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Electromagnetic theory miscellaneous

Electromagnetic Theory

  1. Rectangular waveguide is designed to propagate the dominant mode TE10 at a frequency of 5 MHz. The cutoff frequency is 0.8 of signal frequency the raito of the waveguide height to width is 2 the dimension of the waveguide are—
    1. 3.75 cm, 1.875 cm
    2. 4 cm, 2 cm
    3. 8 cm, 4 cm
    4. 2.54 cm, 1.27 cm
Correct Option: A

We find the relation of waveguide that the cut-off frequency is given by relation

fc =
1
m∏
2+
n∏
2
2∏√με a a

for TE10 mode m = 1 and n = 0 and b =
1
a

fc × 2∏
=
⇒ a = 3.75 cm
ca

and
b =
a
= 1.875 cm
2



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