Electromagnetic communication and radar

Learning objectives

Outcome
Calculate the wavelength or carrier frequency using $\lambda = c/f$ for a given communication band.
Outcome
Determine target range from the echo delay time using $d = c\,t/2$ .
Outcome
Explain how modulation (AM or FM) carries information on a carrier and describe the role of the antenna.
Outcome
Calculate the speed of a moving target from the Doppler frequency shift of a radar signal.

Understand the idea

Electromagnetic communication and radar: principles and applications

Text

Electromagnetic communication and radar both rest on the same fact: an electromagnetic wave travels through vacuum at $c \approx 3\times10^8$ m/s, and its wavelength is fixed by the carrier frequency through $\lambda = c/f$ . A transmitter generates a high-frequency carrier, and information is imprinted on it by modulation — varying the amplitude (AM) or the frequency (FM) — so that the receiver can recover the message. An antenna converts the guided current in a circuit into a freely propagating wave and back again, which is why antenna size scales with wavelength. Radar uses the same waves in a pulse-echo mode: a short pulse is emitted, and the echo returning from a target is timed. Because the wave makes a round trip, the range is $d = c\,t/2$ , and a moving target shifts the echo frequency through the Doppler effect.

Worked examples

Direct

Problem

An FM radio station broadcasts at a frequency $f = 100$ MHz. Find the wavelength in vacuum.

Solution

Use the relation between wavelength and frequency:

\lambda = \frac{c}{f} = \frac{3.00\times10^8\ \text{m/s}}{1.00\times10^8\ \text{Hz}} = 3.0\ \text{m}.

This is a VHF band; a quarter-wave antenna would be about $0.75$ m long, which is convenient for a car receiver.

Application

Problem

A radar emits a pulse and receives the echo from an aircraft after $t = 100$ µs. Find the range to the aircraft.

Solution

The wave travels out and back, so the range is half of the total path:

d = \frac{c\,t}{2} = \frac{(3.00\times10^8\ \text{m/s})(1.00\times10^{-4}\ \text{s})}{2} = 1.5\times10^4\ \text{m} = 15\ \text{km}.

Unit check: $\text{m/s}\cdot\text{s} = \text{m}$ .

Transfer

Problem

A traffic radar operates at $f_0 = 24.15$ GHz and measures a Doppler beat frequency $f_d = 2.00$ kHz from an approaching car. Find the car's speed.

Solution

In radar the wave makes a round trip, so the Doppler shift is doubled:

f_d = \frac{2 v f_0}{c}.

Solve for the speed:

v = \frac{f_d\,c}{2 f_0} = \frac{(2.00\times10^3)(3.00\times10^8)}{2(24.15\times10^9)} = 12.4\ \text{m/s} \approx 45\ \text{km/h}.

An approaching target gives a positive upward frequency shift.

Practice

Text

Simulation plan (safe, no radiation hazard). Work only with a free online pulse-echo simulator or a spreadsheet, not with a real transmitter.

Set the wave speed to $c = 3.00\times10^8$ m/s.
For ranges $d = 5, 10, 20, 40$ km, compute the echo delay $t = 2d/c$ and plot $d$ versus $t$ .
Check that the points lie on a straight line with slope $c/2$ .
Add a moving target with speed $v$ and compute the Doppler shift $f_d = 2 v f_0/c$ for $f_0 = 10$ GHz; plot $f_d$ versus $v$ .

Do not build or point any real transmitting antenna: keep all work inside the simulator or spreadsheet.

Visual reference

Purpose

Show how the round-trip time of a pulse determines range and how the same wave carries a modulated message.

Representation

A two-part diagram: (a) a transmitter, a target at distance $d$ , and the returning echo, with the full path marked $2d$ ; (b) a graph of received signal amplitude versus time, with the transmitted pulse at $t = 0$ and the echo pulse at $t = 2d/c$ .

Real-world examples

Cellular communication: carrier frequency and wavelength

Scenario

A smartphone on an LTE network transmits at $800\ \text{MHz}$ . The speed of light is $c = 3.00\cdot10^8\ \text{m/s}$ , so the carrier wavelength is $\lambda = c/f = 3.00\cdot10^8 / 800\cdot10^6 = 0.375\ \text{m} = 37.5\ \text{cm}$ . The phone antenna is about half a wavelength — roughly $19\ \text{cm}$ — and radiates a power of $0.2\ \text{W}$ . A base station $5\ \text{km}$ away receives the signal about $5\cdot10^3 / 3\cdot10^8 \approx 17\ \mu\text{s}$ after it is sent.

Connection

The scenario uses the fixed link between frequency and wavelength in the electromagnetic spectrum ( $\lambda f = c$ ): the carrier at $800\ \text{MHz}$ dictates the antenna size and the propagation delay at speed $c$ . Modulation and demodulation are the parts of communication theory that imprint and recover information on this carrier.

Speed radar: the Doppler effect

Scenario

A police radar emits a wave at $24.125\ \text{GHz}$ (K-band) toward a car moving at $30\ \text{m/s}$ ( $108\ \text{km/h}$ ). The wave reflected from the car has a Doppler shift $\Delta f = 2 v f / c = 2\cdot30\cdot24.125\cdot10^9 / 3.00\cdot10^8 \approx 4.8\ \text{kHz}$ . The receiver measures this frequency difference and computes the speed.

Connection

This is a direct application of the Doppler effect for electromagnetic waves: the moving car changes the frequency of the reflected wave in proportion to its speed. The radar finds the distance to the target from the round-trip time at speed $c$ .

Weather radar: pulse and range

Scenario

A weather radar operates at $5.6\ \text{GHz}$ (C-band) and sends pulses of duration $1\ \mu\text{s}$ . The pulse occupies $c\tau = 3.00\cdot10^8 \cdot 1\cdot10^{-6} = 300\ \text{m}$ in space, so the range resolution is $\Delta R = c\tau/2 = 150\ \text{m}$ . A raindrop at $60\ \text{km}$ gives an echo $2\cdot60\cdot10^3 / 3\cdot10^8 = 400\ \mu\text{s}$ after the pulse is emitted. From the delay time the radar builds a map of precipitation.

Connection

This uses time-of-flight ranging at the speed of light: the echo delay gives the distance, and the pulse duration limits the range resolution. Backscatter from raindrops turns the radio wave into a measurable echo signal.