Chapter 1. Electromagnetic Theory Sept. 1 st, 2008.
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Transcript of Chapter 1. Electromagnetic Theory Sept. 1 st, 2008.
Chapter 1. Electromagnetic Theory
Sept. 1st, 2008
2
1.1 Introduction to Microwave Engineering
• Microwaves: 300 MHz ~ 300 GHz (1 mm ≤ λ ≤ 1 m) • 1mm ≤ λ ≤ 10mm Millimeter waves• Because of the high frequency (short wavelength),
standard circuit theory generally cannot be used directly to solve microwave network problems.
• Microwave components: distributed elements (the phase of a voltage or current changes significantly over the physical extent of the device)
• Optical engineering• Quasioptical
3
Figure 1.1 (p. 2)The electromagnetic spectrum.
4
Applications of Microwave Engineering
• Microwave engineering 이 필요한 이유– Antenna gain 은 Antenna 의 electrical 크기에
비례 .– 높은 주파수에서는 Bandwidth 가 더 커짐 .– Microwave signal 은 Line of Sight 로 Travel.– Effective reflection area (radar cross section) 는
Target 의 electrical 크기에 비례 .– 분자 , 원자 및 핵공명이 Microwave
주파수에서 일어남 .
5
• Microwave 기술의 응용– Communication systems– Radar systems– Remote sensing– Medical systems
6
A Short History of Microwave Engineering
• Modern electromagnetic theory 의 수립역사1. 1873 년 James Clerk Maxwell 이 이론적으로
EM wave 와 , 빛은 EM energy 의 형태임을 주장
2. 1885 – 1887, Oliver Heaviside 가 Maxwell이론의 수학적 복잡성을 대부분 제거 , Vector 표기 제안 및 Guided wave 와 Transmission line의 응용 제안
3. 1887-1891, Heinrich Hertz, 실험으로 증명
7
• Microwave 기술의 발전1. 1940 년대 , 2 차대전에서 Radar 의 발명으로 큰
발전2. MIT 에서 Radiation Lab 을 세움
– N. Marcuvitz, I.I.Rabi, J.S.Schwinger, H.A.Bethe, E.M.Purcell, C.G.Montgomery, R.H.Dicke…..
– The classic 28-volume Radiation Laboratory Series of books
8
Figure 1.2 (p. 4, see next slide for photograph)Original aparatus used by Hertz for his electromagnetics experiments. (1) 50 MHz transmitter spark gap and loaded dipole antenna. (2) Parallel wire grid for polarization experiments. (3) Vacuum apparatus for cathode ray experiments. (4) Hot-wire galvanometer. (5) Reiss or Knochenhauer spirals. (6) Rolled-paper galvanometer. (7) metal sphere probe. (8) Reiss spark micrometer. (9) Coaxial transmission line. (10-12) Equipment to demonstrate dielectric polarization effects. (13) Mercury induction coil interrupter. (14) Meidinger cell. (15) Vacuum bell jar. (16) High-voltage induction coil. (17) Bunsen cells. (18) Large-area conductor for charge storage. (19) Circular loop receiving antenna. (20) Eight-sided receiver detector. (21) Rotating mirror and mercury interrupter. (22) Square loop receiving antenna. (23) Equipment for refraction and dielectric constant measurement. (24) Two square loop receiving antennas. (25) Square loop receiving antenna. (26) Transmitter dipole. (27) High-voltage induction coil. (28) Coaxial line. (29) High-voltage discharger. (30) Cylindrical parabolic reflector/receiver. (31) Cylindrical parabolic reflector/transmitter. (32) Circular loop receiving antenna. (33) Planar reflector. (34, 35) Battery of accumulators. Photographed on October 1, 1913 at the Bavarian Academy of Science, Munich, Germany, with Hertz’s assistant, Julius Amman. Photograph and identification courtesy of J. H. Bryant, University of Michigan.
9
Figure 1.2 (p. 4)
10
1.2 Maxwell’s Equations
• Maxwell’s work was based on a large body of empirical and theoretical knowledge developed by Gauss, Ampere, Faraday, and others.
• The general form of time-varying Maxwell’s equations in differential form.
0
BE M
tD
H Jt
D
B
What are the sources of electromagnetic fields?
11
Faraday’s law
• Spatially varying field• Moving loop in uniform B-field• Electric generator
12
Corrected Ampere’s circuital law
• James Clerk Maxwell conceived of displacement current as a polarization current which he used to model the magnetic field hydrodynamically and mechanically. He added this displacement current to Ampère's circuital law in his 1861 paper.
13
Gauss’s law
• Gauss's Law is that the [net] electric flux through any closed surface is equal to the charge inside that surface divided by this constant ε.
14
• In free space,
• Since
• The continuity eq. can be derived by taking the Div of (1.1b)
Charge is conserved, or current is continuous since is the outflow of current at a point.
0
0
B H
D E
0 ( )E B Mt
0B
0Jt
J
15
Integral form
0
S V
S
D ds dv Q
B ds
S S S
S S S S
E dl B ds M dst
H dl D ds J ds D ds It t
A sinusoidal E-field in the x direction of the form
ˆ( , , , ) ( , , ) cos( )E x y z t xA x y z t
At t = 0 ˆ( , , ) ( , , ) jE x y z xA x y z e
16
Figure 1.3 (p. 7)The closed contour C and surface S associated with Faraday’s law.
17
( , , , ) Re[ ( , , ) ]j tE x y z t E x y z e
Power and energy
1 1
2 2 3 3
ˆ ( , , ) cos( )
ˆ ˆ( , , ) cos( ) ( , , ) cos( )
E xE x y z t
yE x y z t zE x y z t
In phasor form
31 21 2 3ˆ ˆ ˆ jj jE xE e yE e zE e
2
0
2 2 2 2 2 21 1 2 2 3 30
22 2 21 2 3
1
1[ cos ( ) cos ( ) cos ( )
1 1 1( )
2 2 2
T
T
E E EdtT
E t E t E t dtT
E E E E E E
18
• Assuming an ejωt time dependence,
0
E j B M
H j D J
D
B
19
Figure 1.4a/b (p. 9)Arbitrary volume, surface, and line currents. (a) Arbitrary electric and magnetic volume current densities. (b) Arbitrary electric and magnetic surface current densities in the z = z0 plane.
20
Figure 1.4c/d (p. 9)Arbitrary volume, surface, and line currents. (c) Arbitrary electric and magnetic line currents. (d) Infinitesimal electric and magnetic dipoles parallel to the x-axis.
21
1.3 Fields in Media and Boundary Conditions
• For a dielectric material, an applied E electric dipole moment increase D
where• The ε″ accounts for loss in the medium (heat) due to
damping of the vibrating dipole moment.• The loss of a dielectric material may also be considered
as an equivalent conductor loss.
0
0 0
0 (1 )
e
e
e
D E P
E E
E
E
0 (1 )ej
22
• J = σE Ohm’s law from an EM field point of view
• Loss tangent
( )
H j D J
j E E
j E E
j j j E
tan
0(1 tan ) (1 tan )rj j j
23
• Isotropic material: Pe in the same direction as E.
• Some materials are anisotropic (still linear): crystal structure and ionized gases
• For isotropic material: diagonal ε matrix.
x xx xy xz x x
y yx yy yz y y
z zx zy zz z z
D E E
D E E
D E E
24
For magnetic materials
• Magnetic polarization Pm,
where
Imaginary part of χm or μ: loss due to damping force
• No magnetic conductivity no real magnetic current• Magnetic materials may be anisotropic
0
0
( )
( )m
m
B H P
H H
H
0 (1 )mj
25
If linear media are assumed (ε, μ not depending on E or H)
0
E j H M
H j E J
D
B
D E
B H
ε, μ may be complex and may be tensors possible phase shift between D and E, or B and H.
어떻게 하면 풀수
있을까요 ?
Boundary condition
26
Fields at a General Material Interface
27
Figure 1.5 (p. 12)Fields, currents, and surface charge at a general interface between two media.
28
Figure 1.6 (p. 12)Closed surface S for equation (1.29).
29
S VD ds dv As h 0 2 1
2 1
2 1ˆ ( )
n n s
n n s
s
SD SD S
D D
n D D
Similarly2 1ˆ ˆn B n B
30
Figure 1.7 (p. 13)Closed contour C for Equation (1.33).
31
S V SE dl j B ds M ds
If a magnetic surface current density MS exists on the surface,
( )SM M h
1 2
1 2
2 1 ˆ( )
t t S
t t S
S
lE lE lM
E E M
E E n M
Similarly2 1ˆ ( ) Sn H H J
32
Fields at a Dielectric Interface
1 2
1 2
1 2
1 2
ˆ ˆ
ˆ ˆ
ˆ ˆ
ˆ ˆ
n D n D
n B n B
n E n E
n H n H
Fields at the Interface with a Perfect Conductor (Electric Wall) (MS = 0 assumed)
1
1
ˆ
ˆ 0
ˆ 0
ˆ
S
S
n D
n B
n E
n H J
33
The Magnetic Wall Boundary Condition (JS = 0 assumed)
1
1
ˆ 0
ˆ 0
ˆ
ˆ 0S
n D
n B
n E M
n H
34
• HW1
1. Divergence Theorem 을 증명하고 물리적 의미를 설명하시오 .
2. Stokes’ Theorem 을 증명하고 물리적 의미를 설명하시오 .
3. Ampere 의 법칙과 Gauss 의 법칙에 대하여 설명하시오 .
35
1.4 The Wave Equation and Basic Plane Wave Solutions
The Helmholtz Equation• In a source-free, linear, isotropic, homogeneous region,
2
2 2
2 2
( ) ( 0)
0
E j H
H j E
E j H E
E E E E
E E
Similarly 2 2 0H H
Wave equation or Helmholtz equation
k Wavenumber or propagation constant
36
Plane waves in a lossless medium
• ε and μ: real k: real• Consider only x component and uniform (no variation)
in the x and y direction ( )
• In the time domain, • Phase velocity
• In free-space, vp = c = 2.998x108 m/s
/ / 0x y 2
22
0xx
Ek E
z
( ) ?xE z
( , ) cos(....xE z t E
. 1p
dz d t constv
dt dt k k
37
• The wavelength, λ is defined as the distance between 2 successive maxima on the wave, at a fixed instant of time.
• Magnetic field, (1.41a)
The wave impedance for the plane
(the ratio of the E and H fields)
Ex 1.1
[ ] [ ( )] 2
22 p p
t kz t k z
v v
k f
1[ ]jkz jkz
yH E e E e
/ /k
38
Plane waves in a general lossy medium
E j H
H j E E
2 2 1 0E j E
Define a complex propagation constant
1j j j
22
20x
x
EE
z
Solution?
39
(1 tan ) ( =0)j jk j j
Associated magnetic field
[ ]
1
z zxy
z zy
Ej jH E e E e
z
j
H E e E e
40
Plane waves in a good conductor
• Conductor current >> displacement current (σ >> ωε)
• The skin depth, or characteristic depth of penetration
• Ex 1.2 • The most of the current flow in a good conductor
occurs in an extremely thin region near the surface of the conductor.
(1 )2
j j jj
1 2S
41
• The wave impedance inside a good conductor
1.5 General plane wave solutions
1(1 ) (1 )
2 S
jj j
2 2 22 2 2
0 02 2 20
E E EE k E k E
x y z
2 2 2202 2 2
0i i ii
E E Ek E
x y z
, ,or i x y z
( , , ) ( ) ( ) ( )xE x y z f x g y h z
42
20 0
f g hk
f g h
2 2 2; ;x y z
f g hk k k
f g h
2 2 22 2 2
2 2 20; 0; 0x y z
d f d g d hk f k g k h
dx dy dz
2 2 2 20x y zk k k k
( )( , , ) x y zjk k x k y k zxE x y z Ae
Define a wavenumber vector k
0ˆ ˆ ˆˆx y zk k x k y k z k n
n̂ a unit vector in the direction of propagation
43
• Also define a position vector,
• If
ˆ ˆ ˆr xx yy zz
( , , ) jk rxE x y z Ae
( , , )
( , , )
jk ry
jk rz
E x y z Be
E x y z Ce
0
0
0 0 0
0 0
ˆ ˆ ˆ
( )
0
0
jk r
jk r jk r jk r
jk r jk r
E Ax By Cz
E E e
E E e E e e E
E e jk E e
k E
fA A f f A
44
0x y zk A k B k C
If x y zk k k 0A B C
2 of 3 components can be chosen independently.
0k E
Magnetic field can be found from Maxwell’s eq.
0
00 0
0 00 0
00 0
0 0 0
( )
( )
1 1ˆ ˆ ˆ
jk r
jk r jk r
jk r jk r
E j H
j jH E E e
j jE e E jk
kn E n E n E
45
Figure 1.8 (p. 22)Orientation of the vectors for a general plane wave.0, , H and k k n
46
• The time-domain expression for the E-field can be found as
• Ex 1.3
0
0
( , , , ) Re{ ( , , )
Re{ }
cos( )
j t
jk r j t
E x y z t E x y z e
E e e
E k r t
47
Circularly Polarized Plane Waves
• Plane waves having E-field vector pointing in a fixed direction are called linearly polarized waves.
• Consider the superposition of an x linearly polarized with amplitude E1 and a y linearly polarized with E2.
- If E1 ≠ 0, E2 = 0 linearly polarized in x direction
- If E1 = 0, E2 ≠ 0 linearly polarized in y direction
- If E1 ≠ 0, E2 ≠ 0 (both real), linearly polarized at the angle
01 2ˆ ˆ( ) jk zE E x E y e
1 2
1
tanE
E
48
• If E1 = E2 = E0,
• If E1 = jE2 = E0,
The E-field vector changes with time or equivalently with distance along the z-axis.
• Pick a position (z = 0)
RHCP
00 ˆ ˆ( ) jk zE E x y e
00
0 0 0
ˆ ˆ( )
ˆ ˆ( , ) { cos( ) cos( / 2)}
jk zE E x jy e
E z t E x t k z y t k z
1 sintan
cos
tt
t
49
• Similarly,
LHCP
00 ˆ ˆ( ) jk zE E x jy e
50
Figure 1.9 (p. 24)Electric field polarization for (a) RHCP and (b) LHCP plane waves.
51
1.6 Energy and Power
• In the sinusoidal steady-state case, the time-average stored energy in a volume V is given by, (ε: real)
• Similarly,
• From the Maxwell’s eq.
1Re
4 4e V VW E D dv E E dv
4m VW H H dv
2
2
2 2
( ) | |
( ) | |
| | | |
SH E j H H M
E H E J j E
E J E j E
52
2 2 2
2
2
2 2
( ) ( ) ( )
| | ( | | | | )
( )
( ) | |
( ) | |
| | | |
S
S
E H H E E H
E j E H
E J H M
H E j H H M
E H E J j E
E J E j E
2 2 2
( )
| | ( | | | | )
( )
V S
V V
SV
E H dv E H ds
E dv j E H dv
E J H M dv
,j j
53
2
2 2 2 2
1 1( ) | |
2 2 2
( | | | | ) ( | | | | )2 2
SV S V
V V
E J H M dv E H ds E dv
E H dv j H E dv
Poynting’s theorem
2 ( )S o l m eP P P j W W
where
2 2 2
1( )
21 1
2 2
| | ( | | | | )2 2
S SV
o S S
l V V
P E J H M dv
P E H ds S ds
P E dv E H dv
Poynting vector
54
Figure 1.10 (p. 25)A volume V, enclosed by the closed surface S, containing fields and current sources ss M J
H, E
55
Power absorbed by a Good conductor
• To calculate attenuation and loss due to an imperfect conductor, take the interface between a lossless medium and a good conductor.
• A field is incident from z < 0, and the field penetrates into the conducting region z > 0.
• The real average power entering the conductor volume:
(E & H: fields at the interface)
• The contribution to the integral from S can be made 0 by proper selection of this surface.
0
1ˆRe
2av S SP E H nds
56
Figure 1.11 (p. 26)An interface between a lossless medium and a good conductor with a closed surface S0 + S for computing the power dissipated in the conductor.
57
0
1ˆRe
2ˆ ˆ( ) ( )
av SP E H zds
z E H z E H H H
0
2| |2
1where Re( ) Re (1 )
2 2
Sav S
S
RP H ds
j
RS: the surface resistivity of the conductor
58
1.7 Plane Wave Reflection from a Media Interface
Figure 1.12 (p. 27)Plane wave reflection from a lossy medium; normal incidence.
59
General Medium
• For z < 0,
• In z > 0,
0 00 0
0
1ˆ ˆ,jk z jk z
i iE xE e H y E e
0 00 0
0
ˆ ˆ,jk z jk zr iE x E e H y E e
0 00
ˆ ˆ,z zt i
TE xTE e H y E e
T: Transmission coeff.
1 /
j
j j j
60
• Since Et must be continuous at z = 0,
0
11 ,
TT
0
0 0
2, 1T
61
Lossless Medium
• If the region z > 0 is a lossless dielectric, σ = 0, μ & ε are real. η, Γ, T: real
0
0 0 0 0where
r rj j jk
k
0
0
2 2
1r r
p
r r
r
r
cv
j
62
• For power conservation, in z < 0
• For z > 0,
0 0 0 0
0 0
20
0
2 22 20
0
2 20 0
0
( ) ( )
1ˆ | | ( )( )
1ˆ | | (1 | | )
1ˆ | | (1 | | 2 sin 2 )
i r i r
jk z jk z jk z jk z
jk z jk z
S E H E E H H
z E e e e e
z E e e
z E j k z
2 20
2 2 20 02
0 0
| | | |ˆ
4 1ˆ ˆ| | | | (1 | | )
( )
t t
E TS E H z
z E z E
63
• At z = 0, S- = S+, complex power flow is conserved across the interface.
• For z < 0, the time-average power flow through a 1-m2 cross section,
• Fore z > 0,
2 20
0
1 1 1ˆRe( ) | | (1 | | )
2 2P S z E
2 20
0
1 1 1ˆRe( ) | | (1 | | )
2 2P S z E P
20
0
2 20
0
| |ˆ
| | | |ˆ
i i i
r r r
ES E H z
E TS E H z
i rS S S
64
Good Conductor
• If the region z > 0 is a good conductor,
E and H will be 45o out of phase.• z < 0,
1(1 ) (1 )
2
1(1 ) (1 )
2
s
s
j j j
j j
2 20
0
1ˆ( 0) | | (1 | | )S z z E
65
• z > 0,
• So, at the interface at z = 0, S- = S+, and complex power is conserved.
• Observe
2 220
2 20 2
0
2 2 20
0
| | | |ˆ
4ˆ | |
( )
1ˆ | | (1 | | )
zt t
z
z
E TS E H z e
z E e
z E e
20
0
2 20
0
| |ˆ
| | | |ˆ
i i i
r r r
ES E H z
E TS E H z
66
• The time-average power flows are
power balance at z = 0.• In addition,
2 20
0
1 1 1ˆRe( ) | | (1 | | )
2 2P S z E
2 2 20
0
1 1 1ˆRe( ) | | (1 | | )
2 2zP S z E e
2 2 20 0
0 0
| | | | | |,
2 2i r
E EP P
i rP P P
67
• The electric volume current density flowing in the conducting region is given as:
• The average power dissipated in a 1 m2 cross-sectional volume of the conductor
20ˆ /z
t tJ E x E Te A m
1 1
0 00 0 0
2 22 2 02
0 0
1
21
ˆ ˆ( ) ( )2
1
2 4
tt tV
z z
x y z
z
z
P E J dv
xE Te x E Te dzdydx
E TE T e dz
68
Perfect Conductor
• Assume z > 0 contains a perfect conductor.
• The fields for z > 0 decay infinitely fast, and are identically 0 in the perfect conductor.
• The perfect conductor can be thought of as “shorting out” the incident E-field.
• For z < 0,
; ; 0; 0; 0; 1s T
0 0
0 0
0 0 0
0 0 00 0
ˆ ˆ( ) 2 sin
1 2ˆ ˆ( ) cos
jk z jk zi r
jk z jk zi r
E E E xE e e x jE k z
H H H y E e e y E k z
69
• At z = 0,
• The Poynting vector for z < 0 is
Zero real part: no real power is delivered to the perfect conductor.
• The volume current density
00
2ˆ0,E H y E
20 0 0
0
4ˆ | | sin cosS E H zj E k z k z
0 0 00 00
2 2ˆ ˆ ˆˆ coss
z
J n H z y E k z x E
70
The Surface Impedance Concept
• Particularly where the effect of attenuation or conductor loss is needed, the presence of an imperfect conductor must be taken into account.
• Consider a good conductor in z > 0.• 3 ways to compute the power.
1. Joule’s law (η << η0)
where
2 2 2 2
0 02 2 2
0 00
22
4st s
s
E T E RP
1 1Re( ) Re
2ss s
jR
71
2. Poynting vector
3. Effective surface current density and the surface impedance
so the total current flow per unit width in the x direction
2 2
0 02 20
00
2 Re( ) 21ˆRe( )
2s
z
E E RP S z
20ˆ /z
t tJ E x E Te A m
2000 0
ˆˆ /z
s t
x TEJ J dz x TE e dz A m
72
• For large σ,
• Now replace the exponentially decaying volume current with a uniform volume current extending a distance of one skin depth. Let
0 0 0
2 2(1 ) 2
(1 ) ( ) (1 )s s
s
T j
j j
000 0
0 0
21ˆ ˆ ˆˆ ( ) (1 )s i rz z
EJ n H z H H xE x
/ for 0
0 for s s s
ts
J zJ
z
73
• Find the power lost:
• Power loss can be accurately and simply calculated as
• Ex 1.4
2 22 0
2 200
21
2 2
s s st ssS z S
s
J E RRP dzds J ds
2 2
2 2t s s
s tS S
R RP J ds H ds
74
Figure 1.13 (p. 35)Geometry for a plane wave obliquely incident at the interface between two dielectric regions.
75
Figure 1.14 (p. 38)Reflection coefficient magnitude for parallel and perpendicular polarizations of a plane wave obliquely incident on a dielectric half-space.
76
1.9 Some Useful Theorems
• The Reciprocity Theorem
77
Figure 1.15 (p. 40)Geometry for the Lorentz reciprocity theorem.
78
Image Theory
• In many problems, a current source is located in the vicinity of a conducting ground plane.
• Image theory permits the removal of the ground plane by placing a virtual image source on the other side of the ground plane.
• Consider the surface current density
• A standing wave field in 0 < z < d• Positively traveling wave for z > d
0 ˆS sJ J x
79
Figure 1.16 (p. 42)Illustration of image theory as applied to an electric current source next to a ground plane. (a) An electric surface current density parallel to a ground plane. (b) The ground plane of (a) replaced with image current at z = –d.
80
0 0
0 0
0
( )
( )
jk z jk zsx
jk z jk zsy
E A e e
AH e e
For 0 < z < d,
For z > d, 0
0
0
jk zx
jk zy
E Be
BH e
Boundary Condition Ex = 0 at z = 0
Since MS = 0, sx xz d z d
E E
We have
ˆˆ ( )sS y y z d
J z y H H
002 sin jk djA k d Be
00 0
0 0
2cosjk d
s
B AJ e k d
81
00 00 0 0, sin
2jk ds
s
JA e B jJ k d
Total fields are 0
0
0
0
0 0 0
0 0
0 0 0
0 0
sin
cos
sin
sin
jk dsx s
jk dsy s
jk zx s
jk zy s
E jJ e k z
H J e k z
E jJ k de
H jJ k de
Now apply the image theory to this problem.
By superposition, the total fields for z > 0 can be found by combining the fields from the 2 sources individually.
82
• Fields due to source at z = d,
0
0
0
0
( )0 0
( )0 0
( )0
( )0
for 2
for 2
for 2
for 2
jk z ds
xjk z ds
jk z ds
xjk z ds
Je z d
EJ
e z d
Je z d
HJ
e z d
83
• Fields due to source at z = -d,
0
0
0
0
( )0 0
( )0 0
( )0
( )0
for 2
for 2
for 2
for 2
jk z ds
xjk z ds
jk z ds
xjk z ds
Je z d
EJ
e z d
Je z d
HJ
e z d
84
Figure 1.17 (p. 44)Electric and magnetic current images. (a) An electric current parallel to a ground plane. (b) An electric current normal to a ground plane. (c) A magnetic current parallel to a ground plane. (d) A magnetic current normal to a ground plane.
