Radar Pulse Delay Through Air and Water: 04-BS-9, May 2013, Q8

Question 8: Radar Pulse Delay Through Air and Water (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
Object height above water surface	300 m
Transmitter/receiver depth below surface (case 2)	10 m
Water relative permittivity $\varepsilon_r$	81
Speed of light in vacuum $c$	$3\times10^8$ m/s

Find. Round-trip pulse delay for (1) unit at the water surface, and (2) unit 10 m below the surface.

Round-trip path for each unit position: entirely in air (case 1), or partly in water then air (case 2).

Approach. Find the wave speed in water from $\varepsilon_r$ (nonmagnetic medium, $\mu_r=1$), then sum travel time over each medium the round-trip path crosses.

Speed of light in water. $$n=\sqrt{\varepsilon_r\mu_r}=\sqrt{81}=9, \qquad v_{\text{water}}=\frac{c}{n}=\frac{3\times10^8}{9}=3.33\times10^7\ \text{m/s}.$$
Case 1 — unit at the surface. The whole round trip (up 300 m, reflect, back down 300 m) is in air: $$t_1=\frac{2(300)}{c}=\frac{600}{3\times10^8}$$ $$t_1=\boxed{2.00\ \mu\text{s}}$$
Case 2 — unit 10 m below the surface. Each one-way leg crosses 10 m of water then 300 m of air, so round trip: $$t_2=\frac{2(10)}{v_{\text{water}}}+\frac{2(300)}{c}=\frac{20}{3.33\times10^7}+\frac{600}{3\times10^8}$$ $$t_2=\boxed{2.60\ \mu\text{s}}$$
Comparison. The extra 20 m round-trip path through water (at 1/9 the speed) adds $\Delta t=t_2-t_1=0.60\ \mu\text{s}$ — nearly a third of the total case-1 delay, even though the extra water path is a small fraction of the total distance, because water is traversed 9 times slower than air.

Quantity	Result
$v_{\text{water}}$	$3.33\times10^7$ m/s
Delay, unit at surface ($t_1$)	2.00 µs
Delay, unit 10 m below surface ($t_2$)	2.60 µs
Difference $\Delta t$	0.60 µs

Concept & Theory

Topic. Electromagnetic wave speed in a dielectric medium — radar/sonar-style delay calculations.

Key relations. $v=c/n$, $n=\sqrt{\varepsilon_r\mu_r}$; delay $=$ distance/speed, summed per medium along the path.

Why this works. A wave's phase speed inside a linear, nonmagnetic dielectric slows by the factor $n=\sqrt{\varepsilon_r}$ relative to vacuum/air; travel time simply adds across media because each segment's time is independent of what precedes or follows it, provided the wave direction of travel and total distance in each medium are correctly identified.

Common pitfall. Forgetting to double the one-way distances for a round trip (transmit + reflect + receive), or applying the water's slower speed to the 300 m air path as well instead of only to the 10 m water segment.

Source. Sadiku, Elements of Electromagnetics, Ch. 10 (Electromagnetic Wave Propagation).

Question 8 of 8: Radar Pulse Delay Through Air and Water

Question 8: Radar Pulse Delay Through Air and Water (20 marks)