Page 497 - Engineering Electromagnetics, 8th Edition

CHAPTER 13 Guided Waves 479

We solve for H s by dividing both sides of (69) by − jωµ. Performing this operation
on (70), we obtain the two magnetic field components:

β m − jβ m z
H xs =− E 0 sin(κ m x)e (71)
ωµ
κ m
H zs = j E 0 cos(κ m x)e − jβ m z (72)
ωµ
Together, these two components form closed-loop patterns for H s in the x, z plane,
as can be verified using the streamline plotting methods developed in Section 2.6.
It is interesting to consider the magnitude of H s , which is found through

∗
∗
∗
|H s |= H s · H = H xs H + H zs H zs (73)
s
xs
Carrying this out using (71) and (72) results in
E 0 2 2 2 1/2
|H s |= κ + β 2 1/2 sin (κ m x) + cos (κ m x) (74)
m
m
ωµ
2
2
2
2
2
Using the fact that κ + β = k and using the identity sin (κ m x) + cos (κ m x) = 1,
m
m
(74) becomes
√
k ω µ E 0
|H s |= E 0 = = (75)
ωµ ωµ η
√
where η = µ/ . This result is consistent with our understanding of waveguide
modes based on the superposition of plane waves, in which the relation between E s
and H s is through the medium intrinsic impedance, η.
D13.8. Determine the group velocity of the m = 1 (TE or TM) mode in an
air-filled parallel-plate guide with d = 0.5cmat f = (a)30 GHz, (b)60 GHz,
and (c) 100 GHz.
8
8
Ans. 0; 2.6 × 10 m/s; 2.9 × 10 m/s
D13.9. ATE mode in a parallel-plate guide is observed to have three maxima
in its electric field pattern between x = 0 and x = d. What is the value of m?
Ans. 3
13.5 RECTANGULAR WAVEGUIDES
In this section we consider the rectangular waveguide, a structure that is usually used
in the microwave region of the electromagnetic spectrum. The guide is shown in
Figure 13.7. As always, the propagation direction is along the z axis. The guide is
of width a along x and height b along y.We can relate the geometry to that of the
parallel-plate guide of previous sections by thinking of the rectangular guide as two