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