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3.4 Uniform Plane Waves in Time Domain in Free Space

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3.4 Uniform Plane Waves in Time Domain in Free Space Powered By Docstoc
					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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