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NCERT Exemplar Class 12 Physics Solutions Chapter 15 introduces you to the amazing world of communication – how information is transferred from one place to another. We’re constantly receiving and sharing info through different mediums, and this chapter explains how that smooth and efficient communication actually works.
Ever imagined how we send messages, make phone calls, or watch live broadcasts? This chapter answers those questions and helps you understand the science behind it. The solutions provided will not only help you express your ideas better, especially in academics, but also prepare you well for your board exams and competitive exams like JEE and NEET.
NCERT Exemplar Class 12 Physics Solutions Chapter 15 – Communication Systems helps students understand how information is transmitted over distances using modern communication technology.
MCQ I
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) A is transmitted via space wave while B and C are transmitted via sky wave
(b) A is transmitted via ground wave, B via sky wave and C via space wave
(c) B and C are transmitted via ground wave while A is transmitted via sky wave
(d) B is transmitted via ground wave while A and C are transmitted via space wave
Answer:
The correct option is (b) A is transmitted via ground wave, B via skywave and C via space wave. Because the ground wave propagation is only suitable for up to 2MHz frequencies, and the space wave propagation is for the frequencies more than 40 MHz, whereas the skywave propagation is used for frequencies ranging from 30 to 40 MHz.Question:2
100m long antenna is mounted on a 500m tall building. The complex can become a transmission tower for waves with λ
(a) ~ 400 m
(b) ~ 25 m
(c) ~ 150 m
(d) ~ 2400 m
Answer:
It can easily be calculated asQuestion:3
A1 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(p0/pi)]
(a) 900 W (b) 100 W (c) 990 W (d) 1010 W
Answer:
To calculate the power of the signal transmitted,
The rate of attenuation of the signal $= -2(dB/km)$
The given total length of the path$= 5 km$
The formula used $\Rightarrow dB = 10\log(P_o/P_i)$
After putting all he values,
$P_o = 100W$
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
(a) 1.003 MHz and 0.997 MHz
(b) 3001 kHz and 2997 kHz
(c) 1003 kHz and 1000 kHz
(d) 1 MHz and 0.997 MHz
Answer:
The frequencies of the sidebands would be (a) 1.003 MHz and 0.997 MHz
Question:5
(a) $\omega _m$
(b) $\omega _c$
(c) $\frac{\omega _c +\omega _m}{2}$
(d) $\frac{\omega _c -\omega _m}{2}$
Answer:
Since the frequencies of both the carrier wave and the amplitude-modulated wave are similar, the frequency of the AM wave would be (b) ωcQuestion:6
I-V characteristics of four devices are shown in the figure.
Identify devices that can be used for modulation:
(a) ‘i’ and ‘iii’
(b) only ‘iii’
(c) ‘ii’ and some regions of ‘iv’
(d) All the devices can be used.
Answer:
Question:7
A male voice after modulation-transmission sounds like that of a female to the receiver. The problem is due to
(a) poor selection of modulation index (selected 0 < m < 1)
(b) poor bandwidth selection of amplifiers
(c) poor selection of carrier frequency
(d) loss of energy in transmission
Answer:
When the selection of bandwidth for transmission is improper, the frequency of the modulated signal becomes more though the male voice frequency is less than female; thus, the voice sounds feminine. Therefore, the answer is (b) poor bandwidth selection of amplifiersQuestion:8
A basic communication system consists of
(A) transmitter
(B) information source
(C) user of information
(D) channel
(E) receiver
Choose the correct sequence in which these are arranged in a basic communication system:
(a) ABCDE
(b) BADEC
(c) BDACE
(d) BEADC
Answer:
As clearly stated by the order explained in the given below chain –Question:9
Identify the mathematical expression for amplitude modulated wave:
$(a) Ac \sin\left [ \left \{ \omega_c +1k_{1Vm}\left ( t \right ) \right \} t+\phi \right ]$
(b) $A c \sin \omega_c t+\phi+k_{2 V m}(t)$
$(c) \left \{ Ac +k_{2}V_m(t) \right \}\sin\left ( \omega _ct+\phi \right )$
$(d )A_{c\: Vm}(t)\sin(\omega _ct+\phi )$
Answer:
The correct mathematical expression for an amplitude modulated wave is
$(c) \left \{ Ac +k_{2}V_m(t) \right \}\sin\left ( \omega _ct+\phi \right )$
Question:10
An audio signal of 15 kHz frequency cannot be transmitted over long distances without modulation, because
(a) the size of the required antenna would be at least 5 km which is not convenient
(b) the audio signal cannot be transmitted through sky waves
(c) the size of the required antenna would be at least 20 km, which is not convenient
(d) effective power transmitted would be very low, if the size of the antenna is less than 5 km
Answer:
The correct answer is the options (a, b, d)Question:11
Audio sine waves of 3 kHz frequency are used to amplitude modulate a carrier signal of 1.5 MHz Which of the following statements are true?
(a) The sideband frequencies are 1506 kHz and 1494 kHz
(b) The bandwidth required for amplitude modulation is 6kHz
(c) The bandwidth required for amplitude modulation is 3 MHz
(d) The sideband frequencies are 1503 kHz and 1497 kHz.
Answer:
The types of frequencies under the AM wave are of three types, and the sideband frequencies in AM waves are close to the carrier waves. So according to the given criteria,Question:12
A TV transmission tower has a height of 240 m. Signals broadcast from this tower will be received by LOS communication at a distance of (assume the radius of the earth to be 6.4 × 106 m)
(a) 100 km
(b) 24 km
(c) 55 km
(d) 50 km
Answer:
To calculate the maximum signal transmission from any TV towers, the formula is –Question:13
The frequency response curve in the figure for the filter circuit used for production of AM wave should be
(a) (i) followed by(ii)
(b) (ii) followed by (i)
(c) (iii)
(d) (iv)
Answer:
Question:14
In amplitude modulation, the modulation index m is kept less than or equal to 1 because
(a) m > 1, will result in interference between the carrier frequency and message frequency, resulting in distortion
(b) m > 1 will result in overlapping of both sidebands resulting in loss of information
(c) m > 1 will result in a change in phase between the carrier signal and the message signal
(d) m > 1 indicates the amplitude of message signal greater than the amplitude of the carrier signal resulting in distortion
Answer:
Due to the occurrence of overmodulation $u>1$, this leads to higher chances of sideband overlapping and fading (loss of information). If the value of u is approximately equal to 1, it results in a distortion f carrier wave.Question:15
Which of the following would produce analogue signals and which would produce digital signals?
(i) A vibrating tuning forks
(ii) Musical sound due to a vibrating sitar string
(iii) Light pulse
(iv) Output of NAND gate
Answer:
Analogue signals will be produced by option i) and ii), i.e. a vibrating tuning fork and musical sound due to a vibrating sitar string.
Whereas option iii) and iv) will produce digital signals which are light pulse and output of NAND gate.
Question:16
Would sky waves be suitable for transmission of TV signals of 60 MHz frequency?
Answer:
As the frequency of TV signals is beyond 60 MHz, the skywaves are not reliable for transmission of 60 MHz frequency TV signals. The waves preferred for this type of transmission are space waves.Question:17
Answer:
Frequency and refractive index are directly proportional; thus, due to wave B having a higher frequency than wave A, the refractive index of wave B is higher than the refractive index of wave A.Question:18
Answer:
The change in the ratio of the amplitude of the carrier wave to the amplitude of the original carrier wave is defined as the modulation index.The given minimum amplitude of the AM wave is$= Amin = Ac - Am = 3V$
Thus, the modulation index, m, would be$= Am/Ac = 2/3$
Question:19
Answer:
The formula for the frequency of the tuned amplifier is,Question:20
Why is an AM signal likely to be noisier than an FM signal upon transmission through a channel?
Answer:
AM signal has the characteristic feature of being noisy when broadcasted on channels as compared to the FM signal because its carrier waves’ instantaneous voltage varies by the modulating wave voltage in the AM.
Question:21
Answer:
The loss of signal during the transmission $= 2 \: \: dB/km$
The total path covered by the signal $= 5 \: Km$
Collective loss suffered $= -10\: dB$
Total amplifier gain $= 30 \: dB$
Total gain in signal $= 20 \: dB$
Gain in dB $= 2$
Input power $= 1.01 mW$
Thus, output power would be,
$P_o = 1.01 \times 100$
Hence, the output power is $101\: mW.$
Question:22
Answer:
As given in the question,Question:23
Answer:
In order to connect earth with LOS using space waves,Question:24
Answer:
The values given in the question are$F_{1}\: max = 5 MHz\\F_{2} \: max = 8 MHz$
The formula used is Maximum frequency$=> f_{max }= 9(Nmax)^{1/2}$
So, the Nmax | $F_{1 }= 3.086 \times10^{11} m^{3}$
Again, after value substitution,
$N_{max} | F_{2} = 7.9 \times 10^{^{11}}/m^{3 }$
Question:25
Answer:
The overall power radiated because of energy carried by $\omega c,\left ( \omega c-\omega m \right )\: and\:\left ( \omega c+\omega m \right )$
Since the information is only transmitted by the amplitude modulated signals that are $\left ( \omega c-\omega m \right )\: and\:\left ( \omega c+\omega m \right )$ as these sideband frequencies are mainly closer to the carrier frequency, the cost can easily be minimised by transmitting both these signals.
Question:26
The intensity of a light pulse travelling along a communication channel decreases exponentially with distance x according to the relation $I=I_oe^{-ex}$, where I0 is the intensity at $x = 0 \: and\: \alpha$ is the attenuation constant.
(a) Show that the intensity reduces by 75% after a distance of $( In \: 4/\alpha ).$
(b) Attenuation of a signal can be expressed in decibel (dB) according to the relation $dB=10 \log_{_{10}} (I/I_0)$.What is the attenuation in dB/km for an optical fibre in which the intensity falls by 50% over a distance of 50 km?
Answer:
The given components in the question,Question:27
Answer:
The values given in the question are,
Velocity of waves $= 3 \times 10^{8} m/s$
Time to reach a receiver $= 4.04 \times 10^{-3} s$
Height of satellite, $h = 600\: km$
Radius of earth $= 6400\: km$
Size of transmitting antenna$= hT$
By using the formula = distance travelled by wave/time = velocity
The final velocity of waves $= 606\: km$
After using the Pythagoras theorem,
$d = 85.06\: km$
Thus, the distance between source and receiver$= 2d = 170\: km$
So, the maximum distance from the transmitter to the ground would by the EM waves would be
$= h_t = 565\: m$
Question:28
An amplitude modulated wave is as shown in the figure. Calculate
(i) the percentage modulation
(ii) peak carrier voltage and
(iii) peak value of information voltage
Answer:
As shown in the figure,
Maximum voltage, $Vmax = 100/2 = 50 V$
Minimum voltage, $Vmin = 20/2 = 10 V$
i) Percentage modulation $= (Vmax - Vmin)/(Vmax + Vmin) \times 100$
After substituting the values of Vmax and Vmin, the percentage modulation = 66.67%
ii) the formula of peak carrier voltage is $Vc =( Vmax + Vmin)/2$
Thus,$Vc = 30 V$
iii) Peak value of information voltage $= Vm = uVc = 20V$
Question:29
(i) Draw the plot of amplitude versus ω for an amplitude modulated wave whose carrier wave is carrying two modulating signals,
(ii) Is the plot symmetrical about $\omega _c$? Comment especially about plot in region $\left ( \omega <\omega _c \right )$
(iii) Extrapolate and predict the problems one can expect if more waves are to be modulated.
(iv) Suggest solutions to the above problem. In the process can one understand another advantage of modulation in terms of bandwidth?
Answer:
Question:30
An audio signal is modulated by a carrier wave of 20MHz such that the bandwidth required for modulation is 3kHz. Could this wave be demodulated by a diode detector which has the values of R and C as
$\\(i) R = 1 k\Omega , C = 0.01\mu F\\ (ii) R = 10 k\Omega, C = 0.01\mu F\\ (iii) R = 10 k\Omega, C = 0.1\mu F$
Answer:
Carrier wave frequency, $f_c =20 MHz = 20 \times 10^{6} Hz$
Bandwidth $= 2fm = 3 \times 10^3 Hz$
fm $= 1.5 \times 10^3 Hz$
$\\ \frac{1}{f_c} = 0.5 \times 10^{-7}\\ \\ \frac{1}{f_m} = 0.7 \times 10^{-3}\\ i)\frac{1}{f_c}<<RC< \frac{1}{f_m}$
Therefore, it can be demodulated
$ii)\frac{1}{f_c}<<RC<< \frac{1}{f_m}$
Therefore, it can be demodulated
$iii)\frac{1}{f_c}<RC$
Therefore, it cannot be demodulated
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A communication system helps us share ideas or important information with others effectively and easily.
NCERT Exemplar Solutions for Class 12 Physics Chapter 15 takes you through the basics of communication – from simple voice messages between people to sending files and photos over the internet. You’ll see how communication is used in things like mobile phones, computers, telephones, and more.
This chapter also explains different types of communication systems, how signals are sent and changed, and key terms like bandwidth, attenuation, modulation, and receivers in a way that’s easy to understand.
Also, read NCERT Solutions subject-wise
Must read NCERT Notes subject wise
Also, check the NCERT Books and NCERT Syllabus here:
Students can understand the topic and also use the solutions as a way to refer to their own answers while practicing before the exam.
Our guides have solved all 30 questions from the NCERT exemplar Class 12 Physics chapter 15 exercise
One can easily go to the solutions page, click on the link and the solutions will be downloaded in PDF format.
This chapter is one of the base chapters for those who want to become electrical and communication engineers.
Modulation is the process of altering a signal to make it suitable for transmission over long distances. It helps reduce signal loss and improves clarity.
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