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

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 5.6.7

A spherical wave in lake water

(a) Show that $u (r) = A e^{- j k r} / r$ satisfies the scalar Helmholtz equation $\nabla^{2} u + k^{2} u = 0$ for $r > 0$ (use $\nabla^{2} u = \frac{1}{r} \frac{d ^{2} ( r u )}{d r ^{2}}$ for $u = u (r)$ ). (b) A small source radiates such a wave at $f = 100 MHz$ in lake water, $ε_{r}^{'} = 81$ , $σ = 0.01 S/m$ , $μ = μ_{0}$ . Find $tan δ$ , $k$ , $λ$ and $1/ α$ . (c) The amplitude falls by spreading ( $1/ r$ ) and by absorption ( $e^{- α r}$ ). Up to which distance does spreading dominate? (d) Find, in dB, the spreading loss and the absorption loss from $1$ to $10 m$ and from $10$ to $20 m$ .

Try it yourself first.

Strategy

For a function of $r$ alone the Laplacian reduces to $\frac{1}{r} (r u)^{''}$ , and $r u = A e^{- j k r}$ is a one-dimensional wave. The slope of $ln ∣ u ∣$ splits into a spreading part $1/ r$ and an absorption part $α$ .

$r u = A e^{- j k r}$ and $(r u)^{''} = (- j k)^{2} A e^{- j k r} = - k^{2} r u$ , so $\nabla^{2} u = \frac{1}{r} (- k^{2} r u) = - k^{2} u$ and $\nabla^{2} u + k^{2} u = 0$ for $r > 0$ .

$ω ε^{'} = 2 π \times 1 0^{8} \times 81 \times 1 0^{- 9} /36 π = 0.45 S/m$ and $tan δ = 0.01/0.45 = 0.0222$ . From the closed forms of Eq. (5.6.20) with $k_{0} ε_{r}^{'} = \frac{2 π}{3} \times 9 = 18.85 rad/m$ ,

k λ = 18.851 - j 0.2094 m^{- 1}, = \frac{2 π}{β} = 33.3 cm, \frac{1}{α} = 4.77 m .

$∣ u ∣ = ∣ A ∣ e^{- α r} / r$ , so $- \frac{d ln ∣ u ∣}{d r} = \frac{1}{r} + α$ . Spreading dominates while $1/ r > α$ , that is for $r < 1/ α = 4.77 m$ .

Spreading costs $20 lo g_{10} (r_{2} / r_{1})$ , absorption $8.686 α (r_{2} - r_{1})$ with $8.686 α = 1.82 dB/m$ :

	spreading	absorption
$1 \to 10 m$	$20.0 dB$	$16.4 dB$
$10 \to 20 m$	$6.0 dB$	$18.2 dB$

Answer

tan δ = 0.0222, k = 18.85 - j 0.209 m^{- 1}, λ = 33.3 cm, \frac{1}{α} = 4.77 m spreading dominates for r < 4.77 m

Check

The low-loss estimate $α \approx \frac{1}{2} β tan δ = \frac{1}{2} \times 18.85 \times 0.0222 = 0.2094 Np/m$ agrees with the exact value to four digits.

Remark

Near a source the $1/ r$ law costs the most, far from it the exponential always wins: every further metre of lake water costs $1.8 dB$ , while every further metre of spreading costs less and less. A radio link through water is therefore limited by absorption, which grows with frequency (Section 6.2). Submarine links use VLF.

cf. Problem 5.6.7

Exercise 5.6.7: A spherical wave in lake water