Page 274 - Electromagnetics

By (4.228) we know that α is even in ω and β is odd in ω. Since the derivative of an

odd function is an even function, we also know that β is even in ω. We can therefore
write (4.235) as
1 ∞ 1 − jβ(ω 0 ) ˆ k·r − jβ (ω 0 )(ω−ω 0 ) ˆ k·r
−α(ω 0 ) ˆ k·r
˜

ˆ eE(r, t) = ˆ eE 0 e F(ω − ω 0 )e e +
2π −∞ 2

˜

e
+ F(ω + ω 0 )e jβ(ω 0 ) ˆ k·r − jβ (ω 0 )(ω+ω 0 ) ˆ k·r e jωt dω.
Multiplying and dividing by e jω 0 t and rearranging, we have
1 ∞ 1
˜
jφ j(ω−ω 0 )[t−τ]
ˆ eE(r, t) = ˆ eE 0 e −α(ω 0 ) ˆ k·r F(ω − ω 0 )e e +
2π −∞ 2
˜
e
+ F(ω + ω 0 )e − jφ j(ω+ω 0 )[t−τ] dω
where
ˆ
ˆ
φ = ω 0 t − β(ω 0 )k · r, τ = β (ω 0 )k · r.

Setting u = ω − ω 0 in the first term and u = ω + ω 0 in the second term we have
1 ∞ ju(t−τ)
˜
−α(ω 0 ) ˆ k·r
ˆ eE(r, t) = ˆ eE 0 e cos φ F(u)e du.
2π
−∞
Finally, the time-shifting theorem (A.3) gives us the time-domain wave field
ˆ
ˆ

ˆ eE(r, t) = ˆ eE 0 e −α(ω 0 ) ˆ k·r cos ω 0 t − k · r/v p (ω 0 ) f t − k · r/v g (ω 0 ) (4.236)
where
v g (ω) = dω/dβ = [dβ/dω] −1 (4.237)
is called the group velocity and
v p (ω) = ω/β
is called the phase velocity.
To interpret (4.236), we note that at any given time t the field is constant over the
surface described by
ˆ
k · r = C (4.238)
where C is some constant. This surface isa plane, as shown in Figure 4.10, with its
ˆ
normal along k. It is easy to verify that any point r on this plane satisfies (4.238). Let
ˆ
ˆ
r 0 = r 0 k describe the point on the plane with position vector in the direction of k, and
let d be a displacement vector from this point to any other point on the plane. Then
ˆ
ˆ ˆ
ˆ
ˆ
k · r = k · (r 0 + d) = r 0 (k · k) + k · d.
ˆ
But k · d = 0,so
ˆ
k · r = r 0 , (4.239)
which is a fixed distance, so (238) holds.
Let us identify the plane over which the envelope f takes on a certain value, and follow
its motion as time progresses. The value of r 0 associated with this plane must increase
with increasing time in such a way that the argument of f remains constant:
t − r 0 /v g (ω 0 ) = C.

© 2001 by CRC Press LLC