Question 8.15: A right-hand circularly polarized wave, E^i = (Eo ax - j Eo ......
A right-hand circularly polarized wave, \pmb{E}^{i} =(E_{o} \pmb{a}_{x} – j E_{o} \pmb{a}_{y}) e^{-j\beta z}, propagates in free space(z < 0) and impinges normally on the surface of a perfect conductor at z = 0. Determine
(a) \mathscr{E}^r and its polarization state,
(b) surface current density induced on the surface of the perfect conductor, and
(c) instantaneous form of the total electric field intensity in free space.
(a) Using Γ = –1, the electric field phasor of the reflected wave is written as
\pmb{E}^{r} =\Gamma (E_{o} \pmb{a}_{x} – j E_{o} \pmb{a}_{y}) e^{j\beta z} = (-E_{o} \pmb{a}_{x} + j E_{o} \pmb{a}_{y}) e^{j\beta z} (8-128)
The instantaneous form of the reflected wave is
\mathscr{E}^{r} =Re \left[(-E_{o} \pmb{a}_{x} + e^{j\pi /2 } E_{o} \pmb{a}_{y}) e^{j\beta z} e^{j\omega t}\right] \\ \quad =-\pmb{a}_{x} E_{o} \cos (\omega t +\beta z) – \pmb{a}_{y} E_{o}\sin (\omega t +\beta z) (8-129)
At the z = 0 plane, we have
\mathscr{E}^{r} = – \pmb{a}_{x} E_{o} at \omega t= 0 (8-130a) \\ \mathscr{E}^{r} = – \pmb{a}_{y} E_{o} at \omega t = \pi /2 (8-130b)If the left four fingers follow the rotation of \mathscr{E}^{r} in Eq. (8-130), the left thumb points in the direction of propagation of the reflected wave, as shown in Fig. 8.14. Thus, the reflected wave is left-hand circularly polarized. Note that the polarization state of a circularly polarized wave is reversed when the wave is reflected by a perfect conductor.
(b) Magnetic field intensities of the incident and reflected waves are expressed in free space as
\pmb{H}^{i} = \frac{1}{\eta _{o}}(\pmb{a}_{k}\times \pmb{E}^{i}) = \frac{1}{\eta _{o}}\pmb{a}_{z} \times (E_{o} \pmb{a}_{x} – j E_{o} \pmb{a}_{y}) e^{-j\beta z}\\ \quad \quad \quad \quad \quad \quad \quad \quad =\frac{E_{o}}{\eta _{o}}( \pmb{a}_{y} + j \pmb{a}_{x}) e^{-j\beta z} (8-131) \\ \pmb{H}^{r} = \frac{1}{\eta _{o}}(\pmb{a}_{k}\times \pmb{E}^{r}) = \frac{1}{\eta _{o}}(-\pmb{a}_{z}) \times (-E_{o} \pmb{a}_{x} + j E_{o} \pmb{a}_{y}) e^{j\beta z}\\\quad \quad \quad \quad \quad \quad \quad \quad = \frac{E_{o}}{\eta _{o}}( \pmb{a}_{y} + j \pmb{a}_{x}) e^{j\beta z} (8-132)The total magnetic field intensity at the z = 0 plane is obtained from Eqs. (8-131) and (8-132) as
\pmb{H} (0) = \pmb{H}^{i}(o) + \pmb{H}^{r}(0) = \frac{2 E_{o}}{\eta _{o}}( \pmb{a}_{y} + j \pmb{a}_{x}) (8-133)
Using the outward unit normal to the surface, \pmb{a}_{n} = – \pmb{a}_{z} , the surface current density induced on the conductor is
\pmb{J} = \pmb{a}_{n} \times \pmb{H}(0) = (-\pmb{a}_{z}) \times \frac{2 E_{o}}{\eta _{o}}( \pmb{a}_{y} + j \pmb{a}_{x}) \\ \quad \quad \quad \quad \quad \quad \quad = \frac{2 E_{o}}{\eta _{o}}( \pmb{a}_{x} – j \pmb{a}_{y}) (8-134)
(c) Total electric field intensity in free space is written as
\pmb{E} = \pmb{E}^{i} + \pmb{E}^{r} = (E_{o} \pmb{a}_{x} – j E_{o} \pmb{a}_{y}) e^{- j \beta z} + ( – E_{o} \pmb{a}_{x} + j E_{o} \pmb{a}_{y}) e^{ j \beta z} \\ \quad \quad \quad \quad \quad \quad =- \pmb{a}_{x} 2j E_{o} \sin (\beta z) – \pmb{a}_{y} 2 E_{o} \sin (\beta z)The instantaneous form of the total electric field intensity in free space is therefore
\mathscr{E} = Re \left[ \pmb{E} e^{j\omega t} \right] = Re \left[\pmb{a}_{x} 2 E_{o} \sin (\beta z) e^{(j\omega t – \pi / 2) } – \pmb{a}_{y} 2 E_{o} \sin (\beta z) e^{j\omega t }\right] \\ \quad \quad \quad \quad \quad \quad \quad =2 E_{o} \sin (\beta z) [\pmb {a}_{x} \sin (\omega t) – \pmb{a}_{y} \cos (\omega t)]