Practical_Antenna_Handbook_0071639586

460 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
Equation (20.7) assumes that the dielectric inside the waveguide is air. A more generalized
form, which can accommodate other dielectrics, is
1
f c
=
2 µε
⎛ m⎞
⎜ ⎟
⎝ a ⎠
2 2
⎛ n
+ ⎜
⎞ ⎝ b⎠ ⎟
(20.10)
where e = dielectric constant
m = permeability constant
For air dielectrics, m = m 0 and e = e 0 , from which
1
c =
µ
0ε0
(20.11)
To determine the cutoff wavelength, we can rearrange Eq. (20.10) to the form:
λ C
=
⎛ m⎞
⎜ ⎟
⎝ a ⎠
2
2 2
⎛ n
+ ⎜
⎞ ⎝ b⎠ ⎟
(20.12)
One further expression for air-filled waveguide calculates the actual wavelength in
the waveguide from a knowledge of the free-space wavelength and actual operating
frequency:
λ
g
=
1
λ
0
⎛ f ⎞
⎜ ⎟
⎝ f ⎠
–
c
2
(20.13)
where l g = wavelength in waveguide
l 0 = wavelength in free space
f c = waveguide cutoff frequency
f = operating frequency
Example 20.4 A waveguide with a 4.5-GHz cutoff frequency is excited with a 6.7-GHz
signal. Find (a) the wavelength in free space and (b) the wavelength in the waveguide.
Solution
(a)
λ
o
= c / f
8
3×
10 m / s
=
9
10 Hz
6.7 GHz ×
1GHz
×
= 3 10 8
m / s
9
6.7 × 10 Hz
= 0.0448 m