x
(5.14)
Ey = −
k ∂E z jω µ ∂H z , − + z 2 k p ∂x ω µ ∂y
(5.15)
Hx =
jω ε ∂E z k z ∂H z , − k p2 ∂y ω ε ∂x
Hy = −
jω ε ∂E z k z ∂H z . + k p2 ∂x ω ε ∂y
(5.16)
(5.17)
Now putting Ez=0 we will get the transversal components Ex, Ey, Hx, Hy as functions of Hz and we have TE modes. Conversely, putting Hz=0 we will get the transversal components Ex, Ey, Hx, Hy as functions of Ez and we have TM modes.
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Let us now turn our attention to (5.5). Using k = ω µ ε we get the longitudinal propagation constant
k z = ω 2 µ ε − k p2 .
(5.18)
Three cases can be now distinguished. At high frequencies we have
ω 2 µ ε > k p2 and kz is a real number, so the corresponding mode represents a wave propagating along the transmission line. At low frequencies we have
ω 2 µ ε < k 2p and kz is an imaginary number, so the corresponding mode represents an evanescent wave which does not propagate along the transmission line and its amplitude exponentially decreases. The boundary point determines the so called cut-off frequency of the mode fc =
kp 2π µ ε
.
(5.19)
This means that the transmission line behaves the mode does not the mode propagates as a high-pass filter, as it transmits the mode propagate starting from the cut-off frequency, Fig. 5.2. Note that TEM modes have, as will be shown, f 0 fc kp=0 and, consequently, fc=0, and they can propagate from zero frequency. Fig. 5.2 Each transmission line has its dominant mode. This is the mode with the lowest possible cut-off frequency, i.e., with the lowest kp. It is desired to operate the transmission line in the frequency band of the single mode operation. This is the frequency band at which only the dominant mode propagates. This band is limited from above by the cut-off frequency of the nearest higher mode. When the two modes can propagate simultaneously, the transmitted signal is coupled to these two modes. As they have different propagation constants (5.18), they have different phase velocities and the field of these modes arrives at the output port with a different delay. This results in distortion of the transmitted signal. The particular transmission lines and the modes propagating along them will be studied in the following sections.
6. TEM WAVES ON A TRANSMISSION LINE 6.1 Parameters of a TEM wave Let us study a transmission line with a general cross-section, homogeneous along its infinite length, located in the unbounded homogeneous space parallel to the z-axis. Let the line have perfect conductors with infinite conductivity. We will study the propagation of a
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TEM (transversal electromagnetic) wave along this transmission line. The TEM wave has field components only in the transversal plane and Ez=0, Hz=0. On an infinitely long homogeneous line we assume dependence on the longitudinal coordinate in the form e− jk z z . Thus we have
E(x, y, z ) = E T ( x, y ) e − jk z z , H ( x, y, z ) = H T ( x, y ) e − jk z z .
(6.1)
The transversal field components have the form ET = E x x 0 + E y y 0 , H T = H x x 0 + H y y 0 . Inserting these forms into wave equations (1.41) and (1.43) in the space without sources we get ∆ T E T + k p2 ET = 0 ∆ T H T + k 2p H T = 0
where ∆ T represents Laplace’s operator applied according to transversal coordinates x and y, and k 2p = k 2 − k z2 = ω 2 µε − k z2 is a transversal propagation constant. As both the electric and magnetic fields have zero longitudinal components, these fields have lines of E and H lying only in the transversal planes and therefore they must have a constant phase in these planes. This determines kp = 0 ant thus kz = k. As a result we have ∆T HT = 0 ,
(6.2)
∆ T ET = 0 ,
(6.3)
Equations (6.2) and (6.3) are Laplace equations. The electric and magnetic field transversal components are solutions of the Laplace equation, and this electromagnetic field therefore has the character of a stationary field, of course except for its wave character in the longitudinal direction. As the field distribution in the transversal plane is the same as the distribution of the stationary field, the line must be able to conduct a DC current. Such a transmission line must consist of at least two conductors to transmit the DC current. Generally, a TEM wave can propagate only along a transmission line which consists of at least two conductors. As the field has the character of a stationary field we can unambiguously define the voltage and the current at any point on the transmission line. The voltage is defined by the integral along an arbitrary path c1 lying in the transversal plane u ( z , t ) = ∫ E T ⋅ ds c1
.
(6.4)
z = const .
This voltage depends only on the z-coordinate and time t. The electric current is determined by the integral along the closed path c2 which surrounds one of the conductors (Ampere’s law), and lies in the transversal plane i ( z , t ) = ∫ H T ⋅ dl c2
.
(6.5)
z = const .
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This current depends only on the z-coordinate and time t. Now we can express the electric field and the magnetic field using the voltage and the current E T ( x, y , z , t ) = e T ( x, y ) u ( z , t ) ,
(6.6)
H T ( x, y , z , t ) = h T ( x, y ) i ( z , t ) .
(6.7)
Vector function eT(x,y) represents the distribution of the electric field when DC voltage 1 V is connected to the line terminals. Inserting (6.6) into (6.4) we get
∫ e ( x , y ) ⋅ ds = 1 .
(6.8)
T
c1
Vector function hT(x,y) represents the distribution of the magnetic field when DC current 1 A passes the transmission line. This function is normalized to one, as follows from the insertion of (6.7) to (6.5), (6.9) ∫ h T ( x , y ) ⋅ dl = 1 . c2
The problem of calculating the field distribution is thus divided into two parts. The calculation of functions eT(x,y) and hT(x,y) can be done according to techniques known from the electrostatic field and the stationary magnetic field, solving (6.2) and (6.3). The reader is therefore able to solve this part of the problem. The calculation of voltage u(z,t) and current i(z,t) is known from circuit theory. This is described by the well known telegraph equations. Assuming a steady harmonic state, these equations are ∂U = − ( R + jω L ) I , ∂z
(6.10)
∂I = −(G + jωC )U . ∂z
(6.11)
R, L are line series resistance and inductivity per unit length, G and C are line parallel conductance and capacity per unit length. These primary parameters of a transmission line can be calculated by methods known from electromagnetic field theory. We can eliminate either the voltage or the current from the couple of these equations to get one equation of the second order ∂ 2U ∂z 2 ∂2I ∂z 2
− (R + jωL )(G + jωC )U = 0 ,
(6.12)
− (R + jωL )(G + jωC )I = 0 .
(6.13)
These equations are equivalent to the wave equation. The propagation of the TEM wave along the line is characterized by secondary parameters. These are propagation constant γ, characteristic impedance ZC, phase velocity v, and wavelength λ.
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We start by writing the solution of the telegraph equation for voltage U (6.12) ∂ 2U ∂z
2
− γ 2U = 0 ,
(6.14)
where the propagation constant is written from (6.12)
γ =
(R + jωL )(G + jωC )
.
(6.15)
Here we take the notation of the propagation constant, which is known from circuit theory. The lossless transmission line has R=0 and G=0 and the propagation constant will be
γ = − ω 2 LC = jω LC = jβ ,
(6.16)
where β is the phase constant in the sense of (2.11) valid for a plane electromagnetic wave propagating in a homogeneous space filled by a lossless material. Note that in most of the formulas valid for the TEM wave propagating along the transmission line we can substitute L for µ and C for ε , and we get formulas valid for the plane electromagnetic wave propagating in free space. The solution of (6.14) and the corresponding equation valid for the electric current are, assuming a single wave propagating on an infinitely long line in the positive z direction, U = U 0 e −γz ,
(6.17)
I = I 0 e −γz ,
(6.18)
where U0 and I0 are wave amplitudes. Inserting into (6.12) for the voltage and the current we get − γ U 0 e −γz = −(R + jωL ) I 0 e −γz ,
and from this formula we get, using (6.15), the characteristic impedance of the transmission line, which is defined as the ratio of the voltage amplitude and the current amplitude, ZC =
U 0 R + jωL R + jωL = = . I0 γ G + jωC
(6.19)
A lossless line has ZC =
L . C
(6.20)
The voltage and the current are in phase on a lossless line, as the characteristic impedance is a real number. We can again compare (6.20) with the characteristic impedance of a lossless material for a plane electromagnetic wave (2.26).
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The phase velocity is defined and can be determined in the same way as the phase velocity of a plane electromagnetic wave in free space (2.13). Assuming a lossless line we can get v=
1 ω = . β LC
(6.21)
This velocity determines the velocity of the propagation of a constant phase along the line. In the case of a lossless line it also determines the velocity of the energy transmitted along the line, i.e., the group velocity. The wavelength has again the same meaning as the wavelength of a plane electromagnetic wave propagating in an unbounded space (2.15)
λ=
v 2π = . f β
(6.22)
___________________________ Example 6.1: Calculate the secondary parameters of the co-axial transmission line shown in Fig. 6.1, assuming lossless materials. The line parameters are 2r1=0.46 mm, 2r2=1.5 mm, εr=2. Calculate the distribution of the electric -τ field and the magnetic field in the crosssection of this transmission line. From the theory of electromagnetic field we know the line capacity per unit length τ r3 2πε , C= r2 r2 ln r1 εr r1
and the line inductivity L=
µ 0 r2 ln . 2π r1
Fig. 6.1
The characteristic impedance is, according to (6.20),
ZC =
L = C
µ 0 r2 ln 2π r1 r 60 = ln 2 = 50.13 Ω 2πε 0ε r ε r r1 ln
r2 r1
It is evident that the given parameters correspond to the standard 50 Ω co-axial cable. The phase constant (6.19) is
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β = ω LC = ω
µ 0 r2 2πε ln = ω µε = k 0 ε r , 2π r1 r2 ln
r1
where k0 is the phase constant of the wave propagating in a vacuum. The wavelength (6.22) is
λ=
2π
β
=
2π k0 ε r
=
λ0 , εr
where λ0 is the wavelength of a plane electromagnetic wave in a vacuum. The distribution of the electric field and the magnetic field in the cross-section of this transmission line is known from electromagnetic field theory Er =
U τ = , 2πε r r ln (r2 r1 )
Hα =
I 2πr
.
Note that, using the formula for characteristic impedance ZC of this transmission line, ratio E r H α gives a value equal to the characteristic impedance for a plane electromagnetic wave propagating in the space filled by a dielectric with permittivity εr. ________________________________ The results of the above example can be generalized. The TEM wave propagates along a transmission line with a homogeneous dielectric material between the conductors in the same way as a plane electromagnetic wave propagating in the unbounded space filled by the same homogeneous material. This refers to the propagation constant, the wavelength, and the phase velocity. The characteristic impedance is different from the characteristic impedance of the material for a plane wave. The characteristic impedance of a transmission line is defined as the ratio of the amplitudes of the voltage and of the current. So it must depend not only on the material parameters but also on the line transversal dimensions. The characteristic impedance of the material for a plane wave is defined as the ratio of the field amplitudes, so it depends only on the material parameters. ______________________________ Example 6.2: Calculate the secondary parameters of the two-conductor transmission line (twin-lead transmission line) from Fig. 6.2, assuming lossless materials. In electromagnetic field theory we calculated the line capacity and inductivity per unit length C=
L=
πε a ln r
.
µ0 a ln . π r
r
r a
The characteristic impedance (6.20) is Fig. 6.2
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ZC =
L = C
µ0 a ln π r = 120 ln a . πε εr r ln
a r
The phase constant and the wavelength are
β = ω LC = ω µ 0ε 0ε r = k 0 ε r ,
λ=
λ0 . εr
_____________________________ The results of the above example validate the remarks made to Example 6.1. The TEM wave is not the only wave that can propagate along the above mentioned lines. So called waveguide modes, treated in Chapter 7, can propagate along these lines at sufficiently high frequencies above their cut-off frequency. These modes are undesired, and the lines must be designed to prevent the propagation of waveguide modes.
6.2 Transformation of the impedance along the line 6.2.1 An infinitely long line The general solution of the telegraph equations (6.10) and (6.11) on an infinitely long transmission line consists of two waves, one propagating to the right (denoted by subscript +), and the second propagating to the left (denoted by subscript -) U = U + e −γz + U − eγz , I = I + e −γz + I − eγz =
(6.23)
U + −γz U − γz e − e , ZC ZC
(6.24)
assuming the current amplitudes I + = U + Z C , I − = − U − Z C , as the current flowing to the left has the opposite orientation. The impedance at any point along the line is U + e −γz + U − e γz U , Z (z ) = = Z C I U + e −γz − U − eγz
(6.25)
On an infinitely long line with the only one source connected at z = −∞ , we have only one wave propagating to the right, and the above formulas read U = U + e −γz , (6.26) I = I + e −γz ,
(6.27)
The impedance at any point along the line is
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Z (z ) =
U+ = ZC . I+
(6.28)
The impedance at any point on an infinitely long transmission line is thus equal to the characteristic impedance. This gives us the recipe for realizing an infinitely long line. A line of finite length is terminated by an impedance equal to ZC. This ensures no reflection at the line termination, and the line behaves as an infinitely long line.
6.2.2 A line of finite length Let us take a transmission line of finite length l, terminated at the end by impedance ZL. The voltage U 2 = U (l ) and current I 2 = I (l ) at the end are known. Their ratio is ZL =
U2 . I2
(6.29)
To simplify this problem we use a new coordinate s=l-z measured from the line end, Fig. 6.3. As ds=-dz, we can rewrite equations (6.12) and (6.13) to the form ∂ 2U (s ) ∂s
2
∂ 2 I (s ) ∂s
2
− γ 2U (s ) = 0 ,
s 0
l-z
z
(6.30)
0 z l U2, I2
Fig. 6.3
− γ 2 I (s ) = 0 .
(6.31)
The solution of these equations is U = U + e γs + U − e −γs , I=
(
(6.32)
)
1 U + e γs − U − e −γs . ZC
(6.33)
For z=l we now have s=0 and U2 = U+ +U− , I2 =
1 (U + − U − ) . ZC
From these formulas we get the amplitudes of the wave traveling to the right and the wave traveling to the left as functions of the voltage and the current at the line end U+ =
1 (U 2 + Z C I 2 ) , 2
(6.34)
U− =
1 (U 2 − Z C I 2 ) . 2
(6.35)
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Inserting (6.34) and (6.35) into (6.32) and (6.33) we get the voltage and the current at any point on the transmission line U (s ) = U 2 cosh (γs ) + Z C I 2 sinh (γs ) , I (s ) = I 2 cosh (γs ) +
U2 sinh (γs ) . ZC
(6.36) (6.37)
The impedance at any point along the transmission line is Z=
Z + Z C tgh (γ s ) U U 2 cosh (γ s ) + Z C I 2 sinh (γ s ) = = ZC L , U2 Z C + Z L tgh (γ s ) I sinh (γ s ) I 2 cosh (γ s ) + ZC
(6.38)
Formula (6.38) is the basic formula used in the analysis and design of high frequency circuits. It determines how the terminating impedance is transformed to any point along the line. The propagation constant of a lossless line is determined by (6.16), which is a purely imaginary number. The characteristic impedance is a real number (6.20). Equations (6.36) and (6.37) can now read (13.35) and (13.36) U (s ) = U 2 cos(β s ) + jZ C I 2 sin (β s ) , I (s ) = I 2 cos(β s ) + j
U2 sin (β s ) . ZC
(6.39) (6.40)
The impedance at any point along the line is Z=
Z + jZ C tg (β s ) U = ZC L . Z C + jZ L tg (β s ) I
(6.41)
The two formulas (6.38) and (6.39) confirm the fact stated at the end of the preceding paragraph. Terminating the line by an impedance equal to characteristic impedance ZC, we get the impedance at any point equal to the characteristic impedance. The line behaves as an infinitely long line.
6.2.3 A line terminated by a short cut or by an open end A line terminated by a short cut or by an open end is exposed to total reflection, and thus a reflected wave has the same amplitude as the wave incident to the termination. As a result we get a standing wave. This effect is analogous to the perpendicular incidence of a plane electromagnetic wave to the surface of a well conducting material or a material with infinite permeability – a perfect magnetic material. An open end termination represents an infinite impedance, and the current passing through this impedance I 2 = 0 . Formulas (6.39) and (6.40) are now, using β = 2π / λ (2.15), 2π U = U 2 cos(β s ) = U 2 cos λ
s ,
(6.42)
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I= j
U U2 2π s . sin (β s ) = j 2 sin ZC ZC λ
These voltage and current distributions are plotted in Fig. 6.4. Formulas (6.42) and (6.43) describe a standing wave on a transmission line and are analogous to formulas describing the standing wave after the perpendicular incidence of a s plane electromagnetic wave to the surface of a perfect magnetic material. The active power transmitted by this wave is zero, which gives the wave its name. Using a probe sliding along the line we can detect the distribution of the voltage determined by (6.42) with stable maxima and s minima. The distance between the two adjacent maxima or minima is equal to a half wavelength. In this way we can measure the wavelength. The impedance at any point on the line is Z=
(6.43) u (s )
λ
3 λ/4
λ/2
λ/4
λ
3 λ/4
λ/2
λ/4
i (s )
Fig. 6.4
X(s)
U s = − jZ C cot g 2π . I λ (6.44)
3λ/4
A lossless line terminated by an open end behaves as a reactance with a value from minus to plus infinity. This reactance X(s) is plotted in Fig. 6.5. Fig. 6.5 shows the character of the impedance. It changes from the capacitive character to an impedance of a series resonance circuit, to an inductive character, and the impedance of a parallel resonant circuit. Very close to the line end, where s<<λ, the impedance is Z ≈−j
λ
λ/4 λ/2
Fig. 6.5
1 1 L L 1 =−j = , C ω LC s jωCs C βs
(6.45)
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and the line behaves here as a capacitor with a capacity proportional to the line capacity per unit length. The short cut termination represents zero impedance, and the voltage at the line end is equal to zero. Formulas (6.39) and (6.40) are now s U = j Z C I 2 sin (β s ) = j Z C I 2 sin 2π λ
(6.46) u (s )
s I = I 2 cos(β s ) = I 2 cos 2π . λ (6.47)
These voltage and current distributions are plotted in Fig. 6.6. Formulas (6.46) and (6.47) describe a standing wave which does not transport any active power. The reader can compare these formulas with (3.13) and (3.14), which describe the distribution of a standing wave created by the perpendicular incidence of a plane electromagnetic wave on the surface of a perfect conductor. The impedance at any point on the line is Z=
s
s
λ
3 λ/4
λ
3 λ/4
λ/4
λ/2
i (s )
λ/4
Fig. 6.6
U s = jZ C tg 2π . I λ
(6.48)
X(s)
This function is plotted in Fig. 6.7. Choosing a suitable line length, we can realize a line stub with an arbitrary input reactance value. Fig. 6.7 shows the character of this impedance. Very close to the end of a line (s<<λ) we have the impedance s s Z = jZ C tg 2π ≈ jZ C 2π = λ λ L = j ω LC s = jωLs C
λ 3λ/4
λ/2
λ/4
,
(6.49) and the line terminated by a short cut Fig. 6.7 behaves as an inductor with its inductivity proportional to the line inductivity per unit length.
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A line stub with a short cut is applied more frequently than a stub with an open end. The reason is that the former radiates less energy than the latter. The radiation of energy causes a non zero real part of the impedance, as it represents losses.
6.3 Smith chart When designing microwave circuits the designer is interested in the values of voltages U, currents I, impedances Z and reflection coefficient ρ. It would be a tedious work to calculate these quantities, specially for a lossy transmission line, and transform them along a transmission line without a computer and modern CAD tools. There are some graphical tools that can be used to simplify this task. One of them was introduced by Smith in 1939. It is a chart that enables us to make a simple transformation of the impedance along a transmission line and to recalculate the impedance to a reflection coefficient, and vice versa. It is a very useful tool for designing microwave circuits. It is even used in acoustics. As we work with values of voltage and current, this tool is applicable in the case of transmission lines with a TEM wave, where they are uniquely defined. The Smith chart can even be applied in the case of lines with no TEM wave, e.g., waveguides. Here we can use an appropriate scaling of propagating waves to some hypothetic TEM waves described by voltages and currents. Let us take a TEM transmission line, Fig. 6.8, fed from a generator and terminated by a load, with length l d and characteristic impedance ZC. The generator load ZC position along this z line is determined by the z coordinate z1 measured from the z2 terminals of the l generator. The propagation constant is assumed in the Fig. 6.8 notation used in the theory of electric circuits (6.15)
γ = jk = α + jβ ,
(6.50)
where α is the attenuation constant and β is the phase constant. At point z we have an impedance Z = U(z)/I(z), this impedance causes a reflection with a reflection coefficient
ρ (z ) =
Z (z ) − Z C . Z (z ) + Z C
(6.51)
The reflection coefficient is determined in the same way as the reflection coefficient for the TEM wave incident perpendicular to the boundary between two materials (3.6). From (6.51) we get
Z (z ) = Z C
1 + ρ (z ) . 1 − ρ (z )
(6.52)
The standing wave ratio is defined in the same way as (3.20)
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p( z ) =
1 + ρ (z ) 1 − ρ (z )
.
(6.53)
The transformation of the impedance along the transmission line is described by (6.38). Using (6.38) we can recalculate the impedance from point z1 to z2 and back Z ( z1 ) =
Z ( z 2 ) + Z C tgh (γ d ) U ( z1 ) , = ZC I (z1 ) Z C + Z ( z 2 ) tgh (γ d )
(6.54)
Z (z 2 ) =
Z ( z1 ) − Z C tgh (γ d ) U (z 2 ) , = ZC I (z 2 ) Z C − Z ( z1 ) tgh (γd )
(6.55)
where d = z2 – z1, Fig. 6.8. To define the reflection coefficient we have to decompose the voltage at any point to a wave propagating to the right, i. e., in the positive z direction, and a wave propagating to the left
U + ( z ) = U 0 e −γz ,
U − ( z ) = U 0 e γz .
The transformation of the voltage along the line is U + ( z 2 ) = U + ( z1 ) e −γ ( z2 − z1 ) ,
U − ( z 2 ) = U − ( z1 ) e γ ( z2 − z1 ) ,
(6.56)
From (6.56) we have the transformation of the reflection coefficient
ρ (z 2 ) = ρ ( z1 ) =
U − (z 2 ) U
+
(z 2 )
U − ( z1 ) U
+
(z1 )
= ρ ( z1 ) e 2γd ,
(6.57)
= ρ ( z 2 ) e − 2γd .
(6.58)
To get a universal tool for analyzing of microwave circuits we use the normalized impedances z=
Z . ZC
(6.59)
Consequently (6.51), (6.52) and (6.54) have the form
ρ (z ) = z=
z −1 , z +1
(6.60)
1+ ρ , 1− ρ
(6.61)
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z ( z1 ) =
z ( z 2 ) + tgh (γ d ) . 1 + z (z 2 ) tgh (γ d )
(6.62)
Now we have everything necessary for deriving the Smith chart. First we show the way to make a graphic representation in a complex plane of the reflection coefficient and its transformation. The reflection jv 90° coefficient is generally a complex jv ρ |ρ|=1 number with an amplitude lower |ρ| 0.7 than 1. It is plotted into a complex 0.2 ϕ plane u, jv, Fig. 6.9. The lines of ϕ=0 180° constant amplitude are circles with 0 -180° u u their center at the origin, and the lines of constant phase are radial lines, Fig. 6.9. The reflection coefficient is transformed -90° Fig. 6.9 according to (6.58). For a lossless line it is γ = jβ. The transformation of the reflection coefficient is controlled by
ρ ( z1 ) = ρ ( z 2 ) e − j 2 βd .
(6.63)
shift toward jv 0.125 Now we have to distinguish between the two βλ=const directions of the shift along a line. In the case of d > the generator 0 it is z2 > z1, z1 is closer to the generator and the shift is toward the generator. As d > 0 the phase 0.25 0 in (6.63) decreases and the shift toward the 0 0.5 u generator corresponds to the rotation of the reflection coefficient in the complex plane to the right, Fig. 6.10. In the case of d < 0 it is z2 < z1, z1 is shift toward 0.375 closer to the load and the shift is toward the load. the load As d < 0 the phase in (6.63) increases and the shift toward the load corresponds to the rotation of the Fig. 6.10 reflection coefficient in the complex plane to the left, Fig. 6.10. According to (6.63), the measure of the shift is equal to the product βd. The trip around the circumference of the whole complex plane represents angle 2π and, as β = 2π/λ, this corresponds to 2π = 2 βd = 2
2π
λ
d = 4π
d
λ
=>
d
λ
= 0 .5 .
The angle in the complex plane is not defined in degrees, but in the relative distance measured in the number of wavelengths in the range from 0 to 0.5, see Fig. 6.10. We have to distinguish between the direction toward the load and the direction toward the generator. In the case of a lossless line the transformation is performed along a circle, as the modulus of the reflection coefficient stays unchanged. In the case of a lossy line the modulus of the reflection coefficient decreases exponentially, and the transformation is therefore performed along a spiral. The complex normalized impedance (6.59) and its complex conjugate value are
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z = r + jx =
1+ ρ , 1− ρ
z * = r − jx =
1+ ρ* 1− ρ*
(6.64)
.
(6.65)
By adding and subtracting these two equations we get the real r and imaginary x parts of the normalized impedance r=
1 1 1+ ρ 1+ ρ* , + z + z * = 2 2 1 − ρ 1 − ρ *
jx =
(
)
(6.66)
1 1 1+ ρ 1+ ρ* . − z − z * = 2 2 1 − ρ 1 − ρ *
(
)
(6.67)
From (6.66) and (6.67) we can derive equations which determine images of the lines of the constant values of the real part of normalized impedance r and of the imaginary part of normalized impedance x in the complex plane of the reflection coefficient. The goal is to convert (6.66) and (6.67) into the equation of a general circle in the complex plane ρ = u + jv
(
)
ρ ρ * − (m − jn ) ρ − (m + jn ) ρ * + m 2 + n 2 − R 2 = 0 , where R is the radius of a circle, m and jn are the coordinates of the center of the circle, see Fig. 6.11. After some manipulations, (6.66) can be rewritten into
ρ ρ* −
r r r −1 ρ− ρ* + =0 . r +1 r +1 r +1
(6.68)
jv R jn
(6.69) m
Comparing (6.69) with (6.68) we get the coordinates of the center of the circle (6.69) and its radius m=
r , r +1
Now we are able to transform the lines of the constant values of the real part of the normalized impedance from the complex plane z = r + jx, where these lines are lines parallel with the imaginary axis, into
R=
n=0,
u
Fig. 6.11
1 . (6.70) r +1
jv
jx
r=0 r=1 r=2
r r=0
r=1
r= ∞ u
r=2 Fig. 6.12
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the complex plane ρ = u + jv. This transformation is shown in Fig. 6.12. The corresponding quantities are given in Table 6.1. r m n R The right hand side half of the complex plane z = r + jx is in this way transformed into the inside of the circle in the complex 0 0 0 1 1 1/2 0 1/2 plane ρ = u + jv with radius R = 1. The same procedure can be 2 2/3 0 1/3 performed with the imaginary part of the normalized impedance. Equation (6.67) can be rewritten into 1 0 0 ∞
Table 6.1
1 x
1 x
ρ ρ * − 1 − j ρ − 1 + j ρ * + 1 = 0 .
(6.71)
Comparing (6.71) with (6.68) we get the coordinates of the center of the circle (6.71) and its radius n=
m=1,
1 , x
R=
1 . x
(6.72)
Now we are able to transform the lines of the constant values of the imaginary part of the normalized impedance from the complex plane z = r + jx, where these lines are parallel with the real axis, into the complex plane ρ = u + jv. This transformation is shown in Fig. 6.13. The corresponding quantities are given in Table 6.2. jx x=2
x=1 jv
x=1
x=2 x=0
r
x= -1 x= ∞ x= - ∞
x= -2
x ∞ 2 1 0 -1 -2 -∞
m 1 1 1 1 1 1 1
n 0 1/2 1 ∞ -1 -1/2 0
R 0 1/2 1 ∞ 1 1/2 0
x=0 u
x= -2
x= -1 Fig. 6.13
1
Table 6.2
The Smith chart is obtained by combining the chart from Fig. 6.12, representing the lines of constant r, and the chart from Fig. 6.13, representing the lines of constant x. We have to take into account that the real part of the impedance must be positive. This means that the lines of constant x are limited to the right half of the complex plane z = r + jx, and at the same time to the inside of the circle with
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the unit radius in the plane ρ = u + jv. The final version of the Smith chart is shown in Fig. 6.14. Each point in the Smith chart corresponds to an impedance and at the same time to the reflection coefficient (6.60). The scale of the modulus of the reflection coefficient is shown below the chart. The corresponding angle must be measured using a protractor, or the scale of the angles is shown in some versions of the chart along its perimeter. The scale of the standing wave ratio (6.53), shown in some versions of the chart, is a nonlinear rating from 1 to ∞ . The transformation of the reflection coefficient is performed as explained in the text referring to Fig. 6.10. The scale of the relative distance is shown along the perimeter of the chart, its range being between 0 and 0.5. There are the two scales of the length. The first corresponds to the orientation toward the generator, the second toward the load. The scale of the real part of the normalized impedance is on the horizontal axis of the chart. The scale of the imaginary part of the normalized impedance is shown along the perimeter of the chart.
z = 0.4+j0.7
|ρ| =0.588
Fig. 6.14 In the following examples we show some basic operations with the Smith chart. To show the advantages of applying the Smith chart we choose cases where analytical calculation
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can be used. _____________________________________ Example 6.3: Show the normalized impedance z = 0.4+j07 in the Smith chart. This impedance is shown in Fig. 6.14 together with the corresponding value of the reflection coefficient z −1 = −0.143 + j 0.57 = 0.588 e j104° . z +1 _____________________________________ Example 6.4: The lossless transmission line is terminated by a short. Calculate the normalized impedance at distance l = 3λ/8 from the short. As z(z2) = 0, we can apply (6.48) and we get
ρ (z ) =
2π 3 3 z ( z1 ) = j tg (βl ) = j tg λ = j tg π = − j . λ 8 4
Fig. 6.15 In the Smith chart we use the following technique, Fig. 6.15. The short is a zero impedance. From this point we move in the direction toward the generator along the circle of a constant reflection coefficient by the normalized length l/λ = 3/8 = 0.375. At this distance we read the normalized impedance –j. _____________________________________
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Example 6.5: The lossless transmission line is terminated by an open end termination. Calculate the normalized impedance at distance l = λ/4 from the short. The normalized impedance at any point along the open end terminated transmission line is given by (6.44), so we get π z ( z1 ) = − j cotg (βl ) = − j cotg = 0 . 2
The open end transforms as a short. In the Smith chart we perform the transformation according to Fig. 6.16. The open end termination represents the infinite normalized impedance. From this point we move in the direction toward the generator along the circle of a constant reflection coefficient by the normalized length l/λ = 1/8 = 0.25. At this distance we read the normalized impedance 0.
Fig. 6.16 _____________________________________ Example 6.6: The lossless transmission line is terminated by the normalized impedance zL = 0.8+j. Calculate the normalized impedance at distance l = 0.2λ from the end of the line. Using (6.62) applied to the lossless line and the product βl we get
βl=
2π
λ
z ( z1 ) =
0 .2 λ = 0 .4 π
z L + j tg (β l ) 0.8 + j + j tg (0.4 π ) = = 0.807 − j 1.0061 1 + j z L tg (β l ) 1 + (0.8 + j ) j tg (0.4 π )
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In the Smith chart, Fig. 6.17, we first determine the point corresponding to the loading impedance zL = (0.8+j). We draw a radial line of the constant value of βl through this point and read the value of the relative distance 0.15 on the perimeter of the chart. From this point we move toward the generator by the relative distance 0.2, i.e., to the point 0.15+0.2 = 0.35. Here we read on the circle representing the same value of the modulus of the reflection coefficient as at point zL the normalized impedance z(z1) = 0.8 – j, which is approximately the same value as calculated.
l/λ=0.2
0.8+j
0.8-j
Fig. 6.17 ___________________________________ Example 6.7: The lossy transmission line with attenuation α’ = 0.5 dB is terminated by normalized impedance zL = 0.8+j2.2. Calculate the normalized impedance and the standing wave ratio at distance l = 20 m from the end of the line. The wavelength is 0.44 m. From the attenuation in dB we determine attenuation constant α and from the wavelength phase constant β
( )
α ' = 20 log e −α = −α 20 log(e ) => α = −
α'
20 log(e )
= 0.05756 m-1,
β=
2π
λ
= 14.27 m-1
Using (6.62) we get
z ( z1 ) =
z L + tgh (γ l ) = 1.02 + j 0.156 1 + z L tgh (γ l )
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In the Smith chart, Fig. 6.18, we first determine the point corresponding to the loading impedance zL = (0.8+j2.2). We draw the radial line of the constant value of βl through this point and read the value of the relative distance 0.188 on the perimeter of the chart. From this point we move toward the generator by the relative distance l/λ = 20/λ = 45.454 so we get to the point 0.188+45.454 = 45.642. This means that we go round the chart 91 times and stop at the point 0.142. This is the position of the impedance we are looking for. Now we have to determine the value of the reflection coefficient. The reflection coefficient transforms according to (6.58) so we get
ρ1 = ρ 2 e −2αl = ρ 2 0.1 In the Smith chart in Fig. 6.18 we read the normalized impedance z(z1) = 1.02+j0.15, and using (6.60) and (6.53) we get the standing wave ratio p(z1) = 1.16. 0.142
0.188
zL=0.8+j2.2
|ρ2|
|ρ1| zL=1.02+j0.15
Fig. 6.18 _______________________________________ The normalized impedance could be calculated simply by using transformation formulas (644), (6.48) and (6.62) in examples 6.4 to 6.7. The Smith chart however simplifies the work. In example 6.8 we perform impedance matching, i.e., we make corrections in a circuit to reduce the reflection at the given frequency to zero. This cannot be done simply, as it includes the need to solve a transcendent equation. The Smith chart is a very efficient tool in such a case. ___________________________________
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Example 6.8: The lossless transmission line with characteristic impedance ZC = 300 Ω is terminated by l2 ZC impedance ZL = (420-j180) Ω. Match this impedance by the stub of the same line terminated by a short end connected in series to the line so as to get a zero ZC ZL reflection coefficient, Fig. 6.19. The wavelength is 5 m. First we must determine distance l1 at which we l1 connect the stub. The loading impedance normalized to the characteristic impedance zL = ZL/ZC = 1.4-j0.6 must Fig. 6.19 be transformed within distance l1 to impedance z1 = 1+jx. Here we connect the stub with input impedance z2 = 0-jx. The resulting impedance at this point is then z = z1+z2 = 1, and therefore the reflection coefficient is zero.
l2/λ=0.086
0+j0.6
zL z1
l1/λ=0.043
Fig. 6.20 The solution of this problem in the Smith chart is simple, see Fig. 6.20. Impedance zL is transformed along the circle of the constant value of the reflection coefficient to the point at which the impedance lies on the circle where the real part of the normalized impedance is 1. This gives z1 = 1-j0.6, Fig. 6.20. Corresponding length l1/λ = 0.043. Using the given value of λ we have l1 = 0.215 m. Let us now form the stub. At its input it must have the normalized impedance equal to 0+j0.6. This stub is terminated by a short that is zero impedance. The length of this stub l2 is determined by transforming zero impedance to the point corresponding to the impedance 0+j0.6. From the Smith chart in Fig. 6.20 it follows that the necessary length is l2/λ = 0.086,
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which gives l2 = 0.43 m. The second solution can be obtained by transforming zL to the point with impedance z1 = 1+j0.6. Then the stub must have input impedance 0-j0.6. This solution is not optimal as the corresponding lengths l1 and l2 are longer than those obtained above. _______________________________________
6.4 Problems 6.1 A longitudinally homogeneous transmission line transmits a TEM wave with frequency f=1 kHz. The amplitude of the voltage on the input port is Um=60 V. The end terminals of the line are connected to an impedance equal to the line characteristic impedance. The line parameters are R=4 Ω/km, G=0.5 µS/km, L=2 mH/km, C=6000 pF/km. Calculate the phasor of the voltage on the line end. U2=-24.75+j33.93 6.2 A transmission line is 50 km long, and its parameters are R=6.46 Ω/km, G=1 µS/km, L=4 mH/km, C=12.2 nF/km. The line is loaded by a resistor R L=1000 Ω. The voltage on the load is U2=10 V. The frequency is 0.8 kHz. Calculate the voltage and the current on the input of the line. U1 = 9.033 e j1.748 = −1.5921 + j8.892 V I1 = 20.77 e j1.848 = −569 + j 20 mA
6.3 Calculate the input impedance of a lossless transmission line of the length l=0.4 m. The characteristic impedance is 75 Ω, and the frequency is 0.6 GHz. The line has a shortcut at the end. Z1=230.8 Ω, L1=0.12 µH 6.4 Calculate the input impedance of a lossless transmission line of length l=0.4 m. The characteristic impedance is 75 Ω, and the frequency is 0.6 GHz. The line has an open end. Z1=-24.4 Ω, C1=21.76 pF 6.5 A lossless very short transmission line (l=125 mm) has the input impedance at frequency f=60 MHz Z1S=j8850 Ω when terminated by a short circuit, and Z1O=-j35.5 Ω when terminated by an open circuit. Calculate the inductivity per unit length and the capacity per unit length. L=75 µH/m, C=0.56 pF/m 6.6 Match the transmission line with characteristic impedance 50 Ω terminated by a series combination of the resistor with resistivity R = 100 Ω and a capacitor with capacity C = 1.5 pF, Fig. 6.21, at frequency 5 GHz, which corresponds to the wavelength on the line λ = 42.5 mm.
ZC λ
R C
The short circuited stub of Fig. 6.21 the same transmission line of length l2 = 4.6 mm must be connected in series at distance l1 = 3 mm from the load, Fig.6.19.
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7. WAVEGUIDES WITH METALLIC WALLS In this section we will study transmission lines with metallic walls. Such transmission lines are known as waveguides. The field distribution is determined by solving the wave equation for the longitudinal components of the electric and magnetic fields, as described in section 5. Our task is to solve equation (5.3) with the proper boundary conditions. The tangential component of the electric field has to be equal to zero on the metallic walls. This technique will be directly applied to a parallel plate waveguide and to a waveguide with the rectangular cross-section. The field distribution is described in a rectangular coordinate system in the case of these two lines. In the case of a waveguide with a circular cross-section we have to use a circular coordinate system. The parallel plate waveguide consists of the two separate conducting plates, so it can conduct DC current and it can guide the TEM mode. Together with this TEM mode we will describe the modes known as the waveguide modes. These modes have nonzero cut-off frequency. Waveguides with rectangular and circular cross-sections consist of only one conductor, so they are not able to conduct the TEM mode.
7.1 Parallel plate waveguide The cross-section of the parallel plate waveguide is shown in Fig. 7.1. It is created by the two parallel infinitely wide conducting plates which are located at a distance a. The space between these plates is filled by an ideal a dielectric material. We will describe separately the field of the TE and TM modes. The calculation procedure is simplified by the fact that the field does not depend on the y coordinate. So we put ∂ ∂y = 0 in (5.14) to (5.17). The TE modes thus have Ey, Hx and Hz components. Setting Ez=0 in (5.14) to (5.17) we get
Ey =
σ =∞ ε, µ
σ =∞
jω µ ∂H z , k 2p ∂x
Hx = − j
x
y
z
Fig. 7.1
(7.1)
k z ∂H z . k p2 ∂x
(7.2)
Function H0z representing the transversal field distribution analogous to (5.1) depends only on the x coordinate, so we can put k p = kx ,
(7.3)
and the H0z component is determined by solving the wave equation dH 0 z + k x2 H 0 z = 0 . dx
(7.4)
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The solution is in the form H 0 z = A sin (k x x ) + B cos(k x x ) .
The boundary conditions cannot be fulfilled by this component. We have to use E0y, which is tangent to both conducting plates. At x=0 we have E0y(0)=0, Fig. 7.1, and, consequently, A=0. The second boundary is at x=a, where E0y(a)=0 and sin(kxa)=0 and we get the propagation constant in the x direction kx k xm =
mπ . a
(7.5)
Consequently, the solution of the wave equation (7.4) is mπ H 0 z = B cos a
x .
(7.6)
(7.1) and (7.2) now have the form E0 y = −
jω µ mπ B sin k xm a
H 0x = j
k zm mπ B sin k xm a
x ,
(7.7)
x .
(7.8)
To remove kxm from the denominators of (7.7) and (7.8) we introduce a new constant jω µ and we get the field distribution of the TE modes B' = − kx mπ − jk zm z , E y = B' sin x e a
Hx =
(7.9)
mπ − jk zm z B' sin x e , ωµ a k zm
Hz = j
mπ − jk zm z B' cos xe . ωµ a k xm
(7.10)
(7.11)
The TM modes have nonzero field components Ex, Ez and Hy. The transversal components are from (5.14) to (5.17) Ex = − j
Hy = −j
k zm ∂E z , k p2 ∂x
ω ε ∂E z k p2 ∂x
(7.12)
.
(7.13)
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The equation for determining the transversal distribution of the longitudinal component of the electric field is dE 0 z + k x2 E 0 z = 0 . dx
(7.14)
The boundary conditions are met directly by Ez, which is tangent to both plates. Applying a similar procedure as for the TE modes we get the field distribution mπ H y = C ' cos a
Ex =
Ez =
x e − jk zm z ,
mπ C ' cos ωε a
k zm
(7.15)
x e − jk zm z ,
mπ − jk zm z C ' sin x e . ωε a
jk xm
(7.16)
(7.17)
The propagation constant in the x-direction has the form (7.5) as in the case of the TE modes, and (7.3) is again valid. Modal number m determines the x form of the field distribution in the waveguide cross section. This is shown in m=3 m=0 m=1 m=2 Fig. 7.2. m determines how many halfperiods of the sinus function across the a space between the plates the distribution has. Fig. 7.2 shows the distribution of Ez for the TM modes, or Ey for the TE modes. The longitudinal propagation Ez, Ey constant kz is given by (5.18). Using (7.3) Fig. 7.2 and (7.5) we have for the m-th mode mπ k zm = ω 2 µε − a
2
.
(7.18)
Now we have from (5.19) the cut-off frequency and the cut-off wavelength of both the TE and TM modes with modal number m in the parallel plate waveguide f cm =
λcm =
m 2a µε
,
(7.19)
2a . m
(7.20)
The parallel plate waveguide behaves like the high-pass filter, as the modes can propagate starting from their cut-off frequency. From the longitudinal propagation constant we can determine the parameters describing the propagation of the modes along the parallel plate waveguide. These are the
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wavelength, the phase velocity, and the group velocity. The phase velocity of the m-th mode is 1 ω ω = = = v pm = 2 2 k zm π m mπ ω 2 µε − µε 1 − 2 ω µε a . (7.21) v v = = 2 2 f cm λ 1 − 1 − λ f cm Similarly we can get formulas for the group velocity, the wavelength and the longitudinal propagation constant
λ gm =
v pm f
=
λ f 1 − cm f
2
=
λ λ 1 − λ cm
2
2
v gm
λ f dω = = v 1 − cm = v 1 − dk zm f λcm
k zm
λ f = k 1 − cm = k 1 − f λcm
2
,
(7.22)
2
,
(7.23)
2
.
(7.24)
The plots of these functions vpm/v m=1 m=2 are shown in Fig. 7.3. It is λ / λ gm seen from Fig. 7.3 that as the frequency approaches from above the cut-off frequency, the 1 corresponding mode stops vgm/v propagating as its propagation constant and kzm/k m=1 m=2 group velocity tend to zero. At the same time its phase velocity and wavelength fc2 fc3 fc1 tend to infinity. Below the cut-off frequency these Fig. 7.3 quantities are imaginary numbers and the mode is the evanescent mode, the amplitude of which decreases exponentially along the line, and this mode does not propagate. The field distributions of the TE and TM modes are different, so their wave impedances are different. They are defined as the ratio of the transversal components of the electric and magnetic fields. Using (7.9) and (7.10) we get for the TE modes
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Z mTE = −
Ey Hx
=
ωµ k zm
=
Z0 f 1 − cm f
.
2
(7.25)
For the TM modes we get from (7.15) and (7.16) Z mTM
E k f = x = zm = Z 0 1 − cm H y ωµ f
2
.
(7.26)
Z0 is the wave impedance of the free space. Let us now look for the dominant mode of the parallel plate waveguide. This is the mode with the lowest possible modal number m. It follows from the distribution of the electromagnetic field of the TE modes, (7.9), (7.10), (7.11) that these modes cannot have m = 0, as this case gives a zero field. Consequently, the lowest TE mode is the mode TE1. For the TM modes we can allow m=0, as from (7.15-17) we get H y = C ' e − jkz , Ex =
k
ωε
(7.27)
C ' e − jkz ,
(7.28)
Ez = 0 .
(7.29)
This TM0 mode has only the transversal components of the electric and magnetic fields. Therefore it is the TEM wave. Its cut-off frequency is zero and this mode can propagate from zero frequency and kz0 = k. This confirms the fact that the parallel plate waveguide consists of two separate conductors and is therefore able to conduct the DC current. So the dominant mode of the parallel plate waveguide is the TEM mode, which can propagate from zero frequency. The field distribution of the two lowest TE modes on the parallel plate waveguide is shown in Fig. 7.4. Fig. 7.5 shows the field distribution of the three lowest TM modes. H E
TE1
Fig. 7.4
TE2
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E H
TM0 = TEM E H
TM1
TM2
Fig. 7.5
The explanation of wave propagation based on geometrical optics is shown in Fig. 7.6. The wave is coupled between the plates under angle Θ and bounces up and down due to reflections on the conducting plates. The wavelength in the direction along the line is the Θ λ projection of the free space wavelength, so it must be longer. We have
λg =
λ
cos Θ
,
λg Fig. 7.6
consequently f cos Θ = 1 − cm f
2
.
(7.30)
This angle of establishment determines the conditions of the wave propagation. It corresponds to the relative frequency distance of the wave from the cut-off frequency. The lower this angle is, the further the wave is from the cut-off. Θ = 90°corresponds to the cut-off frequency and it is obvious that the wave does not propagate. __________________________ Example 7.1: Determine the modes which can propagate in the parallel plate waveguide at the frequency 10 GHz. The distance between the plates is 40 mm and the space is filled by air. The mode can propagate if the wavelength at 10 GHz is lower than the cut-off wavelength of this mode. The wavelength at 10 GHz is
λ=
c = 30 mm f
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The cut-off wavelength of the m-th mode is determined by (7.20). So we have the mode can propagate m=0 λc0 = ∞ the mode can propagate m=1 λc1 = 80 mm m=2 λc2 = 40 mm the mode can propagate the mode cannot propagate m=3 λc1 = 26.7 mm The parallel plate waveguide can transmit the TM0, TM1, TE1, TM2 and TE2 modes at the frequency 10 GHz. ________________________________ Example 7.2: Determine the distance between the two conducting planes to get the parallel plate waveguide along which the only TE1 mode propagates at 10 GHz and the mode TE2 is attenuated by –120 dB/m. Do not consider the TM modes. The attenuation of the TE2 mode is E 0 e − jk z 2d 20 log E0
= −120 dB ,
e − jk z 2d = 10 −6 ,
d=1m
-jkz2d = -13.8 ,
kz2 = -j13.8
As kz2 is an imaginary number the TE2 mode does not propagate and is an evanescent mode. From the determined value of kz2 we will calculate the transversal propagation constant ky2 k y2 =
mπ 2π 2 = = k 2 − k z22 = ω 2 µε − (− j13.8) = 210 a a
⇒
a = 29 mm
The cut-off wavelength of the TE1 mode is
λc1 = 2a = 58 mm . The TE1 mode can propagate. ___________________________________
7.2 Waveguide with a rectangular cross-section The parallel plate waveguide is not suitable for practical applications, as it is open from the sides and it can radiate energy to the y sides. To get a practical line we have to confine it by two conducting planes from sides. Thus we get a waveguide of rectangular b cross-section. This waveguide is completely shielded. It is shown in Fig. 7.7. The inner dimensions are a in the direction of the x axis, and b in the direction of the y axis. We will assume an infinitely long waveguide with a 0 x ideally conducting walls filled by a lossless dielectric. z The process of solving the wave equation is shown in chapter 5. In the case of Fig. 7.7
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the TM modes we solve the wave equation (5.3) for E0z and the transversal components of the electric and magnetic fields are calculated from (5.14-17) putting H0z = 0. For the TE modes we determine H0z from the equation analogous to (5.3) and put E0z = 0. The longitudinal component of the magnetic field is
)(
(
H 0 z = A e jk x x + B e − jk x x C e
jk y y
+ De
− jk y y
),
(7.31)
where the separation constants, which represent the propagation constants in the x and y directions, are coupled with kp by k 2p = k x2 + k y2 .
(7.32)
Unknown constants A, B, C, D and the propagation constants must be determined applying the boundary conditions. To meet these boundary conditions we have to know the transversal components of the electric field. From (5.14) and (5.15) we get E0 x = −
E0 y =
(
)(
)
(
)(
)
jωµ ∂H 0 z ω µ k y jk y − jk y = A e jk x x + B e − jk x x C e y − D e y 2 2 k p ∂y kp
ω µ kx jωµ ∂H 0 z jk y − jk y =− A e jk x x − B e − jk x x C e y + D e y . 2 2 k p ∂x kp
The boundary conditions are
(
jk y y
−e
− jk y y
) = 2 jCsin (k
C–D=0
=>
Ce
x = 0, Ey = 0 =>
A–B=0
=>
A e jk x x − e − jk x x = 2 jAsin (k x x ) ,
x = a, Ey = 0 =>
sin (k x a ) = 0 =>
y = b, Ex = 0 =>
sin k y b = 0
( )
=>
(
yy
),
y = 0, Ex = 0 =>
)
kx =
mπ , a
(7.33)
ky =
nπ , b
(7.34)
Now setting a new constant M = 4AC, which determines the field amplitude, we have the field distribution mπ nπ − jk zmn z , H z = M cos x cos ye a b
Hx =
jk zmn mπ mπ nπ − jk zmn z M sin x cos ye , 2 kp a a b
Hy = −
jk zmn nπ mπ nπ − jk zmn z M cos x sin ye , 2 kp b a b
(7.35)
(7.36)
(7.37)
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Ex =
jωµ nπ mπ nπ − jk zmn z M cos x sin ye , k 2p b a b
Ey = −
(7.38)
jωµ mπ mπ nπ − jk zmn z M sin x cos ye , 2 kp a a b
(7.39)
where 2
k 2p
2
mπ nπ = + , a b
(7.40)
A similar procedure can be followed in the case of TM modes. Now we solve the wave equation (5.3). Its solution is
)(
(
E 0 z = A e jk x x + B e − jk x x C e
jk y y
+ De
− jk y y
).
(7.41)
The longitudinal component of the electric field is tangent to all walls of the waveguide. This simplifies the solution. The boundary conditions are x = 0, Ez = 0
=>
A+B=0 =>
y = 0, Ez = 0
=>
C+D=0
x = a, Ez = 0
=>
sin (k x a ) = 0 =>
y = b, Ez = 0
=>
sin k y b = 0
( )
=>
( )
E 0 z = M sin (k x x ) sin k y y
kx =
mπ , a
(7.42)
ky =
nπ . b
(7.43)
The particular components of the electric and magnetic field are derived from (7.41) using (5.14-7) mπ nπ − jk zmn z , ye x sin E z = M sin a b
(7.44)
Ex = −
jk zmn mπ mπ nπ − jk zmn z M cos x sin ye , 2 kp a a b
(7.45)
Ey = −
jk zmn nπ mπ nπ − jk zmn z M sin x cos ye , 2 kp b a b
(7.46)
Hx =
jωε nπ mπ nπ − jk zmn z ye . x cos M sin 2 kp b a b
(7.47)
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Hy = −
jωε mπ mπ nπ − jk zmn z M cos x sin ye . k p2 a a b
(7.48)
kp is determined by (7.40) and kzmn by (5.18). Phase constants kp, kx, ky, kz are equal for the TE and TM modes. This means that all parameters describing the propagation of these modes along the waveguide with a rectangular cross-section are equal for the TE and TM modes. Each mode is described by the two modal numbers m and n. They determine the field distribution in the corresponding direction, as shown in Fig. 7.2. The longitudinal propagation constant is, as follows from (5.18), 2
2
k z = ω µε −
k 2p
mπ nπ = ω µε − − a b 2
2
.
(7.49)
From the condition of zero value of kz we get the cut-off frequency (5.19) of the mode with modal numbers m and n f cmn =
2
1 2π µε
mπ nπ + a b
2
.
(7.50)
The cut-off wavelength is
λcmn =
v
2π
=
. (7.51) 2 2 mπ nπ + a b The reader can now compare relations (7.49), (7.50) and (7.51) valid for a waveguide with a rectangular cross-section with those valid for a parallel plate waveguide (7.18), (7.19) and (7.20). Instead of mπ/a we now have (7.40). It follows that the same relations can be derived for the wavelength along the waveguide, the phase velocity, the group velocity and the longitudinal propagation constant as (7.21), (7.22), (7.23) and (7.24). The wave impedance must be defined separately for the TE and TM modes, as their field distribution is different. It is defined as the ratio of the transversal components f cmn
Z TE ,TM =
Ey Ex =− . Hy Hx
These quantities are described as in the parallel plate waveguide by (7.25) and (7.26). ____________________________ Example 7.3: Determine the modes which can propagate along a rectangular waveguide with the internal dimensions a = 40 mm, b = 25 mm at the frequency f = 10 GHz. The propagating modes must have f < fcmn, which corresponds to the condition k2 > kp2. This gives us 2
2
ω µε >
k 2p
2
mπ nπ = + , a b
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and results in the equation 43.681 > m 2 ⋅ 6.162 + n 2 ⋅ 15.775 This inequality is fulfilled for the modes: TE10, TE01, TE20, TE11, TM11, TE21, TM21 ________________________________ Example 7.4: Calculate the phase and group velocities, the wavelength, the longitudinal propagation constant and the wave impedance of the TE10 and TM12 modes in a rectangular waveguide filled by air with the internal dimensions a = 22 mm, b = 10 mm. The frequency is f = 10 GHz. First we calculate the cut-off frequencies of these two modes f c10 =
f c12 =
1 2a µε
= 6.82 GHz , 2
1 2π µε
2
π 2π + = 30.76 GHz . a b
The TE10 mode can propagate, but the TM12 mode cannot as its cut-off frequency is greater than 10 GHz. Therefore the required parameters will be computed only for the TE10 mode. We will first calculate the term 2
f 1 − c10 = 0.73 f Consequently we have v p10 =
c = 4.1 ⋅ 10 8 m/s 0.73
v g10 = c ⋅ 0.73 = 2.19 ⋅ 108 m/s
λ g10 =
λ 0.73
= 41 mm
k z10 = ω µε ⋅ 0.73 = 152.57 m-1 Z0 = 516.2 Ω 0.73 ____________________________________ TE Z10 =
The most important mode is the dominant mode. This is the mode with the combination of the lowest possible modal numbers m and n and consequently with the lowest cut-off frequency and the simplest field distribution. From the distribution of the field of the TM modes (7.44-48) it follows that it is not possible to set m = 0 and n = 0 as the field is zero.
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In the case of the TE modes it is possible, see (7.35-39), to put m = 0 or n = 0, but not simultaneously. So we have two possible combinations of m and n (1,0) or (0,1). Which cutoff frequency is lower depends on the relation between the dimensions of the waveguide. Accepting the standard notation a > b, we get as the dominant mode the mode TE10. From (7.50) and (7.51) we will get its cut-off wavelength and its cut-off frequency
λc10 = 2a , f c10 =
(7.52)
1
1 . 2 µε a
(7.53)
Formula (7.52) tells us that the waveguide transmits the TE10 mode starting from the frequency at which a = λ/2. The field distribution of the dominant TE10 mode follows from (7.35-39) Ey = −
jω µ a
π
π M sin x e − jk z z , a
π H z = M cos x e − jk z z , a Hx =
jk z a
π
(7.54)
(7.55)
π M sin x e − jk z z . a
(7.56)
There is only one transversal component of the electric field and one transversal component of the magnetic field. The field does not depend on the y coordinate. The longitudinal propagation constant is from (7.49) π k z = ω µε − a 2
2
.
(7.57)
The electric field of this mode has its maximum at the center of the wider dimension, so it can be simply excited by a probe located at the position of the field maximum, Fig. 7.8. Tangent components of the magnetic field create an electric current passing along the surface of the waveguide walls in the direction perpendicular to this field. This means that the Hx component creates currents Ey flowing along the waveguide. This must be taken into account when parts of the waveguide are mounted together. Their flanges must be well fitted to allow these currents to pass, otherwise the increased resistance coaxial probe between flanges raises the losses. Fig. 7.8 ____________________________ Example 7.5: Determine the frequency band of the single-mode operation of the rectangular waveguide with internal dimensions a = 22.86 mm, b = 10.16 mm. The waveguide is filled by air.
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The frequency band is limited from below by the cut-off frequency of the dominant mode TE10. The cut-off wavelength of this mode is (7.52)
λc10 = 2a = 45.72 mm . The cut-off frequency is f c10 =
c
λc10
= 6.56 GHz .
The frequency band of the single-mode propagation is limited from above by the cut-off frequency of the nearest higher mode. In the rectangular waveguide it could be TE01 or TE20 modes, depending on the ratio between the waveguide dimensions. The cut-off wavelengths of these two modes are
λc 01 = 2b = 20.32 mm,
λc 20 = a = 22.86 mm.
In our case a > 2b and consequently λc20 > λc01. The nearest higher mode is therefore the TE20 mode. The cut-off frequency corresponding to λc20 is f c 20 =
c
λc 20
= 13.12 GHz.
The frequency band of the single-mode operation is from 6.56 to 13.12 GHz. ____________________________________ Example 7.6: Design the dimensions of the rectangular waveguide a, b as it transmits at the frequency 10 GHz the only dominant mode TE10 and the two nearest higher modes TE01 and TE20 are equally attenuated. The two modes TE01 and TE20 will be equally attenuated if their longitudinal propagation constants are equal kz01 = kz20
k 2 − k p2 01 = k 2 − k 2p 20
=>
=>
k p 01 =
π b
= k p 20 =
2π a
Consequently we have a relation that assures equal attenuation of the two modes a = 2b The condition for propagation of the dominant mode TE10 is f > f c10 =
c 2a
=>
a>
c = 15 mm 2f
The condition for the attenuation of the two higher modes is f < f c 01 = f c 20 =
c 2b
=>
b<
c = 15 mm, 2f
a = 2b
=>
a < 30 mm
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The waveguide has the dimensions a = 2b and 15 < a < 30 mm. _____________________________________ The power transmitted by the rectangular waveguide is calculated from the Poynting vector
S av
2
* 1 Ey 1 E y E y 1 1 * * = Re E × H = − Re E y H x z 0 = − Re = z z0 . TE 0 TE 2 2 Z10 2 2 Z 10
(
)
(
)
The total active transmitted power is inserted for Ey (7.54) P = ∫∫ S av S
P=
E m2 ab TE 4Z10
1 ⋅ dS = TE 2 Z10
ab
∫ ∫ Ey
2
dxdy =
00
E m2 b
a
∫ sin
TE 2 Z10 0
2 πx
dx , a
.
(7.58)
The maximum amplitude of the electric field is from (7.54) E m = Mωµ a / π . _________________________________ Example 7.7: Calculate the maximum power that can be transmitted by the dominant mode TE10 in a rectangular waveguide with the internal dimensions a = 22.86 mm, b = 10.16 mm at the frequencies 5 GHz and 10 GHz. The waveguide is filled by air with the electric strength Ep = 30 kV/cm. According to Example 7.5 the cut-off frequency of the dominant TE10 mode in this waveguide is 6.56 GHz. This means that this mode does not propagate at 5 GHz. So we will calculate the power only at 10 GHz. The wave impedance of this mode is TE = Z10
Z0 f 1 − c10 f
2
= 500 Ω ,
consequently the transmitted power (7.58) allows the maximum value of the electric field to be equal to the electric strength P=
E p2 ab TE 4 Z10
= 1.05 MW .
___________________________________ Example 7.8: A waveguide with a rectangular cross-section transmits the TM11 mode at the frequency 6 GHz. The waveguide is filled by air and its internal dimensions are a = 71 mm and b = 35.5 mm. Calculate the magnitude of the total current passing the waveguide walls in the longitudinal direction and the magnitude of the displacement current passing the waveguide cross-section in the longitudinal direction. The field distribution is described by the formulas
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E z = j 3000 sin (πx / a )sin (πy / b ) e − j 78.2 z ,
E x = −1060 cos(πx / a ) sin (πy / b ) e − j 78.2 z , E y = −2120 sin (πx / a ) cos(πy / b ) e − j 78.2 z , H x = 9.04 sin (πx / a ) cos(πy / b ) e − j 78.2 z , H y = −4.52 cos(πx / a ) sin (πy / b ) e − j 78.2 z . The conducting current passing along the surface of the waveguide walls can be calculated from the magnetic field boundary condition (1.16) y
n×H = K .
The current directions are shown in Fig. 7.9. In the upper and bottom walls we have
Hx Hy
Kz2
n
K z1 = 9.04 sin (πx / a ) A/m.
n
z
n
Kz2
Kz1 Hx a
0
K z 2 = 4.52 sin (πy / b ) A/m.
Hy
n
b
In the left and right walls we have
Kz1
x
Fig. 7.9
The total current passing the upper wall is a
a
0
0
I1 = ∫ K z1dx = 9.04 ∫ sin (πx / a )dx = 9.04
2π = 0.409 A a
The total current passing the left wall is b
b
0
0
I 2 = ∫ K z 2 dy = 4.52 ∫ sin (πy / b )dy = 4.52
2π = 0.102 A b
The total current is I = 2 I1 + 2 I 2 = 1.022 A The density of the displacement current is J Dz = ωε 0 E z = 1000 sin (πx / a )sin (πy / b ) The total displacement current passing the waveguide cross-section is
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ab
ab
00
00
I D = ∫ ∫ J Dz dxdy = ∫ ∫ 1000 sin (πx / a )sin (πy / b )dxdy = 1000
4ab
π2
= 1.022 A
The results show that the current continuity is preserved. The total value of the conducting current passing the waveguide walls equals the value of the displacement current passing the waveguide cross-section. ____________________________________ All previous results were derived assuming that all materials are without losses. This is an ideal case. There are two sources of losses in a real waveguide. These are the finite conductivity of the waveguide walls and the finite conductivity of the dielectric filling the waveguide. Other losses are caused by the roughness of the waveguide wall and nonperfect connections between the flanges. All these losses increase with the frequency. They are about 0.1 dB/m for the frequency band from 8 to 12.4 GHz, which is the so called X band. It is not possible to design a rectangular waveguide with arbitrary dimensions. The dimensions of these waveguides are determined by international standards depending on the frequency band. An example is the waveguide for the X band. Its internal dimensions are a = 22.86 mm, b = 10.16 mm. The waveguides are at present used at lower frequencies only in the case of the need to transmit high power. Planar transmission lines such as the microstrip line, Fig. 5.1b, are used for low power applications instead of waveguides. Metallic waveguides are very massy and their production is very expensive, so they are not suitable for mass production. At high frequency bands, above 50 GHz, waveguides are still used, as they have lower losses than planar transmission lines. Planar transmission lines are open transmission lines and lose energy due to radiation at high frequencies.
7.3 Waveguide with a circular cross-section A waveguide with a circular cross-section, Fig. 7.10, is used only in some special applications. An example is a rotating joint which transmits an electromagnetic wave to the feeder of a rotating radar antenna. The problem of these waveguides is that due to their geometry they do not keep the plane of the polarization of the transmitted mode when the waveguide is long. The frequency band of the single mode operation of a circular waveguide is narrower than the same band of a rectangular waveguide. In this section we will show only the basic properties of a waveguide with a circular cross-section, Fig. 7.10. The internal radius of this waveguide is ro, and the longitudinal axis is z. We will assume an infinitely long waveguide with ideally conducting walls filled by a lossless dielectric. Even in a waveguide of this geometry ro we can get results similar to those obtained in section 5. We have to use the cylindrical coordinate system. The field in a circular waveguide can be separated into TE and TM modes, which are treated separately. We will see that unlike in the rectangular waveguide the longitudinal z propagation constants and the cut-off frequencies are different for the TE modes and for the TM modes. Fig. 7.10 First we will treat TM modes. The field distribution of these modes is determined by solving the wave equation (5.3) written for the longitudinal component of the electric field (5.1). The cylindrical coordinate system must be used here, see the mathematical appendix.
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E z (r ,α , z ) = E0 z (r ,α ) e − jk z z ,
(7.59)
where the transversal field distribution is obtained by solving the wave equation (5.3) expressed in the cylindrical coordinate system (13.86) ∂ 2 E0 z ∂r 2
1 ∂E0 z 1 ∂ 2 E0 z + 2 + k p2 E0 z = 0 . + 2 r ∂r r ∂α
(7.60)
The propagation constants kp, kz and k are coupled by (5.5). We will solve (7.60) by the method of separation of variables. Function E0z is assumed in the form E 0 z (r ,α ) = R(r ) ⋅ Α(α ) .
(7.61)
Inserting (7.61) into (7.60) and after some manipulation we will get the equation r2
R' ' R ' Α' ' +r + + r 2 k 2p = 0 . R R Α
(7.62)
Term Α' ' / Α must be independent of r and α, and must be equal to constant m, which gives us the equation for function Α Α' '+ m 2 Α = 0 .
(7.63)
The solution of (7.63) is obtained using a proper origin for reading angle α in the simple form Α = Ccos(mα ) .
(7.64)
The electric field must be unique, therefore E 0 z (α ) = E0 z (α + 2mπ ) . It follows from this that m must be an integer number. Inserting solution (7.64) into (7.62) we will get d 2 R 1 dR 2 m 2 + + k p − 2 R = 0 . r d r 2 r d r
(7.65)
This equation is Bessesl’s equation of the m-th order. The solution is in the form of the sum of the Bessel functions of the m-th order Jm and Ym, see the mathematical appendix,
( )
( )
R = A J m k p r + BYm k p r .
(7.66)
To determine constants A and B we have to apply the boundary conditions. The tangent component of the electric field on the surface of the waveguide wall must be zero and the field must have a finite value at the waveguide center. The second condition gives us B = 0, as Bessel function Ym has an infinite value at the origin, see the mathematical appendix. Component Ez is tangent to the waveguide wall, so the first boundary condition reads R(r = ro ) = 0 =>
(
)
J m k p ro = 0 .
(7.67)
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The product kpro must be equal to the n-th zero point of the Bessel function of the m-th order αmn. These values are listed in Table 13.1 in the mathematical appendix. Consequently the transversal propagation constant of the mode with modal numbers m and n is TM k pmn =
α mn
.
ro
(7.68)
From (5.18) and (5.19) we get the longitudinal propagation constant and the cut-off frequency of the TM modes in the circular waveguide TM k zmn
2
= ω µε −
TM f cmn =
1
α = ω µε − mn ro
2 k pmn
α mn
2π µε ro
2
2
,
(7.69)
.
(7.70)
According to Table 13.1 we have the order of the lowest modes TM01, TM11, TM21, TM02 and TM31, as the lowest numbers αmn are α01 = 2.40482, α11 = 3.83171, α21 = 5.13562, α02 = 5.52007, α312 = 6.38016. The field distribution of particular modes can be determined in a similar way as the distribution of TM modes in a rectangular waveguide. The cylindrical coordinate system must be applied. Longitudinal component Ez has the form, using (7.64) and (7.66),
( )
E z = E 0 J m k p r cos(mα ) e − jk zmn z ,
(7.71)
where E0 is an unknown amplitude. The transversal components are Er, Eα, Hr and Hα. The field distribution of the lowest modes is shown in Fig. 7.11. magnetic field
electric field
TE11
TM01
Fig. 7.11
TE01
TM11
The field distribution of the TE modes is determined by solving the wave equation written for the longitudinal component of the magnetic field. Using the same procedure as in the case of the TM modes we will get the distribution of Hz
( )
H z = H 0 J m k p r cos(mα ) e − jk zmn z ,
(7.72)
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where H0 is an amplitude. The TE modes have the Eα component of the electric field tangent to the waveguide wall. This component is proportional to ∂H z / ∂r , so the boundary condition is ∂H z = 0 => J m' k p ro = 0 . (7.73) ∂r r = ro
(
)
The product kpro must be equal to the n-th zero point of the derivative of the Bessel function ' . These values are listed in Table 13.2 in the mathematical appendix. of the m-th order α mn Consequently the transversal propagation constant of the mode with modal numbers m and n is TE k pmn =
' α mn
ro
.
(7.74)
From (5.18) and (5.19) we get the longitudinal propagation constant and the cut-off frequency of the TE modes in a circular waveguide TE k zmn
2
= ω µε −
TE f cmn =
1
2 k pmn
' α mn
2π µε ro
α ' = ω µε − mn ro 2
.
2
,
(7.69)
(7.70)
According to Table 13.2 we have the order of the lowest modes TE11, TE21, TE01, TE31 ' ' ' ' ' are α 11 = 1.84118, α 21 = 3.05424, α 01 = 3.83170, α 31 and TE41, as the lowest numbers α mn ' = 4.20119, α 41 = 5.31755. The transversal propagation constants and the cut-off frequencies of the TE modes are different from those of the TM modes. The field distribution of the lowest modes over the waveguide cross-section is shown in Fig. 7.11. For the circular waveguide the same relations as (7.21), (7.22), (7.23) and (7,24) can be derived for the wavelength along the waveguide, the phase velocity, the group velocity, and the longitudinal propagation constant, of course separately for the TE and TM modes. The wavec impedance is defined as the ratio of the transversal components
Z TE ,TM =
E Er =− α . Hα Hr
The dominant mode of the circular waveguide is the TE11 mode, as its cut-off ' = 1.841. The frequency has the lowest value. The value of the corresponding parameter α 11 cut-off frequency of this mode is TE f c11 =
1
1.841 . 2π µε ro
(7.71)
The cut-off wavelength of the dominant TE11 mode is
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λTE cmn =
2π ε r ro 1.841
= 3.4 ε r ro .
(7.72)
___________________________________ Example 7.9: Determine the frequency band of the single mode propagation of a circular waveguide filled by a dielectric with relative permittivity εr with radius ro = 8.2 mm. The dominant mode of the circular waveguide is the TE11 mode with the cut-off wavelength (7.72). The nearest higher mode is the TM01 mode with the cut-off wavelength
λTc 01M =
v f cTM 01
=
2π ε r ro 2.40482
= 2.6 ε r ro .
For the waveguide radius ro = 8.2 mm we get the band of the single mode operation 21.32 < λ < 27.88 mm 10.76 < f < 14.07 GHz ____________________________________
7.4 Problems Calculate for the modes propagating between the two parallel plates from Example 7 7.1 the wavelength, the phase velocity, the group velocity, the longitudinal propagation constant and the wave impedance. The frequency is 10 GHz. mode vpm kzm vgm ZTE ZTM λgm -1 (m/s) (m/s) (m ) (mm) (Ω) (Ω) 8 8 0th 3 10 3.10 30 209 376.6 8 8 1st 3.24 10 2.78 10 32.4 193.7 349.1 406.3 2nd 4.55 108 1.98 108 45.5 137.9 248.6 570.6
7.2 Calculate the frequency band of the propagation of only the TE1 mode in a parallel plate waveguide filled by air. The distance of the plates is 40 mm. Do not consider the TM modes. fc1 = 3.75 < f < fc2 = 7.5 GHz 7.3 How does the result of Problem 7.2 change if we replace the air by a dielectric with permittivity εr = 9? fc1 = 1.25 < f < fc2 = 2.5 GHz 7.4 Determine the points at the cross-section of the rectangular waveguide at which the magnetic field of modes TE10, TE01, TE20, TE11 and TM11 is zero. 7.5 Determine the points at the cross-section of the rectangular waveguide at which the electric field of modes TE10, TE01, TE20, TE11 and TM11 is zero. 7.6 The maximum value of the amplitude of the electric field in the rectangular waveguide is Em = 600 V/m. The dominant mode TE10 propagates in this waveguide at the frequency 10
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GHz. The waveguide is filled by air and its internal dimensions are a = 22.86 mm and b = 10.16 mm. Calculate the amplitudes of all field components. Eym = -600 V/m Hxm = 1.19 A/m Hzm = 1.04 A/m 7.7 Calculate the electric field intensity at the point with the coordinates x = 5.5 mm and y = 5.1 mm in the cross section of a rectangular waveguide filled by air. The waveguide transmits the dominant mode TE10 at the frequency 15 GHz. The transmitted power is 10 W. The internal dimensions of the waveguide are a = 22.9 mm, b = 10.2 mm. E y = − j 575 e − j 282 z V/m 7.8 Determine which modes can propagate along a circular waveguide filled by air with radius ro = 5.2 mm at frequency f = 30 GHz. TE11, TM01, TE21 7.9 Calculate for the mode TE11 propagating along the waveguide from problem 7.8 at frequency of 30 GHz the longitudinal propagation constant and the wavelength. -1 k zTE 11 = 518.9 m
λTE g11 = 12.1 mm
8. DIELECTRIC WAVEGUIDES In the past, dielectric waveguides were used in microwave systems mainly as dielectric antennas. The wave was guided by a dielectric rod, the cross-section of which was tapered and the wave was gradually radiated out into the space. Dielectric waveguides are nowadays frequently applied in optical communication systems as optical fibers and in the form of planar optical waveguides in the circuits of optical integration systems. The operation of dielectric waveguides is based on total reflection of the wave on the boundary evanescent wave between the two dielectric materials treated in paragraph 3, see Fig. 3.15. The wave is totally vpzv1 boundary. Perpendicular to the boundary the wave in fast wave the first material has the character of a standing wave. The wave propagates as a surface wave in the vpz ε2 evanescent wave couple into this layer the wave as it is incident to the boundary at an angle greater than the critical angle. It Fig. 8.1 is reflected back and is incident to the opposite
108 book - 8
boundary and bounces up and down. According to Fig. 8.1 the wave is in this way coupled to the layer, but it also penetrates into the surrounding material, where its amplitude decreases exponentially. The field distribution across the layer depends on the frequency and on the thickness of the layer. Thus any dielectric structure with a core made from a material with permittivity greater than the permittivity of the surrounding material can behave as a dielectric waveguide. The guiding of the wave is due to the total internal reflection on the surface of the core.
8.1 Dielectric layers A dielectric layer is the dielectric d ε2 guiding structure which can be treated most -a simply. Let us assume a dielectric c y homogeneous layer that is infinitely long and 0 z ε1 wide, and its thickness is 2a, see Fig. 8.2. We a assume that the field does not depend on the y ε2 e ∂ coordinate, therefore = 0 , and that the x ∂y wave propagates in the z axis direction. Thus Fig. 8.2 we can treat the TE and TM modes separately, similarly as in the parallel plate waveguide. According to (5.14-17) the TE modes have non-zero field components Ey, Hx and Hz, and the TM modes have components Hy, Ex and Ez. TE modes Let us first study the TE modes. The H d ε2 mode propagates in the dielectric layer due -a to the internal total reflections, see Fig. 8.3. c y Its electric field has only the Ey component, 0 E z ε1 and the magnetic field must thus have the Hx a and Hz components. The wave equation can ε2 e by now solved for the Ey component of the x electric field. The solution obtained similarly as in paragraph 5 by the method of Fig. 8.3 separation of variables can be written in the particular regions c, d, e, Fig. 8.3, E y1 = [A sin (k x1 x ) + B cos(k x1 x )]e − jk z z ,
x
(8.1)
E y 2 = C e jk x 2 z e − jk z z ,
x < -a ,
(8.2)
E y 3 = D e − jk x 2 z e − jk z z ,
x>a,
(8.3)
where the separation constants are coupled by
k12 =
(
ε r1 k 0 ) = ( ε r1ω ε 0 µ 0 ) = k x21 + k z2 ,
k 22 = k 32 =
2
(
2
ε r 2 k 0 ) = ( ε r 2 ω ε 0 µ 0 ) = k x22 + k z2 . 2
2
(8.4) (8.5)
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Constants kx1 and kx2 are the propagation constants in the x direction in the dielectric layer and in the surrounding material, respectively. We can simplify the treatment of the field by dividing the TE modes into two groups. In the first group we have symmetric, or even modes, which satisfy the condition E y ( x ) = E y (− x ) .
(8.6)
For these modes we put in (8.1) A = 0 and they are described by the cosine function. In the second group we have anti-symmetric, or odd modes which satisfy the condition E y ( x ) = − E y (− x ) .
(8.7)
For these modes we put in (8.1) B = 0 and they are described by the sinus function. The Ey component of the even TE modes in the dielectric layer is
E y1 = B cos(k x1 x ) e − jk z z .
(8.8)
The magnetic field is related to the electric field by Maxwell’s second equation (1.13). Assuming Ex = Ez = 0 we will get
Hz =
1 ∂E y , jωµ ∂x
(8.9)
and the z component of the magnetic field in our three particular areas is H z1 = −
H z2 =
k x1 B sin (k x1 x ) e − jk z z , jωµ
k x2
ωµ
H z3 = −
C e jk x 2 z e − jk z z ,
k x2
ωµ
D e − jk x 2 z e − jk z z ,
x
(8.10)
x < -a ,
(8.11)
x>a.
(8.12)
The field must fulfill the boundary conditions on the layer surfaces at x = a and x = -a. Consequently at x = a and for arbitrary z we have E y1 (a ) = E y 3 (a ) ,
H z1 (a ) = H z 3 (a ) ,
B cos(k x1a ) = D e − jk x 2a , −
(8.13)
k x1 k B sin (k x1a ) = − x 2 D e − jk x 2a . jωµ ωµ
− jk x1 B sin (k x1a ) = k x 2 D e − jk x 2a .
(8.14)
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Dividing (8.14) by (8.13) we get
(k x1a ) tg(k x1a ) = j (k x 2 a ) .
(8.15)
This equation couples propagation constants kx1 and kx2, but to determine them we need to get another equation. This equation is derived by inserting from (8.4) for kz into (8.5) k x21 − k x22 = (ε r1 − ε r 2 )ε 0 µ 0ω 2 .
(8.16)
Now to remove the imaginary unit from (8.15) we introduce a new variable k x' 2 = j k x 2 .
(8.17)
Consequently we have the two equations known as dispersion equations
(k x1a ) tg(k x1a ) = (k x' 2 a ) ,
(8.18)
(k a ) + (k a )
(8.19)
2
x1
' x2
2
= (ε r1 − ε r 2 )ε 0 µ 0ω 2 a 2 .
The Ey component of the odd TE modes is E y1 = A sin (k x1 x ) e − jk z z .
(8.20)
Applying exactly the same procedure for calculating the distribution of the corresponding magnetic field and fulfilling the boundary conditions we will get the dispersion equation for the odd TE modes in the form
(
)
− (k x1a ) cotg(k x1a ) = k x' 2 a .
(8.21)
The second dispersion equation remains in the Ey form (8.19). d The electric field distribution of TE modes in the plane transversal to the m=2 m=0 m=1 propagation direction is described by (8.8) or (8.20) in the layer itself and by (8.2) and (8.3) in c 2a the surrounding material. This distribution is plotted for the three lowest TE modes in Fig. 8.4. The modes are denoted by numbers which determine the number of zeros of the Ey function across the layer. So we have, Fig. 8.4, TE0 TE2 e TE1 the series of modes: even TE0, odd TE1, even even odd even TE2, odd TE3, etc. Solving equations (8.18) and (8.19) or Fig. 8.4 (8.21) and (8.19) we determine the propagation constants kx1 and kx2, and the longitudinal propagation constant is then calculated from (8.4) or (8.5). In this way we can get the dispersion characteristic of the given TE mode propagating along our dielectric layer. The dispersion characteristic represents the dependence of the propagation constant on the frequency. Dispersion equations (8.18) and (8.19) cannot
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be solved analytically. The solution must be performed using a proper numerical method. This problem can be overcome by a simple graphical technique, which gives a clear physical insight of the effect of guiding the waves along the dielectric layer. In the coordinate system ' (k x1a ) and k x 2 a , equation (8.19) represents a circle with its center at the origin. The radius of this circle is
(
k‘x2a M0 M1 M2
)
π/2
(ε r1 − ε r 2 )ε 0 µ 0
R =ωa
TE0
. (8.22)
TE1 TE2
π
TE3
3π/2
kx1a
R
The radius is proportional to the frequency, TE4 the layer thickness and the second root of the difference between the permittivities. M0non The solution of the set of equations (8.18) and (8.19) or (8.21) and (8.19) can be obtained graphically as an intersection of the circle (8.19) with the curve represented Fig. 8.5 by (8.18) or (8.21), Fig. 8.5. From Fig. 8.5 we see that we always have one solution of the dispersion equations, which is represented by the intersection of the circle (8.19) with the first branch of the tangent function on the left hand side of (8.18). This means that the TE0 mode can propagate under any circumstances at any layer at any frequency. This is the dominant mode of this dielectric waveguide. The nearest higher mode is the TE1 mode. So the upper frequency of the frequency band of the single mode propagation is determined by the cut-off frequency of the TE1 mode. The solutions of the dispersion equations lying in the lower half plane of the plane (k x1a ) and k x' 2 a determined by k x' 2 a < 0 are not physical, as they correspond to the field (8.2) and (8.3) which increases exponentially in the surrounding areas d and e. Consequently, the cut-off frequency of the TE1 mode follows from the condition R = π/2, which gives the cut-off frequency of this mode
(
)
f c1 =
(
1 4a
)
1
(ε r1 − ε r 2 )ε 0 µ 0
.
(8.23)
The cut-off frequency of any mode with the modal number m > 0 is determined by the condition R = mπ/2, so generally we have f cm =
m 4a
1
(ε r1 − ε r 2 )ε 0 µ 0
.
(8.24)
Fig. 8.5 shows the plots of the dispersion equations (8.18) for modes TE0, TE2 and TE4, (8.21) for modes TE1 and TE3. There are the three propagating modes TE0, TE1 and TE2 at the given frequency. The corresponding solutions of the dispersion equations are marked in Fig. 8.5 by M0, M1 and M2. The solution marked M2non is non-physical.
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_______________________________________ Example 8.1: Calculate the longitudinal propagation constant of the TE 0 mode propagating along a dielectric layer 5 mm in thickness with the relative permittivity εr = 2.6 at the frequency f = 62.5 GHz. The layer is surrounded by air. According to (8.22) we have R = 4.135. The corresponding dispersion equations are plotted in Fig. 8.5. Our solution is marked by M0. From Fig. 8.5 we can read (k x1a ) = 1.29, k x' 2 a = 3.93 . From these values we have kx1 = 516 m-1, and (8.4) gives us kz = 2042 m-1. ___________________________________ TM modes
(
)
The same procedure can be followed for the TM modes. These modes propagate -a H y in the dielectric layer due to the internal total 0 reflections, see Fig. 8.6. Their magnetic field E has only the Hy component and the electric a field must thus have the Ex and Ez components. The wave equation can now be x solved for the Hy component of the magnetic field. The solution obtained similarly as in Fig. 8.6 paragraph 5 by the method of separation of variables can be written in the particular regions c, d, e, Fig. 8.6, H y1 = [A sin (k x1 x ) + B cos(k x1 x )]e − jk z z ,
x
ε2
d c
ε1 ε2
z
e
(8.25)
H y 2 = C e jk x 2 z e − jk z z ,
x < -a ,
(8.26)
H y 3 = D e − jk x 2 z e − jk z z ,
x>a,
(8.27)
where the separation constants are coupled by (8.4) and (8.5). The distribution of the Hy component of the magnetic field of the TM modes is the same as the distribution of the electric field shown in Fig. 8.4. The rest of the process is the same as in the case of the TE modes. From the Hy component of the magnetic field we calculate the distribution of the Ez electric field by Ez = −
1 ∂H y . jωε ∂x
(8.28)
We divide the field into even and odd modes and we meet the boundary conditions at x = a, which are Hy1(a) = Hy3(a) and Ez1(a) = Ez3(a). As a result we obtain the dispersion equations for the even TM modes
(
)
ε2 (k x1a ) tg(k x1a ) = k x' 2 a , ε1
(8.29)
and for the odd TM modes
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−
(
)
ε2 (k x1a ) cotg(k x1a ) = k x' 2 a . ε1
The equation of the circle (8.19) remains unchanged. The cut-off frequencies of the TM modes are then defined by (8.24) as for the TE modes. The dominant mode is the TM0 mode, which can propagate at any frequency. The frequency band of the single mode operation is confined from above by the cut-off frequency of the TM1 mode (8.23). The only difference between the solutions of the dispersion equations is their shift caused by the term ε2 /ε1, Fig. 8.7. The dispersion characteristic of the modes propagating along the dielectric layer can be obtained by repeating the procedure shown in Example 8.1 for the TE modes, and similarly for the TM modes. The dispersion characteristic of the TM0 and TE1 modes propagating along the dielectric layer defined in Example 8.1 is plotted in Fig. 8.8. Note the cut-off frequency of the TE1 mode.
(8.30) k‘x2a M0 M1 TM0 TM1
TM2
M2 TM3
π/2
π
3π/2
kx1a
TM4 R
M0non
Fig. 8.7
8.2 Dielectric cylinders The dielectric cylinder, the crosssection of which is shown in Fig. 8.9, represents the dielectric waveguide widely used in practical applications. Optical fiber is based on it. In the case of optical fiber the surrounding material with relative permittivity εr2 does not fill the whole space, but creates a cover layer, see Fig. 5.1c. The cylindrical dielectric waveguide is able to transmit the wave due to the internal reflection,
TM0 TE1
fc1
Fig. 8.8
εr2
d
εr1
c
r0
z
Fig. 8.9
assuming that ε1 > ε2. The field distribution can be determined similarly as in the case of a cylindrical waveguide, using the wave equation in the cylindrical coordinate system. However, we have to assume that the field penetrates into the surrounding material d. The field cannot be divided into TE and TM modes due to the boundary conditions on the surface of the dielectric cylinder. All modes propagating along the dielectric cylinder have nonzero components
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Ez and Hz. Only for the cylindrically symmetric modes with the modal number m = 0 do we have the modes TE0n and TM0n. The longitudinal components of the electric and magnetic fields are assumed in the form E z1 (r ,α , z ) = E 0 z1 (r ,α ) e − jk z z ,
r < r0 ,
(8.31)
E z 2 (r ,α , z ) = E0 z 2 (r ,α ) e − jk z z ,
r > r0 .
(8.32)
H z1 (r ,α , z ) = H 0 z1 (r ,α ) e − jk z z ,
r < r0 ,
(8.33)
H z 2 (r , α , z ) = H 0 z 2 (r ,α ) e − jk z z ,
r > r0 .
(8.34)
Functions E0z1, E0z2, H0z1 and H0z2 are solutions of the wave equation (7.60) and a similar equation for the magnetic field. Applying the same procedure as in the case of the metallic waveguide we will obtain the solution
(
)
E z1 = A J m k p1r e jmα e − jk z z ,
(
)
E z 2 = B K m jk p 2 r e jmα e − jk z z ,
(
)
H z1 = C J m k p1r e jmα e − jk z z ,
(
)
H z 2 = D K m jk p 2 r e jmα e − jk z z ,
r < r0 ,
(8.35)
r > r0 ,
(8.36)
r < r0 ,
(8.37)
r > r0 .
(8.38)
Where Km is the modified Bessel function of the second kind, see the mathematical appendix. This function describes the evanescent character of the field outside the cylinder. The electric and magnetic fields (8.35-38) together with Eα1, Eα2, Hα1 and Hα2 have to meet the boundary conditions on the cylinder surface as they are tangent to this surface. From these boundary conditions we can determine the constants B, C and D and the dispersion equations necessary to determine the propagation constants of particular modes.
8.3 Problems 8.1 Determine the frequency band of the single mode operation and the cut-off frequencies of four modes in a plexiglass slab 5 mm in thickness with relative permittivity equal to 2.6. The slab is infinitely wide and long and is surrounded by air. fc1 = 23.7 GHz fc2 = 47.4 GHz fc3 = 71.1 GHz fc4 = 94.8 GHz the single mode operation band is 0 < f < 23.7 GHz 8.2 Determine the maximum thickness of a dielectric layer with relative permittivity εr = 2.28 which ensures propagation of the TE0 mode alone at the wavelength λ0 = 1.3 µm. The dielectric layer is placed between two glass layers with the relative permittivity εr = 2.26. h ≤ 4.6 µm
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8.3 Calculate the phase velocity of the TE0 mode propagating along the dielectric layer from the example 8.1 at the frequency f = 62.5 GHz. vpTE0 = 2.53 108 m/s 8.4 Determine the field distribution and dispersion equations of the TE and TM modes propagating along a dielectric layer of thickness a placed on an ideally conducting plane, Fig. 8.10. The structure in Fig. 8.10 is one half of the structure shown in Fig. 8.2, with the plane of symmetry located at x = 0. Only the x even TM modes and the odd TE modes d ε2 can exist on this structure due to its a symmetry. Apart from this, everything else ε1 c y is the same as was explained for the 0 σ →∞ z dielectric layer. The dominant mode is the TM0 mode, and the TE0 mode does not Fig. 8.10 exist.
9. RESONATORS 9.1 Cavity resonators Microwave resonators are used in microwave circuits similarly as standard resonant LC circuits at lower frequencies. At high frequencies we have to protect the electromagnetic field from radiation outside the resonator. For this reason, resonators are very often used as cavity resonators. These resonators are formed by the volume filled by a dielectric material, mostly air. This volume is shielded by a suitable metallic coating. The electromagnetic field is coupled inside, using a suitable probe. We will treat lossless resonators, assuming that the dielectric material has zero conductivity and on the other hand the metal coating has infinite conductivity. Let us first determine the resonant frequency of a general cavity resonator. We use the power balance for the reactive power described by equation (1.33). Assuming a shielded lossless structure without any radiation, the power balance for the reactive power is
(
(
)
)
1 1 2 2 Im J ⋅ E * dV = ω ∫∫∫ ε E − µ H dV . ∫∫∫ 2 V 2 V Disconnecting the source we have J = 0 and we get the self-oscillation of the field in the volume. The condition for self-oscillation is determined by the equality of energies We = Wm, which is expressed by 2
2
ε ∫∫∫ E dV = µ ∫∫∫ H dV . V
V
The electric field can be determined from Maxwell’s first equation (1.12) assuming σ = 0 and no external current. Then we have
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1
2
E =
rot (H ) . 2
2 2
ω ε
Finally we get the resonant frequency of our cavity
ω 02
=
1
µε
∫∫∫ rot(H )
2
dV
V
∫∫∫
2
H dV
=
1
µε
V
∫∫∫ rot(E)
2
dV
V
∫∫∫
2
E dV
.
(9.1)
V
It follows from (9.1) that the resonant frequency of the cavity depends on the material parameters µ and ε and on the cavity geometry that is hidden in the integrals. This is common with standard LC resonant circuits. However, (9.1) shows that the resonant frequency depends on the distribution of the electric and magnetic fields. Consequently as we can excite an infinite number of particular modes in the cavity, we have an infinite number of resonant frequencies corresponding to these modes. Some modes are known as degenerated. These modes have a different field distribution but an equal resonant frequency. Let us first determine the resonant frequency of the cavity resonator formed by a segment of any waveguide terminated at both ends by ideally conducting planes. The planes are perpendicular to the longitudinal axis of the waveguide. The length of the d z segment is d, Fig. 9.1, so the conducting planes are 0 d located at z = 0 and z = d. It is reasonable to assume that the field distribution of the mode excited in this cavity originates from the distribution of the modes existing in Fig. 9.1 the original waveguide forming the cavity. These modes have transversal and longitudinal propagation constants kpmn and kzmn (5.5) which depend on two transversal modal numbers m and n. The distribution of the transversal components of its electric field is ET0(x,y) as the function of the two transversal coordinates x, y. The electric field in our cavity is now the superposition of the two waves traveling in the positive z direction and in the negative z direction
(
)
ET (u, v, z ) = E T 0 (x, y ) A e − jk zmn z + B e jk zmn z . This field must fulfill the two boundary conditions at z = 0 and z = d. At z = 0 we have ET 0 ( A + B ) = 0
=>
A = -B .
The field distribution is now
(
)
ET (u, v, z ) = E T 0 (u, v ) − B e − jk zmn z + B e jk zmn z = ET 0 (u, v )2 jB sin (k zmn z ) .
(9.2)
At z = d we have
sin (k z d ) = 0 .
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k zmnp =
pπ , d
p = 1, 2, 3, … .
(9.3)
(9.3) tells us that we have the set of discrete values of the longitudinal propagation constants of the modes oscillating in the resonator. These constants depend on modal numbers m, n and p. Now using (5.5) and (2.25) we have the set of resonant frequencies f 0 mnp =
1
2 k pmn
2π µ 0ε 0 ε r
pπ + d
2
.
(9.4)
The particular value of the resonant frequency depends on transversal propagation constant kpmn, which is determined by the type of waveguide from which the resonator is composed. In the case of a waveguide with a rectangular cross section with dimensions a and b, kpmn is determined by (7.40) and the resonant frequency is f 0 mnp =
2
2
mπ nπ pπ + + a b d
1 2π µ 0ε 0 ε r
2
.
(9.5)
This formula is valid for both the TE and TM modes. _____________________________________ Example 9.1: Design a resonator tuned to the resonant frequency 15 GHz for the mode TE101. The resonator is formed by the segment of the rectangular waveguide with the dimensions a = 30 mm, b = 15 mm. The resonator is filled with air. The dominant mode of the rectangular waveguide has the simplest field distribution, so we design the resonator for this mode. Thus for mode TE101 we have the resonant frequency (9.5) f 0101 =
c 1 1 + 2 , 2 2 a d
where c is the speed of light in a vacuum. From the above formula we can determine the necessary length d d=
1 2
1 2 f 0101 − 2 a c
= 10.6 mm.
The resonator must be 10.6 mm long. ___________________________________ In the case of a cylindrical resonator with radius ro we have to distinguish between the TE and TM modes, as their transversal propagation constants are different. For TE modes, kpmn is determined by (7.74), and the resonant frequency is
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f 0 mnp =
2
1 2π µ 0 ε 0ε r
2 ' α mn + pπ . r d o
(9.6)
For TM modes, kpmn is determined by (7.68), and the resonant frequency is f 0 mnp =
2
1 2π µ 0 ε 0ε r
α mn pπ + ro d
2
.
(9.7)
The dominant mode of the cylindrical waveguide is the TE11 mode, so the resonators are ' mostly designed to work with this mode, where α 11 = 1.841 . _____________________________________ Example 9.2: Design a cylindrical resonator tuned to the resonant frequency 14 GHz for the mode TE111. The resonator is filled with air and its radius is ro = 8.2 mm. Applying the same procedure as in example 9.1 we get the resonator length d=
π 2
' 2πf 0101 α 11 − c ro
2
= 16.67 mm.
The resonator must be 16.67 mm long. ___________________________________ A similar procedure can be applied to determine the resonant frequency of a dielectric resonator of either the circular or rectangular cross section terminated by conducting planes. The conducting planes must exceed the cross section of the resonator, as the field is not confined to the dielectric, but it penetrates into the surrounding air. The resonant frequency is then determined by (9.4), where the proper value of the transversal propagation constant must be used. Cavity resonators are used in microwave technology at frequencies above about 5 GHz. At lower frequencies the transversal dimensions of the waveguide, a segment of which forms the resonator, are too big. For this reason, these resonators become impractical at such frequencies. We can use either dielectric resonators or ferrite resonators. Another possibility is to form the resonator from a TEM transmission line. This line can be left either short circuited or open at its ends. A short circuited termination is more convenient, as it does not radiate. A TEM transmission line has kpmn = 0, consequently the resonant frequency is determined directly by the number of periods of the standing wave along the line segment. So we have from (9.4) f0 p =
p 2d µ 0 ε 0 ε r
.
(9.8)
In the case of a real resonator with some losses we define the quality factor of the resonator. The internal quality factor determined by the losses of the resonator itself is defined by
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Q=
ω 0W PL
,
(9.9)
where ω0 is the resonant frequency, W is the averaged value of the total energy of the electromagnetic field inside the cavity, PL is the power lost in the resonator. The losses in the dielectric material filling the cavity can usually be neglected, namely in the case of air. Consequently the power is lost in the cavity walls. The stored energy is W = Weav + Wmav = 2Weav =
ε
E 2 ∫∫∫
2
dV .
(9.10)
V
The power lost in the cavity walls can be calculated from the known distribution of the magnetic field on these walls using (3.26). Dielectric resonators are used namely in microwave integrated systems. They are significantly smaller than cavity resonators, which is their main advantage. These resonators are usually designed as cylinders made of a dielectric material with high permittivity. This significantly reduces the dimensions. The disadvantage is that the field penetrates into the surrounding space, which may increase the losses and reduce the quality factor. On the other hand the electromagnetic field can be simply coupled into such a dielectric resonator by the proximity effect. The resonator is placed in the vicinity of a microstrip line and the field is directly coupled into its volume. Due to the field penetration outside the dielectric resonator, its resonant frequency can in most cases be determined by solving the field distribution numerically. The resonant frequency can be approximately determined by (9.4), as mentioned above, when the dielectric resonator is terminated by conducting planes. Ferrite resonators are produced from mono-crystals of yttrium iron garnet (YIG) in the shape of small spheres. The resonant frequency of these resonators depends on the stationary magnetic field H0 applied to this material. It is, as will be shown in chapter 11,
ω0 = γ H0 ,
(9.11)
where γ is gyromagnetic ratio, see Paragraph 11.1. These resonators can be simply tuned by changing the applied stationary magnetic field.
9.2 Problems 9.1 Calculate the resonant frequency of the cavity resonator formed by a segment of the waveguide with a rectangular cross section with the dimensions a = 25 mm, b = 15 mm, d = 10 mm for the modes TE101 and TE111. The cavity is filled with air. f0101 = 18.14 GHz f0111 = 19 GHz 9.2 Calculate the necessary length d of a cavity resonator of rectangular shape to get the resonator tuned to 10 GHz for the mode TE101. The cavity is filled with air. The transversal dimensions are a = 40 mm, b = 15 mm. d = 16.2 mm 9.3 Calculate the resonant frequency of a resonator with a circular cross section filled with air for the mode TE111. The radius of the cavity is ro = 8.2 mm, the length is d = 25 mm. f0111 = 12.27 GHz
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9.4 Calculate the resonant frequency and the quality factor of a resonator with a rectangular cross section filled with lossless air for the TE101 mode. The dimensions of the cavity are a = 22.86 mm, b = 10.16 mm, d = 22.86 mm. The wall conductivity is σ = 5.7 . 107 S/m. f0101 = 9.28 GHz Q = 7758
10. RADIATION Radiation is usually understood as the transmission of electromagnetic waves in a free space from a source of finite dimensions supplying finite power. We assume an infinite space filled by a homogeneous and isotropic dielectric material without any losses. We have already studied the propagation of waves in a free space, or along transmission lines. We did not take into account the source of the electromagnetic field. For example, a plane electromagnetic wave with surfaces of constant amplitude and phase in the shape of an infinite plane must be excited by a hypothetical source with infinite dimensions that must radiate infinite power. In the case of transmission lines, we just studied the field in the view of possible waves or modes, known as eigen modes, which can propagate along this line. We did not investigate whether these modes would really propagate, assuming that the line were connected to a real source supplying finite power. Radiation is used for the transmission of signals namely in telecommunications, radars, sensors and navigation. Radiation also occurs as a parasitic effect that represents losses, cross-talk and parasitic interference. For example, radiation from open transmission lines and electric circuits. In the first section we showed that electric current and electric charges are sources of an electromagnetic field. However, electric current is the flow of the charge. These two quantities are coupled by a continuity equation (1.8). This means that we can consider only the electric current as the source. The real source is therefore a conducting body of finite dimensions, and the electric current passes through its volume or at high frequencies only along its surface. From the known distribution of this current we can calculate the field distribution by solving the wave equation. The wave equations written directly for field vectors (1.41) and (1.42) have rather complicated functions of the source current and charge. Therefore they are not suitable for solving our problem. To remove this inconvenience we have introduced vector and scalar potentials, which represent the electric and magnetic fields by (1.46) and (1.52). Inserting these formulas into Maxwell’s equations we have derived wave equations for potentials with simple functions of the source current and charge (1.54) and (1.55). Their solution is in the form of the superposition of spherical waves (1.57) and (1.58). In this textbook we will calculate the electromagnetic field distribution only in the cases of simple sources – antennas, where we know in advance the distribution of the electric current. First we will treat simple elementary electric and magnetic dipoles. An electric dipole is represented by a very short straight conductor. The magnetic dipole is represented by a conducting loop of infinitely small dimensions. We will show how to calculate the field radiated by sources of dimensions comparable with the wavelength. We will define the parameters of antennas. Finally, we will mention antenna arrays and receiving antennas.
10.1 Elementary electric dipole An elementary electric dipole consists of two point electric charges q and -q placed at distance d. These charges harmonically change their values. The dipole moment is
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p = Qd =
Id , jω
(10.1)
as the charge is an integral of the electric current. So the dipole can be represented by a straight conductor with passing current I, Fig. 10.1. The conductor must be much shorter than the wavelength, as the amplitude and phase of the current are constant along the conductor. The cross-section of the conductor is negligible. Then the vector potential is simply calculated using (1.58) A = z0
µ 4π
∫∫∫ V
− jkr
Je r
z
A
r dS
ϑ
d
y
J
x
α
dV ,
Fig. 10.1
Current density J has according to Fig. 10.1 the direction determined by unit vector z0, and is constant within the conductor volume. The integral can be omitted, as J is constant within the volume of the dipole and we get e − jkr µ A = z0 Id . 4π r
(10.2)
The magnetic field is calculated from the known distribution of the vector potential using (1.46) and (1.12), assuming that the space is filled with a non-conducting material. First we transform the vector potential from the Cartesian coordinate system to the spherical coordinate system. Using Fig. 10.2 and using the fact that the field does not depend on angle α we get A = Ar r0 + Aϑϑ0 = Az cos(ϑ ) r0 + Az sin (ϑ )ϑ0 . The magnetic field can be obtained using (1.46), taking into account that Aϕ = 0 and ∂ / ∂α = 0 , and using the expression of the rotation operator in the spherical coordinate system (13.89) H=
1
µ
rot (A ) =
α0 ∂ (rAϑ ) − ∂Ar . µr ∂r ∂ϑ
(10.3)
z
(10.4)
Inserting (10.3) into (10.4) we get H=
Aϑ Az
ϑ
Ar
ϑ Fig. 10.2
k 2Id j 1 − jkr sin (ϑ )α 0 . + e 4π kr (kr )2
(10.5)
The electric field follows from Maxwell’s first equation (1.12) using (13.89)
E=
1 jωε
rot (H ) =
1 r0 ∂ [sin (ϑ ) H α ] − ϑ0 ∂ (rH α ) . jωε r sin (ϑ ) ∂ϑ r ∂r
(10.6)
Inserting (10.5) into (10.6) we get the components of the electric field vector
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Er = − j
Eϑ = j
Idωµ j 1 − jkr + cos(ϑ ) , e 2 λ (kr ) (kr )3
Idωµ 2λ
1 j 1 − jkr − sin (ϑ ) , − e 2 (kr )3 kr (kr )
(10.7)
(10.8)
The vector of the magnetic field (10.5) has only one component Hα =
k 2Id 4π
j 1 − jkr e sin (ϑ ) . + 2 kr (kr )
(10.9)
The electric field has two components, Er and Eϑ , and the magnetic field has only one component Hα. This is natural, as the lines of vector H must be of the shape of a circle drawn around the conductor carrying an electric current. Let us now discuss the character of formulas (10.7) to (10.9). The first fraction of these formulas characterizes the source by its current and length, and the frequency and material filling the space. The second term in rectangular brackets defines the dependence of the radiated field on the distance in the form of the normalized distance r/λ as k = 2π/λ. The third term determines the change of the field phase with distance. The last term describes the dependence of the radiated field on angle ϑ. 8 The dependence on distance contains three 1 powers of product kr or relative distance r/λ: (kr )3 6 1 1 1 and . These functions are (kr ) (kr )2 (kr )3 4 1 1 plotted in Fig. 10.3. They differ by the speed of (kr )2 their increase when the argument tends to zero 2 (kr ) and by the speed of their decrease when the argument increases to infinity. Consequently, we 0 can distinguish three areas around the dipole at 0 1 2 3 which the field has a particular behaviour. The kr well known near field zone, which is very close Fig. 10.3 to the dipole, is defined by condition r/λ << 1. Here field components with the term 1/(kr)3 prevail. This term is present in the formulas describing the electric field (10.7) and (10.8). Assuming r/λ << 1, we can simplify (10.7) and (10.8) to Er =
Idωµ 1 cos(ϑ ) , jλ (kr )3
(10.10)
Idωµ 1 Eϑ = sin (ϑ ) . 2 jλ (kr )3
Inserting I = jωq into (10.10) we get the well-known distribution of the electrostatic field excited by a dipole. For this reason, this zone is called the static zone. The magnetic field is of negligible value in this zone.
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At slightly longer distances, term 1/(kr)2 starts to have an important value, and the magnetic field must be taken into consideration. In this area it can be expressed as k 2Id 1 − jkr Id sin (ϑ ) , e sin (ϑ ) 0 ≈ 2 4π (kr ) 4πr 2
Hα =
(10.11)
which, according to the Biot-Savart law, represents the magnetic field excited by the element of a conductor with electric current I, d in length. This area is known as the inductive zone. The most important zone is the far-field zone. This zone is defined by r >> λ. Only terms 1/(kr) are of importance in the formulas describing the field distribution. The field contains only two components
µ e − jkr sin (ϑ ) , Id ε r
Eϑ =
j 2λ
Hα =
j e − jkr Id sin (ϑ ) . r 2λ
(10.12)
(10.13)
These components of the electric and magnetic fields are in phase, are mutually perpendicular, and are perpendicular to the direction of propagation, which is determined by unit vector r0. Their ratio Eϑ = Hα
µ = Zw . ε
equals the wave impedance of the free space. Thus the field has the character of a TEM plane wave propagating in the free space. In fact the field is represented by a spherical wave modified in its direction of propagation by function sin(ϑ). Only components Eϑ and Hα participate in the transmission of power from the dipole. For this reason, the area is known as the radiation zone. Function sin(ϑ) determines the radiation pattern of the dipole. The radiation pattern is generally a function dependent on two angles, ϑ and α, and is defined by F (ϑ ,α ) =
Eϑ (r = const ,ϑ ,α ) , Eϑ max
The radiation pattern is usually α = 0 plotted as its cuts in characteristic planes. For our dipole F (ϑ ,α ) = sin (ϑ ) , and this radiation pattern is shown in Fig. 10.4. The radiation pattern is omnidirectional in the horizontal plane ϑ = 90°. The power radiated by the dipole to the whole space is calculated using Poynting’s vector, calculated in the far field zone
(10.14)
z
ϑ = 90°
ϑ
z
x
y
x Fig. 10.4
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S av =
1 1 1 j Re E × H * = Re Eϑ H α* r0 = Re 2 2 2 2λ
(
)
(
µ I d − jkr − j I d − jkr e sin (ϑ ) e sin (ϑ ) r0 = ε r 2λ r
)
2
µ I d2 = sin 2 (ϑ ) r0 ε 8λ2 r 2
(10.15)
The total power is calculated integrating (10.15) over the surface of a sphere with radius r much greater than the wavelength. An element of this surface in the spherical coordinate system is dS = r2sin(ϑ)dαdϑ
µ I 2d 2 P= ε 8λ2
sin 2 (ϑ ) 2 ∫ ∫ r 2 r sin (ϑ ) dϑdα = 00
2ππ
π
µ I 2d 2 µ I 2d 2 4 3 ( ) sin d = π ϑ ϑ π ε 4λ2 ∫0 ε 4λ 2 3 2
µ I 2 d 2π 40 I 2 d 2π 2 40 2 πd I . = = ε 3λ2 εr λ ε r λ2
P=
(10.16)
The dipole radiating this power represents a load for the source feeding it. So we can substitute the dipole by a radiating resistor with resistance Rr. This is a resistor in which the same power is consumed. The power is P=
1 Rr I 2 , 2
so we get 2P Rr = 2 = I
2
µ d 2 2π d = 80 π 2 . 2 ε 3λ λ
(10.17)
As d << λ, the value of Rr is very small and it is practically impossible to match this elementary dipole to any generator. ____________________________________ Example 10.1: Calculate the power radiated by a dipole radiating into a space filled with air, and also the radiation resistor of this dipole. The dipole is d = 2 m in length. The passing current has amplitude 3 A and frequency 1 MHz. The radiating power is 2 πd
2
2
40 2 πdf P= I = I = 0.16 W εr λ εr c 40
From the radiated power and the passing current we get Rr =
2P = 0.035 Ω I2
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____________________________________ Example 10.2: What power must be radiated by the electric dipole to get the value of an electric field Em = 25 mV/m at the distance from the antenna r = 100 km determined by angle ϑ = 90°. The antenna is much shorter than the wavelength, which is 450 m. The space is filled with air. We are at a distance r=105 >> λ = 450, so we are in the far field zone. The value of the electric field is here, assuming ϑ = 90°, 1 2λ
Eϑ = E m =
µ0 1 Id , ε0 r
and from this formula we get the product Id characterizing the antenna Id =
E m 2λr
µ0 ε0
= 5864 Am .
The radiated power is (10.16) 2
πId P = 40 = 69.5 kW. λ ________________________________________
z I
10.2 Elementary magnetic dipole An elementary magnetic dipole is represented by a current loop with radius a and passing current I. The magnetic dipole is shown in Fig. 10.5. The radius of this loop is much lower than the wavelength, and consequently we can assume that the current has a constant amplitude and phase along the loop. The moment of this dipole is m = I dS ,
dS a
x
y
Fig. 10.5 (10.18)
where dS = π a 2 . The excited field has rotational symmetry, so we can look for the field distribution only in the plane xz. The current densities in the loop at any position x can be decomposed into two components. The x components have opposite directions, so the field excited by them vanishes. The radiated field is thus excited by only the y components of the current density, Fig. 10.6a. Using (1.58), we get the vector potential describing the field excited at point P, Fig. 10.6b, in the form
µ A(P ) = 4π
∫∫∫ V
J (Q ) e rQP
− jkrQP
d VQ
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P y
z
J
ϑ
Jy = Jcos(α) Q
Jx = Jsin(α)
a
x
Jy
α
O
rsin(ϑ)
acos(α)
Jx
rcos(ϑ)
r
y
Q
α
rQP
a
X
x
b Fig. 10.6
where dVQ = S c a dα , Sc is the cross-section of the loop conductor. The vector potential has the direction of the exciting current. This is the y component in the plane xz, but in the whole space it is the α component. Taking into account only the y component Jy = Jcos(α) of the current density and assuming that the integral of J over cross-section Sc equals current I, which is constant along the loop, we get
µ Aα (P ) = 4π =
2π
e
∫ ∫∫
rQP
0 Sc
µ aI 4π
− jkrQP
2π
∫ 0
e
2π
− jkr
e QP µ J cos(α )dS c a dα = a∫ dα ∫∫ J cos(α )dS c = 4π 0 rQP S c
− jkrQP
rQP
cos(α )dα =
π
− jkrQP
e µ aI∫ 2π rQP 0
(10.19)
cos(α )dα
The result of the particular integration in (10.19) is precise. The final integral cannot be simply solved, this can be done only numerically. To get an analytical result, we have to simplify the expression for distance rQP, Fig. 10.6b. We assume r >> a. rQP =
(QX )2 + ( XP )2
=
[a
2
]
+ (OX )2 − 2a(OX ) cos(α ) + [r cos ϑ ]2 =
= a 2 + r 2 sin 2 (ϑ ) − 2ar sin (ϑ ) cos(α ) + r 2 cos 2 (ϑ ) = a 2 + r 2 − 2ar sin (ϑ ) cos(α ) ≈ a a ≈ r 2 − 2ar sin (ϑ ) cos(α ) = r 1 − 2 sin (ϑ ) cos(α ) = r 1 − sin (ϑ ) cos(α ) r r rQP = r − a sin (ϑ ) cos(α ) .
(10.20)
Assuming that cos[ka sin (ϑ ) cos(α )] ≈ 1 and sin[ka sin (ϑ ) cos(α )] ≈ ka sin (ϑ ) cos(α ) as a << λ we can simplify e
− jkrQP
= e − jkr e jka sin (ϑ ) cos (α ) = e − jkr {cos[ka sin (ϑ ) cos(α )] + j sin[ka sin (ϑ ) cos(α )]} ≈ = e − jkr [1 + jka sin (ϑ ) cos(α )]
,
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1 rQP
=
1 a r 1 − sin (ϑ ) cos(α ) r
≈
1 a 1 a 1 + sin (ϑ ) cos(α ) = + 2 sin (ϑ ) cos(α ) . r r r r
Now we can simply express (10.19) as − jkr
π
e QP µ Aα (P ) = aI∫ cos(α )dα = 2π r QP 0 =
π
µ 1 a a I ∫ e − jkr [1 + jka sin (ϑ ) cos(α )] + 2 sin (ϑ ) cos(α ) cos(α ) dα 2π r r 0
After some manipulations and after integration we get Aα (P ) =
µ a 2 I jk 4
1 − jkr sin (ϑ ) . + 2 e r r
(10.21)
From this distribution of the vector potential we calculate the magnetic field using H = rot (A ) µ , and from the magnetic field we calculate the electric field using E = rot (H ) ( jωε ) . The procedure is the same as that applied in the case of the electric dipole. Operator rot must be used in the spherical coordinate system (13.89). As a result we obtain two components of the magnetic field Hr and Hϑ and one component of the electric field Eα Hr =
π (ka )2 I λ
j 1 − jkr + cos(ϑ ) , e 2 (ka )3 (ka )
π (ka )2 I Hϑ = − 2λ Eα =
1 j 1 − jkr − sin (ϑ ) , − e 2 (kr )3 kr (kr )
ωµ (ka )2 I 1 4
j − jkr e sin (ϑ ) . − 2 kr (kr )
(10.22)
(10.23)
(10.24)
Comparing these formulas with (10.7), (10.8) and (10.9) we can see that the field excited by the current loop is dual to the field excited by the electric dipole. For this reason we will not repeat here the discussion from the previous section. It is obvious that, due to this duality, the field excited by the magnetic dipole can be obtained from the formulas (10.7), (10.8) and (10.9), using the transformation: H is exchanged by –E, E is exchanged by H, ε is exchanged by µ, and the dipole moment p = Id/(jω) is exchanged by µm = µπa2I. _________________________________ Example 10.3: Compare the power radiated by the electric dipole to the power radiated by the magnetic dipole. Assume the same current and the same length of the conductor. This means that the length of electric dipole d equals the perimeter of the current loop, i.e., d = 2πa. The power radiated by the electric dipole is determined by (10.16). The power radiated by the magnetic dipole can be calculated in the same way that (10.16) was derived. Another
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way is to use the above mentioned duality and transformation. To use this transformation we have to modify (10.16) to Pe =
µ I 2 d 2π p 2ω 4 = . ε 3λ2 12πε c 3
Inserting here µm instead of p, µ instead of ε and using d = 2πa we get the power radiated by the magnetic dipole
( )
2
ω 4 µ 2 m 2 ω 4 I 2 πa 2 µ Pm = = . 12πµ c 3 12π c 3 Now we can express the ratio of these powers 2
Pe I 2 d 2ω 2 12π c 3 λ = = . 3 4 2 2 4 Pm 12πε c ω I π a µ πa As λ >> a, the power radiated by an electric dipole is much greater than the power radiated by a magnetic dipole. For this reason, only an electric dipole is used as a transmitting antenna. The radiation of the magnetic dipole has very low efficiency. This antenna is used only rarely and only as a receiving antenna. _______________________________________
10.3 Radiation of sources with dimensions comparable with the wavelength In previous sections we calculated the field radiated by electric and magnetic dipoles. We assumed that these dipoles have dimensions much smaller than the wavelength, and the current density is therefore constant within the volume of the dipole. This assumption simplified the calculation of the integral in (1.58). In the case of sources with dimensions comparable with the wavelength we cannot neglect the current variation, and we have to consider that distance R in (1.58) varies within the volume of the source and is different from the radius vector, which defines the position of the point at which the field is calculated. Practical antenna problems are very complicated, as we do not know the current distribution in advance. This distribution however depends on the radiated field via the boundary conditions valid on the antenna surface. The radiated field is determined by the current distribution. This vicious circle must be overcome by a proper numerical method, which enables us to calculate the unknown current distribution applying the proper boundary conditions. Such a problem is however beyond the scope of this textbook. Let us turn our attention to the calculation of the radiated field in the far field zone, and let us assume the known distribution of a current passing our source. We are looking for the field at point P in the far field zone determined by radius vector r, Fig.10.7. The source has volume V. Let us put an element of the current passing the source volume at point Q defined by radius vector rQ, Fig. 10.7. The dimensions of the source are comparable with the wavelength. Assuming that point P is in the far field zone, vectors r and rQP can be considered to be parallel. Thus we can express the distance between points Q and P as
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rQP ≈ r − rQ cos(α ) , (10.25)
P z
and we can assume an even coarser approximation in the denominator of the expression for a vector potential 1 rQP
r rQcos(α)
0
1 ≈ . r
α
J
rQ x
V
Using (1.58) we get the vector potential at point P
y
rQP
Q
Fig. 10.7
µ A= 4π
∫∫∫ V
J (Q ) e rQP
− jkrQP
dV ≈
µ e − jkr 4π r
∫∫∫ J(Q)e
jkrQ cos (α )
dV ,
V
Let us denote a radiation vector
N = ∫∫∫ J (Q ) e
jkrQ cos (α )
dV ,
(10.26)
V
which defines the properties of the source. Then the vector potential is − jkr
µ e A= 4π r
N .
(10.27)
The vector of the electric field is calculated from the vector and scalar potentials by (1.52). The scalar potential is coupled with the vector potential by the Lorentz calibration condition (1.53). Inserting this condition into (1.52) we get
E = − jω A −
j
ωµε
grad div(A ) .
(10.28)
The vector potential (10.27) is inserted into (10.28), taking into account that radiation vector N does not depend on r. We get E=
1 ω µ − jkr (Nϑϑ0 + Nα α 0 ) + 12 [ L ] + L , e −j 4π r r
(10.29)
Nϑ and Nα are the components of N in the spherical coordinate system. We are looking for the field in the far field zone. In this zone only functions with the dependence on distance r of the order of –1 can be taken into account, and the value of the terms of higher orders can be neglected. This means that only the first term in (10.29) can be taken. This gives us a physically reasonable result. The electric field has the form of a spherical wave with a zero longitudinal component and two nonzero transversal components
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Er = 0 , Eϑ = − j
ω µ e − jkr Z e − jkr Nϑ , Nϑ = − j 4π r 2λ r
Eα = − j
ω µ e − jkr Z e − jkr Nα . Nα = − j 4π r 2λ r
(10.30)
As this spherical wave can be locally approximated by a plane TEM wave, we can determine the magnetic field using the wave impedance of the free space (2.26) Hα =
Eϑ , Zw
Hϑ = −
Eα , Zw
(10.31)
The field in the far field zone has locally the character of a TEM wave with elliptical polarization, because the particular transversal components Eϑ, and Eα, have generally different amplitudes and phase. z rQP P _________________________________ Example 10.4: Calculate the electric field radiated by a symmetric dipole of finite length, Fig. 10.8, located at r the origin of the coordinate system. Current passing Q L dipole conductors is in the form of a standing wave. ϑ Let us first calculate the radiation vector, which has the direction of the passing current. Using (10.26) x y we get L I(z) L L jkrQ cos (α ) jkz cos (ϑ ) N = z 0 ∫ ∫∫ J z e dSdz = z 0 ∫ I ( z ) e dz −L S
−L
The dipole is represented by a transmission line with an Fig. 10.8 open end termination at both ends. So the current must fulfill here the boundary condition I(-L) = I(L) = 0, which gives a current distribution in the form of a standing wave I = I m sin [k (L − z )] . The ϑ component of the radiation vector is Nϑ = -Nzsin(ϑ). The electric field has the only component Z e − jkr Z e − jkr Nϑ = j N z sin (ϑ ) = 2λ r 2λ r L 0 Z e − jkr = j sin (ϑ ) ∫ sin [k (L + z )]e jkz cos (ϑ ) dz + ∫ sin[k (L − z )]e jkz cos (ϑ ) dz 2λ r − L 0
Eϑ = − j
After some rearrangements, this formula reads
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Eϑ = j 60 I m
e − jkr cos[kL cos(ϑ )] − cos(kL ) . sin (ϑ ) r
(10.32)
The dipole radiation pattern is F (ϑ , α ) =
cos[kL cos(ϑ )] − cos(kL ) . sin (ϑ )
(10.33)
1 (kL )2 , cos[kL cos(ϑ )] ≈ 1 − 1 (kL )2 cos 2 (ϑ ) , 2 2 and the radiation pattern is described by the simplified formula
For a very short dipole (L << λ) cos(kL ) ≈ 1 −
F (ϑ , α ) =
(kL )2 sin (ϑ ) , 2
which is the radiation pattern of the elemental electric dipole. Fig. 10.9 shows the radiation patterns of dipoles with particular lengths: L → 0, L = λ / 4 (kL = π/2), L = λ/2 (kL = π), L = λ3/4 (kL = π3/2), . L = λ (kL = 2π), L = 2λ (kL = 4π). Increasing further the dipole length we L = λ/4
L→0
L = λ/2
ϑ
L = λ3/4
L=λ
L = 2λ Fig. 10.9
finally get the radiation pattern with two main beams directed along the conductors. This tells us that the power is transmitted only along the conductors. _______________________________________
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The power radiated by our source into the whole space is calculated from Poynting’s vector S av
2
Eϑ + Eα 1 1 = Re(E × H ) = r0 Eϑ H α* − Eα H ϑ* = r0 2 2 240π
(
)
2
= S rav r0 .
(10.34)
Integrating (10.34) over the surface of a sphere with radius r >> λ we get the radiated power π 2π
Pr = ∫ ∫ S rav r 2 sin (ϑ ) dϑ dα .
(10.35)
0 0
Quantity W = Sravr2 is known as the radiation intensity. It represents the power radiated into a unit space angle.
10.4 Antenna parameters
F (dB)
We have already spoken about the antenna radiation pattern (10.14). The radiation pattern determines the dependence of the radiated field on directional angles ϑ and α. It is usually normalized to the maximum of the radiated field. It is generally a complex function. Most often we use the amplitude radiation pattern, but the phase radiation pattern can be also applied. The power radiation pattern corresponds to the second power of (10.14). The amplitude radiation pattern is expressed in absolute units or in dB. Fig. 10.10 shows the radiation pattern of 0 an antenna plotted in dB normalized to FWHP -3 dB 0. It consists of the -5 main lobe and a SLL ϑ = 0° number of side -1 0 lobes. We are in most cases -1 5 interested in the main lobe, at which -2 0 the maximum of power is radiated. The main lobe is -2 5 determined by its width, usually -3 0 defined as the width measured at the -3 5 level –3 dB below -1 8 0 -1 5 0 -1 2 0 -9 0 -6 0 -3 0 0 30 60 90 120 150 180 the main maximum α (° ) of the electric field Fig. 10.10 amplitude, i.e., at one half of the maximum power – the full width at half power (FWHP). We are also interested in the direction of the main lobe. Next we define the side lobe level (SLL), usually in dB, which determines the separation of the maximum lobe of the side lobes from the level of the field in the main lobe. The radiation pattern defines the antenna directivity. The maximum directivity is defined as the ratio of the maximum of the radiation intensity and its average value
133 book - 101
Dmax
4πr 2 S rav max Wmax S rav max r 2 = = = π 2π = Pr Wav 2 r ∫ ∫ S rav sin (ϑ )dϑ dα 4π 0 0
=
4π π 2π
∫ ∫ F (ϑ ,α )
2
4π π 2π
S rav
∫ ∫ S rav max sin (ϑ )dϑ dα 0 0
=
,
(10.36)
sin (ϑ )dϑ dα
0 0
The directivity in the general direction then is D = Dmax F (ϑ , α ) . 2
(10.37)
The antenna directivity evaluates only the directional distribution of the energy flow. It does not define the efficiency of the radiation η. This efficiency is included into an antenna gain. The antenna gain incorporates the efficiency of the conversion of the power supplied from the source feeding the antenna Ps to radiated power Pr. The gain is defined as G=
Wmax 4πWmax = = ηDmax . Ps Pr η 4π
(10.38)
Then antenna input impedance is defined as ZA =
UA , IA
(10.39)
where UA and IA are the voltage and the current at the antenna terminals. It is usually difficult to determine the current distribution over the antenna that defines its input current. In many cases we thus substitute the real part of the antenna input impedance by the radiating resistance (10.17). It is Rr =
2 Pr IA
2
=
2r 2 IA
2
π 2π
∫ ∫ S r sin (ϑ ) dϑ dα
.
(10.40)
0 0
10.5 Antenna arrays We have studied up to now only single radiators. The field radiated by these radiators is calculated as the superposition of the spherical waves coming out from partial elements of the radiator. These particular waves depend on the local amplitude and phase of the current passing this element. The possible ways of modifying the antenna radiation pattern and gain are therefore limited, as we cannot simply change the current distribution over the antenna body. This is the reason why antenna arrays are used. Such an array consists of a number of partial radiating elements. Each element has its own feeding source. Now we have huge possibilities to change the array radiation pattern and its gain. This can be done by changing the amplitudes and phases of the feeding currents and changing the positions of these elements. Antenna arrays can be one dimensional, two dimensional or three dimensional.
134 book - 101
Generally, when we increase the number of particular elements we reduce the width of the radiation pattern. Let us show the tremendous possibilities of antenna arrays through the following simple case, Fig. 10.11. There are two dipoles of total length λ/2 located in parallel with the z axis at distance d symmetrically to the origin, as shown in Fig. 10.11. These dipoles are fed by different currents I1 and I2. The field radiated by a particular dipole was calculated in Example 10.4 (10.32). The field radiated by the couple of dipoles is the superposition of the particular fields (10.32), where kL = π/2
Eϑ = Eϑ1 + Eϑ 2
π cos cos(ϑ ) − jkr − jkr 2 I e 1 + I e 2 = j 60 2 1 r sin (ϑ ) r2 1
P
assuming that the mutual coupling between the dipoles is neglected. Looking for the field in the far field zone, we can express distances r1 and r2, using r as in (10.20) r1 = r +
d sin (ϑ ) cos(α ) , 2
r2 = r −
d sin (ϑ ) cos(α ) . 2
r1
r
r2
z
ϑ 0
λ/2
Consequently we get the field radiated by the couple of dipoles in the form
e − jkr Eϑ = j 60 F1 (ϑ ) r
,
I1
I2
y
d x
α Fig. 10.11
d − jk d sin (ϑ )cos (α ) jk sin (ϑ ) cos (α ) e − jkr 2 F2 (ϑ ,α ) , + I1 e 2 = j 60 F1 (ϑ ) I1 e r (10.41)
where F1 is the radiation pattern of one dipole and F2 is a function known as the array factor. Expression (10.41) can be used to describe the radiated field in the far field zone for any array of equal radiators. We get the radiation pattern of such an array in the simple form of a product
F (ϑ , α ) = F1 (ϑ ,α ) F2 (ϑ , α ) .
(10.42)
Let us return to our problem. (10.41) can be simplified assuming currents of the same amplitude I0 and symmetric phase I 1 = I 0 e jψ ,
I 2 = I 0 e − jψ .
Now the array factor is
135 book - 102
kd F2 (ϑ , α ) = 2 I 0 cos sin (ϑ ) cos(α ) − ψ . 2
(10.43)
This function is plotted for several values of d and ψ in Fig. 10.12 in the plane xy, i.e., for ϑ = π 2 which gives F1 = 1. Fig. 10.12 thus shows the radiation pattern of the couple of dipoles in the plane xy. It is evident that we have really tremendous possibilities in changing the shape of the array radiation pattern even in the case of two radiators. d = λ/8
d = λ/4
d = λ3/8
d = λ/2
d = λ3/4
d=λ
2ψ = 0°
2ψ = 45°
2ψ = 90°
2ψ = 135°
2ψ = 180°
Fig. 10.12 ______________________________ Example 10.5: Calculate the radiation pattern of the array of N dipoles λ/2 in length in the plane xy. The dipoles are parallel with mkdcos(α) the z axis. Their centers are equidistantly located on the x axis with separation d = λ/2, Fig. α 10.13. The dipoles are fed by 0 1 2 equal current I0. The dipoles radiate in the xy plane homogeneously in all
r
rm
d
m
N-1
x
Fig. 10.13
136 book - 102
directions, so F1(π/2,α) = 1. Using Fig. 10.13 and details of calculation from the derivation of (10.41), we get Eϑ =
N −1
∑ A I0
m=0
e − jkrm e − jkr = A I0 rm r
N −1
∑ e jkmd cos(α ) = A I 0
m =0
e − jkr F2 (α ) . r
Substituting q = e jkmd cos (α ) we can read the expression for the array factor normalized to 1 as a geometric sequence of complex terms qm
F2 N
=
N
1 1 q −1 1 = 1 + q 2 + q3 + L q N = N N q −1 N
N −1 q 2
N q2
q
12
−q
−
N 2
− q −1 2
N sin kd cos(α ) 1 2 = N 1 sin kd cos(α ) 2
The array factors computed for N = 1,2,3,4,10 are plotted in Fig. 10.14. These functions are normalized by dividing their values by N. For N = 1 the radiation goes homogeneously in all directions – one dipole. Increasing the number of radiators, we increase the amplitude of the main lobe of F2 at α = 0, and side lobes appear, as was mentioned above. The width of the main beam decreases with the number of dipoles.
F2
Nπ sin cos(α ) 1 4 = N π sin cos(α ) 4
1.2
N=1
1 0.8
N=2
N=4 0.6
N = 10
N=3
0.4 0.2 0 0
30
60
90
120
150
180
α (°)
Fig. 10.14
_______________________________________
10.6 Receiving antennas In the preceding paragraphs we studied the particular sources of an electromagnetic field which are known as transmitting antennas. Here we show how to calculate the power received by an antenna. We define the effective length of an antenna and its equivalent circuit. Let us assume an antenna irradiated by a plane TEM wave, Fig. 10.15. Impedance Zp represents the input impedance of a receiver. The wave described by vectors Ei, Hi and by propagation vector ki is incident to the conducting surface of the antenna. Let us assume that
137 book - 102
this wave is not distorted by this incidence. Consequently it does not fulfill the boundary conditions on the antenna surface. To meet these conditions and to account for the distortion of the field we have to add the field known as scattered, with vectors Er and Hr. Now the total field E = Ei + E r ,
n
Ei
H = Hi + Hr ,
must fulfill the boundary condition, and it models the precise distortion of the field by the antenna. Note that the scattered field must vanish far from the antenna, as the field here must correspond only to the incident wave. We will calculate the power received by the antenna and fed to the input of the receiver as the total power which crosses closed surface S, fully covering the antenna. Let us assume that this surface closely copies the antenna conducting surface. This power is P=
(
[
)
ki Hi
Zp
S Fig. 10.15
)]
(
1 1 E × H * ⋅ dS = ∫∫ (E i + E r ) × H *i + H *r ⋅ dS = ∫∫ 2 S 2 S
[
]
(
)
(
)
1 1 1 = ∫∫ (E i + E r ) × H *i ⋅ dS + ∫∫ E i × H *r ⋅ dS + ∫∫ E r × H *r ⋅ dS 2 S 2 S 2 S
,
(10.44)
In the first integral we can reorder vectors in the mixed product and we can get
[(E
i
]
[
]
+ E r ) × H *i ⋅ dS = (E i + E r ) × H *i ⋅ n dS = [n × (E i + E r )] ⋅ H *i dS = 0
due to the fact that n × (E i + E r ) = 0 , as this vector product determines the tangent component of the total electric field to the conducting surface. The last integral in (10.44) represents the power radiated by the antenna back in the form of a scattered field. This power can be represented as
(
)
1 1 E r × H *r ⋅ dS = Z A I A ∫∫ 2 S 2
2
,
where ZA is the antenna input impedance and IA is the antenna input current. The second integral in (10.44) can be rewritten
(
)
(
)
1 1 1 1 E i × H *r ⋅ dS = ∫∫ E i × H *r ⋅ n dS = − E i ⋅ ∫∫ n × H *r dS = − − E i ⋅ ∫∫ K *A dS = ∫∫ 2 S 2 S 2 2 S S 1 1 = − E i ⋅ i 0 ∫ i *A dl = − E i ⋅ d eff I *A 2 2 d
,
138 book - 102
where i0 is a unit vector in the direction along the antenna, IA is the current passing along the antenna, d is its length and deff represents the antenna effective length. This quantity is defined as the length which, multiplied by the value of the incident electric field, offers the internal voltage received by the antenna. The received power is 1 1 1 2 P = U A I *A = I A Z A − E i ⋅ d eff I *A , 2 2 2
(10.45)
and the voltage induced on the antenna terminals is U A = Z A I A − E i ⋅ d eff .
(10.46)
(10.45) and (10.46) define the receiving antenna equivalent circuit, Fig. 10.16. The input impedance of antenna ZA represents the back radiation which in the receiving antenna is considered as loss. Similarly as the antenna effective length we can define the antenna effective aperture by PR = S av Aeff ,
(10.46)
where PR is the power supplied by the antenna to its load, and Sav is the magnitude of Poynting’s vector of the incident plane electromagnetic wave.
ZA
Zp
UA
IA
E i ⋅ d eff
_____________________________________ Example 10.6: Calculate the effective aperture of the elementary electric dipole d in length, located in air. Assume that the dipole is Fig. 10.16 matched to the load, Fig. 10.16. According to Fig. 10.16, the current supplied by the receiving antenna to the input impedance of the receiver and the corresponding power are IA =
U , ZA + Zp
PR =
1 2 I A Rr , 2
where Rr is radiating resistance of the dipole (10.17), which represents the real part of antenna impedance ZA, and U = Eincd is the voltage received by the antenna. Assuming the impedance matching Z p = Z *A and the current is U . 2 Rr Then the received power is IA =
2
E2 d 2 1 U PR = = inc . 2 4 Rr 8 Rr
139 book - 102
From (10.46), inserting for the received power, for Poynting’s vector S av
2 2 E inc Einc and = = 2 Z 0 240π
for the radiating resistance (10.17), we get the effective aperture Aeff =
2 Einc d 2 240π = 2 8 Rr Einc
d 2 30π
3 = λ2 . 8 d 80π 2 λ 2
Using (10.36) we can calculate the directivity of the elementary electric dipole Dmax =
4π π 2π
∫ ∫ F (ϑ ,α )
2
sin (ϑ )dϑ dα
0 0
=
4π π
2π ∫ sin 3 (ϑ )dϑ
=
3 . 2
0
Combining this result for the directivity with the formula for the effective aperture we get Aeff = D
λ2 . 4π
Although this formula has been derived for the elementary electric dipole, it can be applied for the wide class of antennas. _________________________________________
10.7 Problems 10.1 A short conductor of length l radiates an electromagnetic wave at the frequency f = 100 kHz, Fig. 10.17. At the distance r1 = 1000 m S from this antenna we can measure the electric δ field amplitude Em = 10 V/m. There is a l I r2 r1 rectangular conducting loop with the surface S = 2 0.5 m located at the distance r2 = 1200 m from S the antenna. Calculate the angle δ at which the I loop receives the maximum voltage, and also the r1 value of this voltage. δ = 0° Fig. 10.17 U = 0.0105 V 10.2 Calculate the average value of the power density radiated by an electric dipole at the distance r = 20 km in the directions determined by ϑ = 0°, 45°, 90°, 135°, 180°. The amplitude of the passing current is Im = 50 A, the frequency is f = 500 kHz. ϑ = 0°, 180° S av = 0 W/m2
ϑ = 45°, 135°
S av = 2.04 ⋅ 10 −8 W/m2
ϑ = 90°
S av = 4.08 ⋅ 10 −8 W/m2
10.3 A conductor 50 km in length conducts a current with the amplitude Im = 100 A. The frequency is f = 50 Hz. Calculate the power radiated into the whole space.
140 book - 102
P = 274 W 10.4 Calculate the power radiated to the whole space by a circular loop of radius a = 5 cm. The passing current has the amplitude Im = 5 A, and the frequency is f = 500 kHz. Calculate the radiating resistance of this radiator. P = 0.22 10-10 W Rr = 1.76 10-12 Ω
11. WAVE PROPAGATION IN NON-ISOTROPIC MEDIA We have studied the propagation of electromagnetic waves in isotropic materials in previous chapters. The relations between the field vectors in these materials are described by (1.9), (1.10) and (1.11), where conductivity σ, permeability µ, and permittivity ε are scalar quantities. In non-isotropic materials these formulas have the form (1.26) D=ε E ,
(11.1)
B=µH .
(11.2)
Vectors D, E, and B, H are not parallel (1.27). Tensors of permittivity (1.28) and permeability can be expressed ε xx ε = ε yx ε zx µ xx µ = µ yx µ zx
ε xy ε xz ε yy ε yz , ε zy ε zz µ xy µ yy µ zy
(11.3)
µ xz µ yz . µ zz
(11.4)
Taking into account the non-isotropy of the materials we are able to clarify a number of effects occurring in nature. The number of elements applied in microwave technology is based on the effect of propagation in a non-isotropic material. Such elements are called nonreciprocal, as a wave propagates through them differently in different directions. Nonisotropic behaviour of materials can be either natural or induced. Some mono-crystals in which, e.g., birefringence (double refraction) known from optics, can be observed are materials with natural non-isotropic behaviour. Plasma (ionized gas) and ferrite material are examples with induced non-isotropic behaviour. The non-isotropy is induced in these materials by applying an external magnetic field. Plasma occurs in the upper layers of then atmosphere, and the non-isotropy is induced by the earth’s magnetic field. We will pay attention to ferrite materials, e.g., they are widely used in microwave technology in production of non-reciprocal devices, as isolators, gyrators, hybrids, phase shifters, etc., or ferrite resonators. Ferrites are mixtures of oxides of ferromagnetic materials (Fe, Ni, Cd, Li, Mg) sintered at high temperatures. Ferrites must have very low conductivity (10-4 – 10-6 S/m) to allow an electromagnetic field to propagate through them and to interact with them.
141 book - 11
11.1 Tensor of permeability of a magnetized ferrite Any medium consists of a system of atoms. Each atom consists of a positively charged core and a number of electrons bearing a negative charge. These electrons move along certain orbital paths, and at the same time spin around their axes. This movement and rotation of the charged electrons represents the flow of an electric current. The flowing current excites a magnetic field which is perpendicular to the plane of the current loop. Consequently we can define the orbital magnetic moment and the spin magnetic moment of the electron – mo and msp. The electron has a certain mass. Its movement causes the orbital mechanical moment Lo and the spin mechanical moment Lsp. These moments are coupled by the relations m o = −µ 0
e Lo , 2m
(11.5)
m sp = − µ 0
e L sp = −γ L sp , m
(11.6)
where e and m are the charge and the mass of the electron, and γ = µ 0 e m is known as the gyromagnetic constant. The total moment of the whole Ho atom is the sum of these particular moments. The z moments of the core are negligible in comparison with the moment of the electron. It has been experimentally proved that the orbital moments are of much lower value than the spin moments. Therefore we will treat msp only the spin moments. Let us now study the behaviour of an electron with magnetic spin moment msp and with mechanical spin moment Lsp located in the external magnetic field y electron parallel to the z axis Ho = Hoz0 which is not parallel to msp. Under the external field each single magnetic dipole in the medium rotates. This movement is known Lsp as precession. The end points of vectors msp and Lsp move along a circle, Fig. 11.1. Torque T affecting the x dipole is e T = m sp × H o = − µ 0 L sp × H o . m
Fig. 11.1
It is known from mechanics that the change of Lsp due to T is dL sp dt
= T = m sp × H o .
Inserting for Lsp from (11.6) we get dm sp dt
= −µ 0
e m sp × H o = −γ m sp × H o . m
(11.7)
This vector relation can be rewritten into three scalar equations
142 book - 11
dm spx dt dm spy dt dm spz dt
+ γ H o mspy = 0 ,
(11.8)
− γ H o m spx = 0 ,
(11.9)
=0 .
(11.10)
The solution of (11.8) and (11.9) describes the precession movement of the end point of msp m spx = m∞ cos(ω o t ) ,
(11.11)
m spy = m∞ sin (ω o t ) ,
(11.12)
where ω o = γ H o is known as the frequency of free precession – the Larmor precession frequency. This frequency is proportional to the magnetizing field. In the case of a ferrite without losses the precession is not damped, and it would last to infinity. In a real lossy medium, the end point of msp moves along a spiral and it finally approaches the direction of the external magnetic field. Instead of the magnetic dipole we can follow the vector of magnetization as for a unit volume including N particles with the magnetic moments organized by the external magnetic field into the same direction, and the magnetization is M = N msp. Equation (11.7) now reads dM = −γ M × H o . dt
(11.13)
The solution of this equation is M x = M ∞ cos(ω o t ) ,
(11.14)
M y = M ∞ sin (ω o t ) .
(11.15)
Let us assume that the ferrite material is exposed to the superposition of a DC magnetic field Ho and a high frequency field Hm, the amplitude of which is Hm << Ho. The directions of these fields are different
H c = H o + H m e j ωt = H o z 0 + H m e j ω t .
(11.16)
The total magnetic field thus has a time varying value and a time varying direction. Consequently the precession movement of the magnetization does not have a constant direction but will follow the varying magnetic field, and its frequency will be equal to the frequency of the varying magnetic field. This is know as forced precession. Similarly as the magnetic field we can express the magnetization M c = M o + M m e j ωt = M o z 0 + M m e j ω t ,
(11.17)
143 book - 11
where Mm << Mo. Now we insert the total fields (11.16) and (11.17) into (11.13) and we get jω M m e j ω t = −γ M c × H c =
( ≈ −γ (z
)
= −γ M o × H m e j ω t + M m × H o e j ω t + M m × H m e 2 j ω t + M o × H m e j ω t + M o × H o ≈ 0
× H m M oe jω t + M m × z 0 H oe jω t
)
jω M m = −γ M o z 0 × H m + γ H o z 0 × M m . This vector equation can be expressed using vector components in the following three equations jω M mx = ω M H my − ω o M my , jω M my = −ω M H mx + ω o M mx ,
(11.18)
jω M mz = 0 , where ω M = γ M o is the magnetizing constant. Solving the set of equations (11.18) we get M mx = −
ω oω M ω ωM − H j H my , mx ω 2 − ω o2 ω 2 − ω o2
M mx = j
ω ω ω ωM H mx − 2o M 2 H my , 2 2 ω − ωo ω − ωo
(11.19)
M mz = 0 . The magnetic induction of the high frequency field is B m = µ 0 (H m + M m ) . Particular components of Bm using (11.19) are read ωω ωω Bmx = µ 0 1 − 2o M 2 H mx − jµ 0 2 M 2 H my , ω − ωo ω − ωo Bmx = jµ 0
ω ω ω ωM H mx + µ 0 1 − 2o M 2 H my , 2 2 ω − ωo ω − ωo
(11.20)
Bmz = µ 0 H mz . Consequently we have the tensor of permeability
144 book - 11
µ µ = jµ a 0
− jµ a
0 0 , µ 0
µ 0
(11.21)
where
µ = µ 0 1 −
µa = µ0
ω o ω M , ω 2 − ω o2
(11.22)
ω ωM . ω 2 − ω o2
(11.23)
It is important to remember that the tensor of permeability (11.21) describes the behaviour of the ferrite material under magnetization by the field (11.16). The DC field has the direction parallel to the z axis and the tensor of permeability (11.21) couples together the induction and the intensity of the high frequency magnetic field
Bm = µ Hm .
(11.24)
We noted that non-isotropic behaviour is also displayed by plasma magnetized by a DC magnetic field directed parallel to the z axis. In this case we can express the tensor of permittivity as ε ε = jε a 0
− jε a
ε 0
0 0 , ε 0
(11.25)
and the vectors of the high frequency field are coupled by
Dm = ε Em .
(11.26)
_______________________________________ Example 11.1: Calculate magnetic field Ho to magnetize a ferrite to magnetic resonance at the frequency fr = 15 GHz. The mass of an electron is m = 9.109.10-31 kg, the charge of an electron is e = 1.602.10-19 C. The gyromagnetic constant is
γ = µ0
e = 2.21 ⋅ 10 5 m/C . m
The frequency of the free precession is
ωo = γ Ho
=>
Ho =
ωo = 4.26 ⋅ 10 5 A/m γ
_______________________________________
145 book - 11
Example 11.2: Calculate the tensor of permeability of a ferrite material with ωο = 1.105 1010 s-1 and ωM = 1.547 1010 s-1 (see Problem 11.1) at the frequency f = 300 MHz. Applying formulas (11.22) and (11.23) we get
µ = µ 0 1 −
µa = µ0
ω o ω M = 2.44 µ 0 , ω 2 − ω o2
ωωM = 0.246µ 0 , ω 2 − ω o2
and the tensor of permeability (11.21) is − jµ a 0 µ 2.44 − j 0.246 0 0 = µ 0 j 0.246 2.44 0 µ = jµ a µ 0 0 0 0 1 µ 0 _________________________________________
11.2 Longitudinal propagation of a plane electromagnetic wave in a magnetized ferrite
x Let us assume an unbound space filled homogeneously by a ferrite plane of a constant phase material considered to be lossless. This Ho material is magnetized in the z direction by a static magnetic field Ho = Hoz0. A uniform plane electromagnetic wave propagates in this space in the direction of the z axis, i.e., in the direction parallel y k to the magnetizing field, Fig. 11.2. Both the electric field and the magnetic field of Fig. 11.2 this wave do not depend on the x and y ∂ = 0 and coordinates. Assuming ∂x ∂ = 0 we can rewrite Maxwell’s first and second equations (1.12) and (1.13) into ∂y −
∂H y ∂z
= jωε E x ,
z
(11.27)
∂H x = jωε E y , ∂z
(11.28)
0 = jωε E z ,
(11.29)
−
∂E y ∂z
(
)
= − j ω µH x − jµ a H y ,
(11.30)
146 book - 11
∂E x = − j ω j µ a H x + µH y , ∂z
(11.31)
0 = − jωµ 0 H z ,
(11.32)
(
)
Equations (11.29) and (11.32) tell us that the propagating wave is a transversal electromagnetic wave, as the longitudinal field components are zero. We are looking for the solution to above equations in the form of a plane TEM wave H x = H mx e − jk z z ,
(11.33)
H y = H my e − jk z z ,
(11.34)
E x = E mx e − jk z z = Z xy H my e − jk z z ,
(11.35)
E y = E my e − jk z z = − Z yx H mx e − jk z z ,
(11.36)
where the wave impedances are defined Z xy =
E mx , H my
Z yx = −
E my H mx
(11.37)
,
(11.37)
and can be determined together with propagation constant kz by solving equations (11.27) – (11.32). We insert the form of field (11.33), (11.34), (11.35) and (11.36) into (11.27) – (11.32) jk z H my e − jk z z = − jωε E mx e − jk z z = − jωε Z xy H my e − jk z z , jk z H my == − jωε Z xy H my ,
(11.38)
− jk z H mx e − jk z z = − jωε E my e − jk z z = jωε Z yx H mx e − jk z z , − jk z H mx = jωε Z yx H mx .
(11.39)
Similarly from (11.30) and (11.31) we get
(
)
(11.40)
(
)
(11.41)
k z H mx Z yx = ω µH mx − jµ a H my , k z H my Z xy = ω jµ a H mx + µH my .
From (11.38) and (11.39) we have the wave impedances
147 book - 11
Z xy = Z yx =
kz
ωε
,
(11.42)
From (11.40) and (11.41) we exclude the wave impedance by (11.42). We get the set of two equations for Hmx and Hmy
(k
2 z
)
− ω 2 µε H mx + jω 2ε µ a H my = 0 ,
(
)
− jω 2ε µ a H mx + k z2 − ω 2 µε H my = 0 .
(11.43) (11.44)
Propagation constant kz can be determined from the condition of solubility of the set of the equations (11.43) and (11.44), which requires that the determinant of the system matrix is equal to zero
(k
2 z
− ω 2 µε
) + ( jω 2
2
ε µa
)
2
=0 .
This equation has two solutions defining the propagation constant of a wave propagating through the space filled by a ferrite material k zL, R = ω ε (µ ± µ a ) .
(11.45)
Inserting the propagation constants (11.45) into (11.43) and (11.44) we get H myL = j H mxL ,
(11.46)
H myR = − j H mxR .
(11.47)
The propagating wave can thus be decomposed as the superposition of two partial electromagnetic waves. These waves can be expressed
H L = (x 0 + j y 0 ) H mxL e − jk zL z ,
(11.48)
H R = (x 0 − j y 0 ) H mxR e − jk zR z ,
(11.49)
These two waves are circularly polarized waves. The wave denoted by index L is a circularly polarized wave with left-handed rotation. The wave denoted by index R is a circularly polarized wave with right-handed rotation. The wave propagating in the ferrite consequently behaves as the superposition of the two circularly polarized waves with left-handed and righthanded polarization. Their propagation constants (11.45) are different, and so are the phase velocities and the wave impedances vL =
vR =
ω k zL
ω k zR
=
1 , ε (µ + µ a )
(11.50)
=
1 , ε (µ − µ a )
(11.51)
148 book - 11
ZL =
ZR =
k zL
ωε k zR
ωε
=
µ + µa , ε
(11.52)
=
µ − µa . ε
(11.53)
We substitute for the term (µ ± µ a ) , where µ is defined by (11.22) and µa by (11.23), the effective permeability µef
µ ef = µ ± µ a = µ 0 1 ±
ωM . ω ± ω o (11.54)
The dependence of relative effective permeability µ ef µ 0 and of normalized
10
µef/µ0
8 6
µefR
4
µefL
2 0
ω/ωo 1 vc on phase velocity -2 µefR normalized frequency ω ω o is -4 plotted in Fig. 11.3. There is a -6 frequency band in which the permeability of the wave with -8 right-handed polarization is -10 negative. The corresponding phase velocity (11.51) together 3 with the wave impedance (11.53) are imaginary in this v/c frequency range. This means that the wave does not propagate. 2 There is a sharp resonance in the plots in Fig. 11.3 at ω = ωo. It is vR known as ferromagnetic resonance. At this frequency, 1 vL equal to Larmor precession vR frequency ωo, vector Hm performs the precession 0 movement, which is not damped ω/ωo 1 in the case of a lossless ferrite. The wave with right-handed Fig. 11.3 polarization is intensively attenuated at this resonance. The propagation of the wave with left-handed polarization is not influenced by the ferromagnetic resonance, as the direction of the magnetic field vector rotation is opposite to the precession movement. It follows from this that in practical applications of the ferrite material we have to choose a frequency fairly different from the Larmor precession frequency ωo. This removes the strong wave attenuation together with the severe dependence of all wave parameters on frequency. Note that all this concerns the case of a wave propagating in the direction parallel to DC magnetizing field Ho. The phase velocities of the partial waves with right-handed and left-handed polarization (11.50) and (11.51) are different. Let us now assume that a plane TEM wave with
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linear polarization propagates in the ferrite material. This wave is the superposition of the two above defined waves which can be expressed using (11.48) and (11.49)
H = H R + H L = (x 0 + jy 0 )H mL + (x 0 − jy 0 )H mR . Passing the distance d along the z axis, the two particular waves change their phase differently, as they have different phase velocities and the wave can consequently be expressed assuming that the amplitudes are equal HmL = HmR = Hmx H (d ) = H L e − jk zL d + H R e − jk zR d =
[ (
)
(
)]
= H mx x 0 e − jk zR d + e − jk zL d − jy 0 e − jk zR d − e − jk zL d =
,
(11.55)
k − k zL k − k zL = 2 H mx e − jka d x 0 cos zR d − y 0 sin zR d 2 2
where ka =
k zL + k zR . 2
Equation (11.55) shows that the x component and the y component of the magnetic field of the propagating wave stay in phase. This means that the polarization remains linear. The plane of the polarization is twisted by angle ψ tg (ψ ) =
Hy Hx
k − k zL k − k zL d = KF d , = tg zR d => ψ = zR 2 2
(11.56)
where KF is the Faraday constant. This effect is known as the Faraday effect. The polarization plane of the wave with linear polarization propagating in the ferrite material in the direction parallel with the DC magnetizing field is gradually rotated. The measure of this rotation depends on frequency f, magnetizing field Ho and the parameters of the ferrite. ________________________________________ Example 11.3: Calculate the Faraday constant determining the rotation of the polarization plane of the wave propagating in the ferrite material from Example 11.2 in parallel with the magnetizing field at the frequency 300 MHz. Calculate the distance through which the polarization plane is rotated by 90°. Using (11.45), we get the propagation constants of the wave with left-handed polarization and the wave with right-handed polarization k zL = ω ε (µ + µ a ) = 10.28 m-1, k zR = ω ε (µ − µ a ) = 9.29 m-1.
µ and µa calculated in Example 11.2 were used. Using (11.56) we get
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KF =
k zR − k zL = −0.495 rad/m 2
KF = -28.5 °/m .
The polarization plane is rotated by 90° through the distance 3.16 m. ________________________________________
11.3 Transversal propagation of a plane electromagnetic wave in a magnetized ferrite Let us assume that a plane electromagnetic wave propagates in a ferrite material perpendicular to a DC magnetizing field Ho = Hoz0. Let us say in the direction of the x axis, Fig. 11.4. Both the electric field and the magnetic field of this wave do not depend on the y and z ∂ coordinates. Assuming = 0 and y ∂y ∂ = 0 we can rewrite Maxwell’s first and ∂z second equations (1.12) and (1.13) similarly as in the case of longitudinal propagation into Ex = 0 , −
∂x
= jωε E z ,
(
plane of a constant phase
z
Fig. 11.4
(11.59)
)
(11.60)
∂E z = jω jµ a H x + µ H y , ∂x
(
∂x
Ho
(11.58)
0 = − jω µ H x − jµ a H y ,
∂E y
k
(11.57)
∂H z = jωε E y , ∂x
∂H y
x
)
(11.61)
= − jω µ 0 H z ,
(11.62)
Assuming dependence on the x coordinate in the form e − jk x x which describes a plane electromagnetic wave propagating in the x direction, the equations (11.57) to (11.62) read E mx = 0 ,
(11.63)
k x H mz = ωε E my ,
(11.64)
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k x H my = −ωε E mz ,
(11.65)
µH mx = jµ a H my ,
(11.66)
(
)
− k x E mz = ω jµ a H mx + µH my ,
(11.67)
k x E my = ω µ 0 H mz ,
(11.68)
The set of equations (11.63) to (11.68) can be divided into two sets. The equations (11.64) and (11.68) define the system of Emy and Hmz. The equations (11.65), (11.66) and (11.67) define the system of Emz, Hmx and Hmy. Dividing equations (11.64) and (11.68) we get the propagation constant k x = k xo = ω µ 0 ε .
(11.69)
Using the propagation constant we can define the phase velocity vo =
ω k xo
1
=
.
(11.70)
µ0 . ε
(11.71)
µ 0ε
The wave impedance is Zo =
E my H mz
=
Equations (11.69), (11.70) and (11.71) define the propagation constant, the phase velocity and the wave impedance of a standard plane TEM wave. This wave is known as an ordinary wave. The ordinary wave propagates perpendicularly to the DC magnetizing field and its magnetic field is parallel with Ho. The propagation of the ordinary wave does not depend on Ho . The second set of equations describes the wave propagating perpendicular to Ho. It has the longitudinal component of magnetic field Hmx shifted by π/2 to transversal component Hmy. Electric field Emz is parallel to Ho. This wave is known as an extraordinary wave. From (11.65), (11.66) and (11.67) we get the propagation constant, the phase velocity and the wave impedance of the extraordinary wave k xe = ω ε
ve =
ω k mx
=
µ 2 − µ a2 , µ
(11.72)
1
µ 2 − µ a2 ε µ
E k Z e = − mz = xe = H my ωε
,
(11.73)
µ 2 − µ a2 . εµ
(11.74)
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These quantities depend on Ho. At µ = 0 we have ve = 0 and the extraordinary wave does not propagate. This happens at the frequency
ω t = ω o (ω o + ω M ) .
(11.75)
This effect is transversal ferromagnetic resonance. The extraordinary wave can be described by the field components H y = H my e − jk xe x , H x = H mx e − jk xe x = j
(11.76)
µa H my e − jk xe x , µ
(11.77)
E z = H mz e − jk xe x = − Z e H my e − jk xe x . The plane TEM wave with linear polarization propagating perpendicularly to the DC magnetizing field can be decomposed into ordinary and extraordinary waves, Fig. 11.5. These waves have different phase velocities and therefore they have different phases. The ratio of the electric fields of these waves is E yo E ze
=
E my E mz
e j (k xe − k xo )x =
E my E mz
(11.78) x
kxe extraordinary wave
Hxe Eze Hye kxo
e jψ . ordinary wave
Hzo This shows that the phase between the two Eyo components of the electric field varies with the x coordinate, and the TEM wave which is the superposition of the ordinary wave and the z Ho extraordinary wave changes its polarization as the wave propagates, Fig. 11.6. The wave has linear y polarization at the points at which ψ = (k xe − k xo )x = nπ . The character of the wave Fig. 11.5 polarization changes regularly with x from linear polarization, to elliptical polarization, circular polarization, again to elliptical polarization, etc., Fig. 11.6. ________________________________________ Example 11.4: A plate TEM wave is incident from the air at the angle ϑi = 45° to the surface of the ferrite material from Example 10.2, the magnetizing field is parallel to the ferrite surface, Fig. 11.7, and the ferrite permittivity is εr = 10. The frequency is f = 300 MHz. The wave has arbitrary orientation of vector E to the plane of incidence. Calculate the difference between refraction angles of an ordinary wave and of an extraordinary wave.
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y
0<ψ<π/2
ψ=π/2
π/2<ψ<π
ψ=π
ψ=2π
π<ψ<3π/2 ψ=3π/2 3π/2<ψ<2π
ψ=0 45°
45°
45°
E x z
Ho Fig. 11.6 The refracted wave propagates in the ferrite perpendicular to the magnetizing field. The incident wave must be expressed as the superposition of the wave with horizontal polarization and of the wave with vertical polarization. The wave with horizontal polarization has vector E perpendicular to the plane of incidence, i.e., parallel with Ho. It represents an extraordinary wave in the ferrite. The wave with vertical polarization has vector E ϑi parallel to the plane of incidence, i.e., perpendicular to Ho. It represents an ordinary wave in the ferrite. The propagation constants of these waves (11.69) (10.72) are ferrite H o k = ω µ ε = 19.83 m-1 , o
ϑo
0
µ 2 − µ a2 ke = ω ε = 9.7 m-1 . µ
ϑe Fig. 10.7
From Snell’s law (3.49) we get refraction angles ϑto = 12.9° and ϑte = 27°. The difference between these angles is 14.1°. _________________________________________
11.4 Applications of non-reciprocal devices There are devices working with a longitudinally magnetized ferrite which make use of the Faraday effect, and devices with a transversally magnetized ferrite. The latter have simpler construction and are more universal. Among nonreciprocal devices, we can mention a gyrator marked in circuits as shown in Fig. 11.8a. The gyrator 1 shifts the phase of a transmitted wave in one direction by 180° and there π is no phase shift in the 2 3 return direction. The insulator insulator, Fig. 11.8b, gyrator circulator transmits a wave only in one direction and does not a b c transmit in the return Fig. 11.8
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direction. The circulator, Fig. 11.8c, transmits a wave only in the direction of an arrow from gate 1 to gate 2, from gate 2 to gate 3, and from gate 3 to gate 1. Controlling circuits are steered by changing the electric current in the winding of an electromagnet magnetizing a ferrite. Ferrite resonators were mentioned in Chapter 9. Fig. 11.9 shows a sketch of an insulator with a longitudinally magnetized ferrite working with the use of the Faraday effect. The wave in the form of the TE10 mode in a rectangular waveguide propagates from the left. It is transformed to the TE11 mode of the waveguide with a circular cross-section. Its electric field, Fig. 11.9, is perpendicular to the resistive plate and is not attenuated by this plate. This mode travels in the ferrite and its plane of polarization is rotated by 45°, as its electric field is again perpendicular to the second resistive plate and is transformed to the TE10 mode of an output rectangular waveguide. The rotation of the polarization plane of the wave traveling back is in the opposite direction, so the electric field is parallel to the resistive plate and, moreover, the TE01 mode which cannot propagate is excited in the output waveguide. The wave is thus highly attenuated.
Ho
TE10
TE11
TE11
TE10
TE01
TE11
TE11
TE10
Fig. 11.9
11.5 Problems 11.1 Calculate the frequency of the ferromagnetic resonance and the magnetizing constant of a ferrite magnetized by the field Ho = 5 104 A/m. The magnetization is Mo = 7 104 A/m. ωο = 1.105 1010 s-1 fr = 1.76 GHz ωM = 1.547 1010 s-1 11.2 Recalculate the tensor of permeability of the ferrite material from Example 11.2 at the frequency 3 GHz. 0.266 − j1.25 0 µ = µ 0 j1.25 0.266 0 0 0 1
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11.3 Calculate for the ferrite material from Example 11.2 the propagation constants of the waves with left-handed polarization and with right-handed polarization propagating in parallel with a magnetizing field at the frequency 3 GHz. kL = 77.2 m-1 kR = j62 m-1 The wave with right-handed polarization does not propagate, as kR is a purely imaginary number.
12. APPLICATIONS OF ELECTROMAGNETIC FIELDS Knowledge of electromagnetic field theory is applied in the analysis and design of systems in all branches of high frequency technology. These are namely: microwave technology, antennas, propagation of electromagnetic waves, and optical communications. In particular paragraphs we will briefly introduce these branches. All these particular branches contribute to the design of, e.g., the communication link shown in Fig. 12.1. Microwave technology covers the design of feeders, the output high transmitting antenna
transmitter (generator)
receiving antenna
feeder
feeder
receiver
propagating wave (guiding medium) Fig. 12.1 frequency part of a transmitter, and the input high frequency part of a receiver. Antennas are treated usually separately, as their analysis, optimization and design are very specific. Similarly we treat the propagation of waves in the atmosphere above ground in a special way. In a communication system antennas can be substituted by proper converters to excite a light signal in an optical fiber. This problem belongs to optoelectronic technology. Proper transmission lines, e.g., coaxial cable, were used in older communication links. Special measurement techniques using special measuring devices and systems are applied to verify all steps in the development process of all high frequency circuits and systems. It is beyond the scope of this textbook to treat these techniques. In the following paragraphs we briefly introduce problems connected with microwave technology, antennas, propagation of waves and optoelectronic technology.
12.1 Introduction to microwave technology The spectrum of electromagnetic waves is shown in Fig. 1.1 together with typical applications. Up to the frequency of about 1 GHz most circuits are constructed using lumped parameter circuit components. In the frequency range from 1 up to 100 GHz, lumped circuit elements are usually replaced by transmission line and waveguide components. Thus the term microwave technology refers generally to the engineering and design of information handling
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systems in the frequency range from 1 to 100 GHz, corresponding to wavelengths as long as 30 cm and as short as 3 mm. At shorter wavelengths we have what can be called optical or quasi-optical technology since many of the techniques used here are derived from classical optical techniques. The characteristic feature of microwave technology is the short wavelengths involved, these being of the same order of magnitude as the circuit elements and devices employed. Conventional low frequency circuit analysis based on Kirchhoff’s laws and voltage current concepts cannot be applied. It is necessary instead to carry out the analysis in terms of a description of electric and magnetic fields. There is no distinct frequency boundary at which lumped parameter circuit elements must be replaced by distributed circuit elements. With modern technological processes it is possible to construct printed circuit inductors and on chip capacitors that are so small that they retain their lumped parameter characteristics at frequencies as high as 10 GHz or even higher. Likewise, optical components, such as parabolic reflectors and lenses, are used to focus waves with wavelengths as long as 1 m or more. The first development of microwave technology was started with the development of radars, as the need for high resolution radar capable of detecting small targets is coupled with the need to raise the frequency. This arises predominantly from the need to have antennas of sufficiently small dimensions that will radiate essentially all the transmitted power into a narrow pencil-like beam. In recent years microwave frequencies have also come into widespread use in communication links. This follows from the demand to increase the amount and the speed of transmitted information. Satellite and mobile communication systems are based on microwave technology. At the present time most communication systems use the transmission of digital signals. Waveguides periodically loaded with shunt susceptance elements support slow waves having velocities much less than the velocity of light, and are used in linear accelerators. These produce high energy beams of charged particles for use in atomic and nuclear research. The slow traveling electromagnetic waves interact very efficiently with charged-particle beams having the same velocity, and thereby give up energy to the beam. Another possibility is for the energy in an electron beam to be given up to an electromagnetic wave, with resultant amplification. This latter device is the traveling-wave tube. Sensitive microwave receivers are used in radio astronomy to detect and study the electromagnetic radiation from the sun and a number of stars that emit radiation in this band. Microwave radiometers are also used to map atmospheric temperature profiles, moisture conditions in soils and crops, and for other remote sensing applications. Molecular, atomic, and nuclear systems exhibit various resonance phenomena under the action of periodic forces arising from an applied electromagnetic field. Many of these resonances occur in the microwave range. We have thus a very powerful experimental probe for studying the basic properties of materials. Out of this research on materials have come many useful devices employing ferrites, see Chapter 11. The development of the laser, a generator of essentially monochromatic (singlefrequency) light waves, together with semiconductor technology, has stimulated great interest in the possibilities of developing communication systems at optical wavelengths. There are plenty of the applications of microwaves in industrial processes – microwave ovens, drying of materials, manufacturing wood and paper products, material curing. Microwave radiation has also found some application for medical hyperthermia or localized heating of tumors. At frequencies where the wavelength is several orders of magnitude larger than the greatest dimensions of the circuit or system being examined, conventional circuit elements such as capacitors, inductors, resistors, transistors or diodes are the basic building blocks for the information transmitting, receiving, and processing circuits used. The description or
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analysis of such circuits may be adequately carried out in terms of loop currents and node voltages without consideration of propagation effects. The time delay between cause and effect at different points in these circuits is so small compared with the period of the applied signal as to be negligible. It might be noted here that an electromagnetic wave propagates a distance of one wavelength in a time interval equal to one period of a sinusoidally time varying applied signal. As a consequence, when the distances involved are short compared with a wavelength, the time delay is not significant. As the frequency is raised to a point where the wavelength is no longer large compared with the circuit dimensions, propagation effects can no longer be ignored. A further effect is the great relative increase in the impedance of the connecting leads, terminals, etc., and the effect of distributed capacitance and inductance. In addition, currents circulating in unshielded circuits comparable in size with a wavelength are very effective in radiating electromagnetic waves. The net effect of all this is to make most conventional low frequency circuit elements and circuits hopelessly inadequate at microwave frequencies. As is well known, lumped circuit elements do not behave in the desired manner at high frequencies. For example, a coil of wire may be an excellent inductor at 1 MHz, but at 50 MHz it may be an equally good capacitor because of the predominating effect of interturn capacitance. Even though practical low frequency resistors, inductors and capacitors do not function in the desired manner at microwave frequencies, this does not mean that such energy dissipating and storage elements cannot be constructed at microwave frequencies. On the contrary, there are many equivalent inductive and capacitive devices for use at microwave frequencies. Their geometrical form is quite different, but they can be and are used for much the same purposes, such as impedance matching, resonant circuits, etc. For low power applications, microwave tubes have been completely replaced by transistors, diodes, and negative resistance diodes. However, for high power applications microwave tubes are still necessary. One of the essential requirements in a microwave circuit is the ability to transfer signal power from one point to another without radiation loss. This requires the transport of electromagnetic energy in the form of a propagating wave. A variety of such structures have been developed that can guide electromagnetic waves from one point to another without radiation loss. The simplest guiding structure, from an analysis point of view, is the transmission line. Several of these, such as the open two-conductor line, the coaxial line, the parallel plate line, and the strip line, illustrated in Fig. 5.1b, are in common use at the lower microwave frequencies. At the higher microwave frequencies hollow-pipe waveguides, as illustrated in Fig. 5.1c, may be used. Today, these waveguides are used predominantly only in special systems, where high power signals are transmitted, or where extremely low losses are required. The development of semiconductor high frequency active devices, such as bipolar transistors, field-effect transistors, and diodes, has had a dramatic impact on the microwave engineering field. With the availability of microwave transistors, the focus on waveguides and waveguide components has changed to a focus on planar transmission line structures, such as microstrip lines and coplanar waveguides. These structures, shown in Fig. 5.1b, can be manufactured using printed circuit techniques. They are compatible with microwave semiconductor devices, and can be miniaturized, they are cheap, and are thus suitable for mass production. Microwave circuits are usually designed as hybrid integrated microwave circuits. In hybrid circuit construction the transmission lines and transmission line components, such as matching elements, are manufactured first and then the semiconductor devices are soldered into place. The current trend is toward the use of monolithic microwave integrated circuits (MMIC) in which both the transmission line circuits and the active devices are fabricated on a single chip.
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A unique property of the transmission line is that a satisfactory analysis of its properties may by carried out by treating it as a network with distributed parameters and solving for the voltage and current waves that may propagate along the line, see Chapter 6. In the case of planar transmission lines this can be done at lower microwave frequencies only. At higher frequencies they must be, similarly as hollow-pipe waveguides, treated as electromagnetic boundary value problems, and a solution for the electromagnetic fields must be determined, see Chapter 7. The reason is that it is not possible to define a unique voltage and current that have the same significance as at low frequencies. However using a proper normalization the graphical means of waveguide analysis, such as the Smith chart, see Chapter 6.3, are commonly used. Associated with transmission lines and waveguides are a number of interesting problems related to methods of exciting fields in guides and methods of coupling energy out. Signals from generators and measuring devices are usually carried with the use of coaxial cables. A number of connectors have been produced which are compatible to connecting planar transmission lines by soldering them directly into their layout, Fig. 12.2. These lines can be simple coupled using the proximity effect, see the microstrip line coupler in Fig. 12.3 1
2
3
4 Fig. 12.3
Fig. 12.2 aperture with ports 1 to 4. Three basic coupling methods are used in the case of waveguides, Fig. 12.4: probe coupling, see Fig. 12.4a and Fig. 7.8, loop c a b coupling, see Fig. 12.4b, Fig. 12.4 and aperture coupling between adjacent guides, see Fig. 12.4c. These coupling devices are actually small antennas that radiate into a waveguide. Inductive and capacitive elements take a variety of forms at microwave frequencies. Perhaps the simplest are short circuited sections of a transmission line and a waveguide, see Chapter 6.2.3. These exhibit a range of susceptance values from minus to plus infinity, depending on the length of the line, and hence may act as either inductive or capacitive elements. They may be connected as either series or shunt elements, as illustrated in Fig. 12.5. They are commonly referred to as stubs and are widely used as impedance matching elements, see Example 6.8. In a rectangular waveguide thin conducting windows, or diaphragms, as illustrated in Fig. 12.6, also act as shunt susceptive elements. Their inductive, Fig. 12.6a, or capacitive, Fig. 12.6b, nature depends on whether there is more a - series stub b – shunt stub magnetic energy or electric energy stored in the local fringing fields. Fig. 12.5
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Inductors and capacitors are designed namely for microwave integrated circuits in forms resembling lumped parameter elements. The example is an interdigital capacitor designed for the coplanar waveguide illustrated in Fig. 12.7. An example of an a b inductor is a spiral inductor designed for a microstrip line technology, shown in Fig. 12.8. Fig. 12.6 Resonant circuits are used both at low frequencies and at microwave frequencies to control the frequency of an oscillator and for frequency filtering. At low frequencies this function is performed by an inductor and a capacitor in a series or parallel combination. Resonance occurs when there are equal average amounts of electric and magnetic energy stored. This energy oscillates back and forth between the magnetic field around the inductor and the electric field between the capacitor plates. At microwave frequencies the LC circuit may be replaced by a closed conducting enclosure, or cavity, see Chapter 9.1. The electric and magnetic energy is stored in the field within the cavity. At an infinite number of the resonant frequencies (9.1) there are equal average amounts of electric and magnetic energy stored in the cavity volume. In the vicinity of any one resonant frequency, the input impedance to the cavity has the same properties as for a conventional LC resonant circuit. One metallization slot significant feature worth noting is the very much larger Q values that may be obtained, these Fig. 12.7 being often in excess of 104, as compared with those obtainable from low frequency LC circuits. Dielectric and ferrite resonators are mentioned in Chapter 9.1. Resonators can be formed even using the planar technology of printed transmission lines. An example is a patch resonator, frequently used as a microstrip patch antenna, Fig. 12.9. The Q factor of these resonators is rather low, due to their radiation. Fig. 12.9 shows a rectangular patch resonator, the resonators of various different shapes are even used. When a number of microwave devices are connected by Fig. 12.8 means of sections of transmission lines or waveguides, we obtain a microwave circuit. An analysis of the behaviour of such circuits is carried out either in terms of the equivalent transmission line voltage and current waves or in terms of the conducting amplitudes of the propagating patch waves. The first approach feeding leads to an equivalent microstrip impedance description, and the second emphasizes the dielectric substrate wave nature of the fields and results in a scattering matrix backward metallization formulation. Fig. 12.9
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Particular microwave circuits perform a variety of particular tasks in complex systems or subsystems. We can include among these circuits: oscillators, amplifiers, modulators, demodulators, mixers, filters, switches, multiplexers, demultiplexers, etc. It is beyond the scope of this textbook to treat these circuits in detail. Scattering parameters: A microwave circuit can be generally represented by an N-port, i.e., a circuit with N ports, Fig. 12.10. The internal structure of this bn b1 Un circuit may be rather complex and it can be U1 an a1 very difficult to find the solution of the Maxwell equations describing the field. Let . us try to describe such a circuit. Let us b2 . assume that each port is connected by a U2 a2 . transmission line to the rest of a system. bN . There is a wave, which is incident to this UN aN . circuit port propagating along this line toward . the port, and there is a wave reflected from the port and propagating back. These waves are determined by amplitudes U n+ and U n− . Fig. 12.10 Let us normalize these amplitudes by
U n+ , an = Z Cn
(12.1)
U n− , Z Cn
(12.2)
bn =
where ZCn is the characteristic impedance of the line connected to the n-th port. The voltage and the current at the n-th port are
U n = U n+ + U n− = Z Cn (a n + bn ) , In =
(
)
1 U n+ − U n− = Z Cn
(12.3)
1 (an − bn ) . Z Cn
(12.4)
The power at the n-th port is Pn = Pn+ − Pn− =
[
]
[
1 1 Re U n I n* = Re a n 2 2
2
− bn
2
(
)]
+ b a n* − a n bn* =
1 an 2
2
−
1 bn 2
2
.
So the normalization was done to be able to express simply the power transmitted by the incident wave and the power transmitted by the reflected wave Pn+ =
1 an 2
2
,
(12.5)
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Pn− =
1 bn 2
2
.
(12.6)
It is reasonable to assume that all incident waves contribute to each reflected wave. In the case of a linear circuit, this dependence must be linear. It reads bn = S n1a1 + S n 2 a 2 + S n3 a 3 + ... + S nj a j + ... + S nN a N , n = 1, 2, 3, ….,N .
(12.7)
In the matrix form we have
b1 S11 b S 2 = 21 . . bN S N 1
S12 S 22 . SN2
. S1N a1 . S 2 N a 2 . . . . . S NN a N
(12.8)
Matrix [S ] is known as a scattering matrix. Particular elements of the scattering matrix are defined as S ij =
bi aj
for
ak = 0
until
k≠ j .
(12.9)
These elements represent the transmission from the j-th port to the i-th port. The main diagonal elements Sii represent the reflection coefficient at the i-th port. This assumes that all other ports are matched and no energy comes to them. The elements of a scattering matrix are generally complex numbers and can be displayed on a Smith chart as functions of frequency. Their amplitudes are often expressed in dB. The scattering matrix can be simply transformed to a transition matrix or to a impedance matrix.
12.2 Antennas An antenna is an element that transforms energy from an electric circuit to energy transmitted by a radiated wave into space, or vice versa. The basic theory of wave radiation, antennas, and antenna arrays was presented in Chapter 10 together with definitions of the basic antenna parameters. These are: radiation pattern, input impedance, antenna directivity and gain, effective length and area of the receiving antenna. All these parameters depend on frequency. An antenna is an integral part of communication links, navigation systems, radars, telescopes, and various sensors. Antennas are often designed as metallic bodies. The basic problem in this case is to determine the distribution of the electric current on the surface of the antenna. Once we know the distribution of this current we can relatively simply calculate the field radiated by an antenna using formulas (10.26) to (10.30) derived in Chapter 10. Let us assume an antenna as a perfectly conducting body with surface S, Fig. 12.11, located in an unbounded space filled by homogeneous lossless material with wave impedance Z0 and propagation constant k. This antenna is irradiated by a plane electromagnetic wave described by vector Ei. This field induces an electric current with density K on the surface of an antenna. This current produces electric field Es radiated, or scattered, back by an antenna. Consequently the total field is the sum of these particular fields
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E = Ei + Es .
(12.10)
Let us assume for the sake of simplicity, the problem that irradiating field Ei remains unchanged by the incidence to the surface of an antenna, and so the disturbance of a field is caused purely by scattered field Es. The total field must fulfill the boundary condition on the antenna surface (1.22)
Ei K
k Hi
n R r r’
n×E = 0 , n × E s = −n × E i .
(12.11)
0
The scattered field is determined by (10.28) E s = − jω A −
j
ωµε
S
Fig. 12.11
grad div(A ) ,
(12.12)
where vector potential A is determined by the surface equivalent of (1.58), Fig. 12.12 µ K (r ') e − jkR A(r ) = dS , (12.13) 4π ∫∫ R S where R = r – r’. Inserting for vector potential (12.13) into (12.12) and consequently for the scattered field into (12.11) we get an integral equation for the unknown distribution of electric current K on the antenna surface
µ n × − jω 4π
µ K e − jkR j grad div dS − ∫∫ R 4π ωµε S
K e − jkR = −n × E i . dS ∫∫ R S
We can exchange the order of derivatives and integration in the second term as we integrate according to r’ and the derivatives included in operators gradient and divergence are calculated according to r. So we have j
K e − jkR K e − jkR 1 µ dS = n × E i . grad div n × ∫∫ jω + 4π ωµε R R S
(12.14)
This integro-differential equation cannot be solved analytically. A proper numerical method must be used to solve (12.14). Simplified methods are used in practice to determine the distribution of current on the surface of antennas. These methods depend on the kind of antenna. The solution of the antenna problem is rather difficult. Therefore we often divide this problem into two parts. Solving an inner antenna problem we in fact solve equation (12.14). The aim is to determine the distribution of currents or charges on the antenna body or on some equivalent surfaces. This problem must often be simplified. In the second step we solve the outer antenna problem, which calculates of the field excited by the source currents. If
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we would like to determine the field in the radiating (far field) zone, we can use methods explained in Chapter 10. The behaviour of transmitting and receiving antennas is similar. A radiation pattern for transmission is equal to a radiation pattern for receiving, and the same is valid for an antenna input impedance. Therefore we analyze antennas mostly as transmitting. In the case of a receiving antenna we determine in addition the effective length or aperture and possibly the noise temperature. It is very difficult to list various kinds of antennas, as there is a great variety of them and it is difficult to categorize them in some cases. We did not mention aperture antennas in Chapter 10. The wave passes through an aperture, e.g., through an open-ended waveguide. Examples of this are horn, reflector and slot antennas. A variety of wire antennas are used in particular applications. Ultra wide band antennas must by applied in modern radars and communication systems. Special types of antennas are lens and traveling wave antennas. Planar antennas are widely used, as they are compatible with planar microwave circuits and microwave integrated circuits. There is a huge variety of these antennas. All kinds of antennas can be combined to create an antenna array, see Chapter 10.5. In the following text we briefly introduce particular types of antennas. 12.2.1 Wire antennas Dipole antennas as examples of wire antennas are widely used. They were analyzed in Example 10.4. Greater directivity and L lower side lobe levels than in the case of a straight dipole can be gained using the vee dipole, Fig. 12.12. It is possible to find the angle γ to get maximum directivity. γ An extremely practical wire antenna is the folded dipole. It consists of two parallel dipoles connected at the ends forming a narrow wire loop, as shown in Fig. 12.13, with dimension d much smaller than L. For L = λ/2 the input impedance of this dipole is four times higher than the impedance of a straight dipole. This Fig. 12.12 antenna is suitable for TV and FM radio receiving, as it is well d matched to 300 Ω twin-lead transmission line. We saw in Chapter 10.5 that array antennas can be used to increase directivity. The arrays we have examined had all elements active, requiring a direct connection to each element by a feed network. The feed L networks for arrays are considerably simplified reflector if only a few elements are fed directly. Such an driven element array is referred to as a parasitic array. The directors elements that are not directly driven, called parasites, receive their excitation by near-field Fig. 12.13 coupling from the driven elements. A parasitic linear array of parallel dipoles is known as a Yagi-Uda antenna. These antennas are simple and have a relatively high gain. The basic unit consists of three elements one is driven and two are parasitic, a director and a reflector. More than one director is Fig. 12.14 usually used. Fig. 12.14 shows a Yagi-Uda antenna with five directors. The current induced in closely spaced parasitic elements has the opposite phase to the current of the driven element. From the radiation patterns shown at the bottom of Fig. 10.12 it is clear that such an array radiates endfire. Lengthening one parasitic element known as the reflector, the radiation dual endfire beam is changed to a more desirable single endfire beam. This is more pronounced in a parasitic element shorter than the driven element, known as a director. An example of the radiation pattern of a Yagi-Uda
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antenna with 5 directors is shown in Fig. 12.15. The radiation pattern was measured in the plane of antenna symmetry perpendicular to the antenna elements. 12.2.2 Aperture antennas Aperture antennas can be solved using the equivalence principle. Let electromagnetic sources be contained in volume V bounded by surface S with outward normal n, Fig. 12.16a. The fields E and H exterior to S can be found by removing the sources in V and placing the following surface current densities on S, see Fig. 12.16b,
Fig. 12.15
K = n × Ht ,
(12.15)
M = Et × n ,
(12.16)
where K is current surface E, H E, H Et M n n density on S, M is equivalent magnetic current S Ht K S no sources density on S, Et and Ht are sources zero field the electric and magnetic V V fields produced by the original sources on surface a b S. Thus, with a knowledge of the tangential fields over Fig. 12.16 a surface due to the original sources, we can find the fields everywhere external to the surface through the use of equivalent surface current densities K and M. These densities define vector potentials, see (1.58), A=
µ 4π
K e − jkr ∫∫ r dS , S
(12.17)
F=
µ 4π
M e − jkr ∫∫ r dS , S
(12.18)
where F is the electric vector potential, which is a quantity analogical to magnetic vector potential A. The magnetic and electric vector potentials determine the field vectors, see (10.28), E = − jω A −
H = − jω F −
j
ωµε j
ωµε
grad div(A ) − grad div (F ) +
1
ε
rot (F ) ,
(12.19)
rot (A ) .
(12.20)
1
µ
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Equations (12.19) and (12.20) follow from the symmetry of Maxwell’s equation gained by introducing equivalent current densities (12.15) and (12.16). The calculation of the field in the radiating zone follows the technique described in Chapter 10.3. The gain of the aperture antenna can be estimated by, see Example 10.6 where the homogeneous field distribution over the antenna aperture was assumed, not exactly fulfilled in most aperture antennas, G=
4π
λ2
Aeff ,
(12.21)
where Aeff = ηAeffm is the effective aperture area of an antenna, Aeffm being the maximum effective aperture (10.46) area, and η is the efficiency of radiation. A typical example of aperture antennas B is a pyramidal horn antenna, shown in Fig. E 12.17. This antenna is fed from the rectangular waveguide. The pyramidal horn antenna is flared in both the E- and H-planes. A This configuration will lead to narrow beamwidths in both principal planes, thus Fig. 12.17 forming a pencil beam. The gain can be simply evaluated by (12.21), where Aeff = ηΑΒ. A horn antenna can also be designed on the basis of a waveguide of circular cross-section. In long-distance radio communication and high-resolution radar applications, antennas with high gain are required. Reflector antennas are aperture reflector perhaps the most widely used high gain antennas, plain which routinely achieve gains far in excess of 30 dB in the microwave region. The simplest reflector radiated ray antenna consists of two components: a large (relative to the wavelength) reflecting surface and a much smaller feeding antenna. The most prominent 2a focal point – example is the parabolic reflector antenna, see its feeding point cross section in Fig. 12.18. The reflector has a paraboloid of revolution shape. A feeder is placed at the focal point. For large reflectors (a>>λ) geometrical optics principles can be applied and radiation from the antenna is analyzed by the ray Fig. 12.18 tracing method, see Chapter 4. The gain can be simply evaluated by (12.21), where Aeffm = radiating slot 4πa2, Fig. 12.18. The radiation pattern of these antennas is very narrow – units of degrees. microstrip There are various feeding the versions of slot antennas. slot from a Radiating slots can be metallization back side made in the walls of a waveguide or can be dielectric substrate designed in a planar Fig. 12.19 version, Fig. 12.19.
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12.2.3 Broadband antennas In many applications an antenna must operate effectively over a wide range of frequencies. An antenna with wide bandwidth is referred to as a broadband, or an ultra wide band (UWB), antenna. The term broadband is a relative measure of bandwidth and varies with circumstances. Let fU and fL be the upper and lower frequencies of operation for which satisfactory performance is obtained. The center frequency is fC. Then the bandwidth as a percentage of the center frequency is
BW =
fU − f L × 100 . fC
(12.22)
Resonant antennas have small bandwidths, while, antennas that have traveling waves on them operate over wider frequency ranges. It is frequently desirable to have the radiation pattern and input impedance of an antenna remaining constant over a very wide range of frequencies, say 10:1 or higher. An antenna of this type is referred to as a frequency independent antenna. This is, for example, the case of an infinite biconical antenna, see Fig. 12.20. This antenna must however be Fig. 12.20 truncated, forming a finite biconical antenna and most of its broadband behaviour disappears. The condition of frequency independence is based on a shape determined fully by angles. The concept of angle emphasis has been exploited in recent years and has led to a family of wide bandwidth antennas. These are spiral antennas and log-periodic antennas. Planar spiral antennas can be simply constructed using printed circuit techniques. Their radiation pattern is bi-directional. To obtain radiation in one direction, the spiral antenna is placed on a conical surface. The shape of a spiral antenna is determined by the equation of the spiral in the polar coordinate system r = r0 e aϕ ,
(12.23)
Fig. 12.21
where r0 is the radius for ϕ = 0 and a is a constant giving the flare rate of the spiral. This curve can be used to make the angular antenna shown in Fig. 12.21, which is referred as a planar spiral antenna. The four edges of the metal each have an equation for their curves of the form (12.23), r1 = r0 e aϕ , r2 = r0 e a (ϕ −π / 2 ) , r3 = r0 e a (ϕ −π ) , r4 = r0 e a (ϕ −π / 2−π ) . Maximal radius R determines a lower frequency R = λ L / 4 and the distance of input ports determines the upper frequency. A log-periodic antenna is an antenna having a structural geometry such that its impedance and radiation characteristics repeat periodically as the logarithm of the frequency. There are several configurations of this antenna. Fig 12.22a shows a planar log-periodic toothed trapezoid antenna, Fig. 12.22b shows a log-periodic trapezoid wire antenna, Fig. 12.22c shows log-periodic dipole array geometry. The wide band operation of these
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antennas is obtained due their linearly varied dimensions. Each part of the antenna is active at different frequency band.
a
b Fig. 12.22
c
dielectric substrate 12.2.4 Planar antennas Planar antennas are designed in a planar radiating technology. Due to this patch they are compatible with planar transmission lines and microwave circuits based on planar technology. They can be directly integrated into these circuits and can air spacer microstrip even be a part of feeding the coupling slot monolithic microwave slot from metallization integrated circuits. back side A planar dielectric substrate microstrip patch antenna is shown in Fig. Fig. 12.23 12.9. The drawback of this antenna is its rather narrow band. A number of modifications of a microstrip patch antenna have been designed to widen its frequency band, to get the antenna radiating a circularly polarized wave, to get dual frequency operation. The simplest way to widen the microstrip patch antenna frequency band is to use a thicker substrate, or possibly to divide this substrate into two layers, a thin dielectric substrate and a thick air (foam) layer. This approach is very often combined with an aperture coupling. This antenna is sketched in Fig. 12.23. The thick substrate reduces the antenna quality factor and in this way widens the frequency band. At the same time, the slot through which energy is coupled from the feeding microstrip line to the patch, together with the open ended microstrip line stub terminating the line, is tuned to a frequency slightly different from the patch resonant frequency. This offset can considerably increase the antenna frequency band. There are a number of other types of planar antennas. Some from them were introduced above. An electric dipole alone or as a part of an Yagi-Uda antenna can be designed in planar geometry. Fig. 12.19 shows a slot antenna. Broad band antennas are often designed in planar technology, see the spiral antenna in Fig. 12.21, or the planar log-periodic toothed trapezoid antenna shown in Fig. 12.22a.
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12.3 Propagation of electromagnetic waves in the atmosphere The problem of propagation of electromagnetic waves deals with the part of a communication link shown in Fig 12.1 between the transmitting and receiving antennas. It includes propagation of electromagnetic waves in the space above the Earth. Wave propagation is influenced mainly by two factors. The first factor includes the parameters of the medium itself in which the wave propagates. The second factor covers the geometry of the whole scene. Here we have to assume the actual profile and cover of the ground, including vegetation. The atmosphere changes its parameters – complex permeability – depending on altitude. This causes, e.g., the bending of wave rays in lower parts of the atmosphere (the troposphere), and wave reflection in the ionosphere. The atmosphere is a time varying medium. The time changes are slow, and are caused by alternating seasons of the year, and by changes between day and night. Rapid variation can be caused by changing weather, above all by hydro-meteors present in the atmosphere – rain, snow, fog, water vapour. They cause attenuation and scattering of waves. The propagation of an electromagnetic wave is thus not the matter of a single ray simulating more or less straight propagation of a wave. The wave received by the antenna is in most cases the sum of the particular waves caused by particular reflections. Some of these effects are frequency very selective, some are not and are active in wide frequency bands. Some of these effects vary rapidly with time, some are stable. We can distinguish several basic types of propagation: surface wave, direct wave, reflected and scattered wave, space wave, tropospheric wave, ionospheric wave, Fig. 12.24. The ground surface is a boundary between two electrically different materials. Such a
ionosphere 50 - 400 km ionospheric wave
troposphere up to 10 km tropospheric wave direct wave surface wave reflected wave Earth Fig. 12.24 boundary is able to guide a surface wave, see Chapter 8. The surface wave follows the bent Earth surface. Its polarization is vertical. This kind of connection can be made via long distances namely in the case of long waves, Fig. 1.1. A direct wave represents a connection over short distances between places with direct visibility and in the case of satellite communication. This can happen only at very high frequencies. In practical cases a wave does not propagate in a straight direction. Due to space variations of refractive index of the atmosphere, the trajectories of rays representing the wave
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propagation are bent, see Chapter 4. In the case of a wave transmitted from elevated positions we have to consider not only the direct wave, but the contribution of waves reflected from the ground or scattered from various obstacles. These are the reflected wave and the scattered wave. These waves are superimposed on the direct wave. This interference can cause unpredictable losses of the received signal. The simultaneous presence of the direct and reflected wave, the complex sum of these waves, creates the total field. This, is known as propagation by the space wave. It is typified namely by the direct visibility between the transmitter and the receiver, ensured by elevated antennas at frequencies above 30 GHz. In this case we have so called multi-path propagation, as there could be a number of reflected and scattered waves approaching the receiving antenna at the same time. The propagation of a tropospheric wave is used in communication over very long distances (thousands of kilometers) for short waves. Propagation of this type uses scattering of a wave on non-homogeneities in the troposphere that have different refractive indices. Only a very small part of the transmitted power reaches the receiver, but a connection can be made at distances far beyond the optical horizon. An electromagnetic wave can propagate to long distances due to reflections from the ionosphere. The layer of ionized air causes intensive continuous bending of the trajectories of the rays as finally the rays are returned toward the Earth’s surface. Waves with a wavelength longer than 10 m can be propagated by an ionospheric wave. At present communication over long distances runs via satellites in the microwave frequency ranges – satellite communication links. Let us now discuss the power balance of a communication channel between two antennas located at distance r in free an unbounded space filled by a lossless material assuming that r is in the far field zone of the two antennas. In this case, we can assume propagation of a spherical wave with its amplitude modified by angular dependence defined by the transmitting antenna radiation pattern – antenna directivity DT. Assuming dependence of the radiated field on distance in the form (2.49) we get the average value of Poynting’s vector at the position of the receiving antenna S av =
Pr DT 4πr
2
=
PT GT 4πr 2
,
(12.24)
where Pr is the power radiated by the transmitting antenna, and PT is the total power supplied to the transmitting antenna. The power received by the receiving antenna is PR = S av Aeffmη R ,
(12.25)
where Aeffm is the receiving antenna maximum effective aperture (10.46) and ηR is the efficiency of the receiving antenna. The effective aperture can be simply calculated for an elementary electric dipole. It has the form Aeffm = D
λ2 , 4π
(12.26)
valid for any antenna. Using (10.38), (12.26) and (12.24) we get from (12.25)
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PT GT G R λ2 PR = S av Aeffmη R = . 4πr 2 4π Consequently we get the power received by the receiving antenna 2
λ PR = GT G R PT . 4πr
(12.27)
This term determines the received power assuming that all kinds of losses are neglected. In spite of this fact the received power is lower than the transmitted power. The reason is that the transmitted power is spread into the whole space, see Chapter 2.4, and consequently the field amplitude decreases as 1/r and Poynting’s vector as 1/r2. The term (λ 4πr )2 is known as the free space loss. The transmission formula (12.27) assumes that antennas are targeted mutually in the directions of the maximum of their radiation. Using (10.38) we can introduce antenna directivities with their dependences on angles ϑ and ϕ 2
λ PR = PT ηT η R DT (ϑT , ϕ T ) DR (ϑ R , ϕ R ) , 4πr
(12.28)
where index R denotes the receiving antenna and index T denotes the transmitting antenna. Equation (12.28) is known as the Friis transmission formula. It is applicable in the case of the general directions of the receiving antenna antennas defined by angles transmitting antenna ϑT , ϕ T ,ϑ R , ϕ R , Fig. 12.25. In practical cases of propagation ϑR, ϕR all kinds of losses including the polarization mismatch ϑT, ϕΤ must be included in (12.28). Moreover, the multi-path propagation must be taken into r account and particular waves must be added with their Fig. 12.25 proper phase and target attenuation. transmitter Radar can be assumed as a special case of a communication channel. receiver An electromagnetic wave is transmitted by a transmitter, r reflected back by a target and received by a receiver, Fig. 12.26. The time delay Fig. 12.26 between transmission and reception of the pulses is proportional to the distance of the target. We will calculate the power received by a radar receiver. To simplify this problem we assume that the transmitting and receiving antennas are pointed such that the radiating pattern maxima are directed toward the target. The power density incident on the target is then
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S inc =
PT 4πr 2
GT =
PT AeffT
λ2 r 2
,
(12.29)
where the gain has been expressed using the transmitting antenna effective aperture (12.21). The power intercepted by the target is proportional to the incident power density (12.29), so Pinc = σ S inc ,
(12.30)
where the proportionality constant σ is the radar cross section and is the equivalent area of the target as if the target reradiated the incident power isotropically. Although power Pinc is not really scattered isotropically, the receiver samples the scattered power in only one direction and we are only concerned about that direction and assume the target scatters isotropically. Because Pinc appears to be scattered isotropically, the power density arriving at the receiver is SR =
Pinc
4πr 2
.
(12.31)
The power available at the receiver is then PR = AeffR S R .
(12.32)
Combining the above four equations gives PR = AeffR
AeffT AeffRσ σS inc = , P T 4πr 2 4πr 4 λ2
(12.33)
which is referred to as the radar equation. Usually the transmitting and receiving antennas are identical, that is, AeffT = AeffR and GT = GR = G. Using (12.21) we can rewrite (12.33) in a convenient form as PR = PT
λ 2 G 2σ . (4π )3 r 4
(12.34)
The combination of (12.30) and (12.31) actually forms the definition of the radar cross section 4πr 2 S R σ = , S inc
(12.35)
which is the ratio of 4π times the radiation intensity, r2SR, in the receiver direction to the incident power density from the transmitter direction. The radar equation enables us to determine the range of the radar supposing we know the transmitted power, the antenna parameters, the least power that must be received by the receiver, and the target radar cross section. Equation (12.33) neglects the attenuation of a wave by the atmosphere.
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12.4 Optoelectronics With increasing demands to transmit more and more of information, transmission channels must use wider and wider frequency bands. This is achieved by increasing the transmission frequency. Huge possibilities are offered by using optical beams as a medium for transmitting information. Here we are in the frequency band of the order 1014 Hz. This gives us an extremely wide frequency band, which can be used to transmit a signal, i.e., the great information capacity of an optical link. Optical signals can be exploited when we master the special technology. This concerns generation, modulation, transmission, demodulation and detection of optical signals. Based on the knowledge gained in the preceding text, in this chapter we will introduce basic problems of optoelectronics. We start with optical waveguides, and will continue with coupling the optical signal to them, followed by optical detectors, optical amplifiers, lasers, optical modulators and, finally, optical sensors. 12.4.1 Optical waveguides A system designer must be aware of the fundamental bandwidth limitations in optical waveguides. The most prominent limitation is dispersion, which represents the dependence of the relative propagation constant k/k0 on frequency, where k0 is the phase constant in a vacuum. Dispersion distorts the shape of the pulses transmitted through the waveguide. The pulses are spread in time when traveling along the waveguide, so that they even cannot be recognized at the output. Temporal spreading effectively establishes the maximum data rate for a communication link. There are three types of dispersion: material dispersion, modal dispersion, and waveguide dispersion. In material dispersion, different wavelengths of light travel at different velocities. Consequently, the pulse effectively spreads out (or disperses) in time and space. Modal dispersion arises in waveguides with more than one propagating mode. Each allowed mode in the waveguide travels at a different group velocity. The pulse energy in a waveguide is distributed among the various allowed modes, either through the initial excitation, or through mode coupling that occurs within the waveguide. The modes arrive at the end of the waveguide slightly delayed relative to each other. This effectively spreads the temporal duration of the pulse, which again limits the bandwidth. Waveguide dispersion is a more subtle effect. The propagation constant depends on the wavelength, so even within a single mode different wavelengths propagate at slightly different speeds. Waveguide dispersion is usually smaller than material and modal dispersion. However, in the vicinity of the so called zero dispersion point for materials, waveguide dispersion can be the dominant effect in a single mode system. Waveguide dispersion can be used to cancel material dispersion, allowing the design of special dispersion shifted waveguides. In Chapter 8 we studied basic dielectric waveguides: the dielectric layer – the dielectric slab waveguide and the dielectric cylinder. In these two waveguides the core has a constant permittivity. Such waveguides are known as step-index waveguides. A very common parameter for characterizing waveguides is the numerical aperture NA. This concept is based on ray tracing and refraction, so it is applicable to multimode waveguides. The geometry of this problem is shown in Fig. 12.27. The numerical aperture is defined as the sinus of the maximum angle θmax n=1 fiber cladding n2 under which the ray is coupled into the waveguide. Let us assume a fiber θ ≥ θc with core refraction index n1, refraction index of the surrounding θ inc ≤ θ max fiber core n1 material n2, and this fiber faces a fiber cladding n2 medium with n = 1. The ray will be guided if it strikes the fiber interface Fig. 12.27 at an angle greater than the critical
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angle (3.68). Applying the Snell laws we get NA = sin (θ max ) = n1 sin (90 − θ c ) = n1 cos(θ c ) = = n1 1 − sin 2 (θ c ) = n1 1 − n22 n12 = n12 − n 22
.
(12.36)
The numerical aperture is a useful parameter for large core multimode waveguides, namely for describing the coupling of light into the fiber. The light to be coupled into the waveguide must be focused in such a way that all incident rays lie within this angle θ max. There are two ways to significantly reduce modal dispersion in a waveguide: by using only single mode waveguides, or by using a graded-index waveguide. The single mode waveguide appears to be the simplest, but it is not always a practical solution. It is very difficult to couple light into single mode waveguides, as their core has small dimensions. The second method for reducing modal dispersion is to use graded-index waveguides. These waveguides can be made with relatively large dimensions, easing the coupling and alignment problems common in single mode devices, and they can dramatically reduce modal dispersion. n(x) In a graded-index waveguide the permittivity is a smooth function of position. At the waveguide center it has the maximum value, and it decreases smoothly with distance away from the central axis, see the sketch in Fig. 12.28. These waveguides can be analyzed as has been described in Chapter 4. The path of the ray propagating in the graded-index waveguide has been determined in Example 4.1. It is described by a sinus function. The modal dispersion arose due to the path differences between the high order rays that followed a 0 x long zigzag down the waveguide and the low order rays Fig. 12.28 that traveled straight. Fig. 12.29a illustrates this case for two extreme modes. In the graded-index structure, a ray traveling near the axis will higher order mode higher order mode spend more time in a high index material, Fig. 12.29b, and will travel more slowly than will a low order mode low order mode ray that is farther from the axis. However, rays far step index graded index from the axis follow a a b longer sinusoidal path. Fig. 12.29 Through optimal adjustment of the index gradient, it is possible to minimize the difference in the group delay between the extreme rays. This will reduce modal dispersion and effectively increase the information capacity of the waveguide. Dispersion of the signal can sometimes be compensated or eliminated through clever design, but attenuation simply leads to a loss of signal. Eventually the energy in the signal becomes so weak that it cannot be distinguished with sufficient reliability from the noise always present in the system. Attenuation therefore determines the maximum distance at which an optical link can be operated without amplification. Attenuation arises from several different physical effects. In an optical waveguide, one must consider: intrinsic material absorption, absorption due to impurities, Rayleigh scattering, bending and waveguide scattering losses, and microbending loss. In terms of priority, intrinsic material absorption
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and Rayleigh scattering are the most serious causes of power loss for long distance systems. Impurity absorption has become less of a problem as improved material processing techniques have been developed. Total optical attenuation is characterized
Pout = Pin e −αz ,
(12.37)
where Pout and Pin are the output and input powers of the optical wave, respectively, and α is the attenuation coefficient. The attenuation coefficient depends strongly on the wavelength of the light and the material system involved. The intrinsic material absorption arises from the atomic, molecular, and vibrational transitions. These transitions can absorb electromagnetic energy from the applied field and store it in the excited state. This energy is eventually dissipated through emission of a photon or through the creation of lattice vibrations, and represents a loss to the electromagnetic field. Sometimes the attenuation due to impurities is more severe than the attenuation of the basic material itself. One of the problems of manufacturing optical glass fibers for long distance communication systems is to provide material with sufficient purity. These losses have been represented by complex permittivity (1.23) in Chapter 1. Similarly we can introduce the complex index of refraction (3.11)
n = n’ – jn’’ .
(12.38)
The imaginary part of n can lead to attenuation or gain, depending on its sign. In the case of a passive medium it leads to attenuation and its value is positive. Inserting for wave propagating constant k = k0n = k0(n’-n’’) we can describe the wave traveling along the waveguide
E ( z ) = E0e − jkz = E0e − jk 0 (n '− jn '')z = E0e − k 0 n '' z e − jk 0 n ' z
(12.39)
Rayleigh scattering is a fundamentally different attenuation mechanism. Instead of light being absorbed and converted into stored energy within a medium, it is simply scattered away from its original direction. This is the scattering of light off the random density fluctuations that exist in a dielectric material. These losses are fundamental and cannot be compensated or eliminated. The present attenuation value of 0.2 dB/km in fused silica is the fundamental limit of the performance of such glass fibers. Optical power can be lost through leakage due to fiber bending. If the fiber is bent, the spatial mode is not appreciably changed in shape compared to the straight fiber. However, the plane wavefronts associated with the mode are now pivoted about the center of curvature of the bend. To keep up with the mode, the phase front on the outside of the bend must travel a little faster than the phase front in the core. At some critical distance from the core of the fiber, the phase front will have to travel faster than the local speed of light, c/nclad. Since this is not possible, the field beyond this critical radius breaks away and enters a radiating mode. The power that breaks away is a loss to the waveguide. Microbending loss is caused by putting an optical fiber in a cable and wrapping it about the central cord. The degree of attenuation depends on the specific cabling geometry. There can be a significant loss in optical connections due to misalignment or due to mismatch between the two devices. Misalignment between a source and a single mode waveguide in the case of dimensions less than 1 µm can cause a coupling loss exceeding 1 dB. Coupling problems are exaggerated by the small dimensions of the typical optical waveguides and sources, which makes alignment a critical and challenging task. The
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calculation of the coupling between two optical waveguides is based on a modal description of the waveguides. The couple depends on alignment, dimension differences, and geometric shapes. In the calculation process we have to include the forward waves of the waveguides, the reflected waves and the radiation modes originating from both the reflected wave and the transmitted field. The crucial problem is to couple the light from a source into a waveguide. This coupling is defined by the radiation pattern of the light source, which is defined by the angular dependence of brightness B(ϕ,θ). Brightness is defined as the optical power radiated into a unit solid angle per unit surface area. It is specified in watts per square centimeter per steradian. Consider the case source radiation in Fig. 12.30, showing an pattern optical source end-fire coupling onto the end of a fiber waveguide. The source optical source core acceptance emits light into a cone, angle which partially overlaps the cladding numerical aperture of the lost power waveguide. Any light falling outside either the Fig. 12.30 numerical aperture or the physical core dimensions obviously will not couple to the waveguide. The total power coupled to the waveguide is, assuming waveguide circular symmetry, given by
P=
rmin 2π 2πθ max
∫ 0
∫ ∫ 00
B ( ϕ , θ ) sin ( θ ) d θ d ϕ r dr dϕ S ∫ 0
(12.40)
where the subscript S refers to the source, rmin is the smaller radius of either the fiber core or the source. Brightness B is here integrated over the acceptance solid angle of the fiber. The maximum acceptance angle θmax is defined through the numerical aperture (12.36). 12.4.2 Optical detectors The minimum received power necessary to achieve the desired quality of information in a communication link is established by noise. There are many ways to detect light. Any process which converts optical energy into another useful form of energy can be considered to be a detector. For optoelectronic systems, the most useful detectors are those that convert optical energy directly into electric current or voltage. There are two fundamental classes of detector: quantum, or photon, detectors, which respond to the number of photons that are absorbed, and thermal detectors, which respond to the energy that is absorbed. The quantum detector measures the number of photons received per second N, which is proportional to the power of the optical signal P
N=
ηP hf
=
ηPλ hc
,
(12.41)
where η is the quantum efficiency of the detector, defined as the probability that a free electron or hole is generated per absorbed photon. h is Planck’s constant, hf is the energy of the photon. The number of received photons increases with the wavelength up to the threshold at which the energy of the photon is not sufficient to excite a free electron or hole.
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An ideal thermal detector responds to the total power, independent of the wavelength. Real detectors have spectral response curves which are limited only by the spectral absorption of their coating. The absorbed optical power is converted to heat. In general, thermal detectors have a much slower response time than most quantum detectors. Optical communication systems almost exclusively use quantum detectors. The quality of a signal is degraded by noise, i.e., random fluctuations that cannot be distinguished from the signal. There are many sources of noise. Shot noise is a fundamental noise that exists in all optical detection processes. Its origin is in the randomness of the arrival of photons in the signal. It is fundamental and, consequently, there is no way to avoid it. The absolute noise value is rarely a concern – normally we are worried about the relative noise compared to the signal. The significant parameter is the signal-to-noise ratio - S/N. Four basic parameters characterize the performance of an optical detector. These are: responsivity, spectral response, detectivity, and time response. Responsivity tells us how much signal is obtained per unit optical power. It is measured in A/W or in V/w. Detectivity tells us the minimum detectable power required to achieve S/N = 1. The quantum detectors used in optical communication systems work on the basis of the internal photoeffect. In this case a photon creates a free charge carrier in the material. The creation of excess carriers leads to a change in conductivity, p-n junction voltage, or junction currents. There are three major categories of optoelectronic detectors: photoconductors, pin diodes, and avalanche photodiodes. Photoconductive detectors employ semiconductor materials with a bandgap suited to the wavelength of the light that is being detected. A photon with sufficient energy can lift an electron from the valence band to the conducting band, and thus increases the conductivity. pn junction detectors are very often used in optical systems. A pair electron-hole is generated by a photon in the depletion layer. It is separated by an internal field and contributes to the current excited in this way by the incident light. Adding an intrinsic layer between the p- and n-layers we get a pin photodiode. The intrinsic layer widens the depleted layer and in this way reduces the diode capacitance and shortens the diode time response. An avalanche photodiode uses the internal multiplication effect to achieve higher detectivity than pin photodiodes. The structure of this diode is designed so that the extra electron-hole pairs excited by the detected light are accelerated through the depletion layer by the intense electric field as they are able to generate extra free electron-hole pairs. These subsequent electrons and holes are themselves accelerated, and can lead to further pair creation. As a result of this impact ionization, the total current in the photodiode is greater than would be produced by photoionization alone. 12.4.3 Optical amplifiers and sources A photon with sufficient energy absorbed in a material can excite an atom from its basic energetic state to the upper excited stage. Spontaneous emission of radiation occurs when an excited atom spontaneously relaxes back to the lower basic state, and in the process emits a single photon. This effect of the spontaneous emission of light radiation is used in standard light emitting diodes. The electromagnetic field of the incident optical wave can induce a transition between the upper and lower state, and the atom gives up one quantum of energy to the field. This is known as stimulated emission of radiation. Optical amplifiers and lasers are based on the stimulated emission of radiation. Let us take a monochromatic optical beam that propagates through a material in the z direction. Its electrical field then varies as E ( z ) = E (0 ) e gz ,
(12.42)
where g is the optical gain. This quantity is proportional to the difference between the populations of atoms in the lower basic state N1 and the atoms in the upper excited state N2
177 book 122
g = σ ( N 2 − N1 ) ,
(12.43)
where σ is a constant. From (12.43) it follows that for N2 > N1 we have g > 0. This condition is known as the inversion population of states, and the optical beam is amplified due to the stimulated emission. The inversion population can be obtained by adding the necessary energy from outside, depending on the kind of active material – pumping the laser or the amplifier. The simplest optical amplifier compatible with optical communication links using optical fibers is the erbium-doped fiber optical amplifier. This is a standard optical fiber with the core doped by erbium. Erbium ions in the glass lattice assure appropriate energy levels. This active material is pumped optically by irradiating, e.g., by a tungsten bulb. The energy of the E3 E2 photons of the pumping light must be higher than the stimulated energy of the photons of the amplified light. The pump emission pumping assures transition of atoms from the basic E1 energy state E0 to state E3, see Fig. 12.31, and the E0 stimulated emission then goes from state E2 to state E1. The spontaneous emission in the optical amplifier Fig. 12.31 is the source of the noise added to the amplified signal. Lasers are optical oscillators. The acronym laser stands for light amplification by the stimulated emission of radiation. Like other oscillators, the laser is an optical amplifier with positive feedback. Amplification is assured by the gain, which is established through population inversion due to pumping an active medium. Positive feedback is accomplished in the optical frequency range using mirrors appropriately aligned to resonate one, or a few, cavity modes. Part of the generated power is in this way selectively returned back to the input of the amplifier. In optoelectronics, lasers tend to be based on waveguide structures which are already one-dimensional in nature. The spatial confinement of the oscillating mode is provided by the structure of the device. The longitudinal mode selection and feedback is accomplished by an open resonator known as the Fabry-Perot resonator, Fig. 12.32. It consists of two mirrors with reflection coefficients R1 and R2, separated by R1 R2 distance d. Only light with an integer number of d half-wavelengths in the cavity will be resonant in the structure. The field incident to the mirrors is partially reflected, assuring the feedback, and mirror mirror partially transmitted, providing the laser output. The wave reflected back is amplified, which Fig. 12.32 maintains the oscillations in the cavity. Modern optoelectronics is based on the application of semiconductor injection lasers. These lasers are compatible with driving electronic circuits. They are efficient, small and cheap. The light emitted by these lasers can be relatively simply coupled to optical fibers. The inversion population of states is achieved by injecting free charge carriers across two heterojunctions into a narrow active layer. For this reason, these lasers are known as double heterostructure semiconductor lasers. Semiconductor physics is beyond the scope of this course. 12.4.4 Optical modulators and sensors There are two common methods for encoding a signal onto an optical beam: either directly modulate the optical source, or externally modulate a continuous wave optical source. Direct modulation is the most widespread method of modulation today, but it introduces
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demanding constraints on semiconductor lasers. For example, it is difficult to directly modulate a semiconductor laser at frequencies above a few GHz. Furthermore, it is difficult to maintain single mode operation of these pulsed lasers. Multiple mode lasers have a larger spectral bandwidth, which leads to increased pulse spreading due to dispersion. External modulators offer several advantages over direct modulation. First, one can use a relatively simple and inexpensive continuous wave laser as the primary optical source. Second, since a modulator can encode information based on a number of externally controlled effects, it is not compromised by the need to maintain population inversion or single mode control. The simplest electrooptic modulator is based on the linear electrooptic effect, or Pockel’s effect. This represents the change in the refractive index due to an externally applied electric field. Devices which directly modulate the phase, the intensity, or the polarization of the light are designed with the use of the Pockel’s effect. The phase modulator is simple. The external electric field changes the refractive index, and in this way the phase velocity of the wave propagating through the modulator. This causes changes in the output signal. The phase modulation introduced by the electrooptic effect can be used to create intensity modulation via changes in polarization and through interferometric effects. Polarization modulation can be achieved using the differential retardation between two orthogonal polarizations of the optical wave. To convert polarization modulation into intensity modulation, it is necessary to run the output through a linear polarizer. Phase modulation can be converted into intensity modulation through constructive interference between two waves. The Fabry-Perot interferometer and the Mach-Zender interferometer represent two examples for converting phase modulation into intensity modulation. Fig. 12.33 shows a schematic Mach-Zender interferometer. The single mode waveguide input is split into two single mode waveguides by a 3 dB Y junction. The split beams 3 dB coupler travel different paths, and then recombine at another Y junction. The relative phase difference of the two input V beams can be electrooptically output controlled by applying a voltage to E the center electrode in the structure shown in Fig. 12.33. Because the change in refractive index n depends on the direction of the applied electric Fig. 12.33 field, index n increases in one arm and decreases in the other arm. This differential change in index is used to alter the relative phase of the recombining fields and thus the amplitude of the output wave. Like electrooptic modulators, acoustooptic modulators control the transmission of light by local changes in the refractive index of the transmission medium. The modulation occurs by means of a traveling sound wave which induces stress related modifications of the local refraction index. Optical wave interaction can be produced by either bulk acoustic waves traveling in the volume of the material, or by surface acoustic waves which propagate on the surface within approximately one acoustic wavelength of the surface. Surface acoustic wave devices are well suited to integrated optics applications because the energy of the acoustic field is concentrated in the region of the optical waveguide. Anything that can perturb the optical beam in a fiber can be exploited to make a sensor. Common interactions are length and refractive index modification through stress, strain, or temperature. The designer must know how to make the fiber interact selectively with the measured quantity of interest. An optical sensor can be based on changes in intensity, polarization, phase, wavelength, and direction of propagation. In extrinsic sensors the light leaves the fiber and is modulated prior to being recoupled onto a fiber. In an intrinsic sensor the light is modulated inside the fiber.
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Three basic effects are used to intensity modulate the optical signal: mechanical, based on movement of an aperture or object which interferes with the optical path; refractive, where the cladding layer of the fiber is modified in such a way as to cause the guided light to leave the core; and absorptive, where the optical signal is absorbed by a substance. The critical weakness of the intensity sensor is that the system has no way of knowing if a change in intensity is in response to a change of the measured quantity, or if it is due to some other cause. This weakness can be reduced by using a phase modulation index based, e.g., on the Mach-Zender interferometer, see Fig. 12.33. The measured physical quantity changes the phase of the optical wave in one arm of the interferometer. This causes a change in the output signal.
13. MATHEMATICAL APPENDIX This section summarizes the basic knowledge of mathematics necessary to understand the text. Computation with vectors: A vector quantity is written as a linear combination of basic vectors x0, y0, z0 A = Ax x 0 + Ay y 0 + Az z 0 .
(13.1)
We add vectors by adding their coordinates A + B = ( Ax + B x ) x 0 + (Ay + B y )y 0 + ( Az + B z ) z 0 .
(13.2)
The modulus of a vector is A = A = Ax2 + Ay2 + Az2 .
(13.3)
The scalar product of two vectors is A ⋅ B = AB cos(α ) = Ax B x + Ay B y + Az B z = B ⋅ A ,
(13.4)
where α is the angle between the two vectors A and B. It follows from definition (13.4) that the scalar product of two vectors equals zero for perpendicular vectors. The scalar product is linear
A ⋅ (B + C) = A ⋅ B + A ⋅ C ,
(13.5)
From (13.4) we can determine the angle between vectors
α = arccos
A⋅B . AB
(13.6)
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According to Fig. 13.1 the scalar product determines the projection of vector A in the direction of vector B. Let n be a unit vector normal to a surface, then it follows from Fig. 13.1 that the component of A normal to the surface is defined by An = n ⋅ A .
A Asin(α)
α
•
Acos(α)
(13.7)
B
Fig. 13.1
The vector product of two vectors is defined A × B = AB sin (α ) i n = −B × A ,
(13.8)
where in is a unit vector perpendicular to the plane in which the two vectors lie. The vector product can be calculated in the rectangular coordinate system as x0
y0
A × B = Ax Bx
Ay By
z0
Az = (Ay B z − Az B y )x 0 + ( Az B x − Ax B z ) y 0 + (Ax B y − Ay B x )z 0 . Bz
If the vector product of two nonzero vectors equals zero, it follows from (13.8) that these two vectors are parallel. From Fig. 13.1 it is clear that the value of the vector product determines the projection of vector A into the direction perpendicular to vector B. Therefore using the vector product we can compute the tangential component of a vector to the plane determined by unit normal vector n
At = n × A .
(13.9)
A×B
A
α B Fig. 13.2
(13.10)
The value of the vector product (13.8) equals the area of the parallelogram in Fig. 13.2. The vector product is linear as the scalar product (see (13.5)). The double vector product is defined as
A × (B × C) = B(A ⋅ C) − C(A ⋅ B ) .
A B
(13.11)
C Fig. 13.3
The mixed product is defined as Ax
A ⋅ (B × C) = B ⋅ (C × A ) = C ⋅ (A × B ) = B x Cx
Ay
Az
By Cy
Bz . Cz
(13.12)
The value of the mixed product equals the volume of the parallelepiped shown in Fig. 13.3.
Complex numbers: The basic knowledge is, see Fig. 13.4 defining particular quantities,
181 book-13
y z = x + jy = ρ e jϕ , ρ = x 2 + y 2 , ϕ = arctg , x
e jϕ = cos(ϕ ) + j sin (ϕ ) .
(13.14)
Goniometric functions: Here is a list of basic formulas
ρ ϕ
(13.15)
x
1 [1 − cos(2α )] , 2
cos 2 (α ) =
z
jy
sin 2 (α ) + cos 2 (α ) = 1 , sin 2 (α ) =
(13.13)
(13.16)
Fig. 13.4
1 [1 + cos(2α )] , 2
(13.17)
sin (α ± β ) = sin (α ) cos(β ) ± cos(α ) sin (β ) ,
(13.18)
cos(α ± β ) = cos(α ) cos(β ) m sin (α ) sin (β ) ,
(13.19)
α ± β sin (α ) ± sin (β ) = 2 sin 2
(13.20)
α m β cos , 2
α + β α − β cos(α ) + cos(β ) = 2 cos cos , 2 2
(13.21)
α + β α − β cos(α ) − cos(β ) = −2 sin sin , 2 2
(13.22)
A sin (ωt ) + B cos(ωt ) = C sin (ωt + ϕ ) , C =
B A 2 + B 2 , ϕ = arctg , A
(13.23)
sin ( x ) =
1 jx ( e − e − jx ) , 2j
(13.24)
cos( x ) =
1 jx e + e − jx . 2
(13.25)
sinh ( x ) =
1 x e − e −x , 2
(13.26)
cosh ( x ) =
1 x e + e−x . 2
(13.27)
(
)
Hyperbolic functions:
(
(
)
)
182 book-13
cosh 2 ( x ) = sinh 2 ( x ) + 1 ,
(13.28)
sinh ( x ± y ) = sinh ( x ) cosh ( y ) ± cosh ( x )sinh ( y ) ,
(13.29)
cosh ( x ± y ) = cosh ( x ) cosh ( y ) ± sinh (x ) sinh ( y ) .
(13.30)
From definitions of hyperbolic functions (13.26) and from the definitions of goniometric functions (13.24) and (13.25) it follows sin ( jx ) = j sinh ( x ) , cos( jx ) = cosh (x ) ,
(13.31)
sin ( x + jy ) = sin ( x ) cosh ( y ) + j cos( x )sinh ( y ) ,
(13.32)
cos( x + jy ) = cos( x ) cosh ( y ) − j sin ( x )sinh ( y ) ,
(13.33)
sin (2 x ) + j sinh (2 y ) , cos(2 x ) + cosh (2 y )
(13.34)
sinh ( x + jy ) = sinh ( x ) cos( y ) + j cosh ( x )sin ( y ) ,
(13.35)
cosh ( x + jy ) = cosh ( x ) cos( y ) + j sinh ( x ) sin ( y ) .
(13.36)
tg ( x + jy ) =
Bessel’s functions: A linear combination of Bessel’s functions Jn(x) and Yn(x) of the n-th order is a solution of the differential equation x 2 y ' '+ x y '+(x 2 − n 2 ) y = 0 ,
(13.37)
y = C1 J n ( x ) + C 2 Yn ( x ) .
(13.38)
Functions Jn(x) are defined by the sum
∞
(− 1)k x
n+2k
2 . k = 0 k!Γ(n + k + 1)
J n (x ) = ∑
(13.39)
Functions Yn(x) follow from Jn(x) by the formula Yn ( x ) =
J n ( x ) cos(nπ ) − J − n ( x ) . sin (nπ )
(13.40)
The integrals of Bessel’s functions are
∫x
n
J n−1 ( x ) dx = x n J n ( x ) ,
∫x
−n
J n +1 ( x ) dx = − x − n J n ( x ) .
(13.41)
183 book-13
1
Y 0, Y 1, Y 2
J0 , J1, J2
Plots of these functions of the three lowest orders are shown in Fig. 13.5 and Fig. 13.6. J0
0.8
J1
0.6
J2
0.4
Y0
0.4
Y1
0.2 0
Y2
-0.2
0.2
-0.4
0
-0.6
-0.2
-0.8
-0.4 0
2
4
6
8
10
x
-1 0
2
4
6
8
10
x
Fig. 13.5
Fig. 13.6
To calculate the propagation constants of modes propagating in a waveguide with a circular cross-section, see paragraph 7.3, we need to know the zero points αmn of the Bessel ' function Jm and the zero points α mn of the derivative of this function. The zero points αmn of ' of the derivative J m' are the Bessel function Jm are listed in Tab. 13.1. The zero points α mn listed in Tab. 13.2.
m n=1 n=2 n=3 n=4 n=5
0 2.40482 5.52007 8.65372 11.79153 14.93091
1 3.83171 7.01559 10.17347 13.32369 16.47063
2 5.13562 8.41724 11.61984 14.79595 17.95980
3 6.38016 9.76102 13.01530 16.22347 19.40942
4 7.58834 11.06471 14.37254 17.61597 20.82693
5 8.77148 12.33860 15.70017 18.98013 22.21780
3 4.20119 8.01524 11.34592 14.58585 17.78875
4 5.31755 9.28240 12.68191 15.96411 19.19603
5 6.41562 10.51986 13.98719 17.31284 20.57551
Tab. 13.1 - αmn m n=1 n=2 n=3 n=4 n=5
0 3.83170 7.01558 10.17346 13.32369 16.47063
1 1.84118 5.33144 8.53632 11.70600 14.86359
2 3.05424 6.70613 9.96947 13.17037 16.34752 ' Tab. 13.2 - α mn
The asymptotic formulas of Bessel’s functions Jn(x) and Yn(x) for x increasing to infinity are J n (x ) =
2 1π cos x − n + , πx 2 2
(13.42)
Yn (x ) =
2 1π sin x − n + , πx 2 2
(13.43)
Hankel’s functions (Bessel’s functions of the 3rd kind):
184 book-13
The solution of equation (13.37) is any linear combination of Bessel’s functions Jn and Yn. This enables us to introduce Hankel’s functions by H n1 ( x) = J n ( x ) + jYn ( x ) ,
(13.44)
H n2 ( x) = J n ( x ) − jYn (x ) .
(13.45)
Their approximate values for a high argument are H n1 ( x ) ≈
2 j ( x −π / 4−πn / 2 ) e , πx
(13.46)
H n2 ( x ) ≈
2 − j ( x −π / 4−πn / 2 ) e . πx
(13.47)
It is evident from (13.44) and (13.45) that the relations between Hankel’s functions and Bessel’s functions Jn and Yn are similar to the relations between the exponential function of a complex argument and goniometric functions. Bessel’s functions Jn and Yn describe solutions of the wave equation cylindrical as standing waves, whereas Hankel’s functions describe a traveling wave. Hankel’s functions of the order ½ are defined by 2 jx e , πx
(13.48)
2 − jx e . πx
(13.48)
H 11/ 2 ( x ) = − j H 12/ 2 ( x ) = j
Modified Bessel’s functions: Modified Bessel’s functions are Bessel’s functions of an imaginary argument. Inserting into (13.37) x = jz we get
(
)
z 2 y ' '+ zy '− z 2 + n 2 y = 0 .
(13.49)
The solution of this equation is
y = A I n (z ) + B K n (z ) .
(13.50)
The modified Bessel’s function of the first kind is defined
I n ( z ) = j − n J n ( jz ) ,
(13.51)
which gives a real value for real z. Similarly K n ( z ) = j n+1
π 2
H n1 ( jz )
(13.52)
185 book-13
is the modified Bessel’s function of the second kind, again real for real z. This function can be defined by function In K n (z ) =
π I −n (z ) − I n (z ) . 2 sin (nπ )
(13.53)
Their approximate values for a high argument are
I n (z ) =
ez , 2πz
π
K n (z ) =
2z
(13.54)
e−z .
(13.55)
30
K0, K1
I0 , I 1
Plots of modified Bessel’s functions of the two lowest orders are shown in Fig. 13.7 and Fig. 13.8.
25 20 I0
15
3 2
I1
10
4
K1
1 K0
5 0
0
1
2
3
4
5
0
6 x
Fig. 13.7
0
1
2
3
4
5
6 x
Fig. 13.8
Coordinate systems: We use the three basic coordinate systems: rectangular, cylindrical and spherical. In the rectangular coordinate system each point is determined by three coordinates x, y, z, which represent the corresponding segments on the axes, see Fig. 13.9. Elemental displacements in the directions of the axes are dx, dy and dz. The volume element is dV=dx.dy.dz. The cylindrical coordinates are r, α, z, see Fig. 13.10. r represents the perpendicular distance from the axis z, α is the angle measured in the plane x y starting from the positive x direction, and z has the same meaning as in the rectangular system. The cylindrical coordinates are bound with the rectangular coordinates by the relations x = r cos(α ) , y = r sin (α ) ,
(13.56)
z=z .
Elemental displacements in the directions of the axes are dr, rdα and dz. The volume element is dV=dr.r.dα.dz. If the integrated function depends only on r, we can use the volume element
186 book-13
in the form dV=2πrldr, which is the volume between two coaxial cylinders with radii r and r+dr of length l. The spherical coordinates are r, α, ϑ, see Fig. 13.11. r represents the distance from the origin, α is the angle measured in the plane x y starting from the positive x direction, and ϑ is the angle measured from the positive z direction. The spherical coordinates are bound with the rectangular coordinates by the relations
x = r cos(α ) sin (ϑ ) , y = r sin (α ) sin (ϑ ) ,
(13.57)
z = r cos(ϑ ) . z
z z0
z z0
A(x,y,z) x0
A(r,α,z) r0
y0
z x
α0
α0
α
y
r
z
ϑ
r
r0
A(r,ϑ,α) ϑ0
α
y
y
y x
x
x Fig. 13.10
Fig. 13.9
Fig. 13.11
Elemental displacements in the directions of the axes are dr, rsin(ϑ)dα and rdϑ. The volume element is dV=r2.sin(ϑ).dr.dα.dϑ. If the integrated function depends only on r, we can use the volume element in the form dV=4πr2dr, which is the volume between two concentric spheres with radii r and r+dr.
The flux of a vector (a surface integral): Let J be the distribution of electric current. The total current passing surface S can be calculated in the following way. We discretize the J surface, Fig. 13.12, to elementary surfaces dS S n determined by vectors dS=dSn, where n is the unit normal vector. The current passing the elemental dS surface dS is dI = J n dS = J ⋅ n dS = J ⋅ dS .
dS
The total current is the summation of dI, which is a surface integral
I = ∫∫ J ⋅ dS .
(13.58)
S
Fig. 13.12
In this way we have presented the physical meaning of the surface integral from the scalar product of a vector and the vector of a surface element. In the case of a closed surface we
187 book-13
have the notation
I = ∫∫ J ⋅ dS . S
Path integral: We demonstrate the physical meaning of this integral by calculating the work performed by force F along path c. We divide the path into elemental arcs of length dl. The work performed along an arch is, see Fig. 13.13,
F dl
t0
dl c
dW = Ft dl = F ⋅ t 0 dl = F ⋅ dl .
Fig. 13.13
The work performed by force F along the whole path c is the summation of dW, which is an integral
W = ∫ F ⋅ dl .
(13.59)
c
In the case of a closed path the integral is known as the circulation of vector F
W = ∫ F ⋅ dl .
(13.60)
c
Gradient of a scalar function: The gradient of a scalar function is a vector which determines the direction in which a function value increases at maximum speed, and the modulus of this vector determines the magnitude of this speed. This vector is perpendicular to the planes of the constant function values. In the rectangular coordinate system the gradient is defined by grad ϕ =
∂ϕ ∂ϕ ∂ϕ x0 + y0 + z0 . ∂x ∂y ∂z
(13.61)
The increase of a scalar function corresponding to a shift by dr can be determined neglecting differentials of higher orders
ϕ (r + dr ) = ϕ ( x + dx, y + dy, z + dz ) = , ∂ϕ ∂ϕ ∂ϕ = ϕ ( x, y , z ) + dx + dy + dz + ... ∂x
∂y
∂z
dϕ = ϕ ( x + dx, y + dy, z + dz ) − ϕ ( x, y, z ) ≈ ≈
∂ϕ ∂ϕ ∂ϕ dx + dy + dz = grad ϕ ⋅ dr = grad ϕ dr cos(Θ ) ∂x ∂y ∂z
So the increase of a function is greatest in the direction of gradϕ. The gradient is defined by partial derivatives, so it is a linear operator
188 book-13
grad(λ1ϕ 1 + λ 2ϕ 2 ) = λ1 grad ϕ1 + λ2 grad ϕ 2 ,
(13.62)
where λ1 and λ2 are two scalar constants. For the product of two functions we have grad(ϕψ ) = ϕ gradψ + ψ grad ϕ .
(13.63)
In the literature we can find the symbolic vector ∇ , known as Hamilton’s “nabla” operator. In the rectangular coordinate system this operator is defined as
∇=
∂ ∂ ∂ x0 + y 0 + z 0 . ∂x ∂y ∂z
(13.64)
Using this operator the gradient can be expressed as grad ϕ = ∇ϕ .
(13.65)
Divergence of a vector function: The divergence of vector function J is defined as the volume density of this vector quantity flux flowing out of a point, so it characterizes the sources of a vector, 1 div J = lim V →0 V
∫∫ J ⋅ dS ,
(13.66)
S
where S is the boundary of volume V. In the case of J being a current density, divJ corresponds to the current flowing out of a unit volume. The Gauss-Ostrogradsky theorem follows from this
∫∫ J ⋅ dS = ∫∫∫ div J dV S
.
(13.67)
V
In the rectangular coordinate system the divergence can be calculated as div J =
∂J x ∂J y ∂J z + + . ∂x ∂y ∂z
(13.68)
This operator is a linear operator, as it is defined by partial derivatives, div(λJ 1 + µJ 2 ) = λ div J 1 + µ div J 2 ,
(13.69)
where λ and µ are scalar constants. The divergence applied to the product of scalar function f and vector function J is div( f J ) = grad f ⋅ J + f div J .
(13.70)
The divergence applied to the vector product of two vectors is div(A × B ) = B ⋅ rot A − A ⋅ rot B .
(13.71)
189 book-13
The divergence applied to the gradient of a scalar function defines Laplace’s operator div(grad ϕ ) = ∆ϕ =
∂ 2ϕ ∂ 2ϕ ∂ 2ϕ + + . ∂x 2 ∂y 2 ∂z 2
(13.72)
The divergence can be expressed using Hamilton’s operator div J = ∇ ⋅ J .
(13.73)
Rotation of a vector function: The rotation of a vector function is defined as the surface density of this vector circulation 1 rot F = r0 max lim ∫ F ⋅ dl , S →0 S c
(13.74)
where c is a curve that describes the boundary of surface S, and r0 is a unit vector. This vector determines the direction of vector dS and is selected to obtain the maximum value of 1 lim ∫ F ⋅ dl . As the rotation represents the surface density of a circulation, Stokes’ S →0 S c theorem follows from the definition of the rotation
∫ F ⋅ dl = ∫∫ rot F ⋅ dS . c
(13.75)
S
In the rectangular coordinate system the operator rotation is defined as a symbolic determinant x0 ∂ rot F = ∂x Fx
y0 ∂ ∂y Fy
z0 ∂ . ∂z Fz
(13.76)
This operator is again a linear operator, as it is defined by partial derivatives, rot (λF1 + µF2 ) = λ rot F1 + µ rot F2 .
(13.77)
The rotation applied to the product of a scalar function and a vector function is rot ( f F ) = f rot F + grad f × F .
(13.78)
There are mathematical identities rot (grad f ) = 0 ,
(13.79)
div(rot F ) = 0 ,
(13.80)
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rot (rot F ) = grad(div F ) − ∇ 2 F . (13.81) In the rectangular coordinate system we have ∇ 2 = ∆ The rotation can be expressed by Hamilton’s operator as `
rot F = ∇ × F ,
(13.82)
which follows from (13.63) and (13.9). Differential operators in the cylindrical coordinate system: 1 ∂ϕ ∂ϕ ∂ϕ r0 + α0 + z0 , ∂r r ∂α ∂z
(13.83)
∂F 1 ∂ (r Fr ) + 1 α + ∂Fz , r ∂r r ∂α ∂z
(13.84)
grad ϕ = div F =
∂F 1∂ 1 ∂Fz ∂Fα ∂F ∂F − rot F = r0 + r − z α 0 + (r Fα ) − r z 0 , (13.85) ∂z ∂r ∂α r ∂r ∂z r ∂α ∆ϕ =
1 ∂ ∂ϕ 1 ∂ 2ϕ ∂ 2ϕ . + r + r ∂r ∂r r 2 ∂α 2 ∂z 2
(13.86)
Differential operators in the spherical coordinate system: grad ϕ =
div F =
∂ϕ 1 ∂ϕ 1 ∂ϕ ϑ0 , r0 + α0 + ∂r r sin ϑ ∂α r ∂ϑ
1 ∂ 2 1 ∂Fα 1 ∂ (sin ϑ Fϑ ) , r Fr + + 2 r sin ϑ ∂α r sin ϑ ∂ϑ r ∂r
(
)
∂Fϑ ∂Fr 1 ∂ 1 ∂ ( ) r ( ) α0 + F r F sin ϑ − + − α 0 ϑ r sin ϑ ∂ϑ r ∂r ∂α ∂ϑ , 1 ∂Fr 1 ∂ (r Fα ) ϑ0 + − r sin ϑ ∂α r ∂r
(13.87)
(13.88)
rot F =
∆ϕ =
∂ 2ϕ ∂ ∂ϕ 1 ∂ 2 ∂ϕ 1 1 + + 2 r sin ϑ . 2 2 2 2 ∂ϑ r ∂r ∂r r sin ϑ ∂α r sin ϑ ∂ϑ
(13.89)
(13.90)
14. BASIC PROBLEMS This section lists the basic problems collected throughout the text. The ability to answer them verifies the student’s knowledge and readiness to sit for the exam. 1
Maxwell’s equations in a differential form for a time-harmonic field expressed by
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2 3 4 5 6 7 8 9 10 11 12 13 14 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43
phasors. Maxwell’s equations in an integral form for time-harmonic field expressed by phasors. Boundary conditions of tangential components of E, H in a nonstationary field. Boundary conditions of normal components of E, H in a nonstationary field. Electromagnetic field on a surface of an ideal conductor. Poynting’s vector. Definition and expression using field vectors. Energy balance of active power. Physical meaning of particular items. Energy balance of reactive power. Physical meaning of particular items. The wave equation for E or H in a general material outside of a source region. Timeharmonic field. Expression using phasors. Continuity equation for harmonic field Description of vectors E and H by potentials A and ϕ in time varying field. Wave equations for potentials, sources assumed. General solution of wave equations for potentials, harmonic field is assumed. The expression of E of a time-harmonic plane wave in a general material. Phasor and instantaneous value. Meaning of particular items. What is a surface of constant amplitude and constant phase in a plane wave? What are a uniform wave and a nonuniform wave? What is a phase velocity? How is it defined? What is a group velocity? How is it defined? Draw the orientation of E, H and k in a plane electromagnetic wave. What is the relation of these vectors? What is the wave impedance of a general material? How is it defined? The expressions for k, vf and Z in an ideal dielectric. The expressions for k, vf and Z in a good conductor. What is a penetration depth? How is it defined? Active power transmitted by a plane electromagnetic wave through a surface 1 m2. What is the polarization of an electromagnetic wave? Which kinds of polarization of an electromagnetic wave are there? Under which conditions can two linearly polarized waves create a wave with linear, circular and elliptical polarization? Equation of eikonal. Differential equation describing the ray. Coefficient of reflection for electric field in a perpendicular incidence, general materials. Coefficient of transmission for electric field in a perpendicular incidence, general materials. Coefficients of reflection and transmission for a perpendicular incidence, lossless dielectrics. What is the standing wave? What is the standing wave ratio, its relation to R? What is λ/4 transformer? Snell’s laws. What is the Brewster’s polarization angle, expression? Draw a plot of R=f(ϕi) for ε1<ε2 for both horizontal and vertical polarization. Draw a plot of R=f(ϕi) for ε1>ε2 for both horizontal and vertical polarization. What is the total reflection, the condition for it? What is the surface wave? What is the character of field in the second material in the case of the total reflection? Write a formula describing the spherical wave. Write a formula describing the cylindrical wave.
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44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88
What transmission line can transmit TEM wave? Cut-off frequency of a mode in a parallel plate waveguide. Cut-off wavelength of a mode in a parallel plate waveguide. Sketch the field distribution of the TEM wave propagating in a parallel plate waveguide. Phase velocity of a wave propagating between two parallel plates. Wavelength of a wave propagating between two parallel plates. Propagation constant of a wave propagating between two parallel plates. Wave impedance of a TM and TE waves propagating between two parallel plates. What mode is dominant in the rectangular waveguide? Cut-off frequency of a mode in the rectangular waveguide. Cut-off wavelength of a mode in the rectangular waveguide. Cut-off frequency of the dominant mode in the rectangular waveguide. Cut-off wavelength of the dominant mode in the rectangular waveguide. Phase velocity of a wave propagating in a rectangular waveguide. Wavelength of a wave propagating in a rectangular waveguide. Propagation constant of a wave propagating in a rectangular waveguide. Wave impedance of a TM and TE waves propagating in a rectangular waveguide. Power carried by the dominant mode in a rectangular waveguide. What transmission line can transmit the TEM wave? Cut-off frequency of a TM mode in the circular waveguide. Cut-off frequency of a TE mode in the circular waveguide. What mode is dominant in the circular waveguide? Phase velocity of a wave propagating in a circular waveguide. Wavelength of a wave propagating in a circular waveguide. Propagation constant of a wave propagating in a circular waveguide. Wave impedance of a TM and TE waves propagating in a circular waveguide. Explain the guiding of waves in a dielectric waveguide. Draw the plot of Ey field as a function of the transversal coordinate in the dielectric layer for the first three TE modes. Which parameters determine the resonant frequency of a cavity resonator? Resonant frequency of a rectangular cavity resonator. The telegraph equations for time-harmonic u and i on a TEM transmission line. The meaning of particular items. The characteristic impedance of a transmission line - the lossy and lossless case. Expression and the definition. The input impedance of a lossy and a lossless line of length L terminated by an impedance ZL. The input impedance of a lossless line of length L with a short and open end. Draw lines of constant amplitude of a reflection coefficient in the Smith Chart. Draw lines of constant phase of a reflection coefficient in the Smith Chart. Draw lines of constant real part or a normalized impedance in the Smith Chart. Draw lines of constant imaginary part or a normalized impedance in the Smith Chart. Characteristic regions around the radiating elementary electric dipole. Which components of electric and magnetic field carry power from the radiating electric dipole? Draw radiation patterns of the radiating elementary electric dipole. What is the radiating resistance of the dipole? What is the antenna directivity? What is the antenna gain? Relation between E and D in anisotropic material.
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89 90 91 92 93 94 95
Relation between B and H in anisotropic material. Tensor of permeability of magnetized ferrite. Tensor of permittivity of a magnetized plasma. What is the ferromagnetic resonance? What is the Faraday effect? What is the ordinary wave? What is the extraordinary wave?
14. LIST OF RECOMMENDED LITERATURE C. A. Balanis: Advanced Engineering Electromagnetics, John Wiley & Sons, Inc., New York, USA, 1989. J. D. Jackson: Classical Electrodynamics, 3rd ed., John Wiley & Sons, Inc., New York, USA, 1998. L. B. Felsen, N. Marcuvitz: Radiation and scattering of waves, IEEE Press, Piscataway, N. J., 1994. R. E. COLIN: Field Theory of Guided Waves, 2nd ed., IEEE Press, Piscataway, N. J., 1991. R. E. COLIN: Foundations for Microwave Engineering, 2nd ed., IEEE Press, Piscataway, N. J., 2001. D. M. Pozar: Microwave engineering, 2nd ed., John Wiley & Sons, Inc., New York, 1998. C. A. Balanis: Antenna theory analysis and design, 2nd ed., John Wiley & Sons, Inc., New York, USA, 1997. W. L. Stutzman, G. A. Thiele: Antenna theory and design, John Wiley & Sons, Inc., New York, USA, 1981. J. Lavergnat, M. Sylvain: Radiowave Propagation, John Wiley & Sons, Inc., New York, USA, 2000. L.W. Barclay (Ed): Propagation of radiwaves, 2nd ed., IEE Books, G.B., 2003. C. R. Pollock: Fundamentals of optoelectronics, Richard D. Irwin, Inc., USA, 1995. C. –L. Chen: Elements of optoelectronics and fiber optics, Richard D. Irwin, Chicago, USA, 1996.
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