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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
How do you express amplitude, frequency and phase modulation? |
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Answer» The equation of a carrier wave may be expressed as ` e = E sin (omegat + phi)` where e is instantaneous voltage at time `t , omega` is angluar frequency and `phi` is the phase w.r.t. some refrence. By varying E, we get amplitude modulation. By varying `omega`, we get frequency modulation and by varying `phi`, we get phase modulation. |
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| 2. |
What is the carrier wave? Why high frequency carrier waves are employed for transmission?A. stationary waveB. carrier waveC. modulated waveD. audio wave |
| Answer» Correct Answer - B | |
| 3. |
The ratio of peak value of modulated signal to peak value of carrier signal isA. refractive indexB. modulation indexC. demodulation indexD. none of these |
| Answer» Correct Answer - B | |
| 4. |
Frequency modulation and phase modulation together is referred asA. amplitude modulationB. angle modulationC. power modulationD. wave modulation |
| Answer» Correct Answer - B | |
| 5. |
The process of superimposing signal frequency (i.e. audio wave) on the carrier wave is known asA. modulationB. demodulationC. amplificationD. rectification |
| Answer» Correct Answer - A | |
| 6. |
To avoid distortion in the signal, modulation index should beA. less than 1B. greater than 1C. equal to 1D. less than or equal to 1 |
| Answer» Correct Answer - D | |
| 7. |
The process in which phase of carrier wave changes in accordance with instantaneous value of modulated signal isA. amplitude modulationB. frequency modulationC. phase modulationD. power modulation |
| Answer» Correct Answer - C | |
| 8. |
The process by which some characteristics i.e. amplitude, phase or frequency of hight frequency carrier wave is varied with information signal isA. demodulationB. modulationC. amplificationD. oscillation |
| Answer» Correct Answer - B | |
| 9. |
The process in which frequencies of carrier wave changes in accordance with instantaneous value of modulated signal isA. amplitude modulationB. frequency modulationC. phase modulationD. power modulation |
| Answer» Correct Answer - B | |
| 10. |
In amplitude modulation, the bandwidth isA. equal to frequency of modulated signalB. double the frequency of modulated signalC. half of frequency of modulated signalD. none of these |
| Answer» Correct Answer - B | |
| 11. |
An electronic device which recovers the original audio signal from amplitude modulated wave isA. modulatorB. rectifierC. amplifierD. detector |
| Answer» Correct Answer - D | |
| 12. |
The information carrying capacity of carrier wave is directly related toA. bandwidthB. amplitudeC. velocityD. phase |
| Answer» Correct Answer - A | |
| 13. |
The process of recovering the audio signal from the modulated wave is known asA. amplificationB. oscillationC. rectificationD. demodulation |
| Answer» Correct Answer - D | |
| 14. |
The device used for addition of high frequency carrier wave and information signal isA. amplifierB. rectifierC. modulatorD. demodulator |
| Answer» Correct Answer - C | |
| 15. |
Recovering information from a carrier is known asA. demultiplexingB. demodulationC. modulationD. carrier recovery |
| Answer» Correct Answer - B | |
| 16. |
In Laser communication there isA. low loss of signalB. loss of signalC. no signal securityD. low band width |
| Answer» Correct Answer - A | |
| 17. |
The ground wave communication cannot be occur above a distance ofA. `200 km`B. ` 10 km `C. ` 400 km`D. `500 km` |
| Answer» Correct Answer - B | |
| 18. |
A receiver of communication system consists ofA. detector onlyB. amplifier onlyC. detector, amplifier and speakerD. detector, amplifier, speaker and modulator |
| Answer» Correct Answer - C | |
| 19. |
Global communication is achieved by usingA. signal geostationary satelliteB. minimum two geostationary satellite `180^(@)` apartC. minimum three geostationary satellite `120^(@)` apartD. minimum four geostationary satellites `90^(@)` apart |
| Answer» Correct Answer - C | |
| 20. |
If `E` and `B` represent electric and magnetic field vectors of the electromagnetic wave, the direction of propagation of eletromagnetic wave is along.A. `vecE`B. `vecB`C. `vecB xx vecE`D. `vecE xx vecB` |
| Answer» Correct Answer - D | |
| 21. |
Electric communication was discovered in which century ?A. SixteenthB. EighteenthC. NineteenthD. Twentieth |
| Answer» Correct Answer - C | |
| 22. |
A modulating signal is a square wave as shown in the following figure. While the carrier wave is given by, e = 4 sin (87`pi`t) volt. What is the modulation index ?A. 0.15B. 0.25C. 0.35D. 0.5 |
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Answer» Correct Answer - B The amplitude of the modulating signal = 1V The amplitude (or peak voltage ) of the carrier wave = 4 V Thus `A_m`= 1V and `A_C` = 4 V `therefore mu=A_m/A_C=1/4=0.25` |
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| 23. |
An audio signal given by `e_1 = 12 sin 2pi`(2000 t) amplitude modulates a sinusoidal carrier wave `e^2 = 50 sin 2pi` ( 100, I00t). What is the upper side band frequency?A. a. 101 KHzB. b. 99 KHzC. c. 100 KHzD. d. 102 KHz |
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Answer» Correct Answer - D Comparing the two equations with the standard equation `e=A sin 2 pint` , we get `f_m`=2000 Hz = 2 KHz = frequency of the modulating signal and `f_C` = 1,00,000 Hz = 100 KHz = frequency of the carrier wave `therefore` Upper side band =`f_C+f_m`= 100 + 2 =102 KHz |
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| 24. |
In a YDSE, the separation between slits is `2 mm` where as the distance of screen from the plane of slits is `2.5 m`. Light of wavelengths in the range `200-800 nm` is allowed to fall on the slits. Find the wavelengths in the visible region that will be present on the screen at `1 mm` from central maximum. Also find the wavelength that will be present at that point of screen in the infrared as well as in the ultraviolet region. |
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Answer» Here, `d=2xx10^(-3)m, D=2.5m x=10^(-3)m` For a wavelenght to be present on the screen, interference must be constructive. `therefore (xd)/(D)=n lambda` `lambda=(xd)/(nD)` Put the values of x,d and D to obtain values of `lambda` for n-1,2,3... |
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| 25. |
Both amplitude modulation (AM) and frequency modulation (FM) are used for radio broadcasting. The amplitude of the high frequency carrier wave is varied or modulated in accordance with the variations in the amplitude of the audio signals that are to be transmitted, in te process of amplitude modulation in frequency modulation te amplitude of the carrier remains constant but its frequency is varied inn accordance withh the audio signal. Reception with AM signals is in general is affected by interference of various kinds ad elaborated equipment is required for FM broadcast. In india the radio signals are boarcast using-A. AM signals onlyB. AM signals and FM. Signals all over the countryC. AM signals at few of the stations and FM Signals at a few selected stationsD. FM signals only |
| Answer» Correct Answer - C | |
| 26. |
Both amplitude modulation (AM) and frequency modulation (FM) are used for radio broadcasting. The amplitude of the high frequency carrier wave is varied or modulated in accordance with the variations in the amplitude of the audio signals that are to be transmitted, in te process of amplitude modulation in frequency modulation te amplitude of the carrier remains constant but its frequency is varied inn accordance withh the audio signal. Reception with AM signals is in general is affected by interference of various kinds ad elaborated equipment is required for FM broadcast. The rangeof a transmitter isA. less low frequeciesB. same for all frequenciesC. less for high frequenciesD. none of these |
| Answer» Correct Answer - C | |
| 27. |
Statement-1:Noise reduction is possible in frequency modulation (FM) while it is not feasible in AM STATEMENT-2 In FM, amplitude of transmitted signal remains constantA. Statement-1 is True. Statement-2 is True Statement-2 is a correct explanation for statement-5B. Statement is True Statement-2 is True Statement-2 is NOT a correct explanation for Statement-5C. Statement-1 Is True Statement-2 is FalseD. Statement-1 is False. Statement 2 is True |
| Answer» Correct Answer - A | |
| 28. |
STATEMENT-1 Communication in optical range of frequency allows a large number of channels to be transmitted simultaneously STATEMENT-2 : The bandwidth availability is directly proportional to frequency of carrier waveA. Statement-1 is True. Statement-2 is True Statement-2 is a correct explanation for statement-6B. Statement is True Statement-2 is True Statement-2 is NOT a correct explanation for Statement-6C. Statement-1 Is True Statement-2 is FalseD. Statement-1 is False. Statement 2 is True |
| Answer» Correct Answer - A | |
| 29. |
A sinusoidal carrier wave of amplitude 80 V and frequency 600 KHz is amplitude modulated by a sinusoid voltage of frequency 30 KHz. What is the ratio of upper and lower side band frequencies ?A. `20/19`B. `21/19`C. `25/18`D. `23/17` |
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Answer» Correct Answer - B `f_C` = 600 KHz and `f_m`=30 KHz `therefore` Upper side band = `f_C +f_m` = 600 +30 = 630 KHz and Lower side band `=f_C-f_m`=600-30 = 570 KHz `therefore "U.S.B"/"L.S.B"=630/570=21/19` |
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| 30. |
A sinusoidal carrier voltage of frequency 1200 kHz is amplitude modulated by a sinusoidal voltage of frequnecy 20 kHz. The maximum and minimum modulated carrier amplitudes of 100 V and 90 V are produced. Calculate the frequency of lower and upper side bands, unmodulated carrier amplitude, modulation index and amplitude of each side band. |
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Answer» Here, `v_(c) = 1200 kHz` `v_(m) = 20 kHz` Lower side band frequnecy = `v_(c) - v_(m)` `1200 - 20= 1180 kHz` Upper side band frequency `= v_(c) + v_(m)` `= 1200 + 20 =1220 kHz` Unmodulated carrier amplitude `A_(C) = (A_(max) + A_(min))/2 = 110+90/2 = 100 V` Modulation index `(mu)` `mu = (A_(max)- A_(min))/(A_(max) +A_(min)) = 110-90/110 +90 = 0.1` Amplitude of each side band `= (mu A_(c))/(2) = (0.1 xx 100 / 2) = 5 V` |
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| 31. |
A modulated carrier wave has maximum and minimum amplitudes of 800 m v and 200 m V. What is the percentage modulation? |
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Answer» Here, `A_(max) = 800 m V`, `A_(min) = 200 m V` `mu_(a) = (A_(max) - A_(min))/(A_(max) + A_(min))= (800-200)/(800+200) xx 100%` = 60%. |
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| 32. |
A transmitting antenna of height 25 m is fitted at the top of a tower of height 20m. What is the height of the receiving antenna from ground level if the maximum distance between transmitting antenna and receiving antenna on ground is 56 km. Radius of earth = 6400 km. |
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Answer» Here, height of transmitting antenna `h_(T) = 25+ 20 = 45m`, Maximum distance between transmitting and receiving antenna. `d = 56km = 56 xx 10^3 m`. Let `h_(R)` be the height of receiving antenna in metre. As, `d = sqrt(2Rh_(T)) + sqrt(2Rh_(R))` `:. 56 xx 10^3` ` = sqrt(2xx(6400 xx 10^3) xx 45) + sqrt(2 xx(6400 xx 10^3) xx h_(R))` or ` 56 xx 10^3 = 24 xx 10^3 + sqrt (2 xx 6400 xx 10^3 xx h_R)` or ` 2 xx 64 xx 10^5 xx h_R = [(56 - 24) xx 10^3]^2` `(32 xx 10^3)^2` `h_(R) = (32 xx 10^3)^2/(2xx64 xx 10^5) = 80m` . |
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| 33. |
Write the full forms of the terms: (i) LAN (ii) WWW. |
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Answer» (i) LAN stands for Local Area Network. (ii) WWW stands for World Wide Web. |
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| 34. |
What is the basic idea of mobile telephony? What is the frequency range used in the operation of mobile phones? |
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Answer» Mobile telephony is the provisions of services to phones, which may be moved around freely, rather than staying fixed in one location, The mobile phones operate typically in Ultra High Frequency (UHF) range of 800 - 950 MHz.` |
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| 35. |
A transmitting antenna at the top of a tower has a height `32 m` and the height of the receiving antenna is `50 m`. What is the maximum distance between them for satisfactory communication in `LOS` mode? Given radius of earth `6.4xx10^(6)m`. |
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Answer» `d_(m)=sqrt(2xx64xx10^(5)xx32)+sqrt(2xx64xx10^(5)xx50)m` `=64xx10^(2)xxsqrt(10)+8xx10^(3)xxsqrt(10)m` `=144xx10^(2)xxsqrt(10)m=45.5km` |
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| 36. |
A transmitting antenna at the top of a tower has a height of `45 m` and height of the receiving antenna is `80 m`. What is the maximum distance between them , for satisfactory communication of LOS mode ? Radius of earth ` = 6400 km`. |
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Answer» Correct Answer - ` 56 km` Here , `h_(T) = 45 m , h_(R) = 80 m`, `R = 6400 km = 6.4 xx 10^(6) m` Maximum distance between transmitting and receiving antennas in LOS mode is ` d_(m) = sqrt( 2 R h_(T)) + sqrt ( 2 R h_(R))` ` = sqrt ( 2 xx ( 6.4 xx 10^(6)) xx 45) + sqrt ( 2 xx ( 6.4 xx 10^(6)) xx 80)` `= 3 xx 8 xx 10^(3) + 8 xx 4 xx 10^(3) = 56 xx 10^(3)` ` = 56 km` |
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| 37. |
A transmitting antenna at the top of a tower 32 m and height of receiving antenna is 200 m . What is the maximum distance between them for satisfactory communication in LOS mode ? Radius of earth `6.4xx10^6` m ? |
| Answer» Correct Answer - `224XX10^2sqrt10` m | |
| 38. |
Find the minimum length of antenna used to transmit a radio signal of frequency of 30 MHz. |
| Answer» Correct Answer - 2.5 m | |
| 39. |
Show that the minimum length of antenna required to transmit a radio signal of frequency 10 MHz is 7.5m. |
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Answer» v=10 MHz=`10^7` Hz, `lambda=c/v=(3xx10^8)/10^7`=30 m Minimum length of antenna =`lambda/4`=7.5 m |
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| 40. |
What should be the length of a half wave dipole antenna required to transmit audio frequency signals?A. a.5 kmB. b.10 kmC. c.15 mD. d.7.5 km |
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Answer» Correct Answer - D Audio signals have a frequency range of 20 Hz to 20 KHz `=20xx10^3` Hz `therefore f=20 xx 10^3` Hz The corresponding wavelength `lambda=c/f=(3xx10^8)/(20xx10^3)=(300xx10^6)/(20xx10^3) = 15xx10^3` m = 15 km `therefore` Length of the antenna =`lambda/2` = 7.5 km |
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| 41. |
What is an analog communicating system? |
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Answer» A communication system which makes use of analog signals is called analog communication system e.g., telex. |
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| 42. |
Name of the type of communication in which the signal is a discrete and binary coded version of the message. |
| Answer» Digital communication. | |
| 43. |
The AM wave is equivalent to the summation ofA. two sinusoidal wavesB. three sinusoidal wavesC. four sinusoidal wavesD. none of these |
| Answer» Correct Answer - B | |
| 44. |
In an `FM` system a `7 kHz` signal modulates `108 MHz` carrier so that frequency deviation is `50 kHz`. The frequency modulation index isA. 7. 143B. 8C. 0.71D. 350 |
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Answer» Correct Answer - A Carrier swing=`"Frequency deviation "/"Modulating frequency"` `=(50xx10^3)/(7xx10^3)=50/7=7.143` |
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| 45. |
A message signal of freuency `omega_(m)` is superposed on a carrier wave of frequency `omega_(c)` to get an amplititude modulated wave (AM). The frequency of the AM wave will beA. `omega_m`B. `omega_c`C. `(omega_c+omega_m)/2`D. `(omega_c-omega_m)/2` |
| Answer» Correct Answer - B | |
| 46. |
The AM wave contans three frequencies viz:A. `f_(c)/2,(f_(c)+f_(s))/2,(f_(c)-f_(s))/2`B. `2f_(c),2(f_(c)+f_(s)),2(f_(c)-f_(s))`C. `f_(c),(f_(c)+f_(s)),(f_(c)-f_(s))`D. `f_(c),f_(c),f_(c)` |
| Answer» Correct Answer - C | |
| 47. |
Which of the following would produce analog signals and which would produce digital signals? |
| Answer» (i) analog (ii) analog (iii) digital (iv) digital. | |
| 48. |
In a detector, output circuit consists of `R = 10kOmega and C = 100 p F.` Calculate the frequency of carrier signal it can detect. |
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Answer» Here, `R = 10kOmega = 10^4Omega` `C = 100 p F = 10^(-10) F` Time constant of the circuit `tau = RC = 10^4 xx 10^(-10) = 10^(-6)`s For demodulation, the condition is `1/v_C lt lt RC or v_c gt gt 1/(RC)` i.e., `v_c gt gt 1/10^(-6) = 10^(6) Hz` `:.` Frequency of carrier signal must be much greater than `10^6 Hz or 1 MHz.` |
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| 49. |
Due to economic reasons, only the upper side band of an AM wave is transmitted, but at the receiving station, there is a facility for generating the carrier. Show that if device is available which can multiply two signals, then it is possible to recover the modulating signal at the receiver station. |
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Answer» Let `omega_c and omega_m` be the angular frequency of carrier waves and message signal wave respectively. For simplicity, let the signal received at the receiving station be `e = E_1 cos (omega_c + omega_m)t` Let the instantaneous voltage of carrier waves `(e_c)` given by `e_c = E_c cos omega_ct ` be avialable at the receiving station, Now, `e xx e_c = E_1E_c cos (omega_c + omega_m) t cos omega_ct` `(E_1E_c)/2 [ cos {(omega_c + omega_m) t + omega_ct} + cos { (omega_c + omega_m) t - omega_ct}]` `= (E_1E_c)/2[cos(2omega_c+omega_m) t + cos omega_mt]` At the receiving station, when the signal is passed through low-by pass filter, it will pass the high frequency signals `(2omega_c + omega_m)` but obstruct the low frequency signal `omega_m`. Therefore, we can record the modulating signal `= (E_1E_c)/2 cos omega_mt`, which is a signal of angular frequency `omega_m`. |
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| 50. |
A 400 W carrier wave is modulated to a depth of 75%. Calculate the total power in the modulated wave. |
| Answer» Correct Answer - 512.5 watt | |