Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Home exercise 6.1.6

Interactive: Doppler Effect and Radar

Home exercise 6.1.6

Doppler on a high-speed train

A train runs at $v = 300 km/h$ on a straight track. A base station transmitting at $f = 2.1 GHz$ stands $d = 50 m$ from the track. Let $x$ be the distance, measured along the track, from the train to the point of the track closest to the station ( $x > 0$ before passing).

Show that the receiver on the train sees $f^{'} - f = \frac{f v}{c} \frac{x}{x ^{2} + d ^{2}}$ (moving observer).
Evaluate the shift at $x = 500$ , $100$ , $50$ , $0$ and $- 500 m$ .
How large is the total frequency swing, and in about how long does most of it happen?

Try it yourself first.

Strategy

The train is a moving observer: $f^{'} = f (1 + \frac{v}{c} cos θ)$ by Eq. (6.1.70), where $θ$ is the angle between its velocity and the line from the train to the station. Only the geometry of $cos θ$ is new.

The line from the train to the station has the components $x$ along the track (in the direction of motion while $x > 0$ ) and $d$ across it, so $cos θ = x / x^{2} + d^{2}$ and

f^{'} - f = \frac{f v}{c} \frac{x}{x ^{2} + d ^{2}} .

$v = 300/3.6 = 83.3 m/s$ and $f v / c = 2.1 \times 1 0^{9} \times 83.3/ (3 \times 1 0^{8}) = 583.3 Hz$ :

$x$ (m)	$f^{'} - f$ (Hz)
$500$	$+ 580.4$
$100$	$+ 521.7$
$50$	$+ 412.5$
$0$	$0$
$- 500$	$- 580.4$

Far away the shift tends to $\pm 583.3 Hz$ , so the total swing is $2 \times 583.3 \approx 1.17 kHz$ . Most of it happens while $∣ x ∣ ≲ d$ , that is in about $2 d / v = 100/83.3 = 1.2 s$ .

Answer

f^{'} - f = 583.3 \frac{x}{x ^{2} + d ^{2}} Hz, swing \approx 1.17 kHz in about 1.2 s

Type $x = 500$ , $100$ , $50$ , $0$ and $- 500 m$ below and compare the shift with the table.

InteractiveLoading simulation…

Check

At $x = d$ , $cos θ = 1/ 2$ and the shift is $583.3/ 2 = 412.5 Hz$ . The curve is odd in $x$ , as the symmetry of the pass requires.

Remark

A jump of more than a kilohertz within about a second is a real problem for an LTE or 5G receiver, whose subcarriers are $15 kHz$ apart: the shift is a sizeable fraction of the spacing and must be tracked by the receiver.

Home exercise 6.1.6: Doppler on a high-speed train