Wave Equation and Solution in a Material Medium.ppt

Edward C. Jordan Memorial Offering of the First
Edward C. Jordan Memorial Offering of the First
Course under the Indo-US Inter-University
Course under the Indo-US Inter-University
Collaborative Initiative in Higher Education and
Collaborative Initiative in Higher Education and
Research: Electromagnetics for Electrical and
Research: Electromagnetics for Electrical and
Computer Engineering
Computer Engineering
by
by
Nannapaneni Narayana Rao
Nannapaneni Narayana Rao
Edward C. Jordan Professor of Electrical and Computer Engineering
Edward C. Jordan Professor of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign
University of Illinois at Urbana-Champaign
Urbana, Illinois, USA
Urbana, Illinois, USA
Amrita Viswa Vidya Peetham, Coimbatore
Amrita Viswa Vidya Peetham, Coimbatore
July 10 – August 11, 2006
July 10 – August 11, 2006

4.4
Wave Equation and
Solution for Material
Medium

4.4-3
Waves in Material Media
   
 
 
 
 
,
,
, ,
,
y
x
y x
x
x
y
y
x x x
H z t
E z t
z t
H z t E z t
E z t
z t
E
j H
z
H
E j E j E
z

 

   



 
 
 
 




   


4.4-4
Combining, we get
 
2
2
x
x
E
j j E
z
  

 

 
j j j
     
   
2
2
2
x
x
E
E
z




Define
Then
Wave equation

4.4-5
Solution:
 
   
 
 
 
, Re
Re
Re
cos
cos
z z
x
j t
x x
z z j t
j z j z j t j z j z j t
z
z
E z Ae Be
E z t E z e
Ae Be e
Ae e e e Be e e e
Ae t z
Be t z
 

  
       


  
  
 


 
 

 
 
  
 
 
 
 
 
 
  
  


4.4-6
 = attenuation constant, Np/m
 = phase constant, rad/m
 
  wave
cos
z
Ae t z

  
 

 

        
attenuation
 
  wave
cos
z
Be t z

  

 

       
1
= propagation constant, m
 
attenuation

4.4-7
   
, cos
z
f z t e t z

 

 
2
t 


4
t 


0
t 
z
2



f
1
- 1
0

4.4-8
   
, cos
z
g z t e t z

 
 
0
t 
g
4
t 


2
t 


- z
1
- 1
0
2




4.4-9
1
1
1
x
y
x
y
z z
z z
E
j H
z
E
H
j z
Ae Be
j z
Ae Be
 
 













 
 
 

 
 
 
where intrinsic impedance of the medium.
j
j


 
 


4.4-10
Summarizing,
conversely,
 
j
j j j
j
e
j

     

 
 
   
 

1
Re
1
Im
j
 






 




4.4-11
Example:
5
0 0
For dry earth, 10 s/m, 5 , and .
    

  
Let us compute , , , , and for 100 kHz.
p
v f
    
Solution:
 
5
8
1
1
2
2 10 5 1 0.36
3 10
j j
j j
j
j j
f
j j
   

 


 
 

 
 
 
 
 
 
 
 



4.4-12
0.004683 1.0628 19.8
j
   
0.004683 1.0309 9.9
j
    
 
0.004683 1.0155 0.1772
j j
 
0.00083 0.004756
j
 
0.00083 Np/m

 
0.004756 rad/m
 

4.4-13
5
8
2 10 1.321 10 m/s
0.004756
p
v  


   
2 2 1321.05 m
0.004756
 


  
j
j


 


1
1
j
j j

  




4.4-14
1
1 j j

  



120 1
5 1 0.36
j



 
161.1 28.1
j
  
1
168.6
1.0309 9.9

 
163.55 9.9
  

Wave Equation and Solution in a Material Medium.ppt

More Related Content

Similar to Wave Equation and Solution in a Material Medium.ppt (20)

More from Ankit988352 (12)

Recently uploaded (20)

Wave Equation and Solution in a Material Medium.ppt