Download Chapter 2 Maxwell`s Equations and Plane EM Waves

Chapter 2 Maxwell’s Equations and Plane EM Waves
2-1 Dielectric and Conductor






Displacement vector: D   0 E  P   E   0 1   e  E   0  r E
nv
P

Polarization vector: P  lim
v 0

k
k 1
v

P a R
2
2
2
V
dv' , R 2  x  x'   y  y '  z  z '
2

v
'
4 0
R
1

    
1 a
'  x
y
z
 '    R2
x'
y '
z '
R R





1 
P

'

P
1
 
V 
P '  dv' 
' dv' 
dv'

v'
v' R


4

R
40 v '
R
 
 
0
 



1

  

1  P a n ' 
 ' P 

dS '  
dv'
v'

40 s ' R
R




Surface charge density: ρps= P a n .

Volume charge density: ρp=    P
 


Total charge: Q=  P a n dS '     P dv'  0
s'
v'

 E 



1
0
   

Define D   0 E  P    D  
p

  
    0 E P   



s
 Dx   11  12
Note: Generally, D    E or  D y    21  22
  
 Dz   31  32

 
 11 0
For biaxial dielectric,    0  22
 0
0

0
0 
 33 

 D d S  Q
 13   E x 
 23    E y  .
 33   E z 
 Dx 
8 2 0  E x 


Eg. For an anisotropic medium characterized by D y   0 2 5 0   E y  ,
 

  
 Dz 
0 0 9  E z 

find the value of the effective relative permittivity for (a) E  zˆE 0 , (b)


E  E 0 ( xˆ  2 yˆ ) , (c) E  E 0 (2 xˆ  yˆ ) .
 Dx 
8 2 0 0
0
0








(Sol.) (a) D y   0 2 5 0  0 E0   0 0 E0  9 0 0 E0 ,  r =9
 

  
 
 
 Dz 
0 0 9 1
9
1
 Dx 
8 2 0  1 
4
1








(b) D y   0 2 5 0   2 E0   0  8 E0  4 0  2 E0 ,  r =4
 

  
 
 
 Dz 
0 0 9  0 
 0 
 0 
 Dx 
8 2 0 2
18
 2








(c) D y   0 2 5 0  1 E0   0 9 E0  9 0 1 E0 ,  r =9
 

  
 
 
 Dz 
0 0 9 0
 0 
0
Hall Effect:


Current density: J  yˆ J 0  Nqv
If the material is a conductor or an n-type semiconductor the charge carrier are
electrons: q < 0

 
Hall field: E h  v  B  ( yˆ v0 )  ( zˆB0 )   xˆv0 B0
d
Hall voltage: Vh    Eh dx  v0 B0 d
0
Hall coefficient: C h 
Ex
1

0
J y B z Nq
If the material is a p-type semiconductor, the charge carries are holes: q > 0

Hall field: E h  xˆv0 B0
Hall voltage: Vh  v0 B0 d
Hall coefficient: Ch  0
2-2 Boundary Conditions of Electromagnetic Fields
Boundary conditions for electric fields:
Eg. Show that Et=0 on the conductor plane.
(Proof) ∵ The E-field inside a conductor is zero,
∴




 E d l
 Et W  0  Et  0
 E d S  En S 
 s S

 En  s
0
0

  
Eg. Show that E1t= E2t and a n 2   D1  D2    s on the interface between two


dielectric.

(Proof)

 E d l
abcda
 E1t W  E 2t W  0 , E1t=E2t





 

D

d
S

D

a

D

a

S

a


1
n
2
2
n
1
n 2  D1  D2 S   s S
s



  
a n 2   D1  D2    s or D1n-D2n=ρs


If ρs=0, then D1n=D2n or ε1E1n=ε2E2n
Eg. Two dielectric media are separated by a charge free boundary. The electric
field intensity in media 1 at the point P1 has a magnitude E1 and makes an angle
with the normal. Determine the magnitude and direction of the electric field
intensity at point P2 in medium 2. [交大電子所]
tan  2  2
(Sol.) E2 s i n 2  E1 s i n1 ,  2 E2 cos 2  1 E1 cos1 

tan 1  1
E2  E22t  E22n 
E2 sin  2 2  E2 cos 2 2
2

 1
 
2
 E1 sin 1    E1 cos 1  

2
 
1/ 2
2

 1
 
2
 E1 sin 1   cos  1  

2
 
1/ 2
Eg. Assume that z=0 plane separates two lossless dielectric regions with εr1=2 and






εr2=3. If E1 in region 1 is x 2 y  y 3x  z 5  z  , find E 2 and D2 at z=0 in
region 2.








(Sol.) E1  x 2 y  y 3x  z 5 , E1t  z  0  E 2t  z  0   x 2 y  y 3x ,




D1n z  0  D2 n z  0  2 E1n z  0  3 E 2 n z  0



 10
2     10
 z 5   z , ∴ E 2 z  0  x 2 y  y 3x  z
3
3 
3

E 2 n z  0 


 10


D2 z  0   x 2 y  y 3x  z 3 0
3

Eg. A lucite sheet (εr=3.2) is introduced perpendicularly in a uniform electric





field E 0  x E 0 in free space. Determine Ei , Di and Pi inside the lucite. [中
央地球物理所]




1

Di  x Di  x D0  x  0 E0
(Sol.)

Ei 
1

Di 

 0 r

Di  x
E0
3.2





1 

Pi  Di   0 Ei  x1 
 0 E 0  x 0.6875 0 E 0
 3.2 
(C / m)
Eg. Dielectric lenses can be used to collimate electromagnetic fields. The left
surface of the lens is that of a circular cylinder, and right surface is a plane. If



E1 at point P(r0,45°,z) in region 1 is ar 5  a 3 , what must be the dielectric

constant of the lens in order that E 3 in region 3 is parallel to the x-axis?



(Sol.) Assume E2  ar E2r  a E2 , ∵ E1t  E 2t  E  E 2  3


For E 3 // x  axis  E 2 // x  axis  E 2   E 2 r




 E2r  3
a n  D1  a n  D2   1 Er1   2 Er 2 ,  0 5   0  r 2 3   r 2 
5
3
Eg. A positive point charge Q is at the center of a spherical dielectric shell of an
inner radius Ri and an outer radius Ro. The dielectric constant of the shell is εr.



Determine E , V , D , and P as functions of the radial distance R. [高考]




(Sol.) P  D   0 E   0  r  1 E
R>Ro:

E  aˆ R

D  aˆ R
Q
40 R
2
, V
Q
40 R

Q
and P  0
2
4R
Ri<R<Ro:

E  aˆ R
V  
Q
40  r R
Ro

R  Ri
Q
40 R 2

R
Q
Ro
4 0  r R 2
dR  
4 0 R 2

R<Ri: E  aˆ R
V V
2
Q



Q
Q
1
ˆ
D

a
,
,
P
 aˆ R 1 
R
2
2
4R
4R
 r
 aˆ R
Q
R
Ri

, D  aˆ R
40 R
2
dR 
dR 
Q 
1
1 
4 0   r
Q 
1
1 
4 0   r
 1 
1

 1 
 Ro   r
Eg. Show that μ1H1n=μ2H2n and aˆ n 2  ( H 1  H 2 )  J .


 B d S  0  B1n S  B2n S  0 , B1n=B2n
 μ1H1n=μ2H2n
 H  dl  I   H  dl  H
1
 w  H 2  (w)  J sww
abcda
 H1t  H 2t  J sw  aˆn 2  ( H 1  H 2 )  J
If J=0, then H1t=H2t
 1
1 



 Ro  r R 

Q
,
P
 0,
4R 2
Boundary conditions for magnetic fields:
(Proof)
 Q

2
 4R
 1 1
  
 Ri R 
Eg. Two magnetic media with permeabilities μ1 and μ2 have a common boundary.
The magnetic field intensity in medium 1 at the point P1 has a magnitude H1 and
makes an angle α1 with the normal. Determine the magnitude and the direction
of the magnetic field intensity at point P2 in medium 2.
 H cos 2  1H1 cos1
tan  2 2
(Sol.)  2 2


tan 1 1
H 2 sin  2  H1 sin 1
  2  tan 1 (
2
tan 1 )
1
H 2  H 22t  H 22n



 ( H 2 sin  2 ) 2  ( H 2 cos 2 ) 2  H1 sin 2 1  ( 1 cos1 ) 2 
2


1
2
Eg. Consider a plane boundary (y=0) between air (region 1, μr1=1) and iron


(region 2, μr2=5000). (a) Assuming B1  0.5 xˆ  10 yˆ (mT), find B 2 and the angle

that B2 makes with the interface. (b) Assuming B2  10 xˆ  0.5 yˆ (mT), find B1

and the angle that B1 makes with the normal to the interface.
(Sol.)


(a) B1  0.5 xˆ  10 yˆ , B2  B2 x xˆ  B2 y yˆ , H 2 x 
B2 x
0.5
 H 1x 
 B2 x  2500
5000 o
o

B

B2 y  B1 y  10  B2  2500 xˆ  10 yˆ , tan  2  2 tan  1  500 1x  250
1
B1 y


B
B
(b) B2  10 xˆ  0.5 yˆ , B1  B1x xˆ  B1 y yˆ , H 1x  1x  H 2 x  2 x
1
 B1x 
1
B2 x 
2

10
 0.002 , B1 y  B2 y  0.5 , ∴ B1  0.002 xˆ  0.5 yˆ ,
5000
 BE
B
0.002
tan  1  1x 
 0.004
B1 y
0.5
Magnetic flux lines round a cylindrical bar magnet:
Eg. A very large slab of material of thickness d lies perpendicularly to uniform

magnetic field intensity H 0  zˆH 0 . Ignoring edge effect, determine the magnetic
field intensity in the slab: (a) if the slab material has a permeability μ, (b) if the
slab permanent magnet having a magnetization vector M i  zˆM i . [台大物研]
(Sol.)
(a) B 2  2 H 2 , B2 z  B1z   H 2  o H o , H 2  zˆH 2  zˆ
o
Ho

(b) B 2   o ( H 2  M i ) , B2 z  B1z   o ( H 2  M i )   o H o  H 2  zˆ( H o  M i )
Eg. Assume that N turns of wire are wound around a toroidal core of a
ferromagnetic material with permeability μ. The core has a mean radius r0, a
circular cross section of radius a (a << r0), and a narrow air gap of length lg, as
shown in Figure. A steady current I0 flows in the wire. Determine (a) the
magnetic flux density Bf in the ferromagnetic core; (b) the magnetic field
intensity Hf in the core; and, (c) the magnetic field intensity Hg in the gap. [台大
電研]
(Sol.)
 H  d l  NI
o
, B f  B g  aˆ  B f ,
C
Bf
 o NI o
 o (2ro  l g )  l g
H g  aˆ 
H
NI o
 o (2ro  l g )  l g
f
Bf

 aˆ 
(2ro  l g ) 
Bf
o
l g  NI o
 o NI o
 o (2ro  l g )  l g
2-3 Steady-state Currents
Differential current: I 

Q Nqv  aˆ n st
 

 Nqv  s
t
t

 


Current density: J  Nqv  v (A/m2), I   J  dS (A)
s
 




Let v  E, J  v  E  E
 : mobility
 : conductivi ty
    e ue   h u h
electrons
holes
Eg. An emf V is applied across a parallel-plate capacitor of area S. The space
between the conducting plates is filled with two different lossy dielectrics of
thicknesses d1 and d2, permittivities ε1 and ε2, and conductivities σ1 and σ2,
respectively. Determine (a) the current density between the plates, (b) the electric
field intensities in both dielectrics. [高考]
(Sol.)
 d
d 
V  R1  R2 I   1  2  I
  1S  2 S 
 1 2V
I
V
J 

S d1  1   d 2  2   2 d1   1 d 2
V  E1 d1  E 2 d 2 , J   1 E1   2 E 2 , E1 
 2V
 2 d1   1 d 2
,
E2 
 1V
 2 d1   1 d 2
Eg. Assume a rectangular conducting sheet of conductivity σ, width a, and height
b. A potential difference is applied to the side edges. Find (a) the potential
distribution, (b) the current density everywhere
within the sheet. [台科大電子所]
(Sol.)
(a) V(x)=Cx, V(a)=Ca=V0  V(x)=
(b) E  V ( x)   xˆ
Vo
x
a
Vo
Vo
 J   E   xˆ
a
a
 

 
dQ
d
Equation of continuity: I   J  dS  
   dv     J dv    J 
0
dt
dt v
t
s



 p
 
 t
 
 0 , ρ=  0 e 
If J  E ,   E 
t
 t
Eg. Lightning strikes a lossy dielectric sphere εr=1.2, σ=10S/m, of radius 0.1m at
time t=0, depositing uniformly in the sphere a total charge 1mC. Determine for
all t for (a) the electric field intensity both inside and outside the sphere, (b) the
current density in the sphere, (c) calculate the time it takes for the charge density
in the sphere to diminish to 1% of its initial value, (d) calculate the charge in the
electrostatic energy stored in the sphere as the charge density diminished from
the initial value to 1% of the value. What happens to this energy? (e) determine
the electrostatic energy stored in the space outside the sphere. Does this energy
charge with time?
(Sol.)  0 
Q
4 3
b
3
 0.239C m  ,    0 e
3


 t
4 3

R 


10
 0 R   t
3
ˆ

a
e
 aˆ R 7.5  10 9 R  e  7.4210 t V m 
a  R  b : E i  aˆ R
R
2
3
4R



Q
Q
 aˆ R 2  10 6 V m 
....R  b : E 0  aˆ R
2
4

R
R

0



10
 7.42101 0 t
b  R  b : J i  E i  aˆ R 7.5  10 Re


....R  b : J 0  0

 t


n100
 0.01  t 
 4.88  10 12 s 
c  e  





0

2

 
 2 t
Wi


2
'
 

 0.01  10  4
d  Wi   E i dv  e
Wi 0
2 v'


 
e  W0  0  E 02 4R 2 dR  4.5  10 3  J 
2 0


Boundary conditions for current densities:
Governing Equations for Steady Current Density
Differential Form

 J  0

J
     0
 

  J  0   J 1n  J 2 n



 J
 J 1t  1
      0  J  
2
 2t
  
Integral Form
 
s J  ds  0
c
1  
J  d  0

Eg. Two lossy dielectric media with permittivities and conductivities (ε1, σ1) and
(ε2, σ2) are in contact. An electric field with a magnitude E 1 is incident from
medium 1 upon the interface at an angle α1 and measured from the common
normal, as in Figure.
(a) Find the magnitude and direction of E 2 in medium 2.
(b) Find the surface charge density at the interface.
(Sol.)
(a) E1t  E 2t => E1 sin 1  E 2 sin  2 , J 1n  J 2n =>  1E1 cos 1   2 E2 cos  2

1
2
2
 E 2  E1 sin  1  ( cos  1 )

2
=> 
2

tan

2 
tan  1

1
 1

(b) D 2 n  D1n  s =>  2 E 2n   1E1n   s , ρs=   2   1  E 1 cos 
 2

Eg. Two conducting media with conductivities σ1 and σ2 are separated by an
interface. The steady current density in medium 1 at point P1 has a magnitude J1
and makes an angle α1 with the normal. Determine the magnitude and direction
of the current density at point P 2 in Medium 2. [台大電研]
(Sol.)
J 1 cos 1  J 2 cos  2 ,  2 J 1 sin 1   1 J 2 sin  2 
J2  J  J
2
2t
2
2n

J 2 sin  2 
2
 J 2 cos  2 
2
tan  2  2

tan 1  1
2
 


2
2



 
J 1 sin 1   J 1 cos 1  
  1




 


 J 1n  J 2n   1 E1n   2 E2n

  s    1 2   2  E2n    1   2 1

D

D




E


E




2n
s
1 1n
2 2n
s
1
2
 1n




 E1n .


1 2
If
 2   1   s   1 E1n  D1n .
2-4 Maxwell’s Equations and Plane EM Waves

D
Note:
is equivalent to a current density, called the displacement current density.
t
Eg. A voltage source V0sin(ωt), is connected across a parallel-plate capacitor C.
Find the displacement current in the capacitor.
dvC
A
 CV0 cos t   V0 cos t
dt
d



V0
D 
A
D  E  D   sin t , i D  
 dS   V0 cos t  iC
d
t
d
A
(Sol.) iC  C

V
Lorentz condition:   A  
=0
t










D

A
V
2 A
2
  B  J  
     A  J   (V  )  (  A)   A  J  ( 
)   2
t
t
t
t
t

2




A

V
  2 A   2   J  (  A  
)
t
t



2 A
If Lorentz Condition holds, we have  A   2   J
t
2





A
∵   D      E       (V  )   2V   (  A)   2V   ( V )   
t
t
t
t

 2V


2

t
Effective permittivity:

 



E
 
H  J 
 E  jE  j ( 
) E  j C E
t
j
∴  2V  
 C    j

  ' j ' '     '' . Similarly,    ' j ' '

 '' 

 ' 
Eg. A sinusoidal electric intensity of amplitude 250V/m and frequency 1GHz
exists in a lossy dielectric medium that has a relative permittivity of 2.5 and loss
tangent of 0.001. Find the average power dissipated in the medium per cubic
meter.
(Sol.)

tan    0.001 
,
 0  r
Loss tangent: tan  C 
10 9
)( 2.5)  1.39  10  4 ( S / m)
36
1
1
1
p  JE  E 2   (1.39  10  4 )  250 2  4.34(W / m 3 )
2
2
2
  0.001(2 10 9 )(
Maxwell’s Equations in the source-free regions:






H
E
,  H  
,  E  0,   H  0
  E  
t
t
3
4
Phasor representations: Eg. xˆAe  j ( z  ) , ( xˆ  yˆ )e  j ( z  ) , etc.
5
5
Instantaneous representations: Eg. xˆA cos(t  z   )  Re[ xˆAe  j ( z  )  e jt ] , etc.





H
jωt
In case E and H are proportional to e , we have   E   
  jH and
t



E
 H  
 jE .
t


Eg. Given that H  yˆ 2 cos(15x) sin( 6 10 9 t  z ) in air, find E and β.

(Sol.) Phasor: H  yˆ 2 cos(15x)e  jz , (15 ) 2   2   2  0  0  400 2    13.2

E

  H  [ xˆ158 cos(15x)  zˆj180 sin( 15x)]e  jz
1
j 0


E ( x, z, t )  Re[ E ( x, z )e jt ]


Eg. Given that E  yˆ 0.1cos(10x) sin( 6 10 9 t  z ) in air, find H and β.

(Sol.) Phasor: E  yˆ 0.1cos(10x)e  jz , (10 ) 2   2   2 0 0  400 2    10 3

H 
1
j 0

E 
j
 0
[ xˆ 0.1 cos(10x )  zˆ0.1(10 ) cos(10x )]e  jz


H ( x, z, t )  Re[ H ( x, z )e jt ]
Eg. The electric field intensity of a spherical wave in free space is

E
E  aˆ R 0 sin  cos(t  kR) . Determine the magnetic field intensity.
R

E
(Sol.) Phasor: E  aˆ R 0 sin   e  jkR
R



E
E
1 
 j0 H    E  aˆ
( RE 0 )  aˆ (  jk ) 0 sin   e  jkR  H  aˆ 0
R R
R
R
0
sin   e  jkR
0
Plane EM waves excited by a current sheet:

Given J (t )   xˆJ (t ) at z=0, the field components of the EM plane wave excited by



z
1
z
the current density are E ( z, t )  xˆ J (t  ) and H ( z, t )   yˆ J (t  ) ,
2
vp
2
vp

respectively. If it is a sinusoidal EM plane wave, J (t )   xˆJ 0 cos(t ) at z=0.


J
J
We have E ( z, t )  xˆ 0 cos(t  kz) , H ( z, t )   yˆ 0 cos(t  kz) .
2
2
Electromagnetic wave spectrum:
2-5 Plane EM waves in a simple, nonconducting and source-free region
In a simple, nonconducting and source-free region:






H
E
,  H  
,  E  0,   H  0
  E  
t
t







2E
2E
2
2
2
    E   (  H )   2  (  E )   E   E   E   2  0 .
t
t
t
1
Velocity of the plane EM wave: v=

In vacuum, μ0=4π×10-7, ε0=
1
1
×10-9  c 
 3  10 8 (m / s) .
36
 0 0
Wave number: k=ω/v=   
2
2

v/ f




Assume E  e jt   2 E  k 2 E  0 (drop ejωt factor)

 


d 2 E( z)
Suppose E  E ( z ) 
 k 2 E  0  E ( z )  E0 e  jkz  E0 e jkz
2
dz
Traveling wave in +z-direction:
E0 ( z, t )  Re[ E0 e  jkz  e jt ]  E0 cos(t  kz)
Let ωt-kz=constant  Phase velocity: vp=
dz 

dt k


If E  xˆE x ( z),   E   j ( xˆH x  yˆH y  zˆH z )
 H x  H z  0, H y ( z )  
1
j
and η0=120π  377Ω in free space.
 ( jk ) E x ( z ) 
1

E x ( z ) , where η=

k


,



TEM waves (Transverse electromagnetic waves): E and H ⊥ direction of
propagation ( â n )
 


 

  jk x jk y  jk z 

E( R)  E( x, y, z)  E0 e x y z  E0 e  jk R  E0 e  jkaˆn R , where R  xˆx  yˆy  zˆz ,

k  aˆ n k , and k x2  k y2  k z2   2 









   E  0    ( E0 e  jkaˆ n R )  e  jkaˆn R   E0  E0  (e  jkaˆn R )  E0  (e  jkaˆn R )




 E0   j ( xˆk x  yˆk y  zˆk z )e  jkaˆ n R   jk ( E0  aˆ n )e  jkaˆn R




aˆ n  E0  0  E0  aˆ n (TE). Similarly,   H  0  H 0  aˆ n (TM)
Relation between E-field and H-field of the plane EM wave:

 
 
 
 
 
1
1

E ( R) 
  H ( R) 
( jk )aˆ n  H ( R)  E ( R)  aˆ n  H ( R) , where η=
k
j
j
 


 
 
 
1
1
1
H ( R)  
  E ( R)  aˆ n  E ( R)  H ( R)  aˆ n  E ( R)  H  aˆ n

j



Eg. The instantaneous expression for the magnetic field intensity of a uniform
plane wave propagating in the +y direction in air is given by


H  zˆ 4  10 6 cos(10 7 t  k 0 y  ) A/m. (a) Determine k0 and the location where
4

H z vanishes at t=3ms. (b) Write the instantaneous expression for E .
(Sol.)   10 7   k 0 


c
10 7 

, aˆ n  yˆ

8
30
3  10


 2n  1
1
(a) cos[( 2n  1) ] =0  10 7   3  10 3  y  
  y  30(3  10 4   n)
2
30
4
2
4





y )
(b) E ( z , t )   0 aˆ n  H ( z, t ) , E ( z, t )   xˆ 480 10 6 cos(10 7 t 
30
4

Eg. A 100MHz uniform plane wave E  xˆE x propagates in the +z direction.
Suppose εr=4, μr=1, σ=0, and it has a maximum value of 10-4V/m at t =0 and


z=0.125m. (a) Write the instantaneous expressions for E and H . (b) Determine

the location where E is a positive maximum when t=10-8sec.
(Sol.) k    0  r  0  r 
4
, aˆ n  zˆ ,  
3
0 r
 60
 0 r

(a) E ( z, t )  xˆE x  xˆ10 4 cos( 2  10 8 t  kz   ) has the maximum in case of
2  10 8 t  kz    0   

4

 E ( z , t )  xˆ10  4 cos( 2 10 8 t 
z ),
6
3
6



1
10 4
4

H ( z, t )  aˆ n  E ( z, t )  yˆ
cos( 2 108 t 
z )

60
3
6
(b) cos( 2n )  1 , 2 10 8 (10 8 ) 
4

13 3n
z max   2n  z max  
3
6
8
2
Polarization of the EM wave: The direction of electric field of the EM wave.
In the following text, we assume all EM waves to be z-propagated if we do not
specify them.
Linear polarizations in the x and the y-direction,


respectively: E  xˆE x e  j ( kz  ) , E  yˆE y e  j ( kz  )
Linear
polarization
in
general
case:

E  xˆE x e  j ( kz  )  yˆE y e  j ( kz  ) , where Ex and Ey are in
phase (we can assume the both to be real).
Right–hand
circular

E  xˆE0 e  j ( kz  )  yˆjE0 e  j ( kz  )
polarization:
Left–hand
circular

E  xˆE0 e  j ( kz  )  yˆjE0 e  j ( kz  )
polarization:
Right–hand
elliptical
polarization:

E  xˆE10 e  j ( kz  )  yˆjE20 e  j ( kz  ) ( E10  E20 )
Left–hand
elliptical
polarization:

E  xˆE10 e  j ( kz  )  yˆ jE20 e  j ( kz  ) ( E10  E20 )
We can receive/transmit linearly-polarized EM waves by a linear
dipole antenna.
We can receive/transmit circularly-polarized EM waves by a
circular reflector antenna.

Instantaneous Expression for E of right–hand elliptical–polarization (drop
phase factor e-jθ):

E ( z, t )  Re[ xˆE10 e  jkz  yˆjE20 e  jkz ]e jt   xˆE10 cos(t  kz)  yˆE 20 sin( t  kz)
 xˆE1 ( z, t )  yˆE 2 ( z, t )
E (0, t )
E (0, t )
E (0, t ) 2 E2 (0, t ) 2
E (0, t )
, sin( t )  2
 cos(t )  1
[ 1
] [
]  1 , t  tan 1 2
E10
E20
E10
E20
E1 (0, t )
1. xˆE x 
1
1
( xˆE x  yˆjE y )  ( xˆE x  yˆjE y ) : A linearly polarized plane wave can be
2
2
resolved into a right –hand and left–hand elliptically- or circularly-polarized waves.
2. xˆE0  yˆ jE0  ( xˆ
E.0  E1
E  E0
E  E1
E  E1
 yˆ j 1
)  ( xˆ 0
 yˆ j 0
):
2
2
2
2
A circularly–polarized plane
elliptically–polarized waves.
wave
can
be
resolved
into
two
opposite
E.1  E2
E  E2
E  E2
E  E2
 yˆj 1
)  ( xˆ 1
 yˆj 1
):
2
2
2
2
An elliptically–polarized plane wave can be resolved into two opposite
circularly–polarized waves.
3. xˆE1  yˆjE2  ( xˆ

Eg. The E field of a uniform plane wave propagating in a dielectric medium is
z
z
given by E (t , z )  xˆ 2 cos(10 8 t 
)  yˆ sin( 10 8 t 
) V/m. (a) Determine the
3
3
frequency and wavelength of the wave. (b) What is the dielectric constant of the
medium? (c) Describe the polarization of the wave. (d) Find the corresponding

H field.

(Sol.) Phasor: E  xˆ 2e  jz / 3  yˆje jz / 3
(a)   108  f  1.59  10 7 Hz , k 
(b) v 
1
3
 
2
 2 3
k

 3  108  1 /  0 0 r   r  3

(c) It is the left–hand elliptically-polarized wave propagating along +z direction.
(d)  
0
120
, aˆ n  zˆ

 0 r
3
 1
 1
H  aˆ n  E  zˆ  ( xˆ 2e  jz /




 H ( z, t )  Re[ H ( z )e jt ] 
3
 yˆje  jz / 3 ) 
3
( yˆ 2e  jz /
120
3
 xˆje  jz / 3 )
3
z
z
[ xˆ sin( 108 t 
)  yˆ cos(108 t 
)]
120
3
3
Eg. Write down the instantaneous expression for the electric- and magnetic-field
intensities of sinusoidal time-varying uniform plane wave propagating in free
space and having the following characteristics: (1) f=10GHz; (2) direction of
propagation is the +z direction; (3) left-hand circular polarization; (4) the initial
condition is the electric field in the z=0 plane and t=0 having an x-component
equal to E0 and a y-component equal to √3E0. [台大電研]

2
 10 2
(Sol.)   2  1010 , v   c  3  10 8  k 
k
3

Phasor: E  xˆAe  j ( kz  )  yˆjAe j ( kz  ) for the left-hand circular polarization

 E ( z, t )  Re[ xˆAe ( jkz  )  yˆjAe j ( kz  ) ]e jt  xˆA cos(t  kz   )  yˆA sin( t  kz   )

z=0 and t=0, E (0,0)  xˆA cos( )  yˆA sin(  ) = xˆE0  yˆ 3E0  θ=tan-1(- √3), A=2E0

2
2
E ( z, t )  xˆ 2 E 0 cos[ 2 1010 t  10 2 z  tan 1 ( 3 )]  yˆ 2 E 0 sin( 2 1010 t  10 2 z  tan 1 ( 3 )]
3
3

 1
2 E0
1
H
aˆ n  E 
zˆ  [ xˆAe  j ( kz  )  yˆjAe j ( kz  ) ] 
[ yˆe  j ( kz  )  xˆje  j ( kz  ) ]
0
0
120


jt
 H ( z, t )  Re[ H ( z )e ]
Application of polarization: Liquid Crystal Display (LCD)
The polarizations of incident lights are synchronized by the rotations of molecules of
liquid crystal, which were controlled by an AC voltage. And then the output polarizer
can block the orthogonally-polarized lights to control the output optical intensities.
  
Poynting vector: P  E  H




 




 B  D  

B ,
  D



(
E

H
)

H

(


E
)

E

(


H
)


H


E

EJ
 E  
 H  J 
t
t
t
t


 ( H )  (E )  
2
 1 2
 1 2
 H 
E
 E  J   (  H )  (  E )  E
t
t
t 2
t 2

 
 
2

 2   2
(
E

H
)

d
S
    ( E  H )dv    ( E  H )dv    E dv
s
t v 2
2
v
v
  
 P  E  H is the electromagnetic power flow per unit area.
∴



Instantaneous power density: P( z, t )  Re[ E ( z )e jt ]  Re[ H ( z )e jt ]



E
1
 j ( z  )
Set E ( z )  xˆE x ( z )  xˆE0 e (  j ) z  H ( z )  [aˆ n  E ( z )]  yˆ 0 e z  e
,




∴ E ( z, t )  Re[ E ( z )e jt ]  xˆE 0 e z cos(t  z )

E
and H ( z, t )  Re[ H ( z )e jt ]  yˆ 0 e z cos(t  z   )






 P( z, t )  E ( z, t )  H ( z, t )  Re[ E ( z)e jt ]  Re[ H ( z)e jt ]
 zˆ
E0
2
2
e 2z [cos   cos( 2t  2z   )]  E0
2

 
1
Average power density: Pav  Re( E  H * )
2
2

E0 2z
1 T
Pav   P( z, t )dt  zˆ
e cos  , where T is the period. And it can be proved that
0
T
2
 1  
Pav  Re( E  H * ) .
2

Eg. Show that P( z, t ) of a circularly–polarized plane wave propagating in a
lossless medium is a constant.
(Sol.) Assuming right–hand circularly–polarized plane wave, aˆ n  zˆ

E ( z, t )  E0 [ xˆ cos(t  z )  yˆ sin( t  z )]

 E
1
H ( z, t )  (aˆ n  E )  0 [ xˆ sin( t  z )  yˆ cos(t  z )]


2



E0
ˆ
P( z , t )  E ( z , t )  H ( z , t )  z

Eg. The radiation electric field intensity of an antenna system is

E  aˆ E  aˆ E , find the expression for the average outward power flow per
unit area.
 1

E
E
(Sol.) aˆ n  aˆ r , H  (aˆ n  E )  (aˆ
 aˆ  )



*

 * 1
E *
2
E
1
1
2
Pav  Re( E  H )  Re[( aˆ E  aˆ E )  (aˆ
 aˆ
)] 
aˆ r ( E  E )
2
2


2

Eg. Find P on the surface of a long, straight conducting wire of radius b and
conductivity σ that carries a direct current I. Verify Poynting’s theorem.


 J

  
I
I
I
I2
(Sol.) J  zˆ 2  E   zˆ
, H  aˆ
 P  E  H  aˆ r

2b
b
b 2
2 2 b 3
 

  P  dS    P  aˆ r dS 
s
s
I2
2 b
2
2
 2b  I 2 (

)  I 2R
2
b
2-6 Plane EM Wave in a Lossy Media






 
  H  J  jE  E  jE  j (  j ) E  j c E ,  c    j   '  j '' .


Similarly,  c   '  j ''
Complex wave number: k c    c . Loss tangent: tan  c   ''  ' 
Propagation constant:   jkc  j  c    j  j  (1 


 12
)
j
E  e z  e  jkc z  e z  e  jz
If the medium is lossless, α=0; else if the medium is lossy, α>0.
Phase constant:  
2

1
1
 2


)  1] 2 ,   
[ 1  ( ) 2  1] 2
2

2


1 
 
 1   
Case 1 Low-loss Dielectric:
,     [(1  ( ) 2 ]

8 
2 
 

[ 1 (


(1  j
)

2
Intrinsic impedance:  c 
Phase velocity: v p 

1
1
1 


[1  ( ) 2 ]

8 
 c

Case 2 Good Conductor:
and c 


 1     
 f ,

2


j
f
 (1  j )
(
)  (1  j )

c


Phase velocity: v p 

2



Skin Depth (depth of penetration):  
For a good conductor,  
1


1



2
1


1
f
.

Eg. E (t , z )  xˆ100 cos(10 7 t ) V/m at z=0 in seawater: εr=72, μr=1, σ=4S/m. (a)
Determine α, β, vp, and ηc. (b) Find the distance at which the amplitude of E is
1% of its value at z=0. (c) Write E(z,t) and H(z,t) at z=0.8m, suppose it propagates
in the +z direction.
(Sol.)   107  , f=5×106Hz, σ/ωε0εr=200>>1, ∴ Seawater is a good conductor in
this case.
(a)   f  8.89 Np / m   ,  c  (1  j )
vp 
f

1

2
 3.53  10 6 m / s ,  
 0.707m ,    0.112m



(b) e z  0.01  z 
1
ln( 100)  0.518m

(c) E ( z, t )  Re[ E ( z )e jt ]  xˆ100e z cos(t  z )
z  0.8m  E (0.8, t )  xˆ100e 0.8 cos(t  0.8 )  xˆ 0.082 cos(10 7 t  7.11)


E (0.8) jt
1
e ]  yˆ 0.026 cos(10 7 t  1.61)
H (0.8, t )  aˆ n  E (0.8, t ) , H (0.8, t )  yˆ Re[ x
c

Eg. The magnetic field intensity of a linearly polarized uniform plane wave
propagating in the +y direction in seawater εr=80, μr=1, σ=4S/m is


H  xˆ 0.1sin( 1010 t  ) A/m. (a) Determine the attenuation constant, the phase
3
constant, the intrinsic impedance, the phase velocity, the wavelength, and the
skin depth. (b) Find the location at which the amplitude of H is 0.01 A/m. (c)
Write the expressions for E(y,t) and H(y,t) at y=0.5m as function of t.
(Sol.) (a) σ/ωε=0.18<<1, ∴ Seawater is a low-loss dielectric in this case.



 83.96 Np / m
c 
(1  j
)  41.8e j 0.0283
2 

2
1 
1

    [(1  ( ) 2 ]  300 , v p   3.33  107 m / s ,    1.19  10  2 m ,
8 


2

 6.67  10 3 m

 

(b) e y 
0.01
1
 y  ln 10  2.74  10  2 m
0.1

(c) H ( y, t )  xˆ 0.1e y sin( 1010 t  y 

3
) , y  0.5,   300


 H (0.5, t )  xˆ5.75  10  20 sin( 1010 t  )
3



aˆ n  yˆ  E (0.5, t )  c aˆ n  H (0.5, t )  zˆ 2.41  10 18 sin( 1010 t   0.0283 )
3
Eg. Given that the skin depth for graphite at 100 MHz is 0.16mm, determine (a)
the conductivity of graphite, and (b) the distance that a 1GHz wave travels in
graphite such that its field intensity is reduced by 30dB.
1
(Sol.) (a)  
 0.16  10 3    0.99  10 5 S / m
f
(b) At f=109Hz,   f  1.98 10 4 Np / m
 30(dB)  20 log 10 e z  z 
1.5
 1.75  10  4 m
 log 10 e
Eg. Determine and compare the intrinsic impedance, attenuation constant, and
skin depth of copper σcu=5.8×107S/m, silver σag=6.15×107S/m, and brass
σbr=1.59×107S/m at following frequencies: 60Hz and 1GHz.
(Sol.)   f ,  
1

, f    ,  c  (1  j )


Copper: 60Hz   c  2.02(1  j )  10 6  ,   1.17  10 2 Np / m ,   8.53  10 3 m
1GHz   c  8.25(1  j )  10 3  ,   4.79  10 5 Np / m ,   2.09  10 6 m
Group velocity: v g 
d
1

d d / d

E (t , z )  E0 c o s 
[ (  )t  (    ) z ]  E0 c o s 
[ (  )t  (    ) z ]
 2E0 cos(t  z ) cos(t  z )
Let t  z =constant  v g 
Eg. Show that v g  v p  
dv p
d
dz 
1
d
1




dt   /  d d / d
and v g  v p  
dv p
d
dv p

d
,   v p  , vg 
 vp  

d
d
dv p
2


∵  
,   2 , d  d  0 
, vg  v p  


d
d
d
(Proof) v p 
An example of longitudinal vp>0 but longitudinal vg=0 in barber’s pole.
Eg. A 3GHz, y-polarized uniform plane wave propagates in the +x direction in a
nonmagnetic medium having a dielectric constant 2.5 and a loss tangent 10-2. (a)
Determine the distance over which the amplitude of the propagating wave will be
cut in half. (b) Determine the intrinsic impedance, the wavelength, the phase
velocity, and the group velocity of the wave in the medium. (c) Assuming



E  yˆ 50 sin( 6 10 9 t  ) V/m at x=0, write the instantaneous expression for H
3
for all t and x.
(Sol.) 10  2 

1
 1    10  2  2  3  10 9 
 10 9  2.5  4.166  10 3

36
1 
It is a low–loss dielectric material:     [1  ( ) 2 ]  99.34rad / m
8 
c 


(1  j
) = 2380.29  

2
1

=0.497, e 0.497d   d  1.395m
2
2 

d
1
(b) v p   1.8973  10 8 m / s , v g 

 1.8975  10 8 m / s

d (d / d )
(a)  




 1

j
j (  0.0016 )
(c) E  yˆ 50e 0.497 x  e 3  H  aˆ n  E  zˆ 0.21e 0.497 x  e 3
t

 H ( x, t )  zˆ0.21e 0.497x  sin( 6 10 9 t  31.6x  0.332 ) A/m
Plasma: Ionized gasses with equal electron and ion densities.
Ionosphere: 50~500 Km in altitude

Simple model of plasma: An electron of charge –e, mass m, position x


d 2x
e 


  e2 
2
 eE  m 2  m x  x 
E  Electric dipole p  ex 
E
dt
m 2
m 2

Ne 2 

∴ Total electric dipole moment: P  Np  
E
m 2

 
p2 
Ne 2 
D   0 E  P   0 (1 
) E   0 (1  2 ) E , where  p 
m 2 0

Ne 2
is the plasma
m 0
angular frequency.
   0 (1 
p2
2
)   0 (1 
fp
2
f2
).
Propagation constant:   j  0  1  (
Intrinsic impedance of the plasma:  c 
fp
f
2
2
)
0
1 (
fp
f
where  0  120 ()
)2
Case 1 f<fp: γ is real,  c is pure imaginary  Attenuation  EM wave is in cutoff.
Case 2 f>fp: γ is pure imaginary,  c is real  EM wave can propagate through the
plasma.