Question 10.3: A rectangular waveguide has interior dimensions a = 1.5[cm] ......
A rectangular waveguide has interior dimensions a = 1.5[cm] and b = 1[cm] . It is filled with a nonmagnetic, lossless dielectric( μ = \mu _{o}, ε_{r} = 1.96), and excited with an electromagnetic field( f = 15[GHz] ), which propagates as the TM _{11}mode in the waveguide . Find (a) λ, (b) f_{c(11)}, and (c) \lambda _{11}.
(a) Wavenumber in an unbounded dielectric
k = \frac{2\pi }{\lambda } = \omega \sqrt{ \mu _{o} ε} = \frac{2 \pi \times 15 \times 10^{9} \times 1.4}{3 \times 10^{8}}Thus, the wavelength is
λ = 14.3[mm].
(b) From Eq. (10-46), the cutoff frequency is
\boxed{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(11)} = \frac{1}{2 \times 1.4 \times \sqrt{ \mu _{o} ε_{o}}} \sqrt{ \left\lgroup \frac{1}{1.5 \times 10^{-2}} \right\rgroup^{2} + \left\lgroup\frac{1}{10^{-2}} \right\rgroup ^{2} } = 12.9 [GHz](c) From Eq. (10-49), the waveguide wavelength is
\boxed{\lambda _{mn} = \frac{2 \pi }{\beta _{mn}} =\frac{\lambda }{\sqrt{1 – (f_{c(mn)} / f)^{2}}}} [m] (10-49)
\lambda _{11} = \frac{\lambda }{\sqrt{1 – (f_{c(11)} / f)^{2}}} =\frac{14.3 \times 10^{-3}}{\sqrt{1 – (12.9 /15)^{2}} } = 28.0 [mm]