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)
