GATE EC 2014 Electromagnetics Solution Q 8

GATE EC 2014 Electromagnetics Solution Q 8

For a rectangular waveguide of internal dimensions a × b (a > b), the cut-off frequency for the TE_{11} mode is the arithmetic mean of the cut-off frequencies for TE10 and TE20 mode. If a=\sqrt5cm, the value of b (in cm) is____.

Ans- (2 cm)

Explanation

Cut-off frequency of a rectangular waveguide is given as-

f_c=\frac{v_o}{2\pi}\sqrt{\left(\frac{m\pi}a\right)^2+\left(\frac{n\pi}b\right)^2} f_c\;for\;\;TE_{10}=\frac{v_o}{2\pi}\sqrt{\left(\frac{1\pi}a\right)^2}=\frac{v_o}{2a} f_c\;for\;\;TE_{20}=\frac{v_o}{2\pi}\sqrt{\left(\frac{2\pi}a\right)^2}=\frac{v_o}{2a} f_c\;for\;\;TE_{11}=\frac{v_o}{2\pi}\sqrt{\left(\frac\pi a\right)^2+\left(\frac\pi b\right)^2} =\frac{v_o}2\sqrt{\left(\frac1a\right)^2+\left(\frac1b\right)^2}

Now, f_{cTE_{11}}=\frac{f_{cTE_{10}}+f_{cTE_{20}}}2

\frac{v_o}2\sqrt{\left(\frac1a\right)^2+\left(\frac1b\right)^2}=\frac{\displaystyle\frac{v_o}{2a}+\frac{v_o}a}2 \sqrt{\left(\frac1a\right)^2+\left(\frac1b\right)^2}=\frac1{2a}+\frac1a+=\frac3{2a} \left(\frac1a\right)^2+\left(\frac1b\right)^2=\frac9{4a^2} \left(\frac1b\right)^2=\frac5{4a^2} b=\frac{2a}{\sqrt5}=\frac{2\times\sqrt5}{\sqrt5}

Putting value of a we get, =2\;cm

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