GATE Electronics and Communications (EC) 2019 Solved Paper

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Question : 55 of 65

Marks: +1, -0

A rectangular waveguide of width

{w}

and height

{h}

has cut-off frequencies for

{TE}_{10}

and

{TE}_{11}

modes in the ratio

1: 2

. The aspect ratio

{w} {/} {h}

, rounded off to two decimal places, is ___________

Your Answer:

Solution:

{f}_{ {c}}=\;{ {v}}/{2} √{(\;{ {m}}/{ {a}})^{2}+(\;{ {n}}/{ {b}})^{2}}

For

{T} ε_{10,} \;\; {m}=1, \;\; {n}=0

{fc}_{1}=\;{ {V}}/{2} √{(\;{1}/{ {a}})^{2}=0}=\;{ {v}}/{2 {a}}

For

{T}_{811} \;\; {m}=1, {n}=1

{f}_{ {c} 2}=\;{ {V}}/{2} √{\;{1}/{ {a}^{2}}+\;{1}/{ {b}^{2}}}

Given

\;{ {f}_{ {c}_{1}}}/{ {f}_{ {c}_{2}}}=\;{1}/{2}

\;{ {V} {/} 2 {a}}/{{ {v}}/{2} √{\;{1}/{ {a}^{2}}+\;{1}/{ {b}^{2}}}}=\;{1}/{2}

\;{\;{1}/{ {a}}}/{√{\;{1}/{ {a}^{2}}+\;{1}/{ {b}^{2}}}}=\;{1}/{2} ⇒ \;{\;{1}/{ {a}}}/{{√{ {a}^{2}+ {b}^{2}}}/{ {ab}}}=\;{1}/{2}

\;{ {b}}/{√{ {a}^{2}+ {b}^{2}}}=\;{1}/{2}

⇒ 4 b^{2}=a^{2}+b^{2}

⇒ 3 b^{2}=a^{2}

⇒ \;{b^{2}}/{a^{2}}=\;{1}/{3}

⇒ \;{b}/{a}=\;{1}/{√{3}}

⇒ \;{a}/{b}=√{3}

\;{\;\text " width "\;}/{\;\text " hight "\;}=√{3}=1.732

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