Transcript lect2-02.ppt

Wave Equation
    E   E  (  E )   j  H 
2
 j ( J  jE )
For a Source free medium :
2 E  k 2 E  0
 H k H 0
2
2
;
;
k 2   2 
k  /v 
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
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Plane Waves
2
2
2

E

E

E
2
2
 E  k 0 E  0  2  2  2  k 02 E
x
y
z
 2 Ei  2 E i  2 Ei
2


 k 0 Ei  0
, i  x, y , z
2
2
2
x
y
z
Solve for E x ( x, y , z ), Using separation of variables 
k 2x  k y2  k z2  k 02
 jks x  jk y y  jkz z
E x  Ae
 


r  ax x  a y y  az z
 


, Let k  a x k x  a y k y  a z k z
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E x  Ae  jk r ,
E  E 0 e  jk  r
Similarly E y  Be  jk r ,
E z  Ce  jk r
Since   E  0  k  E 0  0
The vector E 0 is perpendicu lar to the direction of propagatio n k.
The solution is called plane wave
  E   j 0 H
H 
1
j 0
  E 0 e  jk  r 
1
j 0
E 0   e  jk  r 
1
 0
k  E 0 e  jk  r
0
1

nE 
n  E  Y0 n  E 
nE
 0
 0
0
k0
 0 is the interensic impedance of free space  377
Y0 is the intrinsic admittance of free space.
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z
E
H
n
y
x
H is perpendicular to E and to n. (TEM waves)
E  Re( E0 e  jk r  jt )  E0 cos( k  r  t ), if E0 is real
k 0 0  2 ,
*
1
1
1
*
*
P  Re E  H  n  Re Y0 E  ( n  E )  n  Y0 E0  E0
2
2
2
The time average energy densities in the electric and
magnetic fields of a TEM wave are :
*
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1
*
U e   0 E0  E0
, U m  0 H 0  H 0  U e
2
2
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