Page 355 - Electromagnetics

Note that both ±κ satisfy this equation, allowing waves with phase front propagation in
both the ±z-directions.
We see in (4.426) that even for lossless materials certain values of ω result in cos κL > 1,
causing κL to be imaginary and producing evanescent waves. We refer to the frequency
ranges over which cos κL > 1 as stopbands, and those over which cos κL < 1 as passbands.
This terminology is used in filter analysis and, indeed, waves propagating in periodic
media experience effects similar to those experienced by signals passing through filters.

Field produced by an infinite array of line sources. As a second example, consider
an infinite number of z-directed line sources within a homogeneous medium of complex
c
permittivity ˜ (ω) and permeability ˜µ(ω), aligned along the x-axis with separation L
such that
∞

˜ ˜
J(r,ω) = ˆ zI n δ(y)δ(x − nL).
n=−∞
The current on each element is allowed to show a progressive phase shift and attenua-
tion. (Such progression may result from a particular method of driving primary currents
on successive elements, or, if the currents are secondary, from their excitation by an
impressed field such as a plane wave.) Thus we write
˜ I n = I 0 e − jκnL (4.428)
˜
where κ is a complex constant.
We may represent the field produced by the source array as a superposition of the fields
of individual line sources found earlier. In particular we may use the Hankel function
representation (4.345) or the Fourier transform representation (4.407). Using the latter
we have
 
∞+ j
∞ ˜ e − jk y |y|
˜ − jκnL  ω ˜µI 0 (ω) − jk x (x−nL) 
E z (x, y,ω) = e − e dk x .
2π 2k y
n=−∞
−∞+ j
Interchanging the order of summation and integration we have
∞+ j
˜
∞
ω ˜µI 0 (ω) e − jk y |y|
˜ jn(k x −κ)L − jk x x
E z (x, y,ω) =− e e dk x . (4.429)
2π 2k y
n=−∞
−∞+ j
We can rewrite the sum in this expression using Poisson’s sum formula [142].
∞ ∞
1
jnk 0 x
f (x − nD) = F(nk 0 )e ,
D
n=−∞ n=−∞
where k 0 = 2π/D. Letting f (x) = δ(x − x 0 ) in that expression we have
∞
∞
2π L jnL(x−x 0 )
δ x − x 0 − n = e .
L 2π
n=−∞ n=−∞
Substituting this into (4.429) we have
∞+ j
˜
∞
ω ˜µI 0 (ω) e − jk y |y| 2π 2π
˜ − jk x x
E z (x, y,ω) =− δ k x − κ − n e dk x .
2π 2k y L L
n=−∞
−∞+ j
© 2001 by CRC Press LLC