Section 6.2 · Problems · 1 of 8

Problem 6.2.1: Seawater at 1 MHz and at 30 GHz

Interactive: Plane Wave in a Lossy Medium

Problem 6.2.1

Seawater at 1 MHz and at 30 GHz

Analyse seawater ( $σ = 4 S/m$ , $ε_{r} = 81$ , $μ_{r} = 1$ ) for the frequencies $1 MHz$ and $30 GHz$ : classify it at each frequency with $σ / (ω ε)$ , then find $α$ , $β$ , $λ$ , $v_{p}$ , $η_{c}$ and the skin depth $1/ α$ with the appropriate approximate formulas, and check them against the exact ones.

Adapted from: S. J. Orfanidis, Electromagnetic Waves and Antennas, Example 2.7.1.

Try it yourself first.

Strategy

The loss tangent $p = σ / ω ε$ decides which approximation is allowed. Compute it first, use the matching short formulas, then confirm with the exact $γ = j ω μ (σ + j ω ε)$ .

1. Classify

With $ε = 81 ε_{0}$ and $ε_{0} = 1 0^{- 9} /36 π$ , $ω ε = 2 π f \times 81 \times 1 0^{- 9} /36 π = 4.5 \times 1 0^{- 9} f$ :

p_{1 MHz} = \frac{4}{2 π \times 1 0 ^{6} \times 81 \times \frac{1}{36 π} \times 1 0 ^{- 9}} = \frac{4}{4.5 \times 1 0 ^{- 3}} = 888.9 ≫ 1,

p_{30 GHz} = \frac{4}{6 π \times 1 0 ^{10} \times 81 \times \frac{1}{36 π} \times 1 0 ^{- 9}} = \frac{4}{135} = 0.0296 ≪ 1.

At $1 MHz$ seawater is a good conductor, at $30 GHz$ a low-loss dielectric.

2. 1 MHz, good conductor

α = β \approx π f μ_{0} σ = π \times 1 0^{6} \times 4 π \times 1 0^{- 7} \times 4 = 3.97 m^{- 1}, δ = \frac{1}{α} = 0.252 m,

λ η_{c} = 2 π δ = 1.58 m, v_{p} = ω δ = 1.58 \times 1 0^{6} m/s, \approx (1 + j) \frac{π f μ _{0}}{σ} = 0.993 (1 + j) Ω.

3. 30 GHz, low-loss dielectric

With $μ_{0} /81 ε_{0} = 120 π /9 = 41.9 Ω$ ,

α \frac{1}{α} \approx \frac{σ}{2} \frac{μ _{0}}{81 ε _{0}} = \frac{4}{2} \times \frac{120 π}{9} = 83.8 Np/m, = 1.19 cm,

β \approx ω μ_{0} 81 ε_{0} = \frac{2 π \times 3 \times 1 0 ^{10} \times 9}{3 \times 1 0 ^{8}} = 5655 rad/m,

λ v_{p} η_{c} = \frac{2 π}{β} = 1.11 mm, = \frac{c}{9} = 3.33 \times 1 0^{7} m/s, \approx \frac{120 π}{9} (1 + j 0.0148) = 41.9 + j 0.62 Ω.

4. Exact values

At $1 MHz$ the exact formulas give $α = 3.972 Np/m$ , $β = 3.976 rad/m$ , $η_{c} = 0.994 + j 0.993 Ω$ , and at $30 GHz$ $α = 83.77 Np/m$ , $β = 5655.5 rad/m$ , $η_{c} = 41.87 + j 0.62 Ω$ . Both approximations are good to better than $0.1 %$ .

Answer

1 MHz : α = β = 3.97 m^{- 1}, δ = 0.252 m, λ = 1.58 m, v_{p} = 1.58 \times 1 0^{6} m/s, η_{c} = 0.993 (1 + j) Ω, 30 GHz : α = 83.8 Np/m, β = 5655 rad/m, λ = 1.11 mm, v_{p} = 3.33 \times 1 0^{7} m/s, η_{c} = 41.9 + j 0.62 Ω

Drag $f$ below from $1 MHz$ to $30 GHz$ and see seawater move from a good conductor ( $σ / ω ε = 888.9$ ) to a low-loss dielectric ( $σ / ω ε = 0.0296$ ).

InteractiveLoading simulation…

Check

At $1 MHz$ the angle of $η_{c}$ is $4 5^{\circ}$ and $α = β$ , as for any good conductor. At $30 GHz$ $α / β = 0.0148 = p /2$ , the low-loss relation $α = \frac{β}{2} tan δ_{c}$ . "Low-loss" is per wavelength: at $30 GHz$ the field loses only $α λ = 0.093 Np$ ( $0.8 dB$ ) per $1.1 mm$ wavelength, but that is $728 dB/m$ . At $1 MHz$ , the "good conductor", the loss is $8.686 \times 3.97 = 34.5 dB/m$ , twenty times less.

Problem 6.2.1. The exact \alpha(f) and \beta(f) of the seawater model from 10\,\mathrm{Hz} to 100\,\mathrm{GHz}, with the two approximations. Below about 10\,\mathrm{MHz} \alpha=\beta\propto\sqrt f, a good conductor. Above a few GHz \alpha saturates at 83.8\,\mathrm{Np/m} while \beta\propto f, a low-loss dielectric. Around f_t=\sigma/2\pi\varepsilon=889\,\mathrm{MHz} neither approximation holds. — **Figure 6.2.17.** Problem 6.2.1. The exact $α (f)$ and $β (f)$ of the seawater model from $10 Hz$ to $100 GHz$ , with the two approximations. Below about $10 MHz$ $α = β \propto f$ , a good conductor. Above a few GHz $α$ saturates at $83.8 Np/m$ while $β \propto f$ , a low-loss dielectric. Around $f_{t} = σ /2 π ε = 889 MHz$ neither approximation holds.

Check: the whole spectrum

Figure 6.2.17 joins the two answers. Real water is a polar dielectric, though: its $ε_{r}$ falls and its $ε^{''}$ rises above a few gigahertz (a relaxation near $20 GHz$ ), so the $30 GHz$ numbers are those of the model medium, and the true attenuation is larger still.

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