Communication Systems 1. A basic communication system consists of (1) receiver () information source (3) user of information (4) transmitter (5) channe Choose the correct sequence in which these are arranged in a basic communication system. (a) 1345 (b) 4513 (c) 4531 (d) 5143. Which of the foowing is not a transducer? (a) Microphone (b) Loudspeaker (c) Earphone (d) Ampifier 3. If the maximum ampitude of an ampitude moduated wave is 0 V and the minimum ampitude is 4 V, then the moduation index is (a) 1 3 (b) 3 (c) 3 (d) 1 5 4. Which of the foowing statements is correct? (a) Puse position denotes the time of rise or fa of the puse ampitude. (b) Moduation index m is kept 1 to avoid distortion. (c) The audibe range of frequencies is 0 Hz to 00 MHz. (d) In ampitude moduation, the bandwidth is thrice the moduating frequency. 5. AM is used for broadcasting because (a) it requires ess transmitting power compared with other systems. (b) it is more noise immune than other moduation systems. (c) its use avoids receiver compexity. (d) no other moduation system can provide the necessary bandwidth faithfu transmission. 6. A transmitting antenna of height 0 m and the receiving antenna of height h are separated by a distance of 40 km for satisfactory communication in ine-of-sight mode. Then the vaue of h is (Radius of the earth is 6400 km) (a) 0 m (b) 40 m (c) 45 m (d) 50 m 7. Which of the foowing woud produce anaog signas? (1) Light puse. () Musica sound due to a vibrating sitar string. (3) Output of a NOR gate. (4) A vibrating tuning fork. (a) 1 and 3 (b) and 4 (c) 1, and 3 (d) A 8. Identify the correct match from the foowing. (a) Haf dupex Mobie phone device (b) Fu dupex Moduator and device demoduator (c) Modem Waky taky (d) Fax Transmission and reproduction of documents at a distant pace 1 PHYSICS FOR YOU november 14 Page 1
9. A message signa of frequency 10 khz and peak votage of 10 V is used to moduate a carrier of frequency 1 MHz and peak votage of 0 V. Then the moduation index and the side band frequencies respectivey are (a) 0.05 and (1 ± 0.01) MHz (b) 0.5 and (1 ± 0.01) MHz (c) 0.05 and (1 ± 0.005) MHz (d) 0.5 and (1 ± 0.005) MHz 10. A diode AM detector with the output circuit consisting of R = 1 MW and C = 1 pf woud be more suitabe for detecting a carrier signa of (a) 0.1 MHz (b) 0.5 MHz (c) 1 MHz (d) 10 MHz 11. Of the foowing which is preferred moduation scheme for digita communication? (a) Puse Ampitude Moduation (PAM) (b) Puse Position Moduation (PPM) (c) Puse Width Moduation (PWM) (d) Puse Code Moduation (PCM) 1. Match the List I with List II and seect the correct answer using the codes given beow the ists. List I (Name of Service) List II (Frequency band) P. AM broadcast 1. 3.7-4. GHz Q. FM broadcast. 40-890 MHz R. UHF TV 3. 88-108 MHz S. Downink of sateite 4. 540-1600 khz communication Codes: P Q R S (a) 3 4 1 (b) 4 3 1 (c) 1 3 4 (d) 1 4 3 13. The bock diagram of a detector for AM signa is as shown in the figure. Identify the boxes X and Y. X (a) Ampifier (b) Enveope detector (c) Rectifier (d) Rectifier Y Rectifier Rectifier Enveope detector Ampifier 14. In TV broadcasting both picture and sound are transmitted simutaneousy. In this (a) audio signa is frequency moduated and video signa is ampitude moduated. (b) both audio and video signas are frequency moduated. (c) audio signa is ampitude moduated and video signa is frequency moduated. (d) both audio and video signas are ampitude moduated. 15. Which of the foowing waves are used for sateite communication? (a) Ground (b) Space (c) Sound (d) Sky 16. Which of the foowing is an exampe of pointto-point mode of communication? (a) Internet (b) Radio (c) Teephony (d) TV 17. In an ampitude moduated wave for audio frequency of 1000 Hz, the appropriate carrier frequency wi be (a) 10 Hz (b) 100 Hz (c) 1 khz (d) 1 MHz 18. A 1000 khz carrier wave is moduated by an audio signa of frequency range 500-5000 Hz. Then the bandwidth of the channe is (a) 10 khz (b) 0 khz (c) 5 khz (d) 50 khz 19. The sky wave propagation is suitabe for radiowaves of frequency (a) upto 3 MHz (b) from 3 MHz to 0 MHz (c) from 3 MHz to 30 MHz (d) from 3 MHz to 50 MHz 0. In the daytime, ionosphere consists of (a) D, E, and F 1 ayers ony (b) E, F 1 and F ayers ony (c) E and F ayers ony (d) D, E, F 1 and F ayers PHYSICS FOR YOU november 14 13 Page 13
1. How many AM broadcast stations can be accommodated in a 100 khz bandwidth if the highest frequency moduating a carrier is 5 khz? (a) 5 (b) 10 (c) 15 (d) 0. A radiostation has two channes. One is AM at 100 khz and the other FM at 89.5 MHz. For good resuts you wi use (a) onger antenna for the AM channe and shorter for the FM. (b) shorter antenna for the AM channe and onger for the FM. (c) same ength antenna wi work for both. (d) information given is not enough to say which one to use for which. 3. What shoud be the ength of the dipoe antenna for a carrier wave of frequency 300 MHz? (a) 0.5 m (b) 1 m (c) 1.5 m (d) m 4. A message signa of anguar frequency w m is superposed on a carrier wave of anguar frequency w c to get an ampitude moduated wave (AM). The anguar frequency of the AM wave wi be (a) w m (b) w c wc + wm (c) (d) w c w m 5. The radiating power of a inear antenna of radiating ength for a waveength is proportiona to (a) (b) (c) (d) 6. A ground receiver station is receiving a signa at 5 MHz and transmitted from a ground transmitter at a height of 300 m ocated at a distance of 100 km. The signa is coming via (Radius of earth = 6.4 10 6 m, N max of ionosphere = 10 1 m 3 ) (a) sky wave (b) space wave (c) sateite communication (d) a of these 7. An ampitude moduated wave is as shown in figure. The vaue of peak carrier votage and peak information votage respectivey are (a) 5 V, 15 V (c) 15 V, 10 V (b) 10 V, 15 V (d) 5 V, 5 V 8. A radar has a power of 1 kw and is operating at a frequency of 10 GHz. It is ocated on a mountain top of height 500 m. The maximum distance upto which it can detect object ocated on the surface of the earth is (Radius of earth = 6.4 10 6 m) (a) 80 km (b) 16 km (c) 40 km (d) 64 km 9. An AM radio station operating at 530 khz is permitted to broadcast audio frequencies upto 5 khz. The band pass fiter in its moduation circuit can retain the frequencies (a) 1060 khz, 530 khz (b) 535 khz, 530 khz (c) 5 khz, 530 khz (d) 10 khz, 5 khz 30. Figure shows a communication system. What is the output power when input signa is 1.01 mw? Gain in db = 10 og10 Po P i 14 PHYSICS FOR YOU november 14 Page 14
(a) 90 mw (c) 11 mw Soutions (b) 101 mw (d) 10 mw 1. (b) : The bock diagram of a basic communication system is as shown in the figure. Thus, the correct sequence is 4513.. (d) : Any device that converts one form of energy into another is caed a transducer. Loudspeaker, microphone and earphone are transducers but not an ampifier. 3. (c): Let A c and A m be the ampitude of carrier wave and message signa respectivey. Then Maximum ampitude, A max = A c + A m... (i) Minimum ampitude, A min = A c A m... (ii) On soving eqns (i) and (ii), we get Amax + Amin Amax Amin Ac = and Am = Am Amax Amin Moduation index, m= = Ac Amax + Amin Here, A max = 0 V, A min = 4 V \ m = 0 V 4 V 16 V = = 0 V+ 4 V 4 V 3 4. (a) : Moduation index m 1 to avoid distortion. The audibe range of frequencies is 0 Hz to 0 khz. In ampitude moduation, the bandwidth is twice the moduating frequency. 5. (c): AM is used for broadcasting because its use avoids receiver compexity. Ony a diode and a capacitor are sufficient to separate the audio signa from the AM wave. 6. (c): The maximum ine-of-sight distance d M between two antennas having heights h T and h R is dm = RhT + RhR where R is the radius of the earth, h T and h R are the heights of the transmitting and receiving antennas respectivey. Here, d M = 40 km = 40 10 3 m, h T = 0 m, h R = h, R = 6400 km = 6400 10 3 m 3 3 3 \ 40 10 = 6400 10 0 + 6400 10 h 3 3 5 40 10 = 16 10 + 64 10 h 3 [( 40 16) 10 ] h = = 45 m 5 64 10 7. (b) : A vibrating tuning fork and musica sound due to a vibrating sitar string produce anaog signas. Light puse and output of a NOR gate produce digita signas. 8. (d) : Haf dupex device Waky taky Fu dupex device Mobie phone Modem Moduator and demoduator Fax Transmission and reproduction of documents at a distant pace 9. (b) : Here, u m = 10 khz = 0.01 MHz, A m = 10 V, = 1 MHz, A c = 0 V A Moduation index, m 10 V m= = = 05. Ac 0 V The side band frequencies are u SB = ± u m = (1 ± 0.01) MHz 10. (d) : Here, R = 1 MW = 10 6 W C = 1 pf = 10 1 F \ RC = (10 6 )(10 1 ) = 10 6 s For demoduation, 1 1 << RC or uc >> uc RC where is the frequency of the carrier signa. 1 6 \ >> = 10 Hz = 1 MHz 6 10 Thus, it woud be more suitabe for detecting a carrier signa of 10 MHz. 11. (d) : The preferred moduation scheme for digita communication is puse code moduation. PHYSICS FOR YOU november 14 17 Page 17
1. (b) : Name of service AM broadcast FM broadcast UHF TV Frequency band 540-1600 khz 88-108 MHz 40-890 MHz Downink of sateite 3.7-4. GHz communication P-4, Q-3, R-, S-1 13. (c): The bock diagram of a detector for AM signa is as shown in the figure. The box X is a rectifier and the box Y is an enveope detector. 14. (a) : In TV broadcasting both AM and FM are used. AM is for video (picture) signa and FM is for audio (sound) signa. 15. (b) : Space waves are used for sateite communication. 16. (c): In point-to-point communication mode, communication takes pace over a ink between a singe transmitter and a receiver. Teephony is an exampe of such a mode of communication. 17. (d) : Carrier frequency > Audio frequency 18. (a) : Bandwidth is twice the maximum frequency of moduating signa. \ Bandwidth of the channe = (u m ) max = 5000 Hz = 10000 Hz = 10 khz 19. (c): The radiowaves of frequency 3 MHz to 30 MHz are used in sky wave propagation as they are refected by the ionosphere of earth s atmosphere. 0. (d) : Layers of ionosphere D E Exists during Day ony Day ony F 1 Daytime, merges with F at night F Day and night 1. (b) : Here, tota bandwidth = 100 khz = 100 10 3 Hz Any station being moduated by a 5 khz signa wi produce an upper side frequency 5 khz above its carrier and a ower side frequency 5 khz beow its carrier, thereby requiring a bandwidth of 10 khz. Thus, Number of stations accommodated 3 Tota bandwidth = Bandwidth per station = 100 10 Hz = 10 3 10 10 Hz. (b) : For the communication of AM channe of frequency 100 khz, ground wave propagation is used. For this the antenna need not be very ta. For the communication of FM channe of frequency 89.5 MHz, the space wave propagation is needed. For this, very ta towers are used as antenna. 3. (a) : Here, u = 300 MHz = 300 10 6 Hz = 3 10 8 Hz Length of the dipoe antenna = = c = 8 1 3 10 ms = 05. m u 8 3 10 Hz 4. (b) 5. (b) : Power radiated by the inear antenna is proportiona to (/). 6. (a) : Maximum distance covered by space wave communication = Rh 6 = 6. 4 10 300 = 6 km As the distance between transmitter and receiver is 100 km, so propagation via space wave is not possibe for the signa of frequency 5 MHz. For sky wave propagation, Critica frequency, = 9(N max ) 1/ = 9(10 1 ) 1/ = 9 10 6 Hz = 9 MHz As signa of frequency 5 MHz is ess than, so the propagation of signa of frequency 5 MHz is possibe via sky wave. 18 PHYSICS FOR YOU november 14 Page 18
7. (c) : From figure, 50 V V max = = 5 V 10 V V min = = 5 V Peak carrier votage, V V V c = max + min 5V+ 5V 30 V = = = 15V Peak information votage, V V Vm = max min 5V 5V 0 V = = = 10 V 8. (a) : A d R M O R h R B Let M be the mountain top where a radar has been operating to ocate the objects on earth. Let d (MA = MB) be the maximum distance of radar from earth where object can be detected. Refer figure, d + R = (h + R) = h + hr + R or d = h + hr Since h << R, so h can be negected in comparison to hr. \ d = hr or d= hr Here, h = 500 m, R = 6.4 10 6 m \ d = 500 ( 64. 10 6 ) = 80 10 3 m = 80 km 9. (b) : The band pass fiter retains the frequencies, u m and + u m and rejects the frequencies u m, u m and. Here, = 530 khz, u m = 5 khz \ The frequencies retained by the band pass fiter are = 530 khz u m = (530 5) khz = 55 khz + u m = (530 + 5) khz = 535 khz 30. (b) : FOR MORE DETAILED SOLUTION REFER NOVEMBER ISSUE OF PHYSICS FOR YOU Here, transmission path = 5 km Loss suffered in transmission path = db km 1 5 km = 10 db Tota ampifier gain = 10 db + 0 db = 30 db Overa gain of the signa = 30 db 10 db = 0 db As, gain in db = 10 og10 P \ o 0 = 10 og10 P i Po P i or Po 0 og10 og10 100 Pi = 10 or Po = 100 Pi or P o = P i 100 = 1.01 mw 100 = 101 mw PHYSICS FOR YOU november 14 19 Page 19