Problem 5.5.4: Ground-penetrating radar in a lossy soil

Problem 5.5.4

Ground-penetrating radar in a lossy soil

A ground-penetrating radar sends a short pulse (centre frequency $200 MHz$ ) into moist sandy soil with $ε_{r} = 16$ , $μ_{r} = 1$ , $σ = 2 mS/m$ .

Show that the substitution $E = e^{- t / τ} u$ with $τ = 2 ε / σ$ removes the first time derivative from the source-free wave equation of a conducting medium, and find the equation for $u$ .
Find $v$ , $τ$ and the attenuation constant $α = 1/ (v τ)$ .
Check that the pulse is "short" in the sense of Section 5.5.7.
A pipe lies $3 m$ deep. When does its echo arrive, and by what factor has the conduction loss reduced the amplitude on the way down and back? (Ignore the spreading of the beam and the reflection coefficient.)

Try it yourself first.

Strategy

The term $μ σ \partial E / \partial t$ is a friction term. Write the field as a decaying exponential in time times a new unknown, and choose the decay time so that the first derivative cancels. This turns the equation into a wave equation with a small extra term. For a short pulse the extra term is negligible, $u$ travels unchanged, and $e^{- t / τ}$ damps it.

With $E = e^{- t / τ} u$ (dots are time derivatives),

\frac{\partial E}{\partial t} \frac{\partial ^{2} E}{\partial t ^{2}} = e^{- t / τ} (\dot{u} - \frac{u}{τ}), = e^{- t / τ} (\ddot{u} - \frac{2 u ˙}{τ} + \frac{u}{τ ^{2}}) .

Put these into $\nabla^{2} E - μ ε \partial^{2} E / \partial t^{2} - μ σ \partial E / \partial t = 0$ and cancel the common factor $e^{- t / τ}$ :

\nabla^{2} u - μ ε \ddot{u} + (\frac{2 μ ε}{τ} - μ σ) \dot{u} + (\frac{μ σ}{τ} - \frac{μ ε}{τ ^{2}}) u = 0.

With $τ = 2 ε / σ$ the first bracket is $μ σ - μ σ = 0$ , and the second is $μ σ^{2} / (2 ε) - μ σ^{2} / (4 ε) = μ σ^{2} / (4 ε)$ :

Answer

\nabla^{2} u - \frac{1}{v ^{2}} \ddot{u} + α^{2} u = 0, α^{2} = \frac{μ σ ^{2}}{4 ε}, α = \frac{σ}{2} \frac{μ}{ε} = \frac{1}{v τ}

$v = c / 16 = 7.5 \times 1 0^{7} m/s$ and $ε = 16 ε_{0} = 1.415 \times 1 0^{- 10} F/m$ , so

τ α = \frac{2 ε}{σ} = \frac{2 \times 1.415 \times 1 0 ^{- 10}}{2 \times 1 0 ^{- 3}} = 141 ns, = \frac{σ}{2} \frac{μ _{0}}{16 ε _{0}} = \frac{2 \times 1 0 ^{- 3}}{2} \times \frac{120 π}{4} = 0.0942 Np/m,

and indeed $1/ α = v τ = 7.5 \times 1 0^{7} \times 1.415 \times 1 0^{- 7} = 10.6 m$ .

At $200 MHz$ , $ω τ = 2 π \times 2 \times 1 0^{8} \times 1.415 \times 1 0^{- 7} = 178 ≫ 1$ , equivalently $σ / (ω ε) = 0.011 ≪ 1$ . Compared with $\overset{u}{¨} / v^{2} \sim (ω / v)^{2} u$ , the term $α^{2} u$ is smaller by $(α v / ω)^{2} = (1/ ω τ)^{2} = 3 \times 1 0^{- 5}$ . It can be dropped, $u \approx f (z - v t)$ , and at the pulse ( $t = z / v$ ) $E \approx e^{- α z} f (z - v t)$ , which is Eq. (5.5.33).

The path down and back is $6 m$ :

t = \frac{6}{7.5 \times 1 0 ^{7}} = 80 ns, e^{- α \times 6} = e^{- 0.565} = 0.568.

Answer

t = 80 ns, E_{echo} / E_{0} = 0.568 (- 4.9 dB)

Read the echo on the probe at $s = 6 m$ below, then choose wet clay in the menu and watch the pulse leave a wake behind it.

InteractiveLoading simulation…

Check

The ratio $σ / (ω ε) = 0.011$ makes the soil a good dielectric, and the exact attenuation constant of a sinusoidal wave at $200 MHz$ (Home exercise 5.5.1) is $0.0942 Np/m$ as well. In decibels, $20 lo g_{10} 0.568 = - 4.9$ . Units: $σ μ / ε$ is $(S/m) \cdot Ω = 1/ m$ .

(a) The pulse along its path, s=vt. — (a) The pulse along its path, $s = v t$ .

(b) Decay in sand and in clay. — (a) The pulse along its path, $s = v t$ .

Why radar fails in clay

Wet clay ( $ε_{r} = 25$ , $σ = 20 mS/m$ ) has ten times the conductivity: $α = \frac{0.02}{2} \times \frac{120 π}{5} = 0.754 Np/m$ , and the same $6 m$ path leaves $e^{- 4.52} = 1 %$ of the amplitude. Ground-penetrating radar surveys work in dry sand, ice and rock, and struggle in clay and salty soil.