Question 8.14: A uniform plane wave, E^i = ax 20 cos(3 × 10^9t - 17.3z), tr......
A uniform plane wave, \mathscr{E}^{i} = \pmb{a}_{x} 20 \cos (3 \times 10^{9}t – 17.3z) , travels in a lossless dielectric (z < 0) , and impinges normally on the surface of a lossy dielectric (z ≥ 0), for which \varepsilon _{r} = 4 and σ = 0.5[S/m] . Find Γ, S, and the distance of the first maximum from the interface at z = 0 .
For the incident wave in the region z < 0 , we have
\omega = 3 \times 10^{9} [rad / s] \\ \beta _{1} =\omega \sqrt{\mu _{o} \varepsilon _{o} \varepsilon _{r}} = 17.3 , and thus \sqrt{\varepsilon _{r}} = 1.73 \\ \eta _{1} = \sqrt{\frac{\mu _{o}}{\varepsilon _{o}\varepsilon _{r}} } = \frac{377}{1.73} = 217.92In the region z ≥ 0 , the intrinsic impedance is, from Eq. (8-79),
\boxed{\hat{\eta } = \sqrt{\frac{\mu }{\varepsilon } }\sqrt{\frac{1}{1 – j \sigma /(\omega \varepsilon )} }} [Ω] (8-79)
\eta _{2} = \sqrt{\frac{\mu _{o}}{\varepsilon }}\sqrt{\frac{1}{1 – j\sigma /(\omega \varepsilon )} } = \frac{377}{2}\sqrt{\frac{1}{1 -j 0.5 / (3\times 10^{9}\times 4 \times 8.854 \times 10^{-12})} } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad\quad = 188.50 \sqrt{\frac{1}{1- j 4.71} } = 85.90 e^{j0.68}The reflection coefficient is, from Eq. (8-107a),
\boxed{\Gamma = \frac{E^{r}_{o}}{E^{i}_{o}} = \frac{\eta _{2} – \eta _{1}}{\eta _{2} + \eta _{1}}} (8-107a)
\Gamma = \frac{\eta _{2} – \eta _{1}}{\eta _{2} + \eta _{1}} = \frac{85.90 \cos (0.68) – 217.92 + j 85.90\sin (0.68)}{85.90 \cos (0.68) + 217.92 + j 85.90\sin (0.68)} \\ \quad = \frac{- 151.13 + j 54.01}{284.71 + j 54.01} = \frac{160.49 e^{j2.80}}{289.79 e^{j0.19}} = 0.55 e^{j2.61}The standing wave ratio is, from Eq. (8-119),
\boxed{S = \frac{\left|\hat{E}_{1}\right|_{max} }{\left|\hat{E}_{1}\right|_{min} } = \frac{1 + \left|\Gamma \right| }{1 – \left|\Gamma \right| } } (8-119)
S = \frac{1 + \left|\Gamma \right| }{1 – \left|\Gamma \right| } = \frac{1 + 0.55 }{1 – 0.55 } = 3.44Substituting Φ = 2.61, \beta {1} = 17.3 , and n = 0 into Eq. (8-115b), we get
\boxed{z _{max} = – \frac{1}{2\beta _{1}} (\phi + 2\pi n)} (n = 0, ±1, ±2, …) (8-115b)
z_{max} = – \frac{\phi }{2 \beta {1}} = – \frac{2.61}{2 \times 17.3} = -7.5 [cm]