### No Slide Title

```Fundamentals of Electromagnetics:
A Two-Week, 8-Day, Intensive Course for
Training Faculty in Electrical-, Electronics-,
Communication-, and Computer- Related
Engineering Departments
by
Nannapaneni Narayana Rao
Edward C. Jordan Professor Emeritus
of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign, USA
Distinguished Amrita Professor of Engineering
Amrita Vishwa Vidyapeetham, India
Amrita Viswa Vidya Peetham, Coimbatore
August 11, 12, 13, 14, 18, 19, 20, and 21, 2008
4-1
Module 4
Wave Propagation in
Free Space
Uniform Plane Waves in Time Domain
Sinusoidally Time-Varying Uniform Plane Waves
Polarization
Poynting Vector and Energy Storage
4-2
Instructional Objectives
11. Obtain the electric and magnetic fields due to an infinite
plane current sheet of an arbitrarily time-varying uniform
current density, at a location away from it as a function of
time, and at an instant of time as a function of distance, in
free space
12. Find the parameters, frequency, wavelength, direction of
propagation of the wave, and the associated magnetic (or
electric) field, for a specified sinusoidal uniform plane
wave electric (or magnetic) field in free space
13. Write expressions for the electric and magnetic fields of a
uniform plane wave propagating away from an infinite
plane sheet of a specified sinusoidal current density, in
free space
4-3
Instructional Objectives (Continued)
14. Obtain the expressions for the fields due to an
array of infinite plane sheets of specified
spacings and sinusoidal current densities, in free
space
15. Write the expressions for the fields of a uniform
plane wave in free space, having a specified set
of characteristics, including polarization
16. Find the power flow associated with a set of
electric and magnetic fields
4-4
Uniform Plane Waves
in Time Domain
(FEME, Secs. 4.1, 4.2, 4.4, 4.5; EEE6E, Sec. 3.4)
4-5
Infinite Plane Current Sheet Source:
JS  JS t  ax
for z  0
Example: JS t   JS 0 cos t ax
4-6
B
×E  
t
D
×H  J +
t
For a current distribution having only an x-component of
current density that varies only with z,
ax
ay
0
0
Ex
Ey
az

B

z
t
Ez
ax
ay
0
0
Hx
Hy
az

D
J+
z
t
Hz
4-7
Ey
H y
Bx


z
t
Dx

 Jx 
z
t
B y
Ex

z
t
H x D y

z
t
B z
0
t
The only relevant equations are:
B y
Ex

z
t
Thus, E  Ex  z, t  ax
D z
0
t
H y
Dx

 Jx 
z
t
H  Hy  z, t  ay
4-8
In the free space on either side of the sheet, Jx = 0
B y
H y
Ex

  0
z
t
t
H y
Dx
Ex

 0
z
t
t
2 Ex
  H y 
  0 
Combining, we get

2
z
z  t 
  H y 
  0 

t  z 
Ex 

  0   0

t 
t 
2 Ex
2 Ex
 0 0 2
2
z
t
Wave Equation
4-9
Solution to the Wave Equation:

 

 A      f  t  z   
 B   g t  z   
    Af   t  z     g   t  z    


Ex  z , t   Af t  z 0 0  Bg t  z 0 0
Ex
z
0 0
0 0
0 0
2 Ex
z 2
0 0
2 Ex
 0 0 2
t
0 0
0 0
0 0
4-10


z 
z 
Ex  z , t   Af  t    Bg  t  




v
v
p
p




Where vp 

f t
1
00
 3  108 m/s = c, velocity of light
z vp  represents a traveling wave propagating in the
+z-direction.

g t
z vp  represents a traveling wave propagating in the
–z-direction.
4-11
Examples of Traveling Waves:
f  t  z vp    t  z 5 
2
f
t
t0
1
5
1
25
1
0
1
2
1
vp 
 5 m/s
15
z
4-12
g  t  z vp   e
 2t  z
2 t  z 2
e
g
1
t
3
2
1
2
t0
1
0
1
1
vp 
 2 m/s
12
2
z
4-13
H y
Ex
From
 
,
0 t
z
H y
1 Ex

t
0 z
1

0 vp



z 
z 
 Af   t    Bg   t   
vp 
vp  



1 
H y  z , t    Af
0 
where 0 


z 
z 
 t     Bg  t  
vp  
vp 


0  0  Intrinsic impedance
 120  377 ohms
4-14
Thus, the general solution is


z 
z 
Ex  z , t   Af  t    Bg  t  


vp 
vp 


1 
H y  z, t    Af
0 


z 
z 
 t     Bg  t  
vp  
vp 


For the particular case of the infinite plane current sheet in the
z = 0 plane, there can only be a () wave for z > 0 and a ()
wave for z < 0. Therefore,

 Af  t  z vp  ax for z  0
E  z, t   

 Bg  t  z vp  ax for z  0
 A
  f  t  z vp  a y for z  0

H  z, t    0
 B g  t  z vp  a y for z  0

 0
4-15
Applying Faraday’s law in integral form
to the rectangular closed path abcda is
the limit that the sides bc and da0,
Lim
bc 0
da 0
 b E d l  d E d l 
c
 a

Lim
bc 0
da 0
d
 
B dS
dt abcda
 ab Ex z0   dc  Ex z0
0
Af t   Bgt   say, Ft 
4-16
Therefore,

E  z, t   F  t


z
vp

H  z, t    F  t
0 
1

 ax for z

z
vp
0

 a y for z

0
Now, applying Ampere’s circuital law in integral form to the
rectangular closed path efgha is the limit that the sides fg and
he0,
Lim
fg 0
he 0
 f H d l  h H d l
g
 e


Lim
fg 0
he0
d

 efghe J d S  dt

D
d
S
efghe

4-17
ef H y  z 0  hg H y  z 0  ef  JS t 
 1

F t    F t   J S t 
0
 0

1
F t  
Thus, the solution is
0
2
J S t 

z 
E  z, t  
JS  t
ax for z 0

2 
vp 
1 
z 
H  z, t    JS  t
a y for z 0

2 
vp 
Uniform plane waves propagating away from the sheet
to
either side with velocity vp = c.
0
4-18
x
z
y

z 
E  z, t  
J S  t   ax
2 
vp 

z 
E  z, t  
J S  t   ax
2 
vp 
1 
z 
H  z, t    J S  t   a y
2 
vp 
1 
z 
H  z, t    J S  t   a y
2 
vp 
0
0
JS t 
z=0
4-19
z<0
z>0
JS t 
z
z = 0
x
y
z
4-20
a Ex t for z  300 m
b Hy t for z  450 m
4-21
c Ex  z for t  1s
d Hy  z for t  2.5s
4-22
Sinusoidally Time-Varying
Uniform Plane Waves
Secs. 4.1, 4.2, 4.4, 4.5; EEE6E, Sec. 3.5)
4-23
Sinusoidal function of time
4-24
Sinusoidal Traveling Waves
f  t  z vp 
 cos   t  z vp     
 cos t   z    
g  t  z vp 
 cos   t  z vp     
 cos t   z    
where    vp   00
4-25
f  z, t   cos t   z 

t
4
f
t

2
1
t 0
0


2

1
z
4-26
g  z, t   cos t   z 
t

2
t

4
g
1
t 0
z


2
0

1
4-27
For JS t   JS 0 cos t ax for z  0,
The solution for the electromagnetic field is
E
0 J S 0
=
0 J S 0
2
2
cos w  t
z vp  ax for z
cos t
 z  ax for z
JS 0
H
cos   t
2
JS 0

cos t
2
0
0
z vp  a y for z
 z  a y for z
where   w vp  w 00
0
0
4-28
Parameters and Properties
1. t
 z     Phase, 

2.   radian frequency =
t
 rate of change of phase with time
for a fixed value of z. (movie)

f 
 frequency
2
= number of 2 radians of phase change
per sec.
4-29
3.   phase constant =

z
= magnitude of rate of change of phase with
distance z for a fixed value of t. (still photograph)

4. vp  phase velocity =

 velocity with which a constant phase progresses
along the direction of propagation.
follows from d t
 z   0
5.  = wavelength =
2
4-30

 distance in which the phase changes by 2
for a fixed t.
6. Note that
 2 f
vp  
f
 2 
  in m  f in MHz = 300
Ex
Ex
7. 0     
Hy
Hy
= Ratio of the amplitude of E to the amplitude
of H for either wave.
4-31
8. E × H (Poynting Vector, P)
ax × a y  az for (+) wave

ax × a y  az for () wave
is in the direction of propagation.
x
x
E
E
z
H
y
P
P
y
H
z
4-32
Example:
Consider E  37.7 cos  6  108 t  2 z  a y V m.
Then

  6  10 , f 
 3  108 Hz
2
2
  2 ,  
 1m

8
6  108
vp 
 3  108 m s
2
Direction of propagation is –z.
H  0.1 cos  6  108 t  2 z  ax A m
4-33
Array of Two Infinite Plane Current Sheets
JS1
z0
JS 2
z 4
JS1   JS 0 cos t ax for z  0
JS 2   JS 0 sin t ax for z   4
For JS1 ,
0 J S 0
 2 cos t   z  ax for z  0
E1  
0 J S 0 cos t   z  a for z  0
x
 2
4-34
For JS 2 ,
0 J S 0

 


sin t    z    ax for z 

4 
4

 2

E2  
0 J S 0 sin t    z     a for z  

 x

 2
4
4




0 J S 0



 2 sin  t   z  2  ax for z  4




0 J S 0 sin  t   z    a for z  

 x
 2
2
4


0 J S 0
 2 cos t   z  ax for z  4

 0 J S 0 cos t   z  a for z  
x

2
4
4-35
For both sheets,
E = E1  E2
 E1    E2 
for z   4
z 0
z  4


  E1 z 0   E2 z  4 for 0  z   4

 E1 z 0   E2 z  4 for z   4
0 J S 0 cos t   z  ax for z   4

 0 J S 0 sin t sin  z ax for 0  z   4
0
for z  0

No radiation to one side of the array.
1-36
Polarization
(FEME, Sec. 1.4, 4.5; EEE6E, Sec. 3.6)
1-37
Sinusoidal function of time
1-38
Polarization is the characteristic which describes
how the position of the tip of the vector varies
with time.
Linear Polarization:
Tip of the vector
describes a line.
Circular Polarization:
Tip of the vector
describes a circle.
1-39
Elliptical Polarization:
Tip of the vector
describes an ellipse.
(i) Linear Polarization
F1  F1 cos (t   ) a x
Magnitude varies
sinusoidally with time
Direction remains
along the x axis
 Linearly polarized in the x direction.
1-40
Linear polarization
1-41
F2  F2 cos (t   ) a y
Magnitude varies
sinusoidally with time
Direction remains
along the y axis
 Linearly polarized in the y direction.
If two (or more) component linearly polarized vectors
are in phase, (or in phase opposition), then their sum
vector is also linearly polarized.
Ex: F  F1 cos (t   ) a x  F2 cos (t   ) a y
1-42
Sum of two linearly polarized vectors in phase
is a linearly polarized vector
1-43
y
F2

F

F1
x
F2 cos (t   )
–1
 tan
F1 cos (t   )
F2
–1
 tan
F1
 constant
(ii) Circular Polarization
If two component linearly polarized vectors are
(a) equal to amplitude
(b) differ in direction by 90˚
(c) differ in phase by 90˚,
then their sum vector is circularly polarized.
1-44
4-45
Example:
F  F1 cos t ax  F1 sin t a y
F 
 F1
cos t    F1 sin t 
2
2
 F1 , constant
  tan
 tan
1
1
F1 sin t
F1 cos t
 tan t   t
y
F2
F

F1
x
4-46
(iii) Elliptical Polarization
In the general case in which either of (i) or (ii) is not
satisfied, then the sum of the two component
linearly polarized vectors is an elliptically polarized
vector.
Ex: F  F1 cos t a x  F2 sin t a y
y
F2
F
F1
x
1-47
Example: F  F0 cos t a x  F0 cos (t   4)a y
y
F0
F2
F
/4
F1 F0 x
–F0
–F0
4-48
D3.17
F1  F0 cos  2 108t  2 z  ax
F2  F0 cos  2 108t  3 z  a y
F1 and F2 are equal in amplitude (= F0) and differ in
direction by 90˚. The phase difference (say ) depends
on z in the manner –2z – (–3z) = z.
(a) At (3, 4, 0),  = (0) = 0.
 F1  F2 
is linearly polarized.
(b) At (3, –2, 0.5),  = (0.5) = 0.5 .
F1  F2  is circularly polarized.
4-49
(c) At (–2, 1, 1),  = (1) = .
 F1  F2 
is linearly polarized.
(d) At (–1, –3, 0.2) =  = (0.2) = 0.2.
 F1  F2 
is elliptically polarized.
4-50
Power Flow
and Energy Storage
(FEME, Sec. 4.6; EEE6E, Sec. 3.7)
Power Flow and Energy Storage
  E x H)  H  x E) – E  x H)
A Vector Identity
D
B
– H
  Ex H  –E  J – E 
t
t
For J = J 0  J c  J 0   E,
2   1
2 
1


E  J0   E    E     H   (E x H)
t 2
t 2
2
Define P  E x H
Poynting Vector
41
Poynting’s Theorem
  E  J 0  dv   E dv 
2
V
V
 1 E 2 dv   1 H 2 dv  P  dS



 t  2
 t  2
V
Power dissipation
density
Source
power density
V
S
Magnetic stored
energy density
Electric stored
energy density
42
Interpretation of Poynting’s Theorem
Poynting’s Theorem says that the power delivered to the volume V
by the current source J0 is accounted for by the power dissipated in
the volume due to the conduction current in the medium, plus the time
rates of increase of the energies stored in the electric and magnetic
fields, plus another term, which we must interpret as the power carried
by the electromagnetic field out of the volume V, for conservation
of energy to be satisfied. It then follows that the Poynting vector P has
the meaning of power flow density vector associated with the electromagnetic field. We note that the units of E x H are volts per meter
times amperes per meter, or watts per square meter (W/m2) and do
indeed represent power density.
43
The End
```