# Consider an air-filled rectangular waveguide with a cross-section of 5 cm × 3 cm. For this waveguide, the cut-off frequency (in MHz) of TE21 mode is _

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Consider an air-filled rectangular waveguide with a cross-section of 5 cm × 3 cm. For this waveguide, the cut-off frequency (in MHz) of TE21 mode is _________.

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Concept:

The cut-off frequency for a rectangular waveguide is given by:

${f_{m,\;n}} = \frac{C}{2}\sqrt {{{\left( {\frac{m}{a}} \right)}^2} + {{\left( {\frac{n}{b}} \right)}^2}}$

C = velocity of light in free space.

m, n = mode of operation of the waveguide

a, b = dimensions of the waveguide

Calculation:

For TE2, 1 mode, m = 2 and n = 1

Also, a = 5 cm

b = 3 cm

The cut-off frequency will be:

${f_{c\left( {2,\;1} \right)}} = \frac{{3\; \times \;{{10}^{10}}}}{2}\sqrt {{{\left( {\frac{2}{5}} \right)}^2} + {{\left( {\frac{1}{3}} \right)}^2}}$

$= 1.5 \times {10^{10}}\sqrt {0.16 + 0.11} Hz$

= 1.5 × 1010 × 0.52

fc(2, 1) = 7810.24 MHz