EELE 3332 – Electromagnetic II

Chapter 10

Electromagnetic Wave
Propagation
Islamic University of Gaza
Electrical Engineering Department
Dr. Talal Skaik

2012 1
 Electromagnetic wave propagation

•A changing magnetic field produces an electric field, and a

changing electric field produces a magnetic field.
•Accelerating charges produce electromagnetic waves.
•The existence of EM waves, predicted by Maxwell’s equations, was
first investigated by Heinrich Hertz (sometimes called Hertzian
waves).
In general, waves are means of transporting energy or
information.
Typical examples of EM waves include radio waves, TV signals,
radar,…
2
 WAVES

Mechanical waves Electromagnetic waves

Transverse waves Longitudinal waves Transverse waves

Longitudinal Waves: Vibration is parallel to the direction of

propagation. Sound and pressure waves are longitudinal.

Transverse Waves: the motion of the matter particles is

perpendicular to the direction of propagation of the wave itself.

3
 Mechanical Waves
• A material medium is necessary for the transmission for
mechanical waves. Mechanical waves cannot travel through
vacuum.
• Disturbance is transmitted from one layer to the next through
the medium.

Transverse
Wave
(in string)

Longitudinal
Wave
(in spring)
4
 Electromagnetic Waves
•Material medium is not essential for propagation. Em waves travel
through vacuum.
•Disturbance of electric and magnetic fields travelling through space.
•All electromagnetic waves are transverse waves.

• Electromagnetic waves travel through empty space!!

• All electromagnetic waves travels at c = 3108 m/s in vacuum (speed
of light).
• Electric field, magnetic field, and direction of travel are mutually
perpendicular. 5
 Electromagnetic waves

Goal in this chapter: Solve Maxwell’s equations and describe EM

wave motion in the following media:
1. Free Space ( =0,  = 0 ,  =0 )
2. Lossless Dielectrics ( =0,  = r 0 ,  =r 0 or    )
3. Lossy Dielectrics (  0,  = r 0 ,  =r 0 )
4. Good Conductors ( ,  = 0 ,  =r 0 or    )

where  is the angular frequency of the wave

6
 Electromagnetic waves

The wave E  Asin(t   z ) has the following characteristics:

 It varies with both time and space.

 It is time harmonic.
 The amplitude of the wave is A.
 The phase of the wave (in radians) is the term (ωt-βz), depends
on time t and space variable z.
 ω is the angular frequency in (radians/second).
 β is the phase constant, or wave number in (radians per meter).

7
 E  Asin(t   z )

Wave takes distance λ to Wave takes time T to

repeat itself. repeat itself.
λ is called the wavelength T is called the Period
(in meters) (in seconds)

8
Since it takes time T for the wave to travel distance  at the speed u,
  uT
1
( u speed of the wave, depends on the medium u  )

But T  1 / f , where f is the frequency of the wave in Hertz (Hz).
u  f

Since   2 f , and  =
u
2 f 2
  
f 
2
 (rad/m)

This shows that for every wavelength of distance traveled,
a wave undergoes phase change of 2 radians.
9
 Consider a fixed point P on
the wave.
 Sketch E=A sin(ωt-βz) at
times t=0, T/4, and T/2.
 It is evident that as the wave
advances with time, point P
moves along +z direction.
So, the wave E=A sin(ωt-βz) is
travelling with a velocity u in
the +z direction.

10
Point P is a point of constant phase, therefore

t   z  constant
dz 
z  u
dt 

Notes:
A sin(ωt - βz) is wave propagating in +z direction (forward travelling,
or positive-going wave)
A sin(ωt + βz) is wave propagating in -z direction (backward travelling,
or negative going wave)

11
12
13
 Example 10.1
An Electric field in free space is given by

E  50cos(108 t   x) a y V/m

(a) Find the direction of wave propagation.

(b) Calculate β and the time it takes to travel a distance of λ/2.
(c) Sketch the wave at t=0, T/4, and T/2.

(a ) The wave is propagating along - a x direction.

  108 1
(b) in free space u  c,  =    rad/m
u c 3  10 8
3
If T is the period of the wave, it takes T seconds to travel
a distance  at speed c. Hence to travel a distance  /2 will take:
T 1 2 
t1     8  31.42 ns.
2 2 f 2 10 14
Example 10.1 E  50cos(108 t   x) a y V/m

(c ) At t=0, E y  50cos  x
T 1 2
At t= = = ,
4 4 f 4
 2 
E y  50cos  .   x
 4 
 50cos   x   / 2   50sin  x
T 1 2
At t= = = ,
2 2 f 2
 2 
E y  50cos  .   x
 2 
 50cos   x     50cos  x

Notice that the wave travels along  a x 15

 10.3 Wave Propagation in Lossy Dielectrics
A lossy dielectric is a medium in which an EM wave, as it propagates,
loses power owing to imperfect dielectric. (partially conducting medium
with σ≠0)
Consider a linear, isotropic, homogeneous, LOSSY dielectric medium
that is charge free ( v =0). Maxwell's equations in phasor form are:
  Es  0 (1)
  Hs  0 (2)
  E s   j H s (3)
  H s    j  E s (4)
Taking the curl of both sides of equation (3) gives:
    E s   j    H s  (3)
Since     A  (  A)   2 A
 (  E s )   2 E s   j   j  E s

 2E s   2E s  0 , where  2  j   j 

16
 Wave Propagation in Lossy Dielectrics
2E s   2E s  0 where  2  j   j 
 is called the propagation constant of the medium.
By similar procedure, it can be shown that for the H-Field,  2 H s   2 H s  0
2E s   2E s  0
(Vector Wave Equations)
 Hs   Hs  0
2 2

Since  is a complex quantity, let     j

 2  j   j     j    j   j 
2

  2   2  j 2  j   2 
  2   2   2  , 2   (solve for  and  )

    
2
 1 
     1
2     
 

    
2
 1 
     1
2     
  17
 Wave Propagation in Lossy Dielectrics

• α is attenuation constant (Np/m): defines the rate of decay of the

wave in the medium. measured in Nepers per meter (Np/m).

• α=0 for lossless medium (σ=0)

• An attenuation of 1 neper indicates a reduction of
e-1 of the original value.
• (1 Np=20 log10e=8.686 dB).

• β is phase constant (rad/m) : is a measure of the phase shift per

unit length in radians per meter. (also called wave number)

2 
 
 u

18
 Wave Propagation in Lossy Dielectrics
Assume the wave propagates along  a z and E s has only x  component
, then E s  E xs ( z ) a x
Since  2 E s   2 E s  0 ( 2 E s Vector Laplacian )
 2 Exs ( z )  2 Exs ( z )  2 Exs ( z )
Hence     2
Exs ( z )  0
x 2
y 2
z 2

 2 Exs ( z )
or   2
Exs ( z )  0
z 2

This is a scalar wave equation, a differential equation with solution:

Exs ( z )  Eo e  z  E0' e z (where Eo and E0' are constants)
( Second part is zero since we assumed wave traveling along +a z ).


 E( z , t )  Re  Exs ( z )e jt a x   Re E0e  z e j (t   z ) a x 

 E( z, t )  E0e  z cos(t   z ) a x 19
 Wave Propagation in Lossy Dielectrics

 z
E( z, t )  E0e cos(t   z ) a x
A Sketch of |E| at times t=0 and t=Δt is shown

Notice E has only x-component and it is travelling in the +z direction.

20
H(z , t ) can be obtained as:

 
H( z , t )  Re H 0e  z e j (t   z ) a y , H0 
E0

where  is complex quantity known as intrinsic impedance, in ohms
of the medium.
j j
      e 
  j

 / 
with   , tan2 
     2 1/4 
1    
    

 E0  z j (  t   z )

 H= Re  e e ay 
  e j 
 
E0
or H( z, t )  e  z cos(t   z   ) a y
 21
 E( z, t )  E0e  z cos(t   z ) a x
E0
H( z, t )  e  z cos(t   z   ) a y

• Notice that E and H are out of phase by Ѳη at any instant of time.
Thus , E leads H (or H lags E) by Ѳη.
• The ratio of the magnitude of the conduction current density Jc to
that of the displacement current density Jd in a lossy medium is

J cs  Es 
   tan 
J ds j Es 

or tan  

Where tanѲ is known as the loss tangent and Ѳ is the loss angle of the
medium.
22
tanѲ is used to determine how lossy the medium is:
Good (lossless or perfect) dielectric if tanѲ is vary small (σ<<ωε)
Good conductor if tanѲ is very large (σ>>ωε)
• Behaviour of a medium depends not only on parameters σ, ε, and µ,
but also on the frequency. A medium regarded as a good conductor
at low frequencies may be a good dielectric at high frequencies.
 
Since tan   and tan2     2
 
 j 
Since   H s    j  E s  j 1   E s =j c E s
  
    
where ,  c   1  j     j ,  c   ' j '' with  '  ,  '' 
    
 c is called the complex permittivity of the medium.
Notice that the ratio of  '' to  ' is the loss tangent of the medium.
 '' 
tan  = 
 '  23
10.4 Plane waves in lossless dielectrics
In a lossless dielectric, σ<<ωε
(special case of section 10.3), except that
 0,  = 0 r ,  =0r

 2  j   j  , Since  0   2   2 

  j     j 
   0,    
 1 2
u  , =
  
j
Also, Since   and  0
  j
 o
  0

and thus E and H are in time phase with each other. 24
 10.5 Plane waves in free space
In free space
  0,  = 0 ,  =0

This may be regarded as special case of section 10.4.


   0,    0 0 
c
1 2
u  c, =
0 0 
where c=3  108 m/s, is the speed of light in a vaccum.
0 o
  0  0  120 377 
0
0 is called the intrinsic impedance of free space.
25
 Plane waves in free space
If E=E0cos( t- z) a x
then H=H 0cos( t- z) a y
E0
H cos( t- z) a y

In general, if a E , a H , and a k are
unit vectors along the E field,
H field, and the direction of
ave propagation:
ak  aE  a H
a k  a H  a E
aE  aH  ak
Plots of E and H (a) as functions of z at t  0; and (b)
at z  0. The arrows indicate instantaneous values. 26
 Wave representation
Plane Wave and its representation Circular Wave and its representation

27
Plane waves in free space

28
 Plane waves in free space
 Both E and H fields (or EM waves) are everywhere normal to the
direction of wave propagation, ak.
 They form an EM wave and have no electric or magnetic field
components along the direction of propagation. Such a wave is
called a transverse electromagnetic (TEM) wave.
 A combination of E and H is called a uniform plane wave
because E (or H) has the same magnitude throughout any
transverse plane, defined by z=constant.
 Uniform plane waves serve as approximations to practical waves
such as those from a radio antenna a distance sufficiently far from
radiating sources.
29
 Plane Waves in Good Conductors
A perfect or good conductor is one in which σ>>ωε
 ,  = 0 ,  =0 r

 1 j  
 2  j   j  j    j =    = (1  j )
 2  2

      f 
2

 2 2
u  , =
  
j
Also, Since   and  
  j
j 
   45o Thus E leads H by 45o .
 
30
 Plane Waves in Good Conductors
If E  E0e z cos(t   z) a x
E0
then H e z cos(t   z  45o ) a y


Therefore, as the wave travels in a conducting medium, it
amplitude is attenuated by a factor e-αz.
The distance δ, through which the wave amplitude decreases to a
factor e-1 (about 37% of the original value) is called skin depth or
penetration depth of the medium.
E0 e   E0e 1
1
  (general)

The skin depth is a measure of the depth to which an EM
wave can penetrate the medium. 31
 Plane Waves in Good Conductors

Skin depth
illustration

Since for good conductors  =  f 

11
  
  f 
1  z / z
Since  =   E  E0 e cos(t  ) ax
  32
 Plane Waves in Good Conductors
Skin Depth in Copper
Frequency (Hz) 10 60 100 500 104 108 1010

Skin depth (mm) 20.8 8.6 6.6 2.99 0.66 6.6 x 10-3 6.6 x 10-4

 For copper, σ=5.8x107 S/m, µ= µ0,

  66.1/ f (in mm)

 The skin depth decreases with increasing frequency. Thus, E and H can
hardly propagate through good conductors.

The fields and currents are confined to a very

thin layer (the skin) of the conductor surface.
 For a wire of radius a, it is a good
approximation at high frequencies to assume
that all of the current flows in the circular ring of  << a
thickness δ. (as shown)
33
34
 Plane Waves in Good Conductors

Skin effect : is the tendency for high-frequency currents to flow

on the surface of a conductor.
* The effective conductor cross section decreases and the
conductor resistance increases.

It is used to advantage in many applications:

Since the skin depth in silver is very small, silver plating is often
used to reduce the material cost of waveguide components. (e.g
silver-plating on brass).
Hollow conductors are used instead of solid conductors in
outdoor television antennas, and thus saving weight and cost.
35
 Plane Waves in Good Conductors
l
The dc resistance is given by: Rdc 
S
The surface or skin resistance R s (in ) is given by:
1  f
Rs   (real part of  for a good conductor)
 
The ac resistance R ac is calculated by using the dc formula with S  w,
where w is the width:
l Rs l
Rac  
 w w
For a conductor of radius a, w  2 a. So,
l
Rac  2 a a a
    f 
Rdc l 2 2
 a 2
 At high frequencies, Rac is far greater than Rdc 36
 Example 10.2
A lossy dielectric has an intrinsic impedance of 200300at
a particular radian frequency . If, at that frequency, the plane
wave propagating through the dielectric has the magnetic field
component:
 x  1 
H  10e cos  t  x  a y A/m
 2 
find E and  . Determine the skin depth and wave polarization.

37
 Example 10.2 - Solution
The given wave travels along a x so that
ak  a x ; a H  a y , so
 a E  ak  a H  a x  a y  a z
a E  a z
Eo
H o  10,so    200 30o  200e j /6  Eo  2000e j /6
Ho

Except for the amplitude and phase difference, E and H always

have the same form. Hence
E  Re  2000e j /6e x e jt a E 
 x  x 
E  2e cos  t    a z kV/m
 2 6 38
 Example 10.2 - Solution
Knowing that  =1/2 , we need to determine  . Since

         
2 2
 
   1    1 and    1    1
2 
     2 
    
1/2
   
2 
 1    1
     
 
    
2 
 1    1
    

But  tan 2 =tan 60o  3 . Hence

  2  1 
1/2
1 1
        0.2887 Np/m
  2  1 3 3 2 3
1
= 2 3

The wave has an Ez component; hence it is polarized along the z-direction.
39
 Example 10.3
In a lossless medium for which  = 60 , r =1 and
H =  0.1 cos (t  z) ax + 0.5 sin (t  z)a y A/m,
calculate  r ,  , and E.
Solution
In this case ,  =0 ,  =0 , and  =1, so
o r 120 120 120
=  /    or r    2  r  4
o  r r  60
 2
      o o r r  4
c c
 c 1 3  10 
8

or    1.5  108 rad/s

2 2
40
 Example 10.3 - Solution
From the given H field, E can be calculated in two ways: using the
techniques (based on Maxwell's equations) developed in this chapter
or directly using Maxwell's equations as in Chapter 9.
Method 1 : To use the techniques developed in this chapter, we let
H = H1 + H 2
where
H1 =  0.1 cos ( t  z) a x and H 2 = 0.5 sin ( t  z)a y
and the corresponding electric field

E = E1 + E2
where
E1 =E1o cos (t  z) a E1 and E2 = E 2o sin ( t  z)a E2
41
 Example 10.3 - Solution
Notice that although H has components along a x and a y , it has no component
along the direction of propagation; it is therefore a TEM wave.
For E1
 
a E1   a k  a H1    az  a x   a y
E1o   H1o  60 (0.1)  6
Hence
E1 = 6 cos t  z  a y
For E2
 
a E2   a k  a H 2    a z  a y   a x
E2o   H 2o  60 (0.5)  30
Hence
E2 = 30 sin t  z  a x
Adding E1 and E2 gives E ; that is,
E =94.25 sin (1.5  108 t  z) a x + 18.85 cos (1.5  108 t  z) a y V/m 42
 Example 10.3 - Solution
Method 2 : We may apply Maxwell's equations directly.
E 1
H =E   E     H dt
t 
where   0.
ax ay az
   H y H x
But   H   ax  ay
x x z z z
H x ( z) H x ( z) 0

= 0.5 cos ( t  z) a x  0.1 sin ( t  z) a y

Hence
1 0.5 0.1
E
    Hdt 

sin ( t  z) a
x  sin ( t  z) a y

= 94.25 sin ( t  z) a x + 18.85 cos ( t  z)a y V/m 43

 Example 10.4
A uniform plane wave propagating in a medium has
E  2e  z sin(108 t   z )a y V/m. If the medium is characterized by
 r = 1,r = 20, and  = 3 mhos/m, find  ,  , and H.
Solution
We need to determine the loss tangent to be able to tell whether the medium is
a lossy dielectric or a good conductor.
 3
 9
 3393  1
 10
10  1 
8

36
showing that the medium may be regarded as a good conductor at the frequency
1/2
 4  10  20(10 )(3) 
7 8
of operation. Hence,  = =    61.4
2  2 
  61.4 Np/m ,   61.4 rad/m
44
 Example 10.4 - Solution
1/2
  4  10  20(10 ) 
7 8
800
Also, |  |   
  3  2
 
tan 2   3393    45 
o

 4
Hence
 
 z
H  H o e sin  t   z   a H
 4
a H  a k  a E  a z  a y  a x
and
Eo 3
Ho  2  69.1  103
| | 800
3 61.4 z  8 
Thus H  69.1  10 e sin  10 t  61.42 z   a x mA/m
 4 45
 Example 10.5
A plane wave E = Eocos (t   z) ax is incident on a good
conductor at z = 0. Find the current density in the conductor.
Solution
Since the current density J =  E , we expect J to satisfy the wave equation
(2 E s   2 E s  0), that is,
2 J s   2 J s  0
Also the incident E has only an x-component and varies with z.
Hence J = J x (z, t) a x and
d2
2
J sx   2
J sx  0
dz
which is an ordinary differential equation with solution
J sx  Ae  z  Be  z
46
 Example 10.5 - Solution
The constant B must be zero because J sx is finite as z  .
But in a good conductor , >> so that    =1/ .

Hence  =  j   1  j  
1  j

and
J sx  Ae  z (1 j )/
or
J sx  J sx (0) Ae  z (1 j )/
where J sx (0) is the current density on the conductor surface.

47
 Example 10.6
For the copper coaxial cable shown, let a = 2 mm, b = 6 mm, and t = 1 mm.
Calculate the resistance of 2 m length of the cable at dc and at 100 MHz.

Solution : Let R = Ro + Ri
where Ro and Ri are the resistances of the inner and outer conductors.
At dc,
2
Ri     2.744 m
 S  a 2
5.8  10  (2  10 )
7 3 2

Ro   
S  [b  t ]  b 
2 2
 t 2  2bt 
2
  0.8429 m
5.8  10  1  12   10
7 6

Hence R dc  2.744  0.8429  3.587 m

48
 Example 10.6 - Solution
At f=100 MHz ,
 f
Ri     (S  w)
 S  w  2 a 2 a 
2   108  4  107
  0.41 
2  2  10 3
5.8  10 7

Since  = 6.6  m << t = 1 mm ,  = 2 b for the outer conductor. Hence,

 f 2   108  4  107
Ro    0.1384 
2 b  2  6  10 3
5.8  10 7

Rac  0.41  0.1384  0.5484 

which is about 150 times greater than R dc . Thus, for the same effective
current i, the ohmic loss (i 2 R ) of the cable at 100 MHz is far greater than
the dc power loss by a factor of 150.
49