# How is radio wave propagation modelled in seawater?

Tags:
1. Jul 24, 2015

### H Smith 94

Before I start, I apologise for the information dump that is to follow. I don't expect all questions to be answered or all models to be addressed; I simply feel it is appropriate to provide the community with my current knowledge and stage of research so you may not have to search for it yourselves in order to begin a discussion. It also might help some to find a consolidated resource regarding this subject (I know it would have helped me!)

From the very limited literature I can find regarding radio waves in saline-water solutions (as in seawater), I have been able to find very few corroborating models of radio wave propagation through electrically lossy media. I have found the following descriptions, which I have paraphrased here and adapted.

My questions

My primary questions here are:
1. What models of radio wave propagation currently exist?
2. How do these models apply to propagation and attenuation through salt-water solutions and seawater?
3. What are the errors and successes with the models I have provided?
4. Where does my understanding/knowledge falter?
Current models

Note that I have "boxed" the final equations of each model to ease reading.

Ghasemi's (2012) theoretical model

Power flux density

According to Friis' transmission equation, the received power flux density $P_r$ is given by

$$P_r=P_t G_t G_r \left(\frac{λ}{4πr}\right)^2,$$

where $P_t$ is the transmitted power flux density; $G_t$ and $G_r$ are the gain of the transmission antenna and the receiving antenna, respectively; and $r$ is the distance between the two antenna.

Since $λ=c/f$, where $c$ is the speed of light within the medium and $c=c_0/n$, where $n$ is the refractive index of the medium and $c_0$ is the speed of light in free space. Thus,

$$P_r=P_t G_t G_r \left(\frac{c_0}{4πrnf}\right)^2.$$

If salinity is $S$ and the resulting change in refractive index is $\Delta n$, then it may be empirically determined that $S=0.2\,\Delta n$, so if $n=n_\text{water}+\Delta n$ then $n=n_\text{water}+\frac{S}{0.2}$.

Now, since $n_\text{water} =1.3330$,

$$P_r=P_t G_t G_r \left(\frac{c_0}{4πrf}\frac{0.2}{0.267+S}\right)^2.$$

Using the https://www.itu.int/dms_pubrec/itu-r/rec/v/R-REC-V.573-5-200709-I!!PDF-E.pdf [Broken] (pg. 6) $L$, which states that

$$L = 10 \log\left|\frac{P_t}{P_r}\right|$$

we find that,

$$\begin{split}L &= -10 \log\left|P_r/P_t\right|\\&= -10 \log\left|G_t G_R \left(\frac{c_0}{4πrf}\frac{0.2}{0.267+S}\right)^2 \right|.\end{split}$$

This model is according to Ghasemi's Propagation Engineering in Wireless Communications (2012, pp. 37-39) and has been adapted to incorporate salinity.

Using this further, we find that

$$\boxed{L_\text{total} = \text{FSL}\,[\text{dB}] + L_m\,[\text{dB}] − G_t \,[\text{dB}] − G_r \,[\text{dB}]}.$$

where $\text{FSL}$ is the free space loss given by

$$\text{FSL}\,[\text{dB}] = 20 \log\left|\frac{4πrf}{c_0}\frac{0.267+S}{0.2}\right|$$

and $L_m$ is the loss within the medium (undefined by Ghasemi.)

Depth of penetration

If we define depth of penetration as $x\equiv1/\alpha$, where $\alpha = \sqrt{\pi\mu\sigma f}$ is the attenuation, so

$$\begin{split}x&=\frac{1}{\sqrt{\pi\mu\sigma f}}\\&=\frac{1}{\sqrt{\pi\mu_0}}\cdot\frac{1}{\sqrt{\mu_r\sigma f}}\\&=892.0621\frac{1}{\sqrt{\mu_r\sigma f}}.\end{split}$$

Now, "for all practical purposes the relative permeability of salt water is 1," (Herring, 2007) so:

$$\boxed{x=892.0621\frac{1}{\sqrt{\sigma f}}}.$$

The above is a simulation (using Excel) of the model, showing the penetration depth and absolute transmission loss as a function of frequency. Here we witness the penetration depth holding true to our expectations—if perhaps a little more extreme than expected—and the transmission loss showing the *opposite* of what the penetration depth shows.

Zoksimovski (2012) semi-empirical model

I did find a good experimental report, Underwater Electromagnetic Communications Using
Conduction – Channel Characterization (Zoksimovski et. al., 2012), which detailed three separate approximate models for radio wave attenuation using a channel transfer function $H$, outlining two far-field approximations:

$$|H| = A_0 e^{-\alpha_1\sqrt{f}} \ \ \ \ \ \ \ (1)$$

$$|H| = A_0 e^{-\alpha_2 f} \ \ \ \ \ \ \ \ \ (2)$$

and a near-field approximation:

$$\boxed{|H| = \frac{p_1f^2 + p_2f + p_3}{q_1f^2 + q_2f + q_3}\,e^{-\alpha_3 f}}.$$

I'm interested more-so in the near field due to the high observed attenuation of radio waves in seawater. The data provided by Zoksimovski et. al. was unclear at best, but averages were found to be $p_1 = 0.75\times 10^{-5}$, $p_2 = 2.30\times 10^{-4}$, $p_3 = 17.25\times 10^{-5}$, $q_1 = 0.30$, $q_2 = 0.15$, $q_3 = 0.18$, $\alpha_3 = 0.55$. These were the values used in the following simulation.

As you can see this model produces *tiny* received signals which are killed off within 10 Hz. I know this to not be physically accurate. It also does not factor the distance between antenna anywhere into calculations.

This is the graphs produced by Zoksimovski et. al. which do not show the same trends as my simulation (see the thick blue lines for the model discussed.) Perhaps this is an error in my simulation, however intuition suggests that the low values of the parameters will influence the trend in the way I've simulated.

Cole-Cole Relation

The Cole-Cole relation—and its specialisation, the Debye relaxation model—was brought to my attention by O'Shaugnessy's 2012 brilliant paper, Characterising the Relative Permittivity and Conductivity of Seawater for Electromagnetic Communications in the Radio Band.

The Cole-Cole relation states that

$$\hat{\epsilon}(\omega) = \epsilon_\infty + \frac{\Delta\epsilon}{1+(j\omega\tau)^{1-\alpha}}$$

where $\hat{\epsilon}(\omega)$ is the complex permittivity, $\hat{\epsilon}(\omega) = \epsilon' + j\epsilon''$; $\Delta\epsilon = \epsilon_s- \epsilon_\infty$, where $\epsilon_s$ is the static permittivity and $\epsilon_\infty$ is the infinite frequency permittivity; $\omega = 2\pi f$; $\tau$ is the dielectric relaxation time; $j = \sqrt{-1}$; and $\alpha$ is used to describe the spectral shape:

$\alpha = 0$, Debye model,
$\alpha > 0$, 'stretched' relaxation.
In turn,

$$\epsilon_r = \epsilon_\infty + \frac{\Delta\epsilon}{1 + j\frac{\omega\epsilon}{\sigma}} - j\frac{\sigma}{\omega\epsilon_0},$$

where $\sigma$ is the conductivity of the medium.

I'll be honest, I don't fully understand this model, but I appreciate that it attempts to address the dependence of the relative permittivity on frequency. This seems like an important point.

Could anyone perhaps explain this model? How could this be applied into a model for radio propagation? Also, I'm still finding difficulty understanding the application of complex numbers in electromagnetism (such as in wave phase): how does it translate into real terms? It also appears in the following propagation model.

O'Shaugnessy's (2012) Dielectric Propagation Model

Again, this model is from Characterising the Relative Permittivity and Conductivity of Seawater for Electromagnetic Communications in the Radio Band (O'Shaugnessy's, 2012). It states that:

$$\boxed{E_z = E_{0, z} e^{j\omega t - \gamma z}} \ \ \ \ ,\ \ \ \ \boxed{H_z = H_{0, z} e^{j\omega t - \gamma z}}$$

where $E$ represents the electric field strength, $H$ the magnetic field strength and

$$\gamma_{\text{water}}=\alpha + j\beta = j\omega\sqrt{\mu\left( \epsilon - j \frac{\sigma}{\omega}. \right)}$$

The author then continues to discuss the conduction and dielectric bands of this model depending on the relationship between $\frac{\sigma}{\omega}$ and $\epsilon$.

Could anyone help to explain this?

Last edited by a moderator: May 7, 2017
2. Jul 28, 2015

### tech99

I assume both transmitter and receiver and immersed. I think the simple approach is first to work out the free space path loss. A useful formula is PL=20 log f + 20 log d - 27.55 in dB, MHz and m. Then if you know the skin depth, the fields will be 1/e at this depth, which is 0.37 or - 8.6dB. The sea water attenuation in dB/m can then be obtained from 8.6/(skin depth in m). These two losses should be calculated for the intended communication distance and added together.
Remember that antennas for low frequencies are very inefficient. For a receiving antenna, only the noise considerations will apply, so efficiency will not matter so much. The noise may be arriving from the surface above, whilst the signal is arriving horizontally.
In practice, the sea water skin depth might be unreliable as the conduction is by heavy ions. As a matter of interest, I tried making a dipole antenna for 50 MHz using a tube filled with concentrated saline solution and it was almost completely ineffective. I could detect no resonances between 1 and 100 MHz and the input resistance was always very high.

Last edited by a moderator: Jul 29, 2015
3. Jul 29, 2015

### H Smith 94

Hi tech99, thank you for your response!

This sounds like a very useful approach. Would the addition of losses only be applicable in dB?

That sounds like a very interesting experiment. Although it was unsuccessful, do you have any details regarding the set-up? Do you have any plans to improve on the method in the future or perhaps produce a report/paper?

4. Jul 29, 2015

### sophiecentaur

dB is just a ratio of powers and we assume that the system is linear. How else would you quote loss?
In a transmission channel, there is a dB/m and then there is 'geometric' or spreading loss, which is often inverse square but not for a guided wave - as this could be.

5. Jul 29, 2015

### H Smith 94

I'm not used to using dB, I'll be honest. I am much more familiar with absolute signal values such as power density, field strength or the like (I suppose that's the geometric loss that you mention.) I'm still trying to get to grips with it.

Since dB is based around a logarithmic representation, I was under the impression that if absolute signal values are multiplicative, the corresponding signal ratio in dB is additive: i.e. dB makes it a linear system. So, take an idealised form of equation 1,

$$P_r = P_t \left( \frac{\lambda}{4\pi r} \right)^2 \bigg|_{G_t=1,G_r=1},$$
where $P_r$ and $P_t$ are in W/m, for example, then the loss $L_\text{dB}$ in dB is found by

$$\begin{split}L_\text{dB} &= 10 \log \frac{P_t}{P_r} \\&= 10\log\left|( 4\pi r / \lambda)^2\right| \\&= 20\log\left| 4\pi r / \lambda \right| \\&= 20\log\left| 4\pi r \right| - 20\log\left| \lambda \right|. \end{split}$$

Is this a correct understanding of how dB works?

Also, does dB have to always represent a ratio of the same quantities or units? So, for example, would equation 6
$$L_\text{total}\ \text{[dB]} = \text{FSL [dB]} + L_m\ \text{[dB]} − G_ t\ \text{[dB]}−G_r\ \text{[dB]}$$ become something like $$L_\text{total}\ \text{[W/m]} \approx \frac{\text{FSL [W/m]} L_m\ \text{[W/m]}} {G_ t\ \text{[W/m]}G_r\ \text{[W/m]}}$$ in an absolute model?

6. Jul 29, 2015

### sophiecentaur

Looks fine to me. But I didn't understand why you were asking if the answers would be 'in dB'. The main reasons for using dB are that large ranges of values can be accommodated easily and, as gains and losses are multiplicative, you can add dB values. It's a lot more convenient than using standard form, when accuracy is not a serious issue.

7. Jul 29, 2015

### H Smith 94

Oh okay, I get what you were asking. I was trying to understand the justification for adding separate loss values. I get it now, thank you for your help.

8. Jul 29, 2015

### sophiecentaur

One thing about using dB. If you want to add two signals together, you have to go back to the linear world (antilog) or you get nonsense. (You can't multiply 120mW by 34mW)

9. Jul 29, 2015

### tech99

I did this experiment to see if a conductor which uses heavy charge carriers is a poor radiator (because the acceleration is reduced), and so obtain evidence that charge carriers are necessary for radiation. But the resistance of the solution was much higher than expected so I could not obtain reliable measurements, so it needs more thought.
Regarding decibels, the dissipation of sea water causes the field to fall exponentially, and every metre travelled the field will fall the same number of dB. The same happens with a lossy transmission line or light passing through fog etc. But the spreading loss does not behave this way, and the power falls with the inverse square law. So that the number of dB per metre gets less as the path is extended. As mentioned on another thread, an example is that when the Pluto probe has travelled for another nine years, so it is twice as far away, the signal will drop by only 6 dB.

10. Jul 29, 2015

### sophiecentaur

11. Jul 30, 2015

### sophiecentaur

In condensed matter, why would you suggest that motion of the charge carriers is a direct factor? True, the mobility would affect the complex dielectric constant - be responsible for the absorption, if you want a semi-classical model. But the motion of electrons is small, even in a low density ionised gas (Ionosphere) and the theory doesn't include motion of ions (see Appleton Hartree etc.). I thought the radiation process was normally described in terms of 'current', rather than the actual charge carriers. Is there any reason to use a different analysis? I haven't understood the jump from Propagation to Radiating element properties.

12. Jul 30, 2015

### H Smith 94

Ah, that's interesting. I've used an exponential attenuation model when adapting your original suggestions to my scenario:

$$ The attenuation factor originates from the understanding of the exponential attenuation of a signal $I(r)$ whose initial value is $I(0)$, wherein

$$I(r) = I(0)\,e^{-\alpha r}.$$

Therefore,

$$\ln \frac{I(r)}{I(0)} = -\alpha r$$

and by change of base formula, $\ln \frac{I(r)}{I(0)} = 2.306\,\log_{10} \frac{I(r)}{I(0)}$, so

$$\log\frac{I(r)}{I(0)} = -\frac{\alpha r}{2.3026}. "$$​

Then I've again used the ITU loss factor to get an expression for loss in dB. It seems to be working quite well so far.

What's the difference between the exponential decay approach and the spreading loss approach?

13. Jul 30, 2015

### sophiecentaur

The sums are totally different so you'd expect the result to be different. Go with the Maths.