Problem 6.4.4: Oblique incidence on a perfect conductor

Problem 6.4.4

Oblique incidence on a perfect conductor

A uniform sinusoidal plane wave in air with the following phasor expression for electric intensity

E^{i} (x, z) = 10 e^{- j (6 x + 8 z)} a_{y} V/m

is incident on a perfectly conducting plane at $z = 0$ .

Find the frequency and wavelength of the wave.
Determine the angle of incidence.
Find the reflected electric and magnetic fields.
Find the total field in air and the planes where $E_{1}$ vanishes.
Find the surface current density on the plane.

Adapted from: D. K. Cheng, Field and Wave Electromagnetics, 2nd ed., Problem P.8-22.

Try it yourself first.

Strategy

A plane wave varies as $e^{- j β \cdot r}$ with $β = β a_{n}$ and $β = ω / c$ , and $H = \frac{1}{η _{0}} a_{n} \times E$ . Here the interface is $z = 0$ , so its normal is $a_{z}$ and $θ_{i}$ is measured from $a_{z}$ . The plane of incidence is the $x z$ -plane, and $E ∥ a_{y}$ is perpendicular to it: S polarisation, with $Γ_{⊥} = - 1$ at a perfect conductor.

Matching $β \cdot r = 6 x + 8 z$ :

β^{i} = 6 a_{x} + 8 a_{z} = 10 (0.6 a_{x} + 0.8 a_{z}) = β_{1} a_{n}^{i}, β_{1} = 10 rad/m,

ω f λ = β_{1} c = 3 \times 1 0^{9} rad/s, = \frac{ω}{2 π} = 477 MHz, = \frac{2 π}{β _{1}} = \frac{π}{5} = 0.628 m .

In the time domain $E^{i} (x, z, t) = 10 cos (3 \times 1 0^{9} t - 6 x - 8 z) a_{y} V/m$ .

$a_{n}^{i} = sin θ_{i} a_{x} + cos θ_{i} a_{z} = 0.6 a_{x} + 0.8 a_{z}$ , so $cos θ_{i} = 0.8$ and $θ_{i} = 36.8 7^{\circ}$ .

$θ_{r} = θ_{i}$ keeps the $x$ -part of the direction and reverses the $z$ -part: $a_{n}^{r} = 0.6 a_{x} - 0.8 a_{z}$ . $E_{y} (x, 0) = 0$ for every $x$ requires $Γ_{⊥} = - 1$ :

E^{r} (x, z) = - 10 e^{- j (6 x - 8 z)} a_{y} V/m,

H^{r} = \frac{1}{120 π} (0.6 a_{x} - 0.8 a_{z}) \times (- 10) e^{- j (6 x - 8 z)} a_{y} = - \frac{1}{12 π} (0.8 a_{x} + 0.6 a_{z}) e^{- j (6 x - 8 z)} A/m .

Likewise $H^{i} = \frac{1}{12 π} (- 0.8 a_{x} + 0.6 a_{z}) e^{- j (6 x + 8 z)} A/m$ .

Adding the two waves, with $e^{- j 8 z} - e^{j 8 z} = - 2 j sin 8 z$ and $e^{- j 8 z} + e^{j 8 z} = 2 cos 8 z$ :

E_{1} H_{1} = - j 20 sin (8 z) e^{- j 6 x} a_{y} V/m, = - \frac{2}{15 π} cos (8 z) e^{- j 6 x} a_{x} - j \frac{1}{10 π} sin (8 z) e^{- j 6 x} a_{z} A/m .

In the time domain $E_{1} (x, z, t) = 20 sin (8 z) sin (3 \times 1 0^{9} t - 6 x) a_{y} V/m$ : a standing wave along $z$ and a travelling wave along $x$ . $E_{1} \equiv 0$ on the planes

z = - \frac{mπ}{8} = 0, - 0.393, - 0.785, \dots m .

Along $x$ the phase constant is $β_{x} = 6 rad/m$ : $λ_{x} = 1.047 m$ and $v_{x} = ω / β_{x} = 5 \times 1 0^{8} m/s > c$ .

The normal pointing out of the conductor is $a_{n} = - a_{z}$ , and at $z = 0$ only $H_{x}$ survives:

J_{S} = a_{n} \times H_{1} (x, 0) = - a_{z} \times (- \frac{2}{15 π}) e^{- j 6 x} a_{x} .

Answer

J_{S} = \frac{2}{15 π} e^{- j 6 x} a_{y} = 42.4 e^{- j 6 x} a_{y} mA/m

Check

$H_{z} (x, 0) = 0$ : there is no normal $B$ at a perfect conductor. $∣ J_{S} ∣ = 2 H_{0}^{i} cos θ_{i} = 2 \times \frac{10}{120 π} \times 0.8 = 42.4 mA/m$ . The time-average Poynting vector is $\frac{1}{2} Re {E_{1} \times H_{1}^{*}} = \frac{1}{π} sin^{2} (8 z) a_{x} W/ m^{2}$ : power flows only parallel to the wall, and the $z$ -part of the field is purely reactive.

(a) Snapshot of E_y at t=0. — (a) Snapshot of $E_{y}$ at $t = 0$ .

(b) Power flow along the wall. — (a) Snapshot of $E_{y}$ at $t = 0$ .

What the graphs say

The pattern is a standing wave across the wall ( $z$ ) and a travelling wave along it ( $x$ ): the cells slide upwards at $v_{x} = 5 \times 1 0^{8} m/s$ . Power flows only along $x$ , most strongly halfway between the nodal planes. A second conducting plane at $z = - π /8$ would sit where $E_{1} = 0$ already, so it would disturb nothing: the field between the two plates is a mode of a parallel-plate waveguide, guided along $x$ at the phase velocity $5 \times 1 0^{8} m/s$ . This is the starting point of Section 7.2.