Page 347 - Electromagnetics

hence by the Fourier integral theorem

˜
ω ˜µ 1 I(ω) − jk y1 h
A 1 (k x ,ω) − A 2 (k x ,ω) = e .
2k y1
˜
The boundary condition on the continuity of H x yields similarly
˜ I(ω) k y1 k y2
− e − jk y1 h = A 1 (k x ,ω) + A 2 (k x ,ω).
2 ω ˜µ 1 ω ˜µ 2
We obtain
˜
ω ˜µ 1 I(ω) − jk y1 h
A 1 (k x ,ω) = R TM (k x ,ω)e ,
2k y1
˜
ω ˜µ 2 I(ω) − jk y1 h
A 2 (k x ,ω) =− T TM (k x ,ω)e .
2k y2
Here R TM and T TM = 1 + R TM are reflection and transmission coefficients given by
˜ µ 1 k y2 − ˜µ 2 k y1
R TM (k x ,ω) = ,
˜ µ 1 k y2 + ˜µ 2 k y1
2 ˜µ 1 k y2
T TM (k x ,ω) = .
˜ µ 1 k y2 + ˜µ 2 k y1
These describe the reflection and transmission of each component of the plane-wave
spectrum of the impressed field, and thus depend on the parameter k x . The scattered
fields are
∞+ j
˜
ω ˜µ 1 I(ω) e − jk y1 (y+h)
˜ s − jk x x
E (x, y,ω) = R TM (k x ,ω)e dk x , (4.411)
z1
2π 2k y1
−∞+ j
∞+ j
˜
ω ˜µ 2 I(ω) e jk y2 (y−hk y1 /k y2 )
˜ s − jk x x
E (x, y,ω) =− T TM (k x ,ω)e dk x . (4.412)
z2
2π 2k y2
−∞+ j
We may now obtain the field produced by an electric line source above a perfect

2
2
conductor. Letting ˜σ 2 →∞ we have k y2 = k − k →∞ and
2 x
R TM → 1, T TM → 2.
With these, the scattered fields (4.411) and (4.412) become
∞+ j
˜
ω ˜µ 1 I(ω) e − jk y1 (y+h)
˜ s − jk x x
E (x, y,ω) = e dk x , (4.413)
z1
2π 2k y1
−∞+ j
˜ s
E (x, y,ω) = 0. (4.414)
z2
Comparing (4.413) to (4.410) we see that the scattered field is exactly the same as that
˜
produced by a line source of amplitude −I(ω) located at y =−h. We call this line source
the image of the impressed source, and say that the problem of two line sources located
© 2001 by CRC Press LLC