Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Problem 6.3.1

Normal incidence on a dielectric

A uniform plane wave of frequency $f = 100$ MHz travels in air ( $x < 0$ ) along $+ a_{x}$ with

E^{i} (x) = 12 e^{- j β_{1} x} a_{y} V/m

and impinges normally on a lossless nonmagnetic dielectric with $ε_{r} = 9$ that fills $x > 0$ .

Find $β_{1}$ , $β_{2}$ , $λ_{1}$ , $λ_{2}$ and $η_{2}$ .
Find $Γ$ , $τ$ and the reflected and transmitted $E$ and $H$ .
Find the standing-wave ratio in air and the positions of the maxima and minima of $∣ E_{1} (x) ∣$ .
Find the time-average power densities of the three waves and check the power balance.

Try it yourself first.

Strategy

Both media are lossless, so $Γ$ and $τ$ are real. Find the two intrinsic impedances, then $Γ$ and $τ$ from Eq. (6.3.10). The reflected magnetic field carries the extra minus sign of a wave travelling along $- a_{x}$ . The standing-wave pattern follows from $∣1 + Γ e^{2 j β_{1} x} ∣$ , the power from Eq. (6.3.13).

With $ω = 2 π \times 1 0^{8}$ rad/s,

β_{1} β_{2} = \frac{ω}{c} = \frac{2 π}{3} = 2.094 rad/m, λ_{1} = 3 m, = 9 β_{1} = 2 π rad/m, λ_{2} = 1 m,

η_{2} = \frac{η _{0}}{9} = 40 π = 125.7 Ω.

$Γ = \frac{40 π - 120 π}{40 π + 120 π} = - \frac{1}{2}$ and $τ = 1 + Γ = \frac{1}{2}$ . The incident magnetic field has the amplitude $H_{0}^{i} = 12/ (120 π) = 31.8$ mA/m, and

E^{r} H^{r} E^{t} H^{t} = - 6 e^{j 2.094 x} a_{y} V/m, = - \frac{E _{0}^{r}}{η _{1}} e^{j 2.094 x} a_{z} = 15.9 e^{j 2.094 x} a_{z} mA/m, = 6 e^{- j 2 π x} a_{y} V/m, = \frac{6}{40 π} e^{- j 2 π x} a_{z} = 47.7 e^{- j 2 π x} a_{z} mA/m .

With $E_{0}^{r} < 0$ the reflected magnetic field points along $+ a_{z}$ .

$S = (1 + 0.5) / (1 - 0.5) = 3$ . In air $∣ E_{1} (x) ∣ = 121 - \frac{1}{2} e^{j 2 β_{1} x}$ ( $θ_{Γ} = π$ ). It is smallest, $12 \times \frac{1}{2} = 6$ V/m, where $e^{j 2 β_{1} x} = 1$ , that is at $x = - m λ_{1} /2$ , and largest, $18$ V/m, a quarter wavelength further:

∣ E_{1} ∣_{m i n} ∣ E_{1} ∣_{m a x} = 6 V/m at x = 0, - 1.5, - 3, \dots m, = 18 V/m at x = - 0.75, - 2.25, \dots m .

With $P = ∣ E_{0} ∣^{2} / (2 η)$ ,

P^{i} P^{r} P^{t} = \frac{1 2 ^{2}}{2 \times 120 π} = 0.191 W/ m^{2}, = \frac{6 ^{2}}{2 \times 120 π} = 0.048 W/ m^{2}, = \frac{6 ^{2}}{2 \times 40 π} = 0.143 W/ m^{2} .

Answer

Γ = - \frac{1}{2}, τ = \frac{1}{2}, S = 3, R = 25 %, T = 75 %

Check

At $x = 0$ : $E_{y}$ : $12 - 6 = 6$ V/m $✓$ and $H_{z}$ : $31.8 + 15.9 = 47.7$ mA/m $✓$ . Power: $0.191 - 0.048 = 0.143$ W/m $^{2}$ $✓$ , and $R + T = \frac{1}{4} + \frac{η _{1}}{η _{2}} \cdot \frac{1}{4} = \frac{1}{4} + \frac{3}{4} = 1$ . The transmitted electric field is smaller than the incident one, $6 < 12$ V/m, but the transmitted magnetic field is larger, $47.7 > 31.8$ mA/m: medium 2 has the lower impedance.

(a) E_y(x,t) at \omega t=0 (dark) and 60^\circ (light). — (a) $E_{y} (x, t)$ at $ω t = 0$ (dark) and $6 0^{\circ}$ (light).

(b) The envelopes |E| and \eta_1|H|. — (a) $E_{y} (x, t)$ at $ω t = 0$ (dark) and $6 0^{\circ}$ (light).

Check: reading the graphs

Figure 6.3.9 shows the pattern. The envelope in air oscillates between $6$ and $18$ V/m with period $λ_{1} /2 = 1.5$ m, and the time snapshots stay inside it. In the dielectric the snapshots run between $\pm 6$ V/m with the wavelength $1$ m.

Problem 6.3.1: Normal incidence on a dielectric