Question 10.5: In a hollow rectangular waveguide with interior dimensions a......
In a hollow rectangular waveguide with interior dimensions a = 5[cm] and b = 2.5[cm] , an electromagnetic wave of frequency 8[GHz] propagates in the +z-direction. In the transverse plane at z = 0 , the magnetic field is given by
\pmb{H} = 4 \pmb{a}_{x} \sin (20 \pi x) \sin (60 \pi y) + 3 \pmb{a}_{y} \cos (20 \pi x) \cos (60 \pi y) [A/m] with no z-component. Find
(a) TM modes of propagation,
(b) \alpha _{mn} for the evanescent mode with the lowest cutoff frequency, and
(c) magnetic field in the transverse plane at z = 1[m].
(a) From Eq. (10-46), the cutoff frequencies are
\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 \sqrt{\mu _{o} ε_{o}} } \sqrt{\left\lgroup\frac{1}{0.05} \right\rgroup^{2} + \left\lgroup\frac{1}{0.025} \right\rgroup ^{2} } = 6.7 [GHz] \\ f_{c(21)} = \frac{1}{2 \sqrt{\mu _{o} ε _{o}} } \sqrt{\left\lgroup\frac{2}{0.05} \right\rgroup^{2} + \left\lgroup\frac{1}{0.025} \right\rgroup ^{2} } = 8.5 [GHz] \\ f_{c(12)} = \frac{1}{2 \sqrt{\mu _{o} ε_{o}} } \sqrt{\left\lgroup\frac{1}{0.05} \right\rgroup^{2} + \left\lgroup\frac{2}{0.025} \right\rgroup ^{2} } = 12.4 [GHz]The TM_{21} and TM_{12} modes are the evanescent modes in the waveguide operating at 8[GHz] .
(b) From Eq. (10-43), the attenuation constant for TM_{21} mode is
In view of the large value of \alpha _{21}, the evanescent wave can be neglected at the output end at z = 1[m] of the waveguide.
(c) We obtain \pmb{\bar{H} } _{11} from Eq. (10-62) as
The magnitude of the TM_{11} mode contained in H is obtained as follows:
Multiplying Eq. (10-62) by the factor given by Eq. (10-63) we obtain
From Eq. (10-48) we obtain the phase constant for the TM_{11} mode as
\boxed{\beta_{mn} = k \sqrt{1 – \left\lgroup \frac{f_{c(mn)}}{f} \right\rgroup ^{2} } } [rad/m] (10-48)
\beta _{mn} = k \sqrt{1 – \left\lgroup\frac{f_{c(mn)}}{f} \right\rgroup^{2} } = \frac{2\pi \times 8\times 10^{9}}{3\times 10^{8}} \sqrt{1 – \left\lgroup \frac{6.7}{8} \right\rgroup ^{2}} = 91.6 [rad/m] (10-65)
In the plane at z = 1[m], Eqs.(10-64) and (10-65) are combined together to write the magnetic field phasor, that is,
\pmb{H} = [H_{x}(x, y) \pmb{a}_{x} + H_{y}(x, y) \pmb{a}_{y} ]e^{- j 91.6}