Exercise 5.6.20: A damped wave in an oblique direction

About the further exercises of Section 5.6

Twenty further exercises in the spirit of the problems, the worked examples and the home exercises, graded from easy to harder within two groups. Exercises 5.6.1–5.6.10 belong to the lecture: Exercises 5.6.1–5.6.2 are about phasors and the phase velocity, 5.6.3–5.6.5 about the complex permittivity and the wave number, 5.6.6–5.6.7 about the Helmholtz equation and 5.6.8–5.6.10 about power. Exercises 5.6.11–5.6.20 belong to the exercise lecture: reading time-harmonic fields, loops, mixers and receivers, power, and the medium found from measurements. They use only

u = Re {U e^{j ω t}}, sin θ = cos (θ - \frac{π}{2}), \frac{\partial}{\partial t} \to j ω, v_{p} = \frac{ω}{∣\nabla ϕ ∣}, λ = \frac{2 π}{β}, ε_{c} = ε - j \frac{σ}{ω} = ε^{'} - j ε^{''}, tan δ = \frac{ε ^{''}}{ε ^{'}}, k = ω μ ε_{c} = β - j α, \nabla^{2} E + k^{2} E = 0, \nabla \cdot E = 0 (source-free), \nabla \times E = - j ω μ H, \nabla \times H = j ω ε_{c} E, η = \frac{ω μ}{k} = \frac{μ}{ε _{c}}, emf = - \frac{d Φ}{d t},

and, from Section 6.1, the instantaneous and the complex Poynting vectors $P (t) = E (t) \times H (t)$ and $P_{c} = \frac{1}{2} E \times H^{*}$ , with $P_{a v} = Re {P_{c}}$ , and the time-harmonic Poynting relation

- \oint_{S} P_{c} \cdot d S - \int_{V} \frac{1}{2} E \cdot J_{i}^{*} d V = \int_{V} [\frac{1}{2} σ ∣ E ∣^{2} + j 2 ω (w_{m} - w_{e})] d V,

with the time-average energy densities $w_{e} = \frac{1}{4} ε ∣ E ∣^{2}$ and $w_{m} = \frac{1}{4} μ ∣ H ∣^{2}$ (real $ε$ , $μ$ ). For a wave $E_{0} e^{- j k \cdot r}$ every $\nabla$ becomes $- j k$ , so Faraday's law gives $H = k \times E / ω μ$ . A small loop of $N$ turns and area $A$ in a field $H$ of angular frequency $ω$ picks up an emf of amplitude $N ω μ ∣ H \cdot n ∣ A$ , with $n$ its normal.

Exercise 5.6.20

A damped wave in an oblique direction

In a source-free nonmagnetic medium the electric field of a wave is

E (r; t) = (4 a_{x} - 4 a_{y} + 2 a_{z}) e^{- (2 x + 4 y + 4 z) /3} \times cos (6 \times 1 0^{8} t - 2 x - 4 y - 4 z + \frac{π}{3}) V/m .

(a) Write the phasor as $E_{0} e^{- j k \cdot r}$ with a complex vector $k = (β - j α) a_{k}$ . Find $a_{k}$ , $β$ , $α$ , $λ$ and $v_{p}$ , and check $\nabla \cdot E = 0$ . (b) Find $ε_{r}$ and $σ$ . (c) Find $H (r; t)$ . (d) Find $P_{a v}$ at the origin, the power density crossing the plane $z = 0$ there, and the distance along $a_{k}$ over which the power halves.

Try it yourself first.

Strategy

Both the phase and the decay are proportional to $x + 2 y + 2 z$ : the planes of constant phase are also planes of constant amplitude, and the wave is a lossy plane wave along $a_{k} = (a_{x} + 2 a_{y} + 2 a_{z}) /3$ . With $\nabla \to - j k$ every vector operation becomes algebra: $\nabla \cdot E = - j k \cdot E$ , $H = k \times E / ω μ_{0}$ and $k^{2} = k \cdot k = ω^{2} μ_{0} ε_{c}$ .

$2 x + 4 y + 4 z = 6 a_{k} \cdot r$ and $(2 x + 4 y + 4 z) /3 = 2 a_{k} \cdot r$ , so $β = 6 rad/m$ and $α = 2 Np/m$ along $a_{k} = \frac{1}{3} (a_{x} + 2 a_{y} + 2 a_{z})$ :

E (r) λ = (4 a_{x} - 4 a_{y} + 2 a_{z}) e^{j π /3} \times e^{- j (6 - 2 j) a_{k} \cdot r} V/m, = \frac{2 π}{6} = 1.047 m, v_{p} = \frac{6 \times 1 0 ^{8}}{6} = 1 0^{8} m/s .

$\nabla \cdot E = - j k \cdot E$ , and $(a_{x} + 2 a_{y} + 2 a_{z}) \cdot (4 a_{x} - 4 a_{y} + 2 a_{z}) = 4 - 8 + 4 = 0$ .

$k_{0} = ω / c = 2 rad/m$ and $k^{2} = (6 - 2 j)^{2} = 32 - 24 j$ :

\frac{ε _{c}}{ε _{0}} σ = \frac{32 - 24 j}{4} = 8 - 6 j, ε_{r} = 8, = ω ε_{0} \times 6 = \frac{6 \times 1 0 ^{8} \times 1 0 ^{- 9}}{36 π} \times 6 = \frac{1}{10 π} = 0.0318 S/m (tan δ = 0.75) .

$(a_{x} + 2 a_{y} + 2 a_{z}) \times (4 a_{x} - 4 a_{y} + 2 a_{z}) = 12 a_{x} + 6 a_{y} - 12 a_{z} = 6 (2 a_{x} + a_{y} - 2 a_{z})$ and $ω μ_{0} = 240 π$ :

H = \frac{k \times E}{ω μ _{0}} = \frac{( 6 - 2 j ) e ^{j π /3}}{3 \times 240 π} 6 (2 a_{x} + a_{y} - 2 a_{z}) e^{- j k \cdot r} = \frac{6 - 2 j}{120 π} e^{j π /3} (2 a_{x} + a_{y} - 2 a_{z}) e^{- j k \cdot r} .

$(6 - 2 j) /120 π = 0.01678 e^{- j 18.4 3^{\circ}}$ , so

Answer

H (r; t) = 0.01678 (2 a_{x} + a_{y} - 2 a_{z}) e^{- (2 x + 4 y + 4 z) /3} \times cos (6 \times 1 0^{8} t - 2 x - 4 y - 4 z + 41.5 7^{\circ}) A/m

With $η = ω μ_{0} / k = 240 π / (6 - 2 j) = 6 π (6 + 2 j) = 119.2 e^{j 18.4 3^{\circ}} Ω$ and $∣ E_{0} ∣ = 6 V/m$ ,

P_{c} = \frac{1}{2} E \times H^{*} = \frac{∣ E _{0} ∣ ^{2}}{2 η ^{*}} e^{- 4 a_{k} \cdot r} a_{k} = \frac{36 ( 6 + 2 j )}{2 \times 6 π \times 40} e^{- 4 a_{k} \cdot r} a_{k} = \frac{3 ( 3 + j )}{20 π} e^{- 4 a_{k} \cdot r} a_{k} .

At the origin

P_{a v} = \frac{9}{20 π} a_{k} = 0.1432 a_{k} = (0.0477 a_{x} + 0.0955 a_{y} + 0.0955 a_{z}) W/ m^{2},

and the plane $z = 0$ is crossed by $P_{z} = 0.0955 W/ m^{2}$ . The power halves when $e^{- 2 α s} = \frac{1}{2}$ , that is after $s = ln 2/4 = 0.173 m$ along $a_{k}$ .

Answer

a_{k} = \frac{1}{3} (1, 2, 2), β = 6 rad/m, α = 2 Np/m, λ = 1.047 m, v_{p} = 1 0^{8} m/s ε_{r} = 8, σ = 0.0318 S/m ∣ P_{a v} (0) ∣ = 0.143 W/ m^{2}, P_{z} = 0.0955 W/ m^{2}, s_{1/2} = 17.3 cm

Check

$H \cdot a_{k} \propto 2 + 2 - 4 = 0$ and $H \cdot E_{0} \propto 8 - 4 - 4 = 0$ : $E$ , $H$ and $a_{k}$ are mutually perpendicular, and $∣ H_{0} ∣ = 0.01678 \times 3 = 0.0503 = 6/119.2 = ∣ E_{0} ∣/∣ η ∣$ . A finite-difference evaluation of $\nabla \times E$ and $\nabla \times H$ at a random point reproduces $- j ω μ_{0} H$ and $j ω ε_{c} E$ .

Remark

$H$ lags $E$ by $18. 4^{\circ} = δ /2$ , the angle of $η$ , and $Im {P_{c}} = \frac{3}{20 π} = 0.048 W/ m^{2}$ at the origin is reactive. The plane $z = 0$ is crossed by only $cos θ_{z} = 2/3$ of $∣ P_{a v} ∣$ , as in Home exercise 5.6.6. Here the decay is along $a_{k}$ . A wave that enters a lossy medium obliquely from air decays perpendicular to the boundary instead, and then the planes of constant amplitude and of constant phase differ (Section 6.4).

cf. Problems 5.6.8, 5.6.13