Съдържание

IPhO 2002, Теория — Задача 1. Ground-Penetrating Radar

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Всички материали от IPhO 2002

IPhO2002 — I. Ground-Penetrating Radar

Внимание: Бележка към темата

The official solution prints an extra ω in the denominator of the intermediate speed formula in question 1; question 2 prints a Latin a in δ = 1/a; and question 4 places v inside the square root in one intermediate travel-time formula. These expressions are preserved as printed. The two archived English editions contain the same question and official solution.

Условие

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_0\, e^{-\alpha z} cos(\omega t - \beta z)\ , \qquad (1)

where $E_o$ 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}, \quad \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} \qquad (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 : $\mu_o = 4\pi\ \mathrm{x}10^{-7}$ H/m and $\varepsilon_o = 8.85\ \mathrm{x}10^{-12}$ F/m )

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

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 $\left(\frac{\sigma}{\omega\varepsilon}\right)^2 \ll 1$ , (S=ohm $^{-1}$ ; use $\mu=\mu_0$ ). [2 т.]

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 approximately 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 т.]

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:

Framed sketch: a horizontal arrow at the top labelled 'detector position' and x; a vertical arrow pointing down from its middle labelled t; a blue curve shaped like an inverted parabola (hyperbola) with its top touching the vertical axis below the x-arrow and its branches going down on both sides. — Graph of traveltime t vs detector position x, t_min = 100 ns.

Derive $t$ as a function of $x$ and determine $d$ . [3,5 т.]

Отговори

Решение

1.

Speed of radar signal in the material $v_m$ :

$\omega t - \beta z =$ constant $\rightarrow$ $\beta z =$ -constant + $\omega t$ (0.2 pts)

v_m = \frac{\omega}{\beta}

v_m = \frac{1}{\omega\left\{\frac{\mu\varepsilon}{2}\left[(1+\frac{\sigma^2}{\varepsilon^2\omega^2})^{1/2}+1\right]\right\}^{1/2}}

(0.4 pts)

v_m = \frac{1}{\left\{\frac{\mu\varepsilon}{2}(1+1)\right\}^{1/2}} = \frac{1}{\sqrt{\mu\varepsilon}}

(0.4 pts)

2.

The maximum depth of detection (skin depth, $\delta$ ) of an object in the ground is inversely proportional to the attenuation constant:

(0.5 pts) (0.3 pts) (0.2 pts)

\delta = \frac{1}{a} = \frac{1}{\omega\left\{\frac{\mu\varepsilon}{2}\left[\left(1+\frac{\sigma^2}{\varepsilon^2\omega^2}\right)^{1/2}-1\right]\right\}^{1/2}} = \frac{1}{\omega\left\{\frac{\mu\varepsilon}{2}\left[\left(1+\frac{1}{2}\frac{\sigma^2}{\varepsilon^2\omega^2}\right)-1\right]\right\}^{1/2}} = \frac{1}{\omega\left\{\frac{\mu\varepsilon}{2}.\frac{1}{2}\frac{\sigma^2}{\varepsilon^2\omega^2}\right\}^{1/2}}

\delta = \left(\frac{2}{\sigma}\right)\left(\frac{\varepsilon}{\mu}\right)^{1/2}.

Numerically $\delta = \frac{\left(5.31\sqrt{\varepsilon_r}\right)}{\sigma}$ m, where $\sigma$ is in mS/m. (0.5 pts)

For a medium with conductivity of 1.0 mS/m and relative permittivity of 9, the skin depth

\delta = \frac{\left(5.31\sqrt{9}\right)}{1.0} = 15.93\ \mathrm{m}

(0.3 pts) + (0.2 pts)

3.

Lateral resolution:

Right triangle: vertical leg from 'Antenna' at the top down to a rod, labelled d; horizontal leg between two rods (dark discs, each labelled 'rod'), labelled r; hypotenuse from the antenna to the second rod, labelled d + λ/4.

r^2 + d^2 = (d + \frac{\lambda}{4})^2

r = \left(\frac{\lambda d}{2} + \frac{\lambda^2}{16}\right)^{1/2}

(1.0 pts)

r =0.5 m, d =4 m: $\frac{1}{2} = \left(\frac{4\lambda}{2} + \frac{\lambda^2}{16}\right)^{1/2}$ , $\lambda^2 + 32\lambda - 4 = 0$ (0.5 pts)

The wavelength is $\lambda$ =0.125 m. (0.3 pts) + (0.2 pts)

The propagation speed of the signal in medium is

v_m = \frac{1}{\sqrt{\mu\varepsilon}} = \frac{1}{\sqrt{\mu_o\mu_r\varepsilon_o\varepsilon_r}} = \frac{1}{\sqrt{\mu_o\varepsilon_o}}\frac{1}{\sqrt{\mu_r\varepsilon_r}}

$v_m = \frac{c}{\sqrt{\mu_r\varepsilon_r}} = \frac{0.3}{\sqrt{\varepsilon_r}}$ m/ns , where $c = \frac{1}{\sqrt{\mu_o\varepsilon_o}}$ and $\mu_r = 1$

$v_m = 0.1$ m/ns $= 10^8$ m/s (0.5 pts)

The minimum frequency need to distinguish the two rods as two separate objects is

f_{\min} = \frac{v}{\lambda}

(0.5 pts)

f_{\min} = \frac{\frac{0.3}{\sqrt{9}}}{0.125}\,\mathrm{x}10^9\ \mathrm{Hz} = 800\ \mathrm{MHz}

(0.3 pts) + (0.20 pts)

4.

Path of EM waves for some positions on the ground surface

Ground surface line with three transmitter–receiver boxes marked 'T R' under the heading 'Antenna Positions'; a dotted arrow to the right labelled 'Scanning direction'; medium labelled ε1, σ1. Rays 1, 2 and 3 go from the boxes to a blue buried rod at depth d, labelled 'Buried rod(ε2, σ2)'; the horizontal offset x of the first box from the rod is marked at the top.

The traveltime as function of $x$ is

\left(\frac{t\ \ v}{2}\right)^2 = d^2 + x^2\ ,

(1.0 pts)

t(x) = \sqrt{\frac{4d^2 + 4x^2}{v}}

(1.0 pts)

t(x) = \frac{2\sqrt{\varepsilon_{1r}}}{0.3}\sqrt{d^2 + x^2}

Framed graph titled 'Graph of traveltime, t(x)': horizontal arrow labelled 'Antenna Positions' and x, vertical arrow pointing down labelled t, and a blue curve shaped like an inverted parabola (hyperbola) whose top touches the vertical axis.

For $x$ =0 (1.0 pts)

100 = 2×(3/0.3) $d$

$d$ = 5 m (0.5 pts)

Оригинал в Архива: 1.pdf · официални решения: 1_sol.pdf · допълнителни данни: IPhO_2002_Theoretical Question 1.pdf · допълнителни данни: IPhO_2002_Theoretical Solution 1.pdf