Page 366 - Electromagnetics

˜
˜
source I 0 /2 located at y = h and a line source −I 0 /2 located at y =−h. We solve each
of these problems by exploiting the appropriate symmetry, and superpose the results to
find the solution to the original problem.
For the even-symmetric case, we begin by using (4.407) to represent the impressed
field:
∞+ j
˜ I 0 (ω)
ω ˜µ e − jk y |y−h| + e − jk y |y+h|
˜ i 2 − jk x x
E (x, y,ω) =− e dk x .
z
2π 2k y
−∞+ j
For y > h this becomes
∞+ j
˜ I 0 (ω)
ω ˜µ 2 cos k y h
˜ i 2 − jk y y − jk x x
E (x, y,ω) =− e e dk x .
z
2π 2k y
−∞+ j
The secondary (scattered) field consists of waves propagating in both the ±y-directions:
∞+ j
1
˜ s + − jk y y − jk y y − jk x x
E (x, y,ω) = A (k x ,ω)e + A (k x ,ω)e e dk x . (5.7)
z
2π
−∞+ j
s
i
The impressed field is even about y = 0. Since the total field E z = E + E must be
z
z
even in y (E z is parallel to the plane y = 0), the scattered field must also be even. Thus,
A = A and the total field is for y > h
+
−
∞+ j
˜
1 I 0 (ω) 2 cos k y h
˜ + − jk y y − jk x x
E z (x, y,ω) = 2A (k x ,ω) cos k y y − ω ˜µ e e dk x .
2π 2 2k y
−∞+ j
˜
Nowthe electric field must obey the boundary condition E z = 0 at y =±d. However,
˜
since E z is even the satisfaction of this condition at y = d automatically implies its
satisfaction at y =−d. So we set
∞+ j
1 ˜ I 0 (ω) 2 cos k y h − jk y d − jk x x
+
2A (k x ,ω) cos k y d − ω ˜µ e e dk x = 0
2π 2 2k y
−∞+ j
and invoke the Fourier integral theorem to get
˜ I 0 (ω) cos k y h e − jk y d
+
A (k x ,ω) = ω ˜µ .
2 2k y cos k y d
The total field for this case is
∞+ j
˜ I 0 (ω)
ω ˜µ e − jk y |y−h| + e − jk y |y+h|
˜ 2
E z (x, y,ω) =− −
2π 2k y
−∞+ j
2 cos k y h e − jk y d − jk x x
− cos k y y e dk x .
2k y cos k y d
For the odd-symmetric case the impressed field is
∞+ j
˜ I 0 (ω)
ω ˜µ e − jk y |y−h| − e − jk y |y+h|
˜ i 2 − jk x x
E (x, y,ω) =− e dk x ,
z
2π 2k y
−∞+ j
© 2001 by CRC Press LLC