ELECTRONICS SYSTEM & TECHNOLOGIES**Biomedical Technology** 1. A mammography service examined 327 patients during the third calendar quarter of 1996. 719 films were exposed during this period, eight of which were repeats. What is the repeat rate? a. 45.5% b. code 2.45% c. 1.1% d. 54.52% ANS. 1.1% 2. A type of luminescence where the visible light is emitted only during the stimulation of the phosphor. a. fluorescence b. phosphorescence c. after glow d. any of the above ANS. Fluorescence 3. If a heart measures 12.5 cm from side to side at its widest point, and its image on chest radiograph measures 14.7 cm, what is the magnification factor? a. 0.85 b. 1.18 c. 0.15 d. 2.2 ANS. 1.18 MF= (image size)/(object size) MF= (14.7 cm)/(12.5 cm) MF= 1.18 4. 5. 6. Radiation doses in the range of approximately200 to 1000 rad produce the a. Gastrointestinal (GI) syndrome b. Central Nervous System (CNS) syndrome c. Hematologic syndrome d. Prodomal syndrome ANS. Hematologic syndrome Ratio of incident to transmitted radiation through a grid ratio of patient dose with and without grid a. Bucky factor b. Damadian factor c. Grid ratio d. Roentgen equivalent ANS. Bucky factor A radiographic single-phase unit installed in a private office has a maximum capacity of 100 milliseconds of 120kVp and 500mA. What is the power rating? a. 42 kW b. 60 kW c. 600W d. 24kW ANS 42 Kw Power rating=(0.7)x (mA x kVp)/1000 Power rating=(0.7)x ((500)(120))/1000 Power rating=42 kW 7. Who completes the development of first computed tomographic (CT) scanner (EMI, Ltd.) in 1973? a. Mistretta b. Forssmann c. Kuhl d. Hounsfield ANS. Hounsfield 8. It is the study of the response of an image receptor to x-rays a. sensitometry b. xeroradiography c. xeroradiometry d. spectrometry ANS. Sensitometry 9. Ratio of radiographic contrast with a grid to that without a grid a. grid ratio b. contrast improvement factor c. Collidge factor d. Bucky factor ANS. contrast improvement factor 10. Analysis of persons irradiated therapeutically with superficial x-rays has shown that the skin erythema dose required to affect 50% of persons so irradiated is about a. 200 rad b. 400 rad c. 600 rad d. 800 rad ANS. 600 rad 11. ECG recording requires a bandwidth of 0.05 to ___Hz a. 1000 b. 540 c. 100 d. 40 ANS.100 12. Find the potential generated if blood flowing in a vessel with radius 0.9 cm cuts a magnetic field of 250 G. Assume a volume flow rate of 175 cubic cm per second. a. 206 uV b. 309 uV c. 903 uV d. 260 uV ANS. 309 uV E= QB/50πa E= ((175 cc.s)(250 G))/(50π(0.9cm)) E= 309 uV 13. Period during which heart contracts. a. Diastole b. Systole c. block mode Synapse ANS. Systole 14. Apparatus for measuring blood pressure a. defibrillator b. electrocardiogram c. plethysmograph d. sphygmomanometer ANS d. sphygmomanometer d. 15. Given the energy level of 6.624x10^-18 J imparted to an electron stream by an X-ray device, calculate the frequency in MHz. a. 10^6 MHz b. 45.29 MHz c. 10^10 MHz d. 300 MHz ANS. 10^10 MHz f= E/h f= (6.624x〖10〗^(-18) J)/(6.624x〖10〗^(34) J-s) f= 10^10 MHz 16. It was first noted by Heinrich Hertz in 1887 and won Albert Einstein the Nobel Prize in 1905. It refers to the emission of electrons from a clean metallic surface (phototube) when electromagnetic radiation (light waves or X-rays) falls onto that surface. a. Photoelectric effect b. Compton effect c. Bremsstrahlung d. Mie effect ANS. Photoelectric effect 17. Unit of radiation exposure or amount of X-ray radiation that will produce 2.08x10^9 ion pairs per cubic centimetre of air at standard temperature and pressure (STP). a. Curie b. Roentgen c. Radiation absorbed dose (rad) d. Gray ANS. Roentgen 18. An optical electronic device that measures the color concentration of a substance in solution. a. colorimeter b. flame photometer c. spectrophotometer d. chromatograph ANS.colorimeter 19. Recording of heart sounds a. ultracardiography b. cardioacoustics c. electrocardiogram d. phonocardiography ANS. Phonocardiography 20. Recorder for measuring galvanic skin resistance a. electrodermograph b. electromyograph c. electrocorporealograph d. electrogalvanograph ANS.electrodermograph 20.10 Base 5 15. 4.2 Mbps Throughput= (12 000 x 10 000)/60 = 2 Mbps 12. 4A/5A Encoding ANS. It is used in wireless applications in which stations must be able to share the medium without . An IPv4 has an address space of how many? a. A telephone subscriber line must have an SNRdB above 40.4B/5B Encoding 18. foldover distortion c. 6. 20 code ANS. distortion d. 0. Firewall c.0. What proportion of the maximum output voltage is produced? a. 10 Base 5 ANS. Emile Baudot d. 64 kbps c.63 c.2 dB c. flag c. 2B1Q Encoding c. 6. What is the minimum number of bits? a.65 ANS6. determine the energy per bit-to noise power density ratio C = 10e-12 W fb = 60 kbps N = 1.Baudot Code 14. 4294967296 b. slope overload c. ITU d.51. 0.368 Mbps b. In asynchronous transmission. 2449692769 d. PCM d. 6927694924 c.2 dB b. ASCII Code b. Hamming b. 22.678Vmax ANS.786Vmax b. 1.1. 10 Base-T c. Frameguard d. Samuel Baudot ANS. 5.twice the highest frequency of a signal 2.35 13. 0. Baudot b.CRC For the given parameters. twice the highest frequency of a signal ANS.368 Mbps **Electronics/Communications** 1. equal to the highest frequency of a signal c. 3. The thick coaxial cable implementation of standard Ethernet a.stop bit 20. PPM c. 4294967296 232 = 4294967296 16. A coding scheme that records the change in signal level since the precious sample a.2 x 10e-14`W B = 120 kHz a. CRC d. equal to the lowest frequency of a signal b. A network with bandwidth of 10 Mbps can pass only an average of 12000 frames per minute with each frame carrying an average of 10000 bits.length character code developed for machines rather than for people a.84 Mbps 17. neither a nor b ANS. A signal at the input to a mu-law compressor is positive with its voltage one-half the maximum value. William R. Richard W. What is the overhead (number of extra bits) in the DS3 service? a. Inaccuracies caused by the representation of a continuously varying quantity as one of a number of discrete values. 6. 100 Mbps d. trailer b. Flag b.867Vmax d. 4 Mbps ANS. 1000 Base-T b. The minimum sampling rate according to Nyquist Theorem a.ANSI 19. a.Flag 7. A bit pattern or a character added to the beginning and the end of a frame to separate the frames a.**Digital and Data Communications** 1. Footprint ANS. Hamming The most reliable convolutional coding scheme for error detection a. Gray Code c.876Vmax c. What is the throughput of this network? a. 3 Mbps d. 0. Richard W. 5.76 = 40 n = 6. VRC ANS.56 b. stop bit ANS. unipolar code b. Discrete code b. alias signal b. continuous code d. EBCDIC code d.vocoder The early pioneer in the development of errordetection and correction procedures: a. either a or b d. 4B/5B Encoding d. Distortion created by using too low a sampling rate when coding an analog signal for digital transmission a.024 Mbps ANS. Baudot Code ANS. Circuit for digitizing voice at a low data rate by using knowledge of the way in which voice sounds are produced: a.quantizing errors 8. CCITT ANS. 10 Base 2 d. aliasing b. FCC c.876Vmax 9. 9672969442 ANS.048 Mbps d. 16. LRC c. 3. 1. ADC ANS. checksum b.4 GHz ANS.2 dB d. 1 Mbps b.either a or b 11.22. 14. ANSI b. 3. vocoder b. 2 Mbps c. 2. 2.delta modulation 10. Hamming c.2 dB ANS.480 Mbps c. twice the bandwidth of a signal d. codec c. escape byte d. code 39 c.84 Mbps b. 51. quantizing errors ANS.code 39 The first fixed. A block coding technique in which four bits are encoded into a five bit code a.35 SNRdB = 6. sample-and-hold circuit d. A national standards organization that defines standards in the United States a.2 dB . What is the bit rate of STS – 1? a.02nb + 1. It consists of 36 unique codes representing the 10 digits and 26 uppercase letters a.35 d. 2. delta modulation ANS. one or more bits to indicate the end of transmission a. what is the corresponding dBm of this signal? a. What is the multiplier value? a. An antenna with a noise temperature of 75 Kelvin is connected to a receiver input with a noise temperature of 300 K. 438 W c. alias b. Power Line Communications** 1. A/an _______ is a range of frequency in which the upper frequency is double the lower frequency. bandwidth b.12 Megabits 1. Noise figure ANS.97 dB b. Two-dimensional parity check ANS.spread spectrum 3.6 dB c. modulation b. one b. It is a standard designed by ITU to allow telephones in the public telephone network to talk to computers/terminals connected to the internet a. H. parity ANS. An FM transmitter system is using a 1MHz crystal oscillator to generate a very stable 108MHz final carrier frequency. how many bits are needed to send the complete contents of a screen? a. H.6 dB An address space is the total number of addresses used by the protocol. 264 c. 1. a. harmonics . If each pixel uses 1024 colors. The noise temperature is given by: where: NT – noise temp. A multiplexer combines four 100-kbps channels using a time slot of 2 bits. Spread spectrum ANS. H. Gateway c. 46dBm b. In PCM system. 50s c. a. 72 ANS. what is the total transmitted power? a. What is the address space of IPv6? a.6 dB The total noise temperature of the antenna and receiver is NTtot = 75+ 300 = 375. Microelectronics.Dynamic range 7.20s 8. 3. -23 dBm ANS.23 dBm dBm for 200 mW = 10 log (200mW/1mW) = 23 dBm 2.2 dB **Electronic Communications** 1. 210 = 1024 First Multiplier output: 1MHz x 36 = 36 MHz Mixer output: 36MHz + 34.5MHz (filtered output) 36MHz – 34. 2128 ANS.323 3. 12 Gigabits b. Dynamic range b. Simple-parity check d. It is an undesired shift or width change in digital bits of data due to circuitry action which causes bits to arrive at different times a.Checksum It is a connecting device between two internetworks that use different models a. two c. 4. determine the frame duration.5 MHz = 70.5 MHz (difference fed to the next multiplier) Multiplier stage: Output = 108MHz Output = 1. Multiple access c. 3. A computer monitor has a resolution of 1200 by 1000 pixels. 232 b.gateway 5. 400s ANS. a. 11001) a.5 d.2128 10. 120 Megabits d. 1. The sum output of the mixer is filtered out. 391 W b.5 MHz signal. it is the ratio of the maximum input voltage level to the smallest voltage level that can be quantized a. 3 c. 23 dBm d.jitter 6. jitter d. 20s b. Repeater d. 12 Megabits ANS. 688 W ANS. 100s d. five ANS. Figure of merit c.interception by an eavesdropper and without being subject to jamming from a malicious intruder a. find the noise figure of the system. The modulation index of an AM radio station is 0. 641 W d. such that: NF = 10 log NR = 3. Given the reference temperature T0 = 290 K. such as IPv4 and IPv6. If the receiver is receiving 200 mW. 72 Basic Block Diagram of the FM carrier generator: 1 MHz carrier > x 36 > Mixer > Multiplier > 108MHz carrier 11. router ANS. octave d.321 d.5MHz x n N = 108/1. If the carrier power is 500W.29 The Noise figure is the decibel equivalent of the noise ratio. 108 b. NR = Noise ratio Rearranging the above to find NR gives: NR = (375/290) +1 = 2. Bridge b.641 W 4. impulse c. 2 Gigabits c. 0.5 = 72 **Laws and Ethics. 6. multiple access d.02 dB d.5 MHz = 1. frequency doubler c. Checksum b. Two 10101 XOR 11001 = 01100 ( presence of two 1) 9.123 c. The output of the crystal oscillator is fed to a x36 multiplier circuit then mixed with a 34. The sensitivity of a radio receiver is given in terms of dBm.200 times 1000 pixels x 10 bits = 12 Mb 1024 colors requires 10 bits . An error detection method which uses one’s complement arithmetic a. Determine the Hamming distance for the codewords ( 10101. while the difference is fed to another multiplier in order to generate the final carrier frequency. -46 dBm c.75. 323 ANS. Quality factor d. 7. Analog and Digital Electronics. three ANS. 111 b. H. 296 d. CRC c. d. 47.8 μs d. ½ λ b. capacitive c.000 ( 1 + 0. 8.5KHz and the permitted deviation is 4KHz? a.107. 12 c. a.8 = 0. 26% c.26% Г 2 = Pr/Pi Г = SWR-1 SWR+1 . 2. lossless c.octave 2. dielectric constant d. 97% ANS. standing-wave ratio ANS. autodyne ANS. refractive index c.05 +1 Thus : Г 2 = (0. It is the most common means of overcoming the problems of quantizing error and noise. CAT 5 c. The modulated peak value of a signal is 125V and the unmodulated carrier value is 85V.high-level modulator 15.The signal is difficult to demodulate 14.½ λ A mismatched transmission line is also referred to as a ___________ line. 10 b. series-resonant ANS.1/10λ Determine the time taken for a signal to travel down a 10 m transmission line.05 – 1 = 0. twin-lead wire d. a. 108 degrees b. input to an FM receiver has a S/N of 2. 47. A small variable capacitance in parallel with each section of ganged capacitor is called _______.8 Hz c. CAT 4 b. 3. 23. Which of the following is the most widely used UTP cable that can carry baseband data at rates up to 100 Mbps at a range up to 100m? a. One of the advantages of DSB is the savings of great deal of power.6 ns The maximum voltage standing wave of an RG-11/U foam coaxial cable is 52 V and its minimum voltage is 17 V. 2 λ ANS. trimmer 6.7. How many percent of the incident power is the reflected power? a. de-emphasis c. 1/10λ d. shielded-pair ANS. the system is called a _______________________. 66kW d. What is the modulation index? a. 3% d. 14 d. the short-circuited transmission line behaves as a(an) ___________ circuit.47 m = Em/ Ec Emax = Ec + Em .67 Hz d.000 W Pt = 66kW 13. a. 9.82 /2) Pt = 66. a.0.3 Hz ANS.05 Г = 3.47 12. carrierrecovery d.velocity factor 10. 52 degrees d. a. resonant ANS. 14. What is the frequency deviation caused by the noise if the modulating frequency is 1. converter c. SWR = Vmax/ Vmin = 52/17 = 3. 76. ¼ λ c.547. 0.0. How many wavelengths long are required for a pair of conductors to be considered as a transmission line? a. de-emphasis 17.8. 10 a.47 b. a. CAT 6 ANS. If the input in a transmitter is collector-modulated. stripline b. The S/N input is 3:1. “flat” level? a. 5/8 λ ANS. velocity factor b.32 ANS. 3. . 66W c.7)(3x108 m/s) T= 47. 4.3652 16.8 Hz -1 N/S = sin -1 1/2. 250 degrees ANS.108 degrees Θ = 360 td / T T =1/f = 1 / 4MHz = 250 ns Θ = 360 (75) / 250 = 108 degrees Which of the following is a flat conductor separated by an insulating dielectric from a large ground plane that is usually one-quarter or one-half wavelength long? a.8 Hz b. 51% b. padder d. 1. differentialamplifier modulator d. trimmer b.resonant It is the ratio of the speed of propagation on a line to that of light in free-space.3. the signal occupies a narrow bandwidth b. The total transmitted power of an AM broadcast transmitter with a carrier power of 50kW when modulated 80 percent is: a.68 d. a. 1. the signal is weak at high frequencies d.51)2 = 0.ANS. 1 λ d. Em = Emax – Ec = 125 – 85 = 40 V m = 40V/ 85V = 0. the signal is difficult to demodulate ANS.6 ns ANS. 7.51 3. CAT 5e d.5kW b. What circuit is used to return the frequency response of a signal to its normal. if its velocity factor is 0. 40kW ANS. 4. ¼ λ c. At exactly one-quarter wavelength. Determine the phase-shift represented by a 75 ns delay of a 4 MHz signal to a 75-ft cable with a dielectric constant of 2. 62. 547. lossy b. highlevel modulator c. low –level modulator b. lattice modulator ANS. Which of the following is the reason why it is not widely used? a. microstrip One complete revolution around a Smith Chart represents _________ wavelengths.6 ns T= L/ (Vf x Vc) T= 10 m/ (0. the signal has a low S/N c.8 degrees c. 16 ANS. frequency-multiplier ANS. non-resonant d. . microstrip c.460.CAT 5 11. Determine the worst-case output S/N for a narrowband FM receiver with deviation maximum of 10 kHz and a maximum intelligence frequency of 3 kHz.33 ns b. **EST** 1. a. inductive b.66kW Pt = Pc ( 1 + m2/2) Pt = 50. parallel-resonant d.parallel-resonant 5.52ns c. ½λ b.26 The reflected power is 26 % of the incident power : 4. 0. pre-emphasis b.47 c. 2. antenna grounding d. . The shape of the electromagnetic energy radiated from or received by an antenna is called the a.3 dBi and 4. physical construction is easier b.8 dBi 21. loading coil b. Log-Periodic Dipole Array c. Calculate the gain (relative to an isotropic) of a parabolic antenna that has a diameter of 3 m. an LC network c. multiplexing b. buildings and other structures on the ground c. Both antennas have the same gain d. top hat loading c. 1. the number of nodes decreases d. a loading coil d. skin effect b. Antennas that transmit an equal amount of energy in the horizontal direction are called a. coil d. the number of lobes increases LPDA stands for a. amplifying d. radiation pattern d. a.a. directivity 15. Yagi antenna c. a dipole d. a field-strength meter 10. radio signals reflecting off the ground b. a field-strength meter ANS. measured at the feedpoint. The first antenna has a higher gain. electromagnetic pattern c. Log Power Dipole Array ANS. Which of the following is not a reason why most halfwave antennas are mounted horizontally at low frequencies? a. b. companding c. 90% b. none of the above ANS.5 dBd. ________________ means that the characteristics and performance of an antenna are the same whether the antenna is radiating or intercepting an electromagnetic signal. omnidirectional d. directivity d. 79. antenna polarization c. unilateral ANS. a. half-wave dipole d. An antenna's beamwidth is measured a. tower sway ANS. 0. radio signals reflecting off the ground 19.end effect 7. mounting is easier d.8 dBi b. 39. more cost-effective c.66dBi d. 39. 37. 91% c.53 dBi ANS. Marconi antenna b. drooping radials d.\ A half-wave dipole is sometimes called: a. bi-directional b. radiation pattern 13. a. 93% ANS. Calculate the efficiency. LC antenna ANS. antenna reciprocity 3. active antenna b. none of the choices ANS. horizontal range ANS. anti-aliasing ANS. permittivity c. The ability of an antenna to send or receive signals over a narrow horizontal directional range is referred to as a. from front to back d. A dipole antenna has a radiation resistance of 67 ohms and a loss resistance of 5 ohms. focal factor b. the number of lobes increases b. bi-directional array c. Low-Power Directed Array d. between half-power points c. 93% Efficiency = Rr / Rt = 67 / (67 + 5) = 93% Two antennas have gains of 5. all of the choices As the length of a "long-wire" antenna is increased a. 41. a shorted stub b.8 G = 39. Hertz antenna d. respectively. 92% d. Low-Power Dipole Array b. Cannot be determined ANS. a slotted line b.8 G = 10 log 9474. support is easier ANS. companding 18. more cost-effective 12. c.0752 G= 9474. an efficiency of 60% and operates at a frequency of 4 GHz. "Ground Effects" refers to the effects on an antenna's radiation pattern caused by a. Hertz antenna 9. A basic antenna connected to a transmission line plus one or more additional conductors that are not connected to the transmission line form a a.top hat loading 20. antenna pattern ANS. efficiency decreases c. **Electronics System & Technologies** 8. An antenna can be matched to a feed line using a. antenna reciprocity b. faraday effect d. a. The second antenna has a higher gain. active antenna 11.8 dBi G= 2 D2/ 2 = 3x108/ 4MHz = 0. It is a phenomenon caused by any support insulators used at the ends of the wire antenna. an EIRP meter c. parasitic array . signal shape b. between the minor side-lobes ANS. counterpoise ANS. unidirectional c. from +90° to –90° b. end effect c. Log-Periodic Dipole Array Field strength at a distance from an antenna is measured with a. Which has a greater gain? a. 4. The second antenna has a higher gain. omnidirectional 14. between half-power points 5. miller effect ANS. parasitic array b.075 m 6. 2. A receiving antenna with a built-in preamplifier a. passive antenna c. all of the choices ANS. Which of the following methods is used to lower the resonant frequency of a shortened vertical antenna? a. grounding ANS.94 dBi c. fading d. faulty connection of the feed cable ground ANS. A transmitter and a receiver is 45 km apart.shroud 30. TE20 ANS.53 x 108 m/s d. bass drum ANS. 44. 17. the SWR will be 1:10 c. broadside array b. The wavelength of a wave in a waveguide _________. parasitic array ANS. TE11 b.7 dB and a return loss of 26 dB. When the characteristic impedance of the transmission line matches the output impedance of the transmitter and the impedance of the antenna itself. and the load is resistive with an SWR of 3. broadside array 19. 7. 10. is directly proportional to the group velocity ANSWER:is greater than in free space 31.5 m 22. depends only on the waveguide dimensions and the free space wavelength c. a. Yagi antenna 17. 5. 4.5. but absorbs RF power in the opposite direction. minimum power transfer will take place d. and the load is resistive with an SWR of 3. The most ambitious LEO constellation to date is ______. a. Yagi antenna ANS. A source provides 1 W to the isolator. A signal propagated in a waveguide has a full wave of electric intensity change between two further walls.76 m d. dummy load ANS. wave trap ANS.16.4 km d. 45. a balun b. An isolator has a forward loss of 0. TE10 c. How much power returns to the source? a.5m c. An isolator has a forward loss of 0. a. a Q section d. Yagi ANS. is greater than in free space b. 00 latitude c. 10. Marconi antenna c. A metal wrapped around the parabolic antenna aperture to eliminate sidelobes interfering nearby stations is called a. is inversely proportional to the phase velocity d. causing the Bernoulli effect c. None of the choices ANS.Teledesic 2. 638 mW ANS. TE20 26.45 x 108 m/s b. The mode is a. A satellite-dish owner has a 3 meter dish designed for C-band (4 GHz) operation. collinear antenna d. A source provides 1 W to the isolator. maximum power transfer will take place ANS. By how much should two antennas be separated for space diversity in the 11 GHz band? a. 1 W b. 8.5m d. 690 µW ANS. The distance between the satellite and ground station is changing. Hertz antenna b. A one-quarter wavelength of coaxial or balanced transmission line of a specific impedance connected between a load and a source in order to match impedances is a. Iridium b. NAVSTAR c. By how much must the path between the towers clear the obstacle in order to avoid diffraction at a frequency of 11 GHz? a. an autotransformer c. end-fire array c. shield c. 4. How much power is dissipated in the load? a.5 km 25. 9. Why does the downlink frequency appear to vary by several kHz during a low earth orbit satellite pass? a. 54.52 m b. 00 latitude 32. The distance between the satellite and ground station is changing. S/N b. 5. 900 latitude ANS. A stacked collinear antenna consisting of half-wave dipoles spaced from one another by one-half wavelengths is the a. 750 mW d. shroud d. 11.43 x 108 m/s ANS. 350 µW c. Hertz b. It is a microwave device that allows RF energy to pass through in one direction with very little loss. An antenna made up of a driven element and one or more parasitic elements is generally referred to as a a. The owner wants to use the same dish with a new feedhorn. 851 mW c.isolator 29.51 m 24. Suppose that there is an obstacle midway between the transmitter and receiver.44. and no component of the electric field in the direction of propagation.5 km b. Which antennas usually consist of two or more halfwave dipoles mounted end to end? a. circulator c. isolator d. Geostationary satellites are located at ____ with respect to the equator. What effect will the change in . C/N d. causing the Kepler effect b. multiplexer b.7 dB and a return loss of 26 dB. What is the phase velocity of a rectangular waveguide with a wall separation of 3 cm and a desired frequency of operation of 6 GHz? a. 250 mW b. 450 longitude d. The quality of a space-link is measured in terms of the ______ ratio. 5.5m ANS. wide-bandwidth array d.4 km c. The distance between the satellite and ground station is changing.638 mW 27. 535 µW d.43 x 108 m/s 23.5 m b. causing the Boyle’s Law effect d.52 m ANS. 7. The distance between the satellite and ground station is changing. 3.5. TM22 d. G/T 33. Marconi c. causing the Doppler effect ANS. a.43 x 108 m/s 28. a. EIRP ANS. the SWR will be 10:1 b. How far from the transmitter could a signal be received if the transmitting and receiving antennas where 40 m and 20 m.35 x 108 m/s c. collinear d. maximum power transfer will take place 20. The distance between the satellite and ground station is changing. collinear 18. Globalstar ANS. G/T c. radome b. respectively. causing the Doppler effect 1. above level terrain? a. 00 longitude b. a Q section 21. a. Teledesic d. for Ku-band (12 GHz) satellites.51 m c. 775 V ] = 1. Cut-off frequency d. surface waves c. A number from 0 to 1 representing the ability of a surface material to absorb sound energy is known as a.28) + 8 = 137.56 W d. 99.85 W b. a.775 V ] 26 / 20 = log [ voltage in Volts / 0. 6 GHz b. 4. beamwidth decreases to 1/3 of its former value. 1 b. reflection d. 2 c. An acoustical phenomenon wherein the sound continues to persist after the cause of sound has stopped resulting in repeated reflections is called… a.54 dB. 58.75 V b. A loudspeaker produces an SPL of 85dB-SPL at 1 meter distance and input electrical power of 1 W.65 W ANS. d.54 dB.54 dB. reverberation c. beamwidth decreases to 1/3 of its former value. a. 12 GHz ANS. 384 Mm d. The height above the earth’s surface from which a refracted wave appears to have been reflected… a. Ribbon d. 65 dB-SPL d.68 – 120)/10] W = 58. 82 dB-SPL b. Gain decreases by 9. **Printing** 1. MUF c.Condenser c.5 V V (dBu) = 20 log [ voltage in Volts / 0. The highest frequency that can be used for skywave propagation between two specific points on earth’s surface.68 dB-PWL 137. virtual height The highest frequency that can be used for skywave propagation between two specific points on earth’s surface.3) x 0. 12. fading c. 69 dB-SPL SPL@1m/1W = 85 dB-SPL (given) SPL@20m/1W = 85 . b. 69 dB-SPL c. 3. echo b. 7. 46000 km ANS. Condenser .68 W c. 42400 km d. 31 V d.5 Hz + 26 dBu is how many volts? a. Reflection coefficient d. ANS. fraction of an octave f2 = 21/3 x 25 = 31. average height d. a. 5 ANS. Room Constant ANS. center frequency ANS. Gain decreases by 9.5 Hz c. 4. 75 Hz d.5km b. 5. 79 dB-SPL ANS. sky waves A group of filters has 1/3 octave of spacing. 95. space waves ANS. 15. 58. fading 3. 50 Hz b. Dynamic b. a. Critical Frequency 2. 6400 km b.87 km d. If the initial frequency is 25 Hz. the number of polar-orbit satellites required is a.20 log (20m/1m) = 59 dB-SPL SPL@20m/10W = 59 + 10 log (10W/1W) = 69 dBSPL 7.frequency have on the gain and beamwidth of the antenna? a. 380 Mm b.65 Watts 6. c. Absorption Coefficient b.384 Mm Determine the radio horizon for a transmit antenna that is 200 m high and a receiving antenna that is 100 m high a. a.5 km ANS. virtual height c.5 Hz f2 = 2x f1 where x = 1/3. 31. beamwidth increases to thrice of its former value. 386 Mm ANS.5 V A type of microphone that uses the principle of a capacitor as a means of transduction. 4 GHz c. mid frequency ANS.Absorption Coefficient Electromagnetic waves that are directed above the horizon level.68 = 10 log W + 120 W = log-1 [(137. 55.5 km c. 58. What is the length of the path to a geostationary satellite from an Earth station if the angle of elevation is 300? a. Gain increases by 9. what is the next frequency available for the filter? a. weather loss ANS.65 W PWL = SPL + 20 log D(m) + 8 PWL = 100 + 20 log (100/3.54 dB. a. The moon orbits the earth with a period of approximately 28 days. a.5 Hz ANS. 3. 62 V ANS. 56. The by-stander is 100 ft away from the bank. Critical Frequency b. Critical Frequency 4. Critical Frequency b. 99.reverberation 6. mean height ANS. Variation in signal loss caused by natural weather disturbances.775 V Voltage = 15. ground waves b. How loud is the SPL at distance of 20 meters if this speaker is driven to 10 W of electrical power? a. 382 Mm c. It uses a polarizing voltage of between 9 and 48 V of DC supply applied to its diaphragm by an external power supply. beamwidth decreases to 1/3 of its former value.54 dB.775 Volts ] 26 = 20 log [ voltage in Volts / 0. sky waves d. Gain increases by 9. How far away is it? Assume circular orbit. 39000 km c.39000 km 6.Carbon ANS. MUF c. To cover all inhabited regions of the earth. 96. 5.5 V c. 5. attenuation b. Cut-off frequency d. Sound absorption c. 14 GHz **Communications / Broadcast** 1. 15. spreading loss d.3 Voltage = log-1 (1. beamwidth increases to thrice of its former value.1 2. 31. Determine the sound power in Watts produced by the bank’s alarm if a by-stander heard the alarm at a sound pressure level of 100 dB-SPL. What is the nominal uplink frequency for the Ku band? a. Gain increases by 9. 3 d. diffraction ANS. 14 GHz d. actual height b. 97. 6 km/s d. 526. TDMA d. a. 150 ft c. 15 W C 266. secondary service area c.000 ) = 266.4 x 106 m 4 x1011 v 3. 40 kHz b.786 km 2. daytime period b. frequency allocation of AM Radio Broadcasting is from… a. a. 10 W b. ITU-R S. beam b. primetime period ANS. azimuth & elevation 10.07km /s (36. A point in the orbit of an object orbiting the earth that is located closest to Earth. FDMA b. what is the minimum frequency separation in any service area for adjacent AM radio stations? a. What is the maximum power allowable for remote pick-up stations used as broadcast auxillary services for AM and FM broadcast stations? a.dynamic range ANS.07x103 m / s or 1440 minutes 6. 7. longitude & latitude c. As per Phillipine standard. An AM broadcast service area in which groundwave field of 1mV/m (60dBu) is not subject to objectionable interference or objectionable fading. The location of a satellite is generally specified in terms of a. primary service area b. 535 to 1705 kHz . 1440 sec ANS. a. ITU-R S 009 c. 526. The height above mean sea level of a satellite in a geosynchronous orbit around Earth is: a. 526. horizontal b. SCADA ANS. 7.8.6 km/s 5. 36 kHz 7. a. the required obstruction painting and/or lighting must be imposed on mast or tower more than … from the ground level. 76 m/s c. 9.6 km/s2 ANS. 7. Headroom b. signal-to-noise ratio c.786 miles b. 150 ft 5.004 ANS. spot d. experimental period d. 3. 35.000 km above the earth’s surface if the earth’s radius is 6400 km. primary service area 4. a.000 mi c. true north ANS.35. nighttime period c.000 km d.4 x106 m 24hrs v 3.000 km b.5 to 1605 kHz c. 250 ft ANS. 35. azimuth & elevation b. 36. nighttime period 3. none among these ANS.786 NM c. time a. Part of broadcasting in a day that refers to that period of time between 1000 UTC to 2200 UTC. 36 kHz d.latitude & longitude c. perigee Find the velocity of a satellite in a circular orbit 500 km above the earth’s surface. 7. 39. 39.ITU-R S.1001-2 d.600 sec d. vertical 6. ITU-R M1854 b. 1440 mins b.5 to 1705 kHz ANS. intermittent service area d.000 km Formula : 8. azimuth & elevation b. ITU-R S. propagation pattern c. perigee b. apogee c. a. In antenna mast or tower construction. DAMA c. apex d. An ITU radiocommunication standards for satellite services which provides information on the range of frequencies that can be used by fixed satellite service systems for emergency and disaster relief operations. Find the orbital period of a satellite in a circular orbit 36.786 km ANS. 76 km/s b. 3. ANS. bearing d. 1440 mins Formula : Where : C v T = orbital period C = circumference v = orbital velocity C = 2 r = 2π ( 6400 + 36. 35. 300 kHz ANS. bearing ANS.5 to 1705 kHz 2. a. vertical c. Regarding the design of AM antenna. footprint Calculate the length of the path to a geostationary satellite from an earth station where the angle of satellite above earth is 36 X 103 km) a. 24 days c. UTC stands for Universal Time Coordinates. footprint ANS.000 mi. latitude & longitude true morth d. 535 to 1605 kHz b. circular d. 36. 200 ft d.39. 35. dynamic range **Satellite Communications** 1. what should be the polarization of the radiator? a.1001-2 **Communications / Broadcast Engineering and Acoustics** 1. The difference in dB between the loudest level of sound and the softest passage of sound is known as a. As per Philippine standard. 100 ft b. gain d. TDMA ( Time Division Multiple Access) An earth station antenna look angle is determined by : a.000 6400) T d.786 feet d. subsatellite point ANS. elliptical ANS.6 km/s 4 x1011 Formula : v d 6400 Where : v = velocity in meters per second d = distance above earth’s surface in km v 4 x1011 (500 6400) = 7. d (r h) (rcos) rsin 2 2 A satellite access technique where each earth station transmits a short burst of information during a specific slot. The outline of a communications satellite antenna pattern on the earth is known as: a. 200 kHz c. 4. what is the ERP of the station in kW? a.25 = 187. line frequency = 15734. PAL c. 6 MHz d. 4 MHz b. 15 kHz b. Class C d. 19. visual carrier frequency of channel 9 16.96 m c = 3 x 108 m/sec = 300 x 106 m/sec velocity factor vf = 1 / sqrt (εr) = 1 / sqrt (2) = 0. 525 18.TV Pick-up station 21. 6 MHz 20. low band frequency of channel 9 fVC = 186 + 1. 88 – 108 MHz c. what is the ideal RF bandwidth? a. 1. 3. At NTSC standard. 5 MHz c. 5 kW b.556 x 10-6 = 63.2 MHz b. 64.36 m ANS.25 MHz ANS. What is the pilot subcarrier frequency used in FM stereophonic transmission? a. 0. 0. 485 b.556 usec ANS. 187. 66.96 kW GANT = 6 dB = 4 ERP = (5000 – 10) x 4 = 4990 x 4 = 19.580545 d. 193. Exact value of color subcarrier frequency (MHz) is … a. 35 W d. 535 – 1605 kHz b. using Arithmetic Progression f9LB = 174 + (9 – 7)6 = 174 + (2)6 = 174 + 12 = 186 MHz.264 H = 63. What is the mode of radio wave propagation that utilized ionosphere as a medium of transmission and / or reception of radio signals? a.35 MHz ANS. For a total capacity of 36 Mbps and 64-QAM of digital modulation.264 Hz. 175. 3.25 MHz fNLB = 174 + (N – 7)6. 4. 50 – 15000 Hz d. Determine the visual carrier frequency of TV channel 9 a.96 m d. TV Satellite link station ANS.556 usec d.556 usec b. 525 c. Class B What is the modulation used for the stereophonic subcarrier of FM composite baseband signal? a. AM DSB FC b.5 MHz c. the total number of symbol combinations is 64 symbols and the number of bits to produce this is… n = log2 64 = 6 bits per symbol Therefore the ideal RF bandwidth is… BW = fB / n = 36 Mbps / 6 bits BW = 6 MHz 22. Class D ANS. Class B c. ground wave b.8. the number of lines per frame = ________ lines/frame a.707 = vP / c Therefore the velocity of the radio signal in the coaxial line is … . 88 – 108 MHz 13.25 MHz. Class A b. TV Inter-city Relay station d. all of these ANS. At NTSC standard. If the power loss in transmission line is around 10 W and the antenna has a power gain of 6 dB. What is the maximum allowable ERPs for channels 7 – 13 in Metro Manila and Metro Cebu? a. The output power of an FM transmitter is 5 kW.579545 ANS. 19 kHz c.0 m b. 199. 350 kW d.579455 c.19 kHz 11. 5. 585 ANS.25 MHz b. εr = 2? a.25 MHz d.556 usec fH = 15734.579554 b. 316 kW c.960 W or 19. 38 kHz ANS. NTSC b. 63.96 kW 12. A land mobile station used for the transmission of TV program materials and related communications from the scenes of events occuring at remote points from TV broadcast station studios to TV broadcast station. 3. What is the equivalent line period? a. 100 kW b. 30 kW c.NTSC 14. 200 W ANS.96 kW ANS. 35 W What is the classification of an FM station having an ERP not exceeding 30kW? a. c. 19. 3. SECAM d. 4. 1 MW ANS. 1. TV STL station c. AM DSB SC c. 6 MHz d. none among choices ANS. 3.264 Hz H = 1 / fH = 1 / 15734. 30 kHz d. 625 d.85 m c. TV Pick-up station b. 65.sky wave 23. What is the color TV system adopted by the Philippines? a. 20 kW d. Frequency allocation of FM broadcast in the Philippines… a. Nominal RF bandwidth of NTSC TV channel… a. sky wave c.556 usec 19. 1 MW 17. a. space wave d. AM SSB d. AM DSB SC 10.187. What is the wavelength of a radio signal travelling at a frequency of 220 MHz in a coaxial line having Teflon foam as its dielectric.579545 15. AM VSB ANS. 6 MHz For a 64-QAM. 7 MHz ANS.25 MHz c. 63.556 usec c. 9. 174 – 216 MHz ANS. 0. 12 Ω d.96 m approx. Gauss law ANS. 14 MHz ANS. E c. Therefore it is centimetric wave. 28° d. Frequency Response b.963636… m or 0. F2 ANS. 15 Ω c.58 Ω approx.4 MHz d.25° = 4 / 0.707 (300 x 106) = 212. What is the critical frequency of a layer if the maximum value of electron density is 2 x 106 per cm3? a.01 m = 1 cm SHF band wavelength classification is ranging from 1 cm to 100 cm. An ionosphere layer also termed as “absorption layer” because it absorbs most of the frequencies above 100 kHz is known as the … layer.82 Rloss Rloss = (30 – 24. frequency diversity b.5 MHz ANS.centimetric SHF band: 3 GHz – 30 GHz λmax = 3 x 108 / 3 x 109 = 0.58 Ω ANS. E c.82 = 30/(30 + Rloss) 0.4 MHz approx. 30° ANS. What is a diversity scheme that uses two (2) separate antennas and receivers for each single transmitter? a.13 MHz Nmax = 2 x 106 per cm3 = 2 x 106 x 106 per m3 = 2 x 1012 per m3 fc = 9 x sqrt (Nmax) = 9 x sqrt (2 x 1012) fc = 12. increase antenna height 37. 12 MHz c. V. angle diversity d.85 (9. 8.50) Graph ANS. increase antenna gain b.932 MHz via Secant Law OWF = 0. What should be done to increase the transmission distance at frequencies higher than HF bands? a. quadrature diversity ANS. Radiation Pattern c.6) / 0. 75 Ω b. FCC F(50. increase transmitter power d.58 Ω η = Rd/(Rd + Rloss) 0. 24° b. increase receiver sensitivity ANS. 8.Inverse Square law 32. 30° Using Snell’s Law… Sqrt (εr1) sin θi = Sqrt (εr2) sin θr Sqrt (1) sin 45° = Sqrt (2) sin θr θr = sin-1 [(Sqrt (1) sin 45°) / Sqrt (2)] = sin-1 (0. 11-year sunspot c. F1 d.2 MHz b. space diversity c.5) = 30° 25. frequency diversity b. This law is known as… a.11-year sunspot 36.707c = 0. quadrature diversity ANS.3 MHz c.73 MHz or 13 MHz approx. 13 MHz d. space diversity c.932) = 8. frequency diversity b. 24. a. Smith Chart d. The maximum virtual height of the layer is 110 km.273] = 66.431-6 Tables 1 & 2 Wavelength Classifications.4 MHz For flat terrain analysis. D b. The power density is inversely proportional to the square of the distance from the source.36 m/sec (or 212 x 106 m/sec approx.82 = 6. a. Snell’s law c. at the midpoint of the path and the critical frequency is 4 MHz. Its radiation resistance is 30 Ω. increase antenna height c. centimetric d.6 + 0. 31. F2 ANS. D 34.space diversity 28. Radiation Pattern 30. As per ITU-R Recc.132. a. what is the suitable value for the optimum working frequency? Use flat terrain analysis. A phenomenon on the surface of the sun with appearance and disappearance of dark irregularly shaped areas. E 29.25° angle of incidence MUF = fc x sec i = fc / cos i = 4 / cos 66.) so that the wavelength of the signal is … λ = vP / f = 212 x 106 / 220 x 106 = 0. decimetric c. what is the classification of Super High Frequency? a. A graph which shows the radiation in actual field strength of electromagnetic fields at all points which are at equal distance from the antenna is known as… a. What is the value of its loss resistance? a. angle diversity d. The efficiency of an antenna is 82 %. Inverse Square law d. electrical properties of the terrain b. 33.85 x MUF = 0.4422 MHz or 8. 11 MHz b.1 m = 100 cm λmin = 3 x 108 / 30 x 109 = 0. 8. metric b. 35.6. millimetric ANS. The lowest portion of the ionosphere that is useful for long-distance communication by amateurs. SID b.vP = 0. frequency c. Faraday’s law b. 8. 26° c.0) if its angle of incidence is 45°? a. a. 8. This is often termed as the Kennely-Heavyside layer.40275 = 9.5854 Ω or 6. about 100 to 115 km above the earth is known as the… layer. 6. F1 d. Which of the following factors must be considered in the transmission of a surface wave to reduce attenuation? a.034. A radio communication link is to be established via the ionosphere. Sporadic E-layer d. all choices .82 (30 + Rloss) = 30 = 24. Ionospheric storms ANS. tan i = d/2hv i = tan-1 [d/2hv] = tan-1 [500/2(110)] = tan-1 [2. 27. antenna polarization d. If the distance between the radio stations is 500 km. What is the angle of refraction in a Teflon (εr2 = 2) medium of a radio wave from air (εr1 = 1. D 26. What is a diversity scheme that uses two (2) different frequencies in a simplex path? a. What should the mobile transmitter power be set to as a first approximation.852Hz. 852Hz. a.PR Where: PT = transmitted power in dBm PR = received power in dBm PT = -76 dB . which is within 3 – 30 MHz range designated as HF band 39. 90 volts DC b. is the radiated power GANT = log-1 [2.1336Hz.8 dB ANS.97 mi ANS.HF SOLUTION (IF PROBLEM SOLVING): c = 3 x 108 m/sec λ = 11 m = c / f f = 3 x 108 / 11 = 27 MHz approx. the Horizontal frequencies are a.66 km b. the MTSO increases power level ANS. PD = PR/4πd2 PR = PD x 4πd2 = 1.66 mi c. 250 mW d. none of the above ANS.007 W or 3 W approx.1336Hz. central office codes d. shielded twisted-pair copper wire d.57 x 10-4 W/m2 is measured 50 meters from a test antenna whose directive gain is 2. the call is terminated c.770Hz. 200 mW c. **Wireline and wireless ** 1. The typical voltage across a telephone when on-hook is: a.6 dB d. A telephone signal takes 2ms to reach its destination.770Hz. a.1477Hz. language digits b.15 dB. repeater b. control computer d. In a cellular telephone system.twisted-pair copper wire 13. the unit is “handed off” to a closer cell d. A CDMA mobile measures the signal strength from the base as -100 dBm. 53. considering . 852Hz. 2 W c. call blocking is: a.64 PR = PFED x GANT PFED = PR / GANT = 4.1633Hz b. A radio station operates at 11 meter wavelength. 48 volts DC c. 0. 150 mW b. A power density of 1. What is the total radio horizon distance between an 80 ft transmitting station and a 20 ft receiving station? a.15 dB/10] = 1.93 W approx. Calculate the via net loss required for an acceptable amount of echo.in erlangs 3.1633Hz b. 18.2 X 2 + 0. 1209Hz. trunk lines d. 697Hz.97 mi 40. 679Hz. 18.76 dB at mobile power contol? a. VHF d. 0. HF c. 697Hz.770Hz.1747Hz. 48 volts.. a. In Celluar Telephone Systems. 1209Hz. 697Hz.941Hz c. 1336Hz.2 dB b. area codes c.1633Hz .1633Hz ANS.1747Hz. 20 hertz AC ANS. What is the designated band of station’s frequency? a.repeater 4.the unit is “handed off” to a closer cell 5.4 dB Where: VNL = minimum required via net loss in dB t = time delay in ms for propagation one way along line VNL = 0. TRX d. reverse links c.97 km d. Central offices are connected by: a. 53. forward links b.8 dB 9.ANS.forward links 7. occurs on the local loop when there is an electrical power failure c. with echo cancellers b. 1336Hz.area codes 2. In DTMF.941Hz c.2t + 0. cannot occur in the public telephone network b. 18.cluster 6.all choices 38. 0. coaxial cable b.64 = 3. in terms of the grade of service ANS. the cell site switches antennas b.1477Hz. occurs only on long-distance cables d. Cluster c.852Hz. 941Hz d.250 mW PT = -76 dB . In telephony. both a and b b. the four vertical frequencies are ____.48 volts DC a. Telephone traffic is measured ____ 8. UHF ANS. local loops c. in erlangs c. by the relative congestion d. In DTMF. fiber-optic ANS. site d.PR = -76 dB – (-100dBm) = 24 dBm = 250 mW 10. 1 W b.97 mi SOLUTION (IF PROBLEM SOLVING): dRH (TOT) = sqrt (2 x hT) + sqrt (2 x hR) dRH (TOT) = sqrt (2 x 80) + sqrt (2 x 20) dRH (TOT) = 18. sector b.57 x 10-4 x 4π(50)2 = 4. a. 4 W ANS. a.4 dB = 0.93 / 1. 300 mW ANS. TRA ANS. 941Hz d. access digits ANS.3 W From Inverse Square Law. cell segment ANS.4 dB c.770Hz. 0. In a cellular telephone system. MF b. direct link to a branch exchange ANS. each cell site contains a ____. When the signal from a mobile cellular unit drops below a certain level.2t + 0. 0.occurs when the central office capacity is exceeded 12. 1208Hz. occurs when the central office capacity is exceeded ANS.trunk lines 11. Identical telephone numbers in different parts of a country are distinguished by their ___. touch tone processors c.852Hz.770Hz. 90 volts.8 dB VNL = 0. what action occurs? a.4 = 0.941Hz 14. The cable used for local loops in telephone system is mainly: a. group of cells is called ___. 20 hertz AC d. twisted-pair copper wire c. 1208Hz. 3 W d. How much power was fed into the test antenna? a. 679Hz. these are transmissions from base stations to mobile units (Downlink) a. ANS. Basic Service Contract b. Intermodulation System Interference ANS. Red c. Power consumption d. 50 micron b.5 micron d. direct-sequence modulation d.low group frequencies 18.1633Hz b. top group frequencies ANS. In GSM. mid group frequencies d. Power consumption 10. Waveguide dispersion d. Optical Fiber c.reverse links ANS. prevent "singing" c.1209Hz. bearer channels b. voice channels d. Step index A technique that is used to minimize the pulse dispersion effect is to a. All the above ANS. regional center d.Quanta 12.traffic channels 25. Material dispersion c. Which of the following considerations is important when deciding between using a diode laser or an LED? a. AMPS stand for: a.5 micron The abrupt change in refractive index from core to cladding of fiber-optic cable is called the a. Step index ANS. Light Ray b. Quanta d. TRA ANS.10 Base-FB (Backbone) . 10 Base-FL (Link) b. forward links b. Dispersion d. CDMA b. Maximum Signal Carrier 6.5 micron c. Failure characteristics ANS. Response speed ANS. BSC stands for: a. Ultraviolet ANS.frequency hopping 17.Distributed feedback (DFB) and vertical cavity surface emitting (VCSEL) 8. the Vertical frequencies is also known as a. section center c. Recent laser developments for fiber-optic communication include a. low group frequencies b. top group frequencies ANS. It is the central part of the optical communication system a. Heterojunction d. frequency hopping c.Response time 9. Use plastic cladding c. all of the above ANS. allow signals to be multiplexed d. Advanced Mobile Phone Service ANS. Dark current c. traffic channels c. 1336Hz. the most commonly used fiber(s) are a. Fresnel d.Base Station Controller 22. 5.high group frequencies 24. Which of the following Ethernet fiber optic standard uses synchronous. Rayleigh c.Fresnel 23. allow signals to be multiplexed 19. a. Violet d. Cable dispersion ANS. In Cellular Radio. Snell ANS. Responsitivity b.regional center 20. Light Source b. talking channels ANS. Response time b. 125 micron ANS. reverse links c. In DTMF. In the telecommunications industry. high group frequencies c. Waveguide dispersion 7. 62. GSM uses: a. American Mobile Phone System b. high group frequencies c. 3. Mobile Service Cellular d. Advanced Mobile Phone System d. 10 Base-FB (Backbone) c. Mobile Switching Center c. the correct explanation of diffraction was given by a. Boson c. Distributed feedback (DFB) b. 4. In Mobile Communications. Basic Service Code ANS. Total internal reflection b. In Mobile Communications. 50 and 62. allow lines to be "conditioned" b. these are transmissions from mobile units to base stations (Uplink) a. In Cellular Radio. Numerical aperture c. In Celluar Telephone Systems. Power levels c. Light ANS.Optical Fiber 11. voice channels are called: a. Integrated Mobile Subscriber Identification b. all of the above ANS.International Mobile Subscriber Identification ** Electronics and Communications** 1. The bandwidth of voice-grade signals on a telephone system is restricted in order to: a. primary center b. none of the above ANS. MSC stands for: a. Base Signal Controller d. Minimize the core diameter Which is not an important characteristic of a light detector? a. Photodetector d. bottom group frequencies d. the horizontal frequencies is also known as a. International Mobile Subscriber Identification c. The dispersion of light in fiber-optic cable caused by a portion of the light energy traveling in the cladding is called a. Temperature sensitivity d.Advanced Mobile Phone Service 21. The light energy that is always emitted or absorbed in discrete units a. Infrared b. Minimum Signal Carrier 15. Use a higher frequency light source b. Base Station Controller c. Distributed feedback (DFB) and vertical cavity surface emitting (VCSEL) ANS.1477Hz. In 1815. In DTMF. low group frequencies b. In Cellular Radio. Modal dispersion b. IMSI stands for:: a. Vertical cavity surface emitting (VCSEL) c. Analog Mobile Phone Service C. Minimize the core diameter d. Both a & c ANS. TRX d.Mobile Switching Center 16. Maxwell b. Centralized Clock? a. 10 Base-FP (passive) d. 50 and 62. all of the above ANS. The most common light used in fiber-optic links is a. This is the highest-ranking office in the DDD network in telephony in terms of the size of the geographical area served and the trunking options available. Interim Mobile Subscriber Identification d.Infrared 2. 5 Mbps ANS. Antenna arrays d. An optic fiber is made of glass with a refractive index of 1.51. Broadside action d.6◦ c. Resonant antenna c. FDDI .8). What numerical aperture does the fiber have? a. Repeatability ANS. 3. a. Polarizing ANS. 0.Broadside b.55-1. The ratio comparing the power density generated by a practical antenna in some direction.11). It is the ratio of the output current of a photodiode to the input optical power and has the unit of amperes per watt a. 23. 10Mbps b. and the dipole antennas describe so far have been resonant.316µW P(dBm)= 10log = -35dBm 15. Coordinates c. Note: such reflectors are often used at that frequency as antennas outside broadcast television microwave links. Corresponds to a resonant transmission line. FDDI .852 b. 3.Stub b. determine the maximum digital transmission rates for return-to-zero. a. Reciprocity d.25-dB/km loss.352 By equation (20.25)(100)]/(10)] =(1 x 10-4)(1 x 10-25) =0. 0. Azimuth ANS.235 ANS. Dark current b. Launching takes place from air.Broadside action 10. it is a measure of the conversion efficiency of a photodetector. -35dBm P=0. End-fire array ANS. Transmission line d. 20.50 d.20Mbps 16.Aperture 11. a. -45dBm c. Grounding c.8750 ANS. the numerical aperture is found to be: NA=n1√(2∆) =1. bandwidth d.55 and is clad with another glass with a refractive index of 1. directivity b. Jacket ANS. resolution c.352 c. 6. Feeding ANS. with that due to an isotopic antenna radiating the same total power.Top loading What principle that states that the properties of an antenna are independent of whether it is used for transmission or reception. 22.Resonant antenna 12. None of the above ANS. 1. by inference. 4.Beamwidth 2.1-mW light source. the fractional difference between the indexes is: ∆=(n1-n2)/n1 =(1.352 By equation (20.Principle of reciprocity Calculate the beam width between nulls of a 2-m paraboloid reflector used at 6GHz. Light Sensitivity ANS. a.11).55 =0. 35dBm d. Directive gain c.352=20. the numerical aperture is found to be: NA=n1√(2∆) =1.55-1.10).55√[(2)(0.50 This is often used to cure the problem of great thickness required of lenses used at lower microwave frequencies or for strong curved wavefronts.0258)] = 0.6◦ 14. a.8◦ b. a. the acceptance angle is: Θ0(max)=sin-1NA= sin-10. Footprint d. multiple-service standard to allow the transmission of voice and video over an FDDJ network. a.II b. The ratio of the focal length to the mouth diameter is called __________ of the parabola. and also to make the current distribution more uniform. a. Spectral Response d. For a single-mode optical cable with 0. Antenna c. a.51)/1. 50 Mbps d. 0. In fiber optics.Propagation b.140 b. 0. -30dBm ANS. a.13.6◦ By equation (20. focus b. Top loading d.II 14.55 and is clad with another glass with a refractive index of 1.0258 By equation (20. to have __________. a. Buffer Tube ** Signal Processing and Control Systems ** 1. Splicing b. Bulkhead c.032 d. Cladding d. Any array that is directional at right angles to the plane of the array is said. Broadside array c. the fractional difference between the indexes is: ∆=(n1-n2)/n1 =(1.Reflector b. What is the acceptance angle? a. Responsivity c. Non-resonant antenna b. For an optical fiber 10km long with a pulse-spreading constant of 5ns/km.I c. feed d. a. Launching takes place from air. Hi PPI d.Polarization Placing a metallic array on the antenna effects to increase the current at the base of the antenna.Responsivity . Escon ANS. 0. What is this called? a. Buffer Tube b. It is a mixed. -25dBm b.0258)] = 0. Refers to the direction in space of electric vector of the electromagnetic wave radiated from an antenna and is parallel to the antenna itself. A structure-generally metallic and sometimes very complex-designed to provide an efficient coupling between space and the output of a transmitter or input to a receiver. Curving d.Alternation b.6◦ ANS.55 =0. FDDI . The angular separation between the two half-power points on the power density radiation pattern. It is a layer of plastic that surrounds a fiber or group of fibers a. 18. Zoning c.6◦ d.25)(100)]/(10)} =1 x 10-4 x 10{[(0.51. An optic fiber is made of glass with a refractive index of 1. Polarization d. 70 c.Zoning 7.10). just as in camera lenses.Bandwidth b Beamwidth c. 20 Mbps c.55√[(2)(0.3. determine the optical power 100km from a 0.Directive gain(b) 9. Diplexer c. 5. aperture ANS. waveguide ANS. a.51)/1. Elementary doublet ANS. Azimuth ANS.0258 By equation (20.352 13.1mW x 10-{[(0. 20.antenna 8. hub d. ANSWER: d.Lasing 21. Is a PCM system which uses a single bit PCM code to achieve digital transmission of analog signals 19. Dynamic range a.Data Link Layer b. Peer-to-peer network d. QAM 9. DPSK d. Repeater b. Microbending ANS. TCM 7. Photoelectric effect b. The process of constructing an ILD that is similar to LED except that the ends are highly polished. CSMA/CD Data comms a. All of the above ANSWER: a.Trellis Coding PCM d. lasing d. The theory which states that when visible light or high-frequency electromagnetic radiation illuminates a metallic surface. Rayleigh scattering d. Switch b. Presentation Layer a. Planar diffusion c. Is an access method used primarily with LANs configured in a bus topology. Is a multiport bridge that provides bridging function. Pulse spreading b. Photoemission effect d. Scattering loss d. FSK c. Dedicated client server network b. Is the ratio of the largest possible magnitude to the smallest possible magnitude that can be decoded by the digital-to-analog converter. Splitter d. Checksum c. Optics ANS. Quantization Error d. Delta modulation c.Planck’s law 3.048 Mbps ANSWER: b. Planck’s law c. 32 bits 8.544 Mbps c. Longitudinal redundancy check b. Resolution ANSWER: d. a.84 Mbps 11. Is a device which operates in the Network layer and it makes forwarding decisions on the basis of network addresses a. Control and Status Register d. Switch c. Router ANSWER: c. Session Layer a. Photometry c. 24 bits c. Dispersion ANS. TCM 14. None of the above ANS. Is a redundancy error detection scheme that uses parity to determine if a transmission error has occurred within a message and is sometimes called message parity. Peer-to-peer client/server network 5. ANSWER: a.epitaxially grown b. electrons are emitted. CSMA/CD d. Radiometry b. Switch ANSWER: d. Is an empirical record of a system’s actual bit error performance . Is an OSI Layer which is responsible for providing error-free communications across the physical link connecting primary and secondary stations (nodes) ANSWER: a. OC-1 (STS-1) line rate is a. A phenomenon also called stress corrosion resulting if the glass fiber is exposed to long periods of high humidity? a. a. Status Word register c. 1. A phenomenon in optical fibers communication system that is caused by the difference in the propagation times of light rays that take different paths down the fiber. b. 51. Is used to interface DTEs to digital transmission channels b. Wavelength distortion c. 18. Command registerANSWER: b. Router Channel service unit c. hub 4. Vertical redundancy check c. Router b. Static fatigue b. 3 Mbps ANSWER: a. a. Delta modulation c. CSMA c. MA b.17. 51. IP version 4 address length is 32 bits d. a. Data Link layer c. 40 bits Ans. Is an n-bit data register inside the UART that keeps track of the UART’s transmit and receive buffer registers a. The science of measuring only light waves that are visible to the human eye a.Pulse spreading 1. Optometry d. Dynamic range a. Repeater d. Differential b. Is one in which all computers share their resources with all the other computers on the network. Photometry 20. CSMA/CA 13. 2.84 Mbps d. Modulation which combines encoding and modulation to reduce the probability of error. Channel service unit 6. Router 10. Absorption c. 48 bits b. Data modem b. a. Peer-to-peer client/server network c. QAM ANSWER: d. Longitudinal redundancy check ANSWER: b. Static fatigue 2. Transport Layer d. Step size b. Status Word a. Control Register register 12. Ray Theory of light ANS. Character Parity Check d. 0. directive gain c.10 d.The electric field is perpendicular to the earth’s surface. A measure of mismatch in a transmission line a. The maximum horizontal distance between the transmitter and the receiver for line of sight propagation is known as: a. differential mode c. maximum usable propagating frequency frequency (muf) b. C/m ANSWER:a. wave ANSWER:b. directivity c. V/m ANSWER:c. ER d. V-m d. a. normal d. It is the distance between two wave fronts having the same phase at any given instant. 6. scatter space wave d. V/m ANSWER:c. ANSWER: a. reflection coefficient delay b. are of convenient size the transmitting antennas c. power gain d. Travel in a straight line from the transmitting antenna to the receiving antenna ANSWER: b. MF (Medium Frequency) c. radio horizon 9. V/C a. stub d. cut-off frequency ANSWER: critical frequency a. potential difference a. . d. 3. BER 15. Calculate the reflection coefficient. dipole moment d. VLF waves are for some types of services because ANSWER: d. quarter wave transformer ANSWER:c. standing wave ratio d. 7. EER 5. none of these c. HF c.15 b. Is an error-correcting code used for correcting transmission errors in synchronous data streams.25 c. of low powers required b.35 0. end fire a. common a. Hamming code b. they are very reliable 2. 0. normal d. wavefront b. The electric field intensity is measured in: a. service region ANSWER: a. sky wave c. It is a result of an increase in the charge density in dielectric materials. directivity ANSWER: b. vertical horizontal b. MF c. power density propagation 13. field intensity ANSWER a. stub 4.a. c. BER c. potential b. a. polarization b. Hamming code RADIO WAVE 1. UHF a. radio horizon b. standing wave ratio ANSWER: a. In what major RF band is ground wave basically applied? a. a. Typical mode of radiation of helical antenna is: 10. circular c. vertical c. end fire mode b. terminator b. It refers to the maximum antenna gain a. space wave Bisync Code b. A 50 ohms transmission line is connected to a 30 ohm resistive load. they are very reliable ionosphere easily d. wavelength distance d. range d. 14. ground wave b. if wave is sent vertically upward. EBCDIC c. single hop distance c. BERT b. The highest frequency that can be set back to earth by the ionosphere. the polarization is: 11. all of these b. It is a piece of transmission line which is normally short-circuited at the far end. they penetrate the ANSWER:a. Trellis Code d. wavelength ANSWER polarization 8. VHF 12. a. 0. critical frequency d. 92. Statistical concentration d. Quadrature amplitude modulation (QAM) d. marconi b. Radio teletype d. Eye pattern Constellation pattern Orthogonal Frequency Division Multiplexing b. refraction d. Bandwidth requirements of the channel c. Delta modulation b. ANSWER: 6. CSU/DSU d. dipole d.4 + 20 log F + 20 log Dd. Facsimile 15. Compression a. Capture-division multiple-access systems d. Noise performance b. The type of radio transmission that uses pseudorandomly switched transmissions is known as 20. length radius d. Ionospheric fading characteristics 4. Optional Frequency Division Modulation d. Synthesizing a. Power consumption ANSWER: c.2 + 10 log F + 20 log D d. TDM a.25 5. hertz ANSWER: a. Using radio to transmit gathered data on some particular phenomenon without human monitors is known as a. Channel-division multiple-access systems b. E layer a. rhombic c. conductor spacing b. a. a. The technique that uses the BPSK vector relationship to generate an output with logical 0s and 1s determined by comparing the phase of two successive data bits is ANSWER:Code-division multiple-access systems conductor ANSWER: c. Which is properly terminated antenna? a. Code-division multiple-access systems 1.ANSWER: b. c. FSK PSK FPF FSA . 94. F layer b. input b. 7. 92. b. input 17. The characteristic impedance of a transmission line calculation formula is b. b. 94.4 + 20log F + 20 log D DATA a. space loss a. F layer 18. reflection c. midsection 2. 92. Pulse amplitude modulation (PAM) ANSWER: c. Orthogonal Frequency Division Modulation ANSWER: Orthogonal Frequency Division Multiplexing 8. 0. d. diffraction absorption b. Eye pattern ans: diffraction Ref Coeff = (30 – 50)/(30 + 50) = 0. Characteristic impedance of a transmission line is the impedance measured at the _______ when its length is infinite. Over Frequency Division Multiplexing c. a. CVSD b. length 16. Quadrature amplitude modulation (QAM c.25 a. Radio multiplexing ANSWER: Radio telemetry 9. SOLUTION ( IF PROBLEM -SOLVING) c. Loopback ANSWER:a. The acronym OFDM refers to ________. c. DPSK d. Using an oscilloscope to display overlayed received data bits that provide information on noise. jitter. a. D layer A layer d. 3. Which region of the ionosphere is mainly responsible for long-distance night time communications? a. Radio facsimile b. The acronym CDMA refers to ________. Carrier-division multiple-access systems c. Spread spectrum does not depend upon its c. FSK systems are much superior to two-tone amplitudemodulation systems with respect to ANSWER: d. shorted end of the line c. A special digital modulation technique that achieves high data rates in limited-bandwidth channels is called ANSWER:a. Ionospheric fading characteristics d. Radio telemetry c. The acronym for the form of data transmission in which the modulating wave shifts the output between two predetermined frequencies is _____. Pulse-coded modulation (PCM) c. A microwave communications system.4 + 10 log F + 20 log D b. rhombic 19.2 + 20 log F + 20 log D ANSWER: b. and linearity is called a a. DPSK c. conductor diameter c. Spread spectrum c. Occurs when the radio beam is at point of grazing over an obstacle. output ANSWER: b. a. Aliasing c. ANSWER: codec modem DSP codec ASIC 25. joule d. Yellow b. Flow control d. PSK ANSWER: Modem 20. FSK ANSWER: Line Control 10. c. Foldover distortion and aliasing ANSWER: Fold over distortion and aliasing 18. a. remote d. Nyquist rate d. Acquisition time c. a. Line control b. The NRZ digital signal-encoding format has a ________ component in the waveform. Use 16-bit PCM code and include 24 voice channels b. Blue c. alias c. 14. Telemetry may be defined as _____ metering. Use 8-bit PCM code and include 24 voice channels d. The major difficulty faced by delta modulators is a. b. PPM and PWM are superior to PAM systems in a. 21. a. Flat-top time d. energy c. ______ implies that both analog and digital signals share the same channel bandwidth. d. b. The bit ________ is the amount of power in a digital bit for a given amount of time. re-created d. A(n) ________ is a single LSI chip containing both the ADC and DAC circuitry a. Which is not a type of pulse modulation? a. pseudorandom BSC interleaving . Pulse-frequency modulation (PFM) d. Foldover distortion b. detected c. Excessive noise producing errors Slope overload Insufficient frequency response of the intelligence signal d. Red d. b. codec b. Ans: hybrid AM Hybrid FM RF 12. ANSWER: dc dc sinusoidal harmonic parity Reed–Solomon codes utilize a technique called ________ to rearrange the data into a nonlinear ordering scheme to improve the chance of data correction a. d. minimized ANSWER: re-created 11. The ground wire in a USB cable is ______ a. hp ANSWER: energy 24. c. the time that it must hold the sampled voltage is a. A(n) ________ is used to allow a digital signal to be transmitted on an analog channel. Dmin ANSWER: Aperture time 23. c. Pulse-position modulation (PPM) ANSWER: Pulse-frequency modulation (PFM) a. Use delta modulation and include 24 voice channels ANSWER: Use 8-bit PCM code and include 24 voice channels Noise characteristics Bandwidth characteristics Simplicity in design Frequency response of the intelligence signal ANSWER: Noise characteristics 15. Aperture time b. modem d. a. distance b. d. The relationship for bit rate to ________ bandwidth is defined by the Shannon-Hartley theorem. modem b. c. Sequence control a. Brown ANSWER: Brown 19. c. 16. In an S/H circuit. codec c. b.ANSWER: a. Error signals associated with the sampling process are called a. The AT&T T1 lines a. specialized c. Complexity of design ANSWER: Slope overload 17. channel d. Protocol c. Pulse-width modulation (PWM) c. error b. Pulse-amplitude modulation (PAM) b. b. decoded b. Pseudorandom implies a sequence that can be _____ but has the properties of randomness. A procedure that decides which device has permission to transmit at a given time is called a. c. b. data ANSWER: Remote 13. Use delta modulation and include 48 voice channels c. d. amplifier ANSWER: Channel 22. 0.5 kHz 75 kHz = 30 2. A Satellite receiving system includes a dish antenna ( from rest to v. a.00J/m3 b.5 b. 34. 3Q -15 1. Suppose that energy Q is required to accelerate a car 5.00 J of electric potential energy in a volume of 1.113TV/m ANS. -98dBm c.3 =3 5. -75dBm SOLUTION (IF PROBLEM-SOLVING) ᵟ = φ x fi = 0.15 x 10-12W SOLUTION (IF PROBLEM-SOLVING) Using the energy equation W=KE2 – KE1 PN = KT∆f = 1. the signal power is 1W and the noise power is 30mW.38 x 10-23 J/K (35 + 40 + 52)K (1MHz) = 1.4Q MHz frequency range? a. a. a.5mv2 = 3(0. What is the magnitude of the required electric field? a. d. noise temperature 7. 3. v to 2v? What is the noise power to the receiver’s input over a 1a.4 µV 3. 0.5 = 100 = 33. Equivalent temperature of a passive system having the same noise power output of a given system. condition 1 W = 0. 130dBm c.5mv2 - Answer: VN = = SOLUTION (IF PROBLEM-SOLVING) Suppose you want to store 1. 1.38 x 10-23)(302)(6 x 106)(50) = 2. Determine the worst-case output S/N for a broadcast FM program that has a maximum intelligence frequency of 5 kHz. 2.5mv2) = 3Q 2. find the direction of the reflected ray.4 µV SOLUTION (IF PROBLEM-SOLVING) Noise figure Total temperature Equivalent temperature Noise temperature Q = 0. 2124mV condition 2 W=2mv2 – 0. 0. -134dBm SOLUTION (IF PROBLEM-SOLVING) N = KTB B = 1 x 104 Hz T(kelvin) = 17°C + 273° = 290 K = (1.5mv2 – 0 - 4(1. 25 30 c. d.336MV/m c.75 x 10-15 W 6. -80dBm d. 60o d.5 b. 3.15 x 10 W d. The signal power at the input to an amplifier is 100µW and the noise power is 1 µW. 134dBm b. Q b.03W NF (ratio) = 100 33.E=0. The bandwidth of the receiver is 6MHz.Find the noise voltage applied to the receiver input.For an electronic device operating at a temperature of 17°C with a bandwidth of 10 kHz.2Q c. 3005mV d. If the incident ray makes an angle of 60o with the normal.00m3 = 1. 4.225TV/m d.00 m3 in vacuum. The input S/N is 2 a.75 x 10-12W c. 91. 1. ANS.7 V ANSWER: 4. 1. a. Material a is a water and material b is a glass with index of refraction 1. 134dBm d. 49. a.3o c. a.000) + 15 S = -98 dBm 1.475 MV/m b. c.7o 3 ANSWER: 60o . 4.001 2. 3.52.5 V d.475 MV/m (a) SOLUTION (IF PROBLEM-SOLVING VN = = 4(1. At the output.24µV 8.5 x 5 kHz = 2. quantizing ANSWER: interleaving 1. 3Q d. A 300Ω resistor is connected across the 300Ω antenna input of a television receiver. 6. 0. It is desired to operate a receiver with NF = 8dB at S/N = 15 dB over a 200-KHz bandwidth at ambient temperature.5 kHz S = Sensitivity = -74dBm + NF + 10log∆f + desired S/N ** Signal Processing and Control Systems** S = -174 + 8 + 10log(200.24µV E=0. d.4 V b.75 x 10-15W b. The noise voltage produced across a 50Ω is input resistance at a temperature of 302°C with a bandwidth of 6 MHz is ______. 4 SOLUTION (IF PROBLEM-SOLVING) (S/N)p = 100µW 1µW (S/N)o = 1W 0.5 c.d. 5.475 MV/m u=1.38 x 10-23 J/K)(293 K)(6 x 106 Hz)(300 Ω) = 5.12 V c. 3 d. 2. and the resistor is at room temperature .00J/1.38 x 10-23)(290)(1 x 104) = 4 x 10 -17 W N(dBm) = 10log (4 x 10-17) = -134 dBm 0. What is the amplifier noise figure. 5. b. as a ratio? a.5m(2v)2 – 0.4 x 10-6 V = 5. -84dBm b. Calculate the receiver’s sensitivity a. determine the thermal noise power in dBm . – 130dBm ANSWER: c.15 µV c. . neglecting friction. How much added Teq = 35 K) connected via a coupling network (Teq= 40 K) energy would be required to increase the speed from to a microwave receiver (Teq=52 K referred to its input).3o b. 25 x 108 m/s b.Bernoulli’s equation 8.spherical aberration 7. drag velocity b. 0. Calculate the speed of light in this substance. In electrodynamics. Maxwell’s equation d.345c c. commonly generated by electronic devices. In optics.25 x 108 m/s 5.25 x 108 m/s ANS. what refers to waves.900c b. What do you call this effect? a. this shows that the pressure of a fluid decreases as the speed of the fluid increases. Microwaves c. Snell’s effect ANS. 0.25 x 108 m/s In electromagnetics and wave theory. whose wavelengths ranges from approximately 0.900c fires a robot space probe in the same direction as its motion. 0. Bernoulli’s equation c. A spaceship moving away from Earth with speed of 0.microwaves ANS. a. 3.982c 6. a. Clairut’s equation b. Hyrdofluid equation ANS. this effect refers to the blurring of the image produced on a concave mirror due to the convergence of rays far from the mirror to other points on the principal axis. Radio waves b. what term refers to the average speed of the charge carriers? a. 2. What is the probe’s velocity relative to the Earth? a. spherical aberration b.982c ANS.3m to 10-4 m. parallax error d.drift speed . 0. The wavelength of the red light from a helium-neon sensor is 633 nm in air but 474 nm in the aqueous humor inside our eyeballs. 4.700c relative to the spaceship. In fluid mechanics. 0. with speed of 0. Due to their short wavelengths these are often used in radar systems and for studying the atomic and molecular properties of matter. charge velocity ANS. Radar waves d.700c d.00 x 108 m/s c. molecular-kinetic speed c. the pressure of a fluid decreases as the elevation increases. drift speed d. focal divergence c. AM waves d. 3. a.9. 2. In addition. 1.