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Most antennae are deployed in earth's atmosphere which can be approximated as a vacuum. But what about an antennae for a submarine underneath the ocean? Or how about an antennae deployed in a copper mine? What about the most extreme case where one wants to beam a signal from one side of a copper chunk to another?

How does one adapt an antennae to an electrically conductive medium?

  • the interesting thing is that the effective wavelength in a conductive medium is usually much lower than one would guess it from its permittivity alone, hence an underwater (conductvity of sea water ~4S/m) dipole at 100kHz is only ~3m! – hyportnex Apr 13 '17 at 18:13

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Consider a linearly polarized plane wave travelling along the z-axis in a homogeneous conductive medium: the E-field can be written as $E=E_0 e^{\iota \omega t - \gamma z}$ where the complex propagation factor is $$\gamma = \sqrt{\iota \omega \mu (\sigma + \iota \omega \epsilon)} = \alpha + \iota \beta$$ and $\mu, \epsilon, \sigma$ are the permeability, permittivity and conductivity, resp., of the medium. When $\omega$ is small, $\omega << \frac{\sigma}{\epsilon}$, then one can write $$\gamma \approx \sqrt{\iota \omega \mu \sigma} = \sqrt{\frac{\omega \mu \sigma}{2}}(1+\iota)$$ Therefore we have $\alpha \approx \beta = \frac{2\pi}{\Lambda}\approx \sqrt{\frac{\omega \mu \sigma}{2}}$, and the wavelength in the conductive medium is $${\Lambda}\approx 2\pi\sqrt{\frac{2}{\omega \mu \sigma}}$$ Note that this is very different from the lossless medium wavelength $\lambda = \frac {v}{f} = 2\pi \frac{1}{\omega\sqrt{\mu\epsilon}}$.

The main take-away is that for low frequencies the effective wavelength can be much shorter in the conductive medium than if the medium were lossless. This will allow the use of physically small but electrically large half-wave electric dipoles or magnetic loops and some such that are much easier to match than a short dipole. The downside of such communications is the massive dissipative loss incurred. For example, in sea water at 10kHz the wavelength is about 16m, compare that with free space at 30km, but the attenuation per wavelength (16m) is already $2\pi \approx 55dB$!

hyportnex
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