Calculate the cut off wavelength, the guide wavelength, phase velocity and the characterestic wave impedance for TE110 mode.

emphasized textCalculating the free space wavelenth gives

$\lambda = \frac{v_c}{f} = \frac{3\times10^{10}}{9\times10^{10}}= 3.33 cm$

a. The cutoff wavelenth will be

$\lambda _0 = \frac{2a}{m} = \frac{2 \times 4.5}{1} = 9 cm Calculating p, for convenience, gives$ p = \sqrt {1-(\frac{\lambda}{\lambda _0})^2} = \sqrt {1-(\frac{3.33}{9})^2} = \sqrt {1-0.137}= 0.93$Then the quide wavelength is$ \lambda _p = \frac{\lambda}{p} = \frac{3.33}{0.93} = 3.58 cm$The group and phase velocities are simply found from$v_g = v_cp = 3 \times10^{8}\times0.93 = 2.79 \times10^8 m/sv_p = \frac{v_c}{p} = \frac{3\times10^8}{0.93} = 3.23 \times 10^8 m/s$The characteristic wave impedance is$Z_o = \frac{x}{p} = \frac{120 \pi}{0.93} = 405 \ \Omega $