Question 10.2: A parallel-plate waveguide with an air gap of 1[cm] operates......

(a) From Eq. (10-9), the cutoff frequency for TE_{m} mode is

\boxed{f_{c(m)} \frac{m c }{2 n d}}                              [rad/s]                            (10-9)

Cutoff frequencies for TE_{1}, TE_{2}  and  TE_{3} modes are

We note that f_{c(1)},  and  f_{c(2)}  are above cutoff, while f_{c(3)}  is below cutoff.

(b) From Eq. (10-19) we can obtain the amplitude of the TE_{1} mode, which is one of the component waves comprising E in the z = 0 plane, that is,

\boxed{\pmb{E}_{m} = – j \sqrt{\frac{2}{d} } \pmb{a}_{x} \sin \left\lgroup\frac{m \pi }{d} y \right\rgroup e^{- j k_{z}z}}                        (m = 1,2,3,…)                           (10-19)

\int_{y=0}^{y= d}{\pmb{E} \pmb{\cdot }\pmb{E}^{*}_{1} dy} = j 100 \sqrt{2} \int_{y=0}^{y= 0.01}{\sin^{2} \left\lgroup 200 \pi y \right\rgroup \sin \left\lgroup 100 \pi y \right\rgroup dy} \\ \quad \quad \quad \quad \quad \quad \quad = j 0.48                                                            (10-20a)

The amplitude of the TE_{2} mode is

\int_{y=0}^{y= d}{\pmb{E} \pmb{\cdot }\pmb{E}^{*}_{2} dy} = j 100 \sqrt{2} \int_{y=0}^{y= 0.01}{\sin^{2} \left\lgroup 200 \pi y \right\rgroup \sin \left\lgroup 200 \pi y \right\rgroup dy} \\ \quad \quad \quad \quad \quad \quad \quad = 0                                                            (10-20b)

The amplitude of the TE_{3} mode is

\int_{y=0}^{y= d}{\pmb{E} \pmb{\cdot }\pmb{E}^{*}_{3} dy} = j 100 \sqrt{2} \int_{y=0}^{y= 0.01}{\sin^{2} \left\lgroup 200 \pi y \right\rgroup \sin \left\lgroup 300 \pi y \right\rgroup dy} \\ \quad \quad \quad \quad \quad \quad \quad = j 0.34                                                            (10-20c)

By use of Eqs. (10-19) and (10-20), the electric fields of the three component waves are obtained, in the z = 0 plane, as

(c) The phase constant for the TE_{1} mode is obtained from Eq. (10-7):

\beta_{m} \equiv k_{z} = \pm \sqrt{k^{2} – k^{2}_{y}} \\ \quad \quad \quad \quad = \pm \sqrt{\left\lgroup\frac{n \omega }{c} \right\rgroup^{2} – \left\lgroup\frac{m \pi }{d} \right\rgroup^{2} }                      (m = 1,2,3,…)                  (10-7)

\beta_{1} = \sqrt{\left\lgroup\frac{2 \pi \times 35 \times 10^{9}}{3\times 10^{8}} \right\rgroup^{2} – \left\lgroup\frac{\pi }{0.01} \right\rgroup^{2} } = 662.3 [rad / m] .                        (10-22a)

The attenuation constant for the TE_{3} mode is obtained from Eq. (10-16):

\boxed{\alpha _{m} = \frac{n \omega _{c(m)}}{9} \sqrt{ 1 – \left\lgroup\frac{f }{f_{c(m)}} \right\rgroup^{2} } } \qquad [Np / m]                     (10-16)

\alpha _{3} = \frac{2 \pi \times 45 \times 10^{9}}{3\times 10^{3} } \sqrt{ 1 – \left\lgroup\frac{35 }{45} \right\rgroup^{2} } = 592[Np / m]             (10-22b)

Inserting Eq. (10-21) and Eq. (10-22) into Eq. (10-19) we obtain, in the z = 2[cm] plane,