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The Fourier integral theorem implies that the integrand is zero. Then, expanding out
˜
the ρ derivatives, we find that E z (ρ, ω) obeys the ordinary differential equation
2 ˜ ˜
d E z 1 dE z 2 ˜
+ + k E z = 0
dρ 2 ρ dρ
where k = ω/v. This is merely Bessel’s differential equation (A.124). It is a second-order
equation with two independent solutions chosen from the list
J 0 (kρ), Y 0 (kρ), H (1) (kρ), H (2) (kρ).
0 0
We find that J 0 (kρ) and Y 0 (kρ) are useful for describing standing waves between bound-
(1) (2)
aries while H (kρ) and H (kρ) are useful for describing waves propagating in the
0 0
ρ-direction. Of these, H (1) (kρ) represents waves traveling inward while H (2) (kρ) repre-
0 0
sents waves traveling outward. Concentrating on the outward traveling wave we find
that
π (2)
˜
˜
˜
E z (ρ, ω) = A(ω) − j H 0 (kρ) = A(ω)˜ g(ρ, ω).
2
˜
Here A(t) ↔ A(ω) is the disturbance waveform, assumed to be a real, causal function.
To make E z (ρ, t) real we require that the inverse transform of ˜ g(ρ, ω) be real. This
requires the inclusion of the − jπ/2 factor in ˜ g(ρ, ω). Inverting we have
E z (ρ, t) = A(t) ∗ g(ρ, t) (2.347)
(2)
where g(ρ, t) ↔ (− jπ/2)H (kρ).
0
The inverse transform needed to obtain g(ρ, t) may be found in Campbell [26]:
ρ
π ρ U t −
−1 (2) v
g(ρ, t) = F − j H 0 ω = ,
2 v 2 ρ 2
t −
v 2
where U(t) is the unit step function defined in (A.5). Substituting this into (2.347) and
writing the convolution in integral form we have
∞ U(t − ρ/v)
E z (ρ, t) = A(t − t ) dt .
2
2
t − ρ /v 2
−∞
The change of variable x = t − ρ/v then gives
∞
A(t − x − ρ/v)
E z (ρ, t) = dx. (2.348)
2
0 x + 2xρ/v
Those interested in the details of the inverse transform should see Chew [33].
As an example, consider a lossless medium with µ r = 1,
r = 81, and a waveform
A(t) = E 0 [U(t) − U(t − τ)]
where τ = 2 µs. This situation is the same as that in the plane wave example above,
except that the pulse waveform begins at t = 0. Substituting for A(t) into (2.348) and
using the integral
dx √ √
√ √ = 2ln x + x + a
x x + a
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