Power Coupled into Seawater at 1 GHz: 22-Elec-A7 Electromagnetics, May 2014, Q4

Question 4: Power Coupled into Seawater at 1 GHz (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given data
Quantity	Symbol	Value
Frequency	$f$	1000 MHz
Medium 1	—	free space, $\eta_0 = 376.7\ \Omega$
Seawater relative permittivity	$\epsilon_r$	81
Seawater resistivity	$\rho$	1 $\Omega\cdot$m
Seawater conductivity	$\sigma = 1/\rho$	1 S/m
Incidence	—	normal, on a flat interface

Find. The fraction of the incident power density that crosses the surface into the seawater.

Figure 4.1 — Normal incidence at the air-seawater boundary. What is not reflected is what penetrates.

Approach. Evaluate the loss tangent to decide which form of the intrinsic impedance is legitimate, compute the complex $\eta_2$, form the reflection coefficient, and subtract its power from unity.

Classify the medium first. The loss tangent decides everything that follows:$$\tan\delta = \frac{\sigma}{\omega\epsilon_0\epsilon_r} = \frac{1}{2\pi(10^{9})(8.85\times10^{-12})(81)} = \frac{1}{4.504} = 0.222.$$This is neither $\ll 1$ nor $\gg 1$: seawater at 1 GHz is a lossy dielectric, not a good conductor.
Intrinsic impedance of the seawater. The general expression must be used, complex arithmetic and all:$$\eta_2 = \sqrt{\frac{j\omega\mu_0}{\sigma + j\omega\epsilon_0\epsilon_r}} = \frac{\eta_0}{\sqrt{\epsilon_r}}\frac{1}{\sqrt{1 - j\tan\delta}}.$$With $\eta_0/\sqrt{81} = 41.86\ \Omega$ and $\sqrt{1-j0.222} = 1.0061 - j0.1103$,
Evaluate. Dividing through,$$\eta_2 = 41.11 + j4.51\ \Omega = 41.36\angle 6.26^{\circ}\ \Omega.$$The small positive phase angle is the fingerprint of moderate loss; a good conductor would show $45^{\circ}$, a perfect dielectric $0^{\circ}$.
Reflection coefficient at normal incidence. With $\eta_1 = \eta_0 = 376.7\ \Omega$,$$\Gamma = \frac{\eta_2 - \eta_1}{\eta_2 + \eta_1} = \frac{-335.6 + j4.51}{417.8 + j4.51},\qquad |\Gamma| = 0.803.$$The sign tells the physical story: seawater is the far lower impedance, so the surface behaves much like a short circuit and the reflected electric field is inverted.
Power split. Reflected and transmitted power fractions must sum to one:$$\frac{P_r}{P_i} = |\Gamma|^{2} = 0.645,\qquad \frac{P_t}{P_i} = 1 - |\Gamma|^{2},$$$$\boxed{\frac{P_t}{P_i} = 0.355 \quad\text{(35.5 \% penetrates).}}$$

Check: the transmitted fraction is computed as $1-|\Gamma|^2$, the power that crosses the boundary. It is not the power that survives to any particular depth — with $\alpha \approx 20.8$ Np/m at this frequency the field is down to $1/e$ within about 4.8 cm (the power within 2.4 cm), so about 98 % of that 35.5 % is absorbed in the first 10 cm of water.

It is instructive to bracket the answer. Treating the seawater as a lossless $\epsilon_r = 81$ dielectric (ignoring conductivity altogether) gives $\eta_2 = 41.86\ \Omega$ and 36.0 % transmission — barely half a point away, because a loss tangent of 0.22 perturbs the impedance only slightly. Treating it as a good conductor, on the other hand, would give an impedance near $\sqrt{\pi f\mu_0/\sigma}(1+j) = 62.8(1+j)\ \Omega$ and a materially different answer. The lesson is that the classification step is not a formality: at 1 GHz seawater has stopped behaving like the conductor it is at 1 MHz.

Final Results
Quantity	Symbol	Value
Conductivity	$\sigma$	1 S/m
Loss tangent	$\tan\delta$	0.222 (lossy dielectric)
Intrinsic impedance of seawater	$\eta_2$	$41.1 + j4.5\ \Omega$
Reflection coefficient magnitude	$\|\Gamma\|$	0.803
Reflected power fraction	$\|\Gamma\|^{2}$	0.645 (64.5 %)
Transmitted power fraction	$1-\|\Gamma\|^{2}$	0.355 (35.5 %)

Topic. Normal incidence of a plane wave on a lossy half space, and the classification of a medium by its loss tangent.

Key relations. $\tan\delta = \sigma/\omega\epsilon$; $\eta_2 = \sqrt{j\omega\mu/(\sigma+j\omega\epsilon)}$; $\Gamma = (\eta_2-\eta_1)/(\eta_2+\eta_1)$; $P_t/P_i = 1-|\Gamma|^{2}$.

Why this works. At normal incidence the boundary conditions on tangential E and H reduce to a simple impedance-mismatch problem identical in form to a transmission-line junction, so the entire reflection is governed by the ratio of intrinsic impedances. Loss enters only by making $\eta_2$ complex, which rotates the reflection coefficient slightly in phase without changing its magnitude much when the loss tangent is modest. Conservation of power then hands over the transmitted fraction without any need to compute the transmitted field explicitly.

Common pitfall. The standard trap is reaching for the good-conductor forms — $\eta = \sqrt{\pi f\mu/\sigma}(1+j)$ and $\alpha = \sqrt{\pi f\mu\sigma}$ — because the medium is seawater. Those hold only when $\tan\delta \gg 1$, which at 1 GHz is false by a factor of nearly five. A second slip is quoting $|\Gamma|$ itself as the reflected power fraction rather than $|\Gamma|^{2}$.

Source. Hayt & Buck, Engineering Electromagnetics, 9th ed., §12.1–12.2 (reflection at normal incidence); Sadiku, Elements of Electromagnetics, 7th ed., §10.6.

Question 4 of 8: Power Coupled into Seawater at 1 GHz

Question 4: Power Coupled into Seawater at 1 GHz (20 marks)