Question

Ignoring reflection at the air-soil boundary, if the amplitude of a 3-GHz incident wave is 10 V/rn at the surface of a wet soil medium, at what depth will it be down to 1 mV/m? Wet soil is characterized by µr = 1, εr = 9, and σ = 5 × 10–4S/m.

Answer 1

Answer:

5 multiply by 10 S/M

Answer 2

Based on the calculations, the incident wave would be down to 1 mV/m at a depth of 287.82 meters.

How to calculate the depth of the wet soil medium.

Since the medium is a wet soil, it is characterized by the following:

• Dielectric constant, µr = 1.

• Relative permitivitty, εr = 9.

• Electrical conductivity, σ = 5 × 10⁻⁴ S/m.

Next, we would determine loss tangent, so as to know if the wet soil is a low-loss medium:

For a low-loss medium; ε''/ε' = σ/ωε < 0.01.

$\frac{\sigma}{\omega \epsilon} = \frac{5 \times 10^{-4} \;\times\; 36 \pi}{2\pi \;\times \;3 \times 10^9\; \times\; 10^{-9} \times \;9} \\\\\frac{\sigma}{\omega \epsilon} =3.32 \times 10^{-4}$

Therefore, the wet soil medium is a low-loss dielectric and its absorption (attenuation) coefficient is calculated by using this formula:

Note:

$\alpha = \frac{\sigma}{2}\sqrt{\frac{\mu}{\epsilon}} \\\\\alpha = \frac{\sigma}{2} \times \frac{120\pi}{\sqrt{\epsilon_r} } \\\\\alpha =\frac{5 \times 10^{-4} \times 120\pi}{2 \times \sqrt{9} }$

α = 0.032 Np/m.

Now, we calculate the depth:

$E(z) = E_oe^{-\alpha z} = 10e^{-\alpha z}\\\\10^{-3}=10e^{-0.032 z}\\\\ln(10^{-4})=-0.032 z\\\\-9.21 \times 10^9=-0.032z\\\\z=\frac{-92.1 0}{-0.032}$

z = 287.82 meters.

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