3.4 Uniform Plane Waves in Time Domain in Free Space

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```					Edward C. Jordan Memorial Offering of the First
Course under the Indo-US Inter-University
Collaborative Initiative in Higher Education and
Research: Electromagnetics for Electrical and
Computer Engineering

by

Nannapaneni Narayana Rao
Edward C. Jordan Professor of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign
Urbana, Illinois, USA

Amrita Viswa Vidya Peetham, Coimbatore
July 10 – August 11, 2006
3.4

Uniform Plane Waves in
Time Domain in Free Space
3.4-2

Infinite Plane Current Sheet Source:

JS  JS t  ax
for z  0

Example: JS t   JS 0 cos t ax
3.4-3

B                     D
×E                ×H  J +
t                     t

For a current distribution having only an x-component of
current density that varies only with z,

ax   ay    az            ax    ay    az
     B                       D
0    0                 0    0       J+
z    t                 z     t
Ex   Ey    Ez            Hx    Hy   Hz
3.4-4

Ey   B x                            H y     Dx
                                       Jx 
z     t                            z         t
Ex    B y                           H x D y
                                    
z     t                             z   t
Bz                                  Dz
0                                   0
t                                   t
The only relevant equations are:
Ex    B y               H y       Dx
                        Jx 
z     t                 z         t
Thus, E  Ex  z, t  ax    H  H y  z, t  a y
3.4-5

In the free space on either side of the sheet, Jx = 0
Ex    B y        H y             H y  Dx      Ex
        0                            0
z     t          t               z    t       t
2 Ex         H y 
Combining, we get         0        
z 2
z  t 
  H y 
  0        
t  z 
     Ex 
  0   0     
t     t 
2 Ex        2 Ex
 0 0 2     Wave Equation
z 2
t
3.4-6

Solution to the Wave Equation:

               
Ex  z , t   Af t  z 0 0  Bg t  z 0 0       
Ex
 A      f  t  z   
z
0 0                      0 0

 B   g t  z   
0 0                 0 0

2 Ex
    Af   t  z     g   t  z    
z 2
0 0
                     0 0            0 0

2 Ex
 0 0 2
t
3.4-7

     z          z 
Ex  z, t   Af  t    Bg  t  
    vp         vp 
                  
1
Where vp               3  108 m/s = c, velocity of light
00


f t   z vp  represents a traveling wave propagating in the
+z-direction.


g t   z vp  represents a traveling wave propagating in the
–z-direction.
3.4-8

Examples of Traveling Waves:

f  t  z vp    t  z 5 
2

f

1
t0       t
5

1
25

z
1       0                1     2

1
vp      5 m/s
15
3.4-9

g  t  z vp   e
 2t  z         2 t  z 2
e
g

1
1
t
2
t0

z
3      2       1       0           1          2

1
vp      2 m/s
12
3.4-10

Ex        H y
From              ,
z      0 t
H y     1 Ex

t     0 z
1                 z            z 
          Af   t    Bg   t   

0 vp   
          vp 


    vp  

1           z           z 
H y  z , t    Af    t     Bg  t  
0         vp          vp 
                       
where 0     0  0  Intrinsic impedance
 120  377 ohms
3.4-11

Thus, the general solution is
     z          z 
Ex  z, t   Af  t    Bg  t  
    vp         vp 
                  

1           z           z 
H y  z , t    Af    t     Bg  t  
0         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,
 A
 Af  t  z vp  ax for z  0                       f  t  z vp  a y for z  0
                                                  
H  z, t    0
E  z, t   
 Bg  t  z vp  ax for z  0

 B g  t  z vp  a y for z  0
 0

3.4-12

Applying Faraday’s law in integral form
to the rectangular closed path abcda is
the limit that the sides bc and da0,

Lim
 b E d l  d E d l          Lim     d
 a       c
bc 0                                 bc 0           B dS
da 0                     
          da 0    dt abcda

 ab  Ex z0   dc   Ex z0   0

Af t   Bgt   say, F t 
3.4-13

Therefore,                                  
z
E  z, t   F  t               ax for z     0
         vp     
                
 1              z   
H  z, t    F  t                   a y for z   0
0 
               vp   

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
 f H d l  h H d l
         g
fg  0
he  0  e                  

Lim                     d                   
   fg  0
he  0
 efghe J d S  dt

efghe D d S

3.4-14

ef H y z 0  hg H y z 0  ef  JS t 
                  

1            1      
F t    F t   J S t 
0           0     
0
F t          J S t 
2
Thus, the solution is
0       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.
3.4-15
x

y          z

0           z                          0           z 
E  z, t    J S  t   ax              E  z, t    J S  t   ax
2       vp 
                            2        vp 
       

1         z                             1         z 
H  z, t    J S  t   a y            H  z, t    J S  t   a y
2       vp 

2       vp 

JS  t 

z= 0
3.4-16

z< 0                  z> 0   x

z
JS  t       y

z
z = 0
3.4-17

a Ex t for z  300 m

b Hy t for z  450 m
3.4-18

c Ex  z for t  1s

d Hy  z for t  2.5s

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