Nyquist Required Bandwidth Given Continuous Spectrum From 3khz to 8khz
1) Amplitude modulation is
a. Change in amplitude of the carrier according to modulating signal
b. Change in frequency of the carrier according to modulating signal
c. Change in amplitude of the modulating signal according to carrier signal
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. Sensitivity
b. Selectivity
c. Stability
d. None of the above
ANSWER: (b) Selectivity
3) Emitter modulator amplifier for Amplitude Modulation
a. Operates in class A mode
b. Has a low efficiency
c. Output power is small
d. All of the above
ANSWER: (d) All of the above
4) Super heterodyne receivers
a. Have better sensitivity
b. Have high selectivity
c. Need extra circuitry for frequency conversion
d. All of the above
ANSWER: (d) All of the above
5) The AM spectrum consists of
a. Carrier frequency
b. Upper side band frequency
c. Lower side band frequency
d. All of the above
ANSWER: (d) All of the above
6) Standard intermediate frequency used for AM receiver is
a. 455 MHz
b. 455 KHz
c. 455 Hz
d. None of the above
ANSWER: (b) 455 KHz
7) In the TV receivers, the device used for tuning the receiver to the incoming signal is
a. Varactor diode
b. High pass Filter
c. Zener diode
d. Low pass filter
ANSWER: (a) Varactor diode
8) The modulation technique that uses the minimum channel bandwidth and transmitted power is
a. FM
b. DSB-SC
c. VSB
d. SSB
ANSWER: (d) SSB
9) Calculate the bandwidth occupied by a DSB signal when the modulating frequency lies in the range from 100 Hz to 10KHz.
a. 28 KHz
b. 24.5 KHz
c. 38.6 KHz
d. 19.8 KHz
ANSWER: (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
b. RC > 1/W
c. RC < 1/ω
d. RC > 1/ω
ANSWER: (a) RC < 1/W
11) A modulation index of 0.5 would be same as
a. 0.5 of Modulation Depth
b. 1/2% of Modulation Depth
c. 5% of Modulation Depth
d. 50% of Modulation Depth
ANSWER: (d) 50% of Modulation Depth
12) A 3 GHz carrier is DSB SC modulated by a signal with maximum frequency of 2 MHz. The minimum sampling frequency required for the signal so that the signal is ideally sampled is
a. 4 MHz
b. 6 MHz
c. 6.004 GHz
d. 6 GHz
ANSWER: (c) 6.004 GHz
13) The function of multiplexing is
a. To reduce the bandwidth of the signal to be transmitted
b. To combine multiple data streams over a single data channel
c. To allow multiple data streams over multiple channels in a prescribed format
d. To match the frequencies of the signal at the transmitter as well as the receiver
ANSWER: (b) To combine multiple data streams over a single data channel
14) Aliasing refers to
a. Sampling of signals less than at Nyquist rate
b. Sampling of signals greater than at Nyquist rate
c. Sampling of signals at Nyquist rate
d. None of the above
ANSWER: (a) Sampling of signals less than at Nyquist rate
15) The amount of data transmitted for a given amount of time is called
a. Bandwidth
b. Frequency
c. Noise
d. Signal power
ANSWER: (a) Bandwidth
16) An AM broadcast station transmits modulating frequencies up to 6 kHz. If the AM station is transmitting on a frequency of 894 kHz, the values for maximum and minimum upper and lower sidebands and the total bandwidth occupied by the AM station are:
a. 900 KHz, 888 KHz, 12 KHz
b. 894 KHz, 884 KHz, 12 KHz
c. 894 KHz, 888 KHz, 6 KHz
d. 900 KHz, 888 KHz, 6 KHz
ANSWER: (a) 900 KHz, 888 KHz, 12 KHz
Explanation:
Maximum Frequency fUSB = 894 + 6 = 900 kHz
Minimum Frequency fLSB = 894 – 6 = 888 kHz
Bandwidth BW = fUSB fLSB = 900 888 = 12 kHz OR = 2(6 kHz) = 12 kHz
17) The total power in an Amplitude Modulated signal if the carrier of an AM transmitter is 800 W and it is modulated 50 percent.
a. 850 W
b. 1000.8 KW
c. 750 W
d. 900 W
ANSWER: (d) 900 W
Explanation:
The total power in an Amplitude Modulated wave is given by
PT = PC (1+ m22)
Here, PC = 800W,
m = 0.5
therefore,
PT = 800 (1+ (0.5)2/2) = 900 W
18) An unmodulated AM signal produces a current of 5.4 A. If the modulation is 100 percent,
calculate:
(a) the carrier power,
(b) the total power,
(c) the sideband power when it is transmitted through an antenna having an impedance of 50Ω.
a. 1458 W, 2187.5 W, 729.25 W
b. 278 W, 2187.5 W, 1917.25 W
c. 1438 W, 2187.5 W, 759.25 W
d. 280 W, 2187.5 W, 750.25 W
ANSWER: (a) 1458 W, 2187.5 W, 729.25 W
Explanation:
a) PC=I2R = (5.4)2*50 = 1458W
b) IT = Ic√(1+m2/2) = 5.4√(1+12/2)
=6.614 A
PT = IT 2R
= (6.614)2 * 50
= 2187.25 W
c) PSB = PT – PC
= 2187.25 – 1458 W
= 729.25W (for two bands)
For single band, PSB = 729.25/2
= 364. 625 W
19) Calculate the depth of modulation when a transmitter radiates a signal of 9.8KW after modulation and 8KW without modulation of the signal.
a. 80%
b. 67%
c. 50%
d. 100%
ANSWER: (b) 67%
Explanation:
Ptotal = 9.8KW
Pc = 8KW
Power of the signal (Ptotal) transmitted by a transmitter after modulation is given by
Ptotal = Pc (1+ m2/2)
Where Pc is the power of carrier i.e., without modulation
M is the modulation index
Therefore,
9.8= 8 (1+ m2/2)
9.8/8=1+ m2/2
m=0.67 = 67%
20) When AM signal is of 25KHz, calculate the number of channels required in Medium Frequency (MF) band of 300KHz-3000KHz.
a. 94
b. 69
c. 85
d. 54
ANSWER: (d) 54
Explanation:
Medium Frequency (MF) is the band of frequencies from 300 KHz to 3MHz. The lower portion of the MF band (300to 500 kilohertz) is used for ground-wave transmission for reasonably long distances. The upper and lower ends of the mf band are used for naval purpose.
Frequency available in MF band= 3000 – 300 = 2700 KHz
The bandwidth required by 25 KHz signal = 2 * 25= 50 KHz
Therefore the number of channels available = 2700/ 50 = 54
21) 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.
a. 89.33 W
b. 64.85 W
c. 79.36 W
d. 102 W
ANSWER: (c) 79.36 W
Explanation:
Modulation Index = 0.8
Pc = 124W
Power in sidebands may be calculated as = m2 Pc/4
= (0.8)2 * 124/4
= 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.
a. 0.8544
b. 0.6788
c. 0.9999
d. 0.5545
ANSWER: (a) 0.8544
Explanation:
Here, m1 = 0.8
m2 = 0.3
total modulation index mt = √( m1 2 + m2 2 )
= √( 0.82 + 0.32 )
= √ 0.73
= 0.8544
23) Calculate the frequencies available in the frequency spectrum when a 2MHz carrier is modulated by two sinusoidal signals of 350Hz and 600Hz.
a. 2000.35, 1999.65 and 2000.6, 1999.4
b. 1999.35, 1999.65 and 2000.6, 2000.4
c. 2000.35, 2000.65 and 2000.6, 2000.4
d. 1999.35, 1999.65 and 1999.6, 1999.4
ANSWER: (a) 2000.35, 1999.65 and 2000.6, 1999.4
Explanation:
The frequencies obtained in the spectrum after the amplitude modulation are
fc + fm and fc + fm
therefore,
the available frequencies after modulation by 0.350 KHz are
2000KHz + 0.350 KHz = 2000.35 and 2000KHz – 0.350 KHz = 1999.65
the available frequencies after modulation by 0.6 KHz are
2000KHz + 0 .6 KHz = 2000.6 and 2000KHz – 0.6 KHz = 1999.4
24) If an AM signal is represented by
v = ( 15 + 3 Sin( 2Π * 5 * 103 t) ) * Sin( 2Π * 0.5 * 106 t) volts
i) Calculate the values of the frequencies of carrier and modulating signals.
ii) Calculate the value of modulation index.
iii) Calculate the value of bandwidth of this signal.
a. 1.6 MHz and 8 KHz, 0.6, 16 MHz
b. 1.9 MHz and 18 KHz, 0.2, 16 KHz
c. 2.4 MHz and 18 KHz, 0.2, 16 KHz
d. 1.6 MHz and 8 KHz, 0.2, 16 KHz
ANSWER: (d) 1.6 MHz and 8 KHz, 0.2, 16 KHz
Explanation:
The amplitude modulated wave equation is
v = ( 10 + 2 Sin( 2Π * 8 * 103 t) ) * Sin (2Π * 1.6 * 106 t) volts
Instantaneous value of AM signal is represented by the equation
v = {Vc + Vm Sin ( ωm t )} * Sin (ωc t )
comparing it with the given equation,
Vc = 10 V
Vm = 2V
ωc (= 2Π fc) = 2Π * 1.6 * 106
ωm (= 2Π fm) = 2Π * 8 * 103
(i) The carrier frequency fc is = 1.6 * 106 = 1.6 MHz
The modulating frequency fm is = 8* 103 = 8 kHz
(ii) The modulation index m = Vm/Vc = 2/10 = 0.2
(iii) The bandwidth BW = 2 fm = 16 kHz
25) An AM signal has a total power of 48 Watts with 45% modulation. Calculate the power in the carrier and the sidebands.
a. 39.59 W, 4.505 W
b. 40.59 W, 4.205 W
c. 43.59 W, 2.205 W
d. 31.59 W, 8.205 W
ANSWER: (c) 43.59 W, 2.205 W
Explanation:
Given that Pt = 48 W
Modulation index m = 0.45
The total power in an AM is given by
Pt = Pc ( 1 + m2/2)
= Pc ( 1 +0.452/2)
48 = Pc * 1.10125
Therefore, Pc = 48/ 1.10125
= 43.59 W
The total power in two sidebands is 4843.59 = 4.41 W
So the power in each sideband is 4.41/2 = 2.205 W
26) Calculate the power saved in an Amplitude Modulated wave when it is transmitted with 45% modulation
– Without carrier
– Without carrier and a sideband
a. 90%, 95%
b. 82%, 91%
c. 82%, 18%
d. 68%, 16%
ANSWER: (a) 90%, 95%
Explanation:
i) The total power in an AM is given by
Pt = Pc ( 1 + m2/2)
Given: m = 0.45
Therefore Pt = Pc ( 1 + 0.452/2 )
Pt= Pc *1.10125
Pc/ Pt = 1/1.10125
= 0.908
= 90%
This shows that the carrier occupies 90% of total power. So 90% of total power may be saved if carrier is suppressed in the AM signal.
(ii) If one of the sidebands is also suppressed, half of the remaining power will be saved
i.e., 10/2 = 5 %. So a total of 95% (90% + 5% ) will be saved when carrier and a side band are suppressed.
27) What is the carrier frequency in an AM wave when its highest frequency component is 850Hz and the bandwidth of the signal is 50Hz?
a. 80 Hz
b. 695 Hz
c. 625 Hz
d. 825 Hz
ANSWER: (d) 825 Hz
Explanation:
Upper frequency = 850Hz
Bandwidth = 50Hz
Therefore lower Frequency = 850 – 50= 800 Hz
Carrier Frequency = (850-800)/2
= 825 Hz
28) Noise figure of merit in SSB modulated signal is
a. 1
b. Less than 1
c. Greater than 1
d. None of the above
ANSWER: (a) 1
29) For low level modulation, amplifier used is
a. Class A
b. Class C
c. Class A & C
d. None of the above
ANSWER: (a) Class A
30) The antenna current of the transmitter is 10A. Find the percentage of modulation when the antenna current increases to 10.4A.
a. 32%
b. 28.5%
c. 64%
d. 40%
ANSWER:(b) 28.5%
Explanation:
It = Ic √(1+ m2/2)
10.4= 10 √(1+ m2/2)
√ (1+ m2/2) = 1.04
Therefore m = 0.285
= 28.5%
31) What is the change in the value of transmitted power when the modulation index changes from 0 to 1?
a. 100%
b. Remains unchanged
c. 50%
d. 80%
ANSWER: (c) 50%
Explanation:
Pt = Pc ( 1 + m2/2)
Pt= Pc ( 1 + 02/2) = Pc ..(1)
New total power Pt1= Pc ( 1 + 12/2)
= Pc *3/2 ..(2)
(2) / (1),
We get , Pt1/ Pt= 3/2= 1.5
Pt1= 1.5 Pt
i.e. there is increase in total power by 50%
32) Function of RF mixer is
a. Addition of two signals
b. Multiplication of two signals
c. Rejection of noise
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. Poor selectivity
b. Poor Signal to noise ratio
c. Poor sensitivity
d. None of the above
ANSWER: (a) Poor selectivity
34) Advantage of using a high frequency carrier wave is
a. Signal can be transmitted over very long distances
b. Dissipates very small power
c. Antenna height of the transmitter is reduced
d. All of the above
ANSWER: (d) All of the above
35) Advantage of using VSB transmission is
a. Higher bandwidth than SSB
b. Less power required as compared to DSBSC
c. Both a and b
d. None of the above
ANSWER: (c) Both a and b
36) Modulation is required for
a. Reducing noise while transmission
b. Multiplexing the signals
c. Reduction of Antenna height
d. Reduction in the complexity of circuitry
e. All of the above
ANSWER: (e) All of the above
37) Bandwidth required in SSB-SC signal is (fm is modulating frequency):
a. 2fm
b. < 2fm
c. > 2fm
d. fm
ANSWER: (d) fm
Explanation:
In an amplitude modulated wave, total bandwidth required is from fc + fm to fc – fm
i.e. BW = 2fm where fc is carrier frequency.
In SSB-SC transmission, as the carrier and one of the sidebands are suppressed, the bandwidth remains as fm.
38) For over modulation, the value of modulation index m is
a. m < 1
b. m = 1
c. m > 1
d. Not predetermined
ANSWER: (c) m > 1
39) Demodulation is:
a. Detection
b. Recovering information from modulated signal
c. Both a and b
d. None of the above
ANSWER: (c)Both a and b
40) Calculate the side band power in an SSBSC signal when there is 50% modulation and the carrier power is 50W.
a. 50 W
b. 25 W
c. 6.25 W
d. 12.5 W
ANSWER: (c) 6.25 W
Explanation:
The side band power is given by
Pc m2/2
= 50 * (0.5) 2/2
= 6.25W
41) TRF receiver and super heterodyne receiver are used for
a. Detection of modulating signal
b. Removal of unwanted signal
c. Both a and b
d. None of the above
ANSWER: (c) Both a and b
42) Disadvantage of using a DSB or SSB signal modulation is
a. Difficult to recover information at the receiver
b. Carrier has to be locally generated at receiver
c. Both a and b are correct
d. None of the above
ANSWER: (c) Both a and b are correct
43) Calculate the modulation index when the un modulated carrier power is 15KW, and after modulation, carrier power is 17KW.
a. 68%
b. 51.63%
c. 82.58%
d. 34.66%
ANSWER: (b) 51.63%
Explanation:
The total power in an AM is given by
Pt = Pc ( 1 + m2/2)
17 = 15(1 + m2/2)
m2/2 = 0.134
m = 0.5163
= 51.63%
44) An AM transmitter has an antenna current changing from 5 A un modulated to 5.8 A. What is the percentage of modulation?
a. 38.8%
b. 83.14%
c. 46.8%
d. 25.2%
ANSWER: (b) 83.14%
Explanation:
Modulation index m is given by
m= √ (2{It/Ic}2-1)
= √ (2 (5.8/5)2 -1)
= √ (2 (5.8/5)2 -1)
= 0.8314
= 83.14%
45) Calculate the power in a DSB SC signal when the modulation is 60% with a carrier power of 600W.
a. 600 W
b. 540 W
c. 108 W
d. 300 W
ANSWER: (c) 108 W
Explanation:
The total power in an AM is given by
Pt = Pc (1 + m2/2)
Given: m = 0.6
Therefore DSB power = (m2/2)Pc
= 600* (0.6)2/2
= 108 W
46) Analog communication indicates:
a. Continuous signal with varying amplitude or phase
b. No numerical coding
c. AM or FM signal
d. All of the above
ANSWER:(d) All of the above
47) Types of analog modulation are:
a. Phase modulation
b. Frequency modulation
c. Amplitude modulation
d. All of the above
ANSWER: (d) All 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?
a. 0.1364
b. 0.3856
c. 1.088
d. 0.5
ANSWER: (a) 0.1364
Explanation:
The total power in an AM is given by
Pt = Pc (1 + m2/2)
Where Pc is the carrier power and m is the modulation index.
Therefore,
Pt1 = Pc (1 + 0.82/2) = 1.32 Pc
Pt2 = Pc ( 1 + 12/2) = 1.5 Pc
Increase in power = (1.5 Pc – 1.32 Pc)/ 1.32 Pc
= 0.1364
49) Synchronous detection means
a. Extracting week signal from noise
b. Need a reference signal with predetermined frequency and phase
c. Both a and b
d. None of the above
ANSWER: (c) Both a and b
50) Analog signal may be converted into digital signal by
a. Sampling
b. Amplitude modulation
c. Filtering
d. Mixing
ANSWER: (a) Sampling
51) The minimum antenna height required for transmission in reference to wavelength λ is
a. λ
b. λ/4
c. λ/2
d. 4 λ
ANSWER:(b) λ/4
52) Advantages of analog communication over digital communication are:
a. Data rate is low
b. Less transmission bandwidth is required
c. Synchronization is not needed
d. All of the above
ANSWER: (d) All of the above
53) Radio waves travel through
a. Electromagnetic waves
b. Water
c. Wires
d. Fiber optic cable
ANSWER: (a) Electromagnetic waves
54) AM wave may be represented as E(t) cos ωct where E(t) is
a. Envelope of the AM wave
b. Carrier signal
c. Amplitude of modulating signal
d. None of the above
ANSWER: (a) Envelope of the AM wave
55) USB (Upper Side Band) is the band of frequency
a. Above the carrier frequency
b. Includes the carrier frequency
c. That lies in AM spectrum
d. Both a and c are correct
ANSWER:(d) Both a and c are correct
56) LSB (Lower Side Band) is the band of frequency
a. Below the carrier frequency
b. Includes the carrier frequency
c. That lies in AM spectrum
d. Both a and c are correct
ANSWER: (d) Both a and c are correct
57) Bandwidth (B) of an AM signal is given by
a. B = 2 ωm
b. B = (ωc + ωm) – (ωc – ωm)
c. ωm
d. None of the above
e. Both a and b are correct
ANSWER: (e) Both a and b are correct
58) An oscillator for an AM transmitter has a 100μH coil and a 10nF capacitor. If a modulating frequency of 10 KHz modulates the oscillator, find the frequency range of the side bands.
a. 149 KHz to 169 KHz
b. 184 KHz to 296 KHz
c. 238 KHz to 296 KHz
d. 155 KHz to 166 KHz
ANSWER: (a) 149 KHz to 169 KHz
Explanation:
Carrier frequency fc = 1/2Π√LC
= 1/ 2Π√100 * 10 – 6 * 10 * 10-9
= 1/2Π * 10-6
= 1.59 * 105 Hz
= 159 KHz
The modulating frequency fm is 10KHz
Therefore the range of AM spectrum is given by (fc fm ) to (fc + fm )
= (159 – 10) to (159 + 10)
= 149 KHz to 169 KHz
59) In Low level Amplitude Modulation
a. Modulation is done at lower power of carrier and modulating signal
b. Output power is low
c. Power amplifiers are required to boost the signal
d. All of the above
ANSWER: (d) All of the above
60) In High level Amplitude Modulation
a. Modulation is done at high power of carrier and modulating signal
b. Collector modulation method is High level Amplitude Modulation
c. Power amplifiers are used to boost the carrier and modulating signals before modulation
d. All of the above
ANSWER: (d) All of the above
61) Square law modulators
a. Have non linear current-voltage characteristics
b. Are used for Amplitude Modulation
c. Are used for frequency modulation
d. Both a and b are correct
ANSWER: (d) Both a and b are correct
62) AM demodulation techniques are
a. Square law demodulator
b. Envelope detector
c. PLL detector
d. Both a and b are correct
ANSWER: (d) Both a and b are correct
63) The process of recovering information signal from received carrier is known as
a. Detection
b. Modulation
c. Demultiplexing
d. Sampling
ANSWER: (a) Detection
64) Ring modulator is
a. Is used for DSB SC generation
b. Consists of four diodes connected in the form of ring
c. Is a product modulator
d. All of the above
ANSWER: (d) All of the above
65) What is the maximum transmission efficiency of an AM signal?
a. 64.44%
b. 33.33%
c. 56.66%
d. 75.55%
ANSWER: (b) 33.33%
66) In synchronous detection of AM signal
a. Carrier is locally generated
b. Passed through a low pass filter
c. The original signal is recovered
d. All of the above
ANSWER: (d) All of the above
67) Requirements of synchronous detection of AM signal are:
a. Local generation of carrier
b. The frequency of the locally generated carrier must be identical to that of transmitted carrier
c. The phase of the locally generated carrier must be synchronized to that of transmitted carrier
d. All of the above
ANSWER: (d) All of the above
68) Disadvantages of using synchronous detection of AM signal are:
a. Needs additional system for generation of carrier
b. Needs additional system for synchronization of carrier
c. Receiver is complex and costly
d. All of the above
ANSWER: (d) All of the above
69) Quadrature Amplitude Modulation (QAM) is
a. Have same bandwidth used for two DSB-SC signals
b. Is also known as Bandwidth Conservation scheme
c. Is used in color television
d. All of the above
ANSWER: (d) All of the above
70) Pilot carrier is
a. Used with DSB-SC signal
b. A small carrier transmitted with modulated signal
c. Used for synchronization with local carrier at the receiver
d. All of the above
ANSWER:(d) All of the above
71) Generation of SSB SC signal is done by
a. Phase discrimination method
b. Frequency discrimination method
c. Product modulator
d. Both a and b
ANSWER: (d) Both a and b
72) Limitations of Frequency discrimination method are:
a. Cannot be used for video signals
b. Designing of band pass filter is difficult
c. Both a and b
d. None of the above
ANSWER:(c) Both a and b
73) Phase shift method is
a. Includes two balanced modulators
b. Two phase shifting networks
c. Avoids the use of filters
d. All of the above
ANSWER: (d) All of the above
74) Vestigial side band signals are detected by
a. Filters
b. Synchronous detection
c. Balanced modulator
d. None of the above
ANSWER: (b) Synchronous detection
75) Automatic gain control is
a. Provides controlled signal amplitude at the output
b. Adjusts the input to output gain to a suitable value
c. Is used in AM radio receiver
d. All of the above
ANSWER: (d) All of the above
76) In an Amplitude Modulation
a. Amplitude of the carrier varies
b. Frequency of the carrier remains constant
c. Phase of the carrier remains constant
d. All of the above
ANSWER: )(d) All of the above
77) If modulation index is greater than 1
a. The baseband signal is not preserved in the envelope of the AM signal
b. The recovered signal is distorted
c. It is called over modulation
d. All of the above
ANSWER: (d) All of the above
78) Examples of low level modulation are
a. Square law diode modulation
b. Switching modulation
c. Frequency discrimination method
d. Both a and b
ANSWER: (d) Both a and b
79) Frequency components of an AM wave (m = modulation index) are
a. Carrier frequency (ωc ) with amplitude A
b. Upper side band (ωc + ωm) having amplitude mA/2
c. Lower side band (ωc – ωm) having amplitude mA/2
d. All of the above
ANSWER: (d) All of the above
80) Squelch circuit is
a. Suppresses output audio
b. Works when there is insufficient desired input signal
c. Is used to suppress the unwanted channel noise when there is no reception by the receiver
d. All of the above
ANSWER: (d) All of the above
81) In Automatic gain control of the AM receiver
a. Gain of the receiver is adjusted
b. The gain adjustment depends upon the strength of the received signal
c. The output provided is a DC voltage
d. All of the above
ANSWER: All of the above
82) The factors that determine the sensitivity of super heterodyne receiver are
a. Gain of the IF amplifier
b. Noise figure of the receiver
c. Gain of RF amplifier
d. All of the above
ANSWER: (d) All of the above
83) Selectivity of a receiver:
a. Changes with incoming signal frequency
b. Is poorer at high frequencies
c. Is the rejection of the adjacent channel at the receiver
d. All of the above
ANSWER: (d) All of the above
84) Advantages of using an RF amplifier are:
a. Better selectivity
b. Better sensitivity
c. Improved signal to noise ratio
d. All of the above
ANSWER: (d) All of the above
85) Intermediate frequency (IF) should be carefully chosen as
a. High IF results in poor selectivity
b. High IF results in problems in tracking of signals
c. Image frequency rejection becomes poor at low IF
d. All of the above
ANSWER: (d) All of the above
86) Example of continuous wave analog modulation is
a. PCM
b. DM
c. AM
d. PAM
ANSWER: (c) AM
87) The standard value for Intermediate frequency (IF) in AM receivers is
a. 455 KHz
b. 580 KHz
c. 10.7 MHz
d. 50 MHz
ANSWER: (a) 455 KHz
88) The functions of radio receiver are
a. Receive the Incoming modulated carrier by antenna
b. Select the wanted signal and reject the unwanted signals and noise
c. Detection and amplification of the information signal from the carrier
d. All of the above
ANSWER: (d) All of the above
89) Function of frequency mixer in super heterodyne receiver is
a. Amplification
b. Filtering
c. Multiplication of incoming signal and the locally generated carrier
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. Better sensitivity
b. Improved signal to noise ratio
c. Better selectivity
d. All of the above
ANSWER: (d) All of the above
91) The costas receiver is used for
a. FM signal
b. DSB-SC signal
c. PCM signal
d. DM signal
ANSWER: (b) DSB-SC signal
92) Cross talk is –
a. The disturbance caused in the nearby channel or circuit due to transmitted signal
b. Adjacent frequency rejection
c. Generation of closely lying side bands
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
b. fs + 2fi
c. 2fs+ fi
d. 2( fs + fi)
ANSWER: (b) fs + 2fi
94) In Frequency Modulation –
a. Amplitude of the carrier remains same
b. Frequency of the carrier varies in accordance with the modulating signal
c. The number of side bands are infinite
d. All of the above
ANSWER: (d) All of the above
95) Frequency deviation in FM is
a. Change in carrier frequency to the frequency above and below the centre frequency
b. Formation of side bands
c. The variation of the instantaneous carrier frequency in proportion to the modulating signal
d. All of the above
ANSWER: (d) All of the above
96) Carrier swing is defined as
a. The total variation in frequency from the lowest to the highest point
b. Frequency deviation above or below the carrier frequency
c. Width of the side band
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
b. Carrier frequency
c. Modulating frequency
d. Transmitter amplifier
ANSWER: ( a) Amplitude of the modulating signal
98) Drawbacks of using direct method for generation of FM signal are
a. Does not give high stability to FM signal frequency
b. Distorted FM signal is generated due to harmonics of modulating signal
c. Cannot be used for high power FM generation
d. Both a and b
ANSWER: (d) Both a and b
99) Advantage of using direct method for generation of FM signal is
a. It gives high stability to FM signal frequency
b. Distortion free FM signal is generated
c. High power FM generation is possible
d. None of the above
ANSWER: (c) High power FM generation is possible
100) What are the disadvantages of using balanced slope detector for demodulation of FM signal?
a. The detector operates only for small deviation in frequency
b. Low pass filter of the detector produces distortion in the detection
c. Both a and b
d. None of the above
ANSWER: (c) Both a and b
101) Drawbacks of Tuned Radio Receiver are
a. Oscillate at higher frequencies
b. Selectivity is poor
c. Bandwidth of the TRF receiver varies with incoming frequency
d. All of the above
ANSWER: (d) All of the above
102) Sensitivity is defined as
a. Ability of receiver to amplify weak signals
b. Ability to reject unwanted signals
c. Ability to convert incoming signal into Image Frequency
d. Ability to reject noise
ANSWER: (a) Ability of receiver to amplify weak signals
103) In radio receivers, varactor diodes are used for
a. Tuning
b. Demodulation
c. Mixing
d. None of the above
ANSWER: (a) Tuning
104) The standard value for Intermediate frequency (IF) in double conversion FM receivers is
a. 455 KHz
b. 580 KHz
c. 10.7 MHz
d. 50 MHz
ANSWER: (c) 10.7 MHz
105) Amplitude limiter in FM receivers are used to
a. Remove amplitude variations due to noise
b. Filteration
c. Demodulation
d. Amplification
ANSWER: (a) Remove amplitude variations due to noise
106) Pre emphasis is done
a. For boosting of modulating signal voltage
b. For modulating signals at higher frequencies
c. In FM before modulation
d. All of the above
ANSWER: (d) All of the above
107) De emphasis is
a. is restoring of original signal power
b. is done at the detector output of the receiver
c. is the inverse process of Pre emphasis
d. All of the above
ANSWER: (d) All of the above
108) Pre emphasis is done before
a. Before modulation
b. Before transmission
c. Before detection at receiver
d. After detection at receiver
ANSWER: (a) Before modulation
109) What is the effect on the deviation d of an FM signal when it is passed through a mixer?
a. Doubles
b. Reduces
c. Becomes half
d. Remains unchanged
ANSWER: (d) Remains unchanged
110) Armstrong method is used for the generation of
a. Direct FM
b. Indirect FM
c. SSB-SC
d. DSB-SC
ANSWER: (b) Indirect FM
111) The modulation index of FM is given by
a. μ = frequency deviation/ modulating frequency
b. μ = modulating frequency /frequency deviation
c. μ = modulating frequency/ carrier frequency
d. μ = carrier frequency / modulating frequency
ANSWER:(a) μ = frequency deviation/ modulating frequency
112) Disadvantages of FM over AM are
a. Prone to selective fading
b. Capture effect
c. Poorer signal to noise ratio at high audio frequencies
d. All of the above
ANSWER: (d) All of the above
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.
a. 1 KHz
b. 50 KHz
c. 102 KHz
d. 150 KHz
ANSWER: (c) 102 KHz
Explanation:
According to Carson's rule, bandwidth of FM is given by 2(Δf+ fm) where Δf is the deviation in frequency and fm is the frequency of sinusoidal signal. The required bandwidth is therefore calculated as
2 * (50KHz + 1KHz)
= 2 * 51 KHz
= 102 KHz
114) The audio signal having frequency 500Hz and voltage 2.6V, shows a deviation of 5.2KHz in a Frequency Modulation system. If the audio signal voltage changes to 8.6V, calculate the new deviation obtained.
a. 17.2 KHz
b. 19.6 KHz
c. 25.6 KHz
d. 14.6 KHz
ANSWER: (a) 17.2 KHz
Explanation:
Deviation in FM is given by Δf = kf * Am
Therefore, kf = Δf/ Am
= 5.2/2.6
= 2
When voltage changes to 8.6V = Am
New frequency deviation Δf = kf * Am
= 2* 8.6
= 17.2 KHz
115) According to Carson's rule, Bandwidth B and modulating frequency fm are related as
a. B = 2(Δf + fm) Hz
b. B = fm Hz
c. B < 2fm Hz
d. B > 2fm Hz
ANSWER: (a) B = 2(Δf + fm) Hz
116) What is the change in the bandwidth of the signal in FM when the modulating frequency increases from 12 KHz to 24KHz?
a. 40 Hz
b. 58 Hz
c. 24 Hz
d. Bandwidth remains unaffected
ANSWER: (c) 24 Hz
Explanation:
According to Carson's rule, the bandwidth required is twice the sum of the maximum frequency deviation and the maximum modulating signal frequency. Or,
B=2(Δf +fm) Hz
B= 2(Δf +12) Hz = 2 Δf + 24 Hz (1)
Assuming Δf to be constant,
B = 2 Δf + 48 Hz (2)
(2)-(1),
= 24Hz
Therefore the bandwidth changes by 24Hz.
117) What is the maximum frequency deviation allowed in commercial FM broadcasting?
a. 100 KHz
b. 75 KHz
c. 15 KHz
d. 120 KHz
ANSWER: (b) 75 KHz
118) What is the maximum modulating frequency allowed in commercial FM broadcastings?
a. 40 KHz
b. 75 KHz
c. 15 KHz
d. 120 KHz
ANSWER: (c) 15 KHz
119) The ratio of actual frequency deviation to the maximum allowable frequency deviation is called
a. Multi tone modulation
b. Percentage modulation
c. Phase deviation
d. Modulation index
ANSWER: (b) Percentage modulation
120) The range of modulating frequency for Narrow Band FM is
a. 30 Hz to 15 KHz
b. 30 Hz to 30 KHz
c. 30 Hz to 3 KHz
d. 3 KHz to 30 KHz
ANSWER: (c) 30 Hz to 3 KHz
121) FM is advantageous over AM as
a. The amplitude of FM is constant. So transmitter power remains unchanged in FM but it changes in AM
b. The depth of modulation in FM can be changed to any value by changing the frequency deviation. So the signal is not distorted
c. There is less possibility of adjacent channel interference due to presence of guard bands
d. All of the above
ANSWER: All of the above
122) VCO is used to generate
a. Direct FM
b. Indirect FM
c. SSB-SC
d. DSB-SC
ANSWER: (a) Direct FM
123) Change in instantaneous phase of the carrier with change in amplitude of the modulating signal generates
a. Direct FM
b. Indirect FM
c. SSB-SC
d. DSB-SC
ANSWER: (b) Indirect FM
124) Phase-locked loop can be used as
a. FM demodulator
b. AM demodulator
c. FM receiver
d. AM receiver
ANSWER: (a) FM demodulator
125) The increase or decrease in the frequency around the carrier frequency is termed as
a. Figure factor
b. Frequency deviation
c. Modulation index
d. Frequency pectrum
ANSWER: (b) Frequency deviation
126) Carson's rule is used to calculate
a. Bandwidth of FM signal
b. Signal to noise ratio
c. Modulation index
d. Noise figure
ANSWER: (a) Bandwidth of FM signal
127) The ratio of maximum peak frequency deviation and the maximum modulating signal frequency is termed as
a. Frequency deviation
b. Deviation ratio
c. Signal to noise ratio
d. Frequency spectrum
ANSWER: (b) Deviation ratio
128) The equation v(t) = A cos [ωct + kp Φ(t)] represents the signal as
a. Phase modulated signal
b. SSBSC signal
c. DSB SC signal
d. None of the above
ANSWER: (a) Phase modulated signal
129) Calculate the maximum frequency deviation for the FM signal
v(t) = 10 cos (6000t+ 5sin2200t)
a. 2200 Hz
b. 6000 Hz
c. 1750 Hz
d. 11000 Hz
ANSWER: (c) 1750 Hz
Explanation:
A standard FM signal is represented by
v(t) = Ac cos(2πfct + kfsin2πfmt)
Ac = carrier amplitude
fc = carrier frequency
kf = modulation index
fm= modulating frequency = 2200/2π = 350 Hz
kf = frequency deviation/modulating frequency
5 = freq deviation/ 350
Therefore, deviation = 5 * 350
= 1750Hz
130) Calculate the dissipation in power across 20Ω resistor for the FM signal
v(t)= 20 cos(6600t+ 10sin2100t)
a. 5W
b. 20W
c. 10W
d. 400W
ANSWER: (a) 5W
Explanation:
A standard FM signal is represented by
v(t) = Accos(2πfct + kfsin2πfmt)
Ac = carrier amplitude
fc = carrier frequency
kf = modulation index
fm= modulating frequency
kf= frequency deviation/modulating frequency
the power dissipated across 20Ω resistor is given by
Vrms 2/R
=(20/√2)2/R
= 5W
131) What is the value of carrier frequency in the following equation for the FM signal?
v(t)= 5 cos(6600t+ 12sin2500t)
a. 1150 Hz
b. 6600 Hz
c. 2500 Hz
d. 1050 Hz
ANSWER: (d) 1050 Hz
Explanation:
A standard FM signal is represented by
v(t) = Ac cos(2πfct + kfsin2πfmt)
Ac = carrier amplitude
fc = carrier frequency
kf = modulation index
fm= modulating frequency
kf = frequency deviation/modulating frequency
therefore, fc= 6600/2π
= 1050Hz
132) Calculate the modulation index in an FM signal when fm (modulating frequency) is 250Hz and Δf (frequency deviation) is 5KHz.
a. 20
b. 35
c. 50
d. 75
ANSWER: (a) 20
Explanation:
Modulation index is the measure of how much the modulation parameter changes from its un modulated value. The modulation index of FM is given by
μ = frequency deviation/ modulating frequency
= Δf/ fm
Where Δf is the peak frequency deviation i.e. the deviation in the instantaneous value of the frequency with modulating signal.
fm is the value of modulating frequency
μ = 5000/250
= 20
133) After passing the FM signal through mixer, what is the change in the frequency deviation Δ when the modulating frequency is doubled?
a. Becomes 2 Δ
b. Becomes Δ /2
c. Becomes Δ2
d. Remains unchanged
ANSWER: (d) Remains unchanged
134) In frequency modulation,
a. Armstrong method is used for generation
b. Multiple side bands are generated
c. The FM signal has infinite bandwidth
d. All of the above
ANSWER: (d) All of the above
135) Maximum frequency deviation and the maximum bandwidth allowed for commercial FM broadcast is
a. 80KHz, 160Khz
b. 75KHz, 200Khz
c. 60KHz, 170Khz
d. 75KHz, 250Khz
ANSWER: 7 (b) 5KHz, 200Khz
136) Guard bands are provided in FM signal to
a. Prevent interference from adjacent channels
b. To increase the noise
c. To increase bandwidth
d. None of the above
ANSWER: (a) Prevent interference from adjacent channels
137) For a FM signal v(t) = 15 cos ( 10 * 108t + 10 sin 1220t), calculate
1. Carrier frequency
2. Modulating frequency
a. 159.1MHz, 194.1Hz
b. 185.5MHz, 200.15Hz
c. 350.1MHz, 200.1Hz
d. 159.1Hz, 194.1Hz
ANSWER: 1 (a) 59.1MHz, 194.1Hz
138) For a FM signal v(t) = 25 cos (15 * 108t + 10 sin 1550t), calculate:
- Modulation index
- Maximum frequency deviation
a. 10, 3000.1Hz
b. 20, 1550.9Hz
c. 10, 2465.9Hz
d. 10, 2000.0Hz
ANSWER: (c) 10, 2465.9Hz
Explanation:
Standard expression for FM signal is given by
v(t) = A cos ( ωct + mf sin ωmt)
Comparing with the given equation,
Modulation index mf = 10
Maximum frequency deviation is given by
mf = Δf/fm
Δf = mf * fm
Here fm = 1550/2Π = 246.59 Hz
Δf = 10 * 246.59
= 2465.9Hz
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.
a. 100 Watts
b. 10 Watts
c. 200 Watts
d. 20 Watts
ANSWER: (b) 10 Watts
Explanation:
Standard expression for FM signal is given by
v(t) = A cos ( Ωct + mf sin Ωmt)
Comparing with the given equation,
A = 20
The dissipated power is given by P = V2 rms/R
= (20/√2)2/ 20
= 10Watts
140) A 100MHz carrier is frequency modulated by 5 KHz wave. For a frequency deviation of 100 KHz, calculate the carrier swing of the FM signal.
a. 2000 KHz
b. 100 KHz
c. 105 KHz
d. 200 KHz
ANSWER: (d) 200 KHz
Explanation:
Carrier frequency fc = 100MHz
Modulating frequency fm = 5 KHz
Frequency deviation Δf = 100 KHz
Carrier swing of the FM signal = 2 * Δf
= 2 * 100
= 200 KHz
141) A 100MHz carrier is frequency modulated by 10 KHz wave. For a frequency deviation of 50 KHz, calculate the modulation index of the FM signal.
a. 100
b. 50
c. 70
d. 90
ANSWER: (b) 50
Explanation:
Carrier frequency fc = 100MHz
Modulating frequency fm = 10 KHz
Frequency deviation Δf = 500 KHz
Modulation index of FM signal is given by
mf = Δf/fm
= 500 * 103/ 10 * 103
= 50
142) Narrow band FM has the characteristics:
a. The frequency sensitivity kf is small
b. Bandwidth is narrow
c. Both a and b
d. None of the above
ANSWER: (c) Both a and b
143) Wide band FM has the characteristics:
a. The frequency sensitivity kf is large
b. Bandwidth is wide
c. Both a and b
d. None of the above
ANSWER: (c) Both a and b
144) Determine the Bandwidth of a FM wave when the maximum deviation allowed is 75KHz and the modulating signal has a frequency of 10KHz.
a. 170 KHz
b. 200 KHz
c. 100 KHz
d. 1000 KHz
ANSWER: (a) 170 KHz
Explanation:
Modulating frequency fm = 10 KHz
Frequency deviation Δf = 75 KHz
According to Carson s rule, BW = 2(Δf + fm)
= 2 (75 + 10)
= 170 KHz
145) FM signal is better than AM signal because
a. Less immune to noise
b. Less adjacent channel interference
c. Amplitude limiters are used to avoid amplitude variations
d. All of the above
ANSWER: (d) All of the above
146) FM is disadvantageous over AM signal because
a. much wider channel bandwidth is required
b. FM systems are more complex and costlier
c. Adjacent channel interference is more
d. Both a and b
ANSWER: (d) Both a and b
147) For a three stage cascade amplifier, calculate the overall noise figure when each stage has a gain of 12 DB and noise figure of 8dB.
a. 12
b. 24
c. 13.55
d. 8
ANSWER: (c) 13.55
Explanation:
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. So the final noise figure of the cascaded amplifier is obtained by
FN = F1 + (F2 – 1)/ G1 + (F3 – 1)/ G1G2+……+(FN– 1)/ G1G2G3 GN
F1, F2, F3 .. FN, G1,G2, G3…. GN are the noise figures and the gains respectively of the amplifiers at different stages.
F1 = 12, F2 = 12, F3 = 12
G1 = 8, G2 = 8, G3 = 8
FN= 12 + (12- 1)/ 8+ (12- 1)/ 8 * 8
= 12 + 11/8 + 11/64
= 13.55
148) The Hilbert transform of the signal sinω1t + sinω2t is
a. sinω1t + sinω2t
b. cosω1t + cosω2t
c. sinω2t + cosω2t
d. sinω1t + sinω1t
ANSWER: (b) cosω1t + cosω2t
Explanation:
In Hilbert transform, the signal gets shifted by 900.
So the signal sinω1t+ sinω2t gets shifted by 900
sinω1(t+900)+ sinω2(t+900)
= cosω1t+ cosω2t
149) The noise due to random behaviour of charge carriers is
a. Shot noise
b. Partition noise
c. Industrial noise
d. Flicker noise
ANSWER: (a) Shot noise
150) Transit time noise is
a. Low frequency noise
b. High frequency noise
c. Due to random behavior of carrier charges
d. Due to increase in reverse current in the device
ANSWER: (b) High frequency noise
151) Figure of merit γ is
a. Ratio of output signal to noise ratio to input signal to noise ratio
b. Ratio of input signal to noise ratio to output signal to noise ratio
c. Ratio of output signal to input signal to a system
d. Ratio of input signal to output signal to a system
ANSWER: (a) Ratio of output signal to noise ratio to input signal to noise ratio
152) Signum function sgn(f), for f>0, f=0 and f<0, has the values:
a. -1 to +1
b. +1, 0, -1 respectively
c. -∞ to + ∞
d. 0 always
ANSWER: (b) +1, 0, -1 respectively
Explanation:
The sgn(f) is a signum function that is defined in the frequency domain as
sgn(f) = 1, f> 0
= 0, f = 0
= -1, f< 0
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. +/- π
b. +/- π/4
c. +/- π/2
d. Any angle from 00 to 3600
ANSWER: (c) +/- π/2
154) The noise voltage (Vn) and the signal bandwidth (B) are related as
a. Vn is directly proportional to bandwidth
b. Vn is directly proportional to √bandwidth
c. Vn is inversely proportional to absolute temperature
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. Input noise power (Pni) to output noise power (Pno)
b. Output noise power (Pno) to input noise power (Pni)
c. Output noise power (Pno) to input signal power (Psi)
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)
b. F = 10 log10(NF)
c. NF = 10 (F)
d. F = 10 (NF)
ANSWER: (a) NF = 10 log10(F)
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
b. FN = F1 + (F2 – 1)/ G1 + (F3 – 1)/ (G1+G2)+ ..+(FN – 1)/ (G1+G2+G3+…+GN)
c. FN = F1 + F2/ G1 + F3/ (G1+G2) +…+ FN/ (G1+G2+G3+…+GN)
d. FN = F1 + (F2 – 1)/ G1 + (F3 – 1)/ G1G2+…+(FN – 1)/ G1G2G3GN
ANSWER: (d) FN = F1 + (F2 – 1)/ G1 + (F3 – 1)/ G1G2+…+(FN – 1)/ G1G2G3GN
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. The corresponding values of second amplifier are : 25, 80 KΩ, 12 KΩ, 1.2 MΩ respectively. What is the value of equivalent input noise resistance of the given two stage amplifier?
a. 2609.1Ω
b. 2607.1Ω
c. 107.1Ω
d. 2107.1Ω
ANSWER: (b) 2607.1Ω
Explanation:
R1 = Rin1 + Req1 = 700 + 1800 = 2500Ω
R2 = (Ro1 Rin2)/ (Ro1 + Rin2) + Req2 = 30 * 80/(30 + 80) + 12 = 40.92KΩ
R3 = Ro2 = 1.2MΩ
Equivalent input noise resistance of a two stage amplifier is given by
Req = R1 + R2/ A2 1 + R3/ (A2 1 A2 2)
= 2500 + 40.92 * 103/(20)2 + 1.2 * 106/(20)2(25)2
= 2607.1Ω
159) The noise temperature at a resistor depends upon
a. Resistance value
b. Noise power
c. Both a and b
d. None of the above
ANSWER: (b) Noise power
160) Noise voltage Vn and absolute temperature T are related as
a. Vn = 1/ √(4RKTB)
b. Vn = √(4RK)/ (TB)
c. Vn = √(4RKTB)
d. Vn = √(4KTB)/R
ANSWER: (c) Vn = √(4RKTB)
161) Notch filter is a
a. Band pass filter
b. Band stop filter
c. Low pass filter
d. High pass filter
ANSWER: (b) Band stop filter
162) Noise is added to a signal in a communication system
a. At the receiving end
b. At transmitting antenna
c. In the channel
d. During regeneration of the information
ANSWER: (c) In the channel
163) Noise power at the resistor is affected by the value of the resistor as
a. Directly proportional to the value of the resistor
b. Inversely proportional to the value of the resistor
c. Unaffected by the value of the resistor
d. Becomes half as the resistance value is doubled
ANSWER: (c) Unaffected by the value of the resistor
164) Low frequency noise is
a. Transit time noise
b. Flicker noise
c. Shot noise
d. None of the above
ANSWER: (b) Flicker noise
165) Hilbert transform may be used in
a. Generation of SSB signals
b. Representation of band pass signals
c. Designing of minimum phase type filters
d. All of the above
ANSWER: (d) All of the above
166) At a room temperature of 300K, calculate the thermal noise generated by two resistors of 10KΩ and 20 KΩ when the bandwidth is 10 KHz.
a. 4.071 * 10-6 V, 5.757 * 10-6 V
b. 6.08 * 10-6 V, 15.77 * 10-6 V
c. 16.66 * 10-6 V, 2.356 * 10-6 V
d. 1.66 * 10-6 V, 0.23 * 10-6 V
ANSWER: (a) 4.071 * 10-6 V, 5.757 * 10-6 V
Explanation:
Noise voltage Vn = √(4R KTB)
Where, K = 1.381×10-23 J/K, joules per Kelvin, the Boltzmann constant
B is the bandwidth at which the power Pn is delivered.
T noise temperature
R is the resistance
Noise voltage by individual resistors
Vn1 = √(4R1 KTB)
= √(4 * 10 * 103 * 1.381 * 10-23 * 3000 * 10 * 103)
= √16.572 * 10-12
= 4.071 * 10-6 V
Vn2 = √(4R2 KTB)
= √(4 * 20 * 103 * 1.381 * 10-23 * 3000 * 10 * 103)
= √33.144 * 10-12
= 5.757 * 10-6 V
167) At a room temperature of 293K, calculate the thermal noise generated by two resistors of 20KΩ and 30 KΩ when the bandwidth is 10 KHz and the resistors are connected in series.
a. 300.66 * 10-7
b. 284.48 * 10-7
c. 684.51 * 10-15
d. 106.22 * 10-7
ANSWER: (b) 284.48 * 10-7
Explanation:
Noise voltage Vn = √(4R KTB)
Where, K = 1.381 × 10-23 J/K, joules per Kelvin, the Boltzmann constant
B is the bandwidth at which the power Pn is delivered.
T noise temperature
R is the resistance
Noise voltage by resistors when connected in series is
Vn = √{4(R1 + R2) KTB}
= √{4(20 * 103 + 30 * 103) * 1.381 × 10-23 * 293 * 10 * 103 }
= 284.48 * 10-7
168) At a room temperature of 300K, calculate the thermal noise generated by two resistors of 10KΩ and 30 KΩ when the bandwidth is 10 KHz and the resistors are connected in parallel.
a. 30.15 * 10-3
b. 8.23 * 10-23
c. 11.15 * 10-7
d. 26.85 * 10-7
ANSWER: (c) 11.15 * 10-7
Explanation:
Noise voltage Vn = √(4R KTB)
Where, K = 1.381 × 10-23 J/K, joules per Kelvin, the Boltzmann constant
B is the bandwidth at which the power Pn is delivered.
T noise temperature
R is the resistance
Noise voltage by resistors when connected in parallel is
Vn = √{4R KTB}
Here for resistors to be in parallel,
1/R = 1/R1 + 1/R2
= 1/10K + 1/30K
= 0.1333
R = 7.502KΩ
Vn = √{4 * 7.502 * 103 * 1.381×10-23 * 300 * 10 * 103}
= √124.323 * 10-14
= 11.15 * 10-7
169) A periodic signal is
a. May be represented by g(t) = g(t + T0)
b. Value may be determined at any point
c. Repeats itself at regular intervals
d. All of the above
ANSWER: (d) All of the above
170) Sine wave is a
a. Periodic signal
b. Aperiodic signal
c. Deterministic signal
d. Both a and c
ANSWER: (a) Periodic signal
171) Properties of Hilbert transform are:
a. The signal and its Hilbert transform have same energy density spectrum
b. The signal and its Hilbert transform are mutually diagonal
c. Both a and b are correct
d. None of the above
ANSWER: (c) Both a and b are correct
172) An even function f(x) for all values of x and x holds
a. f(x) = f(-x)
b. f(x) = -f(x)
c. f(x) = f(x)f(-x)
d. None of the above
ANSWER: (a) f(x) = f(-x)
173) Random signals is
a. May be specified in time
b. Occurrence is random
c. Repeat over a period
d. None of the above
ANSWER: (b) Occurrence is random
174) Unit step function is
a. Exists only for positive side
b. Is zero for negative side
c. Discontinuous at time t=0
d. All of the above
ANSWER: (d) All of the above
175) In Unit impulse function
a. Pulse width is zero
b. Area of pulse curve is unity
c. Height of pulse goes to infinity
d. All of the above
ANSWER: (d) All of the above
176) For a Unit ramp function area of pulse curve is unity
a. Discontinuous at time t=0
b. Starts at time t=0 and linearly increases with t
c. Both a and b
d. None of the above
ANSWER: (b) Starts at time t=0 and linearly increases with t
177) Thermal noise is also known as
a. Johnson noise
b. Partition noise
c. Flicker noise
d. Solar noise
ANSWER: (a) Johnson noise
178) Threshold effect is:
a. Reduction in output signal to noise ratio
b. Large noise as compared to input signal to envelope detector
c. Detection of message signal is difficult
d. All of the above
ANSWER: (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
b. Vn is Directly proportional to k
c. Vn is Directly proportional to √k
d. Vn is Directly proportional to k3
ANSWER: (c) Vn is Directly proportional to √k
180) The spectrum of the sampled signal may be obtained without overlapping only if
a. fs ≥ 2fm
b. fs < 2fm
c. fs > fm
d. fs < fm
ANSWER: (a) fs ≥ 2fm
181) The desired signal of maximum frequency wm centered at frequency w=0 may be recovered if
a. The sampled signal is passed through low pass filter
b. Filter has the cut off frequency wm
c. Both a and b
d. None of the above
ANSWER: (c) Both a and b
182) A distorted signal of frequency fm is recovered from a sampled signal if the sampling frequency fs is
a. fs > 2fm
b. fs < 2fm
c. fs = 2fm
d. fs ≥ 2fm
ANSWER: (b) fs < 2fm
183) Calculate the minimum sampling rate to avoid aliasing when a continuous time signal is given by x(t) = 5 cos 400πt
a. 100 Hz
b. 200 Hz
c. 400 Hz
d. 250 Hz
ANSWER: (c) 400 Hz
Explanation:
In the given signal, the highest frequency is given by f = 400 π/ 2π
= 200 Hz
The minimum sampling rate required to avoid aliasing is given by Nyquist rate. The nyquist rate is = 2 * f
= 2 * 200
= 400 Hz.
184) Calculate the Nyquist rate for sampling when a continuous time signal is given by
x(t) = 5 cos 100πt +10 cos 200πt – 15 cos 300πt
a. 300Hz
b. 600Hz
c. 150Hz
d. 200Hz
ANSWER: (a) 300Hz
Explanation:
For the given signal,
f1 = 100π/2π = 50Hz
f2 = 200π/2π = 100Hz
f3= 300π/2π = 150Hz
The highest frequency is 150Hz. Therefore fmax = 150Hz
Nyquist rate = 2 fmax
= 2 * 150
= 300Hz.
185) A low pass filter is
a. Passes the frequencies lower than the specified cut off frequency
b. Rejects higher frequencies
c. Is used to recover signal from sampled signal
d. All of the above
ANSWER: (d) All of the above
186) The techniques used for sampling are
a. Instantaneous sampling
b. Natural sampling
c. Flat top sampling
d. All of the above
ANSWER: (d) All of the above
187) The instantaneous sampling
a. Has a train of impulses
b. Has the pulse width approaching zero value
c. Has the negligible power content
d. All of the above
ANSWER: (d) All of the above
188) The sampling technique having the minimum noise interference is
a. Instantaneous sampling
b. Natural sampling
c. Flat top sampling
d. All of the above
ANSWER: (b) Natural sampling
189) Types of analog pulse modulation systems are
a. Pulse amplitude modulation
b. Pulse time modulation
c. Frequency modulation
d. Both a and b
ANSWER: (d) Both a and b
190) In pulse amplitude modulation,
a. Amplitude of the pulse train is varied
b. Width of the pulse train is varied
c. Frequency of the pulse train is varied
d. None of the above
ANSWER: (a) Amplitude of the pulse train is varied
191) Pulse time modulation (PTM) includes
a. Pulse width modulation
b. Pulse position modulation
c. Pulse amplitude modulation
d. Both a and b
ANSWER: (d) Both a and b
192) Drawback of using PAM method is
a. Bandwidth is very large as compared to modulating signal
b. Varying amplitude of carrier varies the peak power required for transmission
c. Due to varying amplitude of carrier, it is difficult to remove noise at receiver
d. All of the above
ANSWER: (d) All of the above
193) In Pulse time modulation (PTM),
a. Amplitude of the carrier is constant
b. Position or width of the carrier varies with modulating signal
c. Pulse width modulation and pulse position modulation are the types of PTM
d. All of the above
ANSWER: (d) All of the above
194) In different types of Pulse Width Modulation,
a. Leading edge of the pulse is kept constant
b. Tail edge of the pulse is kept constant
c. Centre of the pulse is kept constant
d. All of the above
ANSWER: (d) All of the above
195) In pulse width modulation,
a. Synchronization is not required between transmitter and receiver
b. Amplitude of the carrier pulse is varied
c. Instantaneous power at the transmitter is constant
d. None of the above
ANSWER: (a) Synchronization is not required between transmitter and receiver
196) In PWM signal reception, the Schmitt trigger circuit is used
a. To remove noise
b. To produce ramp signal
c. For synchronization
d. None of the above
ANSWER: (a) To remove noise
197) In Pulse Position Modulation, the drawbacks are
a. Synchronization is required between transmitter and receiver
b. Large bandwidth is required as compared to PAM
c. None of the above
d. Both a and b
ANSWER: (d) Both a and b
Source: https://electronicspost.com/multiple-choice-questions-and-answers-on-analog-communication-part-1/
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