Section 4 4 TRƯỜNG ĐIỆN TỪ

Slide Presentations for ECE 329,
Introduction to Electromagnetic Fields,
to supplement “Elements of Engineering
Electromagnetics, Sixth Edition”
by

Nannapaneni Narayana Rao

Edward C. Jordan Professor of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign, Urbana, Illinois, USA
Distinguished Amrita Professor of Engineering
Amrita Vishwa Vidyapeetham, Coimbatore, Tamil Nadu, India


4.4
Wave Equation and
Solution for Material
Medium


4.4-3

Waves in Material Media
H y  z , t 
Ex  z , t 
 
z
t
H y  z , t 
Ex  z , t 
  Ex  z , t   

z
t
Ex
 j H y
z
H y
  Ex  j Ex    j  Ex
z


4.4-4

Combining, we get
2

 Ex
 j   j  Ex
2
z
Define

   j   j   j 
Then

2 Ex
2
Wave equation


E

x
2
z


4.4-5

Solution:

Ex  z   Ae

 z

z

 Be

Ex  z , t  Re  Ex  z   e jt 

 z
z
jt

Re   Ae  Be  e 

Re  Ae e


j    z  j  z

 z

 Ae

e

e

jt

 Be e e j  z e jt 


cos t   z  

j   z





 Be z cos t   z    


4.4-6

Ae  z cos t   z    
      

attenuation   wave

B e z cos t   z    
       

attenuation   wave
 = attenuation constant, Np/m
 = phase constant, rad/m

 = propagation constant, m 1


4.4-7

f  z , t  e

 z

cos t   z 

f
1
0
-1




t
4
t 0
2



t
2

z


4.4-8

g  z , t  e z cos t   z 
g
t
2

- z

t
4

1
t 0

2





0

-1


4.4-9

Ex
 j H y
z
1 Ex
H y 
j z
1 
 Ae  z  Be  z 

j z
1
  Ae  z  Be  z 

j
where  
 intrinsic impedance of the medium.
  j


4.4-10

Summarizing,

   j   j   j 
j

  e 
  j
j

conversely,

1
  
j

 Re

1

  Im




4.4-11

Example:
For dry earth,  10 5 s/m,  50 , and  0 .
Let us compute  ,  , vp ,  , and  for f 100 kHz.

Solution:

  j   j 




 j  j  1 

j



 j  1  j


2 f 

5
2


10
 5 1  j 0.36
j
3 108


4.4-12

 j 0.004683 1.0628  19.8
 j 0.004683 1.0309  9.9

 j 0.004683 1.0155  j 0.1772 
0.00083  j 0.004756
 0.00083 Np/m

 0.004756 rad/m


4.4-13

5

2


10
8
vp  
1.32110 m/s
 0.004756

  2  2
1321.05 m
 0.004756
j

  j

j
1


j
1   j

4.4-14




1


1  j  j

120
5

168.6

1
1  j 0.36

1
1.0309  9.9

163.559.9
161.1  j 28.1 