Question 10.9: An air-filled rectangular waveguide is 1[m] long, with dimen......

(a) The input power is

P_{in} = P_{out} e^{\alpha \ell }= 1.5 \times 10^{3} e^{0.002 \times 2\times 1} = 1,506[W]            (10-102)

Thus, the power dissipated in the waveguide is 6[W].

(b) Cutoff frequency of the dominant mode is

f_{c(10)} = \frac{c}{2a} = \frac{3 \times 10^{8}}{2 \times 0.04} = 3.75[GHz]                (10-103)

The next higher mode is TE_{01}, having a cutoff frequency 4.5[GHz]. Because of the operating frequency 4[GHz], the waveguide supports only TE_{10}mode.

From Eqs. (10-75) and (10-77b), the transverse field components of the TE_{10} mode are written as

\boxed{\eta _{TE} \sqrt{\frac{\mu }{ε} } \frac{1}{\sqrt{1 – (f_{c(mn)} / f)^{2}} } }\qquad [\Omega]         (10-77b)

E_{y} = – \frac{j\omega \mu }{h^{2}} \left\lgroup\frac{\pi }{a} \right\rgroup H_{o} \sin \left\lgroup\frac{\pi }{a}x \right\rgroup \equiv E^{\prime }_{o} \sin \left\lgroup\frac{\pi }{a} x\right\rgroup             (10-104a)

H_{x} = \frac{j\beta _{10} }{h^{2}} \left\lgroup\frac{\pi }{a} \right\rgroup H_{o} \sin \left\lgroup\frac{\pi }{a}x \right\rgroup = – \frac{E^{\prime }_{o}}{\eta _{TE}} \sin \left\lgroup\frac{\pi }{a} x\right\rgroup             (10-104b)

At the input end of the waveguide( z = 0 ), with the aid of Eq. (10-77b), the total time-average power is

P_{in} = \frac{1}{2}\int_{x=0}^{x=a}\int_{y=0}^{y=b}{\pmb{E\times H^{*} \cdot a}_{z} dx dy} = \frac{ab(E^{\prime }_{o})^{2}}{4} \frac{ \sqrt{1 – (f_{c(10)} / f)^{2}}}{\eta _{o}}              (10-105)

Inserting Eqs. (10-102) and (10-103) into Eq. (10-105) we have

Thus,

E^{\prime }_{o}=73.7 [kV/m].

(c) From Eq. (10-104), we see that the maximum electric field intensity is observed at x = a /2 = 0.02[m] .