# For a rectangular waveguide of internal dimensions a × b (a > b), the cut-off frequency for the TE11 mode is the arithmetic mean of the cut-off freque

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For a rectangular waveguide of internal dimensions a × b (a > b), the cut-off frequency for the TE11 mode is the arithmetic mean of the cut-off frequencies for TE10 mode and TE20 mode. If a = √5 cm, the value of b (in cm) is _______.

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Concept:

The cutoff frequency of TEmn mode in the rectangular waveguide is given as:

${f_c} = \frac{C}{{2\pi }}\sqrt {{{\left( {\frac{{m\pi }}{a}} \right)}^2} + {{\left( {\frac{{n\pi }}{b}} \right)}^2}}$

fc for TE10 will be:

${f_{c\left( {10} \right)}} = \frac{C}{{2\pi }}\sqrt {{{\left( {\frac{{1\pi }}{a}} \right)}^2}} = \frac{C}{{2a}}$      ---(1)

fc for TE20 will be:

${f_{c\left( {20} \right)}} = \frac{C}{{2\pi }}\sqrt {{{\left( {\frac{{2\pi }}{a}} \right)}^2}} = \frac{C}{a}$      ---(2)

fc for TE11 will be:

${f_{c\left( {11} \right)}} = \frac{c}{{2\pi }}\sqrt {{{\left( {\frac{\pi }{a}} \right)}^2} + {{\left( {\frac{\pi }{n}} \right)}^2}}$

$= \frac{c}{2}\sqrt {{{\left( {\frac{1}{a}} \right)}^2} + {{\left( {\frac{1}{b}} \right)}^2}} \;$       ---(3)

Calculation:

Given:

${f_c}\left( {T{E_{11}}} \right) = \frac{{{f_c}\left( {T{E_{10}}} \right) + {f_c}\left( {T{E_{20}}} \right)}}{2}$

From equation 1, 2, 3, we can write:

$\frac{c}{2}\sqrt {{{\left( {\frac{1}{a}} \right)}^2} + {{\left( {\frac{1}{b}} \right)}^2}} = \frac{{\frac{c}{{2a}} + \frac{c}{a}}}{2}$

$\sqrt {{{\left( {\frac{1}{a}} \right)}^2} + {{\left( {\frac{1}{b}} \right)}^2}} = \frac{1}{{2a}} + \frac{1}{a} = \frac{3}{{2a}}$

${\left( {\frac{1}{a}} \right)^2} + {\left( {\frac{1}{b}} \right)^2} = \frac{9}{{4{a^2}}}$

a = √5

${\left( {\frac{1}{b}} \right)^2} = \frac{5}{{4{a^2}}}$

$b = \frac{{2a}}{{\sqrt 5 }}$

b = 2 cm