Concept:
The propagation constant is given by \({\rm{\gamma }} = \sqrt {{{\left( {\frac{{{\rm{m\pi }}}}{{\rm{a}}}} \right)}^2} + {{\left( {\frac{{{\rm{n\pi }}}}{{\rm{b}}}} \right)}^2} - {{\rm{k}}^2}}\)
Application:
\({{\rm{f}}_{{\rm{c}},10}} = \frac{{\rm{c}}}{{2{\rm{a}}}} = 6.56{\rm{\;GHz}}\)
\({{\rm{f}}_{{\rm{c}},01}} = \frac{{\rm{c}}}{{2{\rm{b}}}} = 14.76{\rm{\;GHz}}\)
So, 10 GHz can propagate only in TE10 node.
Now, the free space propagation constant will be:
\({\rm{k}} = \frac{{\rm{\omega }}}{{\rm{c}}} = \frac{{2{\rm{\pi }} \times 100 \times {{10}^8}}}{{3 \times {{10}^8}}} \)
\(k= \frac{{200{\rm{\pi }}}}{3}\)
Now, the propagation constant will be:
\({\rm{\gamma }} = \sqrt {{{\left( {\frac{{{\rm{m\pi }}}}{{\rm{a}}}} \right)}^2} + {{\left( {\frac{{{\rm{n\pi }}}}{{\rm{b}}}} \right)}^2} - {{\rm{k}}^2}}\)
With m = 1 and n = 0, we get:
\( {\rm{\gamma }} = \sqrt {{{\left( {\frac{{\rm{\pi }}}{{2.286 \times {{10}^{ - 2}}}}} \right)}^2} - {{\left( {\frac{{200{\rm{\pi }}}}{3}} \right)}^2}}\)
\({\rm{\gamma }}\) = j158.05 m
-1