Communication Systems — Modulation, Demodulation, Bandwidth

How Information Travels

Communication systems transmit information (voice, data, video) from one place to another using electromagnetic waves. The key challenge: audio signals (20 Hz to 20 kHz) have very long wavelengths and cannot be transmitted directly as radio waves. The solution is modulation — embedding the information signal onto a high-frequency carrier wave.

This chapter is relatively low-weightage in CBSE (2-3 marks) and rarely appears in JEE Main. But for NEET and board exams, the conceptual questions are straightforward and free marks if you know the basics. The entire chapter boils down to three ideas: why we modulate, how we modulate, and how signals propagate from transmitter to receiver.

Key Concepts

Why modulation is necessary

Three reasons why we cannot transmit audio signals directly:

Antenna size: For effective radiation, the antenna length should be comparable to the wavelength. A 20 kHz signal has $\lambda = 15$ km — an impractical antenna. A 1 MHz carrier has $\lambda = 300$ m — much more feasible.
Signal mixing: If everyone transmitted raw audio, all signals in the 20 Hz - 20 kHz band would overlap. Modulating different signals onto different carrier frequencies separates them.
Power efficiency: Power radiated by an antenna is proportional to $(\ell/\lambda)^2$ . Low-frequency signals (long $\lambda$ ) radiate very poorly from practical antennas.

Block diagram of a communication system

\text{Source} \to \text{Transmitter (Modulator)} \to \text{Channel} \to \text{Receiver (Demodulator)} \to \text{Output}

The transmitter modulates the message onto a carrier. The channel (atmosphere, cable, fibre) carries the signal. Noise is added in the channel. The receiver extracts the original message from the noisy modulated signal.

Types of modulation

Type	What varies	Carrier property held constant	Used in
AM (Amplitude Modulation)	Amplitude of carrier	Frequency	AM radio (530-1710 kHz)
FM (Frequency Modulation)	Frequency of carrier	Amplitude	FM radio (88-108 MHz)
PM (Phase Modulation)	Phase of carrier	Amplitude	Digital communication

Amplitude Modulation (AM) in detail

The AM signal is:

y(t) = A_c(1 + m\sin\omega_m t)\sin\omega_c t

where:

$A_c$ = carrier amplitude
$m = A_m/A_c$ = modulation index (keep $m \leq 1$ to avoid distortion)
$\omega_m$ = message frequency, $\omega_c$ = carrier frequency

Bandwidth $= 2f_m$ (twice the highest message frequency)

Sideband frequencies: $f_c + f_m$ (upper) and $f_c - f_m$ (lower)

When $m > 1$ , the signal is overmodulated — the envelope distorts and the demodulated signal is no longer faithful to the original. This is why modulation index control is critical.

Power in AM: Total power = carrier power + sideband power.

P_t = P_c\left(1 + \frac{m^2}{2}\right)

Only the sidebands carry information. The carrier itself carries no message — it is wasted power. This is why AM is inefficient and SSB (single sideband) AM was developed.

Frequency Modulation (FM) advantages

FM is less susceptible to noise than AM because most noise affects amplitude, not frequency. FM provides better audio quality and higher bandwidth. However, FM requires a wider bandwidth per channel (about 200 kHz for FM radio vs 10 kHz for AM), so fewer stations fit in a given frequency band.

Propagation modes

Mode	Frequency range	Mechanism	Range
Ground wave	Below ~2 MHz	Follows Earth’s surface, diffraction around obstacles	Up to ~1000 km (AM radio)
Sky wave	~2-30 MHz	Reflects off the ionosphere	Thousands of km (shortwave)
Space wave (line of sight)	Above ~30 MHz	Direct path + ground reflection	Limited by curvature of Earth (FM, TV)

Ground wave: Attenuates with distance due to energy absorption by the ground. Works best at lower frequencies. AM radio uses this.

Sky wave: The ionosphere (ionised upper atmosphere) reflects radio waves below ~30 MHz. Multiple hops between ionosphere and ground can send signals around the world. This is how shortwave radio works.

Space wave: Above 30 MHz, waves pass through the ionosphere instead of reflecting. They must travel in straight lines. This means the transmitter and receiver need line-of-sight, limited by Earth’s curvature. TV towers and cell towers use this.

d = \sqrt{2Rh_T} + \sqrt{2Rh_R}

where $R$ = Earth’s radius (6.4 × 10 $^6$ m), $h_T$ = transmitter height, $h_R$ = receiver height.

For a single tower with receiver at ground ( $h_R = 0$ ): $d = \sqrt{2Rh_T}$

Satellite communication

Geostationary satellites orbit at ~36,000 km with a period of 24 hours — they stay fixed above one point on Earth. Three geostationary satellites can cover nearly the entire globe. Uplink frequency is typically higher than downlink to minimise interference.

Internet and digital communication

Modern communication is predominantly digital. Analog signals are converted to digital (A/D conversion), transmitted as binary data (0s and 1s), and converted back to analog at the receiver (D/A conversion). Digital signals are more robust against noise because the receiver only needs to distinguish between two levels (0 and 1), not a continuous range.

Optical fibre communication uses total internal reflection to carry light signals through thin glass fibres. Advantages: enormous bandwidth, no electromagnetic interference, very low signal loss, high security.

Solved Examples

Example 1 (Easy — CBSE)

An AM signal has carrier frequency 1 MHz and message frequency 5 kHz. Find the sideband frequencies and bandwidth.

Upper sideband: $f_c + f_m = 1000 + 5 = 1005$ kHz. Lower sideband: $f_c - f_m = 1000 - 5 = 995$ kHz.

Bandwidth $= 2f_m = 2 \times 5 = 10$ kHz.

Example 2 (Medium — CBSE)

A TV tower is 100 m tall. Find the maximum distance it can cover. (Take $R = 6.4 \times 10^6$ m.)

$d = \sqrt{2Rh} = \sqrt{2 \times 6.4 \times 10^6 \times 100} = \sqrt{1.28 \times 10^9} \approx 35.8$ km.

For a taller tower (200 m): $d = \sqrt{2 \times 6.4 \times 10^6 \times 200} = \sqrt{2.56 \times 10^9} \approx 50.6$ km. Doubling the height increases range by a factor of $\sqrt{2}$ , not 2.

Example 3 (Medium — NEET)

A carrier of amplitude 10 V is modulated by a signal of amplitude 4 V. Find $m$ and check for distortion.

$m = A_m/A_c = 4/10 = 0.4$ . Since $m < 1$ , no distortion. The modulated signal amplitude varies between $A_c(1+m) = 14$ V and $A_c(1-m) = 6$ V.

Example 4 (Application)

Cell phones use frequencies in the 800 MHz - 2.4 GHz range — space wave propagation only. The range is limited by line of sight and building obstacles. For a 30 m tower: $d = \sqrt{2 \times 6.4 \times 10^6 \times 30} \approx 19.6$ km in open terrain. In cities, buildings absorb and reflect signals, reducing effective range to 2-5 km. Hence towers are placed every few km.

Common Mistakes to Avoid

Confusing bandwidth with carrier frequency. Bandwidth of AM = $2f_m$ (depends on the message, not the carrier). A 1 MHz carrier with 5 kHz audio has bandwidth 10 kHz, not 1 MHz.

Using $m > 1$ without recognizing distortion. If modulation index exceeds 1, the signal is overmodulated — the envelope clips and the demodulated message is distorted. Always check $m \leq 1$ .

Wrong propagation mode for the frequency range. Below 2 MHz → ground wave. 2-30 MHz → sky wave. Above 30 MHz → space wave. FM radio at 100 MHz is space wave, not sky wave.

Forgetting to square root in the space wave formula. $d = \sqrt{2Rh}$ , not $2Rh$ . This is the most common numerical mistake in this chapter.

Saying FM is always better than AM. FM has better noise rejection and audio quality, but AM can propagate over much longer distances (ground wave, sky wave). For long-distance broadcasting in remote areas, AM is still used precisely because FM is limited to line of sight.

Exam Weightage and Strategy

Communication Systems carries 2-3 marks in CBSE Class 12 boards. NEET may ask 0-1 question. JEE Main rarely tests it. Despite low weightage, the marks are easy — the formulas are simple and the concepts are intuitive. Treat this as a quick-score chapter.

Memorise three formulas: AM signal equation, bandwidth = $2f_m$ , and space wave range $d = \sqrt{2Rh}$ . Know the three propagation modes and their frequency ranges. That covers every possible question on this chapter.

Practice Questions

Q1. What is the modulation index if carrier amplitude is 10 V and message amplitude is 6 V?

$m = A_m/A_c = 6/10 = 0.6$ . Since $m < 1$ , the signal is properly modulated without distortion.

Q2. An antenna is 50 m high. What distance can it cover?

$d = \sqrt{2 \times 6.4 \times 10^6 \times 50} = \sqrt{6.4 \times 10^8} \approx 25.3$ km.

Q3. Why can audio signals not be transmitted directly as electromagnetic waves?

Three reasons: (1) Audio frequencies (20 Hz - 20 kHz) need antennas comparable to their wavelength (15 km - 15,000 km) — impractical. (2) All audio signals would overlap in the same frequency band — no channel separation. (3) Power radiated is proportional to $(\ell/\lambda)^2$ — long wavelengths radiate very poorly from practical antennas.

Q4. What advantage does FM have over AM?

FM is less susceptible to noise because most noise affects amplitude (which FM ignores). FM has better audio quality and higher bandwidth for music. However, FM requires wider bandwidth per channel and is limited to line-of-sight range (space wave), while AM can travel much farther via ground and sky waves.

Q5. Two towers of heights 100 m and 64 m are separated by a distance. What is the maximum line-of-sight distance between them?

$d = \sqrt{2Rh_T} + \sqrt{2Rh_R} = \sqrt{2 \times 6.4 \times 10^6 \times 100} + \sqrt{2 \times 6.4 \times 10^6 \times 64}$ $= \sqrt{1.28 \times 10^9} + \sqrt{8.19 \times 10^8} = 35.8 + 28.6 = \mathbf{64.4}$ km.

FAQs

What is the role of a repeater?

A repeater receives a weak signal, amplifies it, and retransmits. Used when the signal must travel beyond line-of-sight distance. Cell towers act as repeaters. Each tower receives the signal from the previous one and retransmits it to the next, extending coverage over large areas.

Why does TV use space wave propagation?

TV signals are above 30 MHz and are not reflected by the ionosphere — they pass through it. They must travel in straight lines (line of sight), which is why TV towers need to be tall. Taller tower = larger coverage area. Cable and satellite TV bypass this limitation entirely.

What is demodulation?

The reverse of modulation — extracting the original message signal from the modulated carrier. For AM, a simple envelope detector (diode + RC filter) strips the carrier and recovers the audio. For FM, a discriminator circuit converts frequency variations back into amplitude variations.

What is bandwidth and why does it matter?

Bandwidth is the range of frequencies occupied by a signal. A voice call needs about 4 kHz bandwidth. Music needs about 20 kHz. Video needs several MHz. The total bandwidth available in any frequency band is limited, so communication systems must use bandwidth efficiently. This is why digital compression (MP3, H.264) is so important — it reduces the bandwidth needed to transmit the same information.

What is the ionosphere and why is it important?

The ionosphere is a layer of the upper atmosphere (60-1000 km altitude) where gas molecules are ionised by solar radiation. The free electrons and ions make this layer reflective to radio waves below ~30 MHz. This is what makes sky wave propagation possible — shortwave radio signals bounce between the ionosphere and the ground, travelling thousands of kilometres. Above 30 MHz (FM, TV), waves pass through the ionosphere into space, which is why these signals are limited to line-of-sight range.