Please refer to CBSE Class 12 Physics HOTs Communication Systems. Download HOTS questions and answers for Class 12 Physics. Read CBSE Class 12 Physics HOTs for Chapter 15 Communication Systems below and download in pdf. High Order Thinking Skills questions come in exams for Physics in Class 12 and if prepared properly can help you to score more marks. You can refer to more chapter wise Class 12 Physics HOTS Questions with solutions and also get latest topic wise important study material as per NCERT book for Class 12 Physics and all other subjects for free on Studiestoday designed as per latest CBSE, NCERT and KVS syllabus and pattern for Class 12
Chapter 15 Communication Systems Class 12 Physics HOTS
Class 12 Physics students should refer to the following high order thinking skills questions with answers for Chapter 15 Communication Systems in Class 12. These HOTS questions with answers for Class 12 Physics will come in exams and help you to score good marks
HOTS Questions Chapter 15 Communication Systems Class 12 Physics with Answers
COMMUNICATION SYSTEMS
1 Mark:
ONE MARK QUESTIONS:
Question. Name the mode of propagation of radio waves which travel in a straight line from the transmitting antenna to the receiving antenna.
Answer : Space waves
Question. Name the type of communication that uses carrier signal having frequencies in the range of 1012 Hz to 1016 Hz.
Answer : Optical Communication using fibre optics
Question. Define modulation index as referred to AM.
Answer : Modulation index µ = Am /Ac
Where Am = modulating signal
Ac = carrier signal
Question. What should be the frequency of carrier wave ?
Answer : It should be from kHz to GHz.
Question. Why short wave band is used for long distance radio broadcast?
Answer : (i) As it has high frequency hence It can travel long distance without getting absorbed by earth’s atmosphere.
TWO MARKS QUESTIONS:
Question. With the help of diagram show the basic constituents of communication system.
Answer : 
Question. Write two basic modes of communication with atleast one example.
Answer : (a)Point to point –Telephone
(b) broadcast-Radio and Television
Question. Write the function of transmitter and receiver in communication system.
Answer : (i) Transmitter-Processes the incoming message/information signals so as to make it suitable for transmission and transmits it to the receiving end through a channel
(ii) Receiver-recovers the desired original message signals from the received signals at the end of channel.
Question. Write the function of transducer and repeater in communication system.
Answer : . (i) Transducer -It is a device that converts one form of energy to another form.
(ii) Repeater – It increases the range of transmission of signal.
Question. Why is the amplitude of modulating signal kept less than the amplitude of carrier wave?
Answer : To keep modulation index less than 1 so that noise level can be kept small in the signal.
Question. A TV tower has a height of 400 m at a given place. Calculate its coverage range if the radius of the earth is 6400 km.
Answer : Coverage Range dm = √2Rh = 71.5 km
Where R is radius of the earth , h is the height of the tower
THREE MARKS QUESTIONS
Question. Explain the process of amplitude modulation. Draw a schematic sketch showing how amplitude modulated signal is obtained by superposing a modulating signal over a sinusoidal carrier wave.
Answer : In Amplitude modulation, the amplitude of the carrier wave is made to vary with time in accordance with the amplitude of modulating signal. (Image)
Question. (i) Write the factors that prevent a baseband signal of low frequency to be transmitted over long distances.
(ii) What is to be done to overcome these factors?
Answer : (i)(a) Size of the antenna should be comparable to the wavelength
(b)Effective power radiated by the antenna
(c) Mixing up of signals from different transmitters
(ii) modulation
Question. A message signal of frequency 20 KHz and peak voltage of 20 volts is used to modulate a carrier signal of frequency 2 MHz and peak voltage of 40 volts. Determine (i) modulation index, (ii) the side bands produced. Draw the corresponding frequency spectrum of amplitude modulated signal.
Answer : (i) Modulation Index = Am/ Ac = 20/40 = 0.5
The side bands are (2000 + 20) KHz = 2020 KHz and
(2000 - 20) KHz = 1980 KHz
Amplitude versus ω for amplitude modulated signal : 
Question. What does the process of detection of amplitude modulated wave mean? The amplitude modulated carrier wave of angular frequency ωc , contains the frequencies ωc ,(ωc + ωm) and (ωc -- ωm) ,where ωm is the angular frequency of the modulating signal. Discuss ,in brief , with the help of a block diagram the essential details of a simple method used for detecting the modulating signal from this modulated carrier wave.
Answer : Detection is the process of recovery of the modulating signal from the modulated carrier wave. The amplitude modulated carrier wave contains three frequencies ωc , (ωc + ωm) and (ωc -- ωm) . In order to detect the original message signal m(t) of angular frequency ωm , a method is shown by the block diagram (Image)
Question. In the block diagram of a simple modulator for obtaining an AM signal identify the boxes A and B . Write their functions.
Answer : A -square law device ,used to produce output voltage or current that is proportional to its input voltage or current
B -band pass filter, it allows specific range of frequencies to pass to the next stage.
1. At a particular place at a distance of 10km from a transmission station a person can receive signals but not able to receive signals at 100km, suggest a method how he can receive signal at 11 km
2. A device is used to communicate through compute to computer name the device.
3. Why ground wave propagation is not suitable for high frequency?
4. Why microwaves are being used in RADAR?
5. Name the type of communication that uses carrier signals having frequencies in the range 1012 Hz.
6. Why long distance radio broadcasts use short-wave bands?
7. A radio can tune to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?
8. Name the process by which exact reproduction of a document at a distant place can be received.
9. The transmitter A and receiver B are not visible to each other on earth surface even then they communicate to each other name the processes of communication.
10. How does the effective power radiated by an antenna vary with wavelength?
2 Marks
11. It is necessary to use satellites for long distance T.V. transmission.Justify?
12. With the help of necessary diagram make it clear that “taller the antenna, greater the coverage of the Television broadcast”.
13. We do not choose to transmit an audio signal by just directly converting it to an e.m. wave of the same frequency, Give two reasons for the same.
14. Distinguish between ‘point to point’ and ‘broadcast’ communication modes. Give one example of each.
15. A transmitting antenna is 32 m high and the receiving antenna 100 m. Calculate the maximum. Distance between them for satisfactory communication in LOS mode. Assume radius of earth 6.4 x 106 m.
3 Marks
16. What does the term LOS communication mean? Name the types of waves that are used for this Communication which of the two-height oftransmitting antenna and height of receiving antenna - Can affect the range over which this mode of communication remains effective?
17. A schematic arrangement for transmitting a message signal (20 Hz to 20 kHz) is given below:
Give two drawbacks from which this arrangement suffers. Describe briefly with the help of a block diagram the alternative arrangement for the transmission And reception of the message signal.
18. Frequencies higher than 10 MHz are found not to be reflected by the ionosphere on a particular Day at a place, calculate the maximum electron density of the ionosphere.
19. A message signal of frequency 15 kHz and peak voltage of 5 volts is used to modulate a carrier of Frequency 1 MHz and peak voltage of 20 volts, Determine (a) Modulation index (b) The side bands Produced
20. What is meant by ‘detection’ of a modulated carrier wave? Describe briefly the essential steps for detection
ANSWERS
(1) By using antenna
(2) Modem
(3) attenuation/power loss
(4) Linear propagation
(5) Optical fibers
(6)
(7) 40m, 25m
(8) Fax
(9) Satellite communication
(10) p α 1/λ2
(11)
(12)
(13) For transmitting an EM wave signal the minimum size of the antenna For an audio frequency wave the size of antenna will be extra large which is not feasible.
(ii) Effective power radiated by antenna is proportional to square of frequency. For an audio Frequency wave the radiated power will be extremely small.
(iii) If different programmes at audio frequencies are directly transmitted then on account Of limited bandwidth these programmes will get mixed up leading to disturbance.
(14) In point-to-point communication mode, the communication takes place over a link between a single transmitter and a receiver. Telephony is an example of point-to-point communication. In broadcast mode, there are a number of receivers corresponding to a single transmitter. Radio and television are examples of broadcast mode of communication.
(15) 45.5km
(16)
(17) (i) Signals cannot go very far without employing large amount of power, because Modulation is not done. (ii) Bandwidth is very short. Alternative arrangement
(18)
(19) (a) 0.25
(b) 1015 kHz, 985 kHz
(20)
Communication Systems
1. Sky wave transmission of electromagnetic wave cannot be used for TV transmission why?
TV transmission takes place at higher frequencies (approx 18 MHz to 200 MHz). Such higher frequencies do not return back to the earth.
2.In digital communication, PCM (Pulse Code Modulation) is preferred than PAM (Pulse Amplitude Modulation). Justify your comment.
(i) PCM is free from noise in interfering signals. It is also coded electrical signal.
(ii) It permits use of repeater for long distance transmission.
3.In sending a binary data over an analog transmission line, what kind of device does the conversion?
MODEM
4.For what kind of signal would you use for encoding and decoding process?
Pulse Code Modulation -PCM is free from noise in interfering signals.
5.When a teacher lectures in the class, is the classroom a full duplex communication system or a half duplex communication system?
In the full duplex mode, transmissions are possible in both directions simultaneously . A standard telephone system is an example of full-duplex transmission.
6. Your wireless notebook computer works on one desk, but when you move it to a nearby desk, you cannot receive a signal. What may be happening? How night you fix the problem?
By increasing the frequency range.(wide Band width)
7.Express the following I.P address in binary:
128.171.17.13
Hint: 128 is 10000000.
8.Convert the following address in binary to decimal notation:
11110000 10101010 00001111 11100011
Hint: 11110000 is 240 in decimal.
9.A tester shows that a UTP (Unshielded Twisted Pair) cord has too much cross-stalk interference. Why?
A standard UTP cord contains 4 wire pairs. The pairs in the bundle will radiate some of the energy producing electromagnetic (EM) interference in other pairs. This mutual EMI among the wire pairs in UTP cord is called cross-stalk interference. The twisting of each pair normally keeps cross-stalk interference to a reasonable level. Installer must be careful not to untwist the wire more than 1.2cm, when addng connectors. This limit will not eliminate terminal cross-stalk interference but it should limit to a acceptable level.
10.How do we make television broadcasts for larger coverage and for long distance?
By using (i) tall antennas which is familiar landmark in many cities and (ii) using artificial satellites —called geostationary satellites. Since television signals are of high frequency and are not reflected by ionosphere so we use satellites to get them reflected & transmission of TV signals can be used for larger coverage as well as for long distance.
11.As we know parallel transmission is faster than serial transmission. Name the device which converts parallel to serial transmission.
Multiplexer
12.In normal radio operation how does channel bandwidth usually related to bandwidth required to transmit a data stream of a given speed.
If you want to transmit at a given bit rate, you will need a channel whose bandwidth is sufficiently wide. Multiplexer is used to convert serial transmission into parallel.
13.Optical Radio telescopes are built on ground but X-Rays telescopes are possible only from satellites orbiting earth. Why? Atmosphere absorbs X-rays while visible and radio waves can penetrate it.
14.Do you think P2P architectures will be popular in the future .Why?
Peer to peer communication applications are likely to appear in near future offering a much broader spectrum of services.
15. Scientists put x-ray astronomical telescope on the artificial satellite orbiting above the earth's atmosphere whereas they build optical and radio-telescopes on the surface of the earth. Why ?
X-rays have very high frequency and much smaller wavelength. These rays get absorbed by the earth's atmosphere. On the other hand, optical (visible) radiations and radio-waves can pass through the atmosphere. That is why optical and radio telescopes can be installed on the earth's surface.
16. In TV transmission which is used –either A.M or F.M?
In TV transmission FM is used for Sound signal and AM for picture signal.
17. In demodulation stage , how the RF signal is removed from AF?
Rectifier is used to rectify the modulated wave (i.e) negative halt of the modulated wave is eliminated. It separates the audio signal from the carrier wave.
18) In the transmission of television signals sky waves are not used Give reason?
The television signals have frequencies in 100-200 MHz range. As ionosphere cannot reflect radio waves of frequency greater than 40 M back to earth, the sky waves cannot be used in the transmission of TV signals.
19) ‘ F2 layer’ plays an important role in communication . Why?
F2is the top most layer of ionosphere. Its height is 240-400 Km in day time where as at night it falls from 400-300Km. It is very thick layer and most of its molecules are ionized. It is the most useful reflecting layer for high frequency radio waves.
20) The strength of received signal is almost constant in the receiving stage. How can you achieve this?
AGC stands for Automatic Gain Control. It is used in receiver so that the strength of received signal is almost constant.
21) In satellite communication, how transponder helps as to Uplink and down link signals? In satellite communication, uplink signals are transmitted from earth stations.
The transponder receives these signals, convert them to different frequencies and retransmits them to back to earth as downlink signals.
22) Why is delta modulation a convenient method of digital modulation.
Delta modulation involves simple pulse coding and decoding methods. A simple delta modulation uses just one bit per sample i.e. a ‘non-zero’ sample or one per sample. Thus, this method is convenient to use.
23. In an optical fibre why the glass fibres have high refractive index as compare to thin layer of coating on it ?
The optical fibres are normally made of glass of high refractive index coated with a thin layer of glass of low refractive index to provide a suitable boundary. It this is not done the phenomenon of total internal reflection is not possible inside the optical fibre or the light would pass from one fibre to another fibre.
24.What is necessary for full duplex operation using a two wire circuit? Give an example of full-duplex transmission.
In the full duplex mode, transmissions are possible in both directions simultaneously but they must be between the same two stations. A standard telephone system is an example of full-duplex transmission.
25.What is necessary for full duplex operation using a two wire circuit? Give an example of full-duplex transmission.
In the full duplex mode, transmissions are possible in both directions simultaneously but they must be between the same two stations. A standard telephone system is an example of full-duplex transmission.
26) Why does the electrical conductivity of earths atmosphere increase with altitude ?
Atmospheric pressure decreases with in crease in altitude. The high energy particles (ie & rays and cosmic rays) coming from outer space and entering our earths atmosphere cause ionization of the atoms of the gases present there . The ionizing power of these radiation decreases rapidly as they approach the earth. due to decrease in number of collision with the gas atoms . It is due to this reason that the electrical conductivity of earths atmosphere increases with altitude
27) If you want transmit an audio signal for long distance , How it can be made possible?
1 .The length of the antenna required is so large (L = 5000m ) that is practically impossible to set up it.
2 The energy radiated from the antenna in audio frequency range is Practically zero
3 The audio signals transmitted from the different broadcasting stations will get inseparably mixed.
28) MUF helps us to receive signal by the applicataion of ionosphere .How it helps us?
Maximum usable frequency (MUF) is defined as the highest frequency that is returned to the earth by the considered layer of the ionosphere after having been beamed at some angle with the normal
MUF=C.F=C.F.secΘ cosΘ
where C.F. is the critical frequency as defined as the highest frequency that is returned to the earth by the considered layer of the ionosphere after having been sent straight (normally) to it "Θ' is the angle between the normal and the incident ray.
29) Numerical aperature depends upon diameter of the core . Give reason?
Refractive index p of core of core of optical fibre is slightly higher than glass cladding.Light propagates through and along the fibre by the series of bounces caused by internal reflection at the interface of the core and cladding. For total internal reflection angle Θc NA=sinΘc =√(μ12 -μ22)
Numerical aperature depends upon diameter of the core It decreases as the diameter of core decreases vice versa
30. How will you distinguish between critical frequency and critical angle..
Critical frequency is the highest frequency that can be propagated directly upward and still be returned to earth by the ionosphere.
Critical angle: Every frequency has a maximum vertical angle at which it can be propagated and still be refracted back by the ionosphere. This angle is called critical angle.
31. How is speech reproduced in a receiver?
The extracted audio frequency signal from modulator is fed to the audio frequency amplifier where it is amplified. The amplified audio frequency signal is given to the speaker which converts it into original speech (in sound wave)
32.In selecting optical fiber, how do each of the following affect maximum distance span of a fiber run between switches?
a) Fiber Quality b) Wavelength.
Answer: a) Fiber Quality
A typical fiber quality is 160Mhz-km i.e., for distance separation of one kilometer bandwidth of 160MHz can be used and for distance 200m a bandwidth of 80MHz can be used.
33.When a packet arrived from the Internet a firm filters them with an application firewall and then passes them to an IPsec Server operating in tunnel mode. The application firewall is designed to protect against e-mail viruses, what is wrong with this setup?
Trojan Horse software on their machine by operating up e-mail attachments. In general the strongest policies mean nothing if users do not follow them so policy implementations must include sanction for employees who violate security policies.
34. Give representative frequency of uplink and downlink communication.
C- Band satellites uses frequency of above 6GHz for uplink and 4GHz for downlink . KV-Band satellite uses frequency of 14Ghz for uplink and 12GHz for downlink.
Ka-Band has uplink frequency of 30GHz and downlink frequency of 20GHz.
35. Line of sight communication (LOS) is good for only 30-50Km and loosing required LOS connection. Why? How to solve the problem in microwave communication.
Microwave system can travel only a limited distance before problems occur. Signals may grow too weak because of attenuation. The receiver might be so far away that target falls below the horizon. Losing the LOS or their may be obstacles between dishes. Microwave systems use repeaters to solve this problem. They capture and regenerate the signal to remove the propagation effects before passing it to the receiving antenna.
36. In optical fiber refractive index of cladding is less than core. Why?
It helps to undergo multiple total internal reflections through the optical fiber.
37. “A RADAR using a wave length 5cm and having an antenna disc of diameter 10m has an angular resolution smaller than 0.01 radian.” Is this statement correct?
Angular resolution = 5x10-2/10 = 0.005 rad, which is less than 0.01 rad. So given statement is correct.
38.If one bit is “1” another bit is a “1”.What will be the result of XOR operation .What if both bits are “0”?
Draw suitable wave form?
39. How do you think TCP/IP standards will have to change in future to continue to improve internet?
TCP/IP layers govern transmission of data across the internet ensuring that any two host computers can communicate . It is dominant in corporate networking .
40.An AM wave is represented by Cm(t) = 6 (1+0.5cos 12560t) cos 22x105 t. Calculate:
Amplitude and frequency of carrier, Modulation Index , Maximum and Minimum amplitude of the AM wave.
41. Using Internet , How can you transfer a file or software from one computer to another. Explain.
File Transfer Protocol(FTP)
42. For better quality of music transmission FM is preferred .Why?
FM gives better quality transmission and has a larger band with . It is in the form of frequency variation , therefore man made noise do little harm .So it is preferred for transmission of music(FM radio-88 to108 MHz)
43. You are trying to connect your computers over a network. In this how can you use DCE and DTE devices to establish the network?
DCE (Data Communications Equipment) : It creates a physical connection between the source and destination computers over a network.
DTE (Data terminal Equipment) : It connects computers to a WAN with the help of a DCE . DTE is end –user equipment , such as a router or computer = 3919.2 x 1000 = 3919200
44) Modem helps us to connect WAN network ? Explain?
Modem is a device that can connect one computer to another across ordinary telephone lines . The current telephone system cannot carry the voltage changes required for a direct digital connection . A modem overcomes this limitations by (i) modulating digital information into audio tones for transmission across the phone line .(ii) demodulating the audio tones back into digital information upon reception. It is from these actions (modulate and demodulate ) that the name modem is derived.
45) Electronic reproduction of a document at a distant place is possible by FAX.How is coding and decoding made possible using FAX?.
telegraphy or FAX.
A FAX machine
(1) digitizes printed images and transmits the corresponding data over the telephone lines.
(ii) receivers digital image information placed on the network, and reconstruct the received image on the paper thus to send the document through FAX.
Explain the Functions fo FAX.(a) At the sending end, there is some sort of sensor (optical scanner) to read the paper.
(b) There is some standard way to encode the white and black spots that the fax machine observes on the paper, so that they can travel through the phone line.
(c) At the receiving end there is a mechanism that marks the paper with white and black spots to reproduce the document
(d) The printer if the FAX machine prints these digital purses faithfully in the form of dots to build up a fascinate copy of the original
46) With the help of diagrams explain the difference between amplitude modulation and frequency modulation.
Amplitude modulation:- In this type of modulation the amplitude of the carrier signal wave varies in accordance with the modulating signal, but in frequency modulation frequency of carrier wave varies. Diagrams below show the types of modulation.
47. Ground receiver station is receiving a signal at (i) 5MH and (ii 100MHz transmitted from a round transmitter at a height of 300 m, located at a distance of 100 km from the receiver station. Identify whether the signal is coming via space wave or sky wave propagation or satellite transponder. Radius of earth = 6.4 x 106 m. Nmax of the Isosphere = 1012 m3
Maximum distance covered by space wave communication = √2RH=62km
Critical frequency fc =9 x Nmax1/2 = 9 MHz
(i) 5 MHz <fc sky wave propagation (ionospheric propagation).
(ii) 100 MHz > fc satellite mode of communication.
48) Draw a block diagram of data transmission and a data receiver. Explain
49.A schematic arrangement for transmitting a message signal (20 Hz to 20kHz) is given below:
51. An audio signal of 1 kHz is used to modulate a carrier of 500 kHz. Determine
(i) Sideband frequency , (ii) Bandwidth required.
Carrier frequency, fc = 500 kHz
Signal frequency, fs = 1 kHz
(i) The AM wave has sideband frequency of (fc + fs) and (fc - fs).501 kHz and 499 kHz
(ii) Bandwidth required = 499 kHz to 501 kHz = 2 kHz
52. The TV transmission tower at a particular place has a height of 160m. What is its coverage range? By how much should the height be increased to double its coverage range? Given that radius of earth = 6400 km.
Here R = 6400km = 6400 X 103 m
h = 160m
Therefore
d = (2 × 6400 X103 × 160)1/2 = 45255m
Coverage range, d = (2Rh)1/2
For the first case
h2/h1 = d22/d12 = (2)2 = 4
h2 = 4h1 = 4 × 160 = 640m
53.A sinusoidal carrier voltage is amplitude modulated by a sinusoidal voltage of 10 kHz to a depth of 30%. Calculate the frequency and amplitude of the two sidebands if
the carrier frequency is 10 MHz and its amplitude is 40V.
(USB=fc+fm=10.01 MHz ; LSB=fc-fm=9.99 MHz)
An audio signal given by em=30sin(2Π*2500t)volts is used to amplitude modulate a carrier wave given by ec=60sin(2Π+200,000t) volts.find
Percent modulation
Components of modulated wave and
The amplitude of each side band.
(m=Em/Ec×100 ; 50% ; 200 kHz ; 197.5 kHz)
55.An amplitude modulated wave is represented as
E=5(1+0.6cos6280t)sin211×104t,volts
What are 3 the minimum and maximum amplitudes of the am wave?
What frequency components are contained in the modulated wave?
What are the amplitudes of the components?
( (i) min. amp.=Ec-mEc=2V ; max.amp.=Ec+mEc=8 V ;
(ii) fc-fm,fc,fc+fm ; 335kHz,336kHz,337 kHz
mEc/2,Ec,mEc/2 ; 1.5 V, 5V,1.5 V)
Communication Systems
1. Sky wave transmission of electromagnetic wave cannot be used for TV transmission why?
TV transmission takes place at higher frequencies (approx 18 MHz to 200 MHz). Such higher frequencies do not return back to the earth.
2.In digital communication, PCM (Pulse Code Modulation) is preferred than PAM (Pulse Amplitude Modulation). Justify your comment.
(i) PCM is free from noise in interfering signals. It is also coded electrical signal.
(ii) It permits use of repeater for long distance transmission.
3.In sending a binary data over an analog transmission line, what kind of device does the conversion?
MODEM
4.For what kind of signal would you use for encoding and decoding process?
Pulse Code Modulation -PCM is free from noise in interfering signals.
5.When a teacher lectures in the class, is the classroom a full duplex communication system or a half duplex communication system?
In the full duplex mode, transmissions are possible in both directions simultaneously . A standard telephone system is an example of full-duplex transmission.
6. Your wireless notebook computer works on one desk, but when you move it to a nearby desk, you cannot receive a signal. What may be happening? How night you fix the problem?
By increasing the frequency range.(wide Band width)
7.Express the following I.P address in binary:
128.171.17.13
Hint: 128 is 10000000.
8.Convert the following address in binary to decimal notation:
11110000 10101010 00001111 11100011
Hint: 11110000 is 240 in decimal.
9.A tester shows that a UTP (Unshielded Twisted Pair) cord has too much cross-stalk interference. Why?
A standard UTP cord contains 4 wire pairs. The pairs in the bundle will radiate some of the energy producing electromagnetic (EM) interference in other pairs. This mutual EMI among the wire pairs in UTP cord is called cross-stalk interference. The twisting of each pair normally keeps cross-stalk interference to a reasonable level. Installer must be careful not to untwist the wire more than 1.2cm, when addng connectors. This limit will not eliminate terminal cross-stalk interference but it should limit to a acceptable level.
10.How do we make television broadcasts for larger coverage and for long distance?
By using (i) tall antennas which is familiar landmark in many cities and (ii) using artificial satellites —called geostationary satellites. Since television signals are of high frequency and are not reflected by ionosphere so we use satellites to get them reflected & transmission of TV signals can be used for larger coverage as well as for long distance.
11.As we know parallel transmission is faster than serial transmission. Name the device which converts parallel to serial transmission.
Multiplexer
12.In normal radio operation how does channel bandwidth usually related to bandwidth required to transmit a data stream of a given speed.
If you want to transmit at a given bit rate, you will need a channel whose bandwidth is sufficiently wide. Multiplexer is used to convert serial transmission into parallel.
13.Optical Radio telescopes are built on ground but X-Rays telescopes are possible only from satellites orbiting earth. Why? Atmosphere absorbs X-rays while visible and radio waves can penetrate it.
14.Do you think P2P architectures will be popular in the future .Why?
Peer to peer communication applications are likely to appear in near future offering a much broader spectrum of services.
15. Scientists put x-ray astronomical telescope on the artificial satellite orbiting above the earth's atmosphere whereas they build optical and radio-telescopes on the surface of the earth. Why ?
X-rays have very high frequency and much smaller wavelength. These rays get absorbed by the earth's atmosphere. On the other hand, optical (visible) radiations and radio-waves can pass through the atmosphere. That is why optical and radio telescopes can be installed on the earth's surface.
16. In TV transmission which is used –either A.M or F.M?
17. In demodulation stage , how the RF signal is removed from AF?
Rectifier is used to rectify the modulated wave (i.e) negative halt of the modulated wave is eliminated. It separates the audio signal from the carrier wave.
18) In the transmission of television signals sky waves are not used Give reason?
The television signals have frequencies in 100-200 MHz range. As ionosphere cannot reflect radio waves of frequency greater than 40 M back to earth, the sky waves cannot be used in the transmission of TV signals.
19) ‘ F2 layer’ plays an important role in communication . Why?
F2is the top most layer of ionosphere. Its height is 240-400 Km in day time where as at night it falls from 400-300Km. It is very thick layer and most of its molecules are ionized. It is the most useful reflecting layer for high frequency radio waves.
20) The strength of received signal is almost constant in the receiving stage. How can you achieve this?
AGC stands for Automatic Gain Control. It is used in receiver so that the strength of received signal is almost constant.
21) In satellite communication, how transponder helps as to Uplink and down link signals? In satellite communication, uplink signals are transmitted from earth stations.
The transponder receives these signals, convert them to different frequencies and retransmits them to back to earth as downlink signals.
22) Why is delta modulation a convenient method of digital modulation.
Delta modulation involves simple pulse coding and decoding methods. A simple delta modulation uses just one bit per sample i.e. a ‘non-zero’ sample or one per sample. Thus, this method is convenient to use.
23. In an optical fibre why the glass fibres have high refractive index as compare to thin layer of coating on it ?
The optical fibres are normally made of glass of high refractive index coated with a thin layer of glass of low refractive index to provide a suitable boundary. It this is not done the phenomenon of total internal reflection is not possible inside the optical fibre or the light would pass from one fibre to another fibre.
24.What is necessary for full duplex operation using a two wire circuit? Give an example of full-duplex transmission.
In the full duplex mode, transmissions are possible in both directions simultaneously but they must be between the same two stations. A standard telephone system is an example of full-duplex transmission.
25.What is necessary for full duplex operation using a two wire circuit? Give an example of full-duplex transmission.
In the full duplex mode, transmissions are possible in both directions simultaneously but they must be between the same two stations. A standard telephone system is an example of full-duplex transmission.
26) Why does the electrical conductivity of earths atmosphere increase with altitude ?
Atmospheric pressure decreases with in crease in altitude. The high energy particles (ie & rays and cosmic rays) coming from outer space and entering our earths atmosphere cause ionization of the atoms of the gases present there . The ionizing power of these radiation decreases rapidly as they approach the earth. due to decrease in number of collision with the gas atoms . It is due to this reason that the electrical conductivity of earths atmosphere increases with altitude.
27) If you want transmit an audio signal for long distance , How it can be made possible?
1 .The length of the antenna required is so large (L = 5000m ) that is practically impossible to set up it.
2 The energy radiated from the antenna in audio frequency range is Practically zero
3 The audio signals transmitted from the different broadcasting stations will get inseparably mixed.
28) MUF helps us to receive signal by the applicataion of ionosphere .How it helps us?
Maximum usable frequency (MUF) is defined as the highest frequency that is returned to the earth by the considered layer of the ionosphere after having been beamed at some angle with the normal
MUF=C.F=C.F.secΘ cosΘ
where C.F. is the critical frequency as defined as the highest frequency that is returned to the earth by the considered layer of the ionosphere after having been sent straight (normally) to it "Θ' is the angle between the normal and the incident ray.
29) Numerical aperature depends upon diameter of the core . Give reason?
Refractive index p of core of core of optical fibre is slightly higher than glass cladding.Light propagates through and along the fibre by the series of bounces caused by internal reflection at the interface of the core and cladding. For total internal reflection angle Θc NA=sinΘc =√(μ12 -μ22)
Numerical aperature depends upon diameter of the core It decreases as the diameter of core decreases vice versa
30. How will you distinguish between critical frequency and critical angle..
Critical frequency is the highest frequency that can be propagated directly upward and still be returned to earth by the ionosphere.
Critical angle: Every frequency has a maximum vertical angle at which it can be propagated and still be refracted back by the ionosphere. This angle is called critical angle.
31. How is speech reproduced in a receiver?
The extracted audio frequency signal from modulator is fed to the audio frequency amplifier where it is amplified. The amplified audio frequency signal is given to the speaker which converts it into original speech (in sound wave)
32.In selecting optical fiber, how do each of the following affect maximum distance span of a fiber run between switches?
a) Fiber Quality b) Wavelength.
Answer: a) Fiber Quality
A typical fiber quality is 160Mhz-km i.e., for distance separation of one kilometer bandwidth of 160MHz can be used and for distance 200m a bandwidth of 80MHz can be used.
33.When a packet arrived from the Internet a firm filters them with an application firewall and then passes them to an IPsec Server operating in tunnel mode. The application firewall is designed to protect against e-mail viruses, what is wrong with this setup?
Trojan Horse software on their machine by operating up e-mail attachments. In general the strongest policies mean nothing if users do not follow them so policy implementations must include sanction for employees who violate security policies.
34. Give representative frequency of uplink and downlink communication.
C- Band satellites uses frequency of above 6GHz for uplink and 4GHz for downlink . KV-Band satellite uses frequency of 14Ghz for uplink and 12GHz for downlink.
Ka-Band has uplink frequency of 30GHz and downlink frequency of 20GHz.
35. Line of sight communication (LOS) is good for only 30-50Km and loosing required LOS connection. Why? How to solve the problem in microwave communication.
Microwave system can travel only a limited distance before problems occur. Signals may grow too weak because of attenuation. The receiver might be so far away that target falls below the horizon. Losing the LOS or their may be obstacles between dishes. Microwave systems use repeaters to solve this problem. They capture and regenerate the signal to remove the propagation effects before passing it to the receiving antenna.
36. In optical fiber refractive index of cladding is less than core. Why?
It helps to undergo multiple total internal reflections through the optical fiber.
37. “A RADAR using a wave length 5cm and having an antenna disc of diameter 10m has an angular resolution smaller than 0.01 radian.” Is this statement correct?
Angular resolution = 5x10-2/10 = 0.005 rad, which is less than 0.01 rad. So given statement is correct.
38.If one bit is “1” another bit is a “1”.What will be the result of XOR operation .What if both bits are “0”?
Draw suitable wave form?
39. How do you think TCP/IP standards will have to change in future to continue to improve internet?
TCP/IP layers govern transmission of data across the internet ensuring that any two host computers can communicate . It is dominant in corporate networking .
40.An AM wave is represented by Cm(t) = 6 (1+0.5cos 12560t) cos 22x105 t. Calculate:
Amplitude and frequency of carrier, Modulation Index , Maximum and Minimum amplitude of the AM wave.
41. Using Internet , How can you transfer a file or software from one computer to another. Explain.
File Transfer Protocol(FTP)
42. For better quality of music transmission FM is preferred .Why?
FM gives better quality transmission and has a larger band with . It is in the form of frequency variation , therefore man made noise do little harm .So it is preferred for transmission of music(FM radio-88 to108 MHz)
43. You are trying to connect your computers over a network. In this how can you use DCE and DTE devices to establish the network?
DCE (Data Communications Equipment) : It creates a physical connection between the source and destination computers over a network.
DTE (Data terminal Equipment) : It connects computers to a WAN with the help of a DCE . DTE is end –user equipment , such as a router or computer = 3919.2 x 1000 = 3919200
44) Modem helps us to connect WAN network ? Explain?
Modem is a device that can connect one computer to another across ordinary telephone lines . The current telephone system cannot carry the voltage changes required for a direct digital connection . A modem overcomes this limitations by (i) modulating digital information into audio tones for transmission across the phone line .(ii) demodulating the audio tones back into digital information upon reception. It is from these actions (modulate and demodulate ) that the name modem is derived.
45) Electronic reproduction of a document at a distant place is possible by FAX.How is coding and decoding made possible using FAX?.
telegraphy or FAX.
A FAX machine
(1) digitizes printed images and transmits the corresponding data over the telephone lines.
(ii) receivers digital image information placed on the network, and reconstruct the received image on the paper thus to send the document through FAX.
Explain the Functions fo FAX.(a) At the sending end, there is some sort of sensor (optical scanner) to read the paper.
(b) There is some standard way to encode the white and black spots that the fax machine observes on the paper, so that they can travel through the phone line.
(c) At the receiving end there is a mechanism that marks the paper with white and black spots to reproduce the document
(d) The printer if the FAX machine prints these digital purses faithfully in the form of dots to build up a fascinate copy of the original
46) With the help of diagrams explain the difference between amplitude modulation and frequency modulation.
Amplitude modulation:- In this type of modulation the amplitude of the carrier signal wave varies in accordance with the modulating signal, but in frequency modulation frequency of carrier wave varies. Diagrams below show the types of modulation.
47. Ground receiver station is receiving a signal at (i) 5MH and (ii 100MHz transmitted from a round transmitter at a height of 300 m, located at a distance of 100 km from the receiver station. Identify whether the signal is coming via space wave or sky wave propagation or satellite transponder. Radius of earth = 6.4 x 106 m. Nmax of the Isosphere = 1012 m3
Maximum distance covered by space wave communication = √2RH=62km
Critical frequency fc =9 x Nmax1/2 = 9 MHz
(i) 5 MHz <fc sky wave propagation (ionospheric propagation).
(ii) 100 MHz > fc satellite mode of communication.
48) Draw a block diagram of data transmission and a data receiver. Explain.
49.A schematic arrangement for transmitting a message signal (20 Hz to 20kHz) is given below:
51. An audio signal of 1 kHz is used to modulate a carrier of 500 kHz. Determine
(i) Sideband frequency , (ii) Bandwidth required.
Carrier frequency, fc = 500 kHz
Signal frequency, fs = 1 kHz
(i) The AM wave has sideband frequency of (fc + fs) and (fc - fs).501 kHz and 499 kHz
(ii) Bandwidth required = 499 kHz to 501 kHz = 2 kHz
52. The TV transmission tower at a particular place has a height of 160m. What is its coverage range? By how much should the height be increased to double its coverage range? Given that radius of earth = 6400 km.
Here R = 6400km = 6400 X 103 m
h = 160m
Therefore
d = (2 × 6400 X103 × 160)1/2 = 45255m
Coverage range, d = (2Rh)1/2
For the first case
h2/h1 = d22/d12 = (2)2 = 4
h2 = 4h1 = 4 × 160 = 640m
53.A sinusoidal carrier voltage is amplitude modulated by a sinusoidal voltage of 10 kHz to a depth of 30%. Calculate the frequency and amplitude of the two sidebands if
the carrier frequency is 10 MHz and its amplitude is 40V.
(USB=fc+fm=10.01 MHz ; LSB=fc-fm=9.99 MHz)
An audio signal given by em=30sin(2Π*2500t)volts is used to amplitude modulate a carrier wave given by ec=60sin(2Π+200,000t) volts.find
Percent modulation
Components of modulated wave and
The amplitude of each side band.
(m=Em/Ec×100 ; 50% ; 200 kHz ; 197.5 kHz)
55.An amplitude modulated wave is represented as
E=5(1+0.6cos6280t)sin211×104t,volts
What are 3 the minimum and maximum amplitudes of the am wave?
What frequency components are contained in the modulated wave?
What are the amplitudes of the components?
( (i) min. amp.=Ec-mEc=2V ; max.amp.=Ec+mEc=8 V ;
(ii) fc-fm,fc,fc+fm ; 335kHz,336kHz,337 kHz
mEc/2,Ec,mEc/2 ; 1.5 V, 5V,1.5 V)
UNIT-10
Communications System
1) Why are micro wave used in radars?
2) Why sky waves are not used in the transmission of television signals?
3) What should be the desirable characteristic of a diode detector?
4) Give a velocity factor of a line.
5) Why is delta modulation a convenient method of digital modulation.
6) Where the two wire transmission line, Coaxial cable, Optical fiber are employed.
7) what is heterodyning?.
8) What is population inversion? How is it achieved?.
9) Enumerate the various types of Lasers?.
10)Lists some of the applications of Lasers?.
11)Name the prime elements of a telecommunications network.
12) Audio signal cannot be transmitted directly in to the space why?
13) What is pulse modulation?
14) What is precisely meant by the term channel in a communication system “?
15) Why does the electrical conductivity of earths atmosphere increase with altitude ?
16) Explain numerical aperture in fibre optical
17) Differentiate between (i) PAM and (II) PPM .
18)Why the transmission of signal is not possible for frequency greater than 20Mhz .
19)How does the effective power radiated by the antenna vary with wavelength?
20)what should be the length of the dipole antenna for a carrier wave of 5 X 108 hz ?
21)By how much should the height of the antenna be increased to double the coverage range R= 6400 Km.
22) A TV. tower has a height of lOOm . How much population is covered by the TV. broadcaste if the average population density around the tower is 1000/km2
23)Ground receiver station is receiving a signal at (i) 5MH and (ii 100MHz transmitted from a round transmitter at a height of 300 m, located at a distance of 100 km from the receiver station. Identify whether the signal is coming via space wave or sky wave propagation or satellite transponder. Radius of earth = 6.4 x 106 m. Nmax of the Isosphere = 1012 m3
24)A schematic arrangement for transmitting a message signal (20 Hz to 20kHz) is given below:
Give two drawbacks from which this arrangement suffers. Describe briefly with the help of a block diagram the alternative arrangement for the transmission and reception of the message
signal.
25) The maximum peak-to-peak voltage of an AM wave is 16mV and the minimum peak-to peak voltage is 4mV.Calculate the modulation factor.
26) An AM wave is represented by the expression:
v = 5(1+0.6cos6280t) sin 221 X 104t volts
(i) What are the maximum and minimum amplitudes of the AM wave.
(ii) What frequency components are contained in the modulated wave.
27) An audio signal of 1 kHz is used to modulate a carrier of 500 kHz. Determine
(i) Sideband frequency.
(ii) Bandwidth required.
28) The antenna current of an AM transmitter is 8A when only carrier is sent but it increases to 8.93A when the carrier is sinusoid ally modulated. Find the percent-age modulation index.
29) A 100 MHz carrier is modulated by a 12 kHz sine wave so as to cause a frequency swing of +50kHz. Find the modulation index.
30) The TV transmission tower at a particular place has a height of 160m. What is its coverage range? By how much should the height be increased to double its coverage range? Given
that radius of earth = 6400 km.
31) A TV tower has a height of 110m. How much population is covered by the TV broadcast if the average population density around the tower is 1000 km-2? Given that radius of
Earth = 6.4 X 106m.
32)A microwave telephone link operating at the cenral frequency of 10 GHz has been established .If 2 % of this is available for microwave communication channel, then how many telephones channels can be simultaneously granted if each telephone is allotted a band width of 8 KHz .
33)You are given three semiconductors A,B,C with respective band gaps of 3eV, 2eV and 1eV for use in a photodetector to detect λ = 1400nm . Select the suitable semiconductor. Give
reasons.
34) Frequencies higher than 10MHz are found not to be reflected by the ionospere on a particular day at a place. Calculate th maximum electron density of the
ionosphere.
ADDITIONAL QUESTIONS:
35) Draw a block diagram of data transmission and a data receiver. Explain them briefly.
36). Explain how the optical communication system offers the possibility of millions of channels with increased band width. Give an additional advantage of optical communication
system over a system employing a co axial cable.
37) Give the names of the Indian satellites launched from the erstwhile USSR for remote sensing.
38) Expand the following abbreviations: APPLE, INSAT, IRS, ISRO, ASLV.
39) . What is the role played by launch vehicles in space exploration ? How is it different from a space shuttle ? Merition any two facilities a country needs to develop to achieve selfsufficiency
in space programmes.
40) Explain how TV programmes are transmitted to remote areas through communication satellites.
41) Explain (a) Modem Transmission (b) Modem Reception.
42) How does Heterodyning distinguish from CW signals?.
43) In a diode AM detector the output circuit consists of R =108Ω and C = 10 pF. A carrier signal of frequency 105hz is to be detected .Is it good?.
44) On a particular day maximum frequency reflected from the ionosphere is 9Mhz.On another day it was found to increase by one Mhz.What is the ratio of the maximum electron
density of the ionosphere on the two days.Hints: Nmax / nmax = (υc’/υc)2= 1.23
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HOTS for Chapter 15 Communication Systems Physics Class 12
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