Page 300 - Electromagnetics

Reflection of time-domain uniform plane waves. Solution for the fields reflected
and transmitted at an interface shows us the properties of the fields for a certain single
excitation frequencyand allows us to obtain time-domain fields byFourier inversion.
Under certain conditions it is possible to do the inversion analytically, providing physical
insight into the temporal behavior of the fields.
As a simple example, consider a perpendicularly-polarized, uniform plane wave incident
from free space at an angle θ i on the planar surface of a conducting material (Figure 4.18).
The material is assumed to have frequency-independent constitutive parameters ˜µ = µ 0 ,
˜ = , and ˜σ = σ. By(4.285) we have the reflected field
r
˜ r
˜ i
˜
˜ r
E (r,ω) = ˆ y ⊥ (ω)E (ω)e − jk (ω)·r = ˆ yE (ω)e − jω ˆ k r ·r (4.300)
c
⊥ ⊥
˜ ˜ i
˜ r
where E = ⊥ E . We can use the time-shifting theorem (A.3)to invert the transform
⊥
and obtain

ˆ r
k · r
r −1 ˜ r
r

E (r, t) = F E (r,ω) = ˆ yE t − (4.301)
⊥ ⊥
c
where we have bythe convolution theorem (12)
r −1 ˜ r

E (t) = F E (ω) = ⊥ (t) ∗ E ⊥ (t).
Here
−1 ˜ i

E ⊥ (t) = F E (ω)
⊥
is the time waveform of the incident plane wave, while
−1 ˜
⊥ (t) = F ⊥ (ω)
is the time-domain reflection coefficient.
ˆ r
By(4.301) the reflected time-domain field propagates along the direction k at the
speed of light. The time waveform of the field is the convolution of the waveform of
the incident field with the time-domain reflection coefficient ⊥ (t). In the lossless case
(σ = 0), ⊥ (t) is a δ-function and thus the waveforms of the reflected and incident fields
are identical. With the introduction of loss ⊥ (t) broadens and thus the reflected field
waveform becomes a convolution-broadened version of the incident field waveform. To
understand the waveform of the reflected field we must compute ⊥ (t). Note that by
choosing the permittivityof region 2 to exceed that of region 1 we preclude total internal
reflection.
We can specialize the frequency-domain reflection coefficient (4.283) for our problem
bynoting that

√ σ 2
2
k 1z = β 1 cos θ i , k 2z = k − k 2 = ω µ 0 + − 0 sin θ i ,
2 1x
jω
and thus
η 0 η 0
Z 1⊥ = , Z 2⊥ = ,
cos θ i σ 2
r + − sin θ i
jω 0
√
where r = / 0 and η 0 = µ 0 / 0 . We thus obtain
√ √
s − Ds + B
˜
⊥ = √ √ (4.302)
s + Ds + B
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