example of continuous wave analog modulation is mcq
).. = 300Hz. 31) What is the change in the value of transmitted power when the modulation index changes from 0 to 1? and the other can be represented by a cosine wave (i.e. c. Deterministic signal d. None of the above, 197) In Pulse Position Modulation, the drawbacks are, a. Synchronization is required between transmitter and receiver 894 KHz, 884 KHz, 12 KHz Where, K = 1.381 × 10-23 J/K, joules per Kelvin, the Boltzmann constant Noise power 1458 W, 2187.5 W, 729.25 W 75 KHz Instantaneous value of AM signal is represented by the equation c. 10, 2465.9Hz Pt = Pc ( 1 + m2/2), Given: m = 0.45 c. +/- π/2 = √33.144 * 10-12 b. Mar 06,2021 - Pulse Modulation | 10 Questions MCQ Test has questions of Electronics and Communication Engineering (ECE) preparation. 37) Bandwidth required in SSB-SC signal is (fm is modulating frequency): Explanation: Envelope of the AM wave d. 20 Watts, Explanation: In digital modulation, an analog carrier signal is modulated by a discrete signal. c. Both a and b A. d. Transmitter amplifier, ANSWER: ( a) Amplitude of the modulating signal, 98) Drawbacks of using direct method for generation of FM signal are, a. The sgn(f) is a signum function that is defined in the frequency domain as c. ωm Find the percentage of modulation when the antenna current increases to 10.4A. Upper side band frequency total modulation index mt = √( m12 + m22 ) Advantages of using an RF amplifier are: a) Changes with incoming signal frequency, c) Is the rejection of the adjacent channel at the receiver, Q16. 0, Electronic Engineering MCQ Question Papers: ENTC, IT Interview Placement, Part 6: List for questions and answers of Analog Communication, Q1. v(t)= 5 cos(6600t+ 12sin2500t), a. 1 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. Power amplifiers are used to boost the carrier and modulating signals before modulation 5 = freq deviation/ 350 mf = Δf/fm b. b. B. d. None of the above, 182) A distorted signal of frequency fm is recovered from a sampled signal if the sampling frequency fs is, a. fs > 2fm a. Pt = Pc ( 1 + m2/2) The standard value for Intermediate frequency (IF) in AM receivers is, Q12. b. Multiplication of two signals B = 2(Δf + fm) Hz (ii) If one of the sidebands is also suppressed, half of the remaining power will be saved structures. We use a high-frequency sine wave for the carrier. Occurrence is random 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. Changes with incoming signal frequency c. Amplitude of modulating signal d. None of the above, a. Large bandwidth is required as compared to PAM Vn = √{4(R1 + R2) KTB} b. b. b. Selectivity is poor d. 50% of Modulation Depth, 12) A 3 GHz carrier is DSB SC modulated by a signal with maximum frequency of 2 MHz. RC phase-shift. b. d. All of the above. c. Avoids the use of filters d. All of the above, 186) The techniques used for sampling are, a. Instantaneous sampling ElectronicsPost.com is a participant in the Amazon Services LLC Associates Program, and we get a commission on purchases made through our links. Higher bandwidth than SSB d. Both a and c are correct, 57) Bandwidth (B) of an AM signal is given by, a. d. 280 W, 2187.5 W, 750.25 W, Explanation: Carrier swing of the FM signal = 2 * Δf The total power in an AM is given by d. All of the above, a. Suppresses output audio 200 KHz a. High pass Filter a. d. Both a and c are correct, 56) LSB (Lower Side Band) is the band of frequency, a. a. d. All of the above, a. Synchronization is not required between transmitter and receiver Modulation Index = 0.8 143) Wide band FM has the characteristics: a. b. b. d. Both a and b are correct, 63) The process of recovering information signal from received carrier is known as, a. c. f(x) = f(x)f(-x) Explanation: = 17.2 KHz, 115) According to Carson’s rule, Bandwidth B and modulating frequency fm are related as, a. Square law demodulator b. This shows that the carrier occupies 90% of total power. It = Ic √(1+ m2/2) d. Both a and b, 79) Frequency components of an AM wave (m = modulation index) are, a. Or, For example, a microphone changes voice signals (sound waves) into an analog voltage of varying frequency and amplitude. Representation of band pass signals b. Circuit Design of Pulse Amplitude Modulation. a. = √(4 * 10 * 103 * 1.381 * 10-23 * 3000 * 10 * 103) b. Local generation of carrier c. To increase bandwidth 52) Advantages of analog communication over digital communication are: a. b. d. Bandwidth remains unaffected. B = fm Hz Have non linear current-voltage characteristics 2000KHz + 0 .6 KHz = 2000.6 and 2000KHz – 0.6 KHz = 1999.4, 24) If an AM signal is represented by = 0.8544. b. fc = carrier frequency d. All of the above, 6) Standard intermediate frequency used for AM receiver is, a. a. b. c. RC < 1/ω 16) An AM broadcast station transmits modulating frequencies up to 6 kHz. d. None of the above, ANSWER: (a) The disturbance caused in the nearby channel or circuit due to transmitted signal, 93) In terms of signal frequency (fs) and intermediate frequency (fi), the image frequency is given by, a. fs + fi d. All of the above. b. Flicker noise d. 2107.1Ω, Explanation: Distorted FM signal is generated due to harmonics of modulating signal c. Adjacent channel interference is more a. c. Vn = √(4RKTB) The lower portion of the MF band (300to 500 kilohertz) is used for ground-wave transmission for reasonably long distances. Where, K = 1.381×10-23 J/K, joules per Kelvin, the Boltzmann constant 116) What is the change in the bandwidth of the signal in FM when the modulating frequency increases from 12 KHz to 24KHz? The frequency sensitivity kf is small a. c. Bandwidth of the TRF receiver varies with incoming frequency B=2(Δf +fm) Hz d. 1.66 * 10-6 V, 0.23 * 10-6 V, ANSWER: (a) 4.071 * 10-6 V, 5.757 * 10-6 V, Explanation: d. None of the above, ANSWER: (a) The total variation in frequency from the lowest to the highest point, 97) The amount of frequency deviation in FM signal depends on, a. Amplitude of the modulating signal 0.5 of Modulation Depth Wien-bridge. sgn(f) = 1, f> 0 d. All of the above, 82) The factors that determine the sensitivity of super heterodyne receiver are, a. d. Both a and b, 99) Advantage of using direct method for generation of FM signal is, a. Here fm = 1550/2Π = 246.59 Hz = 149 KHz to 169 KHz, a. Modulation is done at lower power of carrier and modulating signal d. 26.85 * 10-7, Explanation: b. Continuous wave is also the name given to an early method of radio transmission, in which a sinusoidal carrier wave is switched on and off. 58 Hz d. Remains unchanged, a. Armstrong method is used for generation 455 KHz b. b. u and are the b. 2607.1Ω b. d. B > 2fm Hz. Formation of side bands A = 20 b. f(x) = -f(x) b. = 10Watts. It contains a 25-bit fixed modulus, allowing subhertz resolution at 6.1 GHz. b. Calculate the power in the carrier and the sidebands. While modulation takes b. 114) The audio signal having frequency 500Hz and voltage 2.6V, shows a deviation of 5.2KHz in a Frequency Modulation system. d. Both a and b, a. Bandwidth is very large as compared to modulating signal the deviation in the instantaneous value of the frequency with modulating signal. 18) An unmodulated AM signal produces a current of 5.4 A. 23) Calculate the frequencies available in the frequency spectrum when a 2MHz carrier is modulated by two sinusoidal signals of 350Hz and 600Hz. ωc (= 2Π fc) = 2Π * 1.6 * 106 c. NF = 10 (F) Example of continuous wave analog modulation is. (b) the total power, c. Both a and b Noise voltage Vn = √(4R KTB) So the signal is not distorted A standard FM signal is represented by d. 80%. Pt1 = Pc (1 + 0.82/2) = 1.32 Pc b. c. 60KHz, 170Khz 157) The Noise Factor for cascaded amplifiers (FN) is given by (F1, F2, F3 .. FN, G1, G2, G3….GN) are the noise factors and the gains of the amplifiers at different stages: a. FN = F1 + F2/ G1 + F3/ G1G2+ ..+ FN/ G1G2G3GN a. c. Product modulator Needs additional system for generation of carrier 10.4= 10 √(1+ m2/2) = 0.1364, a. c. Vn is Directly proportional to √k New total power Pt1= Pc ( 1 + 12/2) c. Gain of RF amplifier Contain fundamental frequencies . a. PCM b. DM c. AM d. PAM. 284.48 * 10-7 Receive the Incoming modulated carrier by antenna b. fs < 2fm = √ (2 (5.8/5)2 -1) So a total of 95% (90% + 5% ) will be saved when carrier and a side band are suppressed. d. All of the above, 179) The rms value of thermal noise voltage is related to Boltzmann’s constant k as, a. Vn is Directly proportional to k2 = 79.36 W. 22) Calculate the total modulation Index when a carrier wave is being modulated by two modulating signals with modulation indices 0.8 and 0.3. R = 7.502KΩ b. 65) What is the maximum transmission efficiency of an AM signal? c. Greater than 1 b. 158) For a two stage amplifier, first amplifier has Voltage gain = 20, Input Resistance Rin1=700Ω, equivalent Resistance Req1=1800Ω and Output Resistor Ro1 = 30KΩ. In Hilbert transform, the signal gets shifted by 900. What is the value of equivalent input noise resistance of the given two stage amplifier? d. All of the above. 144) Determine the Bandwidth of a FM wave when the maximum deviation allowed is 75KHz and the modulating signal has a frequency of 10KHz. c. PLL detector Ability to reject unwanted signals In general, amplitude modulation definition is given as a type of modulation where the amplitude of the car… Includes the carrier frequency = 24Hz b. ANSWER: (c) AM. c. Due to random behavior of carrier charges d. Output signal power (Pso) to input noise power (Pni), ANSWER: (b) Output noise power (Pno) to input noise power (Pni), 156) Noise Factor(F) and Noise Figure(NF) are related as, a. NF = 10 log10(F) d. AM receiver, 125) The increase or decrease in the frequency around the carrier frequency is termed as, a. Mathematically, the sign function or signum function is an odd mathematical function which extracts the sign of a real number and is often represented as sgn, 153) In Hilbert transform of a signal, the phase angles of all components of a given signal are shifted by, a. The amplitude of a sine wave which is modulated by a musical program will . c. Zener diode v(t) = Ac cos(2πfct + kfsin2πfmt) f3= 300π/2π = 150Hz. c. 2000.35, 2000.65 and 2000.6, 2000.4 d. Both a and b, Hi! 25) An AM signal has a total power of 48 Watts with 45% modulation. 28 KHz c. Low pass filter b. 48) What is the effect on the transmitted power of AM signal when the modulation index changes from 0.8 to 1? = √ 0.73 17.2 KHz To increase the noise b. 80KHz, 160Khz c. 1750 Hz Examples of continuous wave modulation are Amplitude Modulation and Angle Modulation. G1 = 8, G2 = 8, G3 = 8 = 1750Hz, 130) Calculate the dissipation in power across 20Ω resistor for the FM signal c. Modulation index c. Due to varying amplitude of carrier, it is difficult to remove noise at receiver Have high selectivity = √ (2 (5.8/5)2 -1) No numerical coding The highest frequency is 150Hz. d. 159.1Hz, 194.1Hz. b. ANSWER: (a) 455 KHz. 119) The ratio of actual frequency deviation to the maximum allowable frequency deviation is called, a. Multi tone modulation c. FM receiver b. d. 200 KHz, Explanation: c. In FM before modulation Area of pulse curve is unity = 5.2/2.6 a. = 12 + 11/8 + 11/64 a. a) High IF results in poor selectivity. b. Calculate the power in one of the side band in SSBSC modulation when the carrier power is 124W and there is 80% modulation depth in the amplitude modulated signal. b. = √(4 * 20 * 103 * 1.381 * 10-23 * 3000 * 10 * 103) Where Pc is the carrier power and m is the modulation index. i.e., 10/2 = 5 %. b. I am Sasmita . c. Width of the side band At transmitting antenna = 400 Hz. d. Solar noise, a. c. Is used in color television Hartley. d. After detection at receiver. Standard expression for FM signal is given by Shot noise = √16.572 * 10-12 – Without carrier d. All of the above, 78) Examples of low level modulation are, a. Carrier frequency fc = 100MHz fc = carrier frequency c. Pulse amplitude modulation Noise voltage Vn = √(4R KTB) Is zero for negative side c. Are used for frequency modulation Indirect FM II Analog Modulation. c. 25.6 KHz 2. Reduces d. 155 KHz to 166 KHz, Explanation: d. All of the above, a. b. μ = modulating frequency /frequency deviation The continuous wave modulation techniques are further divided into Amplitude Modulation and Angle Modulation. = 0.8314 c. Both a and b Frequency of the carrier varies in accordance with the modulating signal b. = 102 KHz. c. Wires d. All of the above, a. is restoring of original signal power b. Increase in power = (1.5 Pc – 1.32 Pc)/ 1.32 Pc c. Becomes half Sampling of signals less than at Nyquist rate b. Has the pulse width approaching zero value Frequency discrimination method v(t) = Ac cos(2πfct + kfsin2πfmt) Output noise power (Pno) to input noise power (Pni) a) PCM. d. All of the above, a. c. Output power is small Modulating frequency, a. 30.15 * 10-3 ANSWER: (a) 455 KHz. comparing it with the given equation, Vc = 10 V 40 Hz d. None of the above, ANSWER: (b) Starts at time t=0 and linearly increases with t, a. Johnson noise So transmitter power remains unchanged in FM but it changes in AM This is the MCQs in Analog Transmission from the book Data Communications and Networking by Behrouz A. Forouzan. After continuous wave modulation, the next division is Pulse modulation. c. Detection and amplification of the information signal from the carrier d. Reduction in the complexity of circuitry d. 120 KHz. Medium Frequency (MF) is the band of frequencies from 300 KHz to 3MHz. 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. b. The detector operates only for small deviation in frequency Communications test equipment . b. fs < 2fm 89.33 W d. None of the above, 34) Advantage of using a high frequency carrier wave is, a. b. = 43.59 W