JEE Mains Physics · Communication Systems

Communication Systems, Bands and Antenna Size

A message travels from transmitter to receiver through a channel; each service has its own frequency band and way of travelling, and an antenna must be at least a quarter of a wavelength long.

All Communication Systems notes JEE Mains Physics strategy

Why this matters

Twenty PYQs, nineteen of them multiple choice: four from 2021, seven from 2022 and nine from 2023. Five are about the parts of a system and the bandwidths of signals and media, six match frequency bands or atmosphere layers to NCERT's tables, and nine are about antenna size. Most are recall, so the marks go to knowing the tables exactly.

Concept 1 of 3: Elements of a communication system and the number of channels

Every system has three parts: a transmitter, a channel and a receiver. The message gets weaker and picks up noise on the way, so it is amplified, and over long distances it is received and sent on again. Each message needs a band of frequencies, and a medium offers a band too. Dividing the medium's band by one message's band tells you how many messages it can carry at once.

Definition

Transducer: converts one form of energy into another (a microphone turns sound into an electrical signal).
Attenuation: loss of strength of a signal as it travels through the medium. Amplification: raising its strength with an amplifier.
Modulation: putting the message onto a high-frequency carrier. Demodulation: taking the message back off the carrier at the receiver.
Repeater: a receiver and a transmitter together; it picks up the signal, amplifies it and sends it on.
Noise: unwanted signals that disturb the message. Range: the largest distance over which the signal is received well enough.
Analog signals vary continuously; a digital signal takes only two levels, so it is a rectangular wave. Facsimile (fax) sends a static image of a document.
Bandwidth of signals (NCERT): speech about 2.8 kHz (300 Hz to 3100 Hz), high-quality music about 20 kHz, a TV signal (picture and sound) about 6 MHz.
Bandwidth of media: twisted pair a few MHz, coaxial cable up to about 750 MHz, optical fibre 1 THz to 1000 THz. A guided medium is a wire, a cable or an optical fibre.

Number of channels

N = \frac{\text{available bandwidth}}{\text{bandwidth of one channel}}, \qquad f = \frac{c}{\lambda}

Worked example

An optical link works at a wavelength of 600 nm, and 1% of its frequency is available as bandwidth. How many channels of 5 kHz each can it carry?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 29 June 2022 · Q20Moderate

Example 1 · Communication Systems · Communication Systems, Bands and Antenna Size

Only

2\%

of the optical source frequency is the available channel bandwidth for an optical communicating system operating at

1000\text{ }nm

. If an audio signal requires a bandwidth of

8kHz

, how many channels can be accommodated for transmission:

Find the frequency before taking the percentage

When a source is given by its wavelength, first find

f = c/\lambda

. The percentage is a share of that frequency, not of the wavelength.

A repeater is a receiver and a transmitter

A repeater does not only amplify. It receives the signal, amplifies it and transmits it again, so it combines both ends of the link.

Attenuation and demodulation are easy to swap

Attenuation is the loss of strength in the medium. Demodulation is taking the message back off the carrier at the receiver. Match lists put the two side by side.

Drill 4 more on elements of a communication system and the number of channels

Concept 2 of 3: Frequency bands, propagation modes and atmosphere layers

How a radio wave travels depends on its frequency. Low frequencies hug the ground. Frequencies of a few MHz up to about 30 MHz bounce off the ionosphere and come back far away. Higher frequencies pass straight through the ionosphere, so they must travel in a straight line from one antenna to the other, or up to a satellite.

Definition

Ground wave: travels along the earth's surface. The ground absorbs it more as the frequency rises, so it is used only up to a few MHz.
Sky wave: reflected back to earth by the ionosphere. It works from a few MHz up to about 30 to 40 MHz; above that the wave passes through.
Space wave: travels in a straight line (line of sight), used above about 40 MHz: FM, TV, mobile phones, microwave links and satellites.
Atmosphere layers, by NCERT's table: troposphere up to about 10 km; D layer (part of the stratosphere) 65 to 75 km; E layer (part of the stratosphere) about 100 km; F1 layer (part of the mesosphere) 170 to 190 km; F2 layer (part of the thermosphere) about 300 km at night and 250 to 400 km by day.
The D layer reflects low frequencies and absorbs some medium and high ones; the F2 layer reflects high-frequency waves best, especially at night.
Satellite links use the higher band for the uplink (earth to satellite) and the lower band for the downlink.

Service	Frequency band	Mode	How the wave travels
AM broadcast	540 to 1600 kHz	Ground wave	Along the earth's surface, for frequencies up to a few MHz
Short-wave radio	A few MHz to about 30 MHz	Sky wave	Reflected back to earth by the ionosphere
FM broadcast	88 to 108 MHz	Space wave	In a straight line from the transmitting antenna to the receiving one
Television	54 to 890 MHz	Space wave	In a straight line; the antenna height sets the range TV is split into sub-bands (54 to 72, 76 to 88, 174 to 216 and 420 to 890 MHz). A frequency like 64 MHz is TV, not FM.
Satellite uplink	5.925 to 6.425 GHz	Space wave	Straight up through the ionosphere to the satellite The uplink is the higher band. The downlink comes back on the lower one.
Satellite downlink	3.7 to 4.2 GHz	Space wave	From the satellite straight down to the earth station

Bands as in NCERT's table. Everything above about 40 MHz travels as a space wave.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 24 Jan 2023 · Q108Moderate

Example 2 · Communication Systems · Communication Systems, Bands and Antenna Size

Match List I with List II

LIST I	LIST II
A. AM Broadcast	I. $88 - 108MHz$
B. FM Broadcast	II. $540 - 1600kHz$
C. Television	III. $3.7 - 4.2GHz$
D. Satellite Communication	IV. $54MHz - 890MHz$

Choose the correct answer from the options given below:

kHz for AM, MHz for FM

AM broadcast is 540 to 1600 kHz; FM broadcast is 88 to 108 MHz. An option that gives AM in MHz or FM in kHz is wrong even if the numbers look right.

Uplink is the higher frequency

In satellite links the uplink uses 5.925 to 6.425 GHz and the downlink 3.7 to 4.2 GHz. The two bands are always offered together, so check which one is asked.

Use NCERT's layer labels

NCERT calls the D and E layers part of the stratosphere, F1 part of the mesosphere and F2 part of the thermosphere. Match-list questions follow these labels, so answer by the table.

Drill 5 more on frequency bands, propagation modes and atmosphere layers

Concept 3 of 3: Antenna length and why a high-frequency carrier is needed

An antenna radiates well only when its length is comparable to the wavelength, at least a quarter of it. An audio signal has a wavelength of kilometres, so its antenna would be kilometres long, and it would radiate very little power. So the message is carried on a high-frequency wave, whose wavelength is short. The antenna is then sized for the carrier.

Definition

Minimum antenna length: $l = \dfrac{\lambda}{4}$ , with $\lambda = \dfrac{c}{f}$ .
In a medium the wave is slower: $v = \dfrac{c}{\sqrt{\varepsilon_r \mu_r}}$ and $\lambda = \dfrac{v}{f}$ .
Power radiated by a linear antenna of length l: $P \propto \left(\dfrac{l}{\lambda}\right)^{2}$ . A long wavelength means low power.
Three reasons a low-frequency (baseband) signal is not sent directly: the antenna would be far too long, the radiated power would be low, and the signals of different transmitters would mix.
An AM wave contains $f_c$ and $f_c \pm f_m$ , all close to $f_c$ . So the wavelength radiated is $\lambda = c/f_c$ , and the antenna is sized by the carrier.
A largest antenna size gives a largest wavelength, $\lambda = 4l$ , and so a lowest frequency.

Antenna length and radiated power

l_{\min} = \frac{\lambda}{4} = \frac{c}{4f}, \qquad P \propto \left(\frac{l}{\lambda}\right)^{2}

Worked example

A 15 kHz audio signal is carried on a 1.5 GHz carrier. Find the minimum antenna length needed, and the length that would be needed to send the audio signal directly.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 14 June 2022 · Q16Moderate

Example 3 · Communication Systems · Communication Systems, Bands and Antenna Size

A baseband signal of

3.5MHz

frequency is modulated with a carrier signal of

3.5GHz

frequency using amplitude modulation method. What should be the minimum size of antenna required to transmit the modulated signal?

A quarter, not a half

The minimum antenna length is

\lambda/4

. Using

\lambda/2

doubles the answer, and that value is always among the options.

Size the antenna for the carrier

The radiated wave is the carrier with its side bands, all near

f_c

. Using the message frequency

f_m

gives an antenna far too long.

Divide c by the refractive index in a medium

In a dielectric the wave travels at

c/\sqrt{\varepsilon_r \mu_r}

, so the wavelength is shorter for the same frequency. Using c gives a frequency that is too high.

Largest antenna, lowest frequency

The largest wavelength an antenna can radiate is

4l

, and the largest wavelength is the lowest frequency. Reading it the other way round inverts the answer.

Drill 8 more on antenna length and why a high-frequency carrier is needed

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Elements of a communication system and the number of channels
Number of channels
$N = \frac{\text{available bandwidth}}{\text{bandwidth of one channel}}, \qquad f = \frac{c}{\lambda}$
Antenna length and why a high-frequency carrier is needed
Antenna length and radiated power
$l_{\min} = \frac{\lambda}{4} = \frac{c}{4f}, \qquad P \propto \left(\frac{l}{\lambda}\right)^{2}$

Reference tables (1)

Frequency bands, propagation modes and atmosphere layers6 rows

Service	Frequency band	Mode	How the wave travels
AM broadcast	540 to 1600 kHz	Ground wave	Along the earth's surface, for frequencies up to a few MHz
Short-wave radio	A few MHz to about 30 MHz	Sky wave	Reflected back to earth by the ionosphere
FM broadcast	88 to 108 MHz	Space wave	In a straight line from the transmitting antenna to the receiving one
Television	54 to 890 MHz	Space wave	In a straight line; the antenna height sets the range TV is split into sub-bands (54 to 72, 76 to 88, 174 to 216 and 420 to 890 MHz). A frequency like 64 MHz is TV, not FM.
Satellite uplink	5.925 to 6.425 GHz	Space wave	Straight up through the ionosphere to the satellite The uplink is the higher band. The downlink comes back on the lower one.
Satellite downlink	3.7 to 4.2 GHz	Space wave	From the satellite straight down to the earth station

Bands as in NCERT's table. Everything above about 40 MHz travels as a space wave.

Watch out for (10)

Find the frequency before taking the percentage
→ Elements of a communication system and the number of channels
A repeater is a receiver and a transmitter
→ Elements of a communication system and the number of channels
Attenuation and demodulation are easy to swap
→ Elements of a communication system and the number of channels
kHz for AM, MHz for FM
→ Frequency bands, propagation modes and atmosphere layers
Uplink is the higher frequency
→ Frequency bands, propagation modes and atmosphere layers
Use NCERT's layer labels
→ Frequency bands, propagation modes and atmosphere layers
A quarter, not a half
→ Antenna length and why a high-frequency carrier is needed
Size the antenna for the carrier
→ Antenna length and why a high-frequency carrier is needed
Divide c by the refractive index in a medium
→ Antenna length and why a high-frequency carrier is needed
Largest antenna, lowest frequency
→ Antenna length and why a high-frequency carrier is needed

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