Problem 6.3.2: Standing wave in front of a conducting wall

Problem 6.3.2

Standing wave in front of a conducting wall

A uniform plane wave of frequency $f = 3$ GHz travels in air along $+ a_{x}$ , $H^{i} (x) = 0.05 e^{- j β x} a_{z}$ A/m, and is incident normally on a perfectly conducting wall at $x = 0$ .

Find $β$ , $λ$ and $E^{i} (x)$ .
Find the reflected fields and the total fields $E_{1} (x, t)$ , $H_{1} (x, t)$ in air.
Within $10$ cm of the wall, where are $E_{1}$ and $H_{1}$ zero at all times?
Find the surface current density on the wall.
Find the time-average and instantaneous Poynting vectors.

Try it yourself first.

Strategy

A perfect conductor has $η_{2} = 0$ , so $Γ = - 1$ and $τ = 0$ . Write the incident $E$ from $H$ , add the reflected wave, and convert the sum into a product of a function of $x$ and a function of $t$ : a pure standing wave, as in Eq. (6.3.23). The wall current follows from the jump of $H_{t a n}$ , the Poynting vectors from the product of the total fields.

$β = 2 π f / c = 20 π = 62.8$ rad/m and $λ = 10$ cm. For a plane wave $E = η_{0} H \times a_{k}$ , with $a_{k} = a_{x}$ the direction of travel (the inverse of Eq. (6.3.5)): $η_{0} (0.05 a_{z}) \times a_{x} = 6 π a_{y}$ , so

E^{i} (x) = 18.85 e^{- j 20 π x} a_{y} V/m .

With $Γ = - 1$ : $E^{r} = - 18.85 e^{j 20 π x} a_{y}$ V/m and $H^{r} = - (E_{0}^{r} / η_{0}) e^{j 20 π x} a_{z} = 0.05 e^{j 20 π x} a_{z}$ A/m. Adding,

E_{1} H_{1} = - j 37.7 sin (20 π x) a_{y}, = 0.1 cos (20 π x) a_{z},

and with $ω = 6 π \times 1 0^{9}$ rad/s and $Re {- j e^{j ω t}} = sin ω t$ :

Answer

E_{1} (x, t) H_{1} (x, t) = 37.7 sin (20 π x) \times sin (6 π \times 1 0^{9} t) a_{y} V/m, = 0.1 cos (20 π x) \times cos (6 π \times 1 0^{9} t) a_{z} A/m

$E_{1} \equiv 0$ where $sin (20 π x) = 0$ : $x = 0, - 5, - 10$ cm. $H_{1} \equiv 0$ where $cos (20 π x) = 0$ : $x = - 2.5, - 7.5$ cm.

The normal pointing out of the conductor is $a_{n} = - a_{x}$ , and inside the PEC $H = 0$ :

J_{S} = a_{n} \times H_{1} (0, t) = - a_{x} \times 0.1 cos ω t a_{z} = 0.1 cos (6 π \times 1 0^{9} t) a_{y} A/m .

$\frac{1}{2} Re {E_{1} \times H_{1}^{*}} = \frac{1}{2} Re {- j 3.77 sin β x cos β x} a_{x} = 0$ , but the instantaneous Poynting vector is not zero:

P (x, t) = E_{1} \times H_{1} = 3.77 sin β x cos β x sin ω t cos ω t a_{x} = 0.942 sin (2 β x) sin (2 ω t) a_{x} W/ m^{2} .

Check

$E_{y} (0, t) = 0$ at the wall, as a PEC requires. $∣ J_{S} ∣ = 0.1$ A/m $= 2 H_{0}^{i}$ , the jump of $H_{z}$ from its wall value to zero inside the metal, in agreement with Eq. (6.3.25). The amplitude $0.942$ W/m $^{2}$ is $(E_{0}^{i})^{2} / η_{0} = 18.8 5^{2} /377$ , as in Eq. (6.3.24).

(a) E_y at \omega t=90^\circ,45^\circ,15^\circ (dark to light). — (a) $E_{y}$ at $ω t = 9 0^{\circ}, 4 5^{\circ}, 1 5^{\circ}$ (dark to light).

(a) H_z at \omega t=0,45^\circ,75^\circ (dark to light). — (a) $E_{y}$ at $ω t = 9 0^{\circ}, 4 5^{\circ}, 1 5^{\circ}$ (dark to light).

Remark

$E$ and $H$ are $9 0^{\circ}$ apart in space and in time. The instantaneous energy flow swings back and forth inside each $λ /4 = 2.5$ cm cell, between an $E$ antinode and an $H$ antinode, twice per period, and the net flow is zero: everything that comes in goes back out (Figure 6.3.10).