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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—
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- 3.75 cm, 1.875 cm
- 4 cm, 2 cm
- 8 cm, 4 cm
- 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 = | √ | ![]() | ![]() | 2 | + | ![]() | ![]() | 2 | ||||
2∏√με | a | a |
for TE10 mode m = 1 and n = 0 and b = | |
a |
= | ⇒ a = 3.75 cm | ||
c | a |
and
b = | = 1.875 cm | |
2 |