65 + Interview Questions in COMMUNITCATION-SYSTEMS IN GENERAL AWARENESS Page 1 InterviewSolution

1.	Which of the following frequencies will be suitable for beyond the horizon communication using sky waves?A. 10KHzB. 10MHzC. 1GHzD. 1000GHz
Answer» Correct Answer - B 10KHz cannot be radiated due to large antenna size, 1 GHz and will penetrate.

Discussion

2.	A male voice after modulation-transmission sounds like that of a female to the receiver. The problem is due toA. poor selection of modulation index (selected `0ltmlt1`)B. poor bandwidth selection of amplitudesC. poor selection of carrier frequencyD. loss of energy in transmission
Answer» Correct Answer - B The frequency of male voice is less than that of a female voice in the given problem, the frequency of modulated signal recived becomes more, which is it is so because bandwidth in amplitude modulation is equal to twice the frequency of modulation signal.

Discussion

3.	Modern is a device which performsA. 1. modulationB. 2. demodulationC. 3. rectificationD. 4. modulation and demodulation
Answer» Correct Answer - D Modern is a device which performs modulation and demodulation of the signal.

Discussion

4.	The device which is a combination of a receiver and a transmitter isA. 1. AmplifierB. 2. RepeaterC. 3. TransducerD. 4. Modulator
Answer» Correct Answer - B A repeater is a combination of reciever and a transmitter.

Discussion

5.	Modulation is the process of superposingA. low frequency audio signal on high frequency radio waves.B. low frequency radio signals on low frequency audio wave.C. high frequency radio signal on low frequency audo signal.D. high frequency audio signal on low frequency radio waves.
Answer» Correct Answer - A Modulation is a process of superposing a low frequency audio signals (called modulating signal) on a high frequency radio wave (called carrier wave).

Discussion

6.	A `100m` long antenna is mounted on a `500m` tall building. The complex can become a transmission tower of waves with `lambda`A. `-400m`B. `-25m`C. `-150m`D. `-2400m`
Answer» Correct Answer - A Length of antenna, I=100m, As `l=(lambda)/(4)` or `lambda=4l=4xx100=400m`.

Discussion

7.	A ` 1 KW ` signal is transmitted using a communication channel which provides attenuatiom at the rate of ` - 2 dB per km` . If the communication channel has a total length of ` 5 km` , the power of the signal received is [ gain in `dB = 10 log ((P_(0))/(P_(i)))]`A. 900WB. 100WC. 990WD. 1010W
Answer» Correct Answer - B Loss suffered in the communication channel `=(-2dB//km)5km=-10dB` `therefore 10"log"(P_(0)//P_(i))=-10, "or log" (P_(0)//P_(i))=-1` or `(P_(0)//P_(i))=10^(-1)=(1)/(10)` or `P_(0)=P_(i)//10=(1kW)/(10)=100W`

Discussion

8.	Assertion: Amplification is necessary to compensate for the attenualtion of the signal in communication system. Reason: Amplification is the process of increasing the amplitdue and consequently the strength of a signal using an electronic circuit.
Answer» Correct Answer - A Amplification is necessary to compensate for the attenuation of the signal to communication systems.

Discussion

9.	The skip zone in radio wave trasmission is that range whereA. there is no reception of either ground wave or sky waveB. the reception of ground wave is maximum b ut that of sky wave is minimumC. the reception of ground wave is minimum, but that of sky wave is maximumD. the reception of both ground and sky wave is maximum.
Answer» Correct Answer - A

Discussion

10.	A fax message is to be sent from Delhi to Washington via a geostationary satellite. The minimum distance between the dispatch and its getting received is (Take height of the geostationary satellite=36000km)A. `72xx10^(3)km`B. `12xx10^(3)km`C. `27xx10^(3)km`D. `18xx10^(3)km`
Answer» Correct Answer - A Minimum distance `s=2h=2xx36000=72000km `

Discussion

11.	A TV transmission tower has a height of 240 m. Signals broadcast from this tower will be received by LOS communication at a distance of (assume the radius of earth to be `6.4 xx 10 ^(6) m`)A. 100kmB. 110kmC. 55kmD. 120km
Answer» Correct Answer - C Here, `h=240m,R=64xx10^(6)m` The maximum distance on earth from the transmitter upto which a signal can be recived is `d=sqrt(2Rh)=sqrt(2xx(6.4xx10^(6))xx240)` `=55.4xx10^(3)m=55.4km`

Discussion

12.	A TV transmission tower of antenna is at a height of 20 m. How much service area can it cover if the receiving antenna is (i) at ground level (ii) at a height of 25 m ? Calculate the percentage increase in area covered in case (ii) relative to case (i).A. 2.48B. `348.9%`C. `150%`D. `360.2%`
Answer» Correct Answer - B Here, `h_(T)=20m,R=6.4xx10^(6)m=64xx10^(5)m` If the anteena is at ground level Range, `d=sqrt(2h_(T)R)=sqrt(2xx20xx(64xx10^(5)))` `=16xx10^(3)m=16km ` Area covered, `A=pid^(2)=(22)/(7)xx(!6)^(2)=804.6km^(2)` If the receiving antenna at a height of 25m. Range `d_(1)=sqrt(2h_(T)R)_sqrt(2h_(R)R)` `=sqrt(2xx20xx(64xx10^(5)))+sqrt(2xx25xx64xx10^(5))` `=16xx10^(3)+17.9xx10^(3)=33.9xx10^(3)m=33. 9km` Area covered `A_(1)=pid_(1)^(2)=(22)/(7)xx(33.9)^(2)=3611.8km^(2)`. percentage increase in area `=(A_(1)-A)/(A)xx100` `=((3611.8-804.6)/(804.6))xx100=348. 9%`

Discussion

13.	A TV transmitting antenna is 81 m tall. How much service area it can cover if the receiving antenna is at the ground level? (Radius of earth=`6.4xx10^(6)m`)A. `3258km^(2)`B. `4180km^(2)`C. `2510km^(2)`D. `1525km^(2)`
Answer» Correct Answer - A Service area it can cover, `A=pid^(2)` `A=pi(2hR)=(22)/(7)xx2xx81xx6.4xx10^(6)m^(2)=3258km^(2)`

Discussion

14.	A transmitting antenna at the top of the tower has a height 18 m and the height of the receiving antenna is 32m. The maximum distance between them for satisfactory communciation in line of sight mode is (Radius of earth`= 6.4xx10^(6)m`)A. 15.15kmB. 21.25kmC. 30.45kmD. 35.42km.
Answer» Correct Answer - D here `h_(t)=18m,h_(R)=32m`, `R6.4xx10^(6)m=64xx10^(5) m` The maximum line or sight distance `d_(m)=sqrt(2Rh_(T))+sqrt(2Rh_(R))` `=sqrt(2xx64xx10^(5)xx18)+sqrt(2xx64xx10^(5)xx32)` `=112xx10^(2)xxsqrt(10) m=35417.5m=35.42km`

Discussion

15.	1-V characteristics of four devices are shown in figure. Identify devices that can be used for modulationA. (i) and (iii)B. Only (iii)C. (iii) and some regions of (iv)D. All the devices can be used
Answer» Correct Answer - C Here, (i) represents ideal modulation related to triode valve for linear plate modulation. (ii) represents square law diode modulation. (iii) represents constant current source. (iv) represents part of it as square law modulation. Thus, for modulation (iii) and some regions of (iv) can be used.

Discussion

16.	Assertion: Microwave communication is preferred over optical communication. Reason: Information carrying capacity is directly proportional to bandwidth.
Answer» Correct Answer - A Microwave communication is preferred over optical communication because microwave provide large number of channels and wide bandwidth compared to proportional to bandwidth, so wider the bandwidth. Greater the information carrying capacity.

Discussion

17.	FM boradcast is preferred over AM broadcast becauseA. it is less noisy.B. reproduction is of much better quality.C. it is more noisy.D. both a and b
Answer» Correct Answer - D FM broadcast is preferred over AM broadcast as the noise is much less and reproduction is of much better quality.

Discussion

18.	Of the following which is preferred modulation scheme for digital communication.A. PCMB. PAMC. PPMD. PTM
Answer» Correct Answer - A Pulse Code Modulation is preferred modulation scheme for digital communication.

Discussion

19.	In an amplitude modulation with modulation index 0.5 the ratio of the carrier wave to that of side band in the modulated wave isA. `4:1`B. `1:4`C. `1:3`D. `2:1`
Answer» Correct Answer - A Here, modulation index `mu=0.5` Amplitude of the carrier wave `=A_(c)` Amplitude of the side band `=mu(A_(c))/(2)` Their ratio `=(2)/(mu)=(2)/(0.5)=(4)/(1)`

Discussion

20.	Which of the following is an example of point to point communication mode?A. RadioB. TelevisionC. TelephonyD. All of these
Answer» Correct Answer - C In poin t to point communication mode, communication takes place over a link between a single transmitter and a reciever. Telephony is an example of point to point mode of communication.

Discussion

21.	Essential elements of a communication system areA. transmitter and receiverB. reciever and communication channelC. transmitter and communication channelD. transmitter, comunication channel and receiver
Answer» Correct Answer - D Every communicaton system has the following three essential elements: (i) Transmitter, (ii) Communication channel, (iii) Receiver

Discussion

22.	In which of the following remote sensing technique is not used?A. Medical treatmentB. PollutionC. Wetland mappingD. Ground water survey
Answer» Correct Answer - A Remote sensing is the technique to collector information about an object in respect of its size, colour, nature, location, temperature etc. without physically touching it There are some areas or location which are inaccesible. So to explore these areas or locations, a technizue known as remote sensing is used. Remoter sensing is done through a satellite.

Discussion

23.	Which of the following is digital modulation technique?A. PCMB. PAMC. PPMD. PTM
Answer» Correct Answer - A Pulse Code Modulation is preferred modulation scheme for digital communication.

Discussion

24.	The essential condition for demodulation isA. `v_(c)lt lt RC`B. `(1)/(v_(c))lt lt RC`C. `(1)/(v_(c))gt gtRC`D. `v_(c)gt gtRC`
Answer» Correct Answer - B For demodulation, `(1)/(v_(c))ltltRC`

Discussion

25.	Identify the incorrect statement from the followingA. AM detection is carried out using a rectifier and an envelope detector.B. Pulse position denotes the time of rise or fall of the pulse amplitude.C. Modulation index `mu` is kept `ge` 1, to avoid distoriton.D. Facsimile (FAX) scans the contents of the document to create electronic signals.
Answer» Correct Answer - C Modulation index `mu` is kept `le` 1 to avoid distotion.

Discussion

26.	Audio signal cannot be transmitted becauseA. the signal has more noise.B. the signal cannot be amplified for distance communication.C. the transmitting antenna length is very small to design.D. the transmitting antenna length is very large and impracticable.
Answer» Correct Answer - D Audio signals cannot be transmitted as such b ecause the length of transmitting antenna required would be too large and impracticable.

Discussion

27.	The modulation in which pulse. Duration varies in accordance with the modulating signal is calledA. PAMB. PPMC. PWMD. PCM
Answer» Correct Answer - C In PWM (Pulse width modulation), which is also called PDM (Pulse Duration Modulation), width of pulses vary with modulating signal.

Discussion

28.	Which of the following is an example of broadcast mode of communication?A. RadioB. TelevisionC. MobileD. Both a and b
Answer» Correct Answer - D In the broadc ast mode of communication there is a larger number of receivers corresponding to a single transmitter. So both radio and television are examples of broadcast mode of communication.

Discussion

29.	The frequency band used in the downlink of s atellite communication isA. 9.5 to 2.5 GHzB. 8 96 to 0901 MHzC. 3.7 to 4.2 GHzD. 840 to 935 MHz
Answer» Correct Answer - C The frequency band used in the downlink of satellite communication is 3.7 to 4.2 GHz.

Discussion

30.	Assertion: Digital signals are continous variations of voltage of curren t. Reason: Digital signals are essentially single valued functions of time.
Answer» Correct Answer - D Signal can be either analog or digital. Analog signals are essentially single valued functions of time. Digital signals are those which can take only descrete stepwise values.

Discussion

31.	The mode of propagation used by short wave broadcast services isA. space waveB. sky waveC. ground waveD. both a and c
Answer» Correct Answer - B

Discussion

32.	The intensity of a light pulse travelling along a communication channel decreases exponetially with distance x according to the relation `I=I_(0)e^(-ax)` where `I_(0)` is the intensity at x=0 and `alpha` is the attenuation constant. The percentage decrease in intensity after a distance of `(("In"4)/(alpha))` isA. 0.7B. 0.75C. 0.8D. 0.85
Answer» Correct Answer - B

Discussion

33.	A sinusoidal voltage amplitude modulates another sinusoidal voltage of amplitude 2 kV resulting in two side bands of amplitude 200 V. Find the modulation index.A. 0.2B. 0.3C. 0.4D. 0.5
Answer» Correct Answer - A Amplitude of each sideband `=(muA_(c))/(2)` `therefore 200 V=(muxx2kV)/(2)=(muxx2000V)/(2)` or `mu=(2xx200)/(2000)=0.2`

Discussion

34.	Which one of the following statements is wrong?A. 1. Ground wave propagation can be sustained at frequencies 500 KHz to 1500 KHz.B. 2. Satellite communication is useful for the frequencies above 30 MHz.C. 3. Sky wave propagation is useful in the range of 30 to 40 MHz.D. 4. Space wave propagation takes place through tropospheric space.
Answer» Correct Answer - C

Discussion

35.	By what percentage will the transmission range of a T.V. tower be affected when the height of the tower is increased by 21% ?A. 0.1B. 0.2C. 0.3D. 0.4
Answer» Correct Answer - A

Discussion

36.	Who invented world wide web?A. J.C.R. LickliderB. Tim Berners-LeeC. Alexander Graham BellD. Samuel F,B. Morse
Answer» Correct Answer - B

Discussion

37.	Assertion : The information contained in our original low frequency baseband signal is to be translated into high or radio frequencies before transmission. Reason: For transmitting a signal, the antenna should have a size comparable to the wave length of the signal.
Answer» Correct Answer - A For transmitting a signal, we need an antenna or an serial. Thus antenna should hav e a size comparable to the wavelength of the signal to the signal. For an electromagentic wave of freqeuncy 20 kHz. The wavelength is 15km. Obviously such a long antenna is not possible to constract and operate Hence direct transmission of such baseband signal is not p ractical. Therefore there is a need of translating the information contained in our original low frequency baseband signal into high or radio frequencies before transmission.

Discussion

38.	Assertion: Lond distance communication between two points on the earth is achieved using sky waves. Reason: sky wave propagation takes place above. The frequency of 30 MHz.
Answer» Correct Answer - C Long distance communication between two points on the earth is achieved through reflection of electromagnetic waves by ionosphere such waves are called sky wave propagation takes place u p to frequency of about 30 MHz.

Discussion

39.	The carrier freqeuncy of a station is 40 MHz. A resistor of 10k `Omega` and capacitor of 12 pF are available in the detector circuit. The possible value of C will beA. 12B. 8.2C. 5.6D. All of these
Answer» Correct Answer - D Here, `R=10kOmega=10xx10^(3)Omega` `C=12pF=12xx10^(-12)F` `v_(c)=40MHz=40xx10^(6)Hz=4xx10^(7)Hz`. Time period of carrier frequency `T_(c)=(1)/(v_(c))-(1)/(4xx10^(7))=2.5xx10^(8)s` time constant of C-R circuit, As `T_(c)lt tau`, therefore, circuit is good enough for detection All the values of C given in the options are possible.

Discussion

40.	Compute `LC` product of a tuned amplifer circuit required to generate a carrier `wave` of `1 MHz` for amplitude modulationA. `1.5xx10^(-14)s`B. `1.2xx10^(-12)s`C. `3.2xx10^(-12)s`D. `2.5xx10^(-14) s`
Answer» Correct Answer - D Frequency of tuned amplifier is `v=(1)/( 2pisqrt(LC))` As per question, `v=1 MHz=10^(6)Hz` or `10^(6)=(1)/(2pisqrt(LC))` or `LC=(1)/(4pi^(2)(10^(6))^(2))=2.5xx10^(-14)s`

Discussion

41.	The output current of an 80% mondulating amplitude modulated generator is 1.8A. To what value will the current rise if the generator is additionally modulated by another audiowave of modulation index 0.6?A. 1.71AB. 1.81AC. 1.91AD. 2.01A
Answer» Correct Answer - C Here, `I_(t)=1.8A A: mu_(1)=(80)/(100)=0.80: mu_(2)=0.6` `I_(t)=I_(c)sqrt(1+mu_(1)^(2)//2)` `I_(c)=(1)/(sqrt(1+mu_(1)^(2)//2))=(1.8)/(sqrt(1+(0.8)^(2)//2))=(1.8)/(sqrt(1.32))` `I_("net total")=I_(c)sqrt(1+(mu_(1)^(2))/(2)+(mu_(2)^(2))/(2))` `=(1.8)/sqrt(1.32)sqrt(1+((0.8)^(2))/(2)+((0.6)^(2))/(2))=1.91A`

Discussion

42.	The frequency response curve (figure) for the filter circuit used for production of AM wave should be A. i followed by iiB. ii followd by iC. iiiD. all of these
Answer» Correct Answer - D Bandwidth of AM wave should be `2omega_(m)=(omega_(c)+omega_(m))-(omega_(c)-omega_(m))` it is applicable for case (i),(ii) and (iii) The (iv) will reject the required signal for AM waves.

Discussion

43.	The frequency response curve (figure) for the filter circuit used for production of AM wave should be A. i followed by iiB. ii followd by iC. iiiD. all of these
Answer» Correct Answer - D Bandwidth of AM wave should be `2omega_(m)=(omega_(c)+omega_(m))-(omega_(c)-omega_(m))` it is applicable for case (i),(ii) and (iii) The (iv) will reject the required signal for AM waves.

Discussion

44.	The LC product of a tuned amplifier circuit require to generate a carrier wave of 1 MHz for amplitude modulation isA. `1.5xx10^(-14)s`B. `1.2xx10^(-12)s`C. `3.2xx10^(-12)s`D. `2.5xx10^(-14) s`
Answer» Correct Answer - D Frequency of tuned amplifier is `v=(1)/( 2pisqrt(LC))` As per question, `v=1 MHz=10^(6)Hz` or `10^(6)=(1)/(2pisqrt(LC))` or `LC=(1)/(4pi^(2)(10^(6))^(2))=2.5xx10^(-14)s`

Discussion

45.	A carrier wave of peak voltage `12 V` is used to transmit a message signal. The peak voltage of the modulating signal in order to have a modulation index of `75%` isA. 6VB. 7VC. 8VD. 9V
Answer» Correct Answer - D Here `A_(c)=12V,mu=75%=0.75` As `m u=(A_(m))/(A_(c))` or `A_(m)=muxxA_(c)=9V`

Discussion

46.	A message signal of frequency 10 kHz and peak voltage 10 volt is used to modulate a carrier of frequency 1 MHz and peak voltage 20 volt. Determine (a) modulation index (b) the side bands produced.A. 1000 KHz, 990KHzB. 1010KHz,990KHzC. 990KHz,1000KHzD. 1010KHz,1000KHz
Answer» Correct Answer - B The side bonds produced are at (1000+10)KHz =1010 kHz and (1000-10)kHz= 990kHz.

Discussion

47.	Which one of the following statement is correct?A. A single geostationary satellite can cover the whole part of the earth for microwave communication.B. Atleast three geotationary satellities in the same orbit around earth can cover the whole part of the earth for microwave communication.C. The first Indian communication satellite is Telstar.D. The satellite communication is not like the line of sight microwave communication.
Answer» Correct Answer - B Atleast three geotationary satellites in the same orbit around the earth can cover the whole part of the earth for microwave communication.

Discussion

48.	Assertion: Space waves are used for line -of -sight communication. Reason: Space wave travels in a straight line from transmitting antenna to the receiving antenna.
Answer» Correct Answer - A A space travels in a straight line from transmitting antenna to the receiving antenna. Space wave are used for line-of-sight communication as well as satellite comunication As frequencies above 40 MHz communication is essentially limited t line-of-sight paths.

Discussion

49.	Three waves A,B and C of frequencies 1600 kHz, 5 MHz and 60 MHz, respectively are to be transmitted from one place to another.Which of the following is the appropriate mode of communication?A. A is transmitted via space wave while B and C are transmitted via sky waveB. A is transmitted via ground wave, B via sky wave and C via space waveC. B and C tranmitted via ground wave while A is trasnmitted via sky waveD. B is transmitted via ground wave while A and C are transmitted via space wave.
Answer» Correct Answer - B A is transmitted via ground wave, B via sky wave and C via space wave.

Discussion

50.	A ground receiving station is receiving a signal at 6MHz transmitted from a ground transmitter at a height of 500m located at a distance of 100km. If radius of earth is `6.4xx10^(6)`m, maxim um number density of electron in ionosphere is `10^(12)m^(-3)`. the signal is coming via:A. ground waveB. spac e waveC. Sky wave propagation is useful in the range of 30 to 40 MHz.D. satellite transponder
Answer» Correct Answer - C Maximum distance covered b y space wave communication is `d=sqrt(2Rh)=sqrt(2xx(6.4xx10^(6))xx300)=80xx10^(3)m=80km` Since the distance between transm itter and receiver is 100km, hence for the given frequency signal of 6 MHz. the propagation is not possible via space propagation. for sky wave propagation the critical frequency. `v_(c)=9(N_("max"))^(1//2)=9xx(10^(12))^(1//2)=9xx10^(6)Hz=9MHz` Since 6 MHz `lt9 MHz`. so the propagation of signal of frequency 6 MHz is possible via sky wave.

Discussion

Explore topic-wise InterviewSolutions in .

Which of the following frequencies will be suitable for beyond the horizon communication using sky waves?A. 10KHzB. 10MHzC. 1GHzD. 1000GHz

A male voice after modulation-transmission sounds like that of a female to the receiver. The problem is due toA. poor selection of modulation index (selected `0ltmlt1`)B. poor bandwidth selection of amplitudesC. poor selection of carrier frequencyD. loss of energy in transmission

Modern is a device which performsA. 1. modulationB. 2. demodulationC. 3. rectificationD. 4. modulation and demodulation

The device which is a combination of a receiver and a transmitter isA. 1. AmplifierB. 2. RepeaterC. 3. TransducerD. 4. Modulator

Modulation is the process of superposingA. low frequency audio signal on high frequency radio waves.B. low frequency radio signals on low frequency audio wave.C. high frequency radio signal on low frequency audo signal.D. high frequency audio signal on low frequency radio waves.

A `100m` long antenna is mounted on a `500m` tall building. The complex can become a transmission tower of waves with `lambda`A. `-400m`B. `-25m`C. `-150m`D. `-2400m`

A ` 1 KW ` signal is transmitted using a communication channel which provides attenuatiom at the rate of ` - 2 dB per km` . If the communication channel has a total length of ` 5 km` , the power of the signal received is [ gain in `dB = 10 log ((P_(0))/(P_(i)))]`A. 900WB. 100WC. 990WD. 1010W

Assertion: Amplification is necessary to compensate for the attenualtion of the signal in communication system. Reason: Amplification is the process of increasing the amplitdue and consequently the strength of a signal using an electronic circuit.

A fax message is to be sent from Delhi to Washington via a geostationary satellite. The minimum distance between the dispatch and its getting received is (Take height of the geostationary satellite=36000km)A. `72xx10^(3)km`B. `12xx10^(3)km`C. `27xx10^(3)km`D. `18xx10^(3)km`

A TV transmission tower has a height of 240 m. Signals broadcast from this tower will be received by LOS communication at a distance of (assume the radius of earth to be `6.4 xx 10 ^(6) m`)A. 100kmB. 110kmC. 55kmD. 120km

A TV transmission tower of antenna is at a height of 20 m. How much service area can it cover if the receiving antenna is (i) at ground level (ii) at a height of 25 m ? Calculate the percentage increase in area covered in case (ii) relative to case (i).A. 2.48B. `348.9%`C. `150%`D. `360.2%`

A TV transmitting antenna is 81 m tall. How much service area it can cover if the receiving antenna is at the ground level? (Radius of earth=`6.4xx10^(6)m`)A. `3258km^(2)`B. `4180km^(2)`C. `2510km^(2)`D. `1525km^(2)`

A transmitting antenna at the top of the tower has a height 18 m and the height of the receiving antenna is 32m. The maximum distance between them for satisfactory communciation in line of sight mode is (Radius of earth`= 6.4xx10^(6)m`)A. 15.15kmB. 21.25kmC. 30.45kmD. 35.42km.

1-V characteristics of four devices are shown in figure. Identify devices that can be used for modulationA. (i) and (iii)B. Only (iii)C. (iii) and some regions of (iv)D. All the devices can be used

Assertion: Microwave communication is preferred over optical communication. Reason: Information carrying capacity is directly proportional to bandwidth.

FM boradcast is preferred over AM broadcast becauseA. it is less noisy.B. reproduction is of much better quality.C. it is more noisy.D. both a and b

Of the following which is preferred modulation scheme for digital communication.A. PCMB. PAMC. PPMD. PTM

In an amplitude modulation with modulation index 0.5 the ratio of the carrier wave to that of side band in the modulated wave isA. `4:1`B. `1:4`C. `1:3`D. `2:1`

Which of the following is an example of point to point communication mode?A. RadioB. TelevisionC. TelephonyD. All of these

Essential elements of a communication system areA. transmitter and receiverB. reciever and communication channelC. transmitter and communication channelD. transmitter, comunication channel and receiver

In which of the following remote sensing technique is not used?A. Medical treatmentB. PollutionC. Wetland mappingD. Ground water survey

Which of the following is digital modulation technique?A. PCMB. PAMC. PPMD. PTM

The essential condition for demodulation isA. `v_(c)lt lt RC`B. `(1)/(v_(c))lt lt RC`C. `(1)/(v_(c))gt gtRC`D. `v_(c)gt gtRC`

Audio signal cannot be transmitted becauseA. the signal has more noise.B. the signal cannot be amplified for distance communication.C. the transmitting antenna length is very small to design.D. the transmitting antenna length is very large and impracticable.

The modulation in which pulse. Duration varies in accordance with the modulating signal is calledA. PAMB. PPMC. PWMD. PCM

Which of the following is an example of broadcast mode of communication?A. RadioB. TelevisionC. MobileD. Both a and b

The frequency band used in the downlink of s atellite communication isA. 9.5 to 2.5 GHzB. 8 96 to 0901 MHzC. 3.7 to 4.2 GHzD. 840 to 935 MHz

Assertion: Digital signals are continous variations of voltage of curren t. Reason: Digital signals are essentially single valued functions of time.

The mode of propagation used by short wave broadcast services isA. space waveB. sky waveC. ground waveD. both a and c

A sinusoidal voltage amplitude modulates another sinusoidal voltage of amplitude 2 kV resulting in two side bands of amplitude 200 V. Find the modulation index.A. 0.2B. 0.3C. 0.4D. 0.5

By what percentage will the transmission range of a T.V. tower be affected when the height of the tower is increased by 21% ?A. 0.1B. 0.2C. 0.3D. 0.4

Who invented world wide web?A. J.C.R. LickliderB. Tim Berners-LeeC. Alexander Graham BellD. Samuel F,B. Morse

Assertion : The information contained in our original low frequency baseband signal is to be translated into high or radio frequencies before transmission. Reason: For transmitting a signal, the antenna should have a size comparable to the wave length of the signal.

Assertion: Lond distance communication between two points on the earth is achieved using sky waves. Reason: sky wave propagation takes place above. The frequency of 30 MHz.

The carrier freqeuncy of a station is 40 MHz. A resistor of 10k `Omega` and capacitor of 12 pF are available in the detector circuit. The possible value of C will beA. 12B. 8.2C. 5.6D. All of these

Compute `LC` product of a tuned amplifer circuit required to generate a carrier `wave` of `1 MHz` for amplitude modulationA. `1.5xx10^(-14)s`B. `1.2xx10^(-12)s`C. `3.2xx10^(-12)s`D. `2.5xx10^(-14) s`

The output current of an 80% mondulating amplitude modulated generator is 1.8A. To what value will the current rise if the generator is additionally modulated by another audiowave of modulation index 0.6?A. 1.71AB. 1.81AC. 1.91AD. 2.01A

The frequency response curve (figure) for the filter circuit used for production of AM wave should be A. i followed by iiB. ii followd by iC. iiiD. all of these

The frequency response curve (figure) for the filter circuit used for production of AM wave should be A. i followed by iiB. ii followd by iC. iiiD. all of these

The LC product of a tuned amplifier circuit require to generate a carrier wave of 1 MHz for amplitude modulation isA. `1.5xx10^(-14)s`B. `1.2xx10^(-12)s`C. `3.2xx10^(-12)s`D. `2.5xx10^(-14) s`

A carrier wave of peak voltage `12 V` is used to transmit a message signal. The peak voltage of the modulating signal in order to have a modulation index of `75%` isA. 6VB. 7VC. 8VD. 9V

A message signal of frequency 10 kHz and peak voltage 10 volt is used to modulate a carrier of frequency 1 MHz and peak voltage 20 volt. Determine (a) modulation index (b) the side bands produced.A. 1000 KHz, 990KHzB. 1010KHz,990KHzC. 990KHz,1000KHzD. 1010KHz,1000KHz

Assertion: Space waves are used for line -of -sight communication. Reason: Space wave travels in a straight line from transmitting antenna to the receiving antenna.