Communication System

1. INTRODUCTION

It is no exaggeration to say that there is no modern life without communication. The humming bees , the crying baby , the barking dogs etc., are all trying to convey something to the world around. We need to understand the language of their communication i.e., the expression and reception of feelings that are involved in this process. It is here, that the technology has made rapid strides and made the task simpler, faster and easier.

Thus, communication is an act of exchange of information between the sender and the receiver. Over decades, methods have been evolved to develop languages, codes, signals etc to make communication effective. Communication through electrical signals has made things much simpler because they can be transmitted over extremely large distances in extremely short time as their speed is 3⨯10⁸ m/s. It is clear that any communication system should have three basic parts

a) the transmitter     b) the medium or channel      c) receiver

The block diagram below describes the process

Some important terms needed to understand the basic elements of communication :
a) Information : It is nothing but , the message to be conveyed . The message may be a symbol , code , group of words etc . Amount of information in message is measured in “bits”

b) Communication Channel : Physical medium through which signals propagate between transmitting and receiving stations is called communication channel.

c) Transmitter: It processess the incoming information and makes it suitable for transmission.

d) Receiver: It receives the signal, amplifies, demodulates and de-codes depending upon the necessity.

e) Transducer: It is a device that changes energy from one form to another.

f) Signal: Information in electrical form such as pulses, suitable for transmision is called signal.

g) Noise: This refers to undesired signals which disturb the transmission and processing of signals.

h) Attenuation: It is the loss of strength of a signal during propagation in a medium.

i) Amplification: It is the process of increasing the strength of the signal (amplitude) using an amplifier.

j) Range: It is the maximum distance from a source upto which the signal is received with suffiecient strength.

k) Repeaters: These are the devices used to increase the range of communication system

l) Band width of signals (speech, T.V and digital data): Band width is the frequnency range over which an equipment operatees.
(or)
It is the portion of the spectrum occupied by the signal.
Communication is basically of two types:
a) point to point:- This takes place between a tansmitter and a receiver. Telephonic conversation between two persons is a good example of it.
b) Broad cast mode:- Here, a large number of receivers receive the information from a single transmitter. Radio and television are good examples of broadcast mode.

Bandwidth of signals:
In a communication system, the message signal may be voice, music, picture or data etc.Each of these signals has a spread of different range of frequencies. Hence, the type of communication system needed depends upon the band of frequencies involved. Speech signal requires the band width of 2800 Hz (3100 Hz to 300Hz). For music, a bandwidth of about 20KHz is required (due to high frequency produced by musical instruments).
The audible range of frequencies extends from 20Hz to 20KHz. Video signals require band width of 4.2 MHz for picture transmission. However, a band width of 6MHz is needed for T.V signals. (as it contains both voice and picture)

Bandwidth of transmission medium :

The most used transmission media are wire, free space, and fibre optic cable. Different transmission media offer different band width. Coaxial cable offers a band width of about 750MHz . Radio wave communication through free space takes place over a wide range of frequencies (from 100kHz – GHz) . This range of frequencies is further classified and allocated for various services as shown below. An optical communication using fibres is performed in the frequency range of 1THz – 1000 THz (microwaves to ultraviolet) Optical fibres can offer transmission bandwidth greater than 100 Ghz.

SOME IMPORTANT WIRELESS COMMUNICATION FREQUENCY BANDS.

Service	Frequency bands	Comments
Standard AM broadcast FM broadcast Television       Cellular Mobile   Satellite Communication	540-1600kHz   88-108 MHZ 54-72 MHz 76-88MHz 174-216MHz 420-890 HMz 896-901MHz 840-935 MHz 5.925-6.425 GHz 3.7-4.2 GHz	VHF (veryhigh frequencies) TV UHF (ultra high frequencies) TV Mobile to base station Base station to mobile Uplink Downlink

Propagation of electro magnetic waves in the atmosphere: In the radio wave communication, an antenna at the transmitter radiates the electro magnetic waves. These travel through the space and reach the receiving antenna. As the waves travel, their strength keeps on decreasing. Many factors influence the path and the progress of E.M wave.
The earth’s atmosphere plays a vital role in the propagation of these waves.
Principle of communication: Consider two friends playing with a ball in a closed room. One friend throws the ball (transmitter) and the other receives the ball (receiver). There are three ways in which the ball can be sent to the receiver.
(i) By rolling it along the ground (ii) Throwing directly and (iii) Throwing towards roof and then reflected towards the receiver. Similarly there are three ways of transmitting information from one place to the other using physical space
around the earth. They are :

(i) Along the ground (ground waves)

(ii) Directly in a straight line through intervening topographic space (space wave, or tropospheric wave or surface wave) and

(iii) Upwards in sky followed by reflection from the ionosphere (sky wave) Hence, the three important methods of radio wave Propagation are :
a) Ground wave propagation
b) Sky wave propagation
c) Space wave propagation
a) Ground wave propagation: In this method, the radio waves are guided along the surface. The wave induces charges on the earth. These charges travel with the wave and this forms a current . Now the earth behaves like a leaky capacitor in carrying the induced current.The wave loses some energy, as energy is spent due to flow of charge through the earth’s resistance.The wave also looses energy due to diffraction as it glides along the ground.The loss of energy increases as the frequency increases. Thus ground propagation is suitable upto 2MHz . As they loose energy
they cannot go to long distances on the ground.Maximum range of the ground wave can be increased by increasing the power of the transmitter.
b) Sky wave propagation: Above 2MHz and upto 30 MHz , long distance communication takes place through ionosphere. The ionosphere reflects the radio waves back to the earth. This method is called sky wave propagation. It is used for shortwave broad casting services. Ionosphere is a thick blanket of 65 km to 400 km above the earth’s surface. UV rays and other higher energy radiation coming from space results in the ionisation of air molecules.The ionosphere is further divided into serval layers as shown in table below.It should be understood that degree of ionisation changes with height. This is because the density of atmosphere decreases with height. At great heights, the radiation is intense, but the molecules available are few. On the other hand , near the earth’s surface the molecular concentration is high but the intensity of radiation is low and thus again the ionisation is low. Logically, the peak of
ionisation density occurs at some intermediate heights.The ionosphere acts as a mirror (reflector) for frequcies of 3-30 MHz. Electromagnetic waves of frequences greater than 30 MHz pass through the atmosphere and skip.

The process of bending of EM waves is similar to total internal reflection in optics. The bending of waves can be easily explained on the basis of variation of refractive index of the ionosphere with change in electron density. Suppose that a radio wave enters the ionosphere from the underlying unionized medium. Since the refractive index of ionosphere decreases from D layer to F2 layer, consequently, the incident ray will move away from the normal drawn at the point of incidence following the ordinary laws of refraction

During the propagation in ionosphere the angle of refraction gradually increases and the ray goes on bending more and more till at some point , the angle of refraction becomes 900 and the wave travels parallel to the earth surface . This point is called point of reflection . Then the ray tends to move in the down ward direction and comes back to earth because of symmetry.Super high frequency (SHF) waves propagate as sky waves taking reflection at satellite.

The sky wave propagation can cover a very long distance and so round the globe communication is possible.
c) Space wave propagation: This method is used for line-of-sight [LOS] communication and also for satellite communication. At frequentcies above 40MHz, communication is mainly by LOS method. At such frequencies, relatively smaller antenna can be errected above the ground. Because of LOS propagation, the direct waves get blocked, at some point due to the curvature of the earth as shown in the figure.

For the signal to be received beyond the horizon, the receiving antenna must be high enough to intercept the LOS waves. If the transmitting antenna is at a height h_T then it can be shown that the distance to the horizon d_T is given by

${\mathrm{d}}_{\mathrm{T}}=\sqrt{2{\mathrm{Rh}}_{\mathrm{r}}}$ where ‘R’ is the radius of earth. Similarly if the receiving antenna is at a height ${\mathrm{h}}_{\mathrm{R}}$, the distance to the horizon d_R is ${\mathrm{d}}_{\mathrm{R}}=\sqrt{2{\mathrm{Rh}}_{\mathrm{R}}}$.

∴ The maximum distance ${\mathrm{d}}_{\mathrm{M}}$ between the two antennas is ${\mathrm{d}}_{\mathrm{M}}=\sqrt{2{\mathrm{Rh}}_{\mathrm{T}}}+\sqrt{2{\mathrm{Rh}}_{\mathrm{R}}}$ where

R = Radius of the earth.

h_T = height of the transmitting antenna and

h_R = height of the receiving antenna.

If the Population density around the tower is given, the number of persons covered by the transmistting tower = (Area covered by the tower) ⨯ Population density.

∴ No. of persons covered = ${\mathrm{\pi d}}^2$ ⨯ Population density (Here d= radius of the area covered by single transmitting tower of height h_T)

Television broadcast, microwave and satellite communications are a few examples of Communication systems that use space wave propagation. The figure below illustrates the various modes of wave propagation.

Range of TV transmission:
As the frequency range of TV signals is 100-200 MHz, such signal transmission via ground waves is not possible. In such situations, we use line of sight transmission.

Let CP be the TV tower on the earth’s surface. It’s antenna is at P . Let PC=h. When TV broadcast is made, the signal can reach the earth upto A to B. There will be no reception of the signal beyond A and B. Arc length CA and CB is the range of TV transmission. If O is the centre of the earth, OA=OB=R is the radius of the earth, from right angled triangle OAP

${\mathrm{OP}}^2={\mathrm{OA}}^2+{\mathrm{PA}}^2$

$(\mathrm{h}+\mathrm{R}{)}^2={\mathrm{R}}^2+{\mathrm{PA}}^2$

PA = PB = d

$(\mathrm{h}+\mathrm{R}{)}^2={\mathrm{R}}^2+{\mathrm{d}}^2$

${\mathrm{h}}^2+{\mathrm{R}}^2+2\mathrm{Rh}={\mathrm{R}}^2+{\mathrm{d}}^2$

As h <<R we can ignore ${\mathrm{h}}^2$

${\mathrm{d}}^2=2\mathrm{Rh}\text{ and }\mathrm{d}=\sqrt{2\mathrm{Rh}}$

ange of TV transmission depends upon the height of the transmission antenna. Broadcasts are made from tall transmitting antenna.
Satellite Communication: Long distance communication beyond 10 to 20 MHz was not possible before 1960 because all the three modes of communication discussed above failed (ground waves due to conduction losses, space wave due to limited line of sight and sky wave due to the penetration of the ionosphere by the high frequencies beyond $f_c$).

[** Note: Ionosphere behaves as a rarer medium by which carrier wave is reflected back if its frequency f$\left(\le f_c\right)$.

where $f_c$ is called a “critical frequency” and is given by ${\mathrm{f}}_{\mathrm{c}}\approx 9{\left({\mathrm{N}}_{\max }\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$.

N_max = maximum electron density.]

Satellite communication made this possible.The basic principle of satellite communication is shown in figure. A communication satellite is a spacecraft placed in an orbit around the earth. The frequencies used in satellite communication lie in UHF/microwave regions. These waves can cross the ionosphere and reach the satellite.

For steady, reliable transmission and reception it is preferred that satellite should be geostationary. A geostationary satellite is one that appears to be stationary relative to the earth. It has a circular orbit lying in the equatorial plane of the earth at an approximate height of 36, 000 km. Its time period is 24 hours. If we use three geostationary satellites placed at the vertices of an equilateral triangle as shown in figure the entire earth can be covered by the communication network.
Modulation : It is clear that low frequency waves, can not travel long distances. Hence, to transmit low frequency wave over long distance, we take the help of high frequency waves called carrier wave. The low frequency wave is superposed over high frequency carrier wave. This process is called the modulation. The low frequency wave is called the modulating wave and the high frequency wave is called the carrier wave, and the resultant wave is called modulated wave. In this section we will discuss in detail about modulation. What is it ? What is the need of modulation or how
modulation is done etc.
No signal in general, is a single frequency but it spreads over a range of frequencies called the signal bandwidth. Suppose we wish to transmit an electronic signal in the audio-frequency (20Hz-20kHz) range over a long distance. Can we do it ? No it cannot because of the following problems.
i) Size of antenna: For transmitting a signal we need an antenna. This antenna should have a size comparable to the wavelength of the signal. For an electromagnetic wave of frequency 20kHz, wave length is 15km. Obviously such a long antenna is not possible and hence direct transmission of such signal is not practical

ii) Effective power radiated by an antenna: Power radiated by an antenna $\alpha \frac{1}{{\left(\lambda \right)}^2}$. Therefore power radiated by large wavelength would be small. For good transmission we require high power and hence high frequency transmission is required.
iii) Mixing up of signal from different transmitters : To avoid intermixing of different signals a band of frequencies is alloted to different users. For telephones , band of frequency alloted is 300Hz to 3.4kHz. Suppose you are transmitting at around 1MHz and you are allowing 10kHz band width to each user. So different users or transmitters can transact at frequencies 1.00$\pm$0.005 MHz, 1.01$\pm$0.005MHz, 0.99$\pm$0.005MHz etc. Therefore many channels get allowed if the transmission frequency is high. This is achived by a process called modulation.

Modulation Types: Different types of modulation exist depending upon the specific characterisitc of the carrier wave which is being varied in accordance with the message signal. We know that a sinusoidal carrier wave can be expressed as:

$E=E_0\sin (\omega t+\varphi )$

The three distinct characterisitcs are :

Amplitude (E₀), angular frequency ($\omega$) and phase angle $\varphi$ . Either of these three characteristics can be varied in accordance with the signal. The three types of mudulation are, amplitude modulation (AM), frequency modulation (FM) and phase modulation (PM).
Similarly the characteristics of a pulse are:
pulse amplitude, pulse duration or pulse width and pulse position (time of rise or fall of the pulse amplitude) can be modulated in accordance with the message signal.

AMPLITUDE MODULATION (AM) :

In amplitude modulation, the amplitude (voltage) of a carrier singal is varied in accordance with the intensity of the signal to be transmitted while there is no change in frequency and phase of the carrier wave. Here the peak amplitude of a high frequency sinusodial carrier wave is varied in propagation to the instantaneous amplitude of the modul at ing wave. Greater the amplitude of modulating wave greater are the fluctuations in the amplitude of the
modulated carrier wave. Figure shows graphically the principle of amplitude modulation. It may be seen that the amplitude of both positive and negative half cycles of carrier wave vary in accordance with the instantaneous amplitude

When the modulating signal increases in the positive sense, the amplitude of the modulated wave also increases but during the negative half cycle of signal, the amplitude of modulated carrier wave decreases. Thus the rate at which fluctuations in the amplitude of moudulated wave vary depends on the frequency of modulating wave. The ratio of amplitude change of carrier wave to the amplitude of normal waves is called modulation factor (or) modulation index (m). It determines the strength and quality of transmitted signal. In India amplitude modulation is used in radio broad
casting and for picture signal transmission in television, while frequency modulation is used for sound signal transmission in T.V.

$m=\frac{\text{ Amplitude change of carrier wave }}{\text{ Amplitude of normal carrier wave }}\left(\text{ or }\right)$

$m=\frac{\text{ Maximum value of signal wave }}{\text{ Maximum value of carrier wave }}=\frac{B}{A}$

It is generally expressed in percentage

Percentage of modulation = $\frac{B}{A}\times 100$

example :

$m=\frac{(A+B)–(A–B)}{(A+B)+(A–B)}\times 100=\frac{B}{A}\times 100$

i) A = 1, B = 0.5, m = $\frac{0.5}{1}\times 100=50\%$

ii) If $A=1, B=1, m=\frac{1}{1}\times 100=100\%$

iii) $A=0.5, B=1, m=\frac{1}{0.5}\times 100=200\%$

When modulation index m>1 (or more than 100%) it is said to be over modulation. Actually if modulation index exceeds 100% it produces several distrotion and interference in the transmitter output. Hence modulation index should never exceed 100%. Greater the value of m, the stronger and clearer will be the signal.
If the instantaneous voltage of carrier wave is

$e_c=E_c\sin {\omega }_{c}t=E_c\sin 2\pi f_{c}t$ and the instataneous voltage of modulating wave (signal) is $e_s=E_s\sin {\omega }_{s}t=E_s\sin 2\pi f_{s}t$

The instantaneous voltage of modulated wave is

$E=E_c\sin {\omega }_{c}t+\frac{m{E}_c}{2}\cos \left({\omega }_c–{\omega }_s\right)t–\frac{m{E}_c}{2}\cos \left({\omega }_c+{\omega }_s\right)t$

It may be concluded that modulated wave is the summation of three sinusoidal waves.

i) First having amplitude $E_c$ and frequency $f_c$

ii) Second having amplitude $\frac{m{E}_e}{2}$ and frequency $\left(f_c–f_s\right)$

iii) Third having amplitude $\frac{m{E}_c}{2}$ and frequency $\left(f_c+f_s\right)$

Therefore the process of modulation does not change orginal frequency fc but processes two new frequencies $\left(f_c+f_s\right)$ amd $\left(f_c–f_s\right)$ which are called upper side and lower side band frequencies. The difference between upper side band frequency and lower side band frquency is called band width or channel width. Band width = $2f_s$

The power carried by the carrier wave is $P_c={\left(E_c\right)}^{2}r.m.s/R=E_c^2/2R$ and that of each side band is

${\mathrm{P}}_{\mathrm{SB}}=\frac{{\left(\frac{{\mathrm{mE}}_{\mathrm{C}}}{2}\right)}^2\mathrm{rms}}{2\mathrm{R}}=\frac{{\mathrm{m}}^2{\mathrm{E}}_{\mathrm{C}}^2}{8\mathrm{R}}$

Where R is the resistance of the antenna.

The total power carried by modulated wave is

$P_t=P_c+P_{SB}$

$P_t=\frac{E_c^2}{2R}\left(1+\frac{m^2}{2}\right)$

$P_t=P_c\left(1+\frac{m^2}{2}\right)$

therefore , the fraction of total power carried by side bands

$\frac{P_{SB}}{P_T}=\frac{m^2}{2+m^2}$.

Hence the useful power is in the side bands which depends upon the value of modulation factor m. Greater the value of m greater is useful power carried by the side bands.

FREQUENCY MODULATION (FM) :

Here the frequency of the carrier wave is changed in accrodance with the intensity of the signal while the amplitude and phase of the carrier wave remain unchanged. The amount by which the carrier frequency is varied from its unmodulated value is called the deviation.

PHASE MODULATION (PM) :

In this the phase of the carrier wave is changed while the amplitude and frequency remain unchanged.

Advantages of FM over AM:

i) FM reception is quite immune to noise as compared to AM reception. In FM receivers, the noise can be reduced by increasing the frequency deviation.

ii) FM transmission is highly efficient as compared to AM transmission. In FM transmission, all the transmitted power is useful, whereas in AM tranmission most of the power goes waste in the transmitted carrier, which contains no useful information.

iii) Due to a large number of side bands, FM transmission can be used for the stereo sound tranmission.

Disadvanges of FM:

i) The bandwidth in FM transmission is about 10 times as large as that needed in AM transmission. As a result, much wider frequency channel is required in FM transmission.

ii) FM reception is limited to line of sight. Due to this, area of reception for FM is much smaller than that for AM.

iii) FM transmitting and receiving equipments are very complex as compared to those employed in AM transmission.

SOLVED EXAMPLES

1. How many AM broadcast stations can be accomodated in a 100 kHz bandwidth if the highest modulating frequency of carrier is 5 kHz ?
Sol. Any station being modulated by a 5 kHz signal will produce an upper side frequency 5 kHz above its carrier and a lower side frequency 5 kHz below its carrrier, thereby requiring a bandwidth of 10 kHz. Thus, Number of stations accommodated

$\frac{\text{ Total bandwidth }}{\text{ Band width per station }}=\frac{100}{10}=10$

2. How many 500 kHz waves can be on a 10 km transmission line simultaneously ?

Sol. Let λ be the wavelength of 500 kHz signal. Then,

$\lambda =\frac{c}{f}=\frac{3.0\times {10}^8}{5.0\times {10}^4}m=600m$

The number of waves on the line can be found from,

$n=\frac{d}{\lambda }=\frac{10\times {10}^3}{600}=16.67$

3. A two wire transmission line has a capacitance of 20 pF/m and a characteristic impedance of 50 $\Omega$

(a). What is the inductance per metre of this cable ?

(b) Determine the impedance of an infinitely long section of such cable.

Sol a) The characteristic impedance.

$\mathrm{Z}=\sqrt{\mathrm{L}/\mathrm{C}}$

$\mathrm{L}=\left({\mathrm{Z}}^2\right)\left(\mathrm{C}\right)$

$=(50{)}^2\left(20\times {10}^{–12}\right)\mathrm{H}$

$=0.05\mu H$

b) The characteristic impedance of a transmission line is the impedance that an infinite length of line would present to a power supply at the input end of the line. Thus, $Z_{\infty }=Z_0=50\Omega$

4. T.V. transmission tower at a particular station has a height of 160 m.

(a) What is the coverage range ?

(b) How much population is covered by transmission, if the average population density around the tower is 1200 per km2 ?

Sol. a) Coverage range $d=\sqrt{2gh}$

$=\sqrt{2\times 6400\times {10}^3\times 160 }\mathrm{m}$

= 45.254 km.

b) Population covered = (population density) ⨯ (area covered)

$=\left(1200\right)\times \left(\pi d^2\right)$

$=\left(2400\pi Rh\right)$

= 2400⨯3.14⨯6.4⨯${10}^3$⨯0.16

= 77.17 lac

c) Coverage range $\propto \sqrt{h}$

Therefore coverage range can be doubled by making height of the tower four times to 640m. So, height of the tower should be increased by 480 m.

5. An audio signal given by $e_s=15\sin 2\pi \left(2000t\right)$ modulates a carrier wave given by $e_s=60\sin 2\pi (100,000t)$.
If calculate (a) Percent modulation (b) Frequency spectrum of the modulated wave.

Sol. (a) Signal Amplitude, B = 15

Carrier amplitude, A=60

$\mathrm{m}=\frac{B}{A}=\frac{15}{60}=0.25$

∴ Percentage modulation = 0.25⨯100 = 25%

(b) By comparing the given equations of signal and carrier with their standard form

$e_s=E_s\sin {\omega }_{s}t=E_s\sin 2\pi f_{s}t$ and

$e_c=E_c\sin {\omega }_{c}t=E_c\sin 2\pi f_{c}t$

we have signal frequency $f_s$ = 2000Hz and carrier

frequency $f_c$=100,000Hz

The frequencies present in modulated wave

(i) $f_c=100,000 \mathrm{Hz}=100\mathrm{kHz}$

(ii) $f_c–f_s=100,000–2000=98\mathrm{kHz}$

(iii) $f_c+f_s=100\mathrm{kHz}+2\mathrm{kHz}=102\mathrm{kHz}$

Therefore, frequency spectrum of modulated wave extends from 98kHz to 102 kHz is called band width.

6. The antenna current of an AM transmitter is 8A when only the carrier is sent but it increases to 8.93A when the carrier is modulated. Find percent modulation.

Sol. The modulated or total power carried by AM wave

${\mathrm{P}}_{\mathrm{T}}={\mathrm{P}}_{\mathrm{C}}\left(1+\frac{{\mathrm{m}}^2}{2}\right)$ If R is load resistance, I_m is the current when carrier is modulated and I_c the current when unmodulated, then

$\frac{{\mathrm{P}}_{\mathrm{T}}}{{\mathrm{P}}_{\mathrm{C}}}=\frac{{\mathrm{I}}_{\mathrm{m}}^2\mathrm{R}}{{\mathrm{I}}_{\mathrm{c}}^2\mathrm{R}}$

$\therefore \mathrm{I}+\frac{{\mathrm{m}}^2}{2}=\frac{{\mathrm{I}}_{\mathrm{m}}\mathrm{R}_2}{{\mathrm{I}}_{\mathrm{c}}\mathrm{R}_2}$

Given ${\mathrm{I}}_{\mathrm{m}}=8.93 \mathrm{A}, {\mathrm{I}}_{\mathrm{c}}=8 \mathrm{A}$

$\therefore {\mathrm{m}}^2=2\left[{\left(\frac{8.93}{8.0}\right)}^2–1\right] \therefore \mathrm{m}=0.7$

Therefore, percentage modulation=70%

7. A sinusoidal carrier voltage of 80 volts amplitude and 1 MHz frequency is amplitude modulated by a sinusoidal voltage of frequency 5kHz producing 50% modulation. Calculate the amplitude and frequency of lower and upper side bands.

Sol. Amplitude of both LSB and USB are equal and given by

$=\frac{{\mathrm{mE}}_{\mathrm{c}}}{2}=\frac{0.5\times 40}{2}=20\text{ volts }$

Now frequency of LSB = $f_c–f_s$

= (1000 – 5) kHz = 995 kHz

Frequency of USB = $f_c+f_s$

= (1000 + 5) kHz = 1005 kHz

8. The load current in the transmitting antenna of an unmodulated AM transmitter is 6 Amp. What will be the antenna current when modulation is 60%.

Sol. Total power carried by AM wave

$P_T=P_C\left(1+\frac{m^2}{2}\right)$…………(1)

where ${\mathrm{P}}_{\mathrm{c}}$ is the power of carrier component and m is the modulation factor. If R is the resistance, Im the antenna load current when modulation is 60% and ${\mathrm{I}}_{\mathrm{c}}$ is the antenna load current when un modulated, then

$\frac{{\mathrm{P}}_{\mathrm{T}}}{{\mathrm{P}}_{\mathrm{C}}}=\frac{{\mathrm{I}}_{\mathrm{m}}\mathrm{R}_2}{{\mathrm{I}}_{\mathrm{c}}\mathrm{R}_2}, \therefore 1+\frac{{\mathrm{m}}_2}{2}=\frac{{{\mathrm{I}}_{\mathrm{m}}}^2}{{\mathrm{I}}_{\mathrm{c}}^2}$   using (1)

or   $I_m=I_c\sqrt{\left\{\left(1+\frac{m^2}{2}\right)\right\}}$

Given ${\mathrm{I}}_{\mathrm{c}}$ = 6Amp, m = 0.6

$I_m=6{\left[1+\frac{(0.6{)}^2}{2}\right]}^{1/2}=6[1.086]=6.52\mathrm{Amp}.$

9. A carrier wave of 1000 W is subjected to 100% modulation. Calculate (i) Power of modulated wave, (ii) Power in USB, (iii) Power in LSB.

Sol. (i) Total power of modulated wave

$P_T=P_C\left(1+\frac{m^2}{2}\right)=1000\left(1+\frac{1^2}{2}\right)=1500\mathrm{watt}$

(ii) Power in USB = $\frac{1}{2}{P}_{SB}$

where power carried by side bands is given by Amplitude Modulation and Detection

$P_{SB}=P_C\left(\frac{m^2}{2}\right)=1000\left(\frac{1^2}{2}\right)=500\mathrm{watt}$

$P_{USB}=\frac{1}{2}{P}_{SB}=\frac{1}{2}\times 500=250\mathrm{watt}$

(iii) Since power in LSB = Power in USB

$P_{LSB}=P_{USB}=250\text{ watt }$