Tiêu đề 15.COMMUNICATION SYSTEMS - Explanations.pdf ✅

1 (c) From Z0 = √ L C , C = L Z0 2 = 0.4 × 10−3 160 × 160 = 15.62 × 10−19F = 15.62 nF 4 (c) Here, fm = 10 kHz ∴ fSB = fc ± fm = (712 ± 10) = kHz = 702 kHz and 722 kHz 5 (d) Here, h = 150 km, v = 300 kHz, vc = 100 kHz Dskip = 2h√( v vc ) 2 − 1 = 2 × 150 × 103√( 300 × 102 100 × 103 ) − 1 = 849 Hz 6 (b) The diameter of optical fiber is 10−4 cm 7 (d) The population inversion required for laser action in solid state laser is optical pumping 8 (b) We uses three geostationary satellites placed at the vertiex of an equilateral triangle, then the entire earth, can be covered by the communication network each satellite covers 1/3 of the globe. 9 (c) μ = √1 − 81.45N v 2 = √1 − 81.45 × 400 × 106 (200 × 103) 2 = 0.43 μ = c v ∴ Phase velocity, v = c μ = 3×108 0.43 = 6.98 × 108ms −1 10 (b) Electric flux φ = pulse power area = 1012 10−4 = 1016 Wcm−1 11 (c) A communication satellite receives the coming modulated microwave signal, amplifies it and return it to earth at a different frequency 12 (d) All the three types of energy losses persist in transmission lines 13 (b) Core of acceptance angle θ = sin−1 √n1 2 − n2 2 14 (d) The space wave communication is utilized in all the three communication ie, television, radar and microwave communication 15 (a) Ozone layer extends from 30 km to nearly 50 km above the earth’s surface in ozone sphere. This layer absorbs the major part of ultraviolet radiations coming from the sun and does not allow them to reach the earth’s surface. The range of ultraviolet radiations is 100 Å to 4000 Å. Thus, it blocks the radiations of wavelength less than 3 × 10−7 m (or 3000 Å). 16 (d) All the three modulation techniques are used for transforming digital data into analog signals 17 (b) Characteristic impedance of a coaxial cable is between 50 Ω to 70 Ω. 18 (d) Few advantages of optical fibres are that the number of signals carried by optical fibres is much more than that carried by the Cu wire or radio waves. Optical fibres are practically free from electromagnetic interference and problem of cross talks whereas ordinary cables and microwave links suffer a lot from it. 19 (b) The relativistic mass M is M = γm Where, m is rest mass and γ = 1 √1− u2 c2 is the Lorentz factor having u the relative velocity between the observer and object and c is the speed of light. ∴ M = m √1 − u 2 c 2 Given, M = 2m ∴ 2m = m √1 − u 2 c 2 ⇒ (√1 − u 2 c 2 ) 2 = ( 1 2 ) 2 ⇒ 1 − u 2 c 2 = 1 4 ⇒ u 2 c 2 = 1 − 1 4 = 3 4 ⇒ u 2 c 2 = 3 4 ⇒ u c = √3 2
⇒ u = √3 2 c = √3 2 × 3 × 108 = 2.59 × 108 ms−1 20 (d) All the four pulse systems are employed in pulse modulation 21 (d) Carrier frequency > audio frequency 22 (a) Optical source frequency f = c λ = 3 × 108 /(800 × 10−9 ) = 3.8 × 1014 Hz Bandwidth of channel (1% of above) = 3.8 × 1012Hz Number of channels = (Total bandwidth of channel)/(Bandwidth needed per channel) = (3.8 × 1012)(8 × 103 ) ∼ 4.8 × 108 23 (d) The radio waves can be sent from one place to another through following propagation mode. (i) space wave propagation, sky wave propagation and ground wave propagation 25 (d) The frequency optical communication v = c λ or v = 3 × 108 1200 × 10−9 v = 25 × 1013 Hz But only 2% of the source frequency is available for TV transmission v ′ = 2.5 × 1014 × 2% Or v ′ = 2.5 × 1014 × 2 100 Or v ′ = 5 × 1012 Hz ∴ Member of channels = v ′ bond width Or No of channels = 5×1012 5×106 = 106 = 1 million 26 (d) The critical frequency of sky wave undergoing reflection from a layer of ionosphere is fc = 8 √Nmax Where N is electron density per m3 . Nmax = fc 2 81 = (10 × 106) 2 81 = 1.2 × 1012 m−3 27 (b) A laser device produces amplification in the ultraviolet or visible region. 28 (c) Ultra high frequency waves are generally normally propagated 30 (b) Venus looks brighter than other stars because it is closer to earth than other stars. 31 (c) Taking refraction from air to core of optical fiber. Then, μ1 μ0 = sin 10° sin7° = 0.1736 0.1219 = 1.424 or μ1 = 1.424 (as μ2 = 1) 32 (d) Hubble’s law is the statement in physical cosmology that the red shift in light coming from distant galaxies is proportional to their distance V = H0D Where V is the recessional velocity due o red shift, H0 is Hubble’s constant, D is the proper distance that the light had travelled from the galaxy in the res frame of the observer. This effect helps us in explaining the fact that galaxies are receding from us. 33 (d) The virtual height and critical frequency of E- layer is 110 km and 4 MHz 34 (b) A laser beam is used for carrying out surgery because, it can be sharply focused due to a very high energy per unit area is involved at a location. 35 (c) An antenna is a metallic structure used to radiate or receive EM waves 36 (b) Given d = 12.8 km, R = 6400 km We have d = √2hR h = d 2 2R = (12.8) 2 2 × 6400 = 12.8 m 37 (a) d = √2hR ⇒ d ∝ h 1/2 38 (b) v. f. = 1 √k = 1 √2.6 = 0.62 39 (b) The normalised fiber frequency, v = 2πa λ0 (μ1 2 − μ2 2 ) 1/2 v = 2πa λ0 × 100 = 1% 40 (a) Characteristic impedance (Z0) of a co-axial wire line is lower ie, from 40 to 15Ω 42 (c) The characteristic impedance of a coaxial cable is of the order of 270 Ω. 43 (a) Working of optical fiber is based on total internal reflection. Hence, μ1 > μ2 44 (c) The power of laser is 0.5 W, let n photon/s are incident by laser pulse on the broken ligament, then n × hv = 0.5 W
⇒ nhc λ = 0.5 ⇒ n = 0.5 × λ hc = 0.5×632×10−9 6.626×10−34×3×108 = 1.59 × 1018 photon/s So, number of photons contained in 5 pulses are, n × 5 × (20 × 10 3) = 1.59 × 1023 45 (b) A LASER (Light Amplification by Stimulated Emission of Radiation) is an optical source that emits photons in a coherent beam. The property of coherent beam is used I many military applications to enhance the locating of distant objects. Since, a laser beam typically has low divergence, the laser light appears as a small spot even at long distances, the user simply places the spot on the desired target and barrel of the gun is aligned. 47 (d) The relativistic kinetic energy of a particle of rest mass m0 is given by K = (m − m0 ) c 2 m = m0 √1−(v 2/c 2) , where m is the mass of the particle moving with velocity v. ∴ K = [ m0 √1 − (v 2/c 2) − m0 ] c 2 z According to problem kinetic energy = rest energy ∴ [ m0 √1 − (v 2/c 2) − m0 ] = m0c 2 or m0c 2 √1−(v 2/c 2) = 2m0c 2 or 1 1−(v 2/c 2) = 4 or 4v 2 c 2 = 3 ∴ v = √3c 2 48 (c) Laser beam is almost perfectly parallel and have low divergence. 50 (b) Maximum Range of the radar is given by Rmax = ( PtA 2S 4πλ 2Pmin) 1/4 Where Pt : peak value of transmitted power A: capture area of the receiving antenna S: radar cross-sectional area λ: wavelength of RADAR wave Pmin: minimum receivable power of the receiver 52 (b) Radiowaves of frequency 30 MHz to 300 MHz belong to very high frequency band 56 (d) For good demodulation, 1 f << RC or , RC >> 1 f 57 (c) If maximum electron density of the ionosphere in Nmax per m3 , then fc = 9(Nmax) 1/2 . Above fc , a wave will penetrate the ionosphere and is not reflected by it. 58 (d) An antenna is a form of tuned circuit consisting of inductance and capacitance, and as a result it has a resonant frequency, at the frequency where the capacitive and inductive reactance’s cancel each other out. At this point the antenna appears purely resistive, the reactance being a combination of the loss resistance and radiation resistance. 60 (a) Eg = hc λ = (6.63 × 10−34) × (3 × 108 ) (1400 × 10−9) × (1.6 × 10−19) = 1 eV 62 (b) Atleast three geostationary satellite in the same orbit around the earth can cover the whole part of the earth for microwave communication 63 (a) From R = ρ l a = ρl πD2/4 , when l = 1 m, D = 2 mm, R = 0.1 Ω 64 (d) In space communication, the information can be passed from one place to another with the speed of light(c = 3 × 108 ms −1 ). Hence, time taken for a distance of 100 km = 100×103 3×108 = 3.3 × 10−4 s 65 (c) Here : Velocity of electromagnetic waves in free space and wavelength v = 3 × 108m/s and λ = 150m The frequency of radio waves is given by = v λ = 3 × 108 150 = 2 × 106 Hz = 2 MHz 66 (a) d = √2hR = √2 × 500 × 6.4 × 106 m = 80,000m = 80 km 67 (d) Frequency range for microwaves is 1 × 109 to 3 × 1011 Hz 68 (b) C = 1 nF = 10−9F, L = 10μH = 10−5H v = 1 2π√LC = 1 2π√10−5 × 10−9 = 107 2π
= 1.592 × 104 Hz = 1592 kHz 69 (a) Some important wireless communication frequency bands Name of Service Frequency bands Remarks Standard AM broadcast 540 – 1600 kHz FM broadcast 88 – 108MHz Television 54 – 72MHz VHF (very high frequencie s) 76 – 88MHz (VHF) TV 174 – 216MHz UHF (ultra high frequencie s) 420 – 890MHz (UHF) TV Cellular Mobile Radio 896 – 901MHz Mobile to base station 840 – 935MHz Base station to mobile Satellite Communi- cation 5.925 – 6.425 GHz Uplink 3.7 – 4.2GHz Downlink 71 (a) Here, λ0 = 393.3 nm λ = 401.8 nm Red shift, z = λ−λ0 λ0 = 401.8−393.3 393.3 = 8.5 393.3 = 0.0216 Now z = v c then v = zc = 0.0216 × 3 × 108 = 6.48 × 106 ms−1 = 6480 km s −1 72 (a) Optical communication is made of communication by which we can transfer the information from one place to another through optical carrier waves. Light frequencies used in optical communication system lie between 1014 Hz to 4 × 1014 Hz (ie, 100,000 to 400,000 GHz.) 74 (c) When electromagnetic waves enter the ionised layer of ionosphere, then the relative permittivity of the ionised layer appears to decrease 75 (a) In the communication system, amplitude modulation is used for broadcasting to avoide receiver complexity. 76 (b) Modulation does not change time log between transmission and reception. 77 (d) When ma > 1 then carrier is said to be over modulated 78 (c) 2000 number of telephonic messages can be carried by a optical fiber at an instant 79 (c) The sky waves are the radiowaves of frequency between 2 MHz to 30 MHz. These waves can propagate through atmosphere but reflected back by the ionosphere of earth’s atmosphere. These waves can propagate from transmitter to receiver through sky, therefore, their propagation is called sky wave propagation. 80 (d) An optical fibre is a device based on total internal reflection by which at light signal can be transferred from one place to the other with a negligible loss of energy. 82 (a) By using fc ≈ 9(Nmax) 1/2 ⇒ fc ≈ 2 MHz 83 (b) Some of the characteristics of an optical fibre are as follows. (i) This works on the principle of total internal reflection. (ii) It consists of core made up of glass/silica/plastic with refractive index n1, which is surrounded by a glass or plastic cladding with refractive index n2 (n2 > n1 ). The refractive index of cladding can be either changing abruptly or gradually changing (Graded index fibre). (iii) There is a very little transmission loss through optical fibres. (iv) There is no interference from stray electric and magnetic fields to the signals through optical fibres. 84 (c)