Question 8.13: A standing wave of S = 4 is formed in free space(z < 0). ......

The distance between two adjacent maxima is equal to a half wavelength:

λ = 2 × 0.5[m] , and thus \beta _{1} = \frac{2 \pi }{\lambda } = 2\pi .                        (8-120)

The first maximum corresponds to n = 0 in Eq. (8-115b):

\boxed{z _{max} = – \frac{1}{2\beta _{1}} (\phi + 2\pi n)}                       (n = 0, ±1, ±2, …)                              (8-115b)
z _{max} = -\frac{1 }{2 \beta _{1} }\phi = – 0.2 [m]                                        (8-121)

Inserting Eq. (8-120) into Eq. (8-121), we get

Φ = 0.8π                                                                    (8-122)

From Eq. (8-119), we get

\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)

\left|\Gamma \right| = \frac{S – 1}{S + 1} = \frac{4 -1}{4 + 1} = 0.6                                                               (8-123)

Combining Eq. (8-122) and Eq. (8-123), the reflection coefficient is

\Gamma = 0.6 e^{j 0.8 \pi }                                                                   (8-124)

Rewriting Eq. (8-107a), we have

\boxed{\Gamma = \frac{E^{r}_{o}}{E^{i}_{o}} = \frac{\eta _{2}  –  \eta _{1}}{\eta _{2} + \eta _{1}}}                                                        (8-107a)

\frac{\eta _{2}}{\eta _{1}} = \frac{1 + \Gamma }{1  –  \Gamma }                                                                   (8-125)

Substituting Eq. (8-124) and  \eta _{1} = \eta _{o}= 377 [Ω] into Eq. (8-125), we get the intrinsic impedance of medium 2 as

\eta _{2}=  377  \frac{1 + 0.6 e^{j 0.8\pi } }{1 – 0.6 e^{j 0.8\pi } } = 154 e^{j 0.83} [\Omega ].