Correct Answer - Option 3 : 8√5 GHz
Concept:
The cut-off frequency of waveguide in free space is given by:
\({f_c} = \frac{c}{{2\pi }}\;\sqrt {{{\left( {\frac{{m\pi }}{a}} \right)}^2} + {{\left( {\frac{{n\pi }}{b}} \right)}^2}\;} \)
Where,
c = speed of light in free space = 3 × 108 m/s
m, n = no of modes that shows no of maxima between guide walls of the waveguide
a, b = waveguide dimensions
Calculation:
Given WK 90 waveguide dimension aspect ratio as:
\(\frac{a}{b} = 2\)
The cutoff frequency for TE20 mode is 16 GHz, i.e.
\({\left( {{f_c}} \right)_{T{E_{20}}}} = \frac{C}{{2\pi }}\sqrt {{{\left( {\frac{{2\pi }}{a}} \right)}^2} + {{\left( {\frac{{0.\pi }}{b}} \right)}^2}} \;\)
\({\left( {{f_c}} \right)_{T{E_{20}}}} = \frac{C}{{2\pi }} \times \frac{{2\pi }}{a}\)
\({\left( {{f_C}} \right)_{T{E_{20}}}} = \frac{c}{a}\) -- (1)
\(\therefore \frac{c}{a} = 16\;GHz\)
Now, the cutoff frequency for TE11 mode will be:
\({\left( {{f_c}} \right)_{T{E_{11}}}} = \frac{C}{{2\pi }}\sqrt {{{\left( {\frac{{1.\pi }}{a}} \right)}^2} + {{\left( {\frac{{1.\pi }}{b}} \right)}^2}} \)
\(= \frac{c}{{2\pi \;}} \times \pi \times \sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}} \)
\(= \frac{C}{{2a}}\sqrt {1 + {{\left( {\frac{a}{b}} \right)}^2}} \)
\(= \frac{{\left( {\frac{C}{a}} \right)}}{2}\sqrt {1 + {{\left( {\frac{a}{b}\;} \right)}^2}}\)
From equation (1), we have:
\({\left( {{f_C}} \right)_{T{E_{11}}}} = \frac{{16\;GHz}}{2} \times \sqrt {1 + {{\left( 2 \right)}^2}} \)
\({\left( {{f_c}} \right)_{T{E_{11}}}} = 8\sqrt 5 \;GHz\)
Important points:
- Cut off frequency of waveguide is the minimum frequency that is required to make a wave in the waveguide.
- If f > fc: than wave propagation will occur in the waveguide
- If f = fc : than wave will oscillate between the walls of the waveguide.
- If f < fc : then the wave will not propagate through the waveguide.
- Dominant mode means a mode having the lowest cutoff frequency.
