IPhO 2002 · Theory problem 1

Ground-Penetrating Radar

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Ground-penetrating radar (GPR) is used to detect and locate underground objects near the surface by means of transmitting electromagnetic waves into the ground and receiving the waves reflected from those objects. The antenna and the detector are directly on the ground and they are located at the same point.

A linearly polarized electromagnetic plane wave of angular frequency \(\omega\) propagating in the \(z\) direction is represented by the following expression for its field:

\[ E=E_0e^{-\alpha z}\cos(\omega t-\beta z), \tag{1} \]

where \(E_0\) is constant, \(\alpha\) is the attenuation coefficient and \(\beta\) is the wave number expressed respectively as follows

\[ \alpha=\omega\left\{ \frac{\mu\varepsilon}{2} \left[ \left(1+\frac{\sigma^2}{\varepsilon^2\omega^2}\right)^{1/2}-1 \right]\right\}^{1/2}, \qquad \beta=\omega\left\{ \frac{\mu\varepsilon}{2} \left[ \left(1+\frac{\sigma^2}{\varepsilon^2\omega^2}\right)^{1/2}+1 \right]\right\}^{1/2} \tag{2} \]

with \(\mu,\varepsilon,\) and \(\sigma\) denoting the magnetic permeability, the electrical permittivity, and the electrical conductivity respectively.

The signal becomes undetected when the amplitude of the radar signal arriving at the object drops below \(1/e\) (\(\approx 37\%\)) of its initial value. An electromagnetic wave of variable frequency (10 MHz – 1000 MHz) is usually used to allow adjustment of range and resolution of detection.

The performance of GPR depends on its resolution. The resolution is given by the minimum separation between the two adjacent reflectors to be detected. The minimum separation should give rise to a minimum phase difference of \(180^\circ\) between the two reflected waves at the detector.

Questions:

(Given : \(\mu_0=4\pi\times10^{-7}\ \mathrm{H/m}\) and \(\varepsilon_0=8.85\times10^{-12}\ \mathrm{F/m}\))

  1. Assume that the ground is non-magnetic (\(\mu=\mu_0\)) satisfying the condition \(\left(\dfrac{\sigma}{\omega\varepsilon}\right)^2\ll1\). Derive the expression of propagation speed \(v\) in terms of \(\mu\) and \(\varepsilon\), using equations (1) and (2) [1.0 pts].

  2. Determine the maximum depth of detection of an object in the ground with conductivity of 1.0 mS/m and permittivity of \(9\varepsilon_0\), satisfying the condition \(\left(\dfrac{\sigma}{\omega\varepsilon}\right)^2\ll1\), (\(\mathrm{S}=\mathrm{ohm}^{-1}\) ; use \(\mu=\mu_0\)). [2.0 pts]

  3. Consider two parallel conducting rods buried horizontally in the ground. The rods are 4 meter deep. The ground is known to have conductivity of 1.0 mS/m and permittivity of \(9\varepsilon_0\). Suppose the GPR measurement is carried out at a position aproximately above one of the rod. Assume point detector is used. Determine the minimum frequency required to get a lateral resolution of 50 cm [3.5 pts].

  4. To determine the depth of a buried rod \(d\) in the same ground, consider the measurements carried out along a line perpendicular to the rod. The result is described by the following figure:

Figure 1

Graph of traveltime \(t\) vs detector position \(x\), \(t_{\min}=100\ \mathrm{ns}\).

Derive \(t\) as a function of \(x\) and determine \(d\) [3.5 pts].

Official English text of the problem.

Answers

  1. In the stated low-loss limit, \[ \boxed{v\simeq\frac{1}{\sqrt{\mu\varepsilon}}.} \]

  2. The maximum detection depth is \[ \boxed{D_{\max}\simeq\frac{2}{\sigma}\sqrt{\frac{\varepsilon}{\mu_0}} =15.9\ \mathrm{m}\approx16\ \mathrm{m}.} \]

  3. With rod depth \(h=4\ \mathrm{m}\) and lateral separation \(s=0.50\ \mathrm{m}\), \[ \boxed{f_{\min}\simeq \frac{v}{4\left(\sqrt{h^2+s^2}-h\right)} \approx8.0\times10^8\ \mathrm{Hz} =800\ \mathrm{MHz}.} \]

  4. Taking \(x=0\) directly above the rod, \[ \boxed{t(x)=\frac{2}{v}\sqrt{d^2+x^2} =\sqrt{t_{\min}^2+\frac{4x^2}{v^2}},\qquad d=\frac{vt_{\min}}{2}\approx5.0\ \mathrm{m}.} \]

Solution

1. Propagation speed

A point of constant phase satisfies \[ \omega t-\beta z=\text{constant}. \] Differentiating gives the phase velocity: \[ v=\frac{dz}{dt}=\frac{\omega}{\beta}. \]

Define the small dimensionless parameter \[ q=\frac{\sigma}{\omega\varepsilon},\qquad q^2\ll1. \] Using the binomial expansion, \[ \sqrt{1+q^2}=1+\frac{q^2}{2}+O(q^4). \] Equation (2) therefore gives \[ \beta =\omega\left[ \frac{\mu\varepsilon}{2} \left(2+\frac{q^2}{2}+O(q^4)\right) \right]^{1/2} =\omega\sqrt{\mu\varepsilon} \left(1+\frac{q^2}{8}+O(q^4)\right). \]

Keeping the leading term, \[ \boxed{v\simeq\frac{1}{\sqrt{\mu\varepsilon}}.} \] For non-magnetic ground, \(\mu=\mu_0\). At this order, \(\beta\) is proportional to \(\omega\), so dispersion is negligible and the group velocity also equals this value.

2. Maximum detection depth

For the attenuation coefficient, retain the leading nonzero term of the same expansion: \[ \alpha \simeq \omega\left[ \frac{\mu\varepsilon}{2}\frac{q^2}{2} \right]^{1/2} =\frac{\sigma}{2}\sqrt{\frac{\mu}{\varepsilon}}. \] Thus, to this order, the attenuation coefficient is independent of frequency.

For an object directly below the antenna at depth \(D\), equation (1) gives the incident amplitude ratio \[ \frac{E_{\mathrm{amp}}(D)}{E_0}=e^{-\alpha D}. \] The problem specifies the threshold for the signal arriving at the object, so the propagation distance entering this condition is \(D\). At the limiting depth, \[ e^{-\alpha D_{\max}}=\frac1e, \] and hence \[ D_{\max}=\frac1\alpha \simeq\frac{2}{\sigma}\sqrt{\frac{\varepsilon}{\mu_0}}. \]

Using \[ \sigma=1.0\times10^{-3}\ \mathrm{S/m}, \qquad \varepsilon=9(8.85\times10^{-12}) =7.965\times10^{-11}\ \mathrm{F/m}, \] we obtain \[ \alpha\simeq \frac{1.0\times10^{-3}}{2} \sqrt{\frac{4\pi\times10^{-7}}{7.965\times10^{-11}}} =6.28\times10^{-2}\ \mathrm{m^{-1}}. \] Therefore, \[ \boxed{D_{\max}=15.9\ \mathrm{m}\approx16\ \mathrm{m}.} \]

3. Minimum frequency for a lateral resolution of 50 cm

Work in the vertical plane perpendicular to the rods. Let their common depth be \(h\), their lateral separation be \(s\), and take the antenna–detector position directly above the first rod.

The one-way distances to the two rods are \[ r_1=h,\qquad r_2=\sqrt{h^2+s^2}. \] Since transmitter and detector are at the same point, the reflected signals travel distances \(2r_1\) and \(2r_2\). Their path difference is therefore \[ \Delta L=2\left(\sqrt{h^2+s^2}-h\right). \]

Treating the conducting rods as equivalent reflectors, their common reflection phase cancels. The phase difference at the detector is \[ \Delta\phi=\beta\Delta L =\frac{4\pi f}{v}\left(\sqrt{h^2+s^2}-h\right). \] The specified resolution criterion is \(\Delta\phi\geq\pi\). The minimum frequency occurs at equality: \[ \boxed{ f_{\min}= \frac{v}{4\left(\sqrt{h^2+s^2}-h\right)}. } \]

The wave speed in this ground is \[ v\simeq\frac1{\sqrt{\mu_0(9\varepsilon_0)}} =9.995\times10^7\ \mathrm{m/s}. \] For \(h=4\ \mathrm{m}\) and \(s=0.50\ \mathrm{m}\), \[ \sqrt{h^2+s^2}-h =\sqrt{4^2+0.50^2}-4 =0.0311289\ \mathrm{m}. \] Thus, \[ f_{\min}\simeq \frac{9.995\times10^7}{4(0.0311289)} =8.027\times10^8\ \mathrm{Hz}, \] or, to sensible precision, \[ \boxed{f_{\min}\approx8.0\times10^8\ \mathrm{Hz} =800\ \mathrm{MHz}.} \]

The small-separation approximation gives the same rounded result: \[ \sqrt{h^2+s^2}-h\simeq\frac{s^2}{2h} \quad\Longrightarrow\quad f_{\min}\simeq\frac{vh}{2s^2}. \]

The required frequency lies within the stated \(10\)–\(1000\ \mathrm{MHz}\) range. Moreover, at the calculated frequency, \[ \left(\frac{\sigma}{2\pi f_{\min}\varepsilon}\right)^2 \simeq6.2\times10^{-6}\ll1, \] so the low-loss approximation is consistent. Both rods also lie within the detection range found in part 2.

4. Travel-time curve and rod depth

Choose the horizontal coordinate \(x=0\) directly above the buried rod. In homogeneous ground, the one-way distance from the antenna to the rod is \[ r(x)=\sqrt{d^2+x^2}. \] The measured travel time includes the outward and return journeys: \[ \boxed{t(x)=\frac{2}{v}\sqrt{d^2+x^2}.} \]

The shortest path occurs at \(x=0\), giving \[ t_{\min}=\frac{2d}{v}. \] Consequently, \[ d=\frac{vt_{\min}}{2} =\frac{(9.995\times10^7)(100\times10^{-9})}{2}\ \mathrm{m} =4.998\ \mathrm{m}. \] Thus, \[ \boxed{d\approx5.0\ \mathrm{m}.} \]

Substituting the minimum travel time into the curve gives \[ \boxed{ t(x)=\sqrt{t_{\min}^2+\frac{4x^2}{v^2}} \simeq(100\ \mathrm{ns}) \sqrt{1+\left(\frac{x}{5.0\ \mathrm{m}}\right)^2}. } \]

This is the positive-time branch of the hyperbola \[ \frac{t^2}{t_{\min}^2}-\frac{x^2}{d^2}=1. \] It is symmetric about \(x=0\), has its minimum travel time of \(100\ \mathrm{ns}\) directly above the rod, and increases with \(|x|\). For \(|x|\gg d\), \[ t(x)\sim\frac{2|x|}{v}. \] Because the figure’s time axis points downward, larger travel times appear lower on the page, as shown.

The solution was checked against the official solution.