Determine the spacing between 2 – holes in a 2 –hole directional coupler made in rectangular waveguide version with inner dimension of 2.286 cm X 1.00 cm at 9 GHz.
Given that b=1 cm, a=2.286 cm and f=9 GHz
We know that
In case of dominant mode \(f_c=\frac{C}{2a}=\frac{3\times{10}^{10}}{2\times2.286}=6.56\) GHz
\(\lambda_g=\frac{\lambda}{\sqrt{1-\left(f_c/f\right)^2}}\)
\(\lambda_g=\frac{\frac{C}{f}}{\sqrt{1-\left(f_c/f\right)^2}}\)
\(\lambda_g=\frac{\frac{3\times{10}^{10}}{9\times{10}^9}}{\sqrt{1-\left(6.56\times{10}^9/9\times{10}^9\right)^2}}\)
\(\lambda_g=\frac{3.33}{\sqrt{1-0.532}}=4.87\) cm
Therefore the distance between two holes in a two hole coupler is \(\frac{\lambda_g}{4}=\frac{4.87}{4}=1.22\) cm
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