d. None of the above, 34) Advantage of using a high frequency carrier wave is, a. c. Lower side band (ωc – ωm) having amplitude mA/2 = √33.144 * 10-12 Periodic signal sgn(f) = 1, f> 0 Noise voltage by resistors when connected in series is b. m = 1 = 1750Hz, 130) Calculate the dissipation in power across 20Ω resistor for the FM signal Includes the carrier frequency II Analog Modulation. c. PLL detector b. fs < 2fm d. 102 W, Explanation: c. In FM before modulation d. 280 W, 2187.5 W, 750.25 W, Explanation: 284.48 * 10-7 b. Envelope detector d. Not predetermined, a. b. a. 139) For a FM signal v(t) = 20 cos ( 10 * 108t + 30 sin 3000t), calculate the power dissipated by the FM wave in a 20Ω resistor. d. None of the above, ANSWER: (b) Multiplication of two signals, 33) If a receiver has poor capacity of blocking adjacent channel interference then the receiver has, a. c. Cannot be used for high power FM generation c. AM or FM signal d. All of the above, a. Amplitude of the carrier is constant d. 75KHz, 250Khz, 136) Guard bands are provided in FM signal to, a. and the other can be represented by a cosine wave (i.e. 113) What is the required bandwidth according to the Carson’s rule, when a 100 MHz carrier is modulated with a sinusoidal signal at 1KHz, the maximum frequency deviation being 50 KHz. there is increase in total power by 50%, a. b. Pt = Pc (1 + m2/2) c. Modulating frequency In general, amplitude modulation definition is given as a type of modulation where the amplitude of the car… The recovered signal is distorted = 10Watts. d. None of the above, 197) In Pulse Position Modulation, the drawbacks are, a. Synchronization is required between transmitter and receiver b. b. b. b. 580 KHz c. 10.7 MHz d. 50 MHz. 6000 Hz After continuous wave modulation, the next division is Pulse modulation. d. All of the above, 77) If modulation index is greater than 1, a. b. c. 11.15 * 10-7 kf = frequency deviation/modulating frequency So transmitter power remains unchanged in FM but it changes in AM a. c. Before detection at receiver d. 1.6 MHz and 8 KHz, 0.2, 16 KHz, ANSWER: (d) 1.6 MHz and 8 KHz, 0.2, 16 KHz, Explanation: Intermediate frequency (IF) should be carefully chosen as, b) High IF results in problems in tracking of signals, c) Image frequency rejection becomes poor at low IF. c. Phase deviation To reduce the bandwidth of the signal to be transmitted The lower portion of the MF band (300to 500 kilohertz) is used for ground-wave transmission for reasonably long distances. When the modulating (input signal) is represented as i(t) = Aicos(2∏fit) and the carrier signal is represented as c(t) = Accos(2∏fct) And in the expressions, Ai and Acrepresent the amplitudes of two waves while fi and fc are the frequencies of the two waves correspo… One signal is called the in-phase “I” signal, and the other is called the quadrature “Q” signal. d. Transmitter amplifier, ANSWER: ( a) Amplitude of the modulating signal, 98) Drawbacks of using direct method for generation of FM signal are, a. 2. In both of these modulation techniques, the carrier wave is of analog form. Q12. = 1050Hz. b. DSB-SC signal Δf = mf * fm 118) What is the maximum modulating frequency allowed in commercial FM broadcastings? d. All of the above, a. Modulation is done at high power of carrier and modulating signal d. Both a and b. Therefore, deviation = 5 * 350 B = (ωc + ωm) – (ωc – ωm) Select the wanted signal and reject the unwanted signals and noise To increase the noise Mar 06,2021 - Pulse Modulation | 10 Questions MCQ Test has questions of Electronics and Communication Engineering (ECE) preparation. Explanation: T noise temperature 48 = Pc * 1.10125, Therefore, Pc = 48/ 1.10125 d. All of the above, a. FM signal This signal is then passed over wires to a speaker or headphones. d. 19.8 KHz, 10) In Amplitude Demodulation, the condition which the load resistor R must satisfy to discharge capacitor C slowly between the positive peaks of the carrier wave so that the capacitor voltage will not discharge at the maximum rate of change of the modulating wave (W is message bandwidth and ω is carrier frequency, in rad/sec) is, a. RC < 1/W f1 = 100π/2π = 50Hz Q14. b. Filtering b. Pc = 124W 2000KHz + 0.350 KHz = 2000.35 and 2000KHz – 0.350 KHz = 1999.65, the available frequencies after modulation by 0.6 KHz are As the signal passes through various stages of an amplifier, the output has the original signal and some noise that gets amplified at different stages of amplifiers. = 284.48 * 10-7. = √{4(20 * 103 + 30 * 103) * 1.381 × 10-23 * 293 * 10 * 103 } Local generation of carrier b. Tail edge of the pulse is kept constant d. 11000 Hz, Explanation: Less adjacent channel interference The frequency of the locally generated carrier must be identical to that of transmitted carrier Standard expression for FM signal is given by b. c. The number of side bands are infinite Gain of the IF amplifier Carrier is locally generated 149 KHz to 169 KHz 194) In different types of Pulse Width Modulation, a. 9.8/8=1+ m2/2 86) Example of continuous wave analog modulation is. The dissipated power is given by P = V2rms/R b. (ii) If one of the sidebands is also suppressed, half of the remaining power will be saved Therefore the range of AM spectrum is given by (fc fm ) to (fc + fm ) d. All of the above, a. In continuous-wave modulation, continuous signals are formed after modulation. 117) What is the maximum frequency deviation allowed in commercial FM broadcasting? = 2* 8.6 d. All of the above, 86) Example of continuous wave analog modulation is, 87) The standard value for Intermediate frequency (IF) in AM receivers is, a. Sampling In terms of signal frequency (fs) and intermediate frequency (fi), the image frequency is given by, a) The disturbance caused in the nearby channel or circuit due to transmitted signal, c) Generation of closely lying side bands, Q8. b. RC > 1/W d. All of the above, a. Synchronization is not required between transmitter and receiver d. None of the above, ANSWER: (c) High power FM generation is possible. d. None of the above, ANSWER: (c) Multiplication of incoming signal and the locally generated carrier, 90) The advantages of using an RF amplifier are, a. d. 1.66 * 10-6 V, 0.23 * 10-6 V, ANSWER: (a) 4.071 * 10-6 V, 5.757 * 10-6 V, Explanation: 894 KHz, 884 KHz, 12 KHz 75KHz, 200Khz 2000.35, 1999.65 and 2000.6, 1999.4 c. Becomes half Explanation: To remove noise Percentage modulation After successful attempts, the modulation technique was established and used in electronic communication. 6.08 * 10-6 V, 15.77 * 10-6 V Changes with incoming signal frequency Operates in class A mode Modulating frequency, a. c. Image frequency rejection becomes poor at low IF b. Demodulation Frequency d. None of the above, 29) For low level modulation, amplifier used is, a. a. This set of Data communication and Networking Multiple Choice Questions and Answers (MCQs) focuses on “Analog to Analog Modulation Techniques- Amplitude Modulation (AM)- Frequency Modulation (FM)- Phase Modulation (PM) ”. b. the deviation in the instantaneous value of the frequency with modulating signal. Removal of unwanted signal Phase discrimination method c. 79.36 W d. All of the above, 74) Vestigial side band signals are detected by, a. Filters Generation of SSB signals BW = 2fm where fc is carrier frequency. View Answer: Answer: Option D. Solution: Questions and Answers in Modulation Series. b. 87) The standard value for Intermediate frequency (IF) in AM receivers is. 17 = 15(1 + m2/2) Explanation: = 50. b. For a frequency deviation of 50 KHz, calculate the modulation index of the FM signal. Explanation: I am an M.Tech in Electronics & Telecommunication Engineering. ElectronicsPost.com is a participant in the Amazon Services LLC Associates Program, and we get a commission on purchases made through our links. d. Both a and b, Hi! c. Adjacent channel interference is more 67) Requirements of synchronous detection of AM signal are: a. GN are the noise figures and the gains respectively of the amplifiers at different stages. ANSWER: (a) 455 KHz. c. None of the above 48) What is the effect on the transmitted power of AM signal when the modulation index changes from 0.8 to 1? If you are looking for a reviewer in datacom, a topic in Electronics Systems and Technologies (Communications Engineering) this will definitely help you before taking the Board Exam.In this particular topic you have learned the two types of conversion: digital-to- analog … f3= 300π/2π = 150Hz. 138) For a FM signal v(t) = 25 cos (15 * 108t + 10 sin 1550t), calculate: a. Ability to reject unwanted signals Find the percentage of modulation when the antenna current increases to 10.4A. ADF4158 is a 6.1 GHz, fractional-N frequency synthesizer with direct modulation and waveform generation capability. Noise voltage by individual resistors b. 18) An unmodulated AM signal produces a current of 5.4 A. Receive the Incoming modulated carrier by antenna b. 1. c. In the channel d. Vn is inversely proportional to bandwidth, ANSWER: (b) Vn is directly proportional to √bandwidth, 155) Noise factor for a system is defined as the ratio of, a. d. Signal power. Explanation: General modulated signal: s(t) =A(t)cos[]ωct +φ(t) ω c : carrier frequency A(t): instantaneous amplitude φ(t): instantaneous phase deviation When A(t) is linearly related to the modulating (message) signal total modulation index mt = √( m12 + m22 ) Vn = √{4R KTB} v = ( 10 + 2 Sin( 2Π * 8 * 103 t) ) * Sin (2Π * 1.6 * 106 t) volts. Frequency modulation Frequency of the carrier varies in accordance with the modulating signal c. Instantaneous power at the transmitter is constant Among the types of modulation techniques, the main classification is Continuous-wave Modulation and Pulse Modulation. Vn2 = √(4R2 KTB) = 11.15 * 10-7, a. c. Vn is inversely proportional to absolute temperature RC phase-shift. High frequency noise b. R is the resistance Q13. Detection of modulating signal Given: m = 0.6 Explanation: 144) Determine the Bandwidth of a FM wave when the maximum deviation allowed is 75KHz and the modulating signal has a frequency of 10KHz. Before transmission c. 30 Hz to 3 KHz a. d. Change in amplitude of the carrier according to modulating signal frequency, ANSWER: (a) Change in amplitude of the carrier according to modulating signal, 2) The ability of the receiver to select the wanted signals among the various incoming signals is termed as, a. fc + fm and fc + fm c. 200 Watts b. v(t) = A cos ( Ωct + mf sin Ωmt) =(20/√2)2/R b. = 2 (75 + 10) Passed through a low pass filter Below the carrier frequency c. Product modulator calculate: (a) the carrier power, Two phase shifting networks b. c. Ratio of output signal to input signal to a system Output noise power (Pno) to input noise power (Pni) a. Nyquist rate = 2 fmax c. ωm b. b. (i) Continuous Wave Modulation: When the carrier wave is continuous in nature, the modulation process is known as continuous wave (CW) modulation or analog modulation. According to Carson s rule, BW = 2(Δf + fm) iii) Calculate the value of bandwidth of this signal. 200 KHz 40) Calculate the side band power in an SSBSC signal when there is 50% modulation and the carrier power is 50W. This is the way the telephone system works. i) Calculate the values of the frequencies of carrier and modulating signals. = 2 * 51 KHz Sampling of signals less than at Nyquist rate