Practical_Antenna_Handbook_0071639586

458 P a r t V I : A n t e n n a s f o r O t h e r F r e q u e n c i e s
Solution
(a) Phase velocity
cc
c
vv
p p= v
p
=
sin sin aa
a
×
= 3 10 8
3 10 8
×
= 3 10 8
m / m / ss/
s
sin sin 42°
42°
×
= 3 10 8
3 10 8
×
= 3 10 8
m / m / ss/
s
0.6991
0.6991
8 8 8
= 4.5 = 4.5 ×
10 10 × 10 m / m / ss/ s
(b) Wavelength in free space
λ
fs fs
= c // ff
88 99
= (3 × 10 m // s) s) //(5.6 ×
10 Hz)
= 0.054 m
(c) Wavelength in waveguide
v
pλo
λ =
c
8
(4.5×
10 m / s) (0.054 m)
= = 0.08 m
8
3×
10 m / s
Comparing, we find that the free-space wavelength is 0.054 m, while the wavelength
inside of the waveguide has increased to 0.08 m.
Cutoff Frequency (f c )
The propagation of signals in a waveguide is a function of the frequency of the applied
signal. As the frequency drops, the angle of incidence increases toward 90 degrees.
Indeed, both phase and group velocities are functions of the angle of incidence.
When the frequency drops to a point where the angle of incidence is 90 degrees, then
group velocity is meaningless and propagation ceases. This occurs at the low frequency
cutoff.