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Find the cut off frequency for $T M_{11}$ mode, given dimensions of the waveguide 2 $\mathrm{cm}$ and $4 \mathrm{~cm}$ respectively.
1 Answer
written 23 months ago by | • modified 23 months ago |
Solution:
$ \lambda_{\mathrm{c}}=\frac{2 a b}{\sqrt{m 2 b 2+n 2 a 2}} \\ $
Here,
$ a=2 \times 10^{-2} \mathrm{~m} ; \quad b=4 \times 10^{-2} \mathrm{~m} \\ $
Thus,
$ \lambda_{\mathrm{c}}=3.5777 \mathrm{~cm}$ and $\mathrm{f}_{\mathrm{c}}=\frac{c}{\lambda c}=8.386 \mathrm{GHz} \\ $
$ \lambda_{\mathrm{c}}=\frac{2 a b}{\sqrt{m 2 b 2+n 2 a 2}}\\ $
Here,
$ a=4 \times 10^{-2}, b=8 \times 10^{-2} \mathrm{~m} \\ $
Thus,
$ \lambda_{\mathrm{c}}=7.155 \mathrm{~cm}$ and $\mathrm{f}_{\mathrm{c}}=\frac{c}{\lambda c}=4.19 \mathrm{GHz} \\ $