Question

A rectangular waveguide with air medium has dimensions a = 22.86 mm and b = 10.6 mm is fed by 3 GHz carrier from a coaxial cable, which of the following is a false statement for TE₀₁ mode?

1. Propagating mode
2. Non-propagating mode
3. Propagating mode in case filled fully with dielectric material of proper dielectric constant
4. None of the above

Answer 1

Correct Answer - Option 1 : Propagating mode

Concept:

The cutoff frequency of waveguide is given by:

\({f_c} = \frac{c}{{2\sqrt {{\mu _b}{\epsilon_b}} }}\sqrt {{{\left( {\frac{m}{a}} \right)}^2} + {{\left( {\frac{n}{b}} \right)}^2}} \)

where c = 3 × 108 m/s

The lowest propagating mode is TE10 cutoff frequency

\({f_c} = \frac{c}{{2a}}\)

\(a = \frac{c}{{2{f_c}}}\)

Calculations:

Given:

a = 22.86 mm and b = 10.6 mm

As a > b so,

\({f_c} = \frac{c}{{2a}}\)

\({f_c} = \frac{3\times10^8}{2\times22.86 \ mm}\)

f_c = 6.5 GHz

As input applied frequency is less than the cut-off frequency, the mode is non-propagating. The mode can be made propagating if dielectric with a proper dielectric constant is inserted because it decreases phase velocity.