Page 490 - Engineering Electromagnetics, 8th Edition

472 ENGINEERING ELECTROMAGNETICS

Note, for example, that in an air-filled guide (n = 1) the wavelength at which the
lowest-order mode first starts to propagate is λ c1 = 2d,or the plate separation is
one-half wavelength. Mode m will propagate whenever ω> ω cm ,or equivalently
whenever λ<λ cm . Use of the cutoff wavelength enables us to construct a second
useful form of Eq. (42):

2πn λ 2
β m = 1 − (44)
λ λ cm

EXAMPLE 13.1
A parallel-plate waveguide has plate separation d = 1cm and is filled with teflon
having dielectric constant = 2.1. Determine the maximum operating frequency

r
such that only the TEM mode will propagate. Also find the range of frequencies
over which the TE 1 and TM 1 (m = 1) modes, and no higher-order modes, will
propagate.
Solution. Using (41), the cutoff frequency for the first waveguide mode (m = 1)
will be
ω c1 2.99 × 10 10 10
f c1 = = √ = 1.03 × 10 Hz = 10.3 GHz
2π 2 2.1
To propagate only TEM waves, we must have f < 10.3 GHz. To allow TE 1 and
TM 1 (along with TEM) only, the frequency range must be ω c1 <ω <ω c2 , where
ω c2 = 2ω c1 , from (41). Thus, the frequencies at which we will have the m = 1 modes
and TEM will be 10.3 GHz < f < 20.6 GHz.

EXAMPLE 13.2
In the parallel-plate guide of Example 13.1, the operating wavelength is λ = 2 mm.
How many waveguide modes will propagate?
Solution. For mode m to propagate, the requirement is λ<λ cm .For the given wave-
guide and wavelength, the inequality becomes, using (43),
√
2 2.1 (10 mm)
2mm <
m
from which
√
2 2.1 (10 mm)
m < = 14.5
2mm
Thus the guide will support modes at the given wavelength up to order m = 14. Since
there will be a TE and a TM mode for each value of m, this gives, not including the
TEM mode, a total of 28 guided modes that are above cutoff.