CBSE Class 12 Physics Principles Of Communication chapter notes and important questions

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PRINCIPLES OF COMMUNICATION

Principles of Communication – gist of the chapter and question bank with answers

GIST

1. Basic terminology used in communication

1. Signal: the electrical form of a message, suitable for transmission and reception.
2. Transducer: a device which converts one form of energy into another, e.g. a microphone, a loudspeaker, a photodetector.
3. Noise: unwanted signals that tend to disturb the transmission and processing of message signals in a communication system.
4. Attenuation: the loss of strength of a signal while it propagates through a medium.
5. Amplification: the process of increasing the amplitude of a signal using an electronic circuit.
6. Bandwidth: the range of frequencies over which equipment operates, or the portion of the spectrum occupied by the signal.
7. Transmitter: its function is to transmit the modulated signal.
8. Receiver: its function is to receive the transmitted signal.
9. Demodulator: it separates the baseband signal from the modulated wave.
10. Repeater: it receives the modulated wave, amplifies it and then retransmits it, so as to increase the range of communication.

2. Block diagram of a generalised communication system

3. Bandwidth of signals: speech signals 2800 Hz; music 20 kHz; video signals 4.2 MHz; TV signals 6 MHz.

4. Bandwidth of transmission media: coaxial cable 750 MHz (operated below 18 GHz); optical fibre 100 GHz; for free space refer to Table 15.2, p. 519 of the NCERT book.

5. Propagation of electromagnetic waves

1. Ground wave: the waves are of frequency range 500 kHz–1510 kHz and propagate along the surface of the earth. As the wave glides along the surface of the earth it induces charges on the surface and is attenuated as a result of absorption of energy by the earth. Drawbacks: (1) it cannot be transmitted over long distances due to attenuation; (2) high frequencies cannot be transmitted because of increased attenuation.
2. Sky wave: this mode of communication is used for the frequency range 30 to 40 MHz. Long-distance communication is achieved by ionospheric reflection of radio waves back towards the earth. Sky-wave communication cannot be used for frequencies above 30 MHz, because higher frequencies penetrate the ionosphere and escape.
3. Space wave: communication along a straight line from the transmitting antenna to the receiving antenna. This is also called line-of-sight (LOS) communication. Microwaves are used in this mode because of their directional property (see Fig. 15.5, p. 521 of the NCERT book).

The total coverage distance (range) between a transmitting antenna of height \(h_T\) and a receiving antenna of height \(h_R\) is

\[d=\sqrt{2Rh_T}+\sqrt{2Rh_R}\]

For a given total height of the transmitting and receiving antennas, the range is maximum when the heights of both antennas are equal.

6. Various modes of communication: refer to Fig. 15.6, p. 522 of the NCERT book.

7. Necessity of modulation

1. Size of the antenna: the size of the antenna should be \(\lambda\), or at least \(\dfrac\lambda4\). Before transmission \(\lambda\) should therefore be small, i.e. the frequency high, so there is a need for modulation.
2. Effective power radiated by an antenna: \(P\propto\left(\dfrac1\lambda\right)^2\). For effective radiation of power \(\lambda\) should be small, so the signal should be mixed with a high-frequency carrier wave.
3. Mixing up of signals: due to overlapping or intermixing of signals we would not be able to distinguish between them.

8. Amplitude modulation. If the amplitude of the carrier wave is changed in accordance with the amplitude of the baseband signal (modulating signal), keeping the frequency constant, the type of modulation is called amplitude modulation (see Fig. 15.8, p. 524 of the NCERT book).

The high-frequency carrier wave is represented by \(e_c=E_c\cos\omega_ct\) and the baseband signal by \(e_m=E_m\cos\omega_mt\). The AM wave is represented as

\[e=(E_c+E_m\cos\omega_mt)\cos\omega_ct=E_c(1+m_a\cos\omega_mt)\cos\omega_ct\]

Here \(\dfrac{E_m}{E_c}=m_a\) is the modulation index, defined as the ratio of the amplitude of the baseband signal to the amplitude of the carrier wave.

\[e=E_c\cos\omega_ct+m_aE_c\cos\omega_mt\cos\omega_ct=E_c\cos\omega_ct+\tfrac12m_aE_c(2\cos\omega_mt\cos\omega_ct)\]

\[e=E_c\cos\omega_ct+\tfrac12m_aE_c\cos(\omega_c+\omega_m)t+\tfrac12m_aE_c\cos(\omega_c-\omega_m)t\]

This is the equation of an AM wave. It consists of three terms:

1. the carrier wave of amplitude \(E_c\) and frequency \(\omega_c\);
2. the upper side band (USB) of amplitude \(\dfrac{m_aE_c}2\) and frequency \(\omega_c+\omega_m\);
3. the lower side band (LSB) of amplitude \(\dfrac{m_aE_c}2\) and frequency \(\omega_c-\omega_m\);
4. the frequency spectrum of the AM wave is shown below.

From the frequency spectrum we note that: (1) if the modulation is 100%, the amplitude of each side band is half of the carrier amplitude; (2) bandwidth \(=(\omega_c+\omega_m)-(\omega_c-\omega_m)=2\omega_m\), i.e. twice the frequency of the modulating signal.

Production of AM waves: see the block diagram, Fig. 15.10, p. 525 of the NCERT book. Band pass filter: the function of the band pass filter is to reject d.c. components, low frequencies and high frequencies. It allows to pass only those frequencies which are centred around \(\omega_c\), i.e. \((\omega_c+\omega_m)\) and \((\omega_c-\omega_m)\).

Block diagram of a transmitter: see Fig. 15.11, p. 526 of the NCERT book.

Frequency modulation: if the frequency of the carrier wave is changed in accordance with the baseband signal, keeping the amplitude constant, the type of modulation is called frequency modulation.

QUESTION BANK

What should be the length of a dipole antenna for a carrier wave of \(5\times10^8\) Hz?

Answer: \(L=0.3\) m.

For a dipole antenna \(L=\dfrac\lambda2\), and \(\lambda=\dfrac c\nu=\dfrac{3\times10^8}{5\times10^8}=0.6\) m, so \(L=\dfrac{0.6}{2}=0.3\) m.

A device X can convert one form of energy into another. Another device Y can be regarded as a combination of a transmitter and a receiver. Name the devices X and Y.

Answer: (a) X is a transducer. (b) Y is a repeater.

Name the type of communication that uses carrier waves having frequencies in the range \(10^{12}\) to \(10^{16}\) Hz.

Answer: Optical communication.

Write the expression for band width in FM.

Answer: Bandwidth \(=2f_m\), i.e. twice the frequency of the modulating signal. (The PDF gives this answer for the question as worded for FM; it is actually the result for AM, since in FM the bandwidth is larger, about \(2(\Delta f+f_m)\).)

What is attenuation?

Answer: Attenuation is the loss of strength of a signal while it propagates through a medium (channel).

What is the role of a band pass filter in a modulation circuit?

Answer: It filters out low and high frequencies and allows only the band of frequencies \((\omega_c-\omega_m)\) to \((\omega_c+\omega_m)\) to pass.

Distinguish between analog and digital communication.

Answer: In analog communication the information is carried by a signal whose amplitude (or frequency) varies continuously with time, e.g. a speech or music signal. In digital communication the information is carried by discrete signals, usually the two levels 0 and 1 (binary), e.g. computer data. Digital signals are less affected by noise and are easier to store and process.

State the facts by which the range of transmission of signals by a TV tower can be increased.

Answer: By increasing the height of the transmitting antenna and by increasing the height of the receiving antenna (range \(d=\sqrt{2Rh_T}+\sqrt{2Rh_R}\)).

What percentage of the AM wave power is carried by the side bands for \(m=1\)?

Answer: One-third (33.3%).

Total power \(P_t=P_c\left(1+\dfrac{m^2}2\right)\) and the side-band power is \(P_c\dfrac{m^2}2\). For \(m=1\), \(P_t=1.5P_c\) and the side bands carry \(\dfrac{0.5P_c}{1.5P_c}=\dfrac13\), i.e. 33.3% of the total power.

Why can the moon not be used as a communication satellite?

Answer: The moon is very far (about \(3.8\times10^5\) km), so the signals reaching it and coming back would be very weak and delayed; its orbital period is about 27 days, so it is not stationary with respect to a place on the earth, and it cannot be controlled or amplified like an artificial satellite.

Explain why medium waves are better carriers of signals than radio waves.

Answer: The source gives only the hint “unidirectional propagation”. The question appears to compare microwaves with ordinary radio waves: microwaves have a very high frequency, so a large bandwidth is available, and they travel in a straight line (unidirectional) in narrow beams that can be sent with small antennas. They are not reflected by or absorbed in the ionosphere and so can be used for line-of-sight and satellite links.

What is the requirement for transmitting microwaves from one place to another on the earth?

Answer: The transmitting and receiving antennas must be in line of sight.

Name the type of radio wave propagation involved when TV signals broadcast by a tall antenna are intercepted directly by the receiving antenna.

Answer: Space wave propagation (line-of-sight communication).

Why are sky waves not used for the transmission of TV signals?

Answer: TV signals have frequencies above 30–40 MHz (typically 54–890 MHz). Such high frequencies are not reflected back by the ionosphere; they penetrate it and escape, so sky-wave propagation cannot be used.

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

Answer: \(d\approx62\) km.

\(d=\sqrt{2Rh}=\sqrt{2\times6400\times10^3\times300}=\sqrt{3.84\times10^9}\approx6.2\times10^4\) m \(=62\) km.

How does the effective power radiated by an antenna vary with wavelength?

Answer: The effective power radiated varies inversely as the square of the wavelength, \(P\propto\dfrac1{\lambda^2}\).

Why is ground wave propagation not suitable for high frequencies? (OR) Why is ground wave propagation restricted to frequencies up to 1500 kHz?

Answer: Radio waves of frequency greater than about 1500 kHz are strongly absorbed by the ground, so they are attenuated within a short distance. (The source hint says “1500 MHz”; the limit is 1500 kHz.)

Why are signals not significantly absorbed by the ionosphere in satellite communication?

Answer: Satellite communication employs high-frequency carriers, i.e. microwaves, which pass through the ionosphere without being absorbed or reflected.

How many geostationary satellites are required to provide a communication link over the entire globe, and how should they be parked?

Answer: Three geostationary satellites are required, parked \(120^\circ\) apart in the equatorial plane (in orbit at about 36,000 km height).

Why is the orbit of a remote sensing satellite called sun-synchronous?

Answer: Because whenever such a satellite passes over a particular area of the earth, the position of the Sun with respect to that area remains the same (the satellite passes the area at the same local solar time each time).

At a particular place, at a distance of 10 km from a transmission station, a person can receive signals but is not able to receive signals at 100 km. Suggest a method by which he can receive the signal at 11 km (a distance within range) by using an antenna.

Answer: The range is limited by the height of the antennas: \(d=\sqrt{2Rh_T}+\sqrt{2Rh_R}\). He can increase the height of his receiving antenna (or use a repeater/relay station) so that the line-of-sight range increases and the signal can be received at the larger distance.

Mention the functions of a transponder.

Answer: A transponder is a device fitted on a satellite which receives the signal (uplink), amplifies it, changes its frequency and retransmits it (downlink) to the earth.

What type of modulation is used for commercial broadcast of voice signals?

Answer: Amplitude modulation (AM) is used for medium-wave and short-wave commercial broadcast of voice; frequency modulation is also used for FM broadcast.

Over-modulation results in distortion of the signal in amplitude modulation. Why?

Answer: When the carrier wave is over-modulated (i.e. \(m_a\gt1\)), the modulated wave is absent at the negative peaks of the modulating signal (the envelope no longer follows the message). This results in distortion of the signal.

What would be the modulation index for an amplitude-modulated wave for which the maximum amplitude is \(a\) while the minimum amplitude is \(b\)?

Answer: \(m_a=\dfrac{a-b}{a+b}\).

Modulation index \(m_a=\dfrac{E_m}{E_c}\) …(1). Maximum amplitude of the modulated wave \(a=E_c+E_m\) …(2). Minimum amplitude \(b=E_c-E_m\) …(3). From (2) and (3), \(E_c=\dfrac{a+b}2\) and \(E_m=\dfrac{a-b}2\). From (1), \(m_a=\dfrac{E_m}{E_c}=\dfrac{(a-b)/2}{(a+b)/2}=\dfrac{a-b}{a+b}\).

A carrier wave of peak voltage 20 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 80%?

Answer: 16 V.

\(m_a=\dfrac{E_m}{E_c}\Rightarrow E_m=m_aE_c=0.8\times20=16\) V.

How does the effective power radiated from a linear antenna depend on the wavelength of the signal to be transmitted?

Answer: The effective power radiated is inversely proportional to the square of the wavelength, \(P\propto\dfrac1{\lambda^2}\). So a short wavelength (high frequency) signal is radiated much more effectively.

Draw a block diagram of a transmitter used in a communication system.

Answer: The message signal from the transducer is processed (amplified, filtered) by the baseband processor, then modulated on a high-frequency carrier produced by an oscillator, power-amplified, and radiated by the transmitting antenna.

What does the term ‘demodulation’ mean in a communication system?

Answer: Demodulation is the process of recovering the original baseband (message) signal from the modulated carrier wave at the receiver.

Distinguish between ‘point-to-point’ and ‘broadcast’ modes of communication.

Answer: In point-to-point communication the message is sent from one transmitter to one particular receiver, e.g. a telephone call. In the broadcast mode a single powerful transmitter sends the signal to a large number of receivers spread over a wide area, e.g. radio and TV.

Why are repeaters used in a communication system?

Answer: A repeater receives the signal, amplifies it and retransmits it. It is used to compensate for attenuation and to extend the range of communication.

Write the function of a transducer in a communication system.

Answer: A transducer converts one form of energy into another; for example, at the transmitter a microphone converts sound into an electrical signal, and at the receiver a loudspeaker converts the electrical signal back into sound.

Distinguish between ‘sky wave’ and ‘space wave’ modes of propagation. Why is the sky wave mode of propagation restricted to frequencies up to 40 MHz?

Answer: Sky wave: the radio wave is reflected from the ionosphere back to the earth, and is used for long-distance communication at frequencies from a few MHz up to 30–40 MHz. Space wave: the wave travels in a straight line from the transmitting antenna to the receiving antenna (line of sight) and is used for frequencies above about 40 MHz, e.g. TV and satellite communication.

Above about 40 MHz the waves are not reflected by the ionosphere; they penetrate it and escape, so sky-wave propagation is restricted to frequencies up to about 40 MHz.

Draw a block diagram of a detector for an AM signal and show, using the necessary processes and waveforms, how the original message signal is detected from the input AM wave.

Answer: The AM wave is passed through a rectifier (a diode), which removes the negative half-cycles, giving a rectified wave. This is then passed through an envelope detector (a low-pass RC filter), which removes the high-frequency carrier and follows the envelope of the rectified wave. The output is the original message (modulating) signal.

Why is the frequency of outgoing and incoming signals different in a mobile phone?

Answer: A mobile phone works in full-duplex mode: it transmits and receives at the same time. Different frequencies are used for the outgoing and the incoming signals so that they do not interfere with each other.

Suppose you wish to transmit an electronic signal in the audio frequency range over a long distance directly. Write briefly the three important factors which prevent you from doing so and explain how you overcome these factors.

Answer: (i) Size of the antenna: an efficient antenna needs a length of about \(\dfrac\lambda4\); audio frequencies (20 Hz–20 kHz) have wavelengths of kilometres, so an impossibly long antenna would be needed. (ii) Effective power radiated: \(P\propto\dfrac1{\lambda^2}\), so a low-frequency (long wavelength) signal is radiated with very little power. (iii) Mixing up of signals: signals from different transmitters in the same audio range would overlap and could not be separated at the receiver.

These factors are overcome by modulation: the audio signal is superimposed on a high-frequency carrier wave (small wavelength, so a small antenna, high radiated power) and each station is given a different carrier frequency, so that signals can be separated.

What is ground wave communication? Explain why this mode cannot be used for long distance communication using high frequencies.

Answer: In ground wave (surface wave) communication the radio waves travel along the surface of the earth from the transmitting antenna to the receiving antenna. This is used for frequencies up to about 1500 kHz. As the wave glides over the surface it induces charges in the ground and loses energy through absorption by the earth, so it is attenuated. The attenuation increases rapidly with frequency, so high frequencies are absorbed within a short distance and cannot be used for long-distance communication.

What are the three basic units in a communication system? Write briefly the function of each of these. Write any three applications of the internet used in communication systems.

Answer: The three basic units are: (1) Transmitter, which converts the message signal into a suitable form (modulated signal) and sends it out; (2) Channel (transmission medium), which carries the signal from the transmitter to the receiver (wire, cable, optical fibre or free space); (3) Receiver, which receives the signal and recovers the original message (demodulation).

Applications of the internet: (i) e-mail; (ii) e-commerce (online shopping, banking); (iii) social networking and chat, video conferencing; (also e-learning).

For an amplitude-modulated wave, the maximum amplitude is found to be 10 V while the minimum amplitude is 2 V. Determine the value of the modulation index \(\mu\). What would be the value of \(\mu\) if the minimum amplitude is zero volt? Why is \(\mu\) generally kept less than 1?

Answer: \(\mu=\dfrac23\approx0.67\); with zero minimum amplitude \(\mu=1\).

\(\mu=\dfrac{A_{max}-A_{min}}{A_{max}+A_{min}}=\dfrac{10-2}{10+2}=\dfrac{8}{12}=0.67\). If \(A_{min}=0\), \(\mu=\dfrac{10-0}{10+0}=1\).

\(\mu\) is kept less than 1 to avoid over-modulation: for \(\mu\gt1\) the envelope of the modulated wave no longer follows the message signal and the signal is distorted.

Write two basic modes of communication. Explain the process of amplitude modulation. Draw a schematic sketch showing how an amplitude-modulated signal is obtained by superposing a modulating signal over a sinusoidal carrier wave.

Answer: The two basic modes of communication are point-to-point communication and broadcast communication.

In amplitude modulation the amplitude of the high-frequency carrier wave \(E_c\cos\omega_ct\) is varied in proportion to the instantaneous amplitude of the low-frequency message signal \(E_m\cos\omega_mt\), while the carrier frequency stays constant. The AM wave is \(e=E_c(1+m_a\cos\omega_mt)\cos\omega_ct\), with modulation index \(m_a=\dfrac{E_m}{E_c}\); its envelope has the shape of the message signal.

Write the functions of the following in communication systems: (i) transmitter, (ii) modulator.

Answer: (i) The transmitter processes the message signal, modulates it onto a carrier and radiates it into the channel. (ii) The modulator superimposes the low-frequency message signal on the high-frequency carrier wave (changing its amplitude, frequency or phase) so that the signal can be transmitted efficiently.

Write the functions of the following in communication systems: (i) receiver, (ii) demodulator.

Answer: (i) The receiver picks up the transmitted signal from the channel, amplifies it and recovers the message. (ii) The demodulator (detector) separates the baseband message signal from the modulated carrier wave.

The carrier wave is given by \(C(t)=2\sin(8\pi t)\) volt. The modulating signal is a square wave as shown. Find the modulation index.

Answer: \(m=0.5\).

The amplitude of the carrier is \(E_c=2\) V and the amplitude of the square-wave modulating signal is \(E_m=1\) V. Modulation index \(m=\dfrac{E_m}{E_c}=\dfrac12=0.5\).

The figure shows the block diagram of a generalised communication system. Identify the element labelled ‘X’ and write its function.

Answer: X is the channel (communication channel or transmission medium), the element between the transmitter and the receiver. Its function is to carry the signal from the transmitter to the receiver (it may be a pair of wires, coaxial cable, optical fibre or free space).

Write three important factors which justify the need for modulating a message signal. Show diagrammatically how an amplitude-modulated wave is obtained when a modulating signal is superimposed on a carrier wave.

Answer: The three factors are: (i) size of the antenna: the antenna length should be about \(\dfrac\lambda4\), which is impractically large for low-frequency signals; (ii) effective power radiated: \(P\propto\dfrac1{\lambda^2}\), so low-frequency signals are radiated poorly; (iii) mixing up of signals from different transmitters, which would be impossible to separate.

The block diagram of a receiver is shown in the figure. (a) Identify ‘X’ and ‘Y’. (b) Write their functions.

Answer: (a) X is the intermediate frequency (IF) stage and Y is the AF (audio-frequency) amplifier. (b) The IF stage changes the carrier frequency of the received signal to a lower, fixed intermediate frequency, which is easier to amplify and process. The AF amplifier amplifies the detected (message) signal to the level required to drive the output device (loudspeaker, etc.).

Distinguish between ‘sky waves’ and ‘space waves’ modes of propagation in a communication system. (a) Why is the sky wave mode of propagation restricted to frequencies up to 40 MHz? (b) Give two examples where the space wave mode of propagation is used.

Answer: Sky wave: the signal is reflected back to the earth by the ionosphere; it is used for long-distance communication at frequencies up to about 40 MHz. Space wave: the signal travels directly in a straight line from the transmitting to the receiving antenna (line of sight) at frequencies above about 40 MHz.

(a) Waves of frequency above about 40 MHz are not reflected by the ionosphere; they pass through it and escape. (b) Television broadcast and satellite communication (also microwave links and radar).

A transmitting antenna at the top of a tower has a height of 45 m and the height of the receiving antenna is 80 m. Calculate the maximum distance between them for satisfactory communication in LOS mode. (Radius of the earth \(=6.4\times10^6\) m)

Answer: 56 km.

\(d=\sqrt{2Rh_T}+\sqrt{2Rh_R}=\sqrt{2\times6.4\times10^6\times45}+\sqrt{2\times6.4\times10^6\times80}=24000+32000\) m \(=56\) km.

Name the type of waves which are used for line of sight (LOS) communication. What is the range of their frequencies? A transmitting antenna at the top of a tower has a height of 20 m and the height of the receiving antenna is 45 m. Calculate the maximum distance between them for satisfactory communication in LOS mode. (Radius of the earth \(=6.4\times10^6\) m)

Answer: Space waves (microwaves), with frequencies of about 40 MHz and above (up to several GHz).

\(d=\sqrt{2\times6.4\times10^6\times20}+\sqrt{2\times6.4\times10^6\times45}=16000+24000=40000\) m \(=40\) km.

In the block diagram of a simple modulator for obtaining an AM signal, shown in the figure, identify the boxes A and B.

Answer: A is a square-law device (a non-linear device such as a diode or transistor), which produces the product term of the message and carrier. B is a band pass filter, which allows only the frequencies \(\omega_c\), \(\omega_c+\omega_m\) and \(\omega_c-\omega_m\) to pass and gives the AM wave.

A transmitting antenna at the top of a tower has a height of 45 m and the receiving antenna is on the ground. Calculate the maximum distance between them for satisfactory communication in LOS mode. (Radius of the earth \(=6.4\times10^6\) m)

Answer: 24 km.

\(d=\sqrt{2Rh_T}=\sqrt{2\times6.4\times10^6\times45}=\sqrt{5.76\times10^8}=24000\) m \(=24\) km.

Why are broadcast frequencies (carrier waves) sufficiently spaced in an amplitude modulated wave?

Answer: An AM wave occupies a bandwidth of \(2f_m\) (carrier \(\pm f_m\)). If the carrier frequencies of two stations are too close, their side bands overlap and the signals interfere. Sufficient spacing between carriers (about 10 kHz for AM broadcast) avoids this overlap.

Give reasons for the following: (i) For ground wave transmission, the size of the antenna should be comparable to the wavelength of the signal. (ii) Audio signals converted into electromagnetic waves are not transmitted as such directly.

Answer: (i) An antenna radiates or receives efficiently only if its length is comparable to the wavelength (about \(\dfrac\lambda4\) or more); otherwise the radiated power is very small. (ii) Audio signals are of low frequency (large wavelength), so they would need an impractically large antenna, would be radiated with very little power (\(P\propto\dfrac1{\lambda^2}\)), and would mix up with signals from other transmitters.

Distinguish between amplitude modulation and frequency modulation. Why is the amplitude of the modulating signal kept less than that of the carrier wave?

Answer: In amplitude modulation the amplitude of the carrier is varied according to the message signal while its frequency is kept constant. In frequency modulation the frequency of the carrier is varied according to the message signal while its amplitude is kept constant. FM is less affected by noise but needs a larger bandwidth.

The amplitude of the modulating signal is kept less than that of the carrier (\(m_a\lt1\)) to avoid over-modulation, which would distort the signal.

Explain, with the help of a suitable diagram, the space wave mode of propagation. Give two examples in communication systems where this mode is used. What is the frequency range of these waves? Give the reason for using this range of frequency.

Answer: In the space wave mode the signal travels in a straight line directly from the transmitting antenna to the receiving antenna (line-of-sight communication); the antennas are placed on tall towers so that the earth’s curvature does not block the path. The range is \(d=\sqrt{2Rh_T}+\sqrt{2Rh_R}\), and the total coverage can be increased by raising the antennas or using repeaters. (The diagram shows a tower with a transmitting antenna, a tower with a receiving antenna and the straight-line path between them, with the earth’s curved surface below.)

Examples: television broadcast (and FM radio) and satellite communication (also microwave links and radar).

Frequency range: about 40 MHz up to several GHz (VHF, UHF and microwaves). The reason for this range: waves of these frequencies are not reflected by the ionosphere (they pass through it) and are not carried by the ground wave, so they must travel along the line of sight; being of short wavelength, they are directional and can be radiated by antennas of small size.

• Optical fiber 100 GHz.

• Free space. Table 15.2 page no 519 NCERT Book

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