NCERT Exemplar Class 12 Physics Solutions Chapter 15 Communication System

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:

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:

The correct answer is the option (a)

Question: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(P_o/P_i)]
(a) 900 W (b) 100 W (c) 990 W (d) 1010 W

Answer:

The correct answer is the option (b)

Given, power of signal transmitted is given $P_i=1 \mathrm{~kW}=1000 \mathrm{~W}$
Rate of attenuation of signal $=-2 \mathrm{~dB} / \mathrm{km}$
Length of total path $=5 \mathrm{~km}$
Thus,

$
\text { gain in } d B=5 \times(-2)=-10 d B
$

Also,

$
\text { gain in } d B=10 \log \left(\frac{P_0}{P_i}\right)
$

Here $P_0$ is the power of the received signal.
Putting the given values in Eq. (i),

$
\begin{array}{ll}
& -10=10 \log \left(\frac{P_0}{P_i}\right)=-10 \log \left(\frac{P_i}{P_0}\right) \\
\Rightarrow & \log \frac{P_i}{P_0}=1 \Rightarrow \log \frac{P_i}{P_0}=\log 10 \\
\Rightarrow & \frac{P_i}{P_0}=10 \Rightarrow 1000 \mathrm{~W}=10 P_0 \\
\Rightarrow & P_0=100 \mathrm{~W}
\end{array}
$

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 correct answer is the option (a) 1.003 MHz and 0.997 MHz

$
\begin{aligned}
& \text { Given, frequency of carrier signal is } \omega_c=1 \mathrm{MHz} \\
& \quad \begin{aligned}
\text { and } \text { frequency of speech signal } & =3 \mathrm{kHz} \\
& =3 \times 10^{-3} \mathrm{MHz} \\
& =0.003 \mathrm{MHz}
\end{aligned}
\end{aligned}
$

Now, we know that,

$
\begin{aligned}
\text { Frequencies of side bands } & =\left(\omega_c \pm \omega_m\right) \\
& =(1 \pm 0.003) \\
& =1.003 \mathrm{MHz} \text { and } 0.997 \mathrm{MHz}
\end{aligned}
$

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

(a) $\omega _m$

(b) $\omega _c$

(c) $\frac{\omega _c +\omega _m}{2}$

(d) $\frac{\omega _c -\omega _m}{2}$

Answer:

The correct answer is the option (c)

Question: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:

The correct answer is the option (c)

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:

The correct answer is the option (b)

Question: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:

The correct answer is the option (b) BADEC

Question:9

Identify the mathematical expression for amplitude modulated wave:

(a) $\quad A_c \sin \left[\left\{\omega_c+k_1 v_m(t)\right\} t+\phi\right]$.
(b) $A_c \sin \left\{\omega_c t+\phi+k_2 v_m(\mathrm{t})\right\}$.
(c) $\quad\left\{A_c+k_2 v_m(\mathrm{t})\right\} \sin \left(\omega_c t+\phi\right)$.
(d) $A_c v_m(\mathrm{t}) \sin \left(\omega_c t+\phi\right)$.

Answer:

The correct answer is the option (c)

Consider a sinusoidal modulating signal represented by

$
m(t)=A_m \sin \omega_m t ------(I)
$

where, $A_m=$ Amplitude of modulating signal $\omega_m=$ Angular frequency $=2 \pi \nu_m=\phi \nu_m$
Also consider a sinusoidal carrier wave represented by $C(t)=A_c \sin \omega_c t -------(ii)$
Thus, modulated wave is given by

$
\begin{aligned}
C_m(t) & =\left(A_c+A_m \sin \omega_m t\right) \sin \omega_c t \\
& =A_c\left[1+\frac{A_m}{A_c} \sin \omega_m t\right] \sin \omega_c t
\end{aligned}
$

$
\begin{array}{ll}
\text { Here, } \frac{A_m}{A_c}=M \\
\Rightarrow C_m(t)=\left(A_c+A_c \times \mu \sin \omega_m t\right) \sin \omega_c t
\end{array}
$

Now, we know that $A_c \times \mu=K$ [wave constant]
and $\quad \sin \omega_m t=v_m \quad$ [wave velocity]
Thus, Eq. (iii) becomes

$
C_m(t)=\left(A_c+K \times v_m\right) \sin \omega_c t
$

Now, consider a change in phase angle by $\phi$ then $\sin \omega_c t \rightarrow \sin \left(\omega_c t+\phi\right)$
Thus,

$
C_m(t)=\left(A_c+K v_m\right)\left(\sin \omega_c+\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:

Given, frequency of the wave to be transmitted is

$
\begin{aligned}
v_m & =15 \mathrm{kHz}=15 \times 10^3 \mathrm{~Hz} \\
\text { Wavelength } \lambda_m & =\frac{c}{v_m}=\frac{3 \times 10^8}{15 \times 10^3}=\frac{1}{5} \times 10^5 \mathrm{~m}
\end{aligned}
$

Size of the antenna required,

$
\begin{aligned}
l & =\frac{\lambda}{4}=\frac{1}{4} \times\left(\frac{1}{5} \times 10^5\right) \\
& =5 \times 10^3 \mathrm{~m}=5 \mathrm{~km}
\end{aligned}
$

The audio signals are of low frequency waves. Thus, they cannot be transmitted through sky waves as they are absorbed by atmosphere.
If the size of the antenna is less than 5 km , the effective power transmission would be very low because of deviation from resonance wavelength of wave and antenna length.

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 correct answers are the options (b) and (d)

Given,

$
\begin{aligned}
\omega_m & =3 \mathrm{kHz} \\
\omega_c & =1.5 \mathrm{MHz}=1500 \mathrm{kHz}
\end{aligned}
$
Now, side band frequencies

$
\begin{aligned}
\omega_c \pm \omega_m & =(1500 \pm 3) \\
& =1503 \mathrm{kHz} \text { and } 1497 \mathrm{kHz}
\end{aligned}
$
Also, bandwidth $=2 \omega_m=2 \times 3=6 \mathrm{kHz}$

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:

The correct answers are the options (b), (c) and (d).

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:

The correct answers are the options (b) and (d)

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:

Analog and digital signals are used to transmit information, usually through electric signals. In both these technologies, the information such as any audio or video is transformed into electric signals. The difference between analog and digital technologies is that in analog technology, information is translated into electric pulses of varying amplitude. In digital technology, translation of information is into binary formal (zero or one) where each bit is representative of two distinct amplitudes. Thus, (a) and (b) would produce analog signal and (c) and (d) would produce digital signals.

Question:16

Would sky waves be suitable for transmission of TV signals of 60 MHz frequency?

Answer:

Question:17

Two waves A and B of frequencies 2 MHz and 3 MHz respectively are beamed in the same direction for communication via sky wave. Which one of these is likely to travel a longer distance in the ionosphere before suffering total internal reflection?

Answer:

Question:18

The maximum amplitude of an A.M. wave is found to be 15 V while its minimum amplitude is found to be 3 V. What is the modulation index?

Answer:

The given minimum amplitude of the AM wave is$= A_min = A_c - A_m = 3V$
Thus, the modulation index, m, would be$= A_m/A_c = 2/3$

Question:19

Compute the LC product of a tuned amplifier circuit required to generate a carrier wave of 1 MHz for amplitude modulation.

Answer:

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.

Short Answer

Question:21

Figure shows a communication system. What is the output power when the input signal is of 1.01mW? (gain in d_B = 10 log₁₀ (P_o /P_i ).

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

A TV transmission tower antenna is at the height of 20 m. How much service area can it cover if the receiving antenna is (i) at ground level, (ii) at the height of 25 m? Calculate the percentage increase in area covered in case (ii) relative to the case (i).

Answer:

Question:23

If the whole earth is to be connected by LOS communication using space waves (no restriction of antenna size or tower height), what is the minimum number of antennas required? Calculate the tower height of these antennas in terms of earths radius?

Answer:

Question:24

The maximum frequency for reflection of sky waves from a certain layer of the ionosphere is found to be f_max= 9(N_max)^1/2, where N_max is the maximum electron density at that layer of the ionosphere.
On a certain day it is observed that signals of frequencies higher than 5 MHz are not received by reflection from the F₁ layer of the ionosphere while signals of frequencies higher than 8 MHz are not received by reflection from the F₂ layer of the ionosphere. Estimate the maximum electron densities of the F₁ and F₂ layers on that day.

Answer:

$F_{1}\: max = 5 MHz\\F_{2} \: max = 8 MHz$
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

On radiating (sending out) and AM modulated signal, the total radiated power is due to the energy carried by $\omega c,\left ( \omega c-\omega m \right )\: and\:\left ( \omega c+\omega m \right )$. Suggest ways to minimise the cost of radiation without compromising on the information.

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:27

A 50 MHz sky wave takes 4.04 ms to reach a receiver via re-transmission from a satellite 600 km above earth’s surface. Assuming re-transmission time by satellite negligible, find the distance between source and receiver. If communication between the two was to be done by Line of Sight (LOS) method, what should size and placement of receiving and transmitting antenna be?

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