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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.
| 51. |
If audio signal is tranmitted directly into space, the length of the transmitting antenna required will beA. extremely smallB. extremely largeC. infinitely largeD. moderate |
| Answer» Correct Answer - B | |
| 52. |
A `1 MHz` carrier signal is modulated by a symmertrical sinusoidal wave for period of `100 mus` in a nonlinear (square lae device).Which of the following frequencies will notbe present in the modulated signalA. `990 kHz`B. `1010kHz`C. `1020kHz`D. `1030kHz` |
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Answer» Correct Answer - C The frequency components of modulation signal are from nonlinear device are `f_(c)`, `f_(c)+-f_(m)`, `f_(c)+-3f_(m)`…… So the frequencies present in the `AM` are `100kHz`, `990 kHz`, `1010 kHz`, `970 kHz`, `1030 kHz` |
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| 53. |
A message signal `m(t)=4cos2000pit` modulates the carrier `C(t)=cos2pif_(c)t` where `f_(c)=1MHz` to produce an `AM` signal. For demodulation using envelope detector the time constant `RC` should satisfyA. `0.5ms lt RC lt 1ms`B. `1mus lt lt RC lt lt 1ms`C. `RC gt gt 1mus`D. `RC gt gt 1mus` |
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Answer» Correct Answer - B `f_(c)=1 MHz` and `f_(m)=2000Hz` `(1)/(f_(c)) lt lt RC lt lt (1)/(f_(m))` |
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| 54. |
Depth of modulation in terms of `E_(max)` and `E_(min)` isA. `m_(a)=(E_(max)+E_(min))/(E_(min))`B. `m_(a)=(E_(max)-E_(min))/(E_(max))`C. `m_(a)=(E_(max)+E_(min))/(E_(max)+E_(min))`D. `m_(a)=(E_(max)+E_(min))/(E_(max)-E_(min))` |
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Answer» Correct Answer - C `A_(max)=A_(c)+A_(m)` `A_(min)=A_(c)-A_(m)implies mu=(A_(m))/(A_(c))=(A_(max)-A_(min))/(A_(max)+A_(min))` |
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| 55. |
A `1000 kHz` carrrier is simultaneously modulated with `300Hz`, `800Hz` and `2 kHz` audio waves. The frequencies present in the output areA. `999.7 kHz`, `1000.3kHz`, `999.2kHz`B. `1000.8 kHz`, `998kHz`, `1002kHz`C. `1002.8 kHz`, `996kHz`, `1106kHz`D. Both (`1`) and (`2`) |
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Answer» Correct Answer - D frequencies in output are `f_(c)-f_(m)` & `f_(c)+f_(m)` |
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| 56. |
Audio sine waves of 3 kHz frequency are used to amplitude modulate a carrier signal of 1.5 MHz. Which of the following statements are true?A. The sideband frequencies are 1506 kHz and 1494 kHz.B. The bandwidth required foe amplitude modulation is 6 kHz.C. The bandwidth required for amplitude modulation is 3 MHzD. The sideband frequencies are 1503 kHz and 1494 kHz |
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Answer» Correct Answer - B::D `LSB = (omega_c) = (omega_m)` `1.5xx 10^6 - 3xx 10^3` `= 1497 kHz` and `USB = (omega_c) +(omega_m)` `= 1503 kHz` Bandwidth required `= USB - LSB =2omega_m` `= 2xx3 kHz = 6 kHz`. |
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| 57. |
A audio signal `20sin2pi(1500t)` amplitude modulates `40sin2pi(10^(5))t`. The depth of modulation isA. `25%`B. `20%`C. `50%`D. `40%` |
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Answer» Correct Answer - C `%mu=(A_(m))/(A_(c))xx100` |
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| 58. |
The audio signal used to modulate `60 sin (2pixx10^(6)t)` is `15 sin 300 pi t`. The depth of modulation isA. `50%`B. `40%`C. `25%`D. `15%` |
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Answer» Correct Answer - C |
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| 59. |
`C(t)` and `m(t)` are used to generate an `AM` signal. The modulation indes of generated `AM` signal is `0.5`. Then the quantity `(P_("Total "SB))/(P_("Carrier"))=`A. `1//8`B. `1//4`C. `2//3`D. `9//8` |
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Answer» Correct Answer - A `(P_(Total SB))/(P_(Carrier))=(P_(c)(m^(2)/2))/(P_(c))=(1)/(8)` |
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| 60. |
A: Satellite communication uses different frequency bands for uplink and downlink B: Banwidth of video signals is `4.2 MHz`A. `A` is true but `B` is falseB. `A` is false but `B` is trueC. `A` and `B` are falseD. `A` and `B` are true |
| Answer» Correct Answer - D | |
| 61. |
If the TV telecast is to cover a radius of 120 km ( Given , the radius of the earth = 6400 km ), the height of the transmitting antenna isA. 1280 mB. 1125 mC. 1560 mD. 79 m |
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Answer» Correct Answer - B As ` d=sqrt(2hR)` `implies h=(d^(2))/(2R) =((120 xx 1000)^(2))/(2 xx 6400xx 1000)=1125m` |
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| 62. |
In a diode `AM`- dectector, the output circuit consist of `R=1 kOmega` and `C=10 pF`. A carrier signal of `1000 MHz` is to be detected. Is it goodA. YesB. NoC. Information is not sufficientD. None of these |
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Answer» Correct Answer - A For demodulation `(1)/(f_(c)) lt lt RC` |
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| 63. |
In a diode AM- detector, the output circuit consist of `R=1 k Omega` and `C=10 pF`. A carrier signal of `100 kHz` is to be detected. Is it good?A. YesB. NoC. Information is not sufficientD. None of these |
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Answer» Correct Answer - B |
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| 64. |
What should be the maximum acceptance angle at the aircore interface of an optical fibre if 1 n and 2 n are the refractive indices of the core and the cladding, respectivelyA. `sin^(-1) (n_(2)//n_(1))`B. `sin^(-1) sqrt(n_(1)^(2) - n_(2)^(2))`C. `[tan^(-1) (n_(2))/(n_(1))]`D. `[tan^(-1) (n_(1))/(n_(2))]` |
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Answer» Correct Answer - B |
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| 65. |
Coaxial cable is an example ofA. Optical fibreB. Free spaceC. Wire mediumD. Sea medium |
| Answer» Correct Answer - C | |
| 66. |
Optical fibre communication is generally preferred over general communication system becauseA. it is more efficientB. of signal securityC. both (`1`) & (`2`)D. it is easily available |
| Answer» Correct Answer - C | |
| 67. |
Mean optical power launched into an `8 km` fibre is `120 muW` and mean output power is `4 muW`, then the overall attenuation is (Given log `30=1.477`)A. `14.77 dB`B. `16.77 dB`C. `3.01 dB`D. None of these |
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Answer» Correct Answer - a |
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| 68. |
An audio signal of amplitude 0.1 V is used is used in amplitude modulation of a carrier wave of amplitude 0.2 V. Calculate the modulation index.A. 1B. 0.5C. 1.5D. 2 |
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Answer» Correct Answer - B Here , ` A_(m)=0.1V , A_(c)=0.2 V , mu=? ` ` mu=(A_(m))/(A_(c))=(0.1)/(0.2)=0.5` |
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| 69. |
A 107.6 MHz carrier signal is frequency modulated by a 7kHz sine wave . The resultant FM signal has a frequency deviation of 50 kHz. Determine the modulation index and the carrier swing of the FM waveA. 7.143 and 100 kHzB. 8.234 and 75 kHzC. 6.289 and 150 kHzD. 5.103 and 14 kHz |
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Answer» Correct Answer - A Given `, f_(c)=107.6 MHz , f_(m)=7kHz and Deltaf=50kHz ` The modulation index of the FM wave canbe given as ` m_(f)=("Frequency deviation ")/("Modulating frequency ")` `=(Deltaf)/( f_(m))=( 50 xx 10^(3))/( 7 xx 10^(3))=7.143` Carrier swing ` = 2 xx Deltaf = 2 xx Delta f ` ` = 2 xx 50 xx 10^(3) =100 kHz ` |
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| 70. |
The process of regaining of information from carrier wave at the receiver is termed asA. demodulationB. modulationC. attenuationD. amplification |
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Answer» Correct Answer - A Demodulation is the process of regaining (retrieval ) of information from carrier wave at the receiver . This is the reverse process of modulation. |
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| 71. |
`AM` is used for broad casting because,A. it is more noise immune than other modulating systemsB. it requires less transmitting power compared with other systemsC. its use avoids receiver complexityD. no other modulation system can provide the necessary bandwidth, faithful transmission. |
| Answer» Correct Answer - C | |
| 72. |
Modulation is not used to :-A. reduce the bandwidthB. to separate the transmission of different usersC. to ensure that intelligence may be transmitted to long distancesD. to allow the uses of practical antenna |
| Answer» Correct Answer - A | |
| 73. |
The process of translating the information contained by the low base band signal to high frequencies is calledA. DetectionB. ModulationC. AmplificationD. demodulation |
| Answer» Correct Answer - B | |
| 74. |
During the process of modulation the `RF` wave is calledA. Modulating waveB. Modulated waveC. Carrier waveD. Audio wave |
| Answer» Correct Answer - C | |
| 75. |
Why is the transmission of signals using ground waves restricted up to a frequency of 1500 kMz? |
| Answer» This is because at frequencies higher than 1500 kHz, there is an increase in the absorption of signal by the ground. | |
| 76. |
Which waves constitude amplitude-modulated band? |
| Answer» Electromagnetic waves of frequency less than 30 MHz constitude amplitude- modulated band. | |
| 77. |
State the two functions performed by a modem. |
| Answer» (i) Modulation (ii) Demodulation. | |
| 78. |
Give the frequency ranges of the following (i) High frequency band (HF) (ii) Very high frequency band (VHF) (iii) Ultra high band (UHF) (iv) Super high frequency band (SHF). |
| Answer» (i) `3 MHz "to" 30 MHz` (ii) `30 MHz "to" 300 MHz` (iii) `300 MHz "to" 3000 MHz` (iv) `3000 MHz "to" 30,000 MHz`. | |
| 79. |
If `2f_(c)` and `3f_(m)` are the frequencies of carrier wave and signal, then the band width isA. `f_(m)`B. `6f_(m)`C. `f_(c)`D. `2f_(c)` |
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Answer» Correct Answer - B Frequencies in output are `2f_(c)-3f_(m)` and `2f_(c)+3f_(m)` |
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| 80. |
The total power content of an AM wave is `900` W. For `100%` modulation, the power transmitted by each side band isA. 50 WB. 100 WC. 150 WD. 200 W |
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Answer» Correct Answer - C |
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| 81. |
The maximum distance between the transmitting and receiving TV towers is `D`. If the heights of both transmitting and receiving towers are doubled then the maximum distance between them becomesA. `2D`B. `D//sqrt(2)`C. `4D`D. `D//2` |
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Answer» Correct Answer - B `D=sqrt(2Rh_(T))+sqrt(2Rh_(R))` |
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| 82. |
The maximum distance between the transmitting and receving TV towers is `64 km`. If the ratio of the heights of the TV transmitting tower to receiving tower is `4 : 9`, the heights of the transmitting and receiving tower areA. `51.2 m`, `80m`B. `1280m`, `2880m`C. `80m`, `125m`D. `25m`, `75m` |
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Answer» Correct Answer - B `d=sqrt(2Rh_(T))+sqrt(2Rh_(R))`, `(h_(T))/(h_(R))=(4)/(9)`, `d=64km` |
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| 83. |
Maximum usable frequency (MUF) in F-region layer is x, when the critical frequency is 60 MHz and the angle of incidence is `70^(@)`, then x isA. `150 MHz`B. `170 MHz`C. `175 MHz`D. `190 MHz` |
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Answer» Correct Answer - C |
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| 84. |
Maximum usable frequency (`MUF`) in `F`-region layer is `x`, when the critical frequency is `60 MHz` and the angle of incidence is `60^(@)`. Then `x` isA. `150 MHz`B. `170 MHz`C. `120 MHz`D. `190 MHz` |
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Answer» Correct Answer - C `MUF=(f_(c))/(costheta)` |
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| 85. |
The transmission media can be:A. guided onlyB. unguided onlyC. both (1) and (2)D. neither (1) nor (2) |
| Answer» Correct Answer - C | |
| 86. |
Statement-1: in high latitude one sees colourful curtains of light hanging down from high altitudes. Statement-2: The high energy charged particles from the sun are deflected to polar regions by the magnetic field.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false. |
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Answer» Correct Answer - A |
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| 87. |
A `400 watt` carrier is modulated to a depth of `80%`. Calculate the total power in the modulated wave.A. `528 W`B. `128 W`C. `256 W`D. `400 W` |
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Answer» Correct Answer - A `P_(Total)=P_(C)(1+(m^(2))/(2))=528 watts` |
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| 88. |
Calculate modulation index if carrier waves is modulated by three signals with modulation indices as `0.6`, `0.3` and `0.4`A. `1.0`B. `0.70`C. `0.78`D. `1.3` |
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Answer» Correct Answer - C `m_(t)=sqrt(m_(1)^(2)+m_(2)^(2)+m_(3)^(2))=0.78` |
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| 89. |
Identify the mathematical expression for amplitude modulated wave:A. `A_(c )sin[{omega_(c )+k_(1)V_(m)(t)}t+phi]`B. `A_(c )sin{omega_(c )t+phi+k_(2)V_(m)(t)}`C. `{A_(c ) + k_(2)V_(m)(t)}sin(omega_(c )t +phi)`D. `A_(c) V_(m)(t) sin (omega_(c ) t+phi)` |
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Answer» Correct Answer - C Consider a sinusoidal modulating signal represented by `A_(c) V_(m)(t) sin (omega_(c ) t+phi)` where, `A_(m)` = Amplitude of modulating signal `omega_(m)` = Angular frequency `= 2pi V_(m) = phi V_(m)` Also consider a sinusoidal carrier wave represented by `C(t) = A_(c ) sin omega_(c )t" .....(ii)"` Thus, modulated wave is given by `C_(m) (t) = (A_(c ) +A_(m)sin omega_(m)t)sin omega_(c ) t` `=A_(c)[1+(A_(m))/(A_(c))sin omega_(m)t)sinomega_(c)t` Here, `(A_(m))/(A_(c ))=M` `implies C_(m)(t) = (A_(c ) +A_(c)xxmusinomega_(m)t)sin omega_(c )t" .....(iii)"` Now, we know thta `A_(c) xx mu = K` [wave constant] and `sin omega_(m)t = V_(m)` [wave velocity] Thus, Eq. (iii) becomes `C_(m) (t) = (A_(c) +K xx V_(m))sin omega_(c ) t` Now, consider a change in phase angle by `phi` then `sin omega_(c ) t rarr sin (omega_(c ) t + phi)` Thus, `C_(m)(t) = (A_(c)+KV_(m))(sin omega_(c ) +phi)` |
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| 90. |
(i) The intensity of a light pulse travelling along a communication channel decreases exponentially with distance x according to the relation `I = I_0 e^(-alphax)`, where `I_0` is the intensity at ` x = 0 ` and `alpha` is the attenuation constant. Show that the intensity reduces by 75 percent after a distance of `(ln 4)/(alpha)` (ii) Attenuation of a signal can be expressed in decibel (dB) according to the relation dB `= 10log_10 (I//I_0).` What is the attenuation in `dB//km` for an optical fibre in which the intensity falls by 50 percent over a distance of 50 km? |
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Answer» (a) Given, the intensity of a light pulse `I=I_(0)e^(-alphax)`, where, `I_(0)` is the intensity at x = 0 and `alpha` is constant. According to the question, I = 25% of `I_(0)=(25)/(100).I_(0)=(I_(0))/(4)` Using the formula mentioned in the question, `I=I_(0)e^(-alphax)` `(I_(0))/(4)=I_(0)e^(-alphax)` or `(1)/(4) = e^(-alphax)` Taking log on both sides, we get `ln1-ln4=-alphaxlne" "(becauselne=1)` `-ln4=-alphax` `x=(ln4)/(alpha)` Therefore, at distance `x=(ln4)/(alpha)`, the intensity is reduced to 75% of initial intensity. (b) Let `alpha` be the attenuation in dB/km. If x is the distance travelled by signal, then `10log_(10)((I)/(I_(0)))=-alphax " ....(i)"` where, `I_(0)` is the intensity initially. According to the question, I = 50 % of `I_(0)=(I_(0))/(2)` and x = 50 km Putting the value of x in Eq. (i), we get `10log_(10).(I_(0))/(2I_(0))=-alphaxx50` `10[log1-log2]=-50 alpha` `(10xx0.3010)/(50)=alpha` `therefore` The attenuation for an optical fibre `alpha = 0.0602` dB/km |
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| 91. |
On radiation (sending out) an AM modulated signal, the total radiated power is due to energy carried by `omega_c, omega_c - omega_m and omega_c + omega_m`. Suggest ways to minimise cost of radiation without compromising on information. |
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Answer» In amplitude signal, only side band frequencies contain information. Thus only `(omega_(c ) + omega_(m))`and `(omega_(c )-omega_(m))` contain information. Now, according to question, the total radiated power is due to energy carried by `omega_(c), (omega_(c)-omega_(m))` and `(omega_(c)+omega_(m))`. Thus to minimise the cost of radiation without compromising on information `omega_(c)` can be left and transmitting. `(omega_(c ) + omega_(m)), (omega_(c )-omega_(m))` or both `(omega_(c ) + omega_(m))` and `(omega_(c )-omega_(m))`. |
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| 92. |
Effective power radiated by an antenna isA. Proportional to the square at the length of the antennaB. inversely proportional to the wavelengthC. inversely proportional to the square of the wavelengthD. proportional to the wavelength |
| Answer» Correct Answer - C | |
| 93. |
Statement A: If the antenna is vertical the vertically polarised `EM` wave is radiated Statement B: The vertically polarised `EM` wave has electrical variations in the vertical planeA. `A`is true but `B` is falseB. `A` is false but `B` is trueC. `A` and `B` are falseD. `A` and `B` are true |
| Answer» Correct Answer - D | |
| 94. |
A TV tower has a height of `150 m`. The area of the region covered by the TV broadcast is (Radius of earth `= 6.4 xx 10^(6) m` )A. `9.6pi xx 10^(8)m^(2)`B. `19.2pi xx 10^(7)m^(2)`C. `19.2pi xx 10^(10)m^(2)`D. `19.2pi xx 10^(3)km^(2)` |
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Answer» Correct Answer - D |
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| 95. |
Space wave propagation is used in (`a`) microwave communication (`b`) satellite communication (`c`) TV transmissionA. television communicationB. radar communicationC. microwave communicationD. All of them |
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Answer» Correct Answer - D |
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| 96. |
In amplitude modulation, the modulation index mu, is kept less than or equal to 1 because.A. `m gt 1`, will result in interference between carrier frequency and message frequency, resulting into distortionB. `m gt 1`, will result in overlapping of both side bands resulting into loss of informationC. `m gt 1`, will result in change in phase between carrier signal and message signalD. `m gt 1`, indicates amplitude of message signal greater than amplitude of carrier signal resulting into distortion |
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Answer» Correct Answer - B::D The modulation index (m) of amplitude modulated wave is `m=("amplitude of message signal"(A_(m)))/("amplitude of carrier signal"(A_(c )))` If `m gt 1`, then `A_(m) gt A_(c )`. In this situation, there will be distortion of the resulting signal of amplitude modulated wave. Maximum modulation frequency `(m_(f))` of `A_(m)` wave is `m_(f) = (Deltav_("max"))/(v_(m)("max"))` `=("frequency deviation")/("maximum frequency value of modulating wave")` If `m_(f) gt 1`, then `Deltav_("max")gt v_(m)`. It means, there will be overlapping of both side bands of modulated wave resulting into loss of information. |
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| 97. |
Digital signalsA. provide continuous set of valuesB. represent values are randomlyC. Utilise Decimal code systemD. Utilise binary code system |
| Answer» Correct Answer - D | |
| 98. |
A digital signalA. is less reliable than analog signalB. is more reliable than analog signalC. is equally reliable as the analog signalD. Not at all reliable |
| Answer» Correct Answer - B | |
| 99. |
A digital signal possessA. continuously varying valuesB. only two discrete valuesC. only four discrete valuesD. constant values |
| Answer» Correct Answer - B | |
| 100. |
The audio signalA. can be sent directly over the air for large distanceB. can not be sent directly over the air for large distanceC. possesses very high frequencyD. possesses very low frequency |
| Answer» Correct Answer - B | |