Question 10.6: The TE10 mode of frequency 3[GHz] propagates in a rectangula......
The TE_{10} mode of frequency 3[GHz] propagates in a rectangular waveguide with dimensions a = 2b = 4[cm] , which is filled with a nonmagnetic, lossless dielectric( n = 1.4 ). Find (a) cutoff frequency, (b) phase constant, (c) phase velocity, and (d) wave impedance.
(a) From Eq. (10-46), the cutoff frequency is
f_{c(mn)} = \frac{1}{2\sqrt{\mu ε } } \sqrt{ \left\lgroup \frac{m}{a} \right\rgroup^{2} + \left\lgroup\frac{n}{b} \right\rgroup^{2} } [Hz] (10-46)
f_{c(10)} = \frac{1}{2\sqrt{\mu ε } } \sqrt{\left\lgroup\frac{1}{a} \right\rgroup^{2} + \left\lgroup\frac{0}{b} \right\rgroup^{2} } = \frac{3\times 10^{8}}{2\times 1.4} \left\lgroup \frac{1}{0.04} \right\rgroup = 2.68 [GHz](b) From Eq. (10-48), the phase constant is
\boxed{\beta_{mn} = k \sqrt{1 – \left\lgroup \frac{f_{c(mn)}}{f} \right\rgroup ^{2} } } [rad/m] (10-48)
\beta_{mn} = \frac{n\omega }{c} \sqrt{1 – \left\lgroup \frac{f_{c(mn)}}{f} \right\rgroup ^{2} } = \frac{1.4 \times 2\pi \times 3 \times 10^{9}}{3\times 10^{8}} \sqrt{1 – \left\lgroup\frac{2.68}{3} \right\rgroup ^{2}} \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = 12.58 \pi [rad /m](c) From Eq. (10-52a), the phase velocity of the TE_{10} mode is
v_{p(mn)} = \frac{\omega }{\beta _{mn}} = f \lambda _{mn} (10-52a)
v_{p(10)} = \frac{\omega }{\beta _{10}} = \frac{2\pi \times 3\times 10^{9}}{12.58\pi } =4.77 \times 10^{8}[m/s](d) From Eq. (10-77b), the wave impedance is
\eta _{TE} =\sqrt{ \frac{\mu }{ε }} \frac{1}{\sqrt{1 – (f_{c(mn)}/f)^{2}} } [Ω] (10-77b)
\eta _{TE} = \frac{377}{1.4} \times \frac{1}{\sqrt{1 – (2.68/3)^{2}} }599.2[\Omega ]