Question 7 of 8: Plane-Wave Attenuation in Seawater
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, December 2013 — 07-Elec-A7 Electromagnetics. Three hours, closed book, one approved calculator. Eight questions of equal value (20 marks each); the paper states that any five constitute a complete paper, but all eight are solved here so the set works as a complete study resource. Aids printed on the paper: $\epsilon_0 = 8.85\times10^{-12}$ F/m and $\mu_0 = 4\pi\times10^{-7}$ H/m.
Reference texts. Sadiku, Elements of Electromagnetics, 7th ed.; Hayt & Buck, Engineering Electromagnetics, 9th ed.; Ulaby & Ravaioli, Fundamentals of Applied Electromagnetics, 8th ed.; Pozar, Microwave Engineering, 4th ed.; Balanis, Antenna Theory: Analysis and Design, 4th ed.
Free-space quantities follow from the printed aids: $c = 1/\sqrt{\mu_0\epsilon_0} \approx 3\times10^{8}$ m/s and $\eta_0 = \sqrt{\mu_0/\epsilon_0} = 376.8\ \Omega$.
Question 7: Plane-Wave Attenuation in Seawater (20 marks)
Given. A 1 GHz plane wave inside seawater, a lossy dielectric whose conduction and displacement currents are of comparable size at this frequency.
Given data
Quantity
Symbol
Value
Frequency
$f$
1 GHz ($\omega = 6.283\times10^{9}$ rad/s)
Relative permittivity
$\epsilon_r$
81
Conductivity
$\sigma$
7 S/m
Permeability
$\mu$
$\mu_0$ (non-magnetic)
Find. The attenuation constant, expressed in decibels per metre.
Figure 7.1 — Amplitude decay of the wave with depth. The 1/e penetration depth is only 8.1 mm.
Approach. Evaluate the loss tangent to see which regime applies, use the exact attenuation formula for a general lossy medium, and convert nepers to decibels.
Loss tangent — which regime? The ratio of conduction to displacement current density is$$\tan\delta = \frac{\sigma}{\omega\epsilon} = \frac{7}{(6.283\times10^{9})(81)(8.85\times10^{-12})} = \frac{7}{4.504} = 1.554 .$$This is of order unity, so seawater at 1 GHz is neither a good conductor nor a low-loss dielectric and neither approximation may be used — the exact expression is required.
Exact attenuation constant. For a medium with $\mu$, $\epsilon$ and $\sigma$,$$\alpha = \omega\sqrt{\frac{\mu\epsilon}{2}}\left[\sqrt{1+\left(\frac{\sigma}{\omega\epsilon}\right)^{2}}-1\right]^{1/2}\ \text{Np/m}.$$Both the permittivity and the conductivity appear, which is precisely why an intermediate loss tangent has to be handled exactly.
Substitute the numbers. With $\epsilon = 81\epsilon_0 = 7.169\times10^{-10}$ F/m,$$\omega\sqrt{\frac{\mu_0\epsilon}{2}} = (6.283\times10^{9})(2.122\times10^{-8}) = 133.3,$$$$\left[\sqrt{1+(1.554)^{2}}-1\right]^{1/2} = \left[1.848-1\right]^{1/2} = 0.9209,$$$$\alpha = 133.3 \times 0.9209 = 122.8\ \text{Np/m}.$$The companion phase constant is $\beta = 225.0$ rad/m, giving a wavelength inside the water of only 27.9 mm.
Convert to decibels per metre. One neper of amplitude decay is $20\log_{10}e = 8.686$ dB, so$$\alpha_{\text{dB}} = 122.8 \times 8.686,$$$$\boxed{\alpha = 1.07\times10^{3}\ \text{dB/m}}$$equivalently $\alpha = 122.8$ Np/m, or about 10.7 dB per centimetre.
Interpretation and cross-check. The penetration depth is $\delta = 1/\alpha = 8.14$ mm: the field falls to 37 % of its value in under a centimetre, which is why radio at GHz frequencies is useless underwater and submarine communication uses VLF instead. As a check, the good-conductor approximation $\alpha \approx \sqrt{\pi f\mu_0\sigma} = 166.2$ Np/m over-estimates the true value by 35 %, confirming that seawater is genuinely in the intermediate regime at 1 GHz and that the exact formula was necessary.
