Radio communication carries information on electromagnetic (EM) waves radiated by an antenna. How a wave travels from transmitter to receiver depends mainly on its frequency. There are three principal modes of propagation: the ground (surface) wave, the sky wave and the space wave.
Ground (surface) wave propagation. At low frequencies (up to about $2\,\text{MHz}$, the medium-wave AM band), the wave travels along the curved surface of the Earth, guided by the ground. The wave induces currents in the Earth and is progressively absorbed, so its energy falls rapidly with distance and with increasing frequency. Ground-wave propagation is used for local AM broadcast over a few hundred kilometres.
Sky wave propagation and the ionosphere. The ionosphere is the upper part of the atmosphere (about $65\,\text{km}$ to $400\,\text{km}$) where solar radiation ionises the gas, producing free electrons. For frequencies in the range roughly $2\,\text{MHz}$ to $30\,\text{MHz}$ (short waves), this ionised layer reflects the wave back to Earth, allowing communication over thousands of kilometres by one or more skips between the ground and the ionosphere. The highest frequency that a given ionospheric layer can reflect when the wave is sent straight up is the critical frequency $f_c = 9\sqrt{N_{\max}}$, where $N_{\max}$ is the maximum electron density (per $\text{m}^3$). Waves above $f_c$ (sent vertically) punch through the ionosphere and escape into space, which is why frequencies above about $30\,\text{MHz}$ are not returned by the ionosphere.
Space wave propagation and line-of-sight (LOS). At very high frequencies (above about $40\,\text{MHz}$ — TV, FM, microwave and satellite links), the wave is neither guided by the ground nor reflected by the ionosphere. It travels in a straight line from the transmitting antenna to the receiving antenna — this is line-of-sight communication. Because the Earth is curved, the two antennas must be high enough to see each other over the horizon.
Range of a single antenna. A transmitting antenna of height $h$ above the ground can reach the horizon at a distance $d=\sqrt{2Rh}$, where $R$ is the radius of the Earth ($R \approx 6.4\times10^{6}\,\text{m}$). This is the geometric tangent distance to the Earth's curved surface.
Area covered. The service area of a TV tower of height $h$ is a circle of radius $d=\sqrt{2Rh}$, so the population covered scales with $A=\pi d^{2}=2\pi R h$, i.e. directly with the tower height. Doubling the height doubles the coverage area.
Range with both antennas raised. If both the transmitting antenna (height $h_T$) and the receiving antenna (height $h_R$) are elevated, the maximum line-of-sight distance between them is $d_M=\sqrt{2Rh_T}+\sqrt{2Rh_R}$. Raising the receiver as well as the transmitter extends the reach, which is why both TV towers and rooftop antennas are mounted high.
Summary of frequency bands:
- Ground wave — up to $\sim 2\,\text{MHz}$ (local AM).
- Sky wave (ionospheric reflection) — $\sim 2$ to $30\,\text{MHz}$ (long-distance short-wave).
- Space wave / LOS — above $\sim 40\,\text{MHz}$ (TV, FM, satellite, microwave).
A TV transmitting antenna is $80\,\text{m}$ high. Taking $R=6.4\times10^{6}\,\text{m}$, find the maximum line-of-sight distance up to which it can be received at ground level.
Solution- Range of a single antenna: $d=\sqrt{2Rh}$.
- Substitute $R=6.4\times10^{6}\,\text{m}$, $h=80\,\text{m}$: $d=\sqrt{2\times6.4\times10^{6}\times80}$.
- $2\times6.4\times10^{6}\times80 = 1.024\times10^{9}\,\text{m}^2$.
- $d=\sqrt{1.024\times10^{9}} \approx 3.2\times10^{4}\,\text{m} = 32\,\text{km}$.
Answer: The maximum line-of-sight range is about $32\,\text{km}$.
Find the area covered by the TV tower in the previous problem, and the population covered if the average density is $1000\,\text{per km}^2$.
Solution- Coverage area $A=\pi d^{2}$ with $d=32\,\text{km}$.
- $A=\pi (32)^{2}=\pi \times 1024 \approx 3217\,\text{km}^2$.
- Population $=$ area $\times$ density $= 3217 \times 1000 \approx 3.22\times10^{6}$.
Answer: Area $\approx 3217\,\text{km}^2$; population covered $\approx 3.2$ million.
A transmitting antenna is $32\,\text{m}$ high and the receiving antenna is $50\,\text{m}$ high. Find the maximum line-of-sight distance between them. Take $R=6.4\times10^{6}\,\text{m}$.
Solution- For two raised antennas: $d_M=\sqrt{2Rh_T}+\sqrt{2Rh_R}$.
- $\sqrt{2\times6.4\times10^{6}\times32}=\sqrt{4.096\times10^{8}} \approx 2.024\times10^{4}\,\text{m} \approx 20.2\,\text{km}$.
- $\sqrt{2\times6.4\times10^{6}\times50}=\sqrt{6.4\times10^{8}} \approx 2.53\times10^{4}\,\text{m} \approx 25.3\,\text{km}$.
- $d_M \approx 20.2 + 25.3 = 45.5\,\text{km}$.
Answer: The maximum line-of-sight distance is about $45.5\,\text{km}$.
The maximum electron density of an ionospheric layer is $N_{\max}=9\times10^{10}\,\text{per m}^3$. Find its critical frequency.
Solution- Critical frequency: $f_c=9\sqrt{N_{\max}}$ (with $N$ in $\text{per m}^3$, $f_c$ in Hz).
- $\sqrt{N_{\max}}=\sqrt{9\times10^{10}}=3\times10^{5}$.
- $f_c=9\times3\times10^{5}=2.7\times10^{6}\,\text{Hz}=2.7\,\text{MHz}$.
Answer: The critical frequency is $2.7\,\text{MHz}$; signals above this (sent vertically) pass through the layer.
To what height must a TV transmitting antenna be raised so that it can serve a circular area of radius $100\,\text{km}$? Take $R=6.4\times10^{6}\,\text{m}$.
Solution- From $d=\sqrt{2Rh}$, square both sides: $d^{2}=2Rh$, so $h=\dfrac{d^{2}}{2R}$.
- $d=100\,\text{km}=10^{5}\,\text{m}$, so $d^{2}=10^{10}\,\text{m}^2$.
- $h=\dfrac{10^{10}}{2\times6.4\times10^{6}}=\dfrac{10^{10}}{1.28\times10^{7}} \approx 781\,\text{m}$.
Answer: The antenna must be about $781\,\text{m}$ high to cover a $100\,\text{km}$ radius.
A radio signal of frequency $25\,\text{MHz}$ is to be broadcast over $1500\,\text{km}$. Which mode of propagation is suitable, and why is $25\,\text{MHz}$ a good choice?
Solution- Ground wave is absorbed quickly at this frequency, so it cannot cover $1500\,\text{km}$.
- $25\,\text{MHz}$ lies in the short-wave band ($2$ to $30\,\text{MHz}$), which the ionosphere reflects.
- Hence sky-wave (ionospheric) propagation carries the signal over $1500\,\text{km}$ via reflection from the ionosphere.
Answer: Sky-wave propagation; $25\,\text{MHz}$ is in the short-wave band that the ionosphere reflects, enabling long-distance coverage.