Important Questions for CBSE Class 12 Physics Chapter 15 Communication Systems 2024-25

1 Mark Questions

Refer to Page 1 for 1 Mark Questions in the PDF:

1. Name the type of the communication system in which the signal is a discrete and binary coded version of message or information?

Ans: Digital communication is that type of the communication system in which the signal is discrete and has binary coded version of message or information.

2. What is the purpose of modulating a signal in transmission?

Ans: Since low frequency signals can’t be transmitted to long distances, modulation is done to increase the transmission range.

3. What is the requirement of transmitting microwaves form one position to another on the earth?

Ans: The transmitting and receiving antenna should be in the same line of sight.

4. A T.V. tower has a height of 300m. What is the maximum distance up to which the T.V. transmission can be received?

Ans: In the question, it is given that the height of the tower is 300m. It is asked to find the maximum distance of transmission. 

So, we will use the formula, \[d=\sqrt{2Rh}\]

where \[d\] is the distance of transmission, \[R\] is the radius of the Earth which is 6400km and \[h\] is the height of the tower.

Substitute \[h=300\]and \[R=6400\times 1000\] in the formula \[d=\sqrt{2Rh}\] and simplify.

$d=\sqrt{2\times 6400\times 1000\times 300}$ 

$=61.96\times 1000$

$\approx 62km$  

The maximum distance upto which TV transmission can be received is 62km.

5. Why ground wave propagation is not suitable for high frequencies?

Ans: Signals that have frequencies greater than 1500 KHz are mostly absorbed by the surface of the Earth and hence, cannot be transmitted. Because of this reason ground wave propagation is not suitable for higher frequencies.

6. What type of modulation is used for commercial broadcast of voice signal?

Ans: For commercial broadcast of voice signals, amplitude modulation is used.

2 Marks Questions

Refer to Page 2-6 for 2 Marks Questions in the PDF:

1. A signal jumps from one level to another instantaneously. What will be its frequency?

Ans: If the signal jumps from one level to another instantly it means that its frequency is infinite.

2. Sky has no limit but sky wave propagation has its limit. Explain why?

Ans: Sky wave propagation has its limit because Sky wave propagation happens because of reflection of radio waves by the ionosphere but the ionosphere absorbs high frequency waves and does not reflect them.

3. A transmitting antenna has a height of 50m. If radius of the earth is taken as 6250 km. Find the area covered by it?

Ans: In the question, it is given that the height of the antenna is 50m. It is asked to find the area covered by it. 

So, we will use the formula, \[d=\sqrt{2Rh}\]

where \[d\] is the distance of transmission, \[R\] is the radius of the Earth which is 6250km and \[h\] is the height of the antenna.

Substitute \[h=50\]and \[R=6250\times 1000\] in the formula \[d=\sqrt{2Rh}\] and simplify.

Substitute the value of \[d\] obtained in the formula of area of a circle, \[\pi {{d}^{2}}\].
\[d=\sqrt{2rh}\]

\[d=\sqrt{2\times 6250\times 50\times {{10}^{3}}}\] 

\[d=2.5\times {{10}^{4}}m\]

\[\text{Area covered}=\pi {{d}^{2}}\]

\[=3.14\times {{\left( 2.5\times {{10}^{4}} \right)}^{2}}\]

\[=1963k{{m}^{2}}\]

4. What is the role of layer in communication? 

Ans: The ionosphere's top layer is called F2. It is a reflecting layer for high frequency radio waves with a height of up to 400 kilometres.

5. A carrier wave of peak voltage 12V is used to transmit a message signal. What should be the peak voltage of the modulating signal in order to have a modulation index of $75%$?

Ans: In the question, we will use the formula $\mu =\frac{{{A}_{m}}}{{{A}_{c}}}$ where $\mu =\frac{75}{100}$ and ${{A}_{c}}=12$. Substitute the values in the formula and simplify to find the value of ${{A}_{m}}$.

$\mu =\frac{{{A}_{m}}}{{{A}_{c}}}$ 

${{A}_{m}}=\mu \times {{A}_{c}}$ 

${{A}_{m}}=\frac{75}{100}\times 12 $

${{A}_{m}}=9V $

6. Give the set up of a basic communication system? 

Ans: An information source, a transmitter, a link, and a receiver make up a basic communication system. In a basic communication system, the transmitter is generally placed in a single place, the receiver is positioned at some other place to differentiate them from the transmitter and the channel is the unique physical medium that establishes the connections between them.

(Image Will Be Updated Soon)

7. Distinguish between analog and digital communication?

Ans:

8. Which of the following frequencies will be suitable for beyond-the-horizon communication using sky waves? 

a) 10 kHz 

b) 10 MHz 

c) 1 GHz 

d) 1000 GHz 

Ans: (b) 10 MHz

The signal waves must travel a long distance in order to communicate beyond the horizon. Because of the antenna size, 10 kHz transmissions cannot be broadcast effectively. The ionosphere is penetrated by high-energy signal waves (1GHz - 1000GHz). 10 MHz frequencies are easily reflected by the ionosphere. As a result, signal waves of such frequencies are acceptable for communication beyond the horizon.

9. Frequencies in the UHF range normally propagate by means of:

(a) Ground waves. 

(b) Sky waves. 

(c) Surface waves.

(d) Space waves.

Ans: (d) Space waves

An ultra high frequency (UHF)wave cannot travel down the ground's trajectory or be reflected by the ionosphere due to its high frequency. Line-of-sight communication, which is nothing more than space wave propagation, is used to transmit UHF signals.

10. Digital signals

i) Do not provide a continuous set of values,

ii) Represent values as discrete steps, 

iii) Can utilize binary system, and 

iv) Can utilize decimal as well as binary systems.

Which of the above statements are true? 

(a) (i) and (ii) only 

(b) (ii) and (iii) only 

(c) (i), (ii) and (iii) but not (iv) 

(d) All of (i), (ii), (iii) and (iv). 

Ans: (c) A digital signal uses the binary (0 and 1) scheme to send message signals. The decimal system cannot be trusted within this type of system (which corresponds to analogue signals). Discrete values make up digital signals.

11. Is it necessary for a transmitting antenna to be at the same heights that of the receiving antenna for line-of-sight communication? A TV transmitting antenna is 81m tall. How much service area can it cover if the receiving antenna is at the ground level? 

Ans: There is no physical impediment between the transmitter and the receiver in line-of-sight communication. In such a communication, it is not essential for the transmitting and receiving antennas to be at the same height.

The antenna's height is specified as,$h=81m$

Radius of earth,$R=6.4\times {{10}^{6}}m$

For range,$d=\left( 2Rh \right)\frac{1}{2}$, the relation that provides the service area of the antenna is given by:

$A=n{{d}^{2}}$

$ =n\left( 2Rh \right)$

$=3.14\times 2\times 6.4\times {{10}^{6}}\times 81 $

$=3255.55\times {{10}^{6}}{{m}^{2}}$

$=3255.55$

$\sim 3256{{m}^{2}} $

12. A carrier wave of peak voltage 12 V is used to transmit a message signal. What should be the peak voltage of the modulating signal in order to have a modulation index of 75%? 

Ans:  Amplitude of the carrier wave, ${{A}_{c}}=12V$

Modulation index, 

$m=75% $ 

$ =0.75$  

Amplitude of the modulating wave = ${{A}_{m}}$

Using the relation for modulation index: $m=\frac{{{A}_{m}}}{{{A}_{c}}}$

${{A}_{m}}=m{{A}_{c}} $

$ =0.75\times 12$ 

$ =9V $

13. For an amplitude modulated wave, the maximum amplitude is found to be 10 V while the minimum amplitude is found to be 2 V. Determine the modulation index. What would be the value of if the minimum amplitude is zero volt?

Ans: Maximum amplitude, ${{A}_{\max }}=10V$

Minimum amplitude, ${{A}_{\min }}=2V$

Modulation index$\mu $, is given by the relation: 

$\mu =\frac{{{A}_{\max }}-{{A}_{\min }}}{{{A}_{\max }}+{{A}_{\min }}}$

Substitute ${{A}_{\max }}=10V$ and ${{A}_{\min }}=2V$ in the formula and simplify.

$\mu =\frac{{{A}_{\max }}-{{A}_{\min }}}{{{A}_{\max }}+{{A}_{\min }}}$

$=\frac{10-2}{10+2}$

$=\frac{8}{12}$

$=0.67$

If ${{A}_{\min }}=0$

Then,

$\mu =\frac{{{A}_{\max }}}{{{A}_{\max }}}$

$=\frac{10}{10} $

$=1$

3 Mark Questions

1. Define the term modulation index for A.M. wave. What would be the modulation index for an A.M. wave for which the maximum amplitude is ‘a’ write the minimum amplitude is ‘b’?

Ans: The ratio of the amplitude of the carried wave to the amplitude of the carries (original) wave is known as the modulation index.

i.e. $\mu =\frac{{{E}_{m}}}{{{E}_{c}}}$

Here, Maximum Amplitude $a={{E}_{c}}+{{E}_{m}}$

Minimum Amplitude $b={{E}_{c}}-{{E}_{m}}$

$\Rightarrow {{E}_{c}}=\frac{a+b}{2}$ and

$\Rightarrow {{E}_{m}}=\frac{a-b}{2}$

$\Rightarrow \mu =\frac{a-b}{a+b}$

2. A T.V. tower has a height of 80m. By how much the height of tower be increased to triple its coverage?

Ans: Here, ${{h}_{1}}=80m$. Use the formula ${{d}_{1}}=\sqrt{2{{h}_{1}}R}$ to find the value of ${{d}_{1}}$.

${{d}_{1}}=\sqrt{2{{h}_{1}}R} $ 

$=\sqrt{2\times 80\times R} $

$=\sqrt{160R}$  

If the coverage is tripled then ${{d}_{1}}$  changes to $3{{d}_{1}}$.

Hence, $\sqrt{2{{h}_{2}}R}=3{{d}_{_{1}}}$.

Solving further we get,

$\sqrt{2{{h}_{2}}R}=3\sqrt{160R}$

${{h}_{2}}=720m$

3. An audio signal of amplitude one half of the carries amplitude is employed in amplitude modulation. What is the modulation index? Hence define amplitude modulation? 

Ans: It is given that ${{E}_{m}}=0.5{{E}_{c}}$.

Hence,

${{E}_{\max }}={{E}_{m}}+0.5{{E}_{c}}$

$=1.5{{E}_{c}} $ 

${{E}_{\min }}={{E}_{m}}-0.5{{E}_{c}} $

$=0.5{{E}_{c}}$

Substitute ${{E}_{\max }}=1.5{{E}_{c}}$ and ${{E}_{\min }}=0.5{{E}_{c}}$ in the formula $\mu =\frac{{{E}_{\max }}-{{E}_{\min }}}{{{E}_{\max }}+{{E}_{\min }}}$.  

$\mu =\frac{{{E}_{\max }}-{{E}_{\min }}}{{{E}_{\max }}+{{E}_{\min }}}$ 

$=\frac{1.5{{E}_{c}}-0.5{{E}_{c}}}{1.5{{E}_{c}}+0.5{{E}_{c}}}$

$=0.5$

4. An audio signal of 32kHz modulates a carrier of frequency 84MHz and produces a frequency deviation of 96kHz. 

Find, 

(a) frequency modulation index 

(b) frequency range of the frequency modulated wave?

Ans: We have,

${f_m}=3.2KHz$ 

${f_c}=34MHz$

$\delta =96KHz$

a) Frequency modulated index:

${m_f}=\frac{\delta }{f_m}$ 

$=\frac{96}{3.2} $

$ =30 $

b) Frequency range of the modulated wave:

$fc\pm fm=84\times {{10}^{3}}\pm 3.2KHz$ 

$=83.997MHz\text{ to }84.003MHz$

5. Define the following terms 

(a) Ground wave propagation 

(b) Space wave propagation 

(c) Sky wave propagation

Ans: (a) Ground wave propagation –Ground waves are radio waves that travel along the surface of the earth and propagate along the surface of the earth. It can only operate at frequencies below $1.5$ MHz.

(b) Space wave propagation –Sky waves are radio waves that are reflected back to the earth by the ionosphere, and sky wave propagation is the method through which they travel.

(c) Space wave propagation –The kind of communication known as line of sight communication uses high frequency waves that cannot be reflected back to the earth via sending antenna to receiving antenna. It's also known as propagation of space waves.

6. Which two communication methods make use of space wave propagation method? If the sum of the heights of transmitting and receiving antenna is line of sight communication is fixed at h, show that the range is maximum when the two antenna have a height h/2 each? 

Ans: Satellite communication and line of sight (LOS) communication make use of space waves. 

Now

${{d}_{1}}=\sqrt{2R{{h}_{1}}}$

${{d}_{2}}=\sqrt{2R{{h}_{2}}}$

For maximum range

${{d}_{m}}=\sqrt{2R{{h}_{1}}}+\sqrt{2R{{h}_{2}}}$ 

${{d}_{m}}={{d}_{1}}+{{d}_{2}} $

$ =d$

Given ${{h}_{1}}+{{h}_{2}}=h$

Let ${{h}_{1}}=x$ then ${{h}_{2}}=h-x$

${{d}_{m}}=\sqrt{2Rx}+\sqrt{2R\left( h-x \right)}$

Differentiating w.r.t.$x$.

$\frac{d{{d}_{m}}}{{{d}_{x}}}-\sqrt{\frac{R}{2x}}=\sqrt{\frac{R}{2\left( h-x \right)}}=0$ i.e., $\frac{1}{2x}=\frac{1}{2\left( h-x \right)}$

$\Rightarrow x=\frac{h}{2}$

$\Rightarrow {{h}_{1}}={{h}_{2}}$

$=\frac{h}{2}$

7. A frequency modulated wave is represented by an equation. 

Find (1) carrier frequency 

(2) modulating signal frequency 

(3) Power dissipated if load resistor is of 100$\Omega $? 

Ans: Given $e=10\sin \left( 5\times {{10}^{3}}++6\sin 1000t \right)$

Compare it with general equation 

$e=E\sin \left( {{w}_{c}}t+{{m}_{f}}\sin {{w}_{m}}t \right)$

Carries frequency 

${{v}_{c}}=\frac{{{w}_{c}}}{2\pi }=\frac{5\times {{10}^{3}}}{2\times 3.14}$ 

${{v}_{c}}=79.62MHz$  

Modulating signal frequency​ ${{v}_{m}}=\frac{{{W}_{m}}}{2\pi }=\frac{100}{2\times 3.14}=15.92Hz$

Power dissipated $P=\frac{{{\left( {{E}_{rms}} \right)}^{2}}}{R}=\frac{{{\left( \frac{10}{\sqrt{2}} \right)}^{2}}}{100}=0.5Watts$

8. A modulating signal is a square wave, as shown in Fig. 15.14.

(Image Will Be Updated Soon)

The carrier wave is given by $c\left( t \right)=2\sin \left( 8\Omega t \right)volts$

i) Sketch the amplitude modulated waveform 

ii) What is the modulation index? 

Ans: It can be seen through the given modulating signal that the amplitude of the modulating signal, ${{A}_{m}}=1V$

The carrier wave $c\left( t \right)=2\sin \left( 8nt \right)$ is given.

Amplitude of the carrier wave is as follows, ${{A}_{c}}=2V$

Time period of the modulating signal is as follows, ${{T}_{m}}=1s$

The modulating signal's angular frequency is computed as:

${{\left( t \right)}_{m}}=\frac{2\pi }{{{T}_{m}}}$

$=2\pi \text{ }rad\text{ }{{s}^{-1}}........................\left( i \right)$

The carrier signal's angular frequency is determined as follows:${{\left( t \right)}_{c}}=2\pi \text{ }rad\text{ }{{s}^{-1}}...................\left( ii \right)$

We can deduce the following from equations I and (ii):

${{\left( t \right)}_{c}}=4{{\left( t \right)}_{m}}$

The modulating signal's amplitude modulated waveform is displayed in the following figure.

ii) Modulation index, $m=\frac{{{A}_{m}}}{{{A}_{c}}}=\frac{1}{2}=0.5$

9. Due to economic reasons, only the upper sideband of an AM wave is transmitted, but at the receiving station, there is a facility for generating the carrier. Show that if a device is available which can multiply two signals, then it is possible to recover the modulating signal at the receiver station.

Ans: Let ${{\omega }_{c}}$and${{\omega }_{s}}$ be the carrier and signal waves' respective frequencies.

The signal received at the receiving station, $V={{V}_{1}}\cos \left( {{\omega }_{c}}+{{\omega }_{s}} \right)t$

The carrier wave's instantaneous voltage,${{V}_{in}}={{V}_{c}}\cos {{\omega }_{c}}t$

$\therefore V{{V}_{in}}={{V}_{1}}\cos \left( {{\omega }_{c}}+{{\omega }_{s}} \right)t\cdot \left( {{V}_{c}}\cos {{\omega }_{c}}t \right)$

${{V}_{1}}{{V}_{c}}=\left[ \cos \left( {{\omega }_{c}}+{{\omega }_{s}} \right)t\cdot \cos {{\omega }_{c}}t \right] $

$\frac{{{V}_{1}}{{V}_{c}}}{2}=\left[ 2\cos \left( {{\omega }_{c}}+{{\omega }_{s}} \right)t\cdot \cos {{\omega }_{c}}t \right]$

$\frac{{{V}_{1}}{{V}_{c}}}{2}=\left[ \cos \left\{ \left( {{\omega }_{c}}+{{\omega }_{s}} \right)t+\cos {{\omega }_{c}}t \right\}+\cos \left\{ \left( {{\omega }_{c}}+{{\omega }_{s}} \right)t-{{\omega }_{c}}t \right\} \right]$

$\frac{{{V}_{1}}{{V}_{c}}}{2}=\left[ \cos \left\{ \left( 2{{\omega }_{c}}+{{\omega }_{s}} \right)t+\cos {{\omega }_{s}}t \right\} \right]$

Only high frequency signals are allowed to pass through the low-pass filter at the receiving station. Low frequency signal ${{\omega }_{s}}$ is obstructed. The modulating signal, which is the signal frequency, can thus be recorded at the receiving station as $\frac{{{V}_{1}}{{V}_{c}}}{2}\cos {{\omega }_{s}}t$.

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