Q5Electromagnetics Waves

Question

Q.5. If the cut-off frequency of an air-filled waveguide is 10 GHz and it supports TE01 mode, then what is its size?

Answer 1

For the TEmn mode of a rectangular waveguide of dimensions a × b, the cutoff frequency is:

f_{c} = \frac{c}{2} (\frac{m}{a})^{2} + (\frac{n}{b})^{2}

For TE01 (m = 0, n = 1), only the b-dimension term survives, so fc = c/(2b), giving:

b = \frac{c}{2 f _{c}} = \frac{3 \times 1 0 ^{8}}{2 \times 10 \times 1 0 ^{9}} = 0.015 m = 1.5 cm

The waveguide's narrow (b) dimension must therefore be 1.5 cm; the broad dimension 'a' is not determined by this single cutoff constraint (any a ≥ b, or specifically a > b for a as the standard broad wall, is consistent with TE01 having exactly this cutoff, though in practice a is chosen relative to b by the desired dominant mode and bandwidth).

Q5Electromagnetics Waves

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