CBSE Class 12 Physics Exemplar Chapter 15: Communication Systems NCERT Solutions

Reviewed by NCERT Help Editorial Team · Updated on 29 Jul 2026

NCERT Solutions PDF Class 12 PDF

CBSE Class 12 Physics Exemplar, Chapter 15, delves into the fascinating world of Communication Systems. This chapter explores the fundamental ways radio waves travel through the atmosphere, including ground waves that follow the Earth's curvature, sky waves that bounce off the ionosphere, and space waves that travel directly from transmitter to receiver. It also examines the crucial concept of Amplitude Modulation (AM), detailing how sideband frequencies are generated and how to determine the frequency of the modulated signal. These NCERT Solutions offer clear, step-by-step explanations to help students grasp these concepts, solve related problems, and build a strong foundation for their board exams.

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Quick info

Board	CBSE
Class	Class 12
Subject	Physics Exemplar
Session	2026
Language	English
Type	NCERT Solutions
Chapter	Chapter 15

Chapter summary

Chapter 15 of the CBSE Class 12 Physics Exemplar, "Communication Systems," focuses on the fundamental principles of transmitting and receiving information. The NCERT Solutions provided here cover multiple-choice questions that test understanding of wave propagation modes based on frequency (ground, sky, and space waves) and the characteristics of amplitude modulation, including carrier and sideband frequencies. These solutions aim to clarify complex topics and aid students in mastering the chapter's core concepts.

Learning outcomes

Understand the different modes of wave propagation: ground, sky, and space waves.
Identify the appropriate mode of communication based on wave frequency.
Calculate the frequencies of sidebands in Amplitude Modulation (AM).
Determine the frequency of an AM wave.
Relate antenna length to the wavelength of transmittable waves.
Analyze signal attenuation in communication channels.

Topics covered

Paper topics

Modes of Communication
Ground Wave Propagation
Sky Wave Propagation
Space Wave Propagation
Amplitude Modulation (AM)
Carrier Frequency
Sideband Frequencies
Antenna Length and Wavelength
Signal Attenuation
Decibel Scale

Important topics

Wave Propagation Modes (Ground, Sky, Space)
Frequency Ranges for Propagation Modes
Amplitude Modulation (AM) Principles
Calculation of Sideband Frequencies
Signal Attenuation and Decibels

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Questions and Solutions

Question 1

Three waves A, B and C of frequencies 1600 kHz, 5 MHz and 60 MHz, respectively are to be transmitted from one place to another. Which of the following is the most appropriate mode of communication?

A is transmitted via space wave while B and C are transmitted via sky wave
A is transmitted via ground wave, B via sky wave and C via space wave
B and C are transmitted via ground wave while A is transmitted via sky wave
B is transmitted via ground wave while A and C are transmitted via space wave

Solution: The most appropriate mode of communication depends on the frequency of the wave. The typical frequency ranges for different propagation modes are:

Ground wave propagation: Approximately 530 kHz to 1710 kHz.

Sky wave propagation: Approximately 1710 kHz to 40 MHz.

Space wave propagation: Approximately 54 MHz to 4.2 GHz.

Given the frequencies:

Wave A: 1600 kHz = 1.6 MHz. This falls within the ground wave propagation range.
Wave B: 5 MHz. This falls within the sky wave propagation range.
Wave C: 60 MHz. This falls within the space wave propagation range.

Therefore, wave A is transmitted via ground wave, wave B via sky wave, and wave C via space wave.

Answer: (b) A is transmitted via ground wave, B via sky wave and C via space wave

Question 2

A 100m long antenna is mounted on a 500m tall building. The complex can become a transmission tower for waves with wavelength $\lambda$

$\sim 400 \text{ m}$
$\sim 25 \text{ m}$
$\sim 150 \text{ m}$
$\sim 2400 \text{ m}$

Solution: The effective height of the transmitting antenna system is crucial for determining the range of wavelengths it can effectively transmit. In many antenna designs, particularly for efficient radiation, the antenna length is related to the wavelength. A common relationship for a simple dipole antenna is that its length is approximately half the wavelength ($L = \lambda/2$), or for some configurations, a quarter wavelength ($L = \lambda/4$).

In this problem, the antenna length is given as $l = 100$ m. The source material suggests a relationship where the wavelength $\lambda$ is approximately 4 times the antenna length. This implies the antenna might be acting as a quarter-wave radiator, or the total effective height (building + antenna) is considered in a way that leads to this relationship, though the solution focuses solely on the antenna length.

Using the relationship $\lambda \sim 4l$:

$\lambda \sim 4 \times 100 \, \text{m} = 400 \, \text{m}$

This wavelength is suitable for transmission by an antenna of this size.

Answer: (a) $\sim 400 \text{ m}$

Question 3

A 1 kW signal is transmitted using a communication channel which provides attenuation at the rate of – 2dB per km. If the communication channel has a total length of 5 km, the power of the signal received is [gain in dB = $10 \log \left( \frac{P_0}{P_i} \right)$]

900 W
100 W
990 W
1010 W

Solution: We are given the initial power of the signal ($P_i$), the rate of attenuation, and the total length of the communication channel.

Initial power, $P_i = 1 \text{ kW} = 1000 \text{ W}$.

Rate of attenuation = -2 dB per km.

Total length of the channel = 5 km.

The total attenuation over the 5 km channel is the rate of attenuation multiplied by the length:

Total Attenuation = (-2 dB/km) $\times$ (5 km) = -10 dB.

The formula for gain (or loss, in this case) in decibels is given as:

Gain in dB = $10 \log_{10} \left( \frac{P_0}{P_i} \right)$

Where $P_0$ is the received power and $P_i$ is the initial power. Substituting the values:

$-10 \text{ dB} = 10 \log_{10} \left( \frac{P_0}{1000 \text{ W}} \right)$

Divide both sides by 10:

$-1 = \log_{10} \left( \frac{P_0}{1000} \right)$

To solve for $P_0$, we convert the logarithmic equation to its exponential form:

$10^{-1} = \frac{P_0}{1000}$

$0.1 = \frac{P_0}{1000}$

Now, solve for $P_0$:

$P_0 = 0.1 \times 1000 \text{ W}$

$P_0 = 100 \text{ W}$

Thus, the power of the signal received after passing through the 5 km channel is 100 W.

Answer: (b) 100 W

Question 4

A speech signal of 3 kHz is used to modulate a carrier signal of frequency 1 MHz, using amplitude modulation. The frequencies of the side bands will be

1.003 MHz and 0.997 MHz
3001 kHz and 2997 kHz
1 MHz and 0.997 MHz
1003 kHz and 1000 kHz

Solution: In Amplitude Modulation (AM), the modulated signal contains the original carrier frequency and two side frequencies, known as sidebands. These sideband frequencies are generated by the interaction of the carrier signal and the modulating signal.

The carrier frequency is given as $f_c = 1 \text{ MHz}$.

The frequency of the speech signal (modulating signal) is given as $f_m = 3 \text{ kHz}$.

It is important to use consistent units. Let's convert kHz to MHz:

$f_m = 3 \text{ kHz} = 3 \times 10^{-3} \text{ MHz} = 0.003 \text{ MHz}$.

The frequencies of the upper and lower sidebands are calculated as follows:

Upper Sideband Frequency ($f_{USB}$) = $f_c + f_m$

Lower Sideband Frequency ($f_{LSB}$) = $f_c - f_m$

Substituting the values:

$f_{USB} = 1 \text{ MHz} + 0.003 \text{ MHz} = 1.003 \text{ MHz}$

$f_{LSB} = 1 \text{ MHz} - 0.003 \text{ MHz} = 0.997 \text{ MHz}$

Therefore, the frequencies of the sidebands are 1.003 MHz and 0.997 MHz.

Answer: (a) 1.003 MHz and 0.997 MHz

Question 5

A message signal of frequency $\omega_m$ is superposed on a carrier wave of frequency $\omega_c$ to get an Amplitude Modulated Wave (AM). The frequency of the AM wave will be

$\omega_m$
$\omega_c$
$\frac{\omega_c + \omega_m}{2}$
$\frac{\omega_c - \omega_m}{2}$

Solution: In Amplitude Modulation (AM), the process involves modifying the amplitude of a high-frequency carrier wave according to the instantaneous amplitude of a lower-frequency message signal. The carrier wave's frequency itself is not changed during this process.

The carrier wave has a frequency of $\omega_c$. The message signal has a frequency of $\omega_m$. When the message signal modulates the carrier wave, the resulting AM wave consists of the original carrier frequency component plus two sideband components at frequencies $\omega_c + \omega_m$ and $\omega_c - \omega_m$.

However, the fundamental frequency of the AM wave, which determines its basic oscillation rate, is the frequency of the carrier wave.

Therefore, the frequency of the AM wave is $\omega_c$.

Answer: (b) $\omega_c$

Common mistakes

Confusing the frequency ranges for ground, sky, and space wave propagation.
Incorrectly calculating sideband frequencies in AM.
Misinterpreting the relationship between antenna length and wavelength.
Errors in applying the dB formula for signal attenuation.

Revision tips

Memorize the frequency ranges for ground, sky, and space wave propagation.
Practice calculating sideband frequencies using the formula $f_c \pm f_m$.
Understand the concept of antenna length and its relation to the transmitted wavelength.
Review the formula for signal attenuation in decibels and practice its application.

Practice MCQs

Q1. Three waves A, B, and C with frequencies 1600 kHz, 5 MHz, and 60 MHz, respectively, are to be transmitted. Which of the following is the most appropriate mode of communication for each wave?

Q2. An antenna of length 100 m is mounted on a building 500 m tall. What is the approximate wavelength of the waves that can be transmitted by this setup?

Q3. A 1 kW signal is transmitted through a communication channel with an attenuation rate of -2 dB per km. If the total length of the channel is 5 km, what is the power of the received signal?

Q4. A speech signal of 3 kHz is used to modulate a 1 MHz carrier signal using amplitude modulation. What are the frequencies of the resulting sidebands?

Q5. When a message signal of frequency $ $\omega$ _m$ is superimposed on a carrier wave of frequency $ $\omega$ _c$ to create an Amplitude Modulated (AM) wave, what is the frequency of the resulting AM wave?

Frequently asked questions

What are the main modes of communication discussed in CBSE Class 12 Physics Chapter 15?

The main modes of communication discussed are ground wave propagation (for lower frequencies), sky wave propagation (for medium frequencies), and space wave propagation (for higher frequencies).

How does frequency determine the mode of communication?

Lower frequencies (like 1600 kHz) are best suited for ground wave propagation. Medium frequencies (like 5 MHz) use sky wave propagation, bouncing off the ionosphere. Very high frequencies (like 60 MHz) travel via space waves, often using line-of-sight paths.

What is Amplitude Modulation (AM)?

Amplitude Modulation (AM) is a technique where the amplitude of a high-frequency carrier wave is varied in accordance with the instantaneous amplitude of the message signal.

How are sideband frequencies calculated in AM?

The sideband frequencies are calculated by adding and subtracting the frequency of the message signal ($f_m$) from the frequency of the carrier wave ($f_c$). The frequencies are $f_c + f_m$ (upper sideband) and $f_c - f_m$ (lower sideband).

What is the frequency of an AM wave?

The frequency of an Amplitude Modulated (AM) wave is the same as the frequency of the original carrier wave ($f_c$).

How is signal attenuation measured?

Signal attenuation is often measured in decibels (dB). A negative dB value indicates a loss of signal power over the communication channel.

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