GATE 2016 EC (EC3) – Question 63

Electromagnetics · Waveguides and Optical Fibers · 2 marks · Multiple choice

For an air-filled rectangular waveguide with a=2.286 cm and b=1.016 cm, the increasing order of cutoff frequencies is

$TE_{01}<TE_{10}<TE_{11}<TE_{20}$
$TE_{20}<TE_{11}<TE_{10}<TE_{01}$
$TE_{10}<TE_{20}<TE_{01}<TE_{11}$
$TE_{10}<TE_{11}<TE_{20}<TE_{01}$

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Correct answer: (C) $TE_{10}<TE_{20}<TE_{01}<TE_{11}$

Explanation

The TE cutoff formula is $f_{c,mn}=(c/2)\sqrt{(m/a)^2+(n/b)^2}$.

The four values are approximately 6.562 GHz for TE10, 13.123 GHz for TE20, 14.764 GHz for TE01 and 16.156 GHz for TE11.

Their ascending order is **TE10, TE20, TE01, TE11, C**. The unequal guide dimensions make TE20 lower than TE01.

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