NEET Physics Class 12 Chapter 5 Principles Of Communication Notes

Principles Of Communication Introduction

Communication is the process of transmission of information. The teacher transfers data to the students.

To be successful, the sender and receiver must understand a common language. The electronic communication system requires a source of information, a transmitting medium (channel), and a receiver.

The modern communication system has progressed in all three basic components. The information processing and sorting before communication, the channels like optical fiber, space (satellites), and processing through computers before being delivered.

Elements Of A Communication System

As pointed out communication system has three essential stages. Transmitter, medium, or channel and receiver.

The information may be a message, speech, picture, audio video, or group. The information has to be transformed into electrical signals. The transmitter processes and encodes the information to make it suitable for transmission through the channel as well as for reception.

This encoding process is modulation. Signal propagating through the channel may get distortion and we say noise is added to it due to channel imperfections. So the receiver receives a corrupt version.

The receiver has to process the corrupt signal to a recognizable form of the original signal for delivery to the user. There are two modes of communication.

Point-to-point communication example telephony where there is one transmitter and one receiver.

Broadcast mode involves one transmitter and a large number of receivers Example: Radio, TV, etc. Also on the basis that the signals used may be analog or digital, it may be mentioned that The signals both digital and analog are usually of low frequency and hence cannot be transmitted as such These signals require higher frequency waves on which these can ride over called carrier waves This process is called modulation.

Basic Terminology Used In Communication Systems

Let us acquaint ourselves with the basic terminology used in communications.

Transducer: It is a device that converts one form of energy into the other. However, in communication systems, we have to convert all types of signals into electrical. So an electric transducer is a device that converts a physical signal (or variable) such as pressure, temperature, and force displacement. Light sound, etc. into a corresponding electrical signal.

Signal: Information in an electric form suitable for transmission is called a signal. If the signal is in the form of a continuous variation of voltage or current, it is called an analog signal as shown in Figure 2. The variation has to be single-valued. Since all periodic functions may be broken into sine and cosine components and hence sine wave is a fundamental analogue signal.

Sound and picture signals in TV are analog. In digital signal Figure 2 the voltage has only two values either low (0) or high (1). Thus a digital signal is a discontinuous signal.

There are several coding schemes suitable for digital communication systems. Binary coded decimal (BCD). American standard code for information Interchange (ASCII) is the most popular digital code to represent numbers, letters, and certain characters.

Noise: The unwanted signals generated inside and outside the system are referred to as noise.

Transmitter: It processes the incoming signal to make it suitable for transmission through the channel.

Receiver: The receiver extracts the relevant information from the received signal.

Attenuation: The loss in the signal power or strength during transmission is called attenuation.

Amplification: Increasing signal strength by converting other energy (DC or other frequency) into signal energy using electronic circuitry is called amplification.

Range: It is the largest distance between the source and receiver where the signal of sufficient strength is received.

Bandwidth (BW): The portion of the spectrum occupied by the signal or frequency range over which equipment operates is called bandwidth.

Modulation: Superimposing low-frequency messages over high-frequency carrier waves is called modulation. It is AM (amplitude modulated), FM (frequency modulated or PM phase modulated) depending on which quantity, amplitudes, frequency, or phase of the carrier wave varies with the signal.

Demodulation: The process of retrieving information from received waves is called demodulation.

Repeater: These are used to increase the range. It receives a signal from the transmitter amplifies it and retransmits after amplification. It may be at different frequencies. It is thus a combination of receiver and transmitter. Figure (3) shows how the range is extended beyond the mountain. A satellite station is also a sort of repeater.

Antenna or Aerial:

Bandwidth Of Signals

1. Analog signals: The communication signals are of varied nature and have a different range of frequencies example speech signals have a frequency range of 300 Hz to 3100 Hz. Thus the bandwidth for speech is (3100 – 300) = 2800 Hz = 2.8 Hz. The audio range is 20 to 20 kHz the video signals have a typical bandwidth of 4.2 MHz. A TV signal has both audio and video so its bandwidth is 6 MHz.
2. Digital Signals: Digital signals are in the form of rectangular waves. However, a rectangular wave may be constructed using harmonic sine and or cosine waves of frequencies v, 2v, 3v. This implies infinite bandwidth but for practical purposes, higher harmonics can be neglected. No doubt the received waves are distorted versions of the original wave but information is not lost and the rectangular signal is more or less received.

Band – Widths of various signals used in communication

Bandwidth Of Transmission Medium

The transmission media also have different bandwidths Example widely used coaxial cable has a bandwidth of ~750 MHz and operates below 18 GHz.

Free space communication has a wide range from 100 kHz to 10 GHz Optical fiber operates at Terahertz frequencies and has quite a wide range from 1 THz to 1000 THz (microwaves to UV). It provides a bandwidth of ~100 GHz.

Bandwidths of Transmission Channel/ Media used in communications

Space Communication: Propagation Of Electromagnetic Waves In Space

In space radio communication, an antenna at the transmitter radiates electromagnetic waves that travel through space and reach the receiving antenna at the other end. During propagation not only does the signal diminish but several factors influence the propagation.

The atmosphere is a highly dynamic system whose properties change not only with elevation but also with seasons. The propagation characteristics also depend on the frequency band. The following frequency bands are used in radio communication.

1. Medium Frequency Band (MF) 300 – 3000 kHz
2. High Frequency Band (HF) 3.0 – 30 MHz
3. Very High Frequency Band (VHF) 30 – 300 MHz
4. Ultra High-Frequency Band (UHF) 300 – 3000 MHz
5. Super High-Frequency Band (SHF) 3.0 – 30 GHz

The following are the modes of communication through space:

Ground Wave Transmission

For efficient signal radiation by antenna, its height should be in multiples of (/4) where  is wavelength of CW used. For high i,e. lower frequencies, the antenna size is large and these have to be located near to the ground.

In standard AM broadcast, ground base towers are used for broadcast. The waves propagate parallel to the ground. The propagation is called ground wave propagation. The wave induces a current in the ground where it passes.

Therefore it gets attenuated. Understandably, the higher is frequency greater the attenuation. Therefore range depends on power and frequency. The used frequency is less than a few MHz.

Sky Wave Transmission

The radio waves having a frequency of 2 to 30 MHz where propagating up the sky are reflected by the ionosphere and return to the earth. These waves used for communication, are known as sky waves.

At a height 65 to 400 km above the earth, the atmospheric gas absorbs ultraviolet and other high energetic cosmic radiations coming from the sky and subsequently ionizes there creating an ionic layer called the ionosphere.

Even in single reflection wave can cover about 4000 km distance, hence sky wave propagation is long-range transmission. Round the globe communication.

Is possible using skywave propagation? A few terms are important in skywave propagation

Virtual Height: The reflection of sky waves is through gradual bending like total reflection in the formation of a mirage. Hence virtual height is the height through which the angle of incidence is calculated to send waves.

Critical Frequency (fc): The maximum frequency reflected when beamed straight towards the layer. Frequency f > FC is not reflected

Maximum Usable Frequency (MUF): Radio waves when sent at an angle θ, the maximum usable frequency (MUF) is MUF = fc/cosθ ….…(1)

Skip Distance: The smallest distance of the receiver from the transmitter where a wave of a given frequency reaches. It follows from Figure 5 that the higher is angle of incidence on the ionic layer more is the skip distance.

Fading: Due to the arrival of many signals at the receiver emitted from the source simultaneously but reaching the receiver in time delay due to taking different paths diminished signal due to destructive interference. This is fading.

Space Wave Propagation

In space wave mode of propagation, the waves travel in a straight line from the transmitter to the receiving antenna i.e., communication is in the line of sight (LOS). The curvature of the earth limits the range.

As pointed out earlier this communication is above 40 MHz. At these frequencies, antennae are smaller and may be placed at heights many wavelengths above the ground.

To increase the range to the desired level, a satellite is used as a repeater which effectively increases antenna height to the height of the satellite. TV, mobile, and radar systems are examples.

The range versus antenna height relation may easily be determined using the geometry of

⇒ \(\begin{aligned}
& \Delta H M O, H O^2=H^2+M^2 \\
& \left(R+h_t\right)^2=d^2+R^2, R=\text { radius of the earth } \\
& R^2+h^2+2 R h_t=d_t^2+R^2 \\
& 2 R h_t+h^2=d_t^2, h<<R
\end{aligned}\)

⇒ \(\mathrm{d}_{\mathrm{T}}=\sqrt{2 R h_{\mathrm{t}}}\)

Similarly \(d_R=\sqrt{2 R h_R}\)……………………….2

Hence range \(d_m=\left(d_t+d_R\right)=\sqrt{2 R h_t}+\sqrt{2 R h_R}\)………………………..3

In Case the Satellite is Located At a height then

⇒ \(d_m=d_R=d=\sqrt{2 R h}\)…………………………….4

Area Covered \(\pi d^2=2 \pi R h\)…………………………………..5

Solved Examples

Example 1. In which frequency range, space waves are normally propagated

1. HF
2. VHF
3. UHF
4. SHF

Answer: 3. UHF

Example 2. An antenna behaves as a resonant circuit only when its length is

1. \(\frac{\lambda}{2}\)
2. \(\frac{\lambda}{4}\)
3. λ
4. \(\frac{\lambda}{2} \text { or integral multiple of } \frac{\lambda}{2}\)

Answer: 4. \(\frac{\lambda}{2} \text { or integral multiple of } \frac{\lambda}{2}\)

Question 3. The process of superimposing signal frequency (i.e. audio wave) on the carrier wave is known as

1. Transmission
2. Reception
3. Modulation
4. Detection

Answer: 3. Carrier + signal → modulation.

Question 4. Long-distance short-wave radio broadcasting uses

1. Ground wave
2. Ionospheric wave
3. Direct wave
4. Skywave

Answer: 3. Direct wave

Question 5. The maximum distance up to which TV transmission from a TV tower of height h can be received is proportional to

1. \(h^{1 / 2}\)
2. h
3. \(h^{3 / 2}\)
4. h2

Answer: \(\mathrm{d}=\sqrt{2 \mathrm{hR}} \Rightarrow \quad \mathrm{d} \propto \mathrm{h}^{1 / 2}\)

Example 6. A transmitting antenna at the top of a tower has a height of 32 m and the height of the receiving antenna is 50 m. What is the maximum distance between them for satisfactory communication in LOS mode? Given the radius of Earth 6.4 × 106 m.
Solution:

⇒ \(\begin{aligned}
& \mathrm{d}_{\mathrm{m}}=\sqrt{2 \times 64 \times 10^5 \times 32}+\sqrt{2 \times 64 \times 10^5 \times 50} \mathrm{~m} \\
& =64 \times 10^2 \times \sqrt{10}+8 \times 10^3 \times \sqrt{10} \mathrm{~m} \\
& =144 \times 10^2 \times \sqrt{10} \mathrm{~m}=45.5 \mathrm{~km}
\end{aligned}\)

Example 7. In a communication system, noise is most likely to affect the signal

1. At the transmitter
2. In the channel or the transmission line
3. In the information source
4. At the receiver

Answer: 2. In the channel or the transmission line

Example 8. The waves used in telecommunication are

1. IR
2. UV
3. Microwave
4. Cosmic rays

Answer: 3. UV In telecommunication microwaves are used.

Example 9. Television signals on earth cannot be received at distances greater than 100 km from the transmission station. The reason behind this is that

1. The receiver antenna is unable to detect the signal at a distance greater than 100 km
2. The TV program consists of both audio and video signals
3. The TV signals are less powerful than radio signals
4. The surface of the earth is curved like a sphere

Answer: 4. The surface of the earth is curved like a sphere

Example 10. AM is used for broadcasting because

1. It is more noise-immune than other modulation systems
2. It requires less transmitting power compared with other systems
3. Its use avoids receiver complexity
4. No other modulation system can provide the necessary bandwidth faithful transmission

Answer: 3. Its use avoids receiver complexity

Example. 11 Range of frequencies allotted for commercial FM radio broadcasts is

1. 88 to 108 MHz
2. 88 to 108 kHz
3. 8 to 88 MHz s
4. 88 to 108 GHz

Answer: 1. 88 to 108 MHz

Example 12. At which of the following frequencies, the communication will not be reliable beyond the horizon

1. 1KHz
2. 1 MHz
3. 10 GHz
4. 100 GHz

Answer: 2. MHz travel is a line of sight.

Example 13. Modulation is used to

1. Reduce the bandwidth used
2. Isolate transmission of different users
3. Ensure transmission of intelligence to long-distance
4. To reduce the size of the antenna to a useable range

Answer: 1. Band width is reduced; 2 FM

Example 14. AM is used for broad costing because

1. The signal-to-noise ratio is low
2. It needs less transmission power
3. The AM receiver system is simple width complexity
4. No other modulation provides faithful bandwidth

Answer: 3. The AM receiver system is simple width complexity

Example 15. UHF frequencies normally propagate via

1. Ground wave
2. Skywave
3. Surface-wave
4. Space waves

Answer: 4. UHF travels as a space wave

Example 16. A microwave link operates at a central frequency of 10 GHz and 2% is used for telephone channels. If the telephone is allotted a bandwidth of 8 kHz the number of channels that can be operated simultaneously is

1. 12.5 × 105
2. 12.5 × 103
3. 2.5 × 107
4. 2.5 × 103

Answer: 3. Band width available \(\frac{2}{100} \times 10^9\) one channel needs 8 kHz No of channels operating \(=\frac{2 \times 10^7}{8.5 \times 10^3}=2.5 \times 10^4\)

Modulation And It’s Necessity

The signals to be transmitted (audio, video, or data) are low-frequency signals and there are inherent difficulties in transmitting these signals directly as discussed below.

So these signals are superimposed in any of the properties of high-frequency waves called carriers. When superimposed in amplitude i.e. resultant wave amplitude varies with the signal and we have amplitude modulated wave (AM).

When superimposed in frequency, the frequency of the resultant wave varies with the signal, we have a frequency-modulated (FM) wave similarly if superimposed in phase the phase of the resultant wave varies with the signal and we have a phase-modulated wave (PM). we now outline difficulties in low-frequency transmission.

Size Of Antenna Or Aerial

As pointed out earlier the minimum size of the antenna / aerial required is \(\frac{\lambda}{4}\) For transmitting a 20 kHz signal we require minimum antenna size \(\ell=\frac{\lambda}{4}=\frac{1}{4} \times \frac{3 \times 10^8}{20 \times 10^3} 3.75 \mathrm{~km}.\) This long antenna is not possible.

If instead the signal (or baseband signal) is modulated with CW of frequency 10 MHz the size required is \(\ell=\frac{1}{4} \times \frac{3 \times 10^8}{20 \times 10^3} 7.5 \mathrm{~m}.\) This is practically possible.

Power Radiated

The power radiated by the antenna is determined by length to wavelength ratio (l/λ). More power is radiated at high frequencies making communication better. The power transmitted varies as frequency square (f²)

Mixing Of Signals From Different Transmitters

If signals are transmitted at the baseband (original frequency band) then the receiver shall receive a signal from many transmitters simultaneously and they get mixed. If these signals are modulated on different carrier frequencies, the mixing is avoided and the receiver shall get the desired signal by tuning at that CW frequency.

In digital communication carrier waves are in the form of pulses the modulation may be pulse amplitude, pulse duration, and pulse width modulation as illustrated in Figure 9 The flow chart of Figure 10 below summarizes various types of modulations.

Properties Of Amplitude Modulation:

Frequency spectrum: In amplitude modulated (AM) wave the amplitude of the carrier wave varies with the signal i.e. signal is superimposed in amplitude. If the carrier wave is C(t) = AC sin ACt and the message signal is m(t) = Am sin ωmt then the modulated signal will be

1. Cm(t) = (VC + Vm sin ωmt) sin ωCt
2. Cm(t) = (VC + Vm sin ωmt) sin ωCt

⇒ \(=V_c\left(1+\frac{V_m}{V_c} \sin \omega_m t\right) \sin \omega_c t\)

Where \(\mu=\frac{V_m}{V_c}\) is called modulation index.

⇒ \(C_m(t)=V_c \sin \omega_m t+\left(\frac{V_m}{V_c}\right) \sin \omega_{c t} V_c \sin \omega_c t\)

= VC sin wCt + wVC sin with sin cwt

It is generally less than 1 ( < 1)

⇒ \(C_m(t)=V_c \sin \omega_c t+V_c\left(\frac{1}{2} \cos \left(\omega_c-\omega_m\right) t-\frac{1}{2} \cos \left(\omega_c+\omega_c\right) t\right.\)

Equation (8) shows three sets of angular frequencies original carrier viz., (ωc), (ωc – wm) known as lower sideband and (ωc + ωm) upper sideband of frequencies. The amplitude-modulated frequency spectrum. (f = w/2π)

It follows that if broadcasted bands are sufficiently separated so that sidebands do not overlap, different stations can operate without any interference.

Modulation Index: The ratio of change in amplitude of the carrier wave to amplitude of the original carrier wave is called modulation index or modulation factor, m a
\(m_a=\frac{k V_m}{V_C} \text {, }\) actor k determines the maximum change in amplitude for the given amplitude of the modulating wave.

On amplitude modulation, the maximum and minimum amplitudes are Amax and Amin then the maximum change in amplitude is (Vmax – VC) and so \(m_{\mathrm{a}}=\frac{V_{\max }-V_C}{V_C}\)

For example if VC = V and \(V_m=\frac{V}{2}\) Then \(m_a=V+\frac{V}{2}-V=V / 2=50 \%\)

⇒ \(m_a=V+\frac{V}{2}-V=V / 2=50 \%\)

If VC = VC and Vm = V then Vmax = 2V

If VC = V, Vm = 3/2V then Vmax = 5/2 V

\(m_a=\frac{2 V-V}{V}=1 \text { or } 100 \% \quad \text { and } m_a=\frac{(5 / 2) V-V}{2}=\frac{3}{2} \text { or } 150 \% V\)

In this case, the carrier is overmodulated> 100% it may be noted that the modulation factor ma, determines the strength and quality of the signal transmitted. An audio signal is generally AM modulated and hence higher is modulation the stronger and clearer will be the signal.

If u = 1 modulation index may be expressed in terms of maximum and minimum amplitude, Vmax and Vmin

\(m_a=\frac{V_{\max }-V_{\text {min }}}{V_{\text {max }}+V_{\text {min }}}\)

If the carrier wave is modulated with many signals then the total modulation index is given by

\(m_t=\sqrt{m_1^2+m_2^2+m_3^2 \ldots \ldots \ldots}\)

Bandwidth Required: The required bandwidth is from the lower sideband to the upper side band hence,

⇒ \(\Delta f=\frac{\left(\omega_{\mathrm{c}}+\omega_{\mathrm{m}}\right)}{2 \pi}-\frac{\omega_{\mathrm{c}}-\omega_{\mathrm{m}}}{2 \pi}=\frac{2 \omega_{\mathrm{m}}}{2 \pi}=2 f_m\)

Power in AM Wave: Power dissipated in any circuit having resistance R and supplied at rms voltage Vrms, is given by \(\left(\frac{V_{\mathrm{ms}}^2}{\mathrm{R}}\right)\) The total power is the sum of the power in side bands plus the power in a carrier wave. Total power transmitted

⇒ \(\begin{aligned}
& \text { = } P_{\text {Total }}=P_{\text {LSB }}+P_{\text {USB }}+P_{c w} \\
& =\left(\frac{m_a V_{\mathrm{C}}}{2 \times \sqrt{2}}\right)^2 \frac{1}{R}+\left(\frac{m_a V_{\mathrm{C}}}{2 \times \sqrt{2}}\right)^2 \frac{1}{R}+\left(\frac{V_{\mathrm{C}}}{\sqrt{2}}\right)^2 \frac{1}{R}=\frac{V_{\mathrm{C}}^2}{2 R}\left(1+\frac{m_{\mathrm{a}}^2}{2}\right)
\end{aligned}\)

The ratio of power transmitted to carrier power

⇒ \(\frac{\frac{P_{\text {Total }}}{P_{\mathrm{CW}}}=\frac{V_{\mathrm{C}}^2}{2 R}\left(1+\frac{m_{\mathrm{a}}^2}{2}\right)}{\left(\frac{V_{\mathrm{C}}^2}{2 R}\right)=\left(1+\frac{m_{\mathrm{a}}^2}{2}\right)}\)

Fraction of power transmitted in the sideband

⇒ \(\frac{\frac{P_{S B}}{P_{\text {Total }}}=\frac{1}{R}\left(\frac{m_a V_{\mathrm{C}}}{2 \sqrt{2}}\right)^2}{\frac{V_{\mathrm{C}}^2}{2 R}\left(1+\frac{m_{\mathrm{a}}^2}{2}\right)=\left(\frac{m_a^2 / 2}{1+m_a^2 / 2}\right)}\)

Distortion-free maximum power transfer. For distortion-free transmission ma = 0. so

Production Of Amplitude Modulated Wave

To add the signal to the amplitude of CW the voltage signal is mixed up in a mixer and sampled by a square law device. The output is filtered by a band pass filter centered around the carrier frequency. The signal is amplified by a power amplifier before transmission.

Detection Of Amplitude Modulated Signal

The signal is received by the receiving antenna. As a received signal is weak due to attenuation in the channel, it has to be amplified. The detection of high frequency is difficult so, it is changed to a low frequency called intermediate frequency, (IF).

This converted signal is detected and amplified. The AM wave is rectified and rejects the lower part of the AM wave. Finally envelope is detected (filtering CW component). The entire process

Limitations of AM wave Transmission

AM communication has the following problems

1. Reception is noisy and becomes audio noise caused by various
2. Efficiency is low, machines get mixed
3. The operating range is small
4. The audio quality is poor.

Solved Examples

Example 17. An AM wave has 1800 watts of total power content, for 100% modulation the carrier should have power content equal to

1. 1000 watt
2. 1200 watt
3. 1500 watt
4. 1600 watt

Answer: \(P_t=P_c\left(1+\frac{m_a^2}{2}\right) ; \text { Here } m_a=1\)

\(1800=P_c\left(1+\frac{(1)^2}{2}\right) \quad \Rightarrow P_c=1200 \mathrm{~W}\)

Question 18. A TV tower has a height of 75m. What is the maximum distance and area up to which this TV transmission can be received? Take the radius of the earth as 6.4 × 10 6 m.
Solution: \(d=\sqrt{2 R h}=\sqrt{2 \times 6.4 \times 10^6 \times 75}=3.1 \times 10^4 \mathrm{~m}=31 \mathrm{~km}\)

Question 19. A TV tower has a height of 100 m. How much population is covered by the TV broadcast if the average population density around the tower is 1000 km –2? Given: radius of earth = 6.37 × 106 m.
Solution: h = 100m, R = 6.37 × 106 m,

Average population density = 1000 km–2 = 1000(103)–2m–2 =10–3m–2 2hR Distance up to which the transmission could be viewed, d =

Total area over which transmission could be viewed = πd2 = 2πhR

Population covered = 10–3 × 2πhR = 10–3 × 2 × 3.14 × 100 × 6.37 × 106 = 40 lakh

Example. 20 What is the modulation index of an overmodulated wave

1. 1
2. Zero
3. < 1
4. > 1

Solution 4. When ma > 1 then the carrier is said to be over-modulated.

Example. 21 Which of the following is the disadvantage of FM over AM

1. Larger bandwidth requirement
2. Larger noise
3. Higher modulation power
4. Low efficiency

Solution: 1. Frequency modulation requires a much wider channel (7 to 15 times) as compared to AM.

Example 22. When the modulating frequency is doubled, the modulation index is halved, and the modulating voltage constant of the modulation system is

1. Amplitude modulation
2. Phase modulation
3. Frequency modulation
4. All of the above

Solution: 3. Frequency modulation

Example 23. Indicate which one of the following systems is digital

1. Pulse position modulation
2. Pulse code modulation
3. Pulse width modulation
4. Pulse amplitude modulation

Solution: 2. Pulse code modulation

Example 24. In an FM system, a 7 kHz signal modulates a 108 MHz carrier so that the frequency deviation is 50 kHz. The carrier swing is

1. 7.143
2. 8
3. 0.71
4. 350

Solution: 1. 7.143

Example 25. The sinusoidal carrier voltage of frequency 1.5 MHz and amplitude 50 V is amplitude modulated by the sinusoidal voltage of frequency 10 kHz producing 50% modulation. The lower and upper sideband frequencies in kHz are

1. 1490,1510
2. 1510,1490
3. \(\frac{1}{1490}, \frac{1}{1510}\)
4. \(\frac{1}{1510}, \frac{1}{1490}\)

Solution: Here, fc = 1.5 MHz = 1500 kHz, fm = 10 kHz

• Lower sideband frequency = fc = fm = 1500 kHz – 10 kHz = 1490 kHz
• Upper sideband frequency = fc + fm = 1500 kHz + 10 kHz = 1510 kHz

Frequency Modulated Wave

Most of the noises affect the amplitude of the signal and hence to noise ratio is greatly improved if the amplitude of CW remains unaffected. This is what is done in frequency modulation where the frequency of CW varies.

Analysis

Consider a voltage signal vm = Vm cos wants to be frequency modulated on a carrier voltage wave vc = Vc cos (ωct + ω0).

Where ωm = 2πfm, ωc = 2ωfc are respectively angular frequencies of the signal and the carrier waves and Vm and Vc are their amplitude ω0 is the initial phase of the carrier. The instantaneous phase of carrier wave (CW).

⇒ \(\phi(t)=\omega \mathrm{ct}+\theta_0\)

The angular frequency of the modulated wave shall be

ω = ωc + k Vm cos ωmt

Where k is the frequency conversion factor which is constant. The phase of the FM wave at any instant shall be

⇒ \(\phi(\mathrm{t})=\int \omega \mathrm{dt}=\int\left(\omega_{\mathrm{c}}+\mathrm{kV} \mathrm{m}_{\mathrm{m}} \cos \omega_{\mathrm{m}} \mathrm{t}\right) \mathrm{dt}\)

⇒  \(\phi(t)=\omega_c t+\frac{k V_m}{\omega_m} \sin \omega_m t\)

Hence equation of FM voltage wave is \(v_{F M}(t)=V_c \sin \left[\omega_c t+\frac{k V_m}{\omega_m} \sin \omega_m t\right]\)

The instantaneous frequency of an FM wave is given by

⇒ \(\frac{1}{2 \pi} \frac{\partial \phi(t)}{d t}\)

Therefore \(f=\frac{1}{2 \pi} \omega_c+\frac{k V_m}{2 \pi} \cos \omega_m t\)

The maximum and minimum frequencies are obviously,

⇒ \(\begin{aligned}
& f_{\max }=f_c+\frac{k V_m}{2 \pi} \\
& f_{\min }=f_c-\frac{k V_m}{2 \pi}
\end{aligned}\)

The maximum change in frequency from the mean value is called frequency deviation

⇒ \(f_d=\left(f_{\max }-f_c\right)=\left(f_c-f_{\min }\right)=\frac{k V_m}{2 \pi}\)

The total variation of frequency from the maximum to the minimum is called carrier swing. It is twice the frequency deviation.

⇒ \(C S=2 f d=\frac{k V_m}{\pi}\)

The frequency modulation index mf is defined as the ratio of frequency deviation to the modulation frequency

⇒ \(m_f=\frac{f_d}{f_m}=\frac{\omega_d}{\omega_m}=\frac{k V_m}{\omega_m}\)

The equation of FM wave becomes vFM = VC sin (ct + mf sin mt) ……….(24) (c) Frequency spectrum

FM Side Bands:

Equation (23) may be expanded and trigonometric manipulations shall show that there are a series of sidebands \(\left(f_c \pm f_m\right),\left(f_c \pm 2 f_m\right),\left(f_c \pm 3 f_m\right)\), etc. with decreasing amplitudes. Side bands are equally spaced on either side of carrier frequency FC

Frequency bands in use

As pointed out frequency modulation (FM) has a better quantity of transmission with large bandwidth. The manmade noises and atmosphere changes do not affect transmission quality. Also, the fidelity is good for the transmission of music. The frequency bands in use are:

1. 88 to 108 MHz FM Radio
2. 47 to 230 MHz VHF TV
3. 470 to 960 MHz UHF TV

Digital Communication: Data Transmission & Retrieval

Digital communication ensures less noise and less error communication. Hear, carrier is a digital pulsating wave in binary codes 0 and 1. The analog signal is digitized. There are many encoding steps: source coding channel coding, etc. A typical digital communication system is shown in

There are normally three steps converting the signal into pulses of the same height and negligible width quantization and Coding quantized pulses following some rule.

Modem

Is short-term used for modulators and demodulators? As seen in Figure 1 5 modulator and demodulator are needed for two-way communication and a single modem unit serves the purpose.

Modulation Used

In digital communication modulation techniques used are shown in Figure 9 which is quite illustrative.

Modern communication systems commonly use frequency-shifting keys.

Optical Communication

Typical optical communication system. It uses optical frequency as a carrier so it has the following advantages

1. There is no electromagnetic interference
2. Enormous channel capacity
3. Requires optical fiber as communication channel
4. Mostly used in LAN (Local area networking)
5. Setup for digital communication (Block diagram)

Optical Fibre

Principle: Light travels in an optical fiber through total internal reflection from opposite walls. For total reflection light must travel from the sensor to the rarer medium and the angle of incidence from the densor to the rarer interface should be greater than the critical angle, \(\theta_c=\sin ^{-1}\left(\frac{\mu_{\text {rarer }}}{\mu_{\text {dens }}}\right)\) u’s are corresponding refractive indices

Optical fiber consists of an inner transparent cylindrical core of refractive index 1 surrounded by a transparent cylindrical cladding of refractive index u2 (< u1).

The core-cladding system is secured against mechanical shocks by multilayered polyester nylons, etc., Let a light ray be an incident making an angle u with the axis of the fiber as shown in

The ray incident on the core face is refracted into core and falls on the core-cladding interface at an angle of u1.

Using Snells law

u0 sin u = 1 sin r, r = angle of refraction ………….(25)

Where u0 is the refractive index of the outer medium for air u0 = 1. For the ray to be reflected from the corecladding interface.

\(\theta_1 \geq \theta_c \text { where } \sin \theta_c=\frac{\mu_2}{\mu_1}[\theta_1 \geq \theta_c \text { where } \sin \theta_c=\frac{\mu_2}{\mu_1}\)

Since r = (90 – θc), hence the value of the angle θ for total reflection at the core the clad interface is

⇒ \(\begin{aligned}
\mu_0 \sin \theta & =\mu_1 \sin \left(90-\theta_c\right)=\mu_1 \cos \theta_c=\mu_1 \sqrt{1-\sin ^2 \theta_c .} \\
\text { or } & \sin \theta=\mu_1 \sqrt{1-\left(\frac{\mu_2}{\mu_1}\right)^2}=\sqrt{\mu_1^2-\mu_2^2}
\end{aligned}\)

When 0 increases, r increases but  decreases therefore value of the angle given by Equation (27) is the maximum permissible value and it is called the maximum angle of acceptance

⇒ \(\theta_{\mathrm{m}}=\sin ^{-1}\left(\sqrt{\mu_1^2-\mu_2^2}\right)\)

The quantity sin gives the light-gathering capacity of the fiber. It is called numerical aperture. (NA)

⇒ \(\begin{aligned}
\quad & N A=\mu_0 \sin \theta_m=\sqrt{\mu_1^2-\mu_2^2} \\
\text { or } \quad \text { NA } & =\sqrt{\left(\mu_1+\mu_2\right)\left(\mu_1-\mu_2\right)}
\end{aligned}\)

Since \(\begin{aligned}
& \left(\mu_1-\mu_2\right) \text { is small so } \mu_2=\mu_1 \\
& =\mu_1 \sqrt{2\left(\frac{\mu_1-\mu_2}{\mu_1}\right)}=\mu_1 \sqrt{2 \Delta} \\
& \Delta=\frac{\mu_1-\mu_2}{\mu_1}
\end{aligned}\) is called fractional change in refractive index.

Satellite Communications

In the age of IT explosion where enormous data need to be transmitted and received, there occurs a need for higher frequency bands and more channels. This is possible only with satellite communication.

As discussed earlier, the signal from the transmitting station is sent to communication satellite equipped with transmitting and receiving systems known as radio Transponder (RT).

The signal transmitted and received by satellite is called up-link whereas as transmitted by satellite and received at the ground is called down the link to avoid confusion the frequencies of up and downlinks are kept different.

The commonly used satellite system consists of three geostationary satellites located on the vertices of an equilateral triangle having verticals on a geostationary orbit to cover entire globe space.

Most of the satellites are in geostationary orbit yet two more orbits are used for communication satellites. These are polar circular orbits near the earth about 1000 km, high, their inclination is 90º.

The other is a highly elliptical orbit inclined at 63º to fulfill the needs of high-altitude regions. Commonly known as the 63º slot Finally figure 18 shows the summary of various communication systems.

Example 26. If f0 and ff represent the carrier wave frequencies for amplitude and frequency modulations respectively, then

1. f0 > ff
2. f0 < ff
3. f0 = ff
4. f0 ≥ ff

Answer: f0 < ff

Example 27. The frequency of an FM transmitter without signal input is called

1. Lower sideband frequency
2. Upper sideband frequency
3. Resting frequency
4. None of these

Solution: 3. Resting frequency

Example 28. What type of modulation is employed in India for radio transmission

1. Amplitude modulation
2. Frequency modulation
3. Pulse modulation
4. None of these

Solution: 1. Amplitude modulation

Example 29. While tuning in a certain broadcast station with a receiver, we are actually

1. Varying the local oscillator frequency
2. Varying the frequency of the radio signal to be picked up
3. Tuning the antenna
4. None of these

Solution: 1. Varying the local oscillator frequency

Example 30. Consider telecommunication through optical fibers. Which of the following statements is not true

1. Optical fibers may have homogeneous core with suitable cladding
2. Optical fibers can be of graded refractive index
3. Optical fibers are subject to electromagnetic interference from outside
4. Optical fibers have extremely low transmission loss

Answer: 3. Optical fibers are subject to electromagnetic interference from outside

Example 31. The phenomenon by which light travels in an optical fiber is

1. Reflection
2. Refraction
3. Total internal reflection
4. Transmission

Answer: 3. Total internal reflection

Example 32. Consider an optical communication system operating at λ = 800 nm. Suppose, only 1% of the optical source frequency is the available channel bandwidth for optical communication. How many channels can be accommodated for transmitting audio signals requiring a bandwidth for transmitting audio signals requiring a bandwidth of 8 kHz?

1. 4.8 × 108
2. 48
3. 6.2 × 108
4. 4.8 × 105

Answer: 2. 48

Example 33. A ground receiver station receives a signal at (i) 5 MHz and is transmitted from a ground transmitter at a height of 300 m, located at a distance of 100 km from the receiver station. The signal is coming via. Radius of earth = 6.4 × 106 m. Nmax of isophere = 1012m3.

1. Space wave
2. Skywave propagation
3. Satellite transponder
4. All of these

Answer: 4. All of these

Example 34. The antenna current of an AM broadcast transmitter modulated by 50% is 11 A. The carrier current is

1. 10.35 A
2. 9.25 A
3. 10 A
4. 5.5 A

Solution: 1. 10.35 A

Example 35. If several sine waves with modulation indices n 1, n2, n3,………modulate a carrier wave, then the total modulation index (n) of the wave is

1. n1 + n2…….. + 2 (n1 + n2…)
2. \(\sqrt{n_1-n_2+n_3 \ldots \ldots}\)
3. \(\sqrt{n_1^2+n_2^2+n_3^2 \ldots \ldots . .}\)
4. None Of These

Answer: 3. \(\sqrt{n_1^2+n_2^2+n_3^2 \ldots \ldots. .}\)

Example 36. A transmitter supplies 9 kW to the when unmodulated. The power radiated when modulated to 40% is

1. 5 kW
2. 9.72 kW
3. 10 kW
4. 12 kW

Solution: 2. 9.72 kW

Example 37. In an FM system, a 7 kHz signal modulates a 108 MHz carrier so that the frequency deviation is 50 kHz. The carrier swing is

1. 7.143
2. 8
3. 0.71
4. 350

Solution: 1. 7.143

Example 38. The modulation index of an FM carrier having a carrier swing of 200 kHz and a modulating signal of 10 kHz is

1. 5
2. 10
3. 20
4. 25

Solution 2. 10

Phase Velocity

It is defined as the velocity with which the peak of a sinusoidal pattern is moving. consider wave: cos [Ψ (tx)] cos (kx – wt)

there is peak at Ψ = 0 or θ = 0, so kx – wt = 0

⇒ \(x-w / k t=0 \quad \Rightarrow \quad V p=\frac{w}{k}\)

So for a particular frequency (w) related to time phase velocity is the rate at which the phase of the wave propagates in space.

In the transmission line group of waves travel out with one particular phase velocity is VP = W/K (depending upon frequency ‘w’ oscillation per sec)

⇒ \(V_{\text {group }}=\frac{\mathrm{uw}}{\mathrm{dK}} \quad \omega=\text { wave’s angular frequency }\)

⇒\(V_F=\frac{\text { Ratio of phase velocity }}{\text { velocity of light }}\)

MUF (maximum usable frequency): It is the maximum frequency for a given angle of incidence which gets reflected from the ionosphere. It depends on the angle of incidence.

⇒ \(\text { MUF }=\frac{\text { critical frequency }}{\cos \theta}\)

Relative Permittivity (Dielectric Constant)

The velocity of propagation of a signal in a transmission line is determined mainly by the permittivity of the dielectric material used to construct the line. Permittivity is a measure of the ability of the dielectric material to maintain a difference in electric charge over a given distance.

⇒ \(\varepsilon_{\mathrm{r}}=\frac{\mathrm{C}^2}{\mathrm{~V}_{\mathrm{P}}^2} \quad \Rightarrow \quad \varepsilon_{\mathrm{r}}=\frac{1}{(\text { velocity factor })^2} \quad \Rightarrow \quad \text { velocity factor }=\frac{1}{\sqrt{\varepsilon_{\mathrm{r}}}}\)

where εr = Relative permittivity (dielectric constant)

C = Velocity of light in free space (3 × 108 m/s)

VP = Velocity of propagation (m/s)

Role Of Ionosphere In Radio – Communication

The ionosphere plays a great role in broadcasting, ship and air-craft communication, and navigation by reflecting the radio signals back to the receivers. However, its effectiveness depends on the frequency of the transmitted signal.

This is critical because the behavior of the ionosphere often shows marked differences between day and night. Moreover, it is known for changing its behavior during different seasons. The ionosphere refractive index as represented by the Appleton – Hartee equation is.

⇒ \(n=\left(1-\frac{f_p^2}{f^2}\right)^{1 / 2} \simeq 1-\frac{1}{2} \frac{f_p^2}{f^2}=1-\frac{40.3 N}{f^2}\)

where

fp: electron plasma frequency in Hz

N: electron number density in m–3 from (1), the magnitude of phase velocity can be derived as

⇒ \(V_p=\frac{w}{k}=\frac{w}{n} \cong C\left(1+\frac{40.3 N}{f^2}\right)\)

fP: Depending on the refractive index and frequency the group velocity can be obtained by the equation

⇒ \(vg=\frac{\partial w}{\partial K}=\frac{C}{\left(\frac{\partial(n f)}{\partial f}\right)} \cong C\left(1-\frac{40.3 \mathrm{~N}}{f^2}\right)\)

when the velocity of the wave varies with the frequency, the medium is known as a dispersive medium. because of this dispersion, the idea of group velocity is introduced to represent the velocity of the crest of a group of interfering waves.

⇒ \(\begin{aligned}
& V_p \text { (phase velocity) }=\frac{\omega}{K} \\
& V g \text { (group velocity) }=\frac{d \omega}{d K}
\end{aligned}\)

⇒ \(\begin{aligned}
& =\frac{d}{d k}\left(v_p k\right)=v_p+k \frac{d v_p}{d k} \\
& =v_p+\frac{2 \pi}{\lambda} \times \frac{d v_p}{d(2 \pi / \lambda)} \\
& =v_p+\frac{1}{\lambda} \times \frac{d v_p}{-\frac{1}{\lambda^2} d \lambda}
\end{aligned}\)

⇒ \(\begin{aligned}
& V_g=V p-\lambda \frac{d v_p}{d \lambda} \\
& \omega=2 \pi v=2 \pi \times \frac{E}{h}=\frac{2 \pi m c^2}{h} \\
& \omega=\frac{2 \pi c^2}{h} m_0 \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}
\end{aligned}\)

V = velocity of the particle

⇒ \(K=\frac{2 \pi}{\lambda}=\frac{2 \pi}{h} m v=\frac{2 \pi v}{h} m_0 \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\)

⇒ \(\begin{aligned}
& V_g=\frac{\omega}{K}=\frac{c^2}{V} \\
& V_p V_g=C^2 \quad\left(V_p>c, V g<c\right) \\
& V_g=\frac{d \omega}{d k}=\frac{d \omega / d v}{d k / d v}=V
\end{aligned}\)

Experimental result \(\mu=\left(1-\frac{f_p^2}{f^2}\right)^{1 / 2}=1-\frac{f_p^2}{2 f^2}=1-\frac{40.3 N}{f^2}\)

⇒ \(V_p=\frac{c}{\mu}=C\left(1-\frac{40.3 N}{f^2}\right)^{-1}\)

⇒ \(V_p=C\left(1+\frac{40.3 N}{f^2}\right)\)

VpVg = c2 (=refractive index, vp = phase velocity, vg = group velocity)

⇒ \(V_g=\frac{c^2}{c\left(1+\frac{40.3 N}{f^2}\right)}\)

⇒ \(V g=c\left(1-\frac{40.3 N}{f^2}\right)\)

fp = (N = electron density, fp = electron plasma frequency)

MUF (most useable frequency) \(=\frac{f_c}{\cos i}\)

(Here fc = critical frequency for normal incidence , i = angle of incidence)

⇒ \(\mu=\sqrt{\varepsilon_{\mathrm{r}} \mu_{\mathrm{r}}}\)

Velocity factor \(=V F=\frac{1}{\mu}=\frac{1}{\sqrt{\varepsilon_{\mathrm{r}} \mu_{\mathrm{r}}}}\)

Usually ur=1

therefore \(V F=\frac{1}{\sqrt{\varepsilon_{\mathrm{r}}}}\)