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So I'm an undergrad electrical engineer but I figured this would be a question more suited to physics. I'm working with high frequencies (30-300GHz) which correspond to EM waves of millimetre wavelengths. I've been told that these waves don't travel well through buildings and are easily absorbed by foliage and rain, so I would like to know why if that isn't too vague of a question? More specifically, I would like to know the mathematical relation between how easily these waves are absorbed and the frequency of the waves. If it's the question is too vague then please let me know.

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Have a look at Wikipedia

Liquid water has a broad absorption spectrum in the microwave region, which has been explained in terms of changes in the hydrogen bond network giving rise to a broad, featureless, microwave spectrum. The absorption (equivalent to dielectric loss) is used in microwave ovens to heat food that contains water molecules. A frequency of 2.45 GHz, wavelength 122 mm, is commonly used.

Also if dielectric losses occur, the Poynting Vector $|S| \propto \exp(-\frac{2}{\delta}z)$ depends on the attenuation coefficient $$\frac{2}{\delta} = k_0 \tan \delta_{\epsilon},$$ where $k_{0}$ is the wavelength in the vacuum, $\tan \delta_{\epsilon}$ is the loss coefficient of the dielectric material and $z$ is the length of propagation. And because $$k_{0} = \frac{2 \pi}{\lambda} $$ as smaller the wavelength gets, the harder it is for the EM wave to penetrate the material.

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  • $\begingroup$ I assume \delta is the penetration depth of the em wave, so as the frequency increases, k_0 increases and so the penetration depth decreases? $\endgroup$ – Aidan Apr 15 at 12:31
  • $\begingroup$ $\frac{2}{\delta}$ is the attenuation coefficient. If you just follow the formula, as higher the frequency goes the poynting vectors gets smaller, so the penetration depth decreases. $\endgroup$ – Maxim Apr 15 at 12:55
  • $\begingroup$ No worries, thanks very much for the help! $\endgroup$ – Aidan Apr 15 at 12:58
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The key quantity is $\frac{\sigma}{\omega \epsilon}$ where $\sigma$ is the conductivity at that frequency, and $\epsilon$ is the dielectric constant of the material. When $\frac{\sigma}{\omega \epsilon}\gg 1$, the material is a good conductor at that frequency, and the signal will attenuate as $e^{-\alpha z}$ with $\alpha =\sqrt{\frac{\omega\mu \sigma}{2}}$, with $\mu$ the magnetic permeability. Hence, the answer depends very much on the material and the frequency. For instance, in human tissue with high water content (muscle, skin), $\sigma$ varies from less than 1 to more than 10 over the range $100\hbox{MHz}\le f\le 10000\hbox{MHz}$, whereas $70\le \epsilon_r\le 40$ over that range. For fat, bone and tissue with low water contents, values of $\sigma$ can range from $20$ to about $500$ depending on the specific material (see this website for additional details).

For instance, sea water at $\omega=2\pi\times 20\times 10^3$ has $\sigma=4$ and $\epsilon=81\epsilon_0$ so it's a very good conductor. With this data $\alpha=0.562$ so basically the signal will travel about 40cm before it's significantly attenuated by absorption. On the other hand, fresh water is not a good conductor at this frequency, so the actual amount of salts and other minerals in the actual raindrop is quite important.

As another example more closely related to your question, $\alpha\approx 6\times 10^5$ for aluminum foil at $\omega=2\pi\times 2450\times 10^6$rad/s, whereas the thickness of standard household aluminum foil is $16\times 10^{-6}$, so the field is attenuated by $\approx e^{-10}$ going through the foil, and thus the energy density is decreased by $\approx e^{-20}$.

This is just absorption losses due to propagation; there is also reflection losses, i.e. not all the signal is transmitted at the interface, and if most is reflected there isn't a whole lot of signal transmitted to the second medium to start with, and this is on top of the absorption due to propagation. For instance, most of the electric going from air to sea water is reflected, which makes this and absorption by seawater very difficult for communication at $f=20\times 10^3$Hz.

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  • $\begingroup$ The website link isn't working for me sorry. I'd like to know more, so if you could suggest any resources I'd be very appreciative. $\endgroup$ – Aidan Apr 16 at 12:51
  • $\begingroup$ @Aidan This is covered in most applied E&M textbooks. You can check out Inan, Inan an Said, Engineering Electromagnetics and waves; or Sadiku, Elements of Electromagnetics or Hayt and Buck, Engineering Electromagnetics. $\endgroup$ – ZeroTheHero Apr 16 at 12:56
  • $\begingroup$ Great, thanks very much $\endgroup$ – Aidan Apr 16 at 13:01

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