Page 493 - Engineering Electromagnetics, 8th Edition

CHAPTER 13 Guided Waves 475

Solution. The group delay difference is expressed as

1 1

t = − (s/cm)
ν g2 ν g1
From (57), along with the results of Example 13.1, we have

c 10.3 2
ν g1 = √ 1 − = 0.63c
2.1 25

c 20.6 2
ν g2 = √ 1 − = 0.39c
2.1 25
Then

1 1 1
t = − = 3.3 × 10 −11 s/cm = 33 ps/cm
c .39 .63
Thiscomputationgivesaroughmeasureofthemodal dispersionintheguide,applying
to the case of having only two modes propagating. A pulse, for example, whose center
frequency is 25 GHz would have its energy divided between the two modes. The pulse
would broaden by approximately 33 ps/cm of propagation distance as the energy in
the modes separates. If, however, we include the TEM mode (as we really must), then
√
the broadening will be even greater. The group velocity for TEM will be c/ 2.1. The
group delay difference of interest will then be between the TEM mode and the m = 2
mode (TE or TM). We would therefore have
1 1
t net = − 1 = 52 ps/cm
c .39

D13.5. Determine the wave angles θ m for the first four modes (m = 1, 2,
3, 4) in a parallel-plate guide with d = 2 cm, = 1, and f = 30 GHz.

r
Ans. 76 ;60 ;41 ;0 ◦
◦
◦
◦
D13.6. A parallel-plate guide has plate spacing d = 5mm and is filled with
glass (n = 1.45). What is the maximum frequency at which the guide will
operate in the TEM mode only?

Ans. 20.7 GHz

D13.7. A parallel-plate guide having d = 1cmis filled with air. Find the
cutoff wavelength for the m = 2 mode (TE or TM).

Ans. 1cm