radio communication television and radar principles

Radio communication rests on the emission and reception of electromagnetic waves. A harmonic carrier $s_0(t)=A_0\cos(2\pi f_0 t)$ is radiated by an antenna, and the message is imposed on it by amplitude, frequency, or phase modulation: $s(t)=A(t)\cos(2\pi f_0 t+\varphi(t))$. An antenna radiates efficiently when its size is comparable with the wavelength $\lambda=c/f_0$, which is why long-wave stations need tall masts while mobile links use centimetre waves. Television transmits picture and sound in one channel: the video signal amplitude-modulates the carrier (in practice with a partly suppressed sideband to save spectrum), while line and frame sync pulses let the receiver rebuild the raster. Radar sends a short pulse, receives the echo, and converts the delay $\Delta t$ into range by $d=c\Delta t/2$, the factor 2 accounting for the round trip. A moving target shifts the echo frequency by the Doppler amount $\Delta f=2vf_0/c$, which yields its radial speed.

A radio station transmits at $f_0 = 100$ MHz. Find the wavelength $\lambda$ and estimate the length of a half-wave dipole receiving antenna.

A pulsed radar sends a probing pulse, and the echo from a target returns after $\Delta t = 40$ $\mu$s. The pulse repetition period is $T = 1{.}0$ ms. Find the target range $d$ and the maximum unambiguous range $d_{max}$ of the radar.

A speed-control radar operates at $f_0 = 10$ GHz. Mixing the transmitted and reflected signals gives a difference (Doppler) frequency $\Delta f = 1{.}0$ kHz. Find the radial speed of the car in m/s and km/h, and explain how the sign of the shift shows whether it is approaching or receding.

An at-home modelling lab (entirely safe, no transmission required).

1. In a spreadsheet, tabulate $\lambda = c/f_0$ for carriers at 88, 100, 108 MHz (FM), 600 MHz (television) and 3 GHz (radar). For each row compute $\lambda/2$ and note which antenna size is realistic.
2. Model a pulsed radar: repetition period $T = 1$ ms, target at 15 km. Compute the echo delay $\Delta t = 2d/c$ (it should be 100 $\mu$s) and sketch a 'transmitted pulse — echo' timing diagram.
3. Place the target at 200 km with the same $T$ and show that the echo arrives only after the next pulse: range ambiguity appears, and the instrument may report a false close target.
4. In a free simulator (e.g. Falstad 'Amplitude Modulation' or a similar online oscilloscope) observe the carrier and envelope for AM while changing the modulation depth; compare with FM, where the amplitude stays constant.

Safety. Do not use real transmitters in service or amateur bands without a licence; all measurements and observations are done in a simulator or offline.

Show that radio communication, television and radar share one and the same chain 'carrier → modulation → antenna → propagation → reception → demodulation', and link the echo time delay to the target range.

Block diagram of a radio link: carrier oscillator → modulator (message input) → transmitting antenna → propagation medium → receiving antenna → demodulator. Beside it, two graphs: above, an oscillogram of an AM signal with envelope $A(t)$ on the carrier $\cos(2\pi f_0 t)$; below, a timing diagram of transmitter pulses (period $T$) and the received echo delayed by $\Delta t$, together with the formula $d = c\Delta t/2$.

You call a friend on a mobile phone. The phone emits radio waves at a frequency of $1{,}8\ \text{GHz}$ ($1{,}8\times 10^9\ \text{Hz}$). The transmitter power is about $0{,}2\ \text{W}$, and the nearest base station is $500\ \text{m}$ away. The wavelength is: $\lambda = \frac{c}{f} = \frac{3{,}0\times 10^8\ \text{m/s}}{1{,}8\times 10^9\ \text{Hz}} \approx 0{,}17\ \text{m}.$ The signal reaches the station in about $t = d/c = 500/3{,}0\times 10^8 \approx 1{,}7\times 10^{-6}\ \text{s} = 1{,}7\ \mu\text{s}$.

The scenario uses emission of electromagnetic waves by the phone antenna and modulation: sound is converted into an electrical signal that is superimposed on the $1{,}8\ \text{GHz}$ carrier. The receiving antenna extracts this signal. The frequency choice relies on the spectrum: in the microwave band, compact antennas work and many channels are available.

An officer points a radar at a car. The radar emits pulses at a frequency $f_0 = 24{,}15\ \text{GHz}$. The car moves at speed $v = 30\ \text{m/s}$ ($108\ \text{km/h}$). The reflected signal returns with a Doppler frequency shift: $\Delta f = \frac{2 v f_0}{c} = \frac{2 \cdot 30 \cdot 24{,}15\times 10^9}{3{,}0\times 10^8} \approx 4{,}8\times 10^3\ \text{Hz} = 4{,}8\ \text{kHz}.$ For a distance $d = 150\ \text{m}$, the echo delay is: $t = \frac{2d}{c} = \frac{2\cdot150}{3{,}0\times10^8} = 1{,}0\times10^{-6}\ \text{s} = 1{,}0\ \mu\text{s}.$

The radar principles work here: directed emission of radio waves, reflection from the car, and reception of the echo. The Doppler shift $\Delta f = 2 v f_0/c$ gives speed, while the delay $t = 2d/c$ gives distance. The frequency $24{,}15\ \text{GHz}$ is chosen from the spectrum: the wavelength $\lambda = c/f_0 \approx 0{,}012\ \text{m}$ allows a narrow beam.

A city receives a television signal from a tower of height $h = 300\ \text{m}$ operating at a frequency $f = 600\ \text{MHz}$ ($6{,}0\times10^8\ \text{Hz}$). The wavelength is: $\lambda = \frac{c}{f} = \frac{3{,}0\times10^8}{6{,}0\times10^8} = 0{,}50\ \text{m}.$ The line-of-sight distance to the horizon from the tower height is: $d \approx \sqrt{2 R h} = \sqrt{2\cdot 6{,}37\times10^6\cdot 300} \approx 6{,}2\times10^4\ \text{m} = 62\ \text{km}.$ Therefore reliable reception is possible within a radius of several tens of kilometers.

The example shows television transmission: the video signal modulates the $600\ \text{MHz}$ carrier (UHF band), and the receiving antenna extracts it. Propagation is almost line-of-sight, so the tower height determines the coverage area; this follows from the properties of electromagnetic waves in the UHF part of the spectrum.