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51.

Assertion: Space waves are used for litre -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
52.

Which of the following modes is used for line of sightA. Ground waveB. Sky wave propagation is useful in the range of 30 to 40 MHz.C. Space waveD. All of these

Answer» Correct Answer - C
Space waves are used for line of sight communication as well as satellite communication.
Discussion
53.

In a video signal for transmission of picture. What value of bandwidth is used in communication system?A. 2.4 MHzB. 4.2MHzC. 24MHzD. 42MHz

Answer» Correct Answer - B
A video signal of bandwidth 4.2 MHz is used for transmission of pictures.
Discussion
54.

Three waves A,B and C of frequencies 1500 KHz. 6MHz and 50 MHz respectively are to be transmitted from one place to another. Which of the following is the most appropriate mode of communciation?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
For ground wave propagation , the frequency range is 530kHz to 1710 kHz. For sky wave propagation wave progation the frequency range is 54 MHz to 4.2 GHz. Thus option (b) is true.
Discussion
55.

The sum of the heights of transmitting and receiving attennas in line of sight of communication is fixed at h, find the height of two antennas when r ange is maximum.A. `h//2`B. 2hC. hD. 4h

Answer» Correct Answer - A
The range of line-of-sight communication between two antennas is given by
`v=sqrt(2Rh_(T))+sqrt(2Rh_(R))`
Given `h_(T)+h_(R)=h` and let `h_(T)=H`, then `h_(R)=h-H`
`therefore r=sqrt(2R)[sqrt(H)+sqrt(h-H)]`
For r to be maximum,
`(dr)/(dH)=sqrt(2R)[(1)/(2sqrt(H))+(1)/(2sqrt(h-H)(-1))]=0`
or `(1)/(2sqrt(H))-(1)/(2sqrt(h-H))=0` or `H=h-H` or `H=(h)/(2)`
Discussion
56.

The maximum line-of -sight distance `d_(M)` between two antennas having heights `h_(T)` and `H_(R)` above the earth isA. `sqrt(R(h_(T)+h_(R)))`B. `sqrt(2R//(h_(T)+h_(R)))`C. `sqrt(Rh_(T))+sqrt(2Rh_(R))`D. `sqrt(2Rh_(T))+sqrt(2Rh_(R))`

Answer» Correct Answer - D
The maximum line-of-sight distance `d_(m)` between the two antennas having heights `h_(t)` and `h_(g)` above the earth is given by
`d_(M)=sqrt(2Rh_(T))+sqrt(Rh_(R))`
Discussion
57.

In a communication system, operating at 1200 mm, only 2% of the source frequency is available for T.V. transmission , having a bandwidth of 5 MHz. The number of T.V. channels that can be transmitted isA. 2 millionB. 10 millionC. 0.1 millionD. 1 million

Answer» Correct Answer - D
Optical source frequency, `v=(c)/(lambda)`
or `v=(3xx10^(8)xxms^(-1))/(1200xx10^(-9)m)=2.5xx10^(14)Hz`
Bandwidth of channel=2% of the source frequency
`=(2)/(100)xx2.5xx10^(14)Hz=5xx10^(12)Hz`.
Number of channels `=("Total bandwidth of channel")/("Bandwidth needed per channel")`
`=(5xx10^(12)Hz)/(5xx10^(6)Hz)=10^(6)`=1 million
Discussion
58.

A microwave telephone link operating at the central frequency of 10GHz has been established if 2% of this is available for microwave communication channel. Then how many telephone channels can be simultaneously granted if each telephone is allotted a bandwidth of 8 KHz?A. `1.5xx10^(3)`B. `3.5xx10^(2)`C. `2.5xx10^(4)`D. `4.5xx10^(6)`

Answer» Correct Answer - C
Microwave frequency used in telephone link
=10GHz `=10xx10^(9)Hz=10^(10)Hz`
Frequency available for microwave communication
`=2% "of" 10 GHz=(2)/(100)xx10^(8)Hz`
Bandwidth of each telephone channel
=8KHz=`8xx10^(3)Hz`
Number of microwave telephone channels
`=(2xx10^(8))/(8xx10^(3))=2.5xx10^(4)`
Discussion
59.

An audio signal is modulated by a carrier wave of 20 MHz such that the bandwidth required for modulation is 3 kHz. Could this wave be demodulated by a diode detector which has the values of R and C as (i) `R = 1 k Omega, C = 0.01 mu F (ii) R = 10 k Omega, C = 0.01 muF (iii) R = 10k Omega , C = 1 mu muF`.A. `R=1kOmega,C=0.01muE`B. `R=10kOmega,C=0.01muF`C. `R=10kOmega,C=0.1pF`D. None of these

Answer» Correct Answer - A
`v_(c)=20MHz=20xx10^(6)Hz`
`(1)/(v_(c))=(1)/(20)xx10^(-6)=0.05xx10^(-6)=0.5xx10^(-7)s`
Band width =`20_(m)=3kHz=3xx10^(3)Hz`
`v_(m)=1.5xx10^(3)Hz`
`(1)/(v_(m))=(1)/(1.5)xx10^(-3)=0.7xx10^(-3)s`.
Diode detector can demodulate when
`(1)/(v_(c))ltltRCltlt(1)/(v_(m))`
(i)RC=`1kOmegaxx0.01muF=10^(3)xx(0.01xx10^(-6))=10^(-5)s`. it can be demodulated.
(ii) RC=`10kOmegaxx1xx10^(-8)=10^(-4)s` It can be demodulated
(iii) RC`=10kOmegaxx1pF=10^(4)xx10^(-12)=10^(-8)s`
It cannot be demodulated.
Discussion
60.

If the sum of the heights of transmitting and receiving antennas in line of sight of communication is fixed at h, show that the range is maximum when the two antennas have a height `h//2` each.A. `h//2`B. 2hC. hD. 4h

Answer» Correct Answer - A
The range of line-of-sight communication between two antennas is given by
`v=sqrt(2Rh_(T))+sqrt(2Rh_(R))`
Given `h_(T)+h_(R)=h` and let `h_(T)=H`, then `h_(R)=h-H`
`therefore r=sqrt(2R)[sqrt(H)+sqrt(h-H)]`
For r to be maximum,
`(dr)/(dH)=sqrt(2R)[(1)/(2sqrt(H))+(1)/(2sqrt(h-H)(-1))]=0`
or `(1)/(2sqrt(H))-(1)/(2sqrt(h-H))=0` or `H=h-H` or `H=(h)/(2)`
Discussion
61.

Estimate the fastest bit rate capable of being carried by light of wavelength `1.3 mu m`. How many phone calls could be carried at this bit rate ? Band width of optical fibre = 2 GHz.A. `2.9xx10^(7)`B. `1.5xx10^(6)`C. `2.3xx10^(5)`D. `1.7xx10^(4)`

Answer» Correct Answer - C
Here, `lambda=1.3xx10^(-6)Hz`
`v=(c)/(lambda)=(3xx10^(9))/(1.3xx10^(-6))=2.3xx10^(14)Hz`
Max bit rate `=2U=2xx2.3xx10^(14)=4.6xx10^(14)Hz`
For optical fibre, band width =2GHz=`2xx10^(2)Hz`
`therefore` No. of phone calls `=(4.6xx10^(14))/(2xx10^(4))=2.3xx10^(3)`
Discussion
62.

In optical communication system operating at 1200nm, only 2% of the source frequency is available for TV t ransmission having a bandwidth of 5 MHz. the number of TV channels that can be transmitted isA. 2 millionB. 10 millionC. 0.1 millionD. 1 million

Answer» Correct Answer - D
Optical source frequency, `v=(c)/(lambda)`
or `v=(3xx10^(8)xxms^(-1))/(1200xx10^(-9)m)=2.5xx10^(14)Hz`
Bandwidth of channel=2% of the source frequency
`=(2)/(100)xx2.5xx10^(14)Hz=5xx10^(12)Hz`.
Number of channels `=("Total bandwidth of channel")/("Bandwidth needed per channel")`
`=(5xx10^(12)Hz)/(5xx10^(6)Hz)=10^(6)`=1 million
Discussion
63.

Optical communication system having an operating wavelength `lambda` metre can use only x % of its source frequency as its channel band width. The system is to be used for transmitting T.V signals requiring a band width of F hertz. How many channels can this system transmit simultaneously? Show graphically how this number of channels varies with operating wavelength of the system.A. `lambdaF//Cx`B. `2_(Cx)//100lambdaF`C. ` cx//50lambdaF`D. `cx//100lambdaF`

Answer» Correct Answer - D
Optical s ource frequency `=(c)/(lambda)Hz`
Bandwidth of channel `=x% "of" (c)/(lambda)=(cx)/(100lambda)`
Number of channels.
`n=("Total bandwidth of channel")/("Bandwidth needed per channel")=(cx//100A)/(F)=(cx)/(100lambdaF)`
Discussion
64.

A modulating signal is a square wave as shown in figure. The carrier wave is given by `c(t)=2 sin(8pit)"volt"`. The modulation index isA. 0.2B. 0.3C. 0.4D. 0.5

Answer» Correct Answer - D
Here, `A_(m)=1V,A_(c)=2V`,
Modulation index, `mu=(A_(m))/(A_(c))=(1)/(2)=0.5`
Discussion
65.

Assertion: The television signals are propagated through sky waves. ltbr. Reason: Television signals have freqeuncy in the range of 1000MHz to 2000MHz range.

Answer» Correct Answer - D
As television signals being of frequency 100 MHz to 200 MHz cannot be reflected by ionosphere but they penetrate it, so they not propagated through sky waves, in fact, television signals are propagated through space wave propagation.
Discussion